Four Tensor Derivatives -- EM Field Lagrangian Density

[teroenza]

Homework Statement

Given the Lagrangian density

$$\Lambda = -\frac{1}{c}j^lA_l - \frac{1}{16 \pi} F^{lm}F_{lm}$$

and the Euler-Lagrange equation for it

$$\frac{\partial }{\partial x^k}\left ( \frac{\partial \Lambda}{\partial A_{i,k}} \right )- \frac{\partial \Lambda}{\partial A_{i}} =0$$

derive the inhomogeneous, manifestly covariant, field equations.

Homework Equations

In class we were told

$$\frac{\partial \Lambda}{\partial A_{i,k}} = -\frac{1}{4 \pi} F^{kl}$$

as a starting point, and I am trying to show this.

The Attempt at a Solution

I have that

$$F_{lm}=\frac{\partial A_m}{\partial x^l}-\frac{\partial A_l}{\partial x^m}$$

and

$$F^{lm}=\frac{\partial A^m}{\partial x_l}-\frac{\partial A^l}{\partial x_m}$$

I think I can see how inseting these into the above, possibly taking one of the field tensors to be constant, and taking derivatives (I think the first term does not depend on $A_{i,k}$) leads to the correct form or we were shown in class. I'm not great at four-tensor manipulation or keeping my indices straight, however.

I am trying to understand how to correctly take these derivatives.

$$\frac{\partial }{\partial A_{i,k}} \left ( F^{lm}F_{lm} \right )=0$$

 

[Orodruin]

Well, to start your equation does not make sense, you need to have the same indices on both sides of the equation.

Did you try simply applying the product rule?

 

[teroenza]

I copied the equation incorrectly, it should be:
$$\frac{\partial \Lambda}{\partial A_{l,m}} = -\frac{1}{4 \pi} F^{ml}$$

Using the product rule, I get:
$$\left (\frac{\partial A^m}{\partial x_l}-\frac{\partial A^l}{\partial x_m}\right ) + \left( \frac{\partial A_m}{\partial x^l}-\frac{\partial A_l}{\partial x^m} \right)\frac{\partial }{\partial A_{m,l}} \left(\frac{\partial A^m}{\partial x_l}-\frac{\partial A^l}{\partial x_m} \right )$$

But am not sure what to do with the derivatives in the second term.

 

[Orodruin]

I copied the equation incorrectly, it should be:
$$\frac{\partial \Lambda}{\partial A_{l,m}} = -\frac{1}{4 \pi} F^{ml}$$

Using the product rule, I get:
$$\left (\frac{\partial A^m}{\partial x_l}-\frac{\partial A^l}{\partial x_m}\right ) + \left( \frac{\partial A_m}{\partial x^l}-\frac{\partial A_l}{\partial x^m} \right)\frac{\partial }{\partial A_{m,l}} \left(\frac{\partial A^m}{\partial x_l}-\frac{\partial A^l}{\partial x_m} \right )$$

But am not sure what to do with the derivatives in the second term.

I suggest taking this one step at a time. Without rewriting F in terms of A, what does the product rule give you? (Note that your expression contains too many m and l to make sense. Each should appear twice or not at all in your result. Your free indices are i and k.)

 

[teroenza]

So the first step is:
$$F^{lm} \frac{\partial F_{lm}}{\partial A_{i,k}} + F_{lm}\frac{\partial F^{lm}}{\partial A_{i,k}}$$

 

[Orodruin]

Correct. Now, both terms are equal, so that gives a factor of two (just raise/lower the indices using the metric). This leaves you with the task of computing $\partial F_{lm}/\partial A_{i,k}$. How do you express $F_{lm}$ in terms of $A$? (Note that $A_{i,k} \equiv \partial_k A_i$)

 

[teroenza]

Like this:

$$F_{lm}=\frac{\partial A_m}{\partial x^l}-\frac{\partial A_l}{\partial x^m}=A_{m,l}-A_{l,m}$$

 

[Orodruin]

Yes, so what is the derivative of this with respect to $A_{i,k}$?

 

[teroenza]

Is $A_{l,m}$ antisymmetric? Then I could switch the indices on the second term, add them for the extra factor of 2, and use

$$\frac{\partial A_{m,l}}{\partial A_{i,k}}=\delta^{i}_{m}\delta^{l}_{k}=1$$

 

[Orodruin]

No, it is not antisymmetric, but what makes you think you cannot use this relation separately on each term?

 

[teroenza]

I can use it on each term. I can only see that leaving me with zero instead of adding for the necessary factor of two.

 

[Orodruin]

How would it leave you with zero? The terms are also not equal. What do you get if you do the computation?

 

[teroenza]

It must be simple, but I'm not making the connection. I think the derivative reduces to:

$$\delta^{i}_{m}\delta^{l}_{k}-\delta^{m}_{l}\delta^{i}_{k}$$

Which can only be 0 or 1.

 

[Orodruin]

The second term is incorrect, check the indices. The same indices must be up/down in both terms!

What happens when you contract this expression with the F that was outside?

 

[teroenza]

I see. The resulting difference ends up as $-4F^{ik}$ after swapping the indices on the (antisym.) field tensor.

 

[Orodruin]

Yes, this is correct. So what are the resulting equations of motion?

 

[teroenza]

They would then be $$\partial_kF^{kl} = -\frac{4 \pi}{c}j^l$$