The Lorentz Transformations and a Few Concerns

[Anamitra]

The derivation of the Lorentz transformations is based on the homogeneity[of space and time] and the isotropy of space.
Could one derive the same transformations wrt space which is not homogeneous or[not] isotropic?
You may consider a few chunks of dielectric strewn here and there. I am assuming for the sake of simplicity that they are at rest in some inertial frame. Such a distribution is not possible without introducing gravitational effects. Running of clocks is affected by gravity.Does the anisotropy of space itself have any effect on them[running of clocks]?

[Incidentally isotropy of space is connected with clocks in the derivation of the Lorentz transformation.Clocks placed symmetrically wrt the x-axis[and lying on the y-z plane as example] should record the same time. Otherwise isotropy of spece gets violated.This idea is commonly used in the derivations.You may consider the one given in "Introduction to Relativity" by Robert Resnick ]
Again the Lorentz transformations are embedded in [present in] Maxwell's equations. But they are the vacuum equations---homogeneity of space[and time] and isotropy are in due consideration.

The Lorentz Transformations are of course correct--only in the context of the homogeneity[of space and time] and isotropy of space. They are extremely useful, like frctionless planes we studied in our childhood days.Frictionless planes helped us in understanding mechanics--but it is extremely difficult to realize them in practice.

 

[Bill_K]

Regardless of what Resnick may or may not have said, the Lorentz transformations in no way depend on the isotropy and homogeneity of spacetime. Nor do they depend on the presence or absence of material objects or EM or gravitational fields. Lorentz invariance is a local property of vacuum. All the equations of physics are local equations and rely on Lorentz invariance at every point for their consistency.

 

[Anamitra]

Regardless of what Resnick may or may not have said, the Lorentz transformations in no way depend on the isotropy and homogeneity of spacetime. Nor do they depend on the presence or absence of material objects or EM or gravitational fields. Lorentz invariance is a local property of vacuum. All the equations of physics are local equations and rely on Lorentz invariance at every point for their consistency.

Could you provide me with a sample derivation of the Lorentz Transformations in the "LOCAL CONTEXT"?

Hope you don't have to go back to Resnick!

 

[Anamitra]

A Basic Issue to be addressed:

Can the anisotropy of space itself affect clock rates[considered apart from gravity]?

[Even if you consider an infinitesimally small region of space[surrounded by matter] you can have milloins of directions emanating from a point in it,providing enough scope for anisotropy]

 

[Anamitra]

Regardless of what Resnick may or may not have said, the Lorentz transformations in no way depend on the isotropy and homogeneity of spacetime.

These are the priceless words of the century--from Bill_K

 

[Fredrik]

One way to find the Lorentz transformations is to consider a family of global coordinate systems $\{x_i:\mathbb R^4\rightarrow\mathbb R^4|i\in I\}$ such that each "coordinate change" function $x_i\circ x_j^{-1}$ takes straight lines to straight lines, and the set $\big\{x_i\circ x_j^{-1}|i,j\in I\big\}$ is a group, with a subgroup that's isomorphic to the translation group, and another that's isomorphic to the rotation group.

To drop the requirement of isotropy or homogeneity would be to drop the requirement that the group of coordinate change functions must have the appropriate subgroups. I don't know what we get if you do. I suppose that we don't want to completely drop them, but rather weaken them somewhat, so that the group almost has the translation and rotation groups as subgroups. I don't even know how to make that statement precise.

 

[Fredrik]

Regardless of what Resnick may or may not have said, the Lorentz transformations in no way depend on the isotropy and homogeneity of spacetime. Nor do they depend on the presence or absence of material objects or EM or gravitational fields. Lorentz invariance is a local property of vacuum. All the equations of physics are local equations and rely on Lorentz invariance at every point for their consistency.

Lorentz transformations are permutations of $\mathbb R^4$ that satisfy a few additional conditions. The presence of matter in the actual universe obviously has no effect on $\mathbb R^4$, and therefore no effect on the Lorentz transformations. But you certainly do have to make assumptions of homogeneity and isotropy to be able to "derive" them from Einstein's postulates. (You can't actually derive them from Einstein's postulates. You derive them from mathematical statements that can be thought of as expressing aspects of Einstein's postulates mathematically. Are homogeneity and isotropy such aspects, or are they separate assumptions? I think that's actually a matter of taste. Einstein's postulates aren't very precise, so you can interpret them in more than one way).

 

[robphy]

In Special Relativity, the Lorentz Transformations appear in two places... in spacetime (transforming events or coordinates of events) and in the tangent-spaces associated with each event (transforming tensors or components of tensors).

 

[JDoolin]

Some questions

(1) What is your concept of the universe?

(a) Space-time curvature embedded in an overall flat space-time, or
(b) tiny little differential sections of flat space-time that are approximately flat, embedded in an overall curved universe?​

(2) Where does Lorentz Transformations apply?

(a) Throughout any flat region of spacetime?
(b) Only along a differential path element of space-time.​

(3) Which is the Lorentz Transformation more similar to?

(a) A transformation from spherical to rectangular coordinates:
$$\begin{align*} x &= r \cos(\phi) \sin(\theta)\\ y &= r \sin(\phi)\sin(\theta)\\ z &= r \cos(\theta) \end{align*}$$
(b) A rotation
$$\begin{align*} x' &= x \cos(\theta)+y \sin(\theta) \\ y' &= -x \sin(\theta)+y \cos(\theta) \\ z' &= z \end{align*}$$
​

(4) What is the radial limitation on a rotation transformation?

(a) Universal: If I turn my reference frame to the right, then even galaxies billions of light years away will move counterclockwise around me.
(b) Local: Objects in my room will move counterclockwise around me, but galaxies billions of light years away will maintain their positions.​

 

[JDoolin]

I'd like to simplify question #3 to two dimensions, and delve into it more deeply.

Let's look at the polar to rectangular transformation:$$\begin{align*} x &= r \cos(\theta)\\ y &= r \sin(\theta) \end{align*}$$

This is a global, but nonlinear expression. There is no way to express these equations in the form:

$$\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} \square & \square \\ \square & \square \end{pmatrix}\begin{pmatrix} r\\ \theta \end{pmatrix}$$

However, if we take the derivatives,

$$\begin{align*} dx &= \cos(\theta)dr - r \sin(\theta) d\theta \\ dy &= \sin(\theta)dr + r \cos(\theta) d\theta \end{align*}$$

this becomes an expression which can be expressed in a form which at least resembles a linear equation:$$\begin{pmatrix} dx\\ dy \end{pmatrix} =\begin{pmatrix} \cos(\theta) & -r \sin(\theta)\\ \sin(\theta) & r \cos(\theta) \end{pmatrix}\begin{pmatrix} dr\\ d\theta \end{pmatrix}$$

However, since r and θ are functions of x and y, there's not any global linearity. The function is only "almost" linear in a "sufficiently small" region of space. One could say that the r,θ coordinates are almost linear in a small enough region.

This should be contrasted with the rotation transformation, which has no such restrictions about small regions.

$$\begin{align*} x'&=x \cos(\theta)-y \sin(\theta)\\ y'&=x \sin(\theta) + y\cos(\theta) \end{align*}$$

Unlike the polar-to-rectangular transformation which is a global but NONlinear transformation, this is a global LINEAR transformation. So when we try to change the form, as follows:

$$\begin{pmatrix} x'\\ y' \end{pmatrix} =\begin{pmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta) \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}$$

There is no difficulty. And if we take the derivatives:

$$\begin{pmatrix} dx'\\ dy' \end{pmatrix} =\begin{pmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta) \end{pmatrix}\begin{pmatrix} dx\\ dy \end{pmatrix}$$

There is also no difficulty.

The key difference here is that in the polar-to-rectangular case, the value of r and θ actually are functions of position. In the rotation case, however, the value of θ is a global parameter that does not vary.

So what I'm trying to convince you of here, is that the rotation transformation is NOT a local transformation but it applies to every particle in the universe. If I turn my head to the left, everything in the universe moves clockwise around my head. If I turn my head to the right, everything in the universe moves counterclockwise around my head. There is no situation where the stuff in my room moves counterclockwise, but the stuff outside my room remains stationary, unless I'm not really rotating but my room is.

There's no way to say that the rotation only applies locally. It is a linear global transformation that affects all particles in the universe. I know some people will laugh at me, saying that my turning my head has no effect on distant particles in the universe, but that is not my argument. I'm not saying that turning my head exerts a force on galaxies billions of light years away causing them to go around in a circle.

When you do a rotation transformation, you are changing the observers' reference frame. And thus everything in the universe moves according to the transformation with respect to the observer.

Finally, the Lorentz Transformation equations are NOT ANALOGOUS to the polar-to-cartesian coordinate transformations. They are analogous to the rotation transformation equations.

They are global linear equations that can be expressed as
$$\begin{align*} ct'&=ct \cosh(\varphi)-x \sinh(\varphi)\\ x'&=-ct \sinh(\varphi) + x\cosh(\varphi) \end{align*}$$
and
$$\begin{pmatrix} c t'\\ x' \end{pmatrix}= \begin{pmatrix} \cosh(\varphi) & -\sinh(\varphi)\\ -\sinh(\varphi) & \cosh(\varphi) \end{pmatrix}\begin{pmatrix} c t\\ x \end{pmatrix}$$

The parameter φ is not a function of t or x; it is an independent global parameter, just like θ in the rotation transformation.

I don't know for sure, but it seems to me, these two types of transformations; global but nonlinear, and GLOBALLY LINEAR transformations are confused. The transformation that makes working with spherical coordinates convenient is nonlinear, and on the scale where the coordinates appear linear, it is local.

The Lorentz transformations, on the other hand, have nothing to do with converting from spherical to polar. They have to do with changing velocities; changing rapidity. The Lorentz Transformation equations are a GLOBAL LINEAR TRANSFORMATION, just like rotation.

 

[PhilDSP]

So what I'm trying to convince you of here, is that the rotation transformation is NOT a local transformation but it applies to every particle in the universe.

Yes, I have to agree with that. If you analyze what the LT does on the basis of its original intent by Voigt and H. A. Lorentz (Voigt's use of the transformation preceded that of Lorentz by several years) then what you say (and more) becomes even clearer. As applied to produce both a change in form of the wave equation and the proper change of initial conditions to accompany the wave equation, it results in initial conditions which can never be reconciled between adjacent points of time or space unless a stretch or shift of all other points is accomplished. But each time that is done the situation remains recursively the same as if you're chasing something that can never be caught. There is no way to make the operation of the LT local unless you freeze the transformation and step back into a Galilean frame of reference.

 

[Anamitra]

If we look from the reverse direction[instead of the deductive side] the issue is apparently becoming much simpler.
The manifold[described by the metric coefficients] may take care of all anisotropies and inhomogeneities.A metric can include charges[eg:Reissner-Nordstrom metric]. If a metric can include charges it may include strong charges ,spin and other properties of matter with their distributions.

One could think in the direction of a generalized unified metric incorporating all known/conceivable properties of matter.

These should have their independent effects on clock rates and change of other physical dimensions. In fact the temporal part of the Reissner-Nordstrom metric contains charge as well as mass.
[The same is true for the spatial part]

One could just imagine reducing the mass and keeping the charge same[in this metric]. The manner in which the clock rate would be affected by charge alone could be examined theoretically. One could try out similar exercises with other properties like spin etc in other types of metrics.

The distribution[redistribution] of mass will change the value of the metric coefficients. Distribution[redistribution] of chargees etc will also alter the manifold properties taking care of inhomogeneties and anisotropies.At the same time,one must note, that the clock rates and the spatial dimensions will depend on the distribution/redistribution of these properties in respect of space and time

So by writing a general type of a metric one should be able to explain refraction of light.
But these metrics will fail to explain properties like the slowing down of light in a medium.Perhaps inclusion of refractive index in the metric could improve the situation.But this[ie,RI] is not a fundamental property like charge or mass. I am very much confused on this issue.

Even if RI is included the metric will not interpret the change of the speed of light [in a medium]if it has a form:
ds^2=g(00)dt^2-g(11)dx1^2- g(22)dx^2-g(33)dx3^2

Again if two frames are in relative motion at a uniform rate in curved space-time full of anisotropies and inhomogenities, should one apply the usual lorentz Transformations as we know them?

 

[Anamitra]

The Local Context:
[On Local Inertial Frames]

We may consider a local region in strong curved spacetime. It may be surrounded by inhomgenities and anisotropies. How do we get the Lorentz Transformations here?

Actually a local transformation to flat spacetime is not a physical transformation.We are doing it on paper--in our imagination. So to that little bit of flat spacetime[we got by transformation] add the rest of it---in your imagination.

