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

quantamagazine.org

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

#81

Post by g15jv2dp »

xanderlewis wrote: post #79
g15jv2dp wrote: post #78

Earlier quoted context omitted.

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)].

Are you even trying to understand my point? Yes, in that direction, explicitly constructing an element of the bidual of V from an element of V is easy. To explicitly find an element of V from the element of the bidual, you need to choose a basis. Just try it, come on! Write down the inverse isomorphism. Let \alpha be an element of V**. Then v \in V such that f(v) = \alpha (where f is the isomorphism you wrote down) is given by...?

I will help you: if (e_i) is a basis of V and (e_i^*) is its dual basis, then v = \sum_i \alpha(e_i^*) e_i. Can you find such a formula without mentioning the word "basis"?

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

#82

Post by g15jv2dp »

btilly wrote: post #45
g15jv2dp wrote: post #39

Earlier quoted context omitted.

There are no manifolds at all in this discussion... What are you even talking about? Just stringing words you vaguely know together??

This comment is in a discussion about an article titled, Tensors, the geometric tool that solved Einstein's relativity problem . Therefore, "tensors are literally the topic that started this discussion." Hopefully that's a hint that you should attempt to figure out what someone might be talking about before going to schoolyard insults.

Schoolyard insult? Uh? Can you quote the part of my comment that would be the "schoolyard insult"?

> This comment is in a discussion about an article titled, Tensors, the geometric tool that solved Einstein's relativity problem. Therefore, "tensors are literally the topic that started this discussion."

Again, are manifold involved in any way in the definition of tensors and their properties? No? Then why are you even mentioning "metric tensors"? (Which aren't even tensors, but tensor fields...)

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

#83

Post by prof-dr-ir »

adrian_b wrote: post #65
prof-dr-ir wrote: post #47

Earlier quoted context omitted.

> 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.

The values of the "putative function" do not depend on the chosen coordinate system. This is the essence of notions like scalar, vector, tensor, that they do not depend on the chosen coordinate system. Only their numeric representations associated with a chosen coordinate system do depend on that system. If you compute some arbitrary functions of the numeric components of a tensor in a certain coordinate system, in m…

In the end tensor fields are sections of a bundle.

I insist that calling them "just functions" is simplistic. In fact, I'd say that the complexity of your elaborations kind of proves my point.

Note that I deliberately use "simplistic" and not "wrong", since a section is a function of sorts.

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

#84

Post by will-burner »

cshimmin wrote: post #17
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…

A simple example of a multilinear function is the inner (a.k.a dot) product : it takes a vector (b), and a dual vector (a^T), and returns a number. In tensor notation it's typically written δ_ij. It's multilinear because it's linear in each of its arguments separately: = c and = c . Another simple but less obvious example is a rotation (orthogonal) matrix. It takes a vector as an input, and returns a vector. But a ve…

The dot product! That's a good example, thank you.

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

#85

Post by g15jv2dp »

senderista wrote: post #46
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...

Why? All I need to prove is that I have a canonical linear map going in one direction, and that this map is a bijection. (Since I already constructed such a bijection, I could just answer "for any z in V**, take the unique element v in V such that for all w in V*, w(v) = z(w)".) Do you disagree that I provided such a map? You are correct that V** is not isomorphic to V when V is infinite-dimensional, but your stateme…

I responded to a sibling comment with an explanation: https://news.ycombinator.com/item?id=41234913 But good on you for the condescension even though you didn't understand my point ;)

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

#87

Post by Koshkin »

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…

> all such physical quantities are used to describe multilinear functions

Why always think of tensors as "functions"? In physics, we often think of them as "quantities" - scalar, vector, etc.

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

#88

Post by Koshkin »

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…

More simply, a tensor is an element of a tensor product of linear spaces. (And a monad is simply a monoid in the category of endofunctors.)

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

#89

Post by Koshkin »

cfgauss2718 wrote: post #70
adrian_b wrote: post #65

Earlier quoted context omitted.

The values of the "putative function" do not depend on the chosen coordinate system. This is the essence of notions like scalar, vector, tensor, that they do not depend on the chosen coordinate system. Only their numeric representations associated with a chosen coordinate system do depend on that system. If you compute some arbitrary functions of the numeric components of a tensor in a certain coordinate system, in m…

Here here! Functions do not depend on your choice of coordinates, only the components of tensors do! I think this is why it’s important to keep covariance and contravariance in mind. While tensor(fields) do not depend on coordinates intrinsically, the way we represent them when doing calculations most certainly does, and this is usefully characterized by co/contravariance.

"Here here!" is indeed topological. "Hear, hear!", on the other hand, is conversational.

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

#90

Post by Koshkin »

cfgauss2718 wrote: post #66
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…

Very well, let’s just agree that in physics (r,s) tensors usually refer to sections of the tensor product of some fixed number of copies of the tangent bundle (r copies) and cotangent bundle (s copies) of a smooth manifold (almost always pseudo-Riemannian) and leave it there. Elementary!

This is perfect! Thanks
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