# Visualising tensors

1. Feb 7, 2007

### nolanp2

Can anyone help me to unerstand what tensors are in physics, a few basic examples would probably do, and how they actually go about giving you the results. the only explanations i've been given of them are through maths which are useless to me since i still cant manage to visualise what a matrix is actually doing in many cases. need help urgently!!

2. Feb 7, 2007

### arunma

I've always had trouble with this too. I took both differential geometry and cosmology, and I still didn't get it. I only started to understand tensors when I got an easier introduction to them in special relativity.

Here's my understanding. Tensors are essentially matrices, except they can exist in more than two dimensions. For example, a tensor with one index is a vector. If you were to write it out as a matrix, it would exist in one dimension. Tensors with two dimensions are just typical square matrices. A tensor of three indices can be thought of as a three-dimensional box of numbers. And it just goes from there.

Of course, I've never quite understood the difference between covarient and contravarient tensors. Let me know if you ever find out!

3. Feb 7, 2007

### cristo

Staff Emeritus
Do you know the difference between a covariant and a contravariant vector?

4. Feb 7, 2007

5. Feb 7, 2007

### arunma

Well, I know that one has raised indices, and the other has lowered indices. And I know that raising or lowering both indices simultaneously means that that tensor becomes its inverse. So I can brainlessly do the operation. But I've never understood why this is the case.

6. Feb 7, 2007

### quasar987

The words covariant and contravariant refer to the way the components of the tensor transform as the basis is transformed (contravariant = in a contrary way and covariant = in a ...well in the same way :P).

I feel I cannot say more without rewriting the whole theory from scratch so I refer you to say Sharipov's ebook: http://uk.arxiv.org/abs/math/0403252/

7. Feb 8, 2007

### mjsd

how do you actually want to visualise it? As apples and oranges ? or as a collection of numbers/symbols/sea shells group together in some meaningful way? and that they follow particular rules when you try to juggle the content....?

I guess you may have a chance to visualise tensors if you limit yourself to tensors with very small ranks and very low dimensionality. Otherwise, it would be as difficult as to visualising a 10-dimensional hypercube for instance.

remember tensors are defined by how they transform, and certainly not all tensors naturally lead to some "everyday" visualizations.

8. Feb 14, 2007

### baryon

A tensor is a collection of values that alters a field.

Last edited: Feb 14, 2007
9. Feb 14, 2007

### CPL.Luke

from my very limited understanding of them they can work as follows

an mXm matrix will take a vector and transform it to produce another vector in the same "space" (ex. R^m)

now what if you had an array of several matrices? if you were to multiply this by a vector such that each component of the vector gave a certain "weight" to each of the component matrices in the tensor, thus producing a new matrix that could then be used to tranform a vector.

I believe that general relativity works in this way.

but most simply put a tensor is an array of matrices, and ghigher dimensional tensors are then arrays of lower dimensional tensors. similar in effect to a hypercube.

NOTE: I have no formal traiing or even informal self-study of tensors, I just gathered this from a couple of brief chapters on general relativity a while back, so what I said could easily be very wrong.

10. Feb 14, 2007

### Magister

The contravariant and convariant vectors are related by the way you get the components of the vector. If you use the paralelogram rule of the vectors you get the contravariants components, for instance if you get the components by tracing a line ortogonal to the axis you get the covariant components. Is because of this that in $R^3$ the contravariants and covariants of a vector are equal.
I suppose that the name covariant cames from the fact that when you change one axis only the components related to this axis change while when working with contravariant components all the components change.

I hope I have made my self clear.

11. Feb 14, 2007

### Hurkyl

Staff Emeritus
IMHO, one thing that would help is to try to understand what a tensor is before you try and understand what a tensor field is.

(much like it's easier to learn about vectors before you try to learn about a vector field)

If you've chosen a basis, the coordinates of a vector form an nx1 array of numbers. The coordinates of a covector form a 1xn array of numbers.

In coordinates, a tensor product is a generalization of a scalar product. For example, in three dimensions, if I tensor the covector [a b c] with something (could be anything: a scalar, a vector, a covector, a matrix, or some other kind of tensor)

$$\left[\begin{array}{ccc}a & b & c\end{array} \right] \otimes \mathbf{v} = \left[\begin{array}{ccc}a\mathbf{v} & b\mathbf{v} & c\mathbf{v}\end{array}\right]$$

If v was, say, the vector with coordinates x, y, and z, this tensor product would be the matrix:

$$\left[\begin{array}{ccc}a & b & c\end{array} \right] \otimes \left[\begin{array}{c}x \\ y \\ z\end{array} \right] = \left[\begin{array}{ccc} a \left[\begin{array}{c}x \\ y \\ z\end{array} \right] & b \left[\begin{array}{c}x \\ y \\ z\end{array} \right] & c \left[\begin{array}{c}x \\ y \\ z\end{array} \right]\end{array} \right] = \left[\begin{array}{c|c|c}ax & bx & cx \\ ay & by & cy \\ az & bz & cz \end{array} \right]$$

Actually, we would drop the partitions in this case: they're more useful for other cases, such as if v happened to be the covector [p q r]:

$$\left[\begin{array}{ccc}a & b & c\end{array} \right] \otimes \left[\begin{array}{ccc}p & q & r\end{array} \right] = \left[\begin{array}{ccc|ccc|ccc} ap & aq & ar & bp & bq & br & cp & cq & cr \end{array}\right]$$

Mind you, this is all in a coordinate representation: just like with vectors, we can do algebra with tensors without resorting to coordinates.