It is something remote from reality--a workspace in our imagination.A freely falling lift in the Earth's field can have sufficient inhomogenities and anisotropies around it and does not seem to match with the workspace.
That describes our "frictionless plane"

We can derive the Lorentz Transformations with sufficient conditions of homogenity of space [and time] and isotropy of space[in the "ideal workspace"] from the Postulates of Special Relativity.

[In this process[local transformation from curved space to flat spacetime] we are considering the transformation of a small region of spacetime in total alienation from its surroundings[in so far as anisotropy is concerned], hoping to get correct interpretations from the reverse transformations, when the surroundings are present]

In the transformed flat spacetime[local transformation] the un-transformed anisotropies[and heterogeneities] of the surroundings [of the original curved space] are neglected/ignored. Rather we ASSUME "a Global Transformation" for the rest of the flat space when we consider the lorentz Transformations

[The gravitational field itself, in general will have an anisotropic and a heterogeneous distribution of the metric coefficients wrt some arbitrary point---one may consider this point in the small transformed space in this context]


The Lorentz transformations are of course correct-----only in the proper context of their application

 

[JDoolin]

The Local Context:
[On Local Inertial Frames]

We may consider a local region in strong curved spacetime. It may be surrounded by inhomgenities and anisotropies. How do we get the Lorentz Transformations here?

I would say the Lorentz Transformations should be applied in exact analogy to the Rotation Transformations. That is, if we have some inhomogeneity or anisotropy in spacetime, and we turn and look the other direction, then the whole thing moves, out of our sight.

What coordinates are used when I do this rotation?

Let's look for a moment at a simple example; the Schwarzschild metric:
$$c^2 d\tau^2 = \left( 1 - \frac{2 G M}{c^2 r} \right)c^2 dt^2 - \left( 1-\frac{2 G M}{c^2 r}\right)^{-1} dr^2 - r^2 (d\theta^2 + sin^2(\theta)d\varphi^2)$$
(Thanks for the help finding the parsing error, Fredrik)
It describes a transformation from dt to dτ. And if written in its negative form, it makes a transformation from ds to a function of dr, dθ, and dφ.

My suggestion would be, instead of treating t, r, θ, and φ as "meaningless" variables, we should treat them as spherical coordinates which can easily map to Minkowski coordinates, and are in fact, the domain and range of the Lorentz Transformation equation.

Again, people will laugh at me, and say that there is no place where matter is not present, so, we have schwarzchild metric embedded in another schwarzschild metric, embedded in another schwarzschild metric. In effect, they will say there is no place in the universe where I can get rid of the curvature altogether. And therefore we might as well not try?

It's a fallacious argument. We should always be aware that there may be masses and motion that we have not accounted for, and so lines we think are straight may be curved after all. But the question is not whether we can know for certain that lines are straight. The question is whether the straight lines exist at all.

As long as we have a concept of straightness. Can we not compare one geodesic path to another geodesic path, and unambiguously declare which of those two paths is straighter? Yes, of course. The straight line is the direction that an object would travel if there were no forces acting on it at all. The geodesic is the direction that an object travels when a force is acting on it.

In real life, we know the difference. But once you start reading a lot of General Relativity books, you start calling geodesics straight lines. And for some reason, you ignore t, r, θ, and φ as "meaningless" variables, and you only look at τ, and say. The object in a geodesic only has τ changing, and is otherwise motionless (with respect to itself).

There are some great things you can do with that dτ; in particular, setting it equal to zero, to see what the path of a photon through your curved space is. But that path is through the t, r, θ, and φ coordinates. Those coordinates are not meaningless. They are the overlying GLOBAL LORENTZ FRAME COORDINATES IN SPHERICAL FORM, COMOVING WITH THE CENTRAL MASS.

If you rotate the coordinate system, r, θ, and φ are affected. If you Lorentz Transform the coordinate system, t, r, θ, and φ are affected.

 

[Fredrik]

(sorry, I can't find the parsing error in the LaTeX.)

There's a ^ followed by a space.

 

[JDoolin]

Some questions

(1) What is your concept of the universe?

(a) Space-time curvature embedded in an overall flat space-time, or
(b) tiny little differential sections of flat space-time that are approximately flat, embedded in an overall curved universe?​

If I'm not mistaken, the General Relativity "textbook" answer to this question is (b). But I want to make the argument that the actual answer is (a).

For the Schwarzschild metric, in particular, let's discuss the meanings of these partial derivatives:

$$\left (\frac{\partial \tau}{\partial t} \right )_{\theta,\phi,r \, \mathrm{const}}$$
$$\left (\frac{\partial s}{\partial r} \right )_{\theta,\phi,t \, \mathrm{const}}$$
$$\left (\frac{\partial s}{\partial \phi} \right )_{\theta,r ,t \, \mathrm{const}}$$
$$\left (\frac{\partial s}{\partial \theta} \right )_{\phi,r ,t \, \mathrm{const}}$$

What are the textbook definitions of of t, r, θ, and φ? And what are the labels for

Typically, I find textbooks tend to slide over the subject with a great deal of ambiguity.

"But inspiration aside, it is important to think of these vectors as being located at a single point, rather than stretching from one point to another. (Although this won't stop us from drawing them as arrows on spacetime diagrams.)"

I don't know exactly what Carroll's point is here, but to contrast with Carroll, I will say it is important to think of these vectors as stretching between one event and another event, rather than located at a single point. If you are doing derivatives, with a dt, a dr, a dθ, and a dφ, you can't PUT that at a single point. You can't have a variation in positional variables t, r, θ, and φ "at a single point."

There is an s-component parallel to the r-direction (space-like interval between events), an s-component parallel to the θ direction (space-like interval between events), and an s-component parallel to the φ direction (space-like interval between events), and another component in the t-direction (time-like interval between events), called τ (tau).

There are two different coordinate systems here, and they overlap. You have the three unnamed s-components in the same direction as r, θ, and φ, and the named tau component, in the same direction as t.

The three unnamed s-components and tau make up the curved coordinate system. But there is a continuous bijection from these curved coordinates to the t, r, θ, and φ coordinates, which are flat. The curved coordinates are embedded in the flat coordinates. You can do things easily with the flat coordinates that perhaps you can't easily do with the curved coordinates. For instance, you can do the rotation transformation.

Carroll goes on to say:

A vector is a perfectly well-dened geometric object, as is a vector field, defined as a set of vectors with exactly one at each point in spacetime.

I say there is not one, but two sets of vectors defined at each point in spacetime.

(1) There are Δt, Δθ, Δφ, and Δr which are the flat coordinates, quite easily mapped into Cartesian coordinates, which can then be readily rotated or Lorentz Transformed to whatever angle and rapidity you like,

(2) then there are the "curved" components

$$\begin{matrix} \Delta \tau_{\left (r,\theta,\phi\;\mathrm{const} \right )}\\ \Delta s_\left ({r,\theta, t\;\mathrm{const}} \right )\\ \Delta s_\left ({r,\phi, t\;\mathrm{const}} \right )\\ \Delta s_\left ({\theta,\phi, t\;\mathrm{const}} \right ) \end{matrix}$$

It is these curved components which control the local acceleration and behavior of matter. It is within these curved components where you can set dτ=0 and find the geodesic of a photon, for instance, and all of the neat cool stuff you can do with General Relativity.

 

[JDoolin]

Yes, I have to agree with that. If you analyze what the LT does on the basis of its original intent by Voigt and H. A. Lorentz (Voigt's use of the transformation preceded that of Lorentz by several years) then what you say (and more) becomes even clearer. As applied to produce both a change in form of the wave equation and the proper change of initial conditions to accompany the wave equation, it results in initial conditions which can never be reconciled between adjacent points of time or space unless a stretch or shift of all other points is accomplished. But each time that is done the situation remains recursively the same as if you're chasing something that can never be caught. There is no way to make the operation of the LT local unless you freeze the transformation and step back into a Galilean frame of reference.

A Galilean Transformation "warps" space-time like this:
[URL]http://www.spoonfedrelativity.com/web_images/galileantransform.gif[/URL]
where the events move along lines of constant t.

whereas a Lorentz Transformation "warps" space-time like this:
[URL]http://www.spoonfedrelativity.com/web_images/lorentztransform.gif[/URL]
and the events move along hyperbolic arcs of constant $c t^2 - x^2$

I suppose the Galilean Transformation could be said to be "local" if by "local" you mean that it does not affect events which are occurring "now." But it has a big effect on events which occur in the far future or the far past.

And the Lorentz Transformation could be said to be "local" if by "local" you mean that it does not effect THE event which occurs "here" and "now".

(Either way calling a transformation local based on the only events that it doesn't affect seems backward.)

Accelerating toward an event in the future causes it to lean toward you. But accelerating toward an event in the past causes it to lean away.

 

[PeterDonis]

It describes a transformation from dt to dτ. And if written in its negative form, it makes a transformation from ds to a function of dr, dθ, and dφ.

There are at least two conceptual errors here.

First, the metric, which you have written as a line element, is *not* a coordinate transformation equation. It is an equation that tells you how to calculate an actual physical interval along a differential segment of a curve in spacetime, if you know the coordinates and their differentials along that segment. If you integrate it along a worldline, it tells you the physical interval along that worldline; for example, along a timelike worldline, it tells you the proper time experienced by an observer following that worldline. But proper time is *not* a coordinate; it's a physical observable. The t, r, θ, and φ coordinates are *not* observables; they're arbitrary numbers that are used to label events.

Second, you can't split up the line element into a "positive" and "negative" part, as though one part calculates dτ and the other part calculates ds. The line element is a single expression for calculating a single interval. Putting dτ vs. ds on the left-hand side is just a naming convention, corresponding to a sign convention for the metric coefficients; typically, if the LHS is written as dτ it means you are using a timelike sign convention, where timelike squared intervals are positive, so the "time" metric coefficient will be positive and the "space" ones will be negative, whereas if the LHS is written as ds it means you are using a spacelike sign convention, where spacelike squared intervals are positive, so the "space" metric coefficients are positive and the "time" one is negative. But "time" and "space" here are in quotes because not all metrics are diagonal and not all metrics have one timelike and three spacelike coordinates. Anyway, the sign convention is just that, a convention; it doesn't affect the physics. Same for the convention of writing the LHS as dτ or ds.

My suggestion would be, instead of treating t, r, θ, and φ as "meaningless" variables, we should treat them as spherical coordinates which can easily map to Minkowski coordinates, and are in fact, the domain and range of the Lorentz Transformation equation.

You can try to treat them this way, but it won't work. Minkowski coordinates, interpreted in the usual way, require a flat spacetime; they simply won't work the way you want them to work if spacetime is curved. Same for Lorentz transformations.

Again, people will laugh at me, and say that there is no place where matter is not present, so, we have schwarzchild metric embedded in another schwarzschild metric, embedded in another schwarzschild metric. In effect, they will say there is no place in the universe where I can get rid of the curvature altogether. And therefore we might as well not try?

Yes, exactly. Gravity is everywhere, therefore curvature is everywhere.

It's a fallacious argument. We should always be aware that there may be masses and motion that we have not accounted for, and so lines we think are straight may be curved after all. But the question is not whether we can know for certain that lines are straight. The question is whether the straight lines exist at all.

And the answer is, in a universe with gravity, no, they don't.

As long as we have a concept of straightness. Can we not compare one geodesic path to another geodesic path, and unambiguously declare which of those two paths is straighter? Yes, of course. The straight line is the direction that an object would travel if there were no forces acting on it at all. The geodesic is the direction that an object travels when a force is acting on it.

No, this is wrong; in fact, again, there are at least two conceptual errors here.

First, you are misusing the term "geodesic". In a curved manifold, a "geodesic" is the closest thing you can get to a straight line. There aren't any curves that are straighter. And all geodesics are equally straight, so it makes no sense to say one is straighter than another. I realize that lots of people don't like calling these curves "straight lines", because they aren't Euclidean straight lines; but that's why the term "geodesic" was invented, so we could have a name for curves in curved manifolds that are the closest analogues to Euclidean straight lines; since in a curved manifold there are *no* Euclidean straight lines at all, geodesics are the best we can do.

