Recommendaton for Clarifying Special Relativity

[PeterDonis]

it's a question of what the Doppler formula means. It means that if you plug in the ordinary speed v (based on the inertial definition), you get the relativistic Doppler shift. That's the non-trivial physical fact that the formula is expressing.

Yes, but that physical fact by itself doesn't tell you whether you should interpret it as telling you that, oh, look, the observed Doppler shift just happens to be exactly equal to what the formula tells you when you plug in speed v; or telling you that, oh, look, the speed v just happens to be exactly what you would expect when you plug the Doppler shift into the (inverted) formula.

Your position seems to be that the meaning of v in the Doppler equation need not have anything to do with the ordinary inertial meaning of speed.

No, my position is that the physical fact the formula is expressing can be interpreted in either direction, so to speak. It's telling you about a connection between two different sets of phenomena. It's not telling you which set of phenomena is "more fundamental". That's a matter of interpretation.

 

[Samshorn]

Yes, but that physical fact by itself doesn't tell you whether you should interpret it as telling you that, oh, look, the observed Doppler shift just happens to be exactly equal to what the formula tells you when you plug in speed v; or telling you that, oh, look, the speed v just happens to be exactly what you would expect when you plug the Doppler shift into the (inverted) formula.

In both of the "alternatives" you mentioned, the equation is empirically valid only if v equals the inertial speed, which is the point I've been making: The formula refers to inertial speed or else it's meaningless.

No, my position is that the physical fact the formula is expressing can be interpreted in either direction, so to speak. It's telling you about a connection between two different sets of phenomena. It's not telling you which set of phenomena is "more fundamental". That's a matter of interpretation.

I don't think it's a question of whether one thing is "more fundamental" than another (whatever that might mean). The point is just that both are physically meaningful, and the task of the theory is to describe how they are related. That's why we shouldn't say special relativity has nothing to do with inertial coordinate systems or inertial measures of space and time (or with the Lorentz transformations that relate them to each other). It has everything to do with these things.

As to why we ordinarily conceive of phenomena as existing in space and time, rather than in some abstract "Doppler realm", I think that's a complicated subject. There are always naive positivists urging us to shed our hide-bound notions of a common space and time, and just focus on the raw uninterpreted sense impressions impinging on our own individual world line. The natural tendency of this line of thought is toward solipsism, but even if you succeed in avoiding that trap, most people find that there are good reasons for retaining the conceptual framework of space and time. (I mentioned some in a previous message.) On the other hand, quantum phenomena for n particles seem to reside in 3n-dimensional phase space, suggesting that our concept of 3+1 dimensional spacetime is not fundamental. Nevertheless, we somehow still make the conversion back to space-time representations for most purposes.

 

[bobc2]

...There are always naive positivists urging us to shed our hide-bound notions of a common space and time, and just focus on the raw uninterpreted sense impressions impinging on our own individual world line. The natural tendency of this line of thought is toward solipsism, but even if you succeed in avoiding that trap, most people find that there are good reasons for retaining the conceptual framework of space and time. (I mentioned some in a previous message.)...

Well said.

 

[PeterDonis]

In both of the "alternatives" you mentioned, the equation is empirically valid only if v equals the inertial speed

Yes, but "inertial speed" means "the actual measured speed according to a given procedure". You can describe that procedure and its results without defining inertial coordinates.

That's why we shouldn't say special relativity has nothing to do with inertial coordinate systems or inertial measures of space and time (or with the Lorentz transformations that relate them to each other). It has everything to do with these things.

Once again, you are conflating inertial coordinates with the physical properties and measurements that make inertial coordinates useful. SR does have everything to do with those physical properties and measurements, yes. But you can describe them without defining inertial coordinates.

As to why we ordinarily conceive of phenomena as existing in space and time, rather than in some abstract "Doppler realm", I think that's a complicated subject.

Yes, it is, because it's not just about physics; it's about our cognitive systems, which are much more complicated than the simple physical systems we're talking about here.

There are always naive positivists urging us to shed our hide-bound notions of a common space and time, and just focus on the raw uninterpreted sense impressions impinging on our own individual world line.

I have not said anything like this. Once again, you're conflating coordinates with physical properties that are conveniently described using coordinates. Spacetime, as a geometric object, is certainly the natural outcome of reconciling our common notions of space and time with the other physical facts we have discussed. But spacetime can be described without coordinates. That's all I'm saying. That's not the same as saying spacetime doesn't exist, only our sense impressions do.

 

[Samshorn]

Yes, but "inertial speed" means "the actual measured speed according to a given procedure". You can describe that procedure and its results without defining inertial coordinates.

