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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

#61

Post by ndriscoll »

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…

I think the people who find this definition to be mysterious are really looking for (borrowing from Ravi Vakil[0]) "why is a tensor" rather than "what is a tensor". In that case, a better answer IMO is that it's the "most generic" way to multiply vectors that's compatible with the linear structure: "v times w" is defined to be the symbol "v⊗w". There is no meaning to that symbol.

But these things are vectors, so you could write e.g. v = a⋅x+b⋅y, and then you want e.g. (a⋅x+b⋅y)⊗w = ax⊗w + by⊗w, and so on.

So in some sense, the quotient space construction[1] gives a better "why". It says

* I want to multiply vectors in V and W. So let's just start by writing down that "v times w" is the symbol "v⊗w", and I want to have a vector space, so take the vector space generated by all of these symbols.

* But I also want that (v_1+v_2)⊗w = v_1⊗w + v_2⊗w

* And I also want that v⊗(w_1+w_2) = v⊗w_1 + v⊗w_2

* And I also want that (sv)⊗w = s(v⊗w) = v⊗(sw)

And that's it. However you want to concretely define tensors, they ought to be "a way to multiply vectors that follows those rules". Quotienting is a generic technique to say "start with this object, and add this additional rule while keeping all of the others".

Another way to say this is that the tensor algebra is the "free associative algebra": it's a way to multiply vectors where the only rules you have to reduce expressions are the ones you needed to have.

[0] https://www.youtube.com/live/mqt1f8owKrU?t=500

[1] https://en.wikipedia.org/wiki/Tensor_product#As_a_quotient_s...

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

#62

Post by adrian_b »

abdullahkhalids wrote: post #51
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…

> 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 physical quantities are used to describe multilinear functions, either in linear anisotropic media, or in non-linear anisotropic media, but in the latter case they are applicable only to relations between small differences, where linear approximations may be used.

The multilinear function is the physical concept that is independent of the coordinate system. The concrete computations with a tensor a.k.a. multilinear function may need the computation of contravariant and/or covariant components in a particular coordinate system and the use of their transformation rules. 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.

There is a unique multilinear function a.k.a. tensor, but it can be expressed by an infinity of different arrays of numbers, corresponding to various combinations of contravariant or covariant components, in various coordinate systems. Their transformation rules can be determined by the condition that they must represent the same function. In the books that do not explain this, the rules appear to be magic and they do not allow an understanding of why the rules are these and not others.

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

#63

Post by cryptonector »

mvaliente2001 wrote: post #59

The idea of tensors as "a matrix of numbers" or the example of a cube with vectors on every face never clicked for me. It was this (NASA paper)[ https://www.grc.nasa.gov/www/k-12/Numbers/Math/documents/Ten... ] what finally brought me clarity. The main idea, as others already commented, is that a tensor or rank n is a function that can be applied up to n vector, reducing its rank by one for each vector it consumes.

> a tensor or rank n is a function that can be applied up to n vector

There seems to be a grammar problem here.

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

#64

Post by qsdf38100 »

mvaliente2001 wrote: post #59

The idea of tensors as "a matrix of numbers" or the example of a cube with vectors on every face never clicked for me. It was this (NASA paper)[ https://www.grc.nasa.gov/www/k-12/Numbers/Math/documents/Ten... ] what finally brought me clarity. The main idea, as others already commented, is that a tensor or rank n is a function that can be applied up to n vector, reducing its rank by one for each vector it consumes.

In your cube example you are using the word "vector" to refer to faces of the cube. Did you mean matrix?

My understanding is that the cube is a rank 3 tensor, the faces (or rather slices) of the cube are rank 2 tensors (aka matrices), and the edges (slices) of the matrices are rank 1 tensors (aka vectors).

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

#65

Post by adrian_b »

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.

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 most cases the array of numbers that composes the result will not be a tensor, precisely because the result will really be different in any other coordinate system, while a tensor must be invariant.

All physical laws are formulated only using various kinds of tensors, including vectors and scalars, precisely because they must be invariant at the choice of the coordinate system.

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

#66

Post by cfgauss2718 »

adrian_b wrote: post #44
mr_mitm wrote: post #38

Earlier quoted context omitted.

Yes, very simple, except that when physicists say "tensor", they mean tensor fields, on smooth, curved manifolds, in at least four dimensions, often with a Lorentz metric. Things stop being simple quickly.

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!

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

#68

Post by senderista »

ndriscoll wrote: post #61
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…

I think the people who find this definition to be mysterious are really looking for (borrowing from Ravi Vakil[0]) "why is a tensor" rather than "what is a tensor". In that case, a better answer IMO is that it's the "most generic" way to multiply vectors that's compatible with the linear structure: "v times w" is defined to be the symbol "v⊗w". There is no meaning to that symbol. But these things are vectors, so you…

That abstract approach tends to be how mathematicians view the tensor product (there is also a categorical construction), but I don't find it very helpful for understanding what tensors do, or why they are useful in physics. With the "multilinear map" definition, taking the tensor product T of tensors U and V just means evaluating U and V respectively on the arguments of T and multiplying their outputs. Extend this definition by linearity and you have the tensor product of spaces of tensors.

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

#69

Post by cfgauss2718 »

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…

I think the point above is that in physics tensor is usually overloaded, and those practicing physicists when they speak of tensors are more often referring to tensor fields, and most often this is in a context with more geometric structure than is required by a tensor space in reference to a vector vector bundle. Typically they (physicists) are dealing with domains where the tensor space is in reference to the tangent bundle of a smooth manifold, with the prototypical example being the metric tensor(field) of space time in general relativity. Another prominent example may include tensor fields defined in reference to the tangent bundle of a group of gauge transformations, as in quantum electrodynamics, quantum chromodynamics, etc.

Obviously these things are not just useful to physics, but are indispensable, and so I think the assertion that only the definition of tensor that is useful to physics is the definition tensor=multilinear map is somewhat out of step. Perhaps it would be better to assert that the concept of multilinear map is essential to every useful definition of tensors in physics.

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

#70

Post by cfgauss2718 »

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…

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.
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