Recommendaton for Clarifying Special Relativity

[WannabeNewton]

A tensor is a purely algebraic object. Let $V$ be a real finite dimensional vector space and let $T:V^{*}\times...\times V^{*}\times V\times...\times V\rightarrow \mathbb{R}$ be a tensor over $V$ (there are $k$ products of $V$ and $l$ products of $V^{*}$). If $\{e_{i}\}$ is a basis for $V$ and $\{e^{i*}\}$ is the dual basis then any tensor over $V$ can be written in terms of the simple tensors formed out of this basis i.e. $T = \sum T^{i_1...i_k}{}{}_{v_1...v_l}e_{i_1}\otimes...\otimes e^{v_l*}$. The $T^{i_1...i_k}{}{}_{v_1...v_l}$ are the components of $T$ with respect to the above basis.

Now let $M$ be a smooth manifold and let $T$ be a smooth tensor field (a section of a tensor bundle) on $M$. We can define a smooth basis field for $M$ (a section of the tangent bundle that assigns a basis to each $T_p M$) with an associated smooth dual basis field. Then the components of the tensor field with respect to this basis field are defined point-wise as above for each $T_p M$. If this is an orthonormal basis field then it is often called a frame field (in 4 dimensions a vierbein).

If we now have a coordinate chart $(U,\varphi)$ on $M$ then we can choose to use the coordinate basis fields $\{\frac{\partial }{\partial x^{1}},...,\frac{\partial }{\partial x^{n}}\}$ so that for each $T_p M$ with $p \in U$, the components of $T(p)$ are with respect to $(\frac{\partial }{\partial x^{i}}|_p)$ and the associated covector field $(dx^{i}|_p)$. This is a special case of a basis field (and in general not an orthonormal one) that is associated with a coordinate chart.

One can however always associate a given orthonormal basis for $T_p M$ with some coordinate basis field evaluated at $p$ (and in general this association will only be valid at p). This is how Riemann normal coordinates are constructed.

 

[PeterDonis]

So, in this case, a coordinate-free test theory would be one which reproduces either Minkowski spacetime or Galilean spacetime with some set of tensors and scalars.

Ah, ok, so it would be something like the PPN formalism, and assuming Minkowski spacetime would be like assuming all the PPN parameters have their GR values, instead of figuring out how to test the parameters experimentally. Fair point.

As far as that goes, there is one variable parameter, so to speak, that has been tested: the speed of light itself, or rather the fact that it is independent of the speed of the source. A PPN-like theory of spacetime would have the invariant speed as a variable parameter; its Galilean value would be infinity, and its Minkowski value would be something finite. However, standard SR gives no way of predicting exactly *which* finite speed should be invariant, so this still isn't quite the same as the PPN tests of GR.

 

[Phy_Man]

A tensor is a purely algebraic object.

That is incorrect. Tensors are geometric objects. MTW explain this on page 49. Schutz explains it on page 36. They are referred to as geometrical objects because they can be defined without referring to a specific coordinate system. A point in spacetime (i.e. an event) is also a geometric object. Vectors and 1-forms (which are examples of tensors) are also geometric objects.

 

[WannabeNewton]

A tensor is algebraic as it only requires a vector space to define. It lies within algebraic categories. The tensors obtained by evaluating a section of a tensor bundle over a smooth manifold fiber by fiber of the tangent bundle are simply special cases as they are with respect to the tangent space, which is just a specific type of vector space, defined in the category of smooth manifolds.

MTW and others' use of the word "geometric object", while warranted, is not a mathematical classification of tensors in the language of categories.

As a side note, MTW is about as accurate with mathematics as Fox News is with actual news ;) Schutz is awesome though, got to love that guy.

 

[Jorriss]

That is incorrect.

He's not incorrect. Tensors are elements of products of vector spaces, that's completely algebraic. They are multilinear maps.

 

[Bill_K]

If you missed that point then don't feel bad. It's no big deal. A lot of people seem to forget what it means to represent a point or speak of its coordinates. I.e. if you wish to speak of coordinates then what you've implied by your question is that you have chosen a point to serve as the origin and then the point is the vector displacement from the reference point (i.e. origin) to the point in question.

Certainly one of us has forgotten. A coordinate system does not have to have an origin, let alone a point origin. What point is the origin of Schwarzschild coordinates?

 

[Phy_Man]

A tensor is algebraic as it only requires a vector space to define.

