The Lorentz Transformations and a Few Concerns

[JDoolin]

I think we are agreed then, on the fundamental point of issue. So let me make that point once more, hopefully with a little more clarity. As concisely as possible, my point is that the Minkowski metric and the Schwarzschild metric overlap. The Schwarzschild metric describes physical observables as they really are (with the central gravitating body actually present), and the Minkowski metric describes how things WOULD BE (if the central gravitating body weren't present).

More precisely still, if I have a world-curve of an accelerating clock defined parametrically (t,r(t), θ(t), φ(t)) I would be able to find the age of that clock by integrating the metric $$ds^2 =- \left ( 1 - \frac{2 G M}{c^2 r} \right )c^2 dt^2 + \left ( 1-\frac{2 G M}{c^2 r} \right )^{-1} dr^2 + r^2 (d\theta^2 + sin^2(\theta)d\varphi^2)$$ over that curve.

I could also find how old that clock WOULD have been without the central gravitating body following the same curve (t,r(t), θ(t), φ(t)) but integrating the metric $$ds^2 =- c^2 dt^2 + dr^2 + r^2 (d\theta^2 + sin^2(\theta)d\varphi^2)$$ This is also the metric you would get simply by setting M=0 in the Schwarzschild metric.

We have these two different metrics; one answering the question of "what would the observable be if the mass weren't there" and one answering "what is the observable with the mass there." They overlap. The coordinates (t,r(t), θ(t), φ(t)) don't change their locations, it is only the question of whether you have a central mass.

Whether the mass is there or not, we can define a straight lines with "priveleged" status, being the path the object would have taken if the mass weren't there. That is not an ambiguous direction in Schwarzschild, but it is well defined whether the central mass is there or not.

 

[Passionflower]

You are comparing apples and oranges as the geodesics of the two metrics are completely different.

 

[PeterDonis]

I think we are agreed then, on the fundamental point of issue. So let me make that point once more, hopefully with a little more clarity. As concisely as possible, my point is that the Minkowski metric and the Schwarzschild metric overlap. The Schwarzschild metric describes physical observables as they really are (with the central gravitating body actually present), and the Minkowski metric describes how things WOULD BE (if the central gravitating body weren't present).

Um, no, we don't agree on this point. We appear to agree on the actual physical observables, since you say the Schwarzschild metric describes them. But we do *not* agree that the Minkowski metric describes "how things WOULD BE" as you appear to be using the term; IMO that statement has no physical meaning at all. I would say that the Minkowski metric describes the physical observables in a *different spacetime*, a flat one. See further comments below.

More precisely still, if I have a world-curve of an accelerating clock defined parametrically (t,r(t), θ(t), φ(t)) I would be able to find the age of that clock by integrating the metric $$ds^2 =- \left ( 1 - \frac{2 G M}{c^2 r} \right )c^2 dt^2 + \left ( 1-\frac{2 G M}{c^2 r} \right )^{-1} dr^2 + r^2 (d\theta^2 + sin^2(\theta)d\varphi^2)$$ over that curve.

Ok so far.

I could also find how old that clock WOULD have been without the central gravitating body following the same curve (t,r(t), θ(t), φ(t)) but integrating the metric $$ds^2 =- c^2 dt^2 + dr^2 + r^2 (d\theta^2 + sin^2(\theta)d\varphi^2)$$

Mathematically, yes, I see what you are doing. You are constructing a distorted version of the spacetime where the same events have the same coordinate labels but the metric is different. Again, mathematically you can do this (at least outside the horizon--see just below), but physically it has no meaning. You can capture all the physics in the actual, curved spacetime without bringing this extra math in at all. Also, there is no physical justification for saying that the same 4-tuple of coordinate values labels "the same event" in both cases, since the metric is changed. See further comments below.