Second, you are misunderstanding the physics. A geodesic is the worldline of an object on which *no* forces are acting, when the term "force" is defined correctly, as something that is actually *felt* by an object (and which can be measured by an accelerometer). You define gravity as a force, but an object moving solely under the influence of gravity is weightless; it feels no force, and an accelerometer attached to it would read zero. You can call the path of this object "curved" if you like, but it's the straightest path there is in a spacetime with gravity present. If you disagree, then please explain, in detail, how we are to *physically* pick out the "real straight lines", the ones compared to which we are supposed to view the path of a freely falling object as curved. Oh, and by the way, saying that "straight lines" are defined by the paths of light rays will *not* work; light is bent by gravity as well. It isn't bent as much because it moves faster than ordinary falling objects, but its path is still bent. (If you ask why light and a falling rock travel on such different paths if they're both geodesics, that's because there are many geodesics through any given event, corresponding to all the possible velocities a freely falling object can have at that event, up to and including the speed of light.)

In real life, we know the difference. But once you start reading a lot of General Relativity books, you start calling geodesics straight lines. And for some reason, you ignore t, r, θ, and φ as "meaningless" variables, and you only look at τ, and say. The object in a geodesic only has τ changing, and is otherwise motionless (with respect to itself).

Well, of course. Are you trying to say an object is moving with respect to itself?

The reason we pick out τ is that it's a physical observable; it's the proper time experienced by an observer traveling on that worldline. You can't observe t, r, θ, and φ directly. Sometimes you can observe things that are very *close* to those coordinates, but that only works in regions where gravity is very weak. You can't generalize to everywhere in a curved spacetime.

There are some great things you can do with that dτ; in particular, setting it equal to zero, to see what the path of a photon through your curved space is. But that path is through the t, r, θ, and φ coordinates. Those coordinates are not meaningless. They are the overlying GLOBAL LORENTZ FRAME COORDINATES IN SPHERICAL FORM, COMOVING WITH THE CENTRAL MASS.

If you really believe this, then please demonstrate how, in a curved spacetime with gravity present, two worldlines with the same constant θ and φ, and values of r that differ by a small amount dr, at a given instant of coordinate time t = 0, will continue to remain separated by that same amount dr for all times t. That is what is physically required for your global Lorentz frame coordinates to be valid. If this requirement is violated (which it is in the presence of gravity), your global Lorentz frame coordinates *will not work* the way you are claiming they do. You can *assign* such coordinates as arbitrary labels, but they will not support the claims about the physics that you appear to be making.

 

[PeterDonis]

If I'm not mistaken, the General Relativity "textbook" answer to this question is (b).

Yes.

For the Schwarzschild metric, in particular, let's discuss the meanings of these partial derivatives

See my last post. The metric is a single expression for a single interval; it is not split up into a piece that calculates dtau and a piece that calculates ds.

What are the textbook definitions of of t, r, θ, and φ?

Depends on the spacetime and the coordinate chart you are using to describe it. There is no single "definition" of any set of coordinates that applies generally in GR. You have to specify the spacetime first, and *then* you have to define how the coordinates label events in the spacetime.

The Wikipedia page actually does a pretty good job of explaining the definitions of the exterior Schwarzschild coordinates, which I think is what you're asking about:

http://en.wikipedia.org/wiki/Schwarzschild_coordinates

I don't know exactly what Carroll's point is here, but to contrast with Carroll, I will say it is important to think of these vectors as stretching between one event and another event, rather than located at a single point.

Carroll's point is that vectors are not objects in spacetime itself; they are abstract objects that "live" in the tangent space at each event.

If you are doing derivatives, with a dt, a dr, a dθ, and a dφ, you can't PUT that at a single point. You can't have a variation in positional variables t, r, θ, and φ "at a single point."

No, but you can have a *rate* of variation at a single point, which is what a derivative describes. More precisely, it describes the limit in the variation between two points as the separation between them goes to zero, divided by the separation--but that means that once the limit is taken, you are effectively "at a single point". Derivatives do not describe change over a finite length; they describe change at a single point.

There is an s-component parallel to the r-direction (space-like interval between events), an s-component parallel to the θ direction (space-like interval between events), and an s-component parallel to the φ direction (space-like interval between events), and another component in the t-direction (time-like interval between events), called τ (tau).

Again, see my previous post. s and tau are not separate. The positive and negative terms in the metric are added *together* to get a single interval, which may be timelike, null, or spacelike.

There are two different coordinate systems here, and they overlap. You have the three unnamed s-components in the same direction as r, θ, and φ, and the named tau component, in the same direction as t.

No, this is wrong. See above.

The three unnamed s-components and tau make up the curved coordinate system. But there is a continuous bijection from these curved coordinates to the t, r, θ, and φ coordinates, which are flat.

Nope. See above. However, there is one other thing worth mentioning here. You *can* construct a continuous bijection between two different sets of coordinates; but if one set is "curved" and the other set is "flat", then the bijection will obviously not preserve distances (or in spacetime, intervals) between events. Since preserving intervals between events is a requirement of a coordinate transformation in physics, any such bijection as you describe will not qualify.

Similar comments to the above on the rest of this post: there are not two sets of coordinates, only one.

 

[PhilDSP]

I suppose the Galilean Transformation could be said to be "local" if by "local" you mean that it does not affect events which are occurring "now." But it has a big effect on events which occur in the far future or the far past.

Without taking the time to search for a more precise definition, I'd have to say that a mathematical model is local if the cause of the change of all parameters is limited in displacement across space to a period of time that is greater than or equal to the minimum propagation time allowed in the model.

A Galilean transformation has no intrinsic rate limitation and would seem to support the instantaneous translation of coordinates. So it would be local so far as the rate change of each parameter is consistent with the particular designation of the limiting velocity of EM propagation, gravity, weak and strong forces separately, for example, which don't need to share a single propagation speed.

Under SR, the propagation or displacement of anything is limited to c. Yet the operation of the Lorentz transformation in performing coordinate translations apparently must occur instantaneously. Maybe that's a moot issue as in thinking about this further, the lack of locality is more related to a general mathematical discontinuity between adjacent points of time and space rather than speed of translation. Whereas adjacent points conforming to a Galilean transformation have a shared linear connection, with the LT they don't share that and rather sort of dance around each other depending on how close you wish to make them.

 

[JDoolin]

There are some great things you can do with that dτ; in particular, setting it equal to zero, to see what the path of a photon through your curved space is. But that path is through the t, r, θ, and φ coordinates. Those coordinates are not meaningless. They are the overlying GLOBAL LORENTZ FRAME COORDINATES IN SPHERICAL FORM, COMOVING WITH THE CENTRAL MASS.

If you really believe this, then please demonstrate how, in a curved spacetime with gravity present, two worldlines with the same constant θ and φ, and values of r that differ by a small amount dr, at a given instant of coordinate time t = 0, will continue to remain separated by that same amount dr for all times t. That is what is physically required for your global Lorentz frame coordinates to be valid. If this requirement is violated (which it is in the presence of gravity), your global Lorentz frame coordinates *will not work* the way you are claiming they do. You can *assign* such coordinates as arbitrary labels, but they will not support the claims about the physics that you appear to be making.

This is obvious. Surely you already know the answer I would give: Upon a non-rotating gravitating sphere, place one object on the floor at radius r. Place another object on a table directly above it at radius r+dr. The two objects will remain on the floor and on the table, maintaining their positions in r, theta, phi, but moving through time.

There are at least two conceptual errors here. First, the metric, which you have written as a line element, is *not* a coordinate transformation equation. It is an equation that tells you how to calculate an actual physical interval along a differential segment of a curve in spacetime, if you know the coordinates and their differentials along that segment. If you integrate it along a worldline, it tells you the physical interval along that worldline; for example, along a timelike worldline, it tells you the proper time experienced by an observer following that worldline. But proper time is *not* a coordinate; it's a physical observable. The t, r, θ, and φ coordinates are *not* observables; they're arbitrary numbers that are used to label events.

In the Schwarzschild coordinates, theta, and phi are "arbitrary" in the following sense: Because the schwarzchild metric is spherically symmetric, you can choose the directions for θ=0 and φ=0 in any direction you like. The r coordinate is zero at the center of mass of the central gravitating body, so in that sense, r is not arbitrary. You might suggest that the scale of r is arbitrary since we can measure this in feet, meters, miles, light-years, whatever. And the time coordinate t, is measured from some arbitrarily chosen event, and you can choose that scale arbitrarily, as well, whether it be seconds, hours, months, or years, whatever.

Illustration of spherical coordinates. The red sphere shows the points with r = 2, the blue cone shows the points with inclination (or elevation) θ = 45°, and the yellow half-plane shows the points with azimuth φ = −60°. The zenith direction is vertical, and the zero-azimuth axis is highlighted in green. The spherical coordinates (2,45°,−60°) determine the point of space where those three surfaces intersect, shown as a black sphere.

Second, you can't split up the line element into a "positive" and "negative" part, as though one part calculates dτ and the other part calculates ds. The line element is a single expression for calculating a single interval. Putting dτ vs. ds on the left-hand side is just a naming convention, corresponding to a sign convention for the metric coefficients; typically, if the LHS is written as dτ it means you are using a timelike sign convention, where timelike squared intervals are positive, so the "time" metric coefficient will be positive and the "space" ones will be negative, whereas if the LHS is written as ds it means you are using a spacelike sign convention, where spacelike squared intervals are positive, so the "space" metric coefficients are positive and the "time" one is negative. But "time" and "space" here are in quotes because not all metrics are diagonal and not all metrics have one timelike and three spacelike coordinates. Anyway, the sign convention is just that, a convention; it doesn't affect the physics. Same for the convention of writing the LHS as dτ or ds.

Let $$ds^2 = -\left ( 1 - \frac{2 G M}{c^2 r} \right )c^2 dt^2 + \left ( 1-\frac{2 G M}{c^2 r} \right )^{-1} dr^2 + r^2 (d\theta^2 + sin^2(\theta)d\varphi^2)$$. If the right-hand-side comes out to be a positive number, then the space-time interval is space-like; meaning there is no way for a clock to move between the two events (t,r,θ,φ) and (t+dt, r+dr, θ+dθ, φ+dφ). But it is possible to stretch a physical object such as a ruler between those two events. On the other hand, if the RHS is negative, then the space-time-interval is time-like, meaning it would be impossible to stretch a ruler between those two events, but it is possible for a clock to be at both of those events at different times. I don't understand what you mean "the sign convention...doesn't affect the physics" If you aren't distinguishing between time-like intervals, and space-like intervals, I can't imagine how you can get any of the other physics right.

You can try to treat them this way, but it won't work. Minkowski coordinates, interpreted in the usual way, require a flat spacetime; they simply won't work the way you want them to work if spacetime is curved. Same for Lorentz transformations.

The t, r, θ, and φ coordinates are a flat spacetime; (well, to be more clear, θ, and φ are obviously curved, but there is no difficulty converting t, r, θ, and φ. of any given event, and converting it to (t,x,y,z) using $$\begin{align*} x &= r cos(\phi) sin(\theta) \\ y &= r sin(\phi) sin(\theta) \\ z &= r cos(\theta) \end{align*}$$ From there, there is no difficulty performing a lorentz transformation on the flat coordinates,

Yes, exactly. Gravity is everywhere, therefore curvature is everywhere.

It is not difficult to establish where gravity is present, and where gravity is less present. If you take any given gravitating body, and move further and further away from it, the geodesics become straighter and straighter. Also, as you increase the velocity of a body the trajectory becomes straighter and straighter. There is, of course a limiting factor that even light is affected by gravity, However, from far away, it is clear that objects deep in a gravitational well do not move in straight lines. Objects that are further from a gravitational well move in much straighter lines. It may be appropriate to say that no body will ever follow a perfect straight line in the universe, because of gravity. It is not appropriate or correct to say straight lines do not exist. What you have is curvature of a SUBgeometry which is present everywhere, but it can easily be mapped one-to-one; event by event onto a flat geometry.

And the answer is, in a universe with gravity, no, they don't.

No, this is wrong; in fact, again, there are at least two conceptual errors here. First, you are misusing the term "geodesic". In a curved manifold, a "geodesic" is the closest thing you can get to a straight line. There aren't any curves that are straighter. And all geodesics are equally straight, so it makes no sense to say one is straighter than another. I realize that lots of people don't like calling these curves "straight lines", because they aren't Euclidean straight lines; but that's why the term "geodesic" was invented, so we could have a name for curves in curved manifolds that are the closest analogues to Euclidean straight lines; since in a curved manifold there are *no* Euclidean straight lines at all, geodesics are the best we can do.

Try to understand that I am not drawing a line ON the curved manifold. Think of it more like drawing a line on the plane in a flat surface "above" the curved manifold. Unless you can find a situation where the curved manifold curves down under itself, there is a one-to-one mapping from the curved coordinate system to an overlying global flat coordinate system.
https://www.physicsforums.com/attachment.php?attachmentid=40656&d=1320321617

It doesn't matter if the underlying surfaces are curved, so long as none of the particles in the underlying surface go faster than the speed of light in the overlying geometry, in which case there would be causality problems.

