Gauge Theory - differentiatin a Lagrangian

[countable]

Hi,

Hope some one can help me with a problem I am working on:

It involves working out:

$$\frac{\delta L}{\delta A_\nu}$$ of the following Lagrangian:

$$L=\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{1}{2}<br /> (D_{\mu} \Psi)^{*} D^{\mu}\Psi$$
The solutions show that this is equal to:

$$\frac{\delta L}{\delta A^\nu}=\frac{1}{2}(iq\psi^*D^\nu\psi - iq\psi D^\nu\phi^* )$$ (1)

[I'm guessing the phi above is just a typo - should it be psi?]

Homework Equations

and

The Attempt at a Solution

Would I be right in saying that:

$$(D_{\mu}\Psi)^{*}=\partial_\mu \psi^* + iqA_\mu \psi^*$$ - is this how to deal with complex conjugation? (where $$D_\mu = \partial_\mu - iqA_\mu$$ - ie. the covariant derivative)

If this is correct,

$$\frac{\delta }{\delta A_\nu} (D_{\mu} \Psi)^{*} D^{\mu}\Psi=\frac{\delta }{\delta A_\nu}(\partial_\mu \psi^* + iqA_\mu \psi^*)(\partial^\mu \psi - iqA^\mu \psi)$$

isn't this simply:

$$iq \psi^*(\partial^\mu \psi - iqA^\mu \psi) = iq \psi^* D^\mu \psi$$

So where does the second term in (1) come from?

Also does it make sense to differentiate tensors of different types ie:

$$A^\mu$$ wrt $$A_\mu$$

 

[dextercioby]

Would I be right in saying that:

$$(D_{\mu}\Psi)^{*}=\partial_\mu \Psi^* + iqA_\mu \Psi^*$$ - is this how to deal with complex conjugation? (where $$D_\mu = \partial_\mu - iqA_\mu$$ - ie. the covariant derivative)

This is correct, if A is real, that is only the scalar field is complex.

If this is correct,

$$\frac{\delta }{\delta A_\nu} (D_{\mu} \Psi)^{*} D^{\mu}\Psi=\frac{\delta}{\delta A_\nu}(\partial_\mu \Psi^* + iqA_\mu \Psi^*)(\partial^\mu \Psi - iqA^\mu \Psi)$$

Don't confuse the operators, what you wrote is wrong, you should have written

$$\frac{\partial}{\partial A_\nu} (D_{\mu} \Psi)^{*} D^{\mu}\Psi=\frac{\partial }{\partial A_\nu}(\partial_\mu \Psi^* + iqA_\mu \Psi^*)(\partial^\mu \Psi - iqA^\mu \Psi)$$

Also does it make sense to differentiate tensors of different types ie:

$$A^\mu$$ wrt $$A_\mu$$

No, pay attention to indices

$$\frac{\partial A^{\mu}}{\partial A_{\nu}} = \eta^{\mu\nu}$$

 

[countable]

thanks dexter - very helpful:

So to get the second term in (1) - ie. $$-iq\psi D^\nu \psi^*$$

I have to work out:

$$\frac{\partial }{\partial A_\nu} (D_\mu \psi)^* (-iqA^\mu \psi) = \frac{\partial }{\partial A_\nu} (D_\mu \psi)^* (-iq \eta^{\mu\nu} A_\nu \psi)= -iq\psi \eta^{\mu\nu} (D_\mu \psi)^* =-iq\psi\eta^{\mu\nu}(\partial_\mu \psi^* +iqA_\mu \psi^*)=-iq\psi(\partial^\nu \psi^* +iqA^\nu \psi^*)$$

but the bit in brackets in the last equality does not equal:

$$D^\nu \psi^*$$

as there is a + rather than a -

can you please tell me what am I doing wrong, if you can?

Thanks

 

[dextercioby]

First of all, you differentiate a product which contains the gauge field twice, so that you'll have

$$\left(\frac{\partial}{\partial A_{\nu}} (D_{\mu}\Psi)^{*}\right) (D^{\mu}\Psi) + (D_{\mu}\Psi )^{*} \left(\frac{\partial}{\partial A_{\nu}} (D^{\mu}\Psi)\right)$$

and now

$$\frac{\partial}{\partial A_{\nu}} (D_{\mu}\Psi)^{*} = +iq \delta^{\nu}_{\mu} \Psi^{*}$$

and

$$\frac{\partial}{\partial A_{\nu}} (D^{\mu}\Psi) = -iq \eta^{\mu\nu} \Psi$$

Can you sum the 2 terms now ?

 

[countable]

Well, I understand how to get the first term

$$\frac{\partial}{\partial A_\nu}(D_\mu \psi)^* D^\mu \psi= +iq \delta^{\nu}_{\mu} \Psi^{*} D^\mu \psi +(D^\mu \psi)^* . -iq \eta^{\mu\nu} \psi = iq\psi^* D^\nu \psi - iq\psi \eta^{\mu\nu} (D_\nu \psi)^*$$

The second term in the last equality above is (from the complex conjugation):

$$-iq\psi(\partial^\nu \psi^* +iqA^\nu \psi^*)$$

which isn't equal to

$$-iq\psi D^\nu \psi^*$$

as it should be. Where am I going wrong?

Sorry if I am missing something obvious, and thanks for bearing with me!

 

[dextercioby]

I don't understand your dilemma: the derivative you're looking for is

$$iq \Psi^{*} D^{\nu} \Psi - iq \Psi \left(D^{\nu}\Psi\right)^{*}$$

Now expand the derivatives and couple the terms alike.

 

[countable]

I don't understand your dilemma: the derivative you're looking for is

$$iq \Psi^{*} D^{\nu} \Psi - iq \Psi \left(D^{\nu}\Psi\right)^{*}$$

Now expand the derivatives and couple the terms alike.

I see where I've gone wrong. Thanks for your help Dexter.