It's worth noting that your correspondence only works above the horizon. At r = 2M the Schwarzschild coordinates are singular so you can't label points on the horizon using these coordinates. And inside the horizon, if you were to set up Schwarzschild coordinates, the "age of the clock" you would calculate by your formula above would be imaginary. (More precisely, the squared interval ds^2 would be positive--spacelike--instead of negative--timelike--for a set of parametrized coordinate values where the "Minkowski" squared interval was negative--timelike.)

It's also worth noting that Schwarzschild spacetime is just one of many curved spacetimes in GR. Your "flat correspondence" scheme would not work in most of the others; in FRW spacetime, for example. Even in Kerr spacetime outside the horizon I'm not sure it would work.

We have these two different metrics; one answering the question of "what would the observable be if the mass weren't there" and one answering "what is the observable with the mass there." They overlap. The coordinates (t,r(t), θ(t), φ(t)) don't change their locations, it is only the question of whether you have a central mass.

No, you can't say that events labeled by the same 4-tuple of coordinate values "don't change their locations", because in the actual, physical spacetime with the central mass present, the metric *changes*, therefore the "locations" of events labeled by the same 4-tuple of coordinates change, since "location" is defined by the metric. You can calculate your "Minkowski" interval, but you can't just call that the "location" of a point by fiat; at least not if you want to do physics. Physically, "location" is determined by the metric.

Whether the mass is there or not, we can define a straight lines with "priveleged" status, being the path the object would have taken if the mass weren't there. That is not an ambiguous direction in Schwarzschild, but it is well defined whether the central mass is there or not.

Mathematically, you can "define" straight lines like this, but the definition has no physical meaning. Again, all the physics is captured by the actual, curved metric. There is *no* additional physics being added by your definition of "straight" lines. So physically, in the actual curved spacetime, those lines have no meaning.

If you disagree, then please tell me what actual physics is left out by the curved spacetime description, but is included in the "flat" description. That should be interesting since you've already agreed that the Schwarzschild metric describes all the actual physical observables.

If, on the other hand, you agree that the actual curved metric captures all the physical observables, then you can talk all you want about your mathematically defined "straight" lines, and I can simply ignore all your talk and compute physical observables using the actual curved metric, just as I always have. So why are you bothering to talk about them? Why should anyone else even pay attention?

(Btw, it's not that I don't see how you are picking the "straight lines" out; I do. You are saying that lines of constant r, theta, phi in Schwarzschild coordinates are "straight lines", even though they are not geodesics--objects following these worldlines are accelerated. But physically, nothing is added to my understanding of these lines by picking them out as the "straight" ones in this way; indeed, adding this definition *complicates* matters, since now I have to remember that your "straight lines" are not geodesics, and furthermore that the proper acceleration along one of your "straight lines" *varies* with radial coordinate r. I don't have to deal with any of this in the standard GR picture. Furthermore, I don't even need your criterion--pick out the lines that "would be straight" with the same coordinate values if the metric were Minkowski--to pick out the lines of constant r, theta, phi in Schwarzschild coordinates as being "special"; they are already picked out by the time translation symmetry of the spacetime with the *curved* metric, again without having to introduce your "flat" metric at all. So again, there is *no* physics that is added by your description.)

 

[DrGreg]

As concisely as possible, my point is that the Minkowski metric and the Schwarzschild metric overlap. The Schwarzschild metric describes physical observables as they really are (with the central gravitating body actually present), and the Minkowski metric describes how things WOULD BE (if the central gravitating body weren't present).

Whether the mass is there or not, we can define a straight lines with "priveleged" status, being the path the object would have taken if the mass weren't there.

Instead of considering 4D spacetime, let's consider a 2D analogy, the curved surface of the Earth. For simplicity, let's assume the Earth is a perfect sphere, and choose units to make the radius of the Earth equal 1. The metric for the Earth's surface in terms of latitude and longitude is$$ds^2 = d\theta^2 + \sin^2 \theta \,\, d\phi^2$$Would you be happy for me to say that I can also use the same latitude and longitude to describe what the Earth would be like if it were flat, with a metric$$ds^2 = d\theta^2 + d\phi^2 \,\, \mbox{?}$$Would you be happy to say that the equation$$\theta = A\phi + B$$describes a geodesic on the flat Earth and therefore the same equation defines a "straight line with 'privileged' status" on the surface of the curved Earth?

