Tensor operations, Maxwell's field equations

[MarkSheffield]

I have been working through a relativistic gravitation book ("Gravitation and Cosmology" by Stephen Weinberg) and decided to circle back to the early tensor work in chapter two and just work out the basic tensor math to make sure that I have a feel for how it all goes together. Right at the beginning of this I'm in trouble. Or maybe I'm not, but I can't tell. This is also my first attempt at doing anything with LaTeX, so if something isn't correct with my presentation of all of this, please let me know.

Starting with one of the basics - the tensor for electrodynamics, from the text:

$$F_{\gamma\delta}=\eta_{\gamma\alpha}\eta_{\delta\beta}F^{\alpha\beta}$$

with

$$F^{\alpha\beta} = \left(\begin{array}{cccc} 0 & E_1 & E_2 & E_3\\ -E_1 & 0 & B_3 & -B_2\\ -E_2 & -B_3 & 0 & B_1\\ -E_3 & B_2 & -B_1 & 0\end{array} \right)$$and

$$\eta_{\gamma\alpha} = \eta_{\delta\beta} = \left(\begin{array}{cccc} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

If I break down the operation into steps, I can perform two binary operations by creating an interim tensor, $$T_\delta^{ \alpha} = \eta_{\delta\beta}F^{\alpha\beta}$$ (I already think there's a problem here) and using this in a final operation $$F_{\gamma\delta}=\eta_{\gamma\alpha}T_\delta^{ \alpha}$$

We find the components of $$T_\delta^{ \alpha}$$ by performing the summation over $$\beta$$:
$$T_0^{ 0} = \eta_{0\beta}F^{0\beta} = 0$$
$$T_0^{ 1} = \eta_{0\beta}F^{1\beta} = E_1$$
$$T_0^{ 2} = \eta_{0\beta}F^{2\beta} = E_2$$And so forth until we get all 16 elements of $$T_\delta^{ \alpha}$$

When all is done, I have $$T_\delta^{ \alpha} = \left(\begin{array}{cccc} 0 & E_1 & E_2 & E_3\\ E_1 & 0 & -B_3 & B_2\\ E_2 & B_3 & 0 & -B_1\\ E_3 & -B_2 & B_1 & 0\end{array} \right)$$
When I take this to the next step, $$F_{\gamma\delta}=\eta_{\gamma\alpha}T_\delta^{ \alpha}$$ I get
$$F_{\gamma\delta} = \left(\begin{array}{cccc} 0 & -E_1 & -E_2 &- E_3\\ E_1 & 0 & B_3 & -B_2\\ E_2 & -B_3 & 0 & B_1\\ E_3 & B_2 & -B_1 & 0\end{array} \right)$$
Now this isn't obviously wrong, it just looks wrong. But maybe it's not. Can someone tell me if the initial formulation is correct (I copied this out of "Gravitation and Cosmology") and if the interim tensor $$T_\gamma^{ \alpha}$$ is correct, or maybe point out what fundamental error I created here?

thanks to all - Mark

 

[Altabeh]

$$F_{\alpha\beta} = \left(\begin{array}{cccc} 0 & E_1 & E_2 & E_3\\ -E_1 & 0 & B_3 & -B_2\\ -E_2 & -B_3 & 0 & B_1\\ -E_3 & B_2 & -B_1 & 0\end{array} \right)$$

This is not the matrix of the covariant electromagnetic tensor $$F_{\gamma\delta}.$$ Actually this is the matrix of the contravariant electromagnetic tensor $$F^{\alpha\beta}.$$ But I'm not worried about this because you're making use of this in the calculations correctly.

$$T_0^{ 0} = \eta_{0\beta}F^{0\beta} = 0$$
$$T_0^{ 1} = \eta_{0\beta}\\F^{1\beta} = E_1$$
$$T_0^{ 2} = \eta_{0\beta}F^{2\beta} = E_2$$

When all is done, I have

$$T_\delta^{ \alpha} = \left(\begin{array}{cccc}0 & E_1 & E_2 & E_3\\E_1 & 0 & -B_3 & B_2\\E_2 & B_3 & 0 & -B_1\\E_3 & -B_2 & B_1 & 0\end{array} \right)$$

These are all seamless.

When I take this to the next step...

$$F_{\gamma\delta} = \left(\begin{array}{cccc}0 & -E_1 & -E_2 &- E_3\\E_1 & 0 & B_3 & -B_2\\E_2 & -B_3 & 0 & B_1\\E_3 & B_2 & -B_1 & 0\end{array} \right)$$

Now this isn't obviously wrong

Unfortunately it is and I bet you'll find out where it all went wrong. Just pay a deeper attention to the formula $$F_{\gamma\delta}=\eta_{\gamma\alpha}T_\delta^{ \alpha}$$ and then it's done!

AB


AB

 

[MarkSheffield]

Altabeh - thanks for your reply.

I somehow have the ordering of rows and columns of the tensor mixed up and I'm not sure how. I'll go back and review some info or text on the mixed co-and-contra-variant forms for the second operation and see if I can get something different in the final result.

Rgds - Mark Sheffield

(I fixed/edited the form of the electromagnetic tensor.)

 

[Ben Niehoff]

The sign of E depends on the metric convention used. I think Altebeh is correct; you started with the wrong convention.

However, your calculation is correct. The difference between the covariant and contravariant forms is that B remains the same, and E flips sign.

Edited to add:

Now that you fixed the signs, it is all correct.

 

[MarkSheffield]

Ben - thank you for your reply

So what I have calculated as

$$F_{\alpha\beta} = \left(\begin{array}{cccc} 0 & E_1 & E_2 & E_3\\ -E_1 & 0 & B_3 & -B_2\\ -E_2 & -B_3 & 0 & B_1\\ -E_3 & B_2 & -B_1 & 0\end{array} \right)$$

and
$$F_{\gamma\delta} = \left(\begin{array}{cccc}0 & -E_1 & -E_2 &- E_3\\E_1 & 0 & B_3 & -B_2\\E_2 & -B_3 & 0 & B_1\\E_3 & B_2 & -B_1 & 0\end{array} \right)$$

is correct?

I'm confused.

$$-M_a^{rk}$$

The sign of E depends on the metric convention used. I think Altebeh is correct; you started with the wrong convention.

However, your calculation is correct. The difference between the covariant and contravariant forms is that B remains the same, and E flips sign.

Edited to add:

Now that you fixed the signs, it is all correct.

 

[Fredrik]

This Wikipedia article might be useful.