Recall that you said that a tensor is only a algebraic object. That is not true. Tensors are clearly geometric objects by definition. I referred to those texts because I assumed that you have them. A more precise definition is found in Differential Geometry by Erwin Kreszig, Dover Pub., page 92. Mind you, this is a definition given in this text and as such it's not wrong. There's really no room for debate on this. I know this as a fact. I suspect you'll have to search to find a text with the definition. I admire that by the way. Nothing wrong with challenging what you have a feeling may be wrong. But that's why I posted the reference to the source of the definition.

Please go to those websites I showed you and download the book and read the definition. I'd post it but I have this kink in my neck that is very painful right now.

Do you really believe that MTW would say those things were geometric objects if they really weren't? It seems as though a lot of people here have zero faith in the textbooks that people are learning this subject from. Please think long and hard before you assume that the experts can't get a simple definition wrong. I've yet to see that happen in my lifetime.

Actually I have to question the wisdom of arguing over definitions. I mean really! Where does that get us? Are these things being questioned because someone doesn't know how they're defined? Is someone saying that how things are defined today shouldn't be defined that way? I'm curious, that's all. Although I don't believe it's worth the effort posting on these subects in cases like this. If you folks are happy with what you believe than please let me know. I'm not trying to be a wisenheimer or anything like that. It's just that I live with chronic pain and all thus typing is getting to be very painful. I can't see the usefulness of posting defintions if the only response to them is denial of the definition. So my question is; what do you think is useful in cases like this? I mean after all, you sure don't need someone to look things up for you. Thank you in advance. I appreciate it. My neck needs a rest like you wouldn't believe.

 

[WannabeNewton]

What is your mathematical definition of geometric in the language of category theory? If you are using physics/math books which don't mathematically define the terms they are using as references then that is of no real use. Again, a tensor is purely algebraic as it only requires a vector space structure to define. Nowhere in its definition is there a requirement of a topological structure, a smooth structure, a Riemannian structure, a symplectic structure, or a metric structure. By your argument a group is geometric just because there are specific examples of groups which pertain to Platonic solids or to smooth manifolds but a group is an over-arching algebraic object.

 

[Phy_Man]

He's not incorrect. Tensors are elements of products of vector spaces, that's completely algebraic. They are multilinear maps.

Oy! I'd like to ask you folks to please read my posts more carefully in the future. It's too painful to have to repeat myself.

Jorriss - I did not say that they weren't algebraic. I explicity stated that they are more than that. They are, by defiition, geometric objects. Multilinear maps are geometric objects. I'd appreciate it if you'd look up the definition of geometric object before you claim that the defintion is wrong or that what I said is the definition is wrong. I'm in too much pain to have to repeat myself anymore in this thread. If there are any questions for me on this subject then please send them to me in PM. Thank you.

I don't mean to come off like I'm irritated. I'm just in a ton of pain from all this typing.

 

[Phy_Man]

What is your mathematical definition of geometric in the language of category theory? If you are using physics/math books which don't mathematically define the terms they are using as references then that is of no real use. Again, a tensor is purely algebraic as it only requires a vector space structure to define. Nowhere in its definition is there a requirement of a topological structure, a smooth structure, a Riemannian structure, a symplectic structure, or a metric structure. By your argument a group is geometric just because there are specific examples of groups which pertain to Platonic solids or to smooth manifolds but a group is an over-arching algebraic object.

Sorry but you have all the information you need to answer these questions yourself. I can no longer stand the pain and I'm not into repeating myself for no apparent reason. If you reject the definition then that's not my problem.

 

[Jorriss]

Jorriss - I did not say that they weren't algebraic. I explicity stated that they are more than that. They are, by defiition, geometric objects. Multilinear maps are geometric objects. I'd appreciate it if you'd look up the definition of geometric object before you claim that the defintion is wrong or that what I said is the definition is wrong. I'm in too much pain to have to repeat myself anymore in this thread. If there are any questions for me on this subject then please send them to me in PM. Thank you.

Fair enough, but the modern definition of a tensor does not require structures that are generally associated with a geometric theory such as an inner product structure or metric structure.

 

[WannabeNewton]

Sorry but you have all the information you need to answer these questions yourself. I can no longer stand the pain and I'm not into repeating myself for no apparent reason. If you reject the definition then that's not my problem.

Your reference was an age old dover book on differential geometry. Tensors are not defined like that anymore. If you want to talk mathematics you must use modern definitions. I didn't reject the definition, the mathematical community did.

 

[WannabeNewton]

By the way, for anyone interested, the most general modern definition of a tensor can be found in e.g. chapter 14 of Roman's "Advanced Linear Algebra" or chapter 13 of Lang's "Linear Algebra".