 

[JDoolin]

Peter, it seems to me we are interpreting the meaning of the Schwarzschild metric differently. Whereas I see the coordinates remaining in place as the central mass becomes bigger, you see them moving around somehow to adjust to the "physical observables.

Let me ask a hypothetical question that may further clarify the difference between my interpretation and yours.

Let's say we have a nonrotating planet with a high gravity. And lots and lots of identical meter-sticks. Tall buildings are constructed on the planet that sretch from the ground all the way out into space, and the meter-sticks are placed end to end from the ground up.

Now a photograph is taken from a long distance away; so far away that the small-angle formula applies, i.e. theta=sin(theta)=tan(theta) (ask me if you don't know what I mean.)

Now the meter-sticks measure physical observables, but in the photograph, will it appear that each meter-stick is the same length, or will the meter-sticks near the surface of the planet appear shorter than the meter-sticks further away, due to the effects of the Schwarzschild metric?


The way I interpret the metric, the meter sticks close to the planet will appear shorter in the photograph, even though each meter-stick is "locally" identical. They all measure off the same "physically observable" one meter. However, every observer in the system will agree (if they make careful eough measurements by sight) that the meter-sticks closer to the planet are indeed shorter than the meter-sticks above.

(Conversely, I think we both agree that if an observer climbs up and down the towers, physically checking the length of each meter stick with one he carries, he will find that all of the meter-sticks are the same length, it's just when he makes the comparison by sight. i.e. nonlocal observation of the metric, where the difference in lengths can be observed.)

It seems to me that your interpretation is different; am I right?

 

[PeterDonis]

Peter, it seems to me we are interpreting the meaning of the Schwarzschild metric differently. Whereas I see the coordinates remaining in place as the central mass becomes bigger, you see them moving around somehow to adjust to the "physical observables.

What does "the coordinates remaining in place" even mean? Coordinates are numbers used to label events. If you mean "an event labeled by a given 4-tuple of coordinates remains in place", what does *that* even mean? The only physical interpretation I can give this is that the "location" of an event labeled by a given 4-tuple is "the same" whether a gravitating body is there or not, and since "location" is determined by the metric, this is obviously false; or perhaps a better way to say it would be that any interpretation of the term "location" that has the "location" of an event labeled by a given 4-tuple being "the same" when the metric changes is not a good interpretation for doing physics. You could salvage it somewhat by steadfastly refusing to draw *any* deductions that you would normally draw from the phrase "the location is the same"--for example, you could refuse to deduce that two "locations" which remain the same will be the same physical distance apart (since they won't if the metric changes). But I don't see the point; the usefulness of the term "location" is completely thrown away, so why use it at all?

Let's say we have a nonrotating planet with a high gravity. And lots and lots of identical meter-sticks. Tall buildings are constructed on the planet that sretch from the ground all the way out into space, and the meter-sticks are placed end to end from the ground up.

Now a photograph is taken from a long distance away; so far away that the small-angle formula applies, i.e. theta=sin(theta)=tan(theta) (ask me if you don't know what I mean.)

Now the meter-sticks measure physical observables, but in the photograph, will it appear that each meter-stick is the same length, or will the meter-sticks near the surface of the planet appear shorter than the meter-sticks further away, due to the effects of the Schwarzschild metric?

There was a *long* thread about this recently. The thread started with a different question, but the "meaning" of the Schwarzschild metric as regards radial measures of length came up, starting more or less with this post:

https://www.physicsforums.com/showpost.php?p=3557020&postcount=138

The way I interpret the metric, the meter sticks close to the planet will appear shorter in the photograph, even though each meter-stick is "locally" identical. They all measure off the same "physically observable" one meter. However, every observer in the system will agree (if they make careful eough measurements by sight) that the meter-sticks closer to the planet are indeed shorter than the meter-sticks above.

