# Tensor gradient and maxwell's equation

1. Mar 12, 2005

### steveurkell

Hi everyone,
I now have difficulties in using formula for gradient of tensor. The following is
tensor for field strength
Fsubscripts_alpha_beta =
[ 0 -Ex -Ey -Ez
Ex 0 Bz -By
Ey -Bz 0 Bx
Ez By -Bx 0 ]
My question is, how do we derive the Maxwell equations div(B) = 0 and curl(E)+dB/dt using gradient of tensor of the above field tensor?
I myself not quite sure how to use formula for tensor gradient, especially due to the additional term apart from the vectors and 1-forms that the tensor has
e.g.
del_zeta (S) = (dS/dx_superscript_delta) zeta_superscript_delta
What does zeta term here do?
I will be indebted if someone here could give a real example of how to find the gradient of a tensor with the tensor given in the example.
Thank you for any help
regards,

2. Mar 12, 2005

### pervect

Staff Emeritus
This borad has "tex" notation, you can click on any of the equations below to see the "tex" source, which makes it a lot easier to communicate.

$$\nabla_a$$

What this does is it takes any tensor of rank p, and produces a tensor of rank p+1. A scalar counts as a tensor of rank 0 for this purpose.

If you think about it this should make sense - the partial derivative of a scalar is a vector. The 'a' represents the extra index in the new tensor, you start with a rank p tensor, you end up with a rank p+1 tensor, this means you add one index to the tensor, this added index is represented by the symbol 'a'.

If you have a nice orthonormal coordinate system, the covariant derivative operator reduces to the ordinary derivative.

Thus

$$\nabla_a f = (\frac{\partial f}{\partial t}, \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})$$

where I have replaced a_0, a_1, a_2, a_3 with t,x,y,z to make the point clearer.

Note that we started out with a scalar, and wound up with a one-form. The index a represents the components of that one form (the scalar didn't have any indices at all).

The covariant derivative is only equivalent to the partial derivative in orthonormal coordinate systems where the Christoffel symbols are all zero though.

I hope I've understood your question properly, I'm not positive I have. I realize I haven't answered your question about Maxwell's equations yet, but I want to see if the notation I'm using is the notation you are using.

Sometimes people use other notations and notions (like the exterior derivative) - but that would be symbolized by

dF

3. Mar 12, 2005

### dextercioby

IIUC,you're trying to get from the covariant formalism to the noncovariant one.The equations:

$$\nabla\cdot\vec{B}=0$$(1)

$$\nabla\times\vec{E}+\frac{\partial\vec{B}}{\partial t}=\vec{0}$$ (2-4)...

can be found by simply giving all 4 values to the subscripts in the field equations

$$\partial_{[\mu}F_{\nu\rho]}=0$$

(Pervect mentioned covariant derivative,but in this "fortunate" case

$$\partial_{[\mu}F_{\nu\rho]}\equiv \nabla_{[\mu}F_{\nu\rho]}=0$$ )

So do it...

Daniel.

P.S.Report any problems.