Last edited: Feb 14, 2007
12. Feb 15, 2007

### Mentz114

This intended to help the fellows above who want to know about covariant and contravariant components. As magister above points out, if the axes of our co-ordinate system are not orthogonal, or are curved, there are two ways of relating a point to an axis. One way is to drop a perpendicular, the other way is to go parallel to another axis. So we have 2 numbers where we had one. These are the contravariant and covariant components. We used to describe a point in 2D with 2 co-ords, x and y, now we have a choice of x (contra), x (covar), y(contra), y (covar).

Now switching notation so super- and subscripts are the dimension ( eg 0=time, 1=x, 2=y etc), if we have 2 points in 2D (X,Y) we have 8 numbers ( n=0,1)

$$x^n, x_n , y^n, y_n$$

Luckily the only quantities we can use in physics are products of contravariant and covariant components, so only two sensible choices exist to work with these numbers,

$$x_m y^m = x_0 y^0 + x_1 y^1$$

and

$$x^n y_n$$

which are the same thing. All the intricacies of tensor analysis follow from the fact that physical quantities are scalars.

I haven't mentioned the metric tensor, to keep it simpler.

Last edited: Feb 15, 2007
13. Feb 15, 2007

### CPL.Luke

hmm I'm intrigued

14. Feb 16, 2007

### Mentz114

Hi CPL - what I've posted is the first step in a journey of a thousand leagues.
B Reimann in the 1840's(roughly) investigated systems where the axes were not straight or orthogonal, realising that concepts like position coords are no longer useful, but that distances could be kept invariant.

I can recommend Stephani's book "General Relativity" as an introduction.

M.

Last edited: Feb 16, 2007
15. Feb 16, 2007

### nolanp2

ok i'm trying to focus on seeing low dimensional tensors in simple instances, such as how tensors would be useful in a force field or something similar. can anybody help me with an example of this?

16. Feb 16, 2007

### robphy

Although it has been raised earlier, it's not clear if you are looking for
a geometric visualization (akin to visualizing a vector as an arrow and vector addition via the parallelogram rule)
or
an organizational typeset visualization (akin to visualizing a vector as a column matrix and vector addition via component-wise addition).

If you are looking for a geometric visualization, did you consult the arxiv.org paper?

17. Feb 16, 2007

### Mentz114

Hi Nolanp2. I don't think visualising tensors helps to understand them.
Thinking about geometry might help. You mention force fields, so I'll give you an example of how and why tensors are used. Consider the equation of motion of a charge in an electric field,

m.a = e(E + v x B ), i.e. mass times acceleration is electric charge times E + v x B, where v is the velocity vector and E and B are electric and magnetic field vectors. x is the vector cross-product.Note that this equation is really 3 equations, one each for x, y and z.
This formula is fine but it is not relativistic. It won't work if you put it in a moving frame of reference.
To make the equation obey special relativity, we go to four dimensions,
t,x,y,z, with stipulation that the time dimension has a negative sign, and
that we multiply all times by c=velocity of light to make them distances.
We can now rewrite the equation in tensor notation,
$$ma^{\mu} = e F^{\mu\nu}v_{\nu}$$
which is four equations since $$\mu$$ can be t,x,y,or z. F is the EM field tensor ( see Wiki for instance).
But the point is that this equation is relativistically covariant, i.e it transforms properly between inertial frames. It's also very elegant.

To do a calculation from this formula, say for x ( mu=1), we expand the tensor to give,
$$ma^{1} = e F^{10}v_{0} + e F^{11}v_{1} + e F^{12}v_{2} + e F^{13}v_{3}$$
Note that all the indexes are numbers, and we can plug in the values of the components to get an algebraic differential equation. The tensors are gone.

Last edited: Feb 16, 2007
18. Feb 16, 2007

### nolanp2

as a specific example i'm looking through small oscillations in s dimensions and it gives the formula U= 1/2 SUM[K_(ik)*x_(i)*x_(k)]. is the constant K a tensor in this case and if so what is it doing? need help badly!

19. Feb 16, 2007

### Mentz114

I think this is just using the tensor summation convention. Expand it over all
values of i and k thus,

U = K(0,0)*x(0)*x(0) + K(0,1)*x(0)*x(1) + K(0,2)*x(0)*x(2) + ....

In fact this just matrix multiplication.

All you need now are the values ( components) of K and some x's.

M

Last edited: Feb 16, 2007
20. Feb 17, 2007

### nolanp2

so do the different k values represent different springs (for example) being stretched when one particle connected to them is moved or does it mean a spring is acting different when pulled in different directions? or does a second degree of freedom mean there's a second partcle and the two are connected via springs? in particular what would k_11 or k_22 represent?