No, they will agree that the meter sticks *look* shorter from far away. But locally, each meter stick still measures one meter; if I am far away from the planet, and I have a meter stick with me, which checks out as being exactly the same length as all other meter sticks, and I take it down to the surface of the planet, it will be the same length as the meter sticks there; I can put it side by side with any meter stick and they will exactly overlap. Or, I can take a meter stick from the surface of the planet, that "looks shorter" as seen from far away, put it in my rocket and bring it far away to the location of the telescope through which that meter stick "looked shorter", and it will be the same length as all the other meter sticks in the vicinity of the telescope.

You know this, because you follow up immediately with:

(Conversely, I think we both agree that if an observer climbs up and down the towers, physically checking the length of each meter stick with one he carries, he will find that all of the meter-sticks are the same length, it's just when he makes the comparison by sight. i.e. nonlocal observation of the metric, where the difference in lengths can be observed.)

I agree with this; as I said, I believe we are in agreement on all of the actual physical observables. The only thing I object to is your contention that, because the meter sticks look shorter from far away if they are closer to the planet, they somehow *are* shorter, in some sense. They aren't, because, as you agree, you can take a meter stick anywhere you like and it will be exactly the same length as all the other meter sticks in its vicinity. So the fact that they "look" shorter from far away is an optical illusion, so to speak; it is not caused by any "real" change in length, it is caused by the way gravity bends light rays.

 

[JDoolin]

What does "the coordinates remaining in place" even mean? Coordinates are numbers used to label events. If you mean "an event labeled by a given 4-tuple of coordinates remains in place", what does *that* even mean? The only physical interpretation I can give this is that the "location" of an event labeled by a given 4-tuple is "the same" whether a gravitating body is there or not, and since "location" is determined by the metric, this is obviously false; or perhaps a better way to say it would be that any interpretation of the term "location" that has the "location" of an event labeled by a given 4-tuple being "the same" when the metric changes is not a good interpretation for doing physics. You could salvage it somewhat by steadfastly refusing to draw *any* deductions that you would normally draw from the phrase "the location is the same"--for example, you could refuse to deduce that two "locations" which remain the same will be the same physical distance apart (since they won't if the metric changes). But I don't see the point; the usefulness of the term "location" is completely thrown away, so why use it at all?

Is this another major point of disagreement, then?

In my thinking, the metric (being ∫ds or ∫dτ along a specific spacetime path) is an observer independent quantity which determines how many meter-sticks fit between two specific events along a specific path, or how many ticks on a clock following a specific path between two events.

But just knowing the length of a path does not convey a sense of its location. The location of the path is represented by its coordinates; not it's length.

I can easily see how changing the mass changes the metric, but I can't see how you can claim that changing the mass changes the coordinates. And of course, you can't see how I can claim that changing the mass WOULDN'T change the coordinates. So this issue may be a stalemate.

Okay, so what DO we agree on here. The Schwarzschild metric gives us a RELATIONSHIP between the metric and the coordinates. My central theme is that both the coordinates and the metric are concretely definable constructs. The metric is the actual measurement, while the coordinates are "what the measurements of distance and time would be (referenced from the location of the central mass) if the central mass were zero."

Another idea would be to say that the metric is concretely definable, but that the coordinates are NOT concretely definable. But that begs the question, if the coordinates are meaningless, or not concretely definable, and the Schwarzschild metric gives you the RELATIONSHIP between the neaningFUL metric and the meaningLESS coordinates, wouldn't that mean that the Schwarzschild metric itself is meaningless?

