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Tensors, the geometric tool that solved Einstein's relativity problem

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Re: Tensors, the geometric tool that solved Einstein's relativity problem

#71

Post by cfgauss2718 »

klodolph wrote: post #40
will-burner wrote: post #6

Earlier quoted context omitted.

The definition may be simple, but it's not very concrete and I'd argue that makes it not strait forward. While examples of vector spaces can be very concrete (think R, R^2, R^30), I struggle to think of a concrete example of a multilinear function from vectors and dual vectors in V to numbers in K. On top of that when working with tensors, you don't usually use the definition os a multilinear function at least as far…

I think part of this is “if you have a linear algebra background”. There are a few different explanations of tensors, and different explanations make sense for different people.

Yes, the “multi linear map” definition is accessible to an undergraduate who has taken linear algebra. However, the more common meaning of tensor in physics, like the metric tensor of spacetime, requires some more sophisticated background to understand (differential geometry, Lie Groups come to mind).

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#72

Post by cfgauss2718 »

g15jv2dp wrote: post #21
senderista wrote: post #2

If you have any linear algebra background, then the definition of a tensor is straightforward: given a vector space V over a field K (in physics, K = R or C ), a tensor T is a multilinear (i.e. linear in each argument) function from vectors and dual vectors in V to numbers in K . That's it! A type (p, q) tensor T takes p vectors and q dual vectors as arguments ( p+q is often called the rank of T but is ambiguous comp…

That's incorrect if V is infinite-dimensional. A (0,1)-tensor is just supposed to be an element of V but with your definition you get an element of the bidual of V. Which is not isomorphic to V when dim V is infinite. And even when dim V is finite, you need to choose a basis of V to find an isomorphism with the bidual. From a math point of view, that's just no good.

Can you provide some examples of important tensors in physics for which the underlying vector space is infinite dimensional? I’m most familiar with the setting of tensor fields on manifolds, in which case the vector bundle consists of finite dimensional vector spaces. Nevertheless, I suppose in the absence of a pseudo-Riemannian metric one lacks a natural isomorphism between vectors/dual vectors. Does this “bidual” distinction arise in that case as well?

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#73

Post by cfgauss2718 »

will-burner wrote: post #6
senderista wrote: post #2

If you have any linear algebra background, then the definition of a tensor is straightforward: given a vector space V over a field K (in physics, K = R or C ), a tensor T is a multilinear (i.e. linear in each argument) function from vectors and dual vectors in V to numbers in K . That's it! A type (p, q) tensor T takes p vectors and q dual vectors as arguments ( p+q is often called the rank of T but is ambiguous comp…

The definition may be simple, but it's not very concrete and I'd argue that makes it not strait forward. While examples of vector spaces can be very concrete (think R, R^2, R^30), I struggle to think of a concrete example of a multilinear function from vectors and dual vectors in V to numbers in K. On top of that when working with tensors, you don't usually use the definition os a multilinear function at least as far…

Here’s maybe a useful example. Consider a scalar potential function F on R^3 that describes some nonlinear spring law. At a point p=(x,y,z), the differential dF can be thought of as a (1,0) tensor measuring the spring force. It acts on a particle at p moving with velocity v to give the instantaneous work of the particle on the spring dF(p)(v). Now, suppose that we want to know how this quantity changes when we vary the x coordinate. The x coordinate is also a function of p, we can represent its differential as dx, which is a co-vector(field). The quantity that captures this change can be thought of as a (1,1) tensor field, which is related to the stiffness of the spring potential in the x direction at each point p. In the usual undergraduate setting, this tensor field is given as the hessian of F, call this H. The action of this tensor looks like the product u^T H(p) v, where in our case, u^T = dx(p) = [1 0 0]. A good giveaway for when a “co-vector” appears in a tensor calculation is whenever there is a “row vector” in a matrix operation (most people identify “column” vectors with proper vectors). It’s helpful in this case that “row” rhymes with “co-“.

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#74

Post by enugu »

prof-dr-ir wrote: post #47
adrian_b wrote: post #44

Earlier quoted context omitted.

It does not matter on what set a tensor field is defined. A tensor field is not a tensor, but the value of a tensor field at any point is a tensor, which satisfies the definition given above, exactly like the value of a vector field at any point is a vector. The "fields" are just functions. There are physics books that do not give the easier to understand definition given above, but they give an equivalent, but more…

> The "fields" are just functions. I think this is far too simplistic, for one because the values of this putative function depend on the chosen coordinate system. So I completely agree with the comment you are replying to: when a physicist says "tensor" they really mean a "tensor field" and the definition of the latter is quite a bit more involved than just specifying a multilinear map at each point of a manifold.

There is a way of defining a vector space without an explicit basis(just as a set with an addition and a scalar multiplication). Similarly, there is a way of defining a vector bundle without choosing an explicit coordinates (as an abstract vector space, as defined above, which varies with the point in the space).

Just as a (p,q) tensor is a multilinear object related to a single vector space, a tensor field is a section of a tensor bundle associated to the vector bundle. (A section is just a function on the underlying space whose value at a point lies in the vector space above the point.)