But if I am 4 light years from Alpha Centauri, and say some black hole passed between Alpha Centauri and here. What would happen? The light from Alpha Centauri would probably be bent in every which direction, and I would know there were a problem. Let's say also, that somehow the black hole warps the space-time so that an extra light-year of rulers (physical observables) could fit between Alpha Centauri and me.

Here is where I think we differ. I would say what you have is a bunch of really short rulers, but the distance is still only four light years. You say you have to follow all of the curves, and so the distance is five light years. I'm saying there is a global flat coordinate system overlying all of your curved space; you do NOT have to follow all of the curves. You're saying there is no global coordinate system, and the only way to measure distance is to place rulers end to end, and look at clocks you are physically carrying, ignoring the presence of distant bodies.

Second, you are misunderstanding the physics. A geodesic is the worldline of an object on which *no* forces are acting, when the term "force" is defined correctly, as something that is actually *felt* by an object (and which can be measured by an accelerometer). You define gravity as a force, but an object moving solely under the influence of gravity is weightless; it feels no force, and an accelerometer attached to it would read zero. You can call the path of this object "curved" if you like, but it's the straightest path there is in a spacetime with gravity present. If you disagree, then please explain, in detail, how we are to *physically* pick out the "real straight lines", the ones compared to which we are supposed to view the path of a freely falling object as curved. Oh, and by the way, saying that "straight lines" are defined by the paths of light rays will *not* work; light is bent by gravity as well. It isn't bent as much because it moves faster than ordinary falling objects, but its path is still bent. (If you ask why light and a falling rock travel on such different paths if they're both geodesics, that's because there are many geodesics through any given event, corresponding to all the possible velocities a freely falling object can have at that event, up to and including the speed of light.)

For the purposes of a Lorentz Transformation, It doesn't matter whether a body "feels" a force. What matters is whether it undergoes a change in velocity. I fully agree that when an object's path curves under a gravitational field, there is little or nothing it experiences internally. But I don't think we should limit our description of physics to beings so obsessed with themselves that they never look anywhere except at the clock on their navel. The fact is, I do not "feel" the motion of my body around the Earth as it revolves. I do not "feel" the motion of the Earth around the sun. I do not "feel" the motion of the sun around the center of the Milky Way. Yet I am aware of all of these motions . I am aware that these motions are not straight lines. The accelerometer reads zero, but it is still obvious that I am changing velocity, and changing my reference frame.


To explain, in detail, how we are to *physically* pick out the "real straight lines", it's not a question of picking out THE real straight line, it is in distinguishing between lines which are obviously not straight and those which are straighter.

For each of these motions, I can draw a picture of the orbital path, and then I can at draw a tangent line that is MUCH straighter than the geodesic. How will I pick out the real straight lines? For one, by using those variables t, r, θ, and φ, converting them to cartesian coordinates by $$\begin{align*} x &= r cos(\phi) sin(\theta) \\ y &= r sin(\phi) sin(\theta) \\ z &= r cos(\theta) \end{align*}$$ and then using the usual method for describing straight lines in Cartesian Coordinates.

Well, of course. Are you trying to say an object is moving with respect to itself?

No. What I'm saying is that putting a falling rock in a box that is also falling does not make it move in a straight line. You think that the closest you can get to a straight line is by throwing a rock. That rock is not going in a straight line. It goes up and falls. Ahh, but now you say "Put the rock in a box. See now, in the box the rock is going in a straight line, relative to the box. Ergo, Hence, Therefore, straight line. QED" or From inside the box, the rock appears stationary. Therefore, the rock is only traveling through time. Hence a geodesic arc is a straight line. So by putting the rock in a box, now you think the rock is going in a straight line? The rock is NOT going in a straight line. It is just now that you have both rock and box moving in an arc, and because it is (roughly) the same geodesic, it looks like a straight line.

All I'm saying is, don't put the rock in the stupid box. Don't ignore the presence of gravity. Don't put me in a box and throw the box, because regardless of how it "feels" like I must be floating in free space, that doesn't change the fact that I'm going to hit the ground hard, and frankly hitting the ground IS a physical observable.

The reason we pick out τ is that it's a physical observable; it's the proper time experienced by an observer traveling on that worldline. You can't observe t, r, θ, and φ directly. Sometimes you can observe things that are very *close* to those coordinates, but that only works in regions where gravity is very weak. You can't generalize to everywhere in a curved spacetime.

This tyranny of "observables" is overrated. You limit your notion of distance and time to the actual readings of clocks, and the lengths of rulers that you can physically touch. I watch someone throw a rock and it follows a geodesic. To me that path looks curved. You claim that it is not curved, because if we put the rock in a box that were traveling along with it, also following a geodesic, and rulers in the box, then it would appear from inside the box that the rock was not moving; or might be following a straight line.
The point is that it does not matter whether t, r, θ, and φ can be measured directly. We don't actually need to put clocks and rulers out there to get a pretty good idea where the sun is, where the moon is, where Jupiter is, where the center of the galaxy is. (or was when the light left it). The point is, any nonlocal measurement of distance (i.e. the estimation of distance by sight or parallax, or luminosity.)


My point is, it does not make sense to constrain your physics to only things you can touch. You need to think about far-away objects as well. When discussing far-away objects, the distances are not determined by following every curve of the gravitational manifold, but are instead determined by an overlying global flat coordinate system.

 

[JDoolin]

Without taking the time to search for a more precise definition, I'd have to say that a mathematical model is local if the cause of the change of all parameters is limited in displacement across space to a period of time that is greater than or equal to the minimum propagation time allowed in the model.

A Galilean transformation has no intrinsic rate limitation and would seem to support the instantaneous translation of coordinates. So it would be local so far as the rate change of each parameter is consistent with the particular designation of the limiting velocity of EM propagation, gravity, weak and strong forces separately, for example, which don't need to share a single propagation speed.

Under SR, the propagation or displacement of anything is limited to c. Yet the operation of the Lorentz transformation in performing coordinate translations apparently must occur instantaneously. Maybe that's a moot issue as in thinking about this further, the lack of locality is more related to a general mathematical discontinuity between adjacent points of time and space rather than speed of translation. Whereas adjacent points conforming to a Galilean transformation have a shared linear connection, with the LT they don't share that and rather sort of dance around each other depending on how close you wish to make them.

Well, whether we call it local or global, let's ask what the rotation transformation does, first of all.

Let's say I turn a little to the right, in such a way that the objects 10 feet in front of me move 1 foot to the left. Objects 1000 miles in front of me move 100 miles to the left. Objects a billion light years in front of me move 100 million light years to the left. The point is, all I did was local; turning to the right; but the transformation affects things in direct proportion to their distance. The further away the thing is, the further it moves.

The same can be said for the Lorentz Transformations. Events which are furthest away from the observer undergoing a velocity change are affected the most by the transformation.

In either case, I'm only changing a single local variable, for rotation the local variable is which direction I'm facing. For Lorentz Transformation, that local variable is my rapidity, or my velocity. However, the transformation of coordinates is global.

 

[PeterDonis]

This is obvious. Surely you already know the answer I would give: Upon a non-rotating gravitating sphere, place one object on the floor at radius r. Place another object on a table directly above it at radius r+dr. The two objects will remain on the floor and on the table, maintaining their positions in r, theta, phi, but moving through time.

Ah, I see I left out a key word: "freely falling". Neither of the objects in question are freely falling.

I don't understand what you mean "the sign convention...doesn't affect the physics" If you aren't distinguishing between time-like intervals, and space-like intervals, I can't imagine how you can get any of the other physics right.

I meant that the sign convention that spacelike squared intervals are positive, and timelike ones are negative, is just a convention; you can flip it around, so that timelike squared intervals are positive and spacelike ones are negative, without affecting the physics, provided you change all the formulas appropriately.

The t, r, θ, and φ coordinates are a flat spacetime

As it stands, this statement is at best wrong, and at worst not even wrong. Flatness or curvature is an invariant feature of the underlying geometry, and it is there and can be talked about without even assigning coordinates. If you do assign coordinates, then you figure out whether the underlying geometry is flat or curved by computing the Riemann curvature tensor in those coordinates, using the proper form of the metric in those coordinates for the manifold in question.

When you do this using standard Schwarzschild coordinates on Schwarzschild spacetime, using the expression for the metric on Schwarzschild spacetime in those coordinates, which you have written down several times now, you find that the Riemann curvature tensor is non-zero: i.e., the spacetime is curved.

If you do this using the same Latin and Greek letters, t, r, θ, and φ, for coordinates on Minkowski spacetime, using the expression for the metric on Minkowski spacetime in those coordinates, you find that the Riemann curvature tensor is zero; the spacetime is flat.

It's important to recognize that "curvature" as I've just described it is *intrinsic* curvature of the manifold, since that's what we physically observe. See further comments below.

It may be appropriate to say that no body will ever follow a perfect straight line in the universe, because of gravity. It is not appropriate or correct to say straight lines do not exist.

Yes, it is, because if no body ever follows a perfect "straight line" as you are using the term, how can such a line possibly exist? There's no physical observable that picks it out.

What you have is curvature of a SUBgeometry which is present everywhere, but it can easily be mapped one-to-one; event by event onto a flat geometry.

Try to understand that I am not drawing a line ON the curved manifold. Think of it more like drawing a line on the plane in a flat surface "above" the curved manifold.

In other words, I am supposed to believe that you can draw lines that go outside our universe, taking the "straight" route between two events instead of the "curved" route inside our universe that physical objects actually take. Can you see why I'm a bit skeptical?

Unless you can find a situation where the curved manifold curves down under itself, there is a one-to-one mapping from the curved coordinate system to an overlying global flat coordinate system.

Does the mapping preserve the metric? No. You admit as much later on when you try to claim that the "real" distance to Alpha Centauri remains the same if a black hole comes between there and Earth, even though every actual physical object has to travel a greater distance because of curvature. If the mapping doesn't preserve the metric, then it's physically meaningless.

It doesn't matter if the underlying surfaces are curved, so long as none of the particles in the underlying surface go faster than the speed of light in the overlying geometry, in which case there would be causality problems.

Check out the "river model" of black holes:

http://arxiv.org/abs/gr-qc/0411060

Inside the horizon, freely falling objects do go faster than light relative to the "river bed", which is basically what you mean by the flat geometry that everything is supposedly mapped to. But as the authors of the paper make clear, the "river bed" is not physical and has no physical meaning; it's only used to help with visualization. Also, the authors make clear that the ability to construct a model with a flat "river bed", even just for visualization, is a special feature of Schwarzschild geometry, and cannot be done in general for a curved spacetime.

I'm saying there is a global flat coordinate system overlying all of your curved space; you do NOT have to follow all of the curves.

Show me a physical object that doesn't have to follow all of the curves.

You're saying there is no global coordinate system, and the only way to measure distance is to place rulers end to end, and look at clocks you are physically carrying, ignoring the presence of distant bodies.

Yes. If you disagree, again, show me a physical object that doesn't have to stay within the curved manifold. The distant bodies do affect the observations indirectly, by affecting the curvature, but we can measure the curvature without even knowing the distant bodies are there, as long as we measure the curvature over a region that's not too small for our measurement accuracy.

For the purposes of a Lorentz Transformation, It doesn't matter whether a body "feels" a force. What matters is whether it undergoes a change in velocity.

Velocity relative to what?

The fact is, I do not "feel" the motion of my body around the Earth as it revolves. I do not "feel" the motion of the Earth around the sun. I do not "feel" the motion of the sun around the center of the Milky Way. Yet I am aware of all of these motions . I am aware that these motions are not straight lines. The accelerometer reads zero, but it is still obvious that I am changing velocity, and changing my reference frame.

Relative to what?

To explain, in detail, how we are to *physically* pick out the "real straight lines", it's not a question of picking out THE real straight line, it is in distinguishing between lines which are obviously not straight and those which are straighter.

Relative to what? You keep saying all these things when you have basically admitted that there is *no* physical observable to refer them to. It just "looks to you" like one set of lines is straighter than another. Whatever this is, it isn't physics.

For each of these motions, I can draw a picture of the orbital path, and then I can at draw a tangent line that is MUCH straighter than the geodesic.

I suppose I should note here, once again, that we are talking about paths in *spacetime*, not space. But you can also "draw a tangent line" in spacetime, so I'm ok with what you are visualizing here at this point. However:

How will I pick out the real straight lines? For one, by using those variables t, r, θ, and φ, converting them to cartesian coordinates by $$\begin{align*} x &= r cos(\phi) sin(\theta) \\ y &= r sin(\phi) sin(\theta) \\ z &= r cos(\theta) \end{align*}$$ and then using the usual method for describing straight lines in Cartesian Coordinates.

