41–50 of 98 posts
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…
Earlier quoted context omitted.
No, the isomorphism between V and V** (for finite-dimensional V) is canonical. The canonical isomorphism T:V->V** is easy to construct: map a vector v in V to the element of V** which takes an element w from V* and applies it to v: T(v)(w) = w(v).
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...
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…
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…
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.
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 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 word "tensor" with the current meaning has been used for the first time by Einstein and he has not given any explanation for this word choice. The theory of tensors that Einstein has learned had not used the word "tensor".
Before Einstein, the word "tensor" (coined by Hamilton) was used in physics with the meaning of "symmetrical matrix", because the geometric (affine) transformation of a body that is determined by the multiplication with a symmetric matrix extends (or compresses) the body towards certain directions (the axes that correspond to a rotation that would diagonalize the symmetric matrix). The word "tensor" in the old sense was applied only to what is called now "symmetric tensor of the second order" (which remains the most important kind of the tensors that are neither vectors nor scalars).
Earlier quoted context omitted.
Yes, I had a typo. As for why I said metric, see https://en.wikipedia.org/wiki/Metric_tensor . Which is technically a concept from differential geometry rather than linear algebra. But then again, tensors are literally the topic that started this. And it is only in differential geometry that I've ever cared about mapping from V to V*.
There are no manifolds at all in this discussion... What are you even talking about? Just stringing words you vaguely know together??
Hopefully that's a hint that you should attempt to figure out what someone might be talking about before going to schoolyard insults.
Earlier quoted context omitted.
No, the isomorphism between V and V** (for finite-dimensional V) is canonical. The canonical isomorphism T:V->V** is easy to construct: map a vector v in V to the element of V** which takes an element w from V* and applies it to v: T(v)(w) = w(v).
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...
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…
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.
Earlier quoted context omitted.
For those without a strong math background but more of a programmers background; You know the matrices you work with in 2D or 3D graphics environments that you can apply to vectors or even other matrices to more easily transform (rotate, translate, scale)? Well tensors are the generalisation of this concept. If you’ve noticed 2D games transformation matrices seem similar (although much simpler) to 3D games transforma…
Is there a difference between a 4x4 matrix and a 4x4 tensor?
Interestingly, that extra information helps us to differentiate between the same matrix being used in different "roles". For instance, if you have a 4x4 matrix A, you might think of it like a linear transformation. Given x: V = R^4 and y = Ax, then y is another vector in V. Alternatively, you might think of it like a quadratic form. Given two vectors x, y: V, the value xAy is a real number.
In linear algebra, we like to represent both of those operations as a matrix. On the other hand, those are different tensors. The first would be a rank-(1,1) tensor, the second a rank-(2, 0) tensor.
Ultimately, we might write down both of those tensors with the same 4x4 array of 16 numbers that we use to represent 4x4 matrices, but in the sort of math where all these subtle differences start to really matter there are additional rules constraining how rank-(1, 1) tensors are distinct from rank-(2, 0) tensors.
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…
A tensor is a multi-dimensional array.
:)
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.