Lagrangian for the electromagnetic field coupled to a scalar field

  • #1
Frostman
114
17
Homework Statement:
Consider the following Lagrangian density for the electromagnetic field ##A_\mu## coupled to a scalar field ##\phi## (complex)
$$L=-\frac14F_{\mu\nu}F^{\mu\nu}+(D_\mu\phi)^*D^\mu\phi$$
Where ##D_\mu = \partial_\mu-iqA_\mu## and ## F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu##
1. Write the equations of motion for the two fields.
2. Applying the Noether theorem following the invariance for phase transformation of the field ##\phi##, write the conserved current.
3. Verify explicitly that the current written in this way is preserved.
Relevant Equations:
Euler-Lagrange equations
Noether theorem
It is the first time that I am faced with a complex field, I would not want to be wrong about how to solve this type of problem.
Usually to solve the equations of motion I apply the Euler Lagrange equations.
$$\partial_\mu\frac{\partial L}{\partial \phi/_\mu}-\frac{\partial L}{\partial \phi}=0$$
But since ##\phi## and ##\phi^*## are not independent of each other I will have to follow another path and the only one that comes to mind is the principle of least action. Should I use this approach?

What I get next will be two equations of motion (##\phi## and ##\phi^*##) plus that of the electromagnetic field (##A_\mu##): so I will have 3 EOM not 2. Or the EOM for ##\phi## and ##\phi^*## can be consider as one?
 

Answers and Replies

  • #2
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,122
13,028
You can vary ##\phi## and ##\phi^*## independently from each other, because what counts as independent field-degrees of freedom are real and imaginary parts, i.e., ##\mathrm{Re} \phi## and ##\mathrm{Im} \phi## are to be varied as independent real field-degrees of freedom in the action principle, but that is equivalent to considering ##\phi## and its conjugate complex, ##\phi^*##, as independent field-degrees of freedom.
 
  • #3
Frostman
114
17
You can vary ##\phi## and ##\phi^*## independently from each other, because what counts as independent field-degrees of freedom are real and imaginary parts, i.e., ##\mathrm{Re} \phi## and ##\mathrm{Im} \phi## are to be varied as independent real field-degrees of freedom in the action principle, but that is equivalent to considering ##\phi## and its conjugate complex, ##\phi^*##, as independent field-degrees of freedom.
Okay, so I can re-write my lagrangian as:
$$ L = -\frac14F_{\mu\nu}F^{\mu\nu}+\big(\partial_\mu\phi^*+iqA_\mu\phi^*\big)\big(\partial^\mu\phi-iqA^\mu\phi\big)$$

##\frac{\partial L}{\partial A_\alpha}=\frac{\partial}{\partial A_\alpha}\big(\partial_\mu\phi^*+iqA_\mu\phi^*\big)\big(\partial^\mu\phi-iqg^{\mu\nu}A_\nu\phi\big)=iq\delta_{\alpha\mu}\phi^*\big(\partial^\mu\phi-iqg^{\mu\nu}A_\nu\phi\big)-iqg^{\mu\nu}\delta_{\alpha\nu}\phi\big(\partial_\mu\phi^*+iqA_\mu\phi^*\big)##

##\frac{\partial L}{\partial A_\alpha}=iq\phi^*\partial^\alpha\phi+q^2A^\alpha\phi^2-iq\phi\partial^\alpha\phi^*+q^2A^\alpha\phi^2=2q^2A^\alpha\phi^2##

Since ##\phi## is a scalar and ##\partial^\alpha## a four-vector, I can afford to take the last step, right? Or is it wrong?

##\frac{\partial F}{\partial A_\alpha/_\beta}=-\frac14\frac{\partial }{\partial A_\alpha/_\beta}\big(F_{\mu\nu}F^{\mu\nu}\big)=-\frac14\big(\frac{\partial F_{\mu\nu}}{\partial A_\alpha/_\beta}F^{\mu\nu}+\frac{\partial F^{\mu\nu}}{\partial A_\alpha/_\beta}F_{\mu\nu}\big)=-\frac14\big(2F^{\mu\nu}\frac{\partial F_{\mu\nu}}{\partial A_\alpha/_\beta}\big)=\dots=-F^{\beta\alpha}##

So, we'll have:

##F^{\beta\alpha}/_\beta=-2q^2\phi^2A^\alpha##

##J^{\alpha}=-2q^2\phi^2A^\alpha##

This is the equation of motion for the electromagnetic field ##A^{\alpha}##

While for ##\phi## and ##\phi^*##:

##\frac{\partial L}{\partial \phi}=\frac{\partial }{\partial \phi}\big[\big(\partial_\mu\phi^*+iqA_\mu\phi^*\big)\big(\partial^\mu\phi-iqA^\mu\phi\big)\big]=-iqA^\mu\big(\partial_\mu\phi^*+iqA_\mu\phi^*\big)=-iqA^\mu\phi^*/_\mu+q^2A^2\phi^*##

##\frac{\partial L}{\partial \phi^*}=\frac{\partial }{\partial \phi*}\big[\big(\partial_\mu\phi^*+iqA_\mu\phi^*\big)\big(\partial^\mu\phi-iqA^\mu\phi\big)\big]=iqA_\mu\big(\partial^\mu\phi-iqA^\mu\phi\big)=iqA_\mu\phi/^\mu+q^2A^2\phi##

##\frac{\partial L}{\partial \phi/_\alpha}=\frac{\partial }{\partial \phi/_\alpha}\big[\big(\partial_\mu\phi^*+iqA_\mu\phi^*\big)\big(\partial^\mu\phi-iqA^\mu\phi\big)\big]=\frac{\partial }{\partial \phi/_\alpha}\big[\big(\partial_\mu\phi^*+iqA_\mu\phi^*\big)\big(g^{\mu\nu}\partial_\nu\phi-iqA^\mu\phi\big)\big]=g^{\mu\nu}\delta_{\alpha\nu}\big(\partial_\mu\phi^*+iqA_\mu\phi^*\big)=\phi^*/^\alpha+iqA^\alpha\phi^*##