And you will find that the lines are *not* "straight", in the sense that they will violate the theorems of Euclidean geometry--specifically, the ones that depend on the parallel postulate. The metric in these coordinates will *not* be the Euclidean (in space) or Minkowski (in spacetime) metric.

You appear to believe that the metric doesn't matter: that you can project the curved manifold into a flat manifold, like projecting the Earth's surface onto a flat projection like a Mercator projection, distorting all the distances (and times, in spacetime), and everything will still be just fine. You are simply ignoring actual physical observables when they don't match up with your mathematical scheme. Whatever this is, it isn't physics.

No. What I'm saying is that putting a falling rock in a box that is also falling does not make it move in a straight line.

I am not saying any of the things you think I am saying. I am saying this: in GR, the theory of *physics*, we *define* "straight" motion as freely falling motion. That is our *physical* standard for mapping out the geometry of spacetime. When we look at the behavior of freely falling geodesic worldlines, we find that the geometry they map out is intrinsically curved, and we have an equation, the Einstein Field Equation, that relates that curvature to the presence of matter-energy. In other words, what we call "gravity" is a manifestation of spacetime curvature, *in this theory of physics*.

If you want to put together an alternate theory of physics that uses a flat background and goes through all the contortions to reproduce the predictions of GR, go ahead. But there's no point in wrangling over the meanings of words. If you don't like putting the labels "flat" or "curved" where I am putting them, fine, substitute your own labels. The physics remains the same. You can't change the physics by asserting that a certain set of coordinates is "flat" because you say so, even though you have to distort all the distances and times to make things fit into those coordinates, so that your model loses all connection to what we actually observe.

All I'm saying is, don't put the rock in the stupid box. Don't ignore the presence of gravity. Don't put me in a box and throw the box, because regardless of how it "feels" like I must be floating in free space, that doesn't change the fact that I'm going to hit the ground hard, and frankly hitting the ground IS a physical observable.

Where have I said the rock won't hit the ground? You are reading an awful lot into what I'm saying, that I haven't said. I've never said the rock isn't moving relative to the ground.

This tyranny of "observables" is overrated.

And you would substitute what? Apparently a bunch of unobservables that throw away the observables.

You limit your notion of distance and time to the actual readings of clocks, and the lengths of rulers that you can physically touch.

As the ultimate *basis* for distances and times, yes. Obviously you can build accounts of distances and times over extended regions on this basis. If I want to know how much proper time will elapse along the Earth's worldline from now until Christmas, I can calculate it based on local clock and ruler readings.

I watch someone throw a rock and it follows a geodesic. To me that path looks curved. You claim that it is not curved, because if we put the rock in a box that were traveling along with it, also following a geodesic, and rulers in the box, then it would appear from inside the box that the rock was not moving; or might be following a straight line.

No, I claim the rock's path is not curved because it feels no force. Again, that is a *definition* I adopt because I am using a particular theory of physics. You are free to adopt a different definition, as long as you contort your model appropriately. But you'll have to expect some skepticism when you explicitly state that the distances and times we actually measure are not the "real" ones.

When discussing far-away objects, the distances are not determined by following every curve of the gravitational manifold, but are instead determined by an overlying global flat coordinate system.

Really? Give some examples, please. I certainly agree there are ways of obtaining observables for distant objects without stretching rulers from here to there. But to relate any of those observables to a "distance" requires a theoretical model that gives you the relationship. The theoretical relationships used in actual cosmology are based on GR, so they are based on distances that "follow every curve" of the manifold.

 

[Passionflower]

JDoolin, the Earth goes around the Sun just as much as the Sun goes around the Earth, it is simply a matter of relativity.

 

[JDoolin]

Yes.



See my last post. The metric is a single expression for a single interval; it is not split up into a piece that calculates dtau and a piece that calculates ds.



Depends on the spacetime and the coordinate chart you are using to describe it. There is no single "definition" of any set of coordinates that applies generally in GR. You have to specify the spacetime first, and *then* you have to define how the coordinates label events in the spacetime.

The Wikipedia page actually does a pretty good job of explaining the definitions of the exterior Schwarzschild coordinates, which I think is what you're asking about:

http://en.wikipedia.org/wiki/Schwarzschild_coordinates

The question I am asking is what are the textbook definitions of t, r, θ, and φ. The article describes a "coordinate singularity" at t=t0, r=r0, θ=0, and φ undefined being the north pole, and t=t0, r=r0, θ=Pi, and φ undefined being the south pole. That's fine. That is precisely what I meant by them. But the variable r extends from 0 to infinity, the variable θ from 0 to Pi, and φ from 0 to 2 Pi. The domain of r goes from 0 to infinity, even though you have a coordinate singularity at r=r0. This is the geometry of a black hole, and there is a certain radius, called the schwarzschild radius or the event horizon where light will fall in. My point is that the (t,r,θ,φ) variables are not meaningless. They represent position and time of a global flat coordinate system in spherical coordinates centered at (t=t0, r=0).

Carroll's point is that vectors are not objects in spacetime itself; they are abstract objects that "live" in the tangent space at each event.



No, but you can have a *rate* of variation at a single point, which is what a derivative describes. More precisely, it describes the limit in the variation between two points as the separation between them goes to zero, divided by the separation--but that means that once the limit is taken, you are effectively "at a single point". Derivatives do not describe change over a finite length; they describe change at a single point.

In order to have a rate of change, you have to allow variables to change. A derivative is not a change at a single point. It is the "limit" as the distance approaches zero. If you actually try to describe the change at a single point, it is 0/0, which is undefined. A "limit" in math means playing the game of whatever number you can give, I can give one smaller, but LARGER than zero. Once you actually say the denominator is dx=zero, then it is a division by zero error.

Again, see my previous post. s and tau are not separate. The positive and negative terms in the metric are added *together* to get a single interval, which may be timelike, null, or spacelike.

I misspoke. Obviously, (t,r, θ, φ) are four parameters while $$s^2 = - c^2 d\tau^2$$ is but a single parameter. So let me clarify and correct so that you may understand what I meant: If the value of s is imaginary, then it represents the time between time-like-separated events. I was thinking of the time on the clock being one of your "physical observables" and thinking there was only one physical observable there. But if we have a clock going between these two events, then the velocity of the clock has a component in the r direction, the θ direction and the φ direction; (I'm not sure if you count these components of velocity in your "physical observables" or not, but I did neglect to think of them when I made my earlier post). If the value s is real, it represents the distance between two events. (The events may have different values of t, in which case they are simultaneous only in some non-comoving reference frame. This is a more complicated problem and we would have to discuss the orientation of the ruler as well as its velocity.) But if the t-values are the same, then the orientation of the ruler has a component in the r direction, the theta direction, and the phi direction. Since you've got a ruler there, and its length is one of your "physical observables", the different components of direction are also "physical observables"

What I am saying is that the (t,r, θ, φ) coordinates overlap with what you call "physical observables" coordinates. See below.

No, this is wrong. See above.



Nope. See above. However, there is one other thing worth mentioning here. You *can* construct a continuous bijection between two different sets of coordinates; but if one set is "curved" and the other set is "flat", then the bijection will obviously not preserve distances (or in spacetime, intervals) between events. Since preserving intervals between events is a requirement of a coordinate transformation in physics, any such bijection as you describe will not qualify.

But the Schwarzschild metric does NOT preserve distances. It does NOT preserve space-time intervals. Are you arguing that the Schwarzschild metric is invalid? The Schwarzschild metric does NOT preserve the speed of light. It makes the light travel slower and slower as it goes down toward the Schwarzschild radius. What Schwarzschild metric DOES do is maintain the local measure of distance and the local measure of time, so that locally it looks like the speed of light is preserved. But globally, the speed of light is not preserved. From the global perspective, the schwarzschild metric causes the light to slow, and turn and bend, not unlike the way a transparent crystal causes light to slow and bend.But the Schwarzschild metric does NOT preserve distances. It does NOT preserve space-time intervals. Are you arguing that the Schwarzschild metric is invalid? The Schwarzschild metric does NOT preserve the speed of light. It makes the light travel slower and slower as it goes down toward the Schwarzschild radius. What Schwarzschild metric DOES do is maintain the local measure of distance and the local measure of time, so that locally it looks like the speed of light is preserved. But globally, the speed of light is not preserved. From the global perspective, the schwarzschild metric causes the light to slow, and turn and bend, not unlike the way a transparent crystal causes light to slow and bend.

Similar comments to the above on the rest of this post: there are not two sets of coordinates, only one.

Which of the two sets of coordinates are you referring to as the only set of coordinates? In the schwarzcshild metric there is a coordinate system composed of (t, r, θ, φ), but then you keep referring to "observables" being the lengths of local rulers, and the times on local clocks. one could construct a sort of pseudo-coordinate system where you take as your measure of time the times on local clocks, and take as your measure of distance the lengths marked off on local rulers. In this pseudo-coordinate system, the local speed of light is constant. This pseudo-coordinate system is curved, since the rulers at different locations are different lengths, and the clocks at different locations have different rates. If you want to construct a coordinate system strictly from "observables" defined as lengths of local rulers, and you get a curved coordinate system.

However, in the schwarzschild metric, we also have (t,r,θ, φ) which we both agree do not mark off the lengths of local rulers or the times on local clocks. I think we can both acknowledge, at least, that the coordinates (t,r, θ, φ) are NOT the physical "observables" that you've been describing, that may help our communication. So to help me understand your point here, explain which coordinate system you mean; the one made of (t,r,θ, φ) or the one made up of local rulers and clocks?

When using these parameters (t, r, θ, φ) it is helpful to discuss many different derivatives and partial derivatives, in sort of a similar way as to when in thermodynamics, functions of (pressure, volume, temperature, chemical potential, and total energy) etc. used. The idea is you can hold certain parameters constant, and allow other parameters to vary, and find helpful formulas for isothermal, isobaric, adiabatic reactions, and many more. In particular, you can say "hold r, θ, φ constant, and find dtau/dt as a function of r." You can set ds^2=0 and hold theta and phi constant, and find dr/dt as a function of r, to find the velocity of a photon falling into the Schwarzschild geometry. You can set ds^2=0 and hold theta =Pi/2, and find a relationship between radial velocity, dr/dt and angular velocity dphi/dt, and this is a good start to working out the arc that a photon will follow in the schwarzschild geometry. I have listed the questions below, along with my approach to getting the answers, so you can understand exactly what I mean.

Exercise 1. Assume a photon is moving around a gravitating mass in the equatorial plane. Use the Schwarzschild metric to find a relationship between radial velocity, $v_r=dr/dt$ and tangential velocity $v_t=r dφ/dt$ .

Answer: Sed ds=0, for a photon, and Set θ=Pi/2 for the equatorial plane, then Schwarzschild metric simplifies to $$0 =\left( 1-\frac{r_s}{r} \right )c^2 dt^2+\left( 1-\frac{r_s}{r} \right )^{-1}dr^2+r^2 d\phi^2$$ Divide by dt^2. $$\begin{align*} 0 &=-\left( 1-\frac{r_s}{r} \right )c^2 +\left( 1-\frac{r_s}{r} \right )^{-1}\left (\frac{dr}{dt} \right )^2+\left (r \frac{d\phi}{dt} \right )^2 \\ 0 &=-\left( 1-\frac{r_s}{r} \right )c^2 +\left( 1-\frac{r_s}{r} \right )^{-1} v_r^2+v_t^2 \\ v_t^2 &=\left( 1-\frac{r_s}{r} \right )c^2 -\left( 1-\frac{r_s}{r} \right )^{-1} v_r^2 \\ v_t &=\pm \sqrt{\left( 1-\frac{r_s}{r} \right )c^2 -\left( 1-\frac{r_s}{r} \right )^{-1} v_r^2} \end{align*}$$

Exercise 2. Find the tangential velocity of a photon as it makes its closest approach to a gravitating body.

Answer: Using the results from previous problem, but dr/dt=0; $$v_t =\pm c \sqrt{1-\frac{r_s}{r} }$$


Exercise 3. Find the rate of time tau of a clock sitting on a non-rotating surface at radius r, as compared to the global coordinate time t.