There was a *long* thread about this recently. The thread started with a different question, but the "meaning" of the Schwarzschild metric as regards radial measures of length came up, starting more or less with this post:

https://www.physicsforums.com/showpost.php?p=3557020&postcount=138No, they will agree that the meter sticks *look* shorter from far away. But locally, each meter stick still measures one meter; if I am far away from the planet, and I have a meter stick with me, which checks out as being exactly the same length as all other meter sticks, and I take it down to the surface of the planet, it will be the same length as the meter sticks there; I can put it side by side with any meter stick and they will exactly overlap. Or, I can take a meter stick from the surface of the planet, that "looks shorter" as seen from far away, put it in my rocket and bring it far away to the location of the telescope through which that meter stick "looked shorter", and it will be the same length as all the other meter sticks in the vicinity of the telescope.

You know this, because you follow up immediately with:
I agree with this; as I said, I believe we are in agreement on all of the actual physical observables. The only thing I object to is your contention that, because the meter sticks look shorter from far away if they are closer to the planet, they somehow *are* shorter, in some sense. They aren't, because, as you agree, you can take a meter stick anywhere you like and it will be exactly the same length as all the other meter sticks in its vicinity. So the fact that they "look" shorter from far away is an optical illusion, so to speak; it is not caused by any "real" change in length, it is caused by the way gravity bends light rays.

You're saying that the meter-sticks "look" shorter, but that is an optical illusion caused by the way gravity bends light rays. I agree that there is going to be an additional optical effect caused by the bending of light rays, but the Schwarzschild metric itself does not include this optical effect.

I generated an image to give further explanation to what I am saying.

The innermost circle has radius r=1.1

The circles beyond that are drawn so that $$\int ds = \int_{1.1}^{r_f}\frac{1}{\sqrt{1-\frac{1}{r}}}dr$$ evaluates to 0.1, 0.2, 0.3, 0.4, ... 6.1, 6.2, 6.3. i.e. the circles are drawn so that the radial distance between each has the same "physical observable" distance.

The axes give values for r=2, r=4, r=6, where r=1 represents 1 schwarzschild radius. The figure is drawn so that the innermost circle is drawn at 1.1 Schwarzschild radius, staying well clear of the singularity.

From the diagram, it is clear that length "AB" looks shorter than length "CD" But I feel it is not just a matter of some optical illusion. From a global point-of-view, the length "AB" really is shorter than length "CD"

Also, from a global point of view, the clocks at A would be going much slower than the clocks at D. The time-slowing effect may be a stronger argument that this is no illusion. A person may climb up to D, and verify that his pocketwatch is going at the same rate as the clock at D. Then he goes down to A and verifies that his pocketwatch is going at the same rate as the one at A. However, as he continues to do this, he will find that when he goes above, the time on his clock really gets ahead of the one below. And/or when he goes below, his clock really gets behind the one above. It's a slightly different phenomenon than with the rulers, which don't have a memory, but with the clocks, they will lose their synchronization, so you can measure locally that there is a difference in the rates of time "globally."

There should still be an additional distortion effect on top of this one, whereby the image is even further distorted, caused by the bending of light rays on their way toward the camera. This distortion is calculatable based on where the camera is positioned.

 

[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.

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.

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.

I thought this particular thread of the argument was interesting. I have been claiming all along that geodesics are NOT straight lines.

But when I offered an example of an actual straight line, you rejected it because it wasn't a geodesic.

Do you have any better argument that the lines of constant r, theta, phi are not straight?

 

[PeterDonis]

In my thinking, the metric (being ∫ds or ∫dτ along a specific spacetime path) is an observer independent quantity which determines how many meter-sticks fit between two specific events along a specific path, or how many ticks on a clock following a specific path between two events.

No problem here.

But just knowing the length of a path does not convey a sense of its location. The location of the path is represented by its coordinates; not it's length.

If you choose to define "location" this way. "Location" is just a word. All the *physics* is conveyed by the statement about the integral of the metric along a specific path. Given the physics, I don't think the definition of "location" you appear to be using is a very useful one; but either way it doesn't change the physics.

