Motivation for tensors in GR

jdstokes

I quite often hear that GR is formulated in terms of tensors because laws of physics expressed in terms of tensor equations are indepedent of choice coordinates because they `transform nicely'.

I thought the motivation for tensors was that since spacetime is curved, we locally linearize it by introducing the tangent space to a point. Then since all physics happens within the tangent space, all quantities of interest can be expressed as multilinear mappings from the tangent/cotangent space to the real numbers. Coordinate invariance then follows because of the same reason that vectors have a coordinate independent existence in the tangent space, tensors have a coordinate independent existence in tensor product space.

I don't see the connection between coordinate invariance and ``transforming nicely'' under a change of coordinates. Ie, what would be the supposed `bad' implication of not transforming according to the tensor transformation law??

CompuChip

Not transforming according to the tensor transformation law would mean that you would get additional terms containing partial derivatives of one set of coordinates with respect to the other, which would not cancel in the end. This means that your answer would become dependent on the coordinates you would choose.

lbrits

I guess you could say that, from a differential geometry point of view, tensors (or something equivalent to tensors), are a unique way of dealing with functions and their derivatives on manifolds. I think uniqueness follows from a choice of defining an inner product, but anyway. So once you've agreed on how to take dot products on flat space, you either end up with tensors and metrics and whatnot on curved space, or you end up with something else. The something else isn't going to be coordinate invariant because of uniqueness.

Now, historically a lot of "reasons" people had turned out to be false but had an important role to play in developing our understanding of GR. I think that few people have been more confused by GR than Einstein himself.

In any event, for a concrete example, lets say I can find a coordinate system in which I can write the laws of physics as [tex]\partial_\mu \partial^\mu \Phi = 0[/tex] where [tex]\Phi[/tex] is just some scalar field. Now, no one in their right mind would claim this has any meaning (what I wrote down is garbage), but bear with me. So I go out and claim that this is a law of nature. Someone else picks a different coordinate system, and they evaluate what [tex]\partial_\mu \partial^\mu \Phi[/tex] is. Now, they take their resulting expression, and write it in terms of my coordinates, and they do not find [tex]0[/tex]. Obviously my coordinates were special (specifically, [tex]\Gamma^\alpha_{\beta\gamma} = 0[/tex]) for me, whereas not for them, and what I thought was a "law of nature" was more like a crude observation which doesn't immediately generalize.

I think it is important to phrase the answer to your question in this way, because there are different meanings of coordinate invariance (to some people, at least). But here it is meant in the sense that "the equation looks the same" rather than "evaluates to the same".

pmb_phy

Tensors are actually used in Newtonian physics as well. 3-vectors and 3-tensors are tensors by definition. They are referred to as Cartesian tensors (aka Affine tensors). They are defined according to how their components transform under an orthogonal transformation. In SR the tensors are Lorentz tensors which are defined according to how their components transform under a Lorentz transformation. In GR tensors can be defined according to how their components transform under an arbitrary coordinate transformation. The reason they transform as such is because their definition is independant of any particular coordinate system and laws of physics must be able to be defined in such an arbitrary fashion.
The motication is that the laws of physics must have the same form in in all coordinate systems, regardless of whether the undelying manifold is curved or not.
"Invariance" refers to the geometric nature of tensors in that they have a coordinate independant meaning. In fact a tensor can be defined as a map from vectors and 1-forms to real numbers (scalars). No coordinate system is used in such a definition. However once a coordinate system is defined it can be shown that it implies the transformation properties that you're referring to given the definition of how the components of vectors and 1-forms transform under coordinate transformation. The "bad" impilication that you're asking about is coordinate dependance. Such a dependance would violate a principle of general relativity.

Pete

jdstokes

Any tensor equation can be written as [latex]T = 0[/latex] where [latex]T[/latex] is some tensor and 0 is the zero tensor e.g. in component notation for a a rank-2 tensor this would read [latex]\sum_{\mu,\nu}T^{\mu\nu}\partial_\mu\otimes\partial_\nu = 0[/latex].