Answer: In this case, dr=dθ =dφ = 0, since the position of the clock is not changing. The Schwarzschild metric simplifies to $$\begin{align*} c^2 d\tau^2&=\left ( 1-\frac{r_s}{r} \right )c^2 dt^2 \\ \frac{d\tau}{dt}&= \pm c \sqrt{ 1-\frac{r_s}{r}} \end{align*}$$


Exercise 4; Assume a photon is moving directly toward a gravitating mass. Use the Schwarzschild metric to find dr/dt, the radial velocity of the photon as a function of r.

Answer: $$ds^2 =- \left ( 1 - \frac{2 G M}{c^2 r} \right )c^2 dt^2 + \left ( 1-\frac{2 G M}{c^2 r} \right )^{-1} dr^2 + r^2 (d\theta^2 + \sin^2(\theta)d\varphi^2)$$ ds=0 because it is a photon. dθ=dφ=0 because it is coming straight in. Let $$r_s=\frac{2 G M}{c^2}$$ Then the equation simplifies to $$\begin{align*} 0 &=-\left ( 1-\frac{r_s}{r} \right )c^2 dt^2+\left( 1-\frac{r_s}{r} \right )^{-1}dr^2 \\ \left ( 1-\frac{r_s}{r} \right )c^2 dt^2 &= \left( 1-\frac{r_s}{r} \right )^{-1}dr^2\\ \left (\frac{dr}{dt} \right )^2 &=\left( 1-\frac{r_s}{r} \right )^2 c^2 \\ \frac{dr}{dt}&= \pm \left ( 1-\frac{r_s}{r} \right )c \end{align*}$$

 

[Passionflower]

But the variable r extends from 0 to infinity, the variable θ from 0 to Pi, and φ from 0 to 2 Pi. The domain of r goes from 0 to infinity, even though you have a coordinate singularity at r=r0.

Show me one credible textbook where the validity of r in the Schwarzschild coordinate chart is stated as 0 to infinity.

My point is that the (t,r,θ,φ) variables are not meaningless. They represent position and time of a global flat coordinate system in spherical coordinates centered at (t=t0, r=0).

They are certainly not meaningless however r in the Schwarzschild coordinate chart does not represent a position. In fact r is a measure of Gaussian curvature.

So let me clarify and correct so that you may understand what I meant: If the value of s is imaginary, then it represents the time between time-like-separated events.

When do you think the value of s becomes imaginary in the Schwarzschild coordinate chart?
Within the valid range of r in the Schwarzschild coordinate chart the value is never imaginary.

 

[PeterDonis]

The question I am asking is what are the textbook definitions of t, r, θ, and φ.

You have the definitions of the angular coordinates correct. The definition of the radial coordinate r is that a 2-sphere at radial coordinate r has physical area $4 \pi r^{2}$; because of the coordinate singularity at r = 2M, the range of r is *not* 0 to infinity. Rather, there are two separate Schwarzschild charts, an exterior one where r > 2M, and an interior one where 0 <= r < 2M. The Schwarzschild coordinates are not valid at r = 2M; to analyze what is happening at r = 2M, you need to either switch charts or take limits as r -> 2M from one side or the other (depending on which chart you are in), if those limits are well-defined. For the rest of this post, I will only talk about the exterior chart, where r > 2M.

The definition of the time coordinate t in the exterior chart is, roughly speaking, that it is the proper time experienced by an observer at r = infinity. (More precisely, it's the limit of the proper time of an observer at r as r -> infinity.)

My point is that the (t,r,θ,φ) variables are not meaningless. They represent position and time of a global flat coordinate system in spherical coordinates centered at (t=t0, r=0).

No, not a flat coordinate system; an asymptotically flat coordinate system. Big difference.

In order to have a rate of change, you have to allow variables to change. A derivative is not a change at a single point. It is the "limit" as the distance approaches zero. If you actually try to describe the change at a single point, it is 0/0, which is undefined. A "limit" in math means playing the game of whatever number you can give, I can give one smaller, but LARGER than zero. Once you actually say the denominator is dx=zero, then it is a division by zero error.

I don't see the point of debating elementary calculus here. When I compute a value for a derivative, such as $dx^{a} / d\tau$, I take a limit as $x^{a}$ and $\tau$ approach the values that they take at a particular event. Once I've taken the limit, I call it the derivative at that event. If you don't like that nomenclature, feel free to invent your own terminology. It doesn't change the math or the physics.

if we have a clock going between these two events, then the velocity of the clock has a component in the r direction, the θ direction and the φ direction...Since you've got a ruler there, and its length is one of your "physical observables", the different components of direction are also "physical observables"

You are misstating how the physical observables are observed. If a clock is going between two events, the physical observables are the clock readings at each event. Those are independent of the velocity of the clock, because the velocity of the clock is frame-dependent and the clock readings are not. If I am moving relative to the clock, I need to know its velocity relative to me to *calculate* what I predict the clock's readings to be at the two events; but that calculation does not affect the actual readings.

Same with the ruler: if I want to measure the distance between two events, I put a ruler between the events; in order to do this at all, the ruler needs to be in the same surface of simultaneity as both events. The readings of the two ruler marks at each event then give the distance, and that observable is frame-independent; but the "velocity" of the ruler is not, and does not affect the readings. If I am moving relative to the ruler, again, I need to know its velocity relative to me to calculate what I predict its readings to be, but that doesn't affect the actual readings.

But the Schwarzschild metric does NOT preserve distances. It does NOT preserve space-time intervals.

Okay, you appear to be extremely confused about what a metric *means*. The metric *is* the expression for space-time intervals; to say that it does not preserve space-time intervals is not even wrong.

The Schwarzschild metric does NOT preserve the speed of light. It makes the light travel slower and slower as it goes down toward the Schwarzschild radius.

Slower and slower relative to the *coordinates*. Not slower and slower relative to observers who are actually there where the light is. To observers who are there where the light is, the light moves at c. You are mistaking an artifact of coordinates for an actual physical observable. Try looking at the velocity of ingoing light in Eddington-Finkelstein or Painleve coordinates; it doesn't go to zero as r -> 2M.

If the light really was slowing down, physically, as it approached r = 2M, then you would be able to find a coordinate-independent quantity that expressed its "velocity" and show that that quantity approached zero as r -> 2M. Can you find such a quantity? If not, then you are simply wrong when you say the light slows down, just because its coordinate velocity in a particular chart goes to zero.

What Schwarzschild metric DOES do is maintain the local measure of distance and the local measure of time, so that locally it looks like the speed of light is preserved...It does NOT preserve space-time intervals.

Again, the second sentence is not even wrong. The first sentence is more or less correct. But in order to even *define* a "space-time interval" in a curved spacetime outside a small local region, you have to first define along what curve you are going to measure the interval. Then you have to integrate the metric along that curve, summing up all the small contributions from each local piece of it, using the line element expression (the Schwarzschild line element in this case), to arrive at the total length of the curve. This is no different than the same calculation along a curve on a curved surface in ordinary geometry. This calculation *defines* what space-time intervals are, so to say that the metric does not preserve them is, again, not even wrong.

Perhaps you are confused by the fact that, in flat spacetime, where SR is globally valid, it is easy to gloss over the fact that spacetime intervals are defined along specific curves, because there are obvious curves that are "privileged" relative to all others: the worldlines of inertial (i.e., freely falling) observers, and the lines (or surfaces, if we are including all three spatial dimensions) of simultaneity orthogonal to those worldlines. A set of such worldlines that are all parallel, and the set of surfaces orthogonal to the entire set of such worldlines, is one way of specifying a global inertial frame in SR. The standard definition of an interval, using the Minkowski metric in differential form...

$$d\tau^{2} = dt^{2} - dr^{2} - r^{2} d\Omega^{2}$$

...then integrates trivially over the "privileged" curves, so that a finite interval can be written (assuming purely radial motion) as:

$$\tau^{2} = t^{2} - r^{2}$$

But this is only justified because the differential line element has the Minkowski form, and because we are using coordinates tied to a global inertial frame. If spacetime is curved, the differential line element cannot in general be made to take the Minkowski form (you can do it, at best, at a single event), so you have to use the more general procedure I described above to obtain actual physical distances and times. (Even in flat spacetime, the differential line element is more complicated if we use non-inertial coordinates, such as Rindler coordinates, so we have to use the more general procedure to compute intervals in those coordinates.)

Which of the two sets of coordinates are you referring to as the only set of coordinates? In the schwarzcshild metric there is a coordinate system composed of (t, r, θ, φ), but then you keep referring to "observables" being the lengths of local rulers, and the times on local clocks.

Coordinates are arbitrary sets of numbers that are used to label events; in GR, as long as the coordinates meet minimal requirements of continuity and differentiability, you can assign them any way you want. Obviously there are practical advantages in picking coordinates that fit well with the particular shape of the geometry you are trying to describe, which is why, for example, we use angular coordinates on 2-spheres instead of Cartesian ones. But that's a calculational convenience only and doesn't affect the physics.

Observables are numbers that are independent of any coordinate system; they describe actual physical observations like the readings on clocks or the lengths between marks on rulers. You can use coordinates to calculate predictions for these observables, but the observables themselves are not coordinates and can be defined independently of all coordinates. You can describe all the physics entirely in terms of observables, without ever mentioning coordinates at all.

one could construct a sort of pseudo-coordinate system where you take as your measure of time the times on local clocks, and take as your measure of distance the lengths marked off on local rulers. In this pseudo-coordinate system, the local speed of light is constant.

Yes, you can construct a local chart this way, and it will cover a small local region. But that's all.

This pseudo-coordinate system is curved, since the rulers at different locations are different lengths, and the clocks at different locations have different rates. If you want to construct a coordinate system strictly from "observables" defined as lengths of local rulers, and you get a curved coordinate system.

Now you are talking about something different, a global coordinate system that tries to "fit together" the various local ones, although how exactly you propose to do the fitting is vague. If you think I have been talking about such a system in any of my posts, then you have been misunderstanding me. The Schwarzschild coordinate system is *not* such a system.

However, in the schwarzschild metric, we also have (t,r,θ, φ) which we both agree do not mark off the lengths of local rulers or the times on local clocks. I think we can both acknowledge, at least, that the coordinates (t,r, θ, φ) are NOT the physical "observables" that you've been describing, that may help our communication.

Yes, you are right, the coordinates are *not* the physical observables.

So to help me understand your point here, explain which coordinate system you mean; the one made of (t,r,θ, φ) or the one made up of local rulers and clocks?

Mostly I have been talking about the Schwarzschild exterior coordinates, (t, r, θ, φ). Sometimes I have mentioned the strictly local system using readings from local rulers and clocks, but I haven't talked at all about trying to "extend" such a system beyond the small local region around a given event in which spacetime curvature effects are small enough to be ignored.

I have listed the questions below, along with my approach to getting the answers, so you can understand exactly what I mean.

As far as I can tell, these calculations are correct. And they provide a good illustration of the difference between a coordinate-dependent quantity and an observable. Exercises 1, 2, and 4 are calculating coordinate-dependent quantities. Exercise 3 is calculating an observable. (I would prefer "gravitational redshift/blueshift for a static observer at radius r" to "rate of time" as the name for this observable, since the redshift/blueshift is how it would actually be measured.) Since this post is getting long, I'll add a separate post discussing the difference in more detail.

 

[PeterDonis]

Okay, we have the four quantities from post #25, and I said in my last post that #1, #2, and #4 are coordinate-dependent, but #3 is a coordinate-independent observable. How do we tell the difference?

First, we note that, as I said at the end of my last post, the way we actually measure #3 is by the gravitational redshift/blueshift for light that is traveling radially between two observers who are "hovering" at a constant radial coordinate r (and are also not changing angular coordinates). So the observable we want to calculate is the observed redshift/blueshift of a light ray emitted at r very large (we'll use the limit as r -> infinity), when it is observed at a finite radius r. We'll see that this comes out to be the same expression as #3.

The law of physics we need is that the gravitational redshift/blueshift is given by the inner product of the light ray's 4-velocity vector with the observer's 4-velocity vector. (Purists might object to the term "4-velocity" in reference to the 4-vector describing a light ray, since such a vector must be null and some definitions of the term "4-velocity" require it to be a unit vector; that's a matter of terminology, not physics, so I'm ignoring it here.) The 4-velocity of the observer static at radius r is (listing the components in order t, r--the angular coordinates can be ignored for this entire problem since all motion is purely radial):

$$u^{a} = \left( u^{t}, u^{r} \right) = \left( \frac{1}{\sqrt{1 - \frac{2M}{r}}}, 0 \right)$$

You may wonder why $u^{t}$ is not simply 1, as it would be for an observer at rest in flat spacetime. For a timelike observer, 4-velocity is a unit vector, as I noted just now, so its norm has to be 1:

$$u = g_{ab} u^{a} u^{b} = 1$$

Since only the t component is nonzero, g_tt is the only metric coefficient that matters in the norm above, and you can see that for the norm to be 1, u^t must be as I have written it.