I can easily see how changing the mass changes the metric, but I can't see how you can claim that changing the mass changes the coordinates. And of course, you can't see how I can claim that changing the mass WOULDN'T change the coordinates. So this issue may be a stalemate.

I don't say that changing the mass changes the coordinates, or that it doesn't. I say that the whole idea of "changing the coordinates" or not, when you're talking about two different spacetimes, is physically meaningless; it's just words.

The coordinates are just numbers used to label events. When you have a mass present, you have a different spacetime, a different geometry, and a different set of events. Saying that an event in the curved spacetime with a given (t, r, theta, phi) is "the same event" as an event in the flat spacetime with those coordinate values is just words; it has no physical meaning. Setting up such a correspondence may help someone (such as you) to visualize what is going on when the mass is present, but that's not the same as it having an actual physical meaning.

The metric is the actual measurement, while the coordinates are "what the measurements of distance and time would be (referenced from the location of the central mass) if the central mass were zero."

And that statement has no physical meaning. It may help you with visualization, but all the physics can be captured without taking it into account at all.

Another idea would be to say that the metric is concretely definable, but that the coordinates are NOT concretely definable. But that begs the question, if the coordinates are meaningless, or not concretely definable, and the Schwarzschild metric gives you the RELATIONSHIP between the neaningFUL metric and the meaningLESS coordinates, wouldn't that mean that the Schwarzschild metric itself is meaningless?

I didn't say the coordinates are meaningless. I said they are arbitrary numbers used to label events. The metric tells you how the coordinate labeling expresses the physics. Pick a different coordinate labeling, and the metric looks different. Look up Painleve coordinates, or Eddington-Finkelstein coordinates, or Kruskal coordinates; all of them describe the *same* geometry, the *same* physics, as Schwarzschild coordinates, they just label events differently, so the metric looks different. Does that mean they're "meaningless"? Does the fact that Cartesian and polar coordinates on a plane label points differently mean those are "meaningless"? The metric looks different for those two cases too, but both describe the same geometry (a flat Euclidean plane).

You're saying that the meter-sticks "look" shorter, but that is an optical illusion caused by the way gravity bends light rays. I agree that there is going to be an additional optical effect caused by the bending of light rays, but the Schwarzschild metric itself does not include this optical effect.

Yes, it does, because the metric describes *all* curves, not just timelike or spacelike ones; it describes null curves too. To find out how a meter stick appears to someone far away, you figure out the paths that light rays follow from the meter stick to the observer. How do you do that? By using the metric to figure out what paths null curves (curves with ds = 0) will follow.

I thought this particular thread of the argument was interesting. I have been claiming all along that geodesics are NOT straight lines.

But when I offered an example of an actual straight line, you rejected it because it wasn't a geodesic.

Do you have any better argument that the lines of constant r, theta, phi are not straight?

The word "straight" has different possible definitions. I was pointing out that according to the definition which is used in GR, geodesics are straight and accelerated worldlines are not. But I was also pointing out that, even according to a *Euclidean* (or "Minkowskian", since we're talking about spacetime) definition of "straight", lines of constant r, theta, phi are not straight in Schwarzschild coordinates. If those lines were straight by the Minkowskian definition of "straight", then objects following them would have to be in free fall, since the lines remain the same distance apart for all time. But objects following those lines are not in free fall. So the lines are not "straight" according to any definition of the word "straight" that is usual in relativity physics.

You are obviously using a different definition of "straight". That's a matter of words, not physics; the physics is the same regardless of which lines we call "straight", just as it's the same regardless of how we define a "location" (see above). I happen to think your choice of words is not a very useful one for studying the physics we're talking about, but that doesn't make your choice of words "wrong", just not very useful IMO. I don't really care about the words; I care about the physics. If any misunderstanding is suspected based on not using words the same way, we can just switch to math terminology, as you did in the first passage I quoted in this post (about the integral of the metric); that expresses the physics involved clearly and concisely, and that's all that's necessary.