If someone defines a speed (for example) using a procedure operationally equivalent to defining an inertial coordinate system and dividing the space difference by the time difference, then I'd count that as using an inertial coordinate system, which is to say, using inertial measures of space, time, speed, angles, etc. For example, the MTW exercise you cited as a coordinate-free "calculation" used an inertial coordinate system by referring to the "ordinary speed v".

Once again, you are conflating inertial coordinates with the physical properties and measurements that make inertial coordinates useful.

No, I'm conflating the use of inertial coordinates with the use of measures of distance, time, speed, angles, etc., based on the concept of inertial coordinates. If, for example, someone says an object is moving at speed v, and they mean this as the ordinary speed in terms of an inertial coordinate system, then I would say they have invoked an inertial coordinate system. (The v in the Doppler formula is just such a speed, and hence it refers to inertial coordinates.)

SR does have everything to do with those physical properties and measurements, yes. But you can describe them without defining inertial coordinates.

If someone asks you what special relativity predicts for the Doppler shift of a light source moving away from us at the speed v, where v is an "ordinary speed" defined in terms of inertial coordinates, how would your application of coordinate-free reasoning erase the fact that what you're doing is explicitly answering a question whose parameters are defined in terms of inertial coordinates?

Spacetime can be described without coordinates. That's all I'm saying.

Ohanian and Ruffini had it right: "We must not forget that the physicist who wishes to measure a tensor has no choice but to set up a coordinate system, and then measure the numerical values of the components. Thus, to carry out the comparison of theory and experiment, the physicist cannot ultimately avoid the language of components; only a pure mathematician can adhere exclusively to the abstract coordinate-free language of differential forms..." You see, this is the point: The epistemological foundations of a physical theory rest entirely on "the comparison of theory and experiment", and this decisive link is provided by coordinates. Without this link, any formal mathematical expressions are devoid of physical content.

 

[Dale]

Sure, just write down all the equations the same way you can in GR, using only coordinate-free tensor expressions, and use the metric of Minkowski spacetime.

No, if you use the metric of Minkowski spacetime then you are already assuming SR and not making a test theory for SR.

Or do you mean by "test theory" something like the Cartan geometric formulation of Newtonian gravity? In other words, casting Galilean spacetime in terms of coordinate-free tensor equations?

In order to test a theory you cannot assume it, so the best approach is to assume a test theory. A test theory is a general theory which has a set of one or more unknown parameters. Various competing theories (such as SR or Newtonian physics) are then reproduced through specific choices of the unknown parameters. You then propose an experiment to put constraints on the unknown parameters and see how closely they match the parameters corresponding to the various theories.

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.

I don't know that that has been done specifically, but I don't see why it couldn't be.

I also don't see why it couldn't be, but as far as I know it has not been done, so I would be reluctant to make claims about it.

 

[Dale]

Ohanian and Ruffini had it right: "We must not forget that the physicist who wishes to measure a tensor has no choice but to set up a coordinate system, and then measure the numerical values of the components. Thus, to carry out the comparison of theory and experiment, the physicist cannot ultimately avoid the language of components; only a pure mathematician can adhere exclusively to the abstract coordinate-free language of differential forms..."

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. 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.

 

[Phy_Man]

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. 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.

Ummmm, nope! I'm afraid that you missed a very fundamental fact here. Ohanian and Ruffini are absolutely correct. They're experts in their field and know precisely what they're talking about. Setting up a system of basis vectors is identically the same thing as setting up a coordinate system. Contracting components on a basis is just another name for measuring components. Dealing with components is identical to dealing with a coordinate system. From your response it appear as if you might have a flawed notion of what a coordinate system is.

 

[Dale]

Setting up a system of basis vectors is identically the same thing as setting up a coordinate system.

See the attached image for a system of basis vectors. What are the coordinates of the blue point?

 

[Phy_Man]

See the attached image for a system of basis vectors. What are the coordinates of the blue point?

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

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? While you're at it please make it clear as to whether you're asking me for the components of the displacement vector (which is what a position vector is) from the origin that the point represents or whether you have something else in mind. Thank you.

When someone sets up a basis then wants to express the position vector in terms of that basis they have to locate the two points which define the displacement vector; one point represents the point represented by the origin and the point of interest. When you tell me that there is a point in there and I should tell you its coordinates then what you're telling me is that you have chosen a point to serve as the origin. If not then you haven't mentioned a displacement vector and I have no response as a result.

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.

Or perhaps you thought you had me? Nope. Sorry. :D

 

[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).