Hello everyone!

I am hoping there are some people on this forum who are familiar with Tensor calculus. I have just found a book on it and it looks very interesting!! I may have some questions as I go through this book and I will start with an easy one. What is a Kronecker Delta? and why do we need it? Also could anyone give me a brief overview of the Einstein Summation Convention.

Thanks heaps!

Ben.

the kronecker delta is just an element of the identity matrix. the identity matrix has 1s on the diagonal, and 0s elsewhere, so the kronecker delta is 1 if the two indices are equal, and 0 otherwise.

the einstein summation convention. well, it just means, don t write the sigma. in a single multiplicative term, if you see a single index twice, once above the line, and once below, then you sum over that index. that index is now contracted, and the resulting sum transforms like the type of tensor you would have if both instances of the index were erased.

you should never see that index more than twice, nor should you see it twice raised, or twice lower. it is acceptable to repeat the index in another term.

added terms should have the same overall tensor type. the should have the same symbols above the line, and the same symbols below the line.

Tensor analysis might be a big leap from where you are. I assume you at least have some background in calculus

If might be a good idea to find a good book or a good course in Differential Geometry which uses tensor notation and provides a bit of insight into other uses of tensor analysis. Relativity mathematics is applied differential geometry, usually using tensor notation.

Originally posted by lethe
added terms should have the same overall tensor type. the should have the same symbols above the line, and the same symbols below the line. To this I just wanted to add, "with the exception of the indices involved in a contraction," since these are independent from term to term.

Some additional thoughts

Tensor notation is often used for matrix algebra. For example, the following is the sum of two matrix products using tensor notation.

M^j_k = A^j_x*B^x_k + C^j_x*D^x_k

In the above, M, A, B, C, D are matrices. j, k, x are integers. Superscripts are row indicators, while subscripts are column indicators.

It is a property of tensor notation that each term must have paired subscripts and superscripts. Note that each of the above three terms has j as a superscript and k as a subscript, while x appears as both a subscript and a superscript for elements to be multiplied.

In the above, the sum of all possible element products is implied by x, which appears as both a subscript and a superscript in the same term.

The pairing property of tensor analysis is illustrated by the following more complex formula.

Zⁱ_jk = Aⁱⁿ_jk*X_n + Bⁱ_ik*Y

A person familiar with tensor notation would know there was an error in the above, and assume that the following was intended.

Zⁱ_jk = Aⁱⁿ_jk*X_n + Bⁱ_jk*Y

Tensors are not matrices, in general. There is a more strict condition on tensors.

Errandir: You sare correct, matrices are not tensors. However, the notation is often used for matrix algebra.

Originally posted by errandir
Tensors are not matrices, in general. There is a more strict condition on tensors.

in general, tensors are not matrices, this is true. however, matrices are tensors, therefore matrix algebra furnishes a very excellent example showing how tensor manipulations work, one which most beginners would be more familiar with.


Originally posted by Dinosaur
Errandir: You sare correct, matrices are not tensors. However, the notation is often used for matrix algebra.

Matrices are coordinate representations of a certain class of tensor, namely, the (1,1) rank tensor, i.e. a linear transformation of a vector space. tensor index notation is a coordinate notation for a more general tensor (i.e. a multilinear mapping from many copies of a vector space and its dual).

therefore it is perfectly acceptable to do matrix manipulations as if you were doing tensor manipulations. you are doing tensor manipulations. however, i only feel comfortable seeing the tensor index notation for matrices if with one raised and one lowered index (since a matrix is a (1,1) rank tensor). however, many physicists do not agree with me and use the notation you have here, Dinosaur.

carry on.

Originally posted by lethe
however, i only feel comfortable seeing the tensor index notation for matrices if with one raised and one lowered index (since a matrix is a (1,1) rank tensor). however, many physicists do not agree with me and use the notation you have here, Dinosaur.[/pedantry]


whoa, dinosaur. i owe you an apology, i was on a computer that was not displaying superscripts correctly. apparently you use the correct notation, the way i like it.