If I locally reparametrize the manifold (ie change coordinates by rotating, Lorentz boost or whatever) then the following will be true

[latex]\sum_{\mu,\nu}T'^{\mu\nu}\partial'_\mu\otimes\partial'_\nu = 0[/latex] where the new components are related to the old ones by the tensor transformation law. This obviously looks the same as the first expression. But what does any of this have to do with cancelation of terms, nothing canceled.

jdstokes

I think the problem I'm having is trying to understand why the form of ANY equation could depend on the coordinates chosen. A concrete example of may assist here.

It seems so obvious that if we only have access to the tangent space then our laws of physics will depend on those vectors. Saying that an equation of physics is has the same form in any coordinate system is merely a restatement of the fact that tensors are built out of tangent and cotangent vectors which are themselves coordinate independent objects.

Fredrik

It seems to me that the real question here is "Why can the laws of physics be expressed in terms of real-valued functions on M and local sections of tensor bundles of M?" I don't see how that question can be answered with anything but "We're just lucky I guess", or a direct derivation from a more fundamental theory.

Edit: Oooooh....ooohhhh...500 posts.

genneth

There are several ways to define things like vectors and tensors. It is an unfortunate state of affairs that physicists generally prefer the coordinate-based method. In essence, you need to distinguish between two things --- the tensor itself and the list of numbers that you choose to use to represent the tensor. The latter requires a coordinate system or frame for it to make sense, but the former is independent of the choice. Pure mathematicians often define vectors spaces as a purely algebraic structure, and tensors as linear spaces on top of that. As you've noticed, this has nothing to do with the way they "transform". Choosing a frame, and then writing down a set of numbers allow you to do concrete calculations with them. Change the frame, and the same tensor now has different numbers --- that is the transform. The fact is that the transformation is always a specific type, and can be calculated independently. Now the amazing thing is that you can also go backwards. So start with lists of numbers, and require that they transform in a certain way when you make a "change of basis", and you can recover the coordinate-free description as the unique result.

jdstokes

Hi Genneth,

I also find the ``physicist way'' of defining tensors to be extremely annoying because of its coordinate dependence, even though I consider myself a physicist.

Let me see if I can run through cartesian tensors the way a mathematician would see them. The first thing you do when you encounter a space [latex]S[/latex] with an inner product is to linearize it by introducing a vector space to each point. This is done by first defining a parametrization of the space and then defining the tangent space to be the set of derivatives of smooth curves passing through the point.

In the case of Euclidean space, there is a well-defined mapping between any tangent space defined by parallel transport, thus it is only necessary to consider the tangent space at the origin.

Note that a choice of basis for the tangent space is determined by a choice of parametrization [latex]F : \mathbb{R}^3 \to \mathbb{E}^3[/latex]. [latex]F[/latex] should be conceputalized as a physical measurement apparatus for assigning numbers to points in space (e.g. using steel rulers).

I think it is important to realize (as many physicists probably overlook this), that one cannot even talk about vectors (abstract algebraic things) in Euclidean space unless one has a notion of how to parametrize Euclidean space. It is through this parametrization that one defines vectors (which turn out to be independent of the choice of parametrization).

What the physicists don't seem to realize is that the choice of parametrization is totally arbitrary. In the same way that I could change parametrization by rotating my steel rulers, I could also change to spherical polar coordinates and THEN rotate my coordinates. In spherical coordinates, the components won't change under rotation the way a physicist would expect them to, so they would arge that the object is not a vector, but they are wrong, because they don't understand what a vetor is.

jdstokes

I'd be interested to see the proof of this. Can suggest a reference please?

jdstokes

Do you know a proof of this?

lbrits

Well, mmm... On a smooth manifold there is a natural correspondence between tangent vectors to curves through a point and directional derivatives of functions on the manifold at that point. So we establish [tex]\vec{v}(f) \mapsto \tfrac{d}{d \lambda}(f)[/tex]. At a point, these span the tangent space [tex]T_p(M)[/tex]. Now be daring and define it's dual space of oneforms like [tex]df[/tex] with the rule [tex]\langle \vec{v}, df\rangle = \tfrac{df}{d\lambda}[/tex], and which span [tex]T^*_p(M)[/tex].