##\frac{\partial L}{\partial \phi^*/_\alpha}=\frac{\partial }{\partial \phi^*/_\alpha}\big[\big(\partial_\mu\phi^*+iqA_\mu\phi^*\big)\big(\partial^\mu\phi-iqA^\mu\phi\big)\big]=\frac{\partial }{\partial \phi/^*_\alpha}\big[\big(\partial_\mu\phi^*+iqA_\mu\phi^*\big)\big(\partial^\mu\phi-iqA^\mu\phi\big)\big]=\delta_{\alpha\mu}\big(\partial^\mu\phi-iqA^\mu\phi\big)=\phi/^\alpha-iqA^\alpha\phi##

And I get:

##\phi^*/^\alpha_\alpha+iqA^\alpha/_\alpha\phi^*+iqA^\alpha\phi^*/_\alpha=-iqA^\alpha\phi^*/_\alpha+q^2A^2\phi^*##

##\phi/^\alpha_\alpha-iqA^\alpha/_\alpha\phi-iqA^\alpha\phi/_\alpha=iqA_\alpha\phi/^\alpha+q^2A^2\phi##

But equations of motion CANNOT be complex. So I made some mistakes, but I can't find it.
In any case, is it right the process that I use to get them?
 
  • #4
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,122
13,028
Why can't they be complex. After all ##\phi## is a complex field, and you also get out that one of the KG equations is the complex conjugated of the other. So this looks pretty good. On the other hand the equation for the em. field is not right. Carfully check at least your calculation of ##\partial L/\partial A_{\alpha}##.

Hint: It's much easier to take the variation of the action than to use the Euler-Lagrange equations. E.g., the variation of ##\phi^*## gives
$$\delta L=(\partial_{\mu} \delta \phi^* + \mathrm{i} q A_{\mu} \delta \phi^*)(\partial^{\mu} \phi-\mathrm{i} q A^{\mu} \phi).$$
Integrating this you get the variation of the action. Doing an integration by parts for the first term in the left bracket this gives
$$\delta S=\int_{\mathbb{R}^4} \mathrm{d}^4 x \delta \phi^* (\mathrm{i} q A_{\mu}-\partial_{\mu}) (\partial^{\mu} \phi-\mathrm{i} q A^{\mu} \phi).$$
Since this has to vanish for arbitrary ##\delta \phi^*## you get the EoM
$$-\Box \phi+\mathrm{i} q \partial_{\mu} (A^{\mu} \phi)+\mathrm{i} q A_{\mu} \partial^{\mu} \phi +A_{\mu} A^{\mu} \phi=0.$$
This is the same as your result for the EoM of ##\phi##.

As I said, check your Maxwell equations for ##A^{\mu}## again!
 
  • #5
Frostman
114
17
Why can't they be complex. After all ##\phi## is a complex field, and you also get out that one of the KG equations is the complex conjugated of the other. So this looks pretty good. On the other hand the equation for the em. field is not right. Carfully check at least your calculation of ##\partial L/\partial A_{\alpha}##.

I obtain the same result. I suppose that:

##iq\phi^*\partial^\alpha\phi+-iq\phi\partial^\alpha\phi^*##

Can't be simplified as 0

Or what I wrote as:

##2q^2A^{\alpha}\phi^2##

Should be written as:

##2q^2A^{\alpha}|\phi|^2##

If it's something else can you give me a hint?

Hint: It's much easier to take the variation of the action than to use the Euler-Lagrange equations. E.g., the variation of ##\phi^*## gives
$$\delta L=(\partial_{\mu} \delta \phi^* + \mathrm{i} q A_{\mu} \delta \phi^*)(\partial^{\mu} \phi-\mathrm{i} q A^{\mu} \phi).$$
Integrating this you get the variation of the action. Doing an integration by parts for the first term in the left bracket this gives
$$\delta S=\int_{\mathbb{R}^4} \mathrm{d}^4 x \delta \phi^* (\mathrm{i} q A_{\mu}-\partial_{\mu}) (\partial^{\mu} \phi-\mathrm{i} q A^{\mu} \phi).$$
Since this has to vanish for arbitrary ##\delta \phi^*## you get the EoM
$$-\Box \phi+\mathrm{i} q \partial_{\mu} (A^{\mu} \phi)+\mathrm{i} q A_{\mu} \partial^{\mu} \phi +A_{\mu} A^{\mu} \phi=0.$$
This is the same as your result for the EoM of ##\phi##.

As I said, check your Maxwell equations for ##A^{\mu}## again!

Oh I see, it's much easier and less calculations.
 
  • #7
Frostman
114
17
I think, you get in the right direction!
I go for

##iq\phi^*\partial^\alpha\phi-iq\phi\partial^\alpha\phi^*##

From the moment that ##\phi## and ##\phi^*## have imaginary part different (sign).

So the solution became

##J^{\alpha}=-iq\phi^*\partial^\alpha\phi+iq\phi\partial^\alpha\phi^*-2q^2\phi^2A^\alpha##
 
  • #8
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,122
13,028
...and also ##\phi^* \phi=|\phi|^2## instead of ##\phi^2## in the last term.
 

Suggested for: Lagrangian for the electromagnetic field coupled to a scalar field

Replies
0
Views
662
Replies
4
Views
134
Replies
24
Views
1K
Replies
2
Views
518
Replies
5
Views
1K
Replies
1
Views
2K
Replies
0
Views
575
Top