[Astronomical]

On wikipedia articles I often get stuck because of the use of this funny notation being used, such as something like this:











Could somebody help me understand the heck the above mean? Perhaps with an elementary example?

[ben-smith]

(Original post by Astronomical)
On wikipedia articles I often get stuck because of the use of this funny notation being used, such as something like this:











Could somebody help me understand the heck the above mean? Perhaps with an elementary example?

These are tensors which are being expressed with the einstein summation convention (ESE).
1) in ESE, if you have two indices with the same letter with one in the upper (contravariant) position and the other in the lower (covariant) position there is an implied summation over that letter. This means that the only free indices are a and c so it's really a rank 2 tensor which is what the expression is saying
2) I'm not that hot on tensor products but I think it's just defining what the outer product of two tensors is in terms of the tensor product.
3)This should read: the covariant derivative of the tensor T^(ab) wrt b is 0. The covariant derivative is a way of differentiating tensor quantities such that the output transforms as a tensor and is thus useful when we're in say riemannian spaces.
4)This is just a tensor of rank 4 with two covariant and two contravariant indices.

I think that's right but take what I say with a pinch of salt.

[Astronomical]

(Original post by ben-smith)
These are tensors which are being expressed with the einstein summation convention (ESE).
1) in ESE, if you have two indices with the same letter with one in the upper (contravariant) position and the other in the lower (covariant) position there is an implied summation over that letter. This means that the only free indices are a and c so it's really a rank 2 tensor which is what the expression is saying
2) I'm not that hot on tensor products but I think it's just defining what the outer product of two tensors is in terms of the tensor product.
3)This should read: the covariant derivative of the tensor T^(ab) wrt b is 0. The covariant derivative is a way of differentiating tensor quantities such that the output transforms as a tensor and is thus useful when we're in say riemannian spaces.
4)This is just a tensor of rank 4 with two covariant and two contravariant indices.

I think that's right but take what I say with a pinch of salt.

Good lord, I think I've fallen in the deep end! Thanks for the reply, however it may as well be in Spanish, such is my understanding of tensors and covariant derivatives.

Are tensor products similar to dot or cross products?

[FireGarden]

(Original post by Astronomical)
Good lord, I think I've fallen in the deep end! Thanks for the reply, however it may as well be in Spanish, such is my understanding of tensors and covariant derivatives.

Are tensor products similar to dot or cross products?

Dot and Cross products are tensors..

[Astronomical]

(Original post by FireGarden)
Dot and Cross products are tensors..

Now I really am lost.

[ben-smith]

(Original post by FireGarden)
Dot and Cross products are tensors..

unless I'm mistaken, the cross product is only a tensor up to a minus sign, it's a pseudo-tensor.
(not that it matters, your point stands)

[FireGarden]

(Original post by Astronomical)
Now I really am lost.

What is it you are trying to study? Tensors are not the easiest of objects to understand (they give me a goddamn headache..).

Tensors are geometrical objects that obey certain transformations rules so that they are coordinate-invariant. Scalars, vectors, and matrices are all types of tensor (0th, 1st, and 2nd rank respectively, where rank denotes the number of indices needed to represent the tensor). However, tensors are more general than simply thinking of these objects, as they are used to describe the linear relations between scalars, vectors, and other tensors, etc. The dot product is a linear relation between two vectors, or a vector with itself, resulting in a scalar which is coordinate-invariant, and transforms linearly - hence classifies as a tensor.

They are needed most significantly when you are working in non-euclidean spaces, as they have this additional weirdness of dealing with 'covariant' and 'contravariant' components (this is why you have indices flying everywhere - you need each index to deal with some component of the space you are in, and where it is (either raised or lowered) tells you how the component changes with a transformation), which are in fact one and the same in euclidean space.

As far as my understanding goes (not really a lot, on tensors), I have no knowledge of tensor products (unless (and I suspect it does, but am hesitant to declare it!) matrix multiplication is a form of it; but I do not understand it from a tensor point of view). I've simply seen them briefly for use in classical mechanics, most notably, the Inertia Tensor.

[agostino981]

@ben-smith I think Example 4 is a 2,3 tensor.

Dot product is a kind of contraction.