 

[Dale]

I'm sorry my good man but I don't see a blue point.

Look harder, it is there. You need to click on the thumbnail to make it full-size.

I assume that you understand that when I said that setting up a coordinate system is identical to settiing up a basis that it can't be taken to mean that a set of basis vectors is a coordinate system, right?

If they are "identically the same thing" (your words) then given a specified set of basis vectors you should be able to determine the coordinates.

Regarding your question, I assume you meant to place a point in that diagram somewhere to represent the position vector which is a displacement vector from a point chosen as the origin?

Nope, I didn't identify any point as the origin. That is one specific problem with your claim that makes it wrong. Setting up a system of basis vectors in no way identifies an origin.

 

[Phy_Man]

Fair enough, but the modern definition of a tensor does not require structures that are generally associated with a geometric theory such as an inner product structure or metric structure.

Since I don't see a question mark I won't respond to this. I have no idea what that's supposed to mean anyway (not that I want to). The modern definition of a tensor is a multilinear map from vectors and 1-forms to reals. The definition of geometrical object is consistent with that fact. As I said, I won't discuss definitions anymore. I'm too old and too tired for such useless conversations. If people don't like the way things are defined I'd wish they'd simply say so and stop wasting time. Thanks.

 

[TrickyDicky]

Ohanian and Ruffini are wrong on this point ("no choice but to set up a coordinate system"). You can set up a set of basis vectors without setting up a coordinate system.

If I may, the set of basis you were thinking of happened to be ordered? There might be some confusion with your statements if you don't specify this. Ohanian and Ruffini are clearly referring to ordered bases.

Components are then contractions with one of the basis vectors, which is still coordinate-free. So just because you are dealing with components doesn't mean that you are dealing with coordinates. Their conclusion ("no choice ...") doesn't follow from their argument ("cannot ultimately avoid ... components"). Coordinates always imply a unique vector field called the coordinate basis, but a basis does not imply a unique coordinate system.

See above. Also, Ohanian and Ruffini are not talking about a "unique" coordinate system in the quoted paragraph but about coordinate systems in general.

 

[WannabeNewton]

DaleSpam's point was that you don't need a coordinate system to talk about the components of tensors. That only applies to the components of a tensor as represented in a coordinate basis. The problem is that there are GR books which use the word "component" as if it unequivocally meant "coordinate basis components". Wald does this too in fact, in the beginning of ch4.

 

[Samshorn]

You can set up a set of basis vectors without setting up a coordinate system... So just because you are dealing with components doesn't mean that you are dealing with coordinates... a basis does not imply a unique coordinate system.

No one is saying it does. Remember, this discussion began with the claim that by using the "coordinate-free" approach we can dispense with "reference frames and coordinate systems and Lorentz transformations". Since a reference frame is an equivalence class of coordinate systems that all share the same measures of spatial distances, temporal intervals, speeds, angles, etc., (and also to sidestep the ambiguous aspects of the word "frame", and also since it is the natural contra-distinction to "coordinate-free") we've been sometimes referring to reference frames informally as coordinate systems - but not with the intent of suggesting uniqueness, which would be absurd. The coordinate system obviously only needs to be specified up to the point of determining all measures of distances, times, speeds, angles, etc. Which basically means we need to specify the frame - or a basis if you prefer.

Note that those saying we can dispense with frames are also saying that components have no physical meanings - not just that coordinates have no physical meanings. This is a fairly standard notion of what the coordinate-free approach entails. Any time you resort to indices on your tensors, and actually quantify the components of a tensor, you are diverging from the coordinate-free precepts, by their own admission, because even they recognize that choosing a basis is tantamount to establishing a (equivalence class of) coordinate system. For example, D'Inverno has this to say

"There are two distinct approaches to the teaching of tensors: the abstract or index-free (coordinate-free) approach and the conventional approach based on indices... The disadvantage [of the coordinate-free approach] is that when one wants to do a real calculation with tensors, as one frequently needs to, then recourse has to be made to indices."

You see, the very use of indices (or quantifying components) is understood to be a transgression against the coordinate-free (index-free) approach. And, again, in order to do actual calculations, this is what we must do. And from the foundational standpoint (which is what this thread is about), since the comparison with observation unavoidably involves this kind of actual calculation, we can't dispense with (equivalence classes of) coordinate systems, or, if you prefer, frames, or basis, or however you prefer to think about it.

 

[robphy]

For example, D'Inverno has this to say

"There are two distinct approaches to the teaching of tensors: the abstract or index-free (coordinate-free) approach and the conventional approach based on indices... The disadvantage [of the coordinate-free approach] is that when one wants to do a real calculation with tensors, as one frequently needs to, then recourse has to be made to indices."