The light ray's 4-vector is more complicated, but fortunately the key calculation is already done; it's simply exercise 4! That exercise tells us how the radial coordinate of the photon changes with coordinate time, which is the r-component of its 4-velocity; and since the norm of the photon's 4-velocity is zero, we can compute the t-component by setting the norm equal to zero and plugging in the r-component and the metric coefficients g_rr and g_tt. The result is:

$$k^{a} = \left( k^{t}, k^{r} \right) = \left( 1, 1 - \frac{2M}{r} \right)$$

The inner product of the two vectors, which is the observable we seek, is then simply:

$$ku = g_{ab} k^{a} u^{b} = g_{tt} k^{t} u^{t} = \sqrt{1 - \frac{2M}{r}}$$

which agrees with the expression given in exercise #3. But note how this expression is written: it is a contraction of the metric with two vectors, i.e., it is a scalar invariant. That means it is independent of the coordinate system; if I calculate the same invariant in any coordinate system, I will get the same answer. For example, try the same calculation in Painleve coordinates; here the metric line element is:

$$ds^{2} = - \left( 1 - \frac{2M}{r} \right) dT^{2} + 2 \sqrt{\frac{2M}{r}} dT dr + dr^{2} + r^{2} d\Omega^{2}$$

The 4-velocity of an observer at constant r is the same, because g_tt is the same; but the radial component of the 4-velocity of the photon is different; a similar procedure to exercise #4 gives:

$$\frac{dr}{dT} = - 1 - \sqrt{\frac{2M}{r}}$$

for an ingoing photon. We can then find the t-component of the photon's 4-velocity as before, by setting the norm equal to zero and substituting the metric coefficients and the r-component. We find that the t-component is again 1, the same as before (the change in the metric is just right to offset the change in the r-component), and that means the inner product of the photon's 4-velocity with the static observer's 4-velocity will again be the same (since it is just g_tt times the two t-components, so all three factors are the same as before).

Now try to do the same thing I just did for the quantities in exercises #1, #2, and #4. You will find that you can't; there is no coordinate-independent expression corresponding to any of these quantities. That is why they are not considered observables, while the quantity in exercise #3 is.

 

[Passionflower]

Slower and slower relative to the *coordinates*. Not slower and slower relative to observers who are actually there where the light is. To observers who are there where the light is, the light moves at c. You are mistaking an artifact of coordinates for an actual physical observable. Try looking at the velocity of ingoing light in Eddington-Finkelstein or Painleve coordinates; it doesn't go to zero as r -> 2M.

It is very easy to demonstrate that if we have for instance two static observers A and B some distance removed that light signals from A to B and B to A take a different amount of time. Furthermore if we know the distance between them we can time how long it takes for light to travel, or alternatively we could use a radar method by placing a mirror in both observers and record the roundtrip times.

That the light speed is not everywhere the same is even simpler to see when you use a PG chart.

And yes, before we get 27 postings, locally, e.g. at a point it is measured at c, always, but measurements, especially in space, tend to span beyond a point.

 

[yoron]

You might want to use the conceptual frame to say that light speed differ comparing A-B to B-A in a accelerating frame, but I don't see it that way. The only thing differing to me is how gravity acts upon their paths, relative the detectors. locally both A-B and B-A will give 'c'. So yes, to me Peter is definitely defining it correctly.

If it wasn't this way, light would not be a constant.

 

[Passionflower]

You might want to use the conceptual frame to say that light speed differ comparing A-B to B-A in a accelerating frame, but I don't see it that way. The only thing differing to me is how gravity acts upon their paths, relative the detectors.

So even if an experiment would demonstrate it you would still believe light speed is constant?

locally both A-B and B-A will give 'c'. So yes, to me Peter is definitely defining it correctly.

If it wasn't this way, light would not be a constant.

What is locally when A and B are for instance 1000 kilometers removed?

Locally the speed of light is c but locally the world is flat as well.

 

[juanrga]

The derivation of the Lorentz transformations is based on the homogeneity[of space and time] and the isotropy of space.
Could one derive the same transformations wrt space which is not homogeneous or[not] isotropic?
You may consider a few chunks of dielectric strewn here and there. I am assuming for the sake of simplicity that they are at rest in some inertial frame. Such a distribution is not possible without introducing gravitational effects. Running of clocks is affected by gravity.Does the anisotropy of space itself have any effect on them[running of clocks]?

[Incidentally isotropy of space is connected with clocks in the derivation of the Lorentz transformation.Clocks placed symmetrically wrt the x-axis[and lying on the y-z plane as example] should record the same time. Otherwise isotropy of spece gets violated.This idea is commonly used in the derivations.You may consider the one given in "Introduction to Relativity" by Robert Resnick ]
Again the Lorentz transformations are embedded in [present in] Maxwell's equations. But they are the vacuum equations---homogeneity of space[and time] and isotropy are in due consideration.

The Lorentz Transformations are of course correct--only in the context of the homogeneity[of space and time] and isotropy of space. They are extremely useful, like frctionless planes we studied in our childhood days.Frictionless planes helped us in understanding mechanics--but it is extremely difficult to realize them in practice.

Effectively the LT is only valid in a specific context, as Robert Resnick remarks. Precisely when you introduce a second body, you cannot consider that homogeneity is valid everywhere. The properties around particle 1 in the direction where particle 2 is placed are not the same than the properties in the direction where the particle 2 is not (charge distribution is not homogeneous for instance).

This has important consequences. As Jackson emphasizes in his book, one cannot built a Lagrangian for a charged two-body system that was truly Lorentz invariant.

 

[yoron]

Well, theoretically it is defined so, flat that is, in the same way we also use 'point particles', but in general there are no straight lines naturally in nature, as I know? And in 'space' you will find 'gravity', everywhere even though we also define a 'flat space'. And, it's not a theoretical definition that you won't be able to measure lights speed in a vacuum as other than 'c' locally, again as I know it, it's a experimental one even though rather counter intuitive. And the last point stands, if you want to define 'c' as a 'variable' you can't simultaneously define it as a constant. So I believe that to be correct.

"If general relativity is correct, then the constancy of the speed of light in inertial frames is a tautology from the geometry of spacetime. The causal structure of the universe is determined by the geometry of "null vectors". Travelling at the speed c means following world-lines tangent to these null vectors. The use of c as a conversion between units of metres and seconds, as in the SI definition of the metre, is fully justified on theoretical grounds as well as practical terms, because c is not merely the speed of light, it is a fundamental feature of the geometry of spacetime"

Is The Speed of Light Constant.

 

[JDoolin]

Sorry for the delay. This was a long post, and took some thought. Thanks.

Ah, I see I left out a key word: "freely falling". Neither of the objects in question are freely falling.

You asked how, in a curved spacetime with gravity present, two worldlines with the same constant θ and φ, and values of r that differ by a small amount dr, at a given instant of coordinate time t = 0, will continue to remain separated by that same amount dr for all times t. I gave the correct answer, and apparently you knew the correct answer. But you are also right; you cannot have freely falling objects maintain the same constant dr for all times t. That's the essential flaw of using geodesics as a coordinate system.

I meant that the sign convention that spacelike squared intervals are positive, and timelike ones are negative, is just a convention; you can flip it around, so that timelike squared intervals are positive and spacelike ones are negative, without affecting the physics, provided you change all the formulas appropriately.

If you like. But regardless of how you do it, $$ds^2=-c^2 d\tau^2$$ and if ds is real then dtau is imaginary. And if dtau is real, then ds is imaginary. There's only one degree of freedom here in the space-time interval, and the interval is either space-like or time-like or zero.

As it stands, this statement is at best wrong, and at worst not even wrong. Flatness or curvature is an invariant feature of the underlying geometry, and it is there and can be talked about without even assigning coordinates. If you do assign coordinates, then you figure out whether the underlying geometry is flat or curved by computing the Riemann curvature tensor in those coordinates, using the proper form of the metric in those coordinates for the manifold in question.

When you do this using standard Schwarzschild coordinates on Schwarzschild spacetime, using the expression for the metric on Schwarzschild spacetime in those coordinates, which you have written down several times now, you find that the Riemann curvature tensor is non-zero: i.e., the spacetime is curved.

If you do this using the same Latin and Greek letters, t, r, θ, and φ, for coordinates on Minkowski spacetime, using the expression for the metric on Minkowski spacetime in those coordinates, you find that the Riemann curvature tensor is zero; the spacetime is flat.

It's important to recognize that "curvature" as I've just described it is *intrinsic* curvature of the manifold, since that's what we physically observe. See further comments below.

Okay. It appears we both agree that , t, r, θ, and φ can represent coordinates on the Minkowski spacetime. And those coordinates are "flat" according to the Riemann curvature tensor. So it appears that where we disagree is that I am claiming that the (t, r, θ, φ) in the Schwarzschild metric have exactly the same global geometric meaning as the (t, r, θ, φ) in the Minkowski spacetime.

Yes, it is, because if no body ever follows a perfect "straight line" as you are using the term, how can such a line possibly exist? There's no physical observable that picks it out.

Right, there is no physical observable that picks it out. But we don't need a physical observable. Regarding straight lines, one should NOT ask "What direction does the light ray actually take around the planet." but rather one SHOULD ask, "What direction would the ray have gone if the mass had not been there." There is a clear and unambiguous answer.

In other words, I am supposed to believe that you can draw lines that go outside our universe, taking the "straight" route between two events instead of the "curved" route inside our universe that physical objects actually take. Can you see why I'm a bit skeptical?

Sure, it is an odd way of saying it. But if you imagine the path of a satellite if the Earth did not exist, you are essentially imagining a line that exists "outside the universe." There is still no difficulty imagining this line and superimposing it inside our universe.

Does the mapping preserve the metric? No. You admit as much later on when you try to claim that the "real" distance to Alpha Centauri remains the same if a black hole comes between there and Earth, even though every actual physical object has to travel a greater distance because of curvature. If the mapping doesn't preserve the metric, then it's physically meaningless.
Check out the "river model" of black holes:

http://arxiv.org/abs/gr-qc/0411060

Inside the horizon, freely falling objects do go faster than light relative to the "river bed", which is basically what you mean by the flat geometry that everything is supposedly mapped to. But as the authors of the paper make clear, the "river bed" is not physical and has no physical meaning; it's only used to help with visualization. Also, the authors make clear that the ability to construct a model with a flat "river bed", even just for visualization, is a special feature of Schwarzschild geometry, and cannot be done in general for a curved spacetime.

I'd like to see the original papers of Gullstrand and Painleve, but I have a few comments. (1) The conservation of energy requirement uses a nonrelativistic calculation of kinetic energy. (2) Space does not flow like a river into a black hole. Objects flow toward it, or in their reference frame, the black hole moves toward them. (3) Whether or not the authors believe the "river bed" has physical meaning, it does represent a global Minkowski Reference frame comoving with the black hole. And like I've been saying earlier, although the black hole prevents objects from maintaining straight lines in that reference frame, nothing prevents us from imagining the straight lines, and superimposing those straight lines in the physical space. (4) Using the negative escape velocity as the "speed of space" certainly does not form a Lorentz Reference frame.

Show me a physical object that doesn't have to follow all of the curves.
Yes. If you disagree, again, show me a physical object that doesn't have to stay within the curved manifold. The distant bodies do affect the observations indirectly, by affecting the curvature, but we can measure the curvature without even knowing the distant bodies are there, as long as we measure the curvature over a region that's not too small for our measurement accuracy.

Again, here is what I was talking about with the "tyranny of physical observables" While all physical objects (that we know of) will follow the geodesic curves, we can also describe the curve that would exist if a certain mass were not there. We can draw a line on a page that is straighter than any geodesic. We can envision a path through space that is straighter than any geodesic. When we KNOW what a straight line looks like, and we KNOW those straight lines can be mapped right into the real space.

Velocity relative to what?
Relative to what?

My velocity is changing relative to everything else in the universe. There is another way of looking at it. From my perspective, I did not change velocity, but everything else in the universe rotated and changed velocity relative to me. In fact, that is precisely what the Lorentz Transformation and rotation transformations do. They take local parameters( my change in angle and speed) and transform the coordinates of every object (event) in the universe. Effectively, as I circle around the planet, there are two ways of looking at it: Either (a) I am continually changing velocity, or (b) I am continually changing my momentarily comoving reference frame. I actually think that the second way may be a more constructive way of looking at it.