The indices indicate the basis of the object. Upper indices are contravariant components while lower indicies are covariant components. Contra- and covariance show their importance when you are working with a curvilinear coordinate, for instance, the spherical coordinate. Contra- and covariance exist all the time and the reason why you don't need those in school maths and physics is because they are naturally identical in euclidean flat space.

There are basically 2 main fundamental theorems. The addition thm and the quotient thm.

In addition thm, if you have 2 tensors with different indices, you can write another tensor which contains all these indices, i.e. Example 2 in the first post.
In quotient thm, if you have 2 tensors with same indices and these 2 indices must be different (1 contravariant and 1 covariant) and you can contract them to reduce the ranks of the tensor, i.e. Example 1 in first post*.
*Usually indicates the metric tensor, which defines the metric of the manifold (the First Fundamental Form if it is a surface), and , which is a kronecker delta, a fancy way to write an identity matrix.

The reason why contra- and covariant components are identical because the metric tensor is an identity matrix while the metric tensor and the inverse metric tensor are used to raise and lower the indices, so nothing is changed.

If I recall, tensor product is just multiplying different tensors, such as a vector another vector will form a 2,0 tensor, i.e. a matrix. The tensor product notation is rarely used in physics and in geometry, they use to indicate the multiplication of vector spaces.

Cross product is an operation among tensor. Cross product is limited on only. A more general case will be using Levi-Civita Symbols and the operation will contains 3 different tensors, a 0,3 Levi-Civita Symbol, and 2 vectors you want to cross with.

As I mentioned before, dot product is a contraction and it is just obeying the quotient thm. If they are contra- or covariant components at the same time, it will become the trace of the matrix the formed.

I guess these are the very basic stuffs you need for tensors. If I have any mistake, please kindly point it out.

EDIT:

Dot product is a contraction between vector and covector. If you want to dot 2 vectors, apply a metric tensor to either one of them first. Again, dot product is just available in , so what you are dotting are actually vector and covector.

In mathematics, covectors have basis while vectors, in the eyes of mathematicians, have basis , here, indices represent contra- or covariance.

EDIT 2:

Covariant Derivative (some will use a semi-colon instead), if I recall, is the directional derivative of a tensor. There are 2 different kinds of derivatives, one is covariant derivative and the other is Lie derivative , where is a vector field.

You need to know what is connection first, a quick look at wikipedia would be a good preview.

If you want to do directional derivative along a vector field on a vector field, the normal directional derivative is meaningless. There are 2 ways to do so, one is to move the specific vector along the vector field you normally want to do directional derivative with and this is covariant derivative, while Lie derivative is to move the entire vector field instead of a specific vector.

To do either one of them, you need to use connections, as operations between vector space on different points are meaningless without using a connection. You may also find out that you can indeed use covariant derivative to represent Lie derivative as well:
, where is an arbitrary vector.

[Snakefingers13]

I'm a complete novice, so bear with me, but aren't tensors just fancy matrices of forces on an object?


This was posted from The Student Room's iPhone/iPad App

[agostino981]

Nope.

Tensor is more than matrices. Scalar is Rank-0 Tensor, Vector is Rank-1 Tensor, Matrix is Rank-2 Tensor, Hypermatrix is Rank-3 Tensor and so on.

In continuum mechanics and rigid mechanics, tensors are mainly used to indicate stresses, inertia, etc. Force can be considered as tensor as well, but tensors aren't forces.

In General Relativity, tensors are used to represent almost everything, such as the curvature of spacetime, the geodesic of objects in a curved spacetime, and a lot more.

[ben-smith]

(Original post by agostino981)
@ben-smith I think Example 4 is a 2,3 tensor.

lol, how good am I at counting?

[agostino981]

(Original post by ben-smith)
lol, how good am I at counting?

That's alright, you rarely need to count when you are doing maths.

[ben-smith]

(Original post by agostino981)

In quotient thm, if you have 2 tensors with same indices and these 2 indices must be different (1 contravariant and 1 covariant) and you can contract them to reduce the ranks of the tensor, i.e. Example 1 in first post*.
*Usually indicates the metric tensor, which defines the metric of the manifold (the First Fundamental Form if it is a surface), and , which is a kronecker delta, a fancy way to write an identity matrix.