You see, the very use of indices (or quantifying components) is understood to be a transgression against the coordinate-free (index-free) approach.

"Coordinate-free" and "index-free" are not synonyms.
(Penrose's) Abstract Index Notation is coordinate-free, but uses indices to label slots of a tensor.
http://en.wikipedia.org/wiki/Abstract_index_notation
(abstract-index does NOT mean, e.g., "$\mu$ stands for $\{ t,x,y,z \}$".)

Classical tensor calculus notation, however, uses indices which refer to particular choices of coordinates or bases. Here, it may be that "$\mu$ stands for $\{ t,x,y,z \}$".

The abstract index notation tries to bridge
the component-based methods of classical tensor calculus
with the coordinate-free methods of modern mathematical treatments.
http://books.google.com/books?id=hQdh3SVgZ8MC&pg=PA56&dq="abstract+index"+penrose

 

[TrickyDicky]

DaleSpam's point was that you don't need a coordinate system to talk about the components of tensors.

I doubt that was Dalespam's point. Such triviality has nothing to do with the Ohanian and Rufinni quote he said was incorrect. You seem to be stuck with the algebraic aspect of tensors and maybe didn't notice that O&R mention: "the physicist who wishes to measure a tensor" suggesting they are referring to the geometric aspect of tensors of a geometric space (such as the Minkowski manifold if we talk about SR).

 

[WannabeNewton]

That's the whole point of a frame field; the integral curves of the frame field represent a family of ideal observers who each carry a measuring apparatus with which they make a complete set of measurements of the components of tensor fields on space-time. These measurements will be independent of whatever coordinate chart you use to represent the frame field.

 

[TrickyDicky]

That's the whole point of a frame field; the integral curves of the frame field represent a family of ideal observers who each carry a measuring apparatus with which they make a complete set of measurements of the components of tensor fields on space-time. These measurements will be independent of whatever coordinate chart you use to represent the frame field.

Mate, I suggest you read again Samshorn's posts in this thread and then consider the apparently innocent phrase " a family of ideal observers who each carry a measuring apparatus". You might find it iluminating.

 

[TrickyDicky]

WN I see you edited your initial post and I'm not disagreeing with your new last line at all.

 

[WannabeNewton]

I'm not disagreeing with you either. I'm just trying to describe the physical/geometric nature of frame fields. If I have a frame field $\{(e_{a})^{\mu}\}$, the integral curves of $(e_0)^{\mu}$ (which is chosen to be time-like by definition of a frame field) will be a congruence of time-like curves, at the least on some proper open subset of space-time. At each event, we think of the orthonormal basis $\{(e_a)^{\mu}|_p\}$ for $T_p M$ as representing three perpendicular meter sticks and a clock, carried by the observer represented by the unique time-like curve in the congruence intersecting that event. I refer again to section 13.1 of Wald.

Notice how I said before that "these measurements will be independent of whatever coordinate chart you use to represent the frame field" but you still need some coordinate chart to actually write down the frame field and make down to Earth computations. You can just be rest assured that it wouldn't have mattered which coordinate chart you chose.

EDIT: By the way, I'm not berating Ohanian's book or anything. I have the book myself (assuming we are talking about "Gravitation and Spacetime") and I absolutely treasure it, especially its devoted focus to experiment instead of theory.

 

[TrickyDicky]

I'm not disagreeing with you either.
Notice how I said before that "these measurements will be independent of whatever coordinate chart you use to represent the frame field" but you still need some coordinate chart to actually write down the frame field and make down to Earth computations. You can just be rest assured that it wouldn't have mattered which coordinate chart you chose.

Totally agreed.
But then I'm not sure what the disagreement regarding the O&R quote is about (or the PeterDonis-Samshorn exchange for that matter). Are you guys making some nuanced distinction between coordinate charts and coordinate systems? I thought they were the same thing.

 

[WannabeNewton]

I wasn't going against or with the quote in any way. I was just trying my best to possibly clear up the various ways in which tensors are used/described in GR texts because there is a noticeable lack of consistency amongst them (which is not to say it's anyone's fault of course).

 

[TrickyDicky]

I was just trying my best to possibly clear up the various ways in which tensors are used/described in GR.

I know, and I always find it informative. Even if the discussion in this case seemed more centered in SR.