Great, thanks for the replies guys.

A couple of you have been mentioning matricies. I would like to know more about this.

Linear Transformations:
I understand that doing tensor calculus requires a sound understanding of linear algebra but more so, the ability to calculate linear transformations.

I have done transformations before in calculus (ie. integration, differentiation, functions, etc...). Is this the same idea but with linear algebra?

Also could someone give me an idea of what I am doing geometrically when I transform.

Thanks

Originally posted by oxymoron
Great, thanks for the replies guys.

A couple of you have been mentioning matricies. I would like to know more about this.

Linear Transformations:
I understand that doing tensor calculus requires a sound understanding of linear algebra but more so, the ability to calculate linear transformations.

I have done transformations before in calculus (ie. integration, differentiation, functions, etc...). Is this the same idea but with linear algebra?

Also could someone give me an idea of what I am doing geometrically when I transform.

Thanks

linear transformations? well it is a little hard to describe a general linear transformation geometrically. it can be a rule that says multiply all the vectors in your space by 2, making them twice as long. it can be a rule that says rotate all your vector by 40 degrees. it could be a rule that says switch 2 of your basis vectors.

it is basically any rule that gives you a new vector for each vector in the space, in a linear way.

Originally posted by oxymoron
I have done transformations before in calculus (ie. integration, differentiation, functions, etc...). Is this the same idea but with linear algebra?I don't think so (in the way I suspect you are considering the issue).

The derivative <i>could</i> be sort of thought of as an infinite dimensional matrix. In order to calculate the derivative of a function, f(x), in this formalism, you would have to take the inner prodect of |x> with |f'> = |Df> = D|f>, where D is the matrix for differentiation. You would have to use an integral to define the inner product, since the index on the matrix and vectors is continuous. I was assuming that this is not what you had in mind.

I just had a thought/question pop up in my head when I wrote this: does this mean that the eigenvectors of D are the exponentials (when projected onto the x basis)?




Originally posted by oxymoron
Also could someone give me an idea of what I am doing geometrically when I transform.It's probably best to get an intuition from 3-D real space (layman geometry). Here, transformations do three things: stretch, rotate, and reflect. The matrix/transformation takes an arrow as an input, and then spits out an arrow as an output. This new arrow may be stretched, rotated, and/or reflected.

If the matrix/transformation is unitary (an important class), then the arrow is not stretched.

If the matrix/transformation is Hermitian (another important class), then there is a set of arrows that it will not rotate that can be used as an orthogonal basis (like the x,y,z basis).

These two matrices/transformations are not mutually exclusive.

There are certain arrows for every matrix/transformation that the matrix transformation will not rotate (it may stretch or reflect, but not rotate). These arrows are called eigenvectors. Keep in mind that they are specific to the matrix/transformation. You can have a set of eigenvectors for one matrix, that another matrix can rotate.

The derivative could be sort of thought of as an infinite dimensional matrix. In order to calculate the derivative of a function, f(x), in this formalism, you would have to take the inner prodect of |x> with |f'> = |Df> = D|f>, where D is the matrix for differentiation. You would have to use an integral to define the inner product, since the index on the matrix and vectors is continuous. I was assuming that this is not what you had in mind.
Actually that is exactly what I meant! :)

I might know more about this is a few days as we are about to start this topic in Linear Algebra. It looks quite interesting and no doubt I will have a few more questions on this soon! Thanks errandir and lethe once again for your efforts.

Cheers. :)

Ben.

Originally posted by oxymoron
Actually that is exactly what I meant!My appologies.:)

I'm a bit surprised, though, that you have gotten into this notion before a sound appreciation of the geometrical interpretation.

Originally posted by errandir
I don't think so (in the way I suspect you are considering the issue).