Hmmm, that's not the quotient theorem I'm familiar unless I'm misunderstanding your post. I thought it was: given arbitrary tensors A,B then
A=CB====> C is a tensor.

I may be misremembering though.

In General Relativity, tensors are used to represent almost everything, such as the curvature of spacetime, the geodesic of objects in a curved spacetime, and a lot more.

As a purely parenthetical point, the geodesic equation in GR actually owes most of it's significance to the non zero christoffel symbols which are definitely not tensors. Pedantic post is pedantic.

[agostino981]

(Original post by ben-smith)
Hmmm, that's not the quotient theorem I'm familiar unless I'm misunderstanding your post. I thought it was: given arbitrary tensors A,B then
A=CB====> C is a tensor.

I may be misremembering though.




As a purely parenthetical point, the geodesic equation in GR actually owes most of it's significance to the non zero christoffel symbols which are definitely not tensors. Pedantic post is pedantic.

Oops, I guess I used the wrong word, lol, it should be contraction thm. Names got blurred when you knew read a bunch of different thms with similar name.

Christoffel symbols, like Levi-Civita Symbols, are definitely not tensors, just arrays of numbers derived from the metric tensor and its derivative. However, you can do tensor algebra upon the christoffel symbols.

The geodesic equation is the parallel transport of the tangent vectors. Although technically Christoffel symbols aren't tensors obviously, but they can be expressed using abstract index notations and obeys the rules.

I didn't really say that Christoffel symbol is a tensor, and there are a lot of 0s in Christoffel symbols in GR.

[Astronomical]

As much as I appreciate the detailed responses in here, guys, I am going to have to let the team down, in that frankly, I don't have a clue what's going on. Contra-variant and co-vector, and countless other terms, are just flying over my head left, right and centre!

Having said that, I can't wait until I do actually get taught this stuff. Are you all maths students, or physics? I'm just starting a theoretical physics degree in October so it'll likely be a few more years until I get into this stuff - I just can't help nosing around on wikipedia getting ahead of myself! It doesn't help that, unless you already know what these things are, the wiki articles might as well be written upside down and back to front - they would make just as much sense that way!

[ben-smith]

(Original post by Astronomical)
As much as I appreciate the detailed responses in here, guys, I am going to have to let the team down, in that frankly, I don't have a clue what's going on. Contra-variant and co-vector, and countless other terms, are just flying over my head left, right and centre!

Having said that, I can't wait until I do actually get taught this stuff. Are you all maths students, or physics? I'm just starting a theoretical physics degree in October so it'll likely be a few more years until I get into this stuff - I just can't help nosing around on wikipedia getting ahead of myself! It doesn't help that, unless you already know what these things are, the wiki articles might as well be written upside down and back to front - they would make just as much sense that way!

It's worth battling through (be it now or later), tensors are just awesome.
Starting maths degree in october.

[Astronomical]

(Original post by ben-smith)
It's worth battling through (be it now or later), tensors are just awesome.
Starting maths degree in october.

Mother of god, you didn't even start yet and you understand this stuff?

Where abouts is your degree? I'm going to guess Cambridge.

[nuodai]

(Original post by Astronomical)
Mother of god, you didn't even start yet and you understand this stuff?

You beat me to the same question! I did a lecture course on vector calculus which included tensors two years ago and even I don't understand those beasts. (And I didn't at the time.)

I'm worried about some of the F38 frequenters of late.

[Astronomical]

(Original post by nuodai)
You beat me to the same question! I did a lecture course on vector calculus which included tensors two years ago and even I don't understand those beasts. (And I didn't at the time.)

I'm worried about some of the F38 frequenters of late.

Well, if you couldn't manage it, I'm surprised anyone can! The question you linked made me want to cry.

Unfortunately, I don't even know what F38 is/stands for. I've been saying that far too much in this thread.

[nuodai]

(Original post by Astronomical)
Unfortunately, I don't even know what F38 is/stands for. I've been saying that far too much in this thread.

Sorry, that's my spent-far-too-much-time-on-TSR-to-claim-to-have-any-other-life-even-though-I-swear-I-actually-do talking; F38 is the Maths section (since it says ?f=38 in the URL).