Lagrangian Field Theory - Maxwell's Equations

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Homework Help Overview

The discussion revolves around the application of Lagrangian field theory to derive a specific derivative of a Lagrangian involving Maxwell's equations. The original poster presents a Lagrangian that includes terms with derivatives of a vector potential and seeks to compute the derivative with respect to one of the derivatives of the vector potential.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to manipulate the Lagrangian and compute the derivative, expressing concerns about potential mistakes in index handling and consistency. Some participants suggest focusing on proper index notation and manipulation rules to avoid confusion. Others propose rewriting the Lagrangian with different indices to clarify the calculations.

Discussion Status

Participants have provided guidance on index manipulation and suggested alternative formulations of the Lagrangian to aid understanding. The original poster acknowledges an error in the Lagrangian's expression and indicates that they ultimately arrived at the correct answer, highlighting the importance of careful index management.

Contextual Notes

The discussion includes concerns about the original expression of the Lagrangian and the implications of using repeated indices in calculations, which may lead to confusion or errors in the derivation process.

Woolyabyss
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Homework Statement


$$ L = -\frac{1}{2} (\partial_{\mu} A_v) (\partial^{\mu} A^v) + \frac{1}{2} (\partial_{\mu} A^v)^2$$

calculate $$\frac{\partial L}{\partial(\partial_{\mu} A_v)}$$

Homework Equations


$$ A^{\mu} = \eta^{\mu v} A_v, \ and \ \partial^{\mu} = \eta^{\mu v} \partial_{v}$$

The Attempt at a Solution


rewrite equation as $$ L = - \frac{1}{2} \eta^{\mu a} \eta^{v b} (\partial_{\mu} A_v) (\partial_{a} A_b) + \frac{1}{2} \eta^{v b} \eta^{v b} (\partial_{\mu} A_b) (\partial_{\mu} A_b )$$

now taking $$\frac{\partial L}{\partial(\partial_{\mu} A_v)}$$

we get $$ - \frac{1}{2} ( \eta^{\mu a} \eta^{v b} \partial_{a} A_b + \eta^{\mu a} \eta^{v b} (\partial_{\mu} A_v) \delta^{\mu v}_{a b} ) + \frac{1}{2} \eta^{v b} \eta^{v b}( \delta_{b}^{v} \partial_{\mu} A_b + \partial_{\mu} A_b \delta_{b}^{v}) $$

This is what I've got so far. I'm pretty certain I've made a number of mistakes here. On the left hand side I've got the mu's and v's repeating 3 times which probably isn't a good sign and on the right hand side the indices of the delta don't seem to be consistent.

The answer is

$$ \frac{\partial L}{\partial(\partial_{\mu} A_v)} = - \partial^{\mu} A^v + (\partial_{\rho} A^{\rho} ) \eta^{\mu v} $$

Any help would be appreciated.
 
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You seem to have some trouble with index manipulations. There are some very simple rules that you should always follow:
  1. You can never have more than two of the same index. If you at some point write down an expression with more than two of the same index you have done something wrong and you need to work around it by using other index manipulations.
  2. If you have two of the same index, one needs to be covariant and the other contravariant. This represents a sum according to the summation convention. This has the following implication:
    • You can rename any index that appears twice as long as you rename it to something that does not appear in your expression.
  3. If you have one of a particular index, it is a free index and needs to appear on both sides of an equality. You are not allowed to rename this index.
As an example, a very common mistake is of the following nature: You have an expression that contains a scalar that can be expressed as a product ##V^2 = V_\mu V^\mu## and you want to compute ##\partial_\mu V^2##. You insert the expression for ##V^2## and get ##\partial_\mu V^2 = \partial_\mu V_\mu V^\mu##. This expression is nonsense as you have three ##\mu## indices. What went wrong is that you inserted ##V^2 = V_\mu V^\mu## in an expression that already had a ##\mu## index. However, what you can do is to use (2) to replace the ##\mu## index in the ##V^2## expression to obtain ##V^2 = V_\nu V^\nu##. Since there are no ##\nu## indices in ##\partial_\mu V^2## you can now insert this into the expression to find that ##\partial_\mu V^2 = \partial_\mu V_\nu V^\nu##, which is a valid expression.

I suggest you go through your expressions with the above in mind. Note that the last term of the Lagrangian you are given a priori violates some of the rules above. However, it is quite common to by the square denote multiplication with the same expression, but with the indices swapped from covariant to contravariant and vice versa. In other words ##(\partial_\mu A^\nu)^2## should be read as ##(\partial_\mu A^\nu)(\partial^\mu A_\nu)##, not as ##(\partial_\mu A^\nu)(\partial_\mu A^\nu)##, which is meaningless.
 
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If you're having trouble, it might be helpful if you do not use the indices \mu, v in your Lagrangian so that you don't get confused with the indices in the derivative that you're taking. So for your case:

<br /> L = -\frac{1}{2} \eta^{a c} \eta^{b d}(\partial_{a} A_b) (\partial_{c} A_d) + \frac{1}{2} \eta^{a b} \eta^{c d}(\partial_{a} A_b) (\partial_{c} A_d)<br />
or
<br /> L = \frac{1}{2}\left(\eta^{a b} \eta^{c d} - \eta^{a c} \eta^{b d}\right) (\partial_{a} A_b) (\partial_{c} A_d) <br />

Then use the relation
<br /> \frac{\partial }{\partial (\partial_\mu A_v)} (\partial_{a} A_b) = \delta^{\mu}_a \delta^{v}_b<br />
repeatedly. Now that you're not reusing indices in different parts of your equations it may eliminate some confusion.
 
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Thanks a lot to the both of you. I made an error when I wrote down the Lagrangian, the term on the right was actually $$ \frac{1}{2} ( \partial_{\mu} A^{\mu} )^2 $$

Regardless, I managed to get the right answer in the end. It never occurred to me that if my derivative had indices matching indices of the Lagrangian I could run into problems.
 

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