Now, if you expand [tex]\vec{v}[/tex] in terms of a basis [tex]\tfrac{\partial}{\partial x^\mu}[/tex] and [tex]df[/tex] in terms of a dual basis [tex]dx^\mu[/tex] you will find that the inner product [tex]\langle\, , \, \rangle[/tex] is the normal Euclidean one, and you can convince yourself that it is coordinate invariant.

Now, to do anything like GR we would actually need to add "geometry" to the manifold in addition to topology, and so we need to know how to take dot products between vectors. A "natural" way to do this is a map from vectors to oneforms, since we already have an induced inner product. This map is the metric, and when you've done this, you get something like the Euclidean dot product between vectors when you have a flat metric.

I should retract my statement that it is a unique way. I don't really know how else you would do it. What I mean is that it is a "natural" way, in the sense that you take what you do in flat space, but do it in a coordinate free manner. When you make the space curved, everything follows from your coordinate invariant inner product [tex]\langle \, , \, \rangle[/tex] and your choice of metric.

As a physicist, I want to address your "don't know what a vector is", but right now I have to make supper =)

Edit: my discussion about tangent spaces is for the benefit of those joining the discussion, of course.

lbrits

Are you referring to active verus passive transformations?

In any case, in spherical coordinates, if you rotate your coordinates then the components of the vector would change exactly as a physicist would expect them to... if not, you're hanging out with the wrong physicists =)

jdstokes

In spherical polar coordinates If I rotate about the z-axis by [latex]\Delta \varphi[/latex] then my coordinates change by [latex](r,\theta,\varphi) \mapsto (r,\theta, \varphi + \Delta\varphi)[/latex]: not what was expected from the definition of a cartesian tensor.

lbrits

In what universe? =)

jdstokes

Why do you question that the components will transform differently, it seems clear to me that if you represent a vector in the curvilinear basis then the components will transform differently under rotation than were you to represent the vector in an orthogonal basis.

jdstokes

Getting back to the motivation for tensors, I think you might be able to at least partially justify it as follows:

Given a tensor [latex]T[/latex], we know from the chain rule that if the component representation wrt the parametrization [latex]x[/latex] is [latex]T^{\mu_1,\ldots,\mu_k}_{\nu_1,\ldots,\nu_k}[/latex], then the components with respect to another parametrization are

[latex]T^{\mu_1',\ldots,\mu_k'}_{\nu_1',\ldots,\nu_k'} = \frac{\partial x^{\mu_1'}}{\partial x_{\mu_1}}\cdots \frac{\partial x^{\mu_k'}}{\partial x_{\mu_k}}\frac{\partial x^{\nu_1'}}{\partial x_{\nu_1}}\cdots \frac{\partial x^{\nu_k'}}{\partial x_{\nu_k}}
T^{\mu_1,\ldots,\mu_k}_{\nu_1,\ldots,\nu_k}[/latex]

Thus, given tensor equation [latex]T=0[/latex] we then have

[latex]T^{\mu_1,\ldots,\mu_k}_{\nu_1,\ldots,\nu_k} = 0[/latex]

and moreover

[latex]T^{\mu_1',\ldots,\mu_k'}_{\nu_1',\ldots,\nu_k'} = \frac{\partial x^{\mu_1'}}{\partial x_{\mu_1}}\cdots \frac{\partial x^{\mu_k'}}{\partial x_{\mu_k}}\frac{\partial x^{\nu_1'}}{\partial x_{\nu_1}}\cdots \frac{\partial x^{\nu_k'}}{\partial x_{\nu_k}}
\times 0 = 0[/latex].

Thus the tensor equation holds in all frames of reference.

Surprisingly simple yet illustrates the point.