Usually, the vector bundle relevant in physics is the tangent bundle of a 4-manifold.

This abstract way of defining tensors and tensor fields is manifestly invariant under coordinate changes, but it takes some machinery to set up. Whereas the 'numbers associated to each coordinate system which transform in a certain way' is more direct, but the rules can seem arbitrary at first sight. Also, maybe this approach can generalize to allow more transformation rules which might take some time to put into an abstract setting.

Standard example is a matrix A which transforms as PAP^-1 (where P is linear coordinate change) vs a matrix T which is a linear map between vector spaces.

The same issue appears in software where you can expose a data structure as a tuple of numbers/string fields and then define functions on them, or you can expose it as an abstract data type where the user of the library can only apply certain functions on them and the implementation author can choose different representations(coordinate changes) in which to easily compute the functions.

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#75

Post by abdullahkhalids »

adrian_b wrote: post #62
abdullahkhalids wrote: post #51

Earlier quoted context omitted.

> There are physics books that do not give the easier to understand definition given above, but they give an equivalent, but more obscure, definition of a tensor, by giving the transformation rules for its contravariant components and for its covariant components at a change of the reference system. The definition of a tensor as linear maps, while simple to understand, has no content that is useful for doing physics.…

No, the definition of a tensor as a linear map, is the only definition that is useful for doing physics. All the physical quantities that are defined to be tensors are quantities used to transform either vectors into other vectors or tensors of higher orders into other tensors of higher orders (for instance the transformation between the electric field vector and the electric polarization vector). Therefore all such…

> On the other hand, the abstract formulation of the physical laws does not need such details, but only the high-level definitions using multi-linear functions, and it is independent of any choice for the coordinate system.

This is exactly the point. Abstract physical laws must be invariant to coordinate transformation. From a pedagogical point of view, perhaps this is less important when discussing anisotropic media, but critical when discussing general relativity. Hence, the first reason why many physicist book writer think it very important that covariance/contravariance of tensors be central to both their definition and their pedagogy as applied to physics. You have to convince the student that tensors are the right mathematical objects to describe reality because they preserve this invariance.

The second reason is just as important. Physics is nothing without validating abstract physical laws by experiment. And that validation can not be done without computing predictions. Which in turn will require the right coordinate system, which will require covariance/contravariance of tensors. You can't just disregard these computations as unimportant or unnecessary from either a pedagogical point of view or a deeper philosophical one.

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#76

Post by enugu »

eigenket wrote: post #55
bjourne wrote: post #49

Earlier quoted context omitted.

Well, I can write a definition that is both easier to understand and shorter than yours: A tensor is a multi-dimensional array. :)

This only works in finite dimensions, which for mathematicians excludes pretty much all of the interesting cases.

Finite dimension tensors is interesting both in physics(ex: mechanics, electromagnetism, general relativity) and mathematics(ex: representation theory, differential geometry). Infinite dimensions is also used in physics(quantum theory) and in mathematics (Operator Algebras, representation theory again).

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#77

Post by eigenheart »

ijidak wrote: post #30

Here is a video series on tensors I've enjoyed: https://youtube.com/playlist?list=PLJHszsWbB6hrkmmq57lX8BV-o... And this series by Dialect: https://youtube.com/playlist?list=PL__fY7tXwodmfntSAAyBDxZ4_...

https://news.ycombinator.com/item?id=17817655

Hackernews user saivan started notes on eigenchris's tensor series videos.

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#78

Post by g15jv2dp »

xanderlewis wrote: post #42
g15jv2dp wrote: post #34

Earlier quoted context omitted.

GP is giving you an element of V**. You want to turn it into a vector. To do that, please make the inverse isomorphism explicit without using a basis. I'll wait...

You said even when dim V is finite you need a basis to find an isomorphism with V**. But that’s not true. You’re right if you mean V*, but not the bidual.

I'm talking about a practical task. Not an abstract existence result.

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#79

Post by xanderlewis »

g15jv2dp wrote: post #78
xanderlewis wrote: post #42

Earlier quoted context omitted.

You said even when dim V is finite you need a basis to find an isomorphism with V**. But that’s not true. You’re right if you mean V*, but not the bidual.

I'm talking about a practical task. Not an abstract existence result.

It’s not that abstract. You can literally write down the isomorphism: v —> [e_v: V* —> F, e_v(f) = f(v)].

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#80

Post by eigenket »

enugu wrote: post #76
eigenket wrote: post #55

Earlier quoted context omitted.

This only works in finite dimensions, which for mathematicians excludes pretty much all of the interesting cases.

Finite dimension tensors is interesting both in physics(ex: mechanics, electromagnetism, general relativity) and mathematics(ex: representation theory, differential geometry). Infinite dimensions is also used in physics(quantum theory) and in mathematics (Operator Algebras, representation theory again).

I guess? Mostly the cool applications in physics and differential geometry are about tensor fields, which are more complicated than bare tensors. You could argue that they're talking about finite dimensional tensors but tensor fields are kinda a different object (at least subjectively, to me).
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