 

[WannabeNewton]

I think even in SR, the fact that the states of the physical theory on Minkowski space-time can be taken to be tensor fields and that its isometry group is the proper Poincare group allows for an instructive use of the same type of formalism i.e. frame fields and the link between the measurements of components of tensor fields made using one frame field and another through special covariance under the proper Poincare transformations.

 

[Dale]

If I may, the set of basis you were thinking of happened to be ordered? There might be some confusion with your statements if you don't specify this. Ohanian and Ruffini are clearly referring to ordered bases.

Yes, I was assuming a standard tetrad, so smooth, orthonormal, and ordered. It still doesn't give you a coordinate system. See the picture I drew earlier. The basis vector field simply doesn't imply a coordinate system.

 

[TrickyDicky]

Yes, I was assuming a standard tetrad, so smooth, orthonormal, and ordered. It still doesn't give you a coordinate system. See the picture I drew earlier. The basis vector field simply doesn't imply a coordinate system.

I think we all agree your abstract example doesn't give you a unique coordinate system. But this misses the point of the O&R quote.

 

[Dale]

No one is saying it does.

Phy_man did.

a reference frame is an equivalence class of coordinate systems that all share the same measures of spatial distances, temporal intervals, speeds, angles, etc.

I have never heard that definition of a reference frame. I have always heard a reference frame described as a smooth set of orthonormal vectors at each point, i.e. a tetrad or vierbien. From the tetrad you could find an equivalence class of coordinate systems which all share the same tetrad as their coordinate basis, but the tetrad itself does not pick out any specific coordinate system, so using the tetrad is not the same as using the coordinate systems.

we've been sometimes referring to reference frames informally as coordinate systems

I do that too . That may be the confusion.

Note that those saying we can dispense with frames are also saying that components have no physical meanings - not just that coordinates have no physical meanings.

I am not saying that. Components are themselves vectors, a given vector can always be expressed as a sum of components. As vectors, components "live" in the tangent space.

Coordinates "live" in the manifold itself, not the tangent space. In general, they do not follow the laws of vector addition. They are not themselves vectors.

You can do physics without coordinates (if you are masochistic), but I don't see any way to do physics without components. Coordinates certainly make things easier, and from a coordinate system it is easy to obtain a tetrad and the associated components whenever needed, but they are indeed not "foundational", so you always have a choice to use them or not.

 

[Dale]

I think we all agree your abstract example doesn't give you a unique coordinate system. But this misses the point of the O&R quote.

That example was a response to Phy_man's assertion that a coordinate system and a set of basis vectors are "identically the same thing". They are clearly not the same thing.

 

[DrGreg]

I think we all agree your abstract example doesn't give you a unique coordinate system. But this misses the point of the O&R quote.

It's not just an issue of uniqueness. There are other examples of tetrads that cannot be associated with any coordinate system.

Consider the frame field associated with a Born-rigid rotating cylinder in flat spacetime. Set the timelike basis vector along the worldlines of all the observers at rest relative to the cylinder and three spacelike vectors orthogonal to that, in the radial, tangential and axial directions. If you choose a single observer you can find a local coordinate system associated with that observer's tetrad only, but you can't find an extended single coordinate system compatible with all of the tetrads everywhere.

 

[TrickyDicky]

It's not just an issue of uniqueness. There are other examples of tetrads that cannot be associated with any coordinate system.

Consider the frame field associated with a Born-rigid rotating cylinder in flat spacetime. Set the timelike basis vector along the worldlines of all the observers at rest relative to the cylinder and three spacelike vectors orthogonal to that, in the radial, tangential and axial directions. If you choose a single observer you can find a local coordinate system associated with that observer's tetrad only, but you can't find an extended single coordinate system compatible with all of the tetrads everywhere.

But of course, they are called nonholonomic or noncoordinate basis and it's an abstract notation frequently used.
But we have been talking here about measuring components, that is, giving numerical values as the mathematical representation of physical tensors(that are geometric objects) and for doing that it seems to me one needs coordinate basis and these components will depend on the coordinate system used. So surely depending on the level of abstraction one can dispense with coordinates but when giving specific numerical values one depends on the specific coordinate system used and as have been commented several times if measuring physical quantities in SR one ultimately is always referring to inertial coordinates.

 

[TrickyDicky]

That example was a response to Phy_man's assertion that a coordinate system and a set of basis vectors are "identically the same thing". They are clearly not the same thing.

Clearly. But you were who equated what O&R said about measuring the components of a physical tensor with setting up a basis when it is not the same thing, the latter is a more abstract and broader concept and doesn't depend on coordinate systems while the former depends on the chosen reference frame(ordered coordinate basis).