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- Thread starter JD_PM
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nrqed

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This was my point about finding the E-L equation. A classical field must obey that equation. So one must find the E-L equation and consider only the field configurations that obey them.

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I'll work it out and post what I get.

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$$\partial_{\mu} \Big( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} A_{\nu})} \Big) - \frac{\partial \mathcal{L}}{\partial A_{\nu}} = 0$$

At #8 I showed that

$$\frac{\partial \mathcal{L}}{\partial(\partial_{\mu} A_{\nu})} = -\partial^{\mu} A^{\nu}$$

The given Lagrangian doesn't depend on ##A_{\nu}## explicitly. Thus:

$$\frac{\partial \mathcal{L}}{\partial A_{\nu}} = 0$$

Thus our equation of motion is

$$-\partial_{\mu} \partial^{\mu} A^{\nu} = -\eta^{\nu \rho} \partial_{\mu} \partial^{\mu} A_{\rho} = 0$$

##\eta^{\nu \rho}## is a non-zero element, so we can always multiply both sides of the above equation by ##\frac{1}{\eta^{\nu \rho}}## to get

$$\partial_{\mu} \partial^{\mu} A_{\rho} = 0$$

Which is what we needed to finish b)

Thank you nrqed!

- #30

nrqed

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Good job!

$$\partial_{\mu} \Big( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} A_{\nu})} \Big) - \frac{\partial \mathcal{L}}{\partial A_{\nu}} = 0$$

At #8 I showed that

$$\frac{\partial \mathcal{L}}{\partial(\partial_{\mu} A_{\nu})} = -\partial^{\mu} A^{\nu}$$

The given Lagrangian doesn't depend on ##A_{\nu}## explicitly. Thus:

$$\frac{\partial \mathcal{L}}{\partial A_{\nu}} = 0$$

Thus our equation of motion is

$$-\partial_{\mu} \partial^{\mu} A^{\nu} = -\eta^{\nu \rho} \partial_{\mu} \partial^{\mu} A_{\rho} = 0$$

##\eta^{\nu \rho}## is a non-zero element, so we can always multiply both sides of the above equation by ##\frac{1}{\eta^{\nu \rho}}## to get

$$\partial_{\mu} \partial^{\mu} A_{\rho} = 0$$

Which is what we needed to finish b)

Thank you nrqed!

Although you cannot divide by ##\eta^{\nu \rho}##. Note that ##\rho## is summed over so it does not make sense to divide by that. Even worse, some of the components are zero. It is better to notice that ##\partial_{\mu} \partial^{\mu} A^{\nu}=0## implies that

$$\eta_{\rho \nu} \partial_{\mu} \partial^{\mu} A^{\nu}= 0 $$ and now we can move the ##\eta## inside the derivative, yielding what you wanted.

- #31

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Ahhh you're right.Even worse, some of the components are zero.

This equation

$$\partial_{\mu} \partial^{\mu} A^{\nu} = \eta^{\nu \rho} \partial_{\mu} \partial^{\mu} A_{\rho} = 0$$

It is equivalent to

$$\partial_{\mu} \partial^{\mu} A_{\nu} = \eta_{\nu \rho} \partial_{\mu} \partial^{\mu} A^{\rho} = \partial_{\mu} \partial^{\mu} A_{\nu} = 0$$

Which is what we want.

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