Relative to what? You keep saying all these things when you have basically admitted that there is *no* physical observable to refer them to. It just "looks to you" like one set of lines is straighter than another. Whatever this is, it isn't physics.
I suppose I should note here, once again, that we are talking about paths in *spacetime*, not space. But you can also "draw a tangent line" in spacetime, so I'm ok with what you are visualizing here at this point. However:

If you can see that the tangent lines of any given geodesic actually have a meaningful distant meaning, and are NOT just "abstract vector spaces" existing only "at" a point, then you are well on your way to appreciating the overlying flat geometry.

And you will find that the lines are *not* "straight", in the sense that they will violate the theorems of Euclidean geometry--specifically, the ones that depend on the parallel postulate. The metric in these coordinates will *not* be the Euclidean (in space) or Minkowski (in spacetime) metric.

Well, let's get specific, and logical, about what I mean by that. I'm saying the overlying geometry represents "What would happen if the masses weren't there" I think we all agree that if no masses were present in the universe, we have a Minkowski metric. But the point is, we can construct a Minkowski metric (or at least we can get closer and closer to it) simply by imagining the tangents to the geodesics.

You appear to believe that the metric doesn't matter: that you can project the curved manifold into a flat manifold, like projecting the Earth's surface onto a flat projection like a Mercator projection, distorting all the distances (and times, in spacetime), and everything will still be just fine. You are simply ignoring actual physical observables when they don't match up with your mathematical scheme. Whatever this is, it isn't physics.

(1) No, I believe the metric does matter. (2) Yes, I do believe you can project the curved manifold onto a flat manifold AND vice versa; you can project the flat manifold onto the curved manifold, not unlike a mercator projection, distorting all the distances and times--essentially that is what the Schwarzschild metric does. dtau and dt are NOT the same. ds and dr are not the same. (3) No, I am not ignoring actual physical observables; I am merely trying to place the physical observables in a global context. (4) You HAVE to deal with the flat coordinates because that's where we can make comparisons. If we actually existed in the curved coordinates, then the clock rates in the valley and the clock rates on the mountain would be the same.

I am not saying any of the things you think I am saying. I am saying this: in GR, the theory of *physics*, we *define* "straight" motion as freely falling motion. That is our *physical* standard for mapping out the geometry of spacetime. When we look at the behavior of freely falling geodesic worldlines, we find that the geometry they map out is intrinsically curved, and we have an equation, the Einstein Field Equation, that relates that curvature to the presence of matter-energy. In other words, what we call "gravity" is a manifestation of spacetime curvature, *in this theory of physics*.

If you want to put together an alternate theory of physics that uses a flat background and goes through all the contortions to reproduce the predictions of GR, go ahead. But there's no point in wrangling over the meanings of words. If you don't like putting the labels "flat" or "curved" where I am putting them, fine, substitute your own labels. The physics remains the same. You can't change the physics by asserting that a certain set of coordinates is "flat" because you say so, even though you have to distort all the distances and times to make things fit into those coordinates, so that your model loses all connection to what we actually observe.

Okay, other words besides flat that could be substituted include "Euclidean" "Cartesian" or "Minkowski." The point is that if you don't acknowledge an overlying flat coordinate system, then how can you have an underlying curved coordinate system? If you can imagine a tangent-line parallel to the curve of a geodesic at a point, that tangent line should go on forever. It should not be a "tangent space defined only at a point." These tangent lines parallel to the curve of the geodesic are paths that the object WOULD take if the gravitating mass weren't there. When we imagine paths of objects in a space where no matter exists, we are imagining a Minkowski space, where parallel lines remain parallel. The Minkowski space is an overlying coordinate system. And specifically, the Schwarzschild metric (t,r,theta,phi) are spherical coordinates in that global flat Minkowski Euclidean space. They are the coordinates that represent straight lines geodesics would follow if the central mass were not there.

Where have I said the rock won't hit the ground? You are reading an awful lot into what I'm saying, that I haven't said. I've never said the rock isn't moving relative to the ground.

Directly above in this very post, you told me that I was not allowed to use a clock sitting on the floor and a clock sitting on a table to describe two objects that maintained a constant distance r. You said you "left out a key word 'freely-falling.'" The ground does not follow a geodesic, nor do objects sitting on the ground.

And you would substitute what? Apparently a bunch of unobservables that throw away the observables.

No, don't throw away the observables, but put them in a global context. I go upstairs and look at the clock and see that it is going at the same rate as my watch, I look at the height and it's the same length as my tie.. Then I go downstairs and see that the clock and my watch are going at the same rate, and its height is the same length as my tie. However, when I look at both clocks at the same time, I find that the downstairs clock is going slower than the upper clock, and the lower clock is actually shorter than the upper clock. You can't get the entire picture of what is going on without looking at nonlocal observables. There is a global reference frame where the upper clock is actually taller and faster than the lower clock. It's not a meaningless set of Greek and Latin letters, but a global flat coordinate system.

As the ultimate *basis* for distances and times, yes. Obviously you can build accounts of distances and times over extended regions on this basis. If I want to know how much proper time will elapse along the Earth's worldline from now until Christmas, I can calculate it based on local clock and ruler readings.

And it makes for pretty good data why? Because the spacetime around Earth is ALMOST flat.

No, I claim the rock's path is not curved because it feels no force. Again, that is a *definition* I adopt because I am using a particular theory of physics. You are free to adopt a different definition, as long as you contort your model appropriately. But you'll have to expect some skepticism when you explicitly state that the distances and times we actually measure are not the "real" ones.

I am trying to give definitions and draw distinctions. I already pointed out that at each point in space, there are two important vector spaces; one made up of lengths and times of the physical observables. And another made up of distance and direction of the overlying flat coordinate space. However, you are arguing that there is only one vector space. Once you acknowledge that there are two vector spaces, then you can also draw the distinctions that I am making, and make sense of what I'm saying.

Really? Give some examples, please. I certainly agree there are ways of obtaining observables for distant objects without stretching rulers from here to there. But to relate any of those observables to a "distance" requires a theoretical model that gives you the relationship. The theoretical relationships used in actual cosmology are based on GR, so they are based on distances that "follow every curve" of the manifold.

Well, the fact is that we don't see very many situations where light is significantly curved. If there is a big enough mass between us and the distant star to deflect the light, then either the star is eclipsed, by the mass in front of it, or there is a visible gravitational lensing effect. Yes, the light coming from there to here does follow every curve in the manifold, but throughout the region in between, the manifold is ALMOST flat, and when I say flat, that is in comparison to the overlying Minkowski coordinates. Only in regions of intense gravitational fields, in regions where the light would be eclipsed anyway, do you find a significant variation from flatness.

 

[PeterDonis]

You asked how, in a curved spacetime with gravity present, two worldlines with the same constant θ and φ, and values of r that differ by a small amount dr, at a given instant of coordinate time t = 0, will continue to remain separated by that same amount dr for all times t. I gave the correct answer, and apparently you knew the correct answer. But you are also right; you cannot have freely falling objects maintain the same constant dr for all times t. That's the essential flaw of using geodesics as a coordinate system.

No, that's the essential flaw in assuming that SR is globally valid, i.e., that spacetime is flat. It isn't flat. If we don't have agreement on this point, then any further discussion is useless unless you can give an actual *physical* observable that picks out your "flat" lines in spacetime instead of the "curved" ones. I emphasize that this isn't just about geodesics; there are no "flat" lines in your sense in a curved spacetime, geodesic or otherwise. None. If you think there are, show me one. It doesn't have to be a geodesic; any "flat" line will do.

So it appears that where we disagree is that I am claiming that the (t, r, θ, φ) in the Schwarzschild metric have exactly the same global geometric meaning as the (t, r, θ, φ) in the Minkowski spacetime. Right, there is no physical observable that picks it out. But we don't need a physical observable. Regarding straight lines, one should NOT ask "What direction does the light ray actually take around the planet." but rather one SHOULD ask, "What direction would the ray have gone if the mass had not been there." There is a clear and unambiguous answer. Sure, it is an odd way of saying it. But if you imagine the path of a satellite if the Earth did not exist, you are essentially imagining a line that exists "outside the universe." There is still no difficulty imagining this line and superimposing it inside our universe.

See further comments below on the "river model". Also, even if there is a (strained) way of doing this in one particular spacetime, that's a far cry from having a way of doing it in *any* spacetime. Finally, as I've said before, and as I point out again below, even if you can construct such a "flat background", when you project the actual, physical, curved spacetime into it, the metric gets distorted, so the laws of physics, as projected into the flat background, are *not* the laws of SR. So even where what you say is possible, it doesn't mean what you think it means.

(2) Space does not flow like a river into a black hole. Objects flow toward it, or in their reference frame, the black hole moves toward them.

In one model, yes. The river model is another model. Since both models make exactly the same physical predictions, they are physically equivalent.

(3) Whether or not the authors believe the "river bed" has physical meaning, it does represent a global Minkowski Reference frame comoving with the black hole.

A "frame" that nevertheless violates the laws of SR, because objects inside the EH move inward faster than light in this "frame".

And like I've been saying earlier, although the black hole prevents objects from maintaining straight lines in that reference frame, nothing prevents us from imagining the straight lines, and superimposing those straight lines in the physical space.

And distorting the metric so that the laws of physics in the "flat background" are not the laws of SR. I just gave one example of that above.

(4) Using the negative escape velocity as the "speed of space" certainly does not form a Lorentz Reference frame.

Yes, exactly. The "flat background" is *not* a global Lorentz Reference frame. In a curved spacetime there is no such thing. There just isn't.

Again, here is what I was talking about with the "tyranny of physical observables" While all physical objects (that we know of) will follow the geodesic curves, we can also describe the curve that would exist if a certain mass were not there. We can draw a line on a page that is straighter than any geodesic. We can envision a path through space that is straighter than any geodesic. When we KNOW what a straight line looks like, and we KNOW those straight lines can be mapped right into the real space.

Only by distorting the metric and thereby distorting the physical laws in this "flat" background. You can complain all you want about a "tyranny of physical observables", but you do remember that we are discussing physics here, right? It's a little weird to come into a physics forum and complain that people want theories to match physical observables.

My velocity is changing relative to everything else in the universe. There is another way of looking at it. From my perspective, I did not change velocity, but everything else in the universe rotated and changed velocity relative to me.

Sure, you can always construct your own "local" reference frame around any event on your worldline. It will only cover a small local patch of spacetime around that event, though. More precisely, the coordinate values in that frame will only obey the laws of SR, to some given accuracy of measurement, within a small local patch around that event.

In fact, that is precisely what the Lorentz Transformation and rotation transformations do.

In that small local patch, within the given accuracy of measurement, yes. Outside that small local patch, no. You can certainly try to extend your local coordinate patch, but you will find that the coordinate values no longer work right--they no longer predict the correct observational results--when you plug them into the SR formulas. The Lorentz Transformation is just one example of that. You can, if you insist, work out what changes in coordinate values a Lorentz Transformation in your local patch around a given event induces in the region outside that patch, but again, you will find that those changes don't obey the laws of SR.

They take local parameters( my change in angle and speed) and transform the coordinates of every object (event) in the universe. Effectively, as I circle around the planet, there are two ways of looking at it: Either (a) I am continually changing velocity, or (b) I am continually changing my momentarily comoving reference frame. I actually think that the second way may be a more constructive way of looking at it.

Yes, I would agree. And if you are in free fall, and yet you are continually changing your MCIF, then you are not obeying the laws of SR. The laws of SR require that an object in free fall, feeling no force, can never change its MCIF.

If you can see that the tangent lines of any given geodesic actually have a meaningful distant meaning, and are NOT just "abstract vector spaces" existing only "at" a point, then you are well on your way to appreciating the overlying flat geometry.

Sorry, this train never left the station, because its driver (you) fails to see that there is no unique way to define the "same" tangent vector at different events in a curved manifold. If I have a tangent vector T at event A, and I want to compare it with a tangent vector V at event B, I have to first "transport" T to event B (or V to event A, but I'll stick with moving T for this example). But what vector at B I end up with when I transport T from A to B depends on the path I take between those events; that's one way of expressing what it means for a manifold to be curved. So there is no unique way to compare T and V, because there is no unique way to transport either one to the other's "location" to compare them.

I won't comment right now on the rest of your post because I think the above captures the fundamental point at issue. I don't think we disagree on the actual math (except that the last point I raised, about the non-uniqueness of transporting vectors from event to event, is very important, so if you're not sure about it, we should resolve it), or the actual physical observables; I think the disagreement is about "interpretation" of an aspect of the math that you think is very important physically, but I think is irrelevant physically, though it can help sometimes with visualization.