The derivative <i>could</i> be sort of thought of as an infinite dimensional matrix. In order to calculate the derivative of a function, f(x), in this formalism, you would have to take the inner prodect of |x> with |f'> = |Df> = D|f>, where D is the matrix for differentiation. You would have to use an integral to define the inner product, since the index on the matrix and vectors is continuous. I was assuming that this is not what you had in mind.
you do not need an inner product to calculate the matrix elements of the derivative operator, nor any linear operator.

I just had a thought/question pop up in my head when I wrote this: does this mean that the eigenvectors of D are the exponentials (when projected onto the x basis)?
yes (you have to include a factor of i, if you want the derivative to be hermitean)


If the matrix/transformation is unitary (an important class), then the arrow is not stretched.

If the matrix/transformation is Hermitian (another important class), then there is a set of arrows that it will not rotate that can be used as an orthogonal basis (like the x,y,z basis).
this is correct, although i think if you want to use layman intuition in R3, you should talk about the real valued analogues of unitarity and hermiticity, namely orthogonality and symmetry. the statements you make about those classes of matrices also apply to these classes (since the reals are a subset of the complexes).

another property of symmetric/hermitean matrices is that they preserve the normal inner product.

also, i should say that the real condition for a vector not being stretched is not orthogonality/unitarity, although they certainly do imply that vectors do not get stretched. the condition for vectors not to stretch is that the matrix be unimodular (i.e. the determinant is 1). in fact, i can be even more general, and only require that the modulus of the determinant is 1. this will allow rotations which also do not stretch vectors.

What is meant by curvilinear coordinates? Is it just the fancy name they give to spherical, cylindrical etc.. coordinates?

What is the Jacobian (you get this from a Jacobian Matrix)? And what is a Jacobian Matrix?

Do these two questions relate and is there any desire to change from one coordinate system to another?

Originally posted by oxymoron
What is meant by curvilinear coordinates? Is it just the fancy name they give to spherical, cylindrical etc.. coordinates?
Pretty much. Technically it is defined to be "a coordinate system composed of intersecting surfaces", which covers cylindrical, spherical and rectangular coordinate systems.


Originally posted by oxymoron
What is the Jacobian (you get this from a Jacobian Matrix)? And what is a Jacobian Matrix?

A good explanation of it is here: http://www.wikipedia.org/wiki/Jacobian, but some MVC is needed to understand it. Post if you have any questions about it.

Originally posted by oxymoron

Do these two questions relate and is there any desire to change from one coordinate system to another?

Sort of. Curvilinear coordinates is just a "fancy" term, while the Jacobian is used to change from one coordinate system to another.

The Jacobian is a matrix of partial derivatives. If you have (say) five functions of five variables, the Jacobian is a 5X5 matrix defined as follows. PartialDerivative(Function^j)/Variable_k, with (j, k) taking on values from 1 to 5.The above notation is made up for this post and might not be standard. The intent is Element^j_k is the partial derivative of function^j with respect to variable_k

I assume that the above is also called a Jacobian matrix, but I am not certain of this.

There is a Jacobi procedure for finding eignevalues of a symmetric matrix using successive approximations which converge to a diagonal matrix of eigenvalues. Perhaps a symmetric matrix is called a Jacobian matrix.

I am not sure, but curvilinear coordinates might refer to a concept a bit more complicated than non-Cartesian systems like spherical or cylindrical coordinates.

In the discipline called Differential (or metric) Geometry, the coordinate system changes at each point on a surface. At each point along a curve, the coordinates system used consists of the tangent to the curve, the principle normal for the curve, and the line perpendicular to both. This might be what is meant by curvilinear coordinates.

Originally posted by Dinosaur
Tensor analysis might be a big leap from where you are. I assume you at least have some background in calculus

If might be a good idea to find a good book or a good course in Differential Geometry which uses tensor notation and provides a bit of insight into other uses of tensor analysis. Relativity mathematics is applied differential geometry, usually using tensor notation.

yes i agree......back round on calculus and pdes are needed. and vector calculus is good to review,