# Lagrangian for the electromagnetic field coupled to a scalar field

• Frostman
In summary, the conversation discusses the use of the Euler-Lagrange equations to solve equations of motion, but due to the dependence of certain fields on each other, the principle of least action must be applied. The resulting equations of motion for the electromagnetic field and the fields ##\phi## and ##\phi^*## are derived, with the latter being found to be complex. However, this is not a problem as the fields are complex and the resulting equations are consistent. The correct process for deriving the equations of motion is also discussed.
Frostman
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?

Delta2
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.

vanhees71 said:
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?

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!

vanhees71 said:
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?

vanhees71 said:
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.

I think, you get in the right direction!

vanhees71 said:
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##

...and also ##\phi^* \phi=|\phi|^2## instead of ##\phi^2## in the last term.

Frostman

## 1. What is the Lagrangian for the electromagnetic field coupled to a scalar field?

The Lagrangian for the electromagnetic field coupled to a scalar field is a mathematical expression that describes the dynamics of the electromagnetic field and the scalar field. It is written as L = -1/4FμνFμν + 1/2(∂μφ)(∂μφ) - V(φ), where Fμν is the electromagnetic field tensor, φ is the scalar field, and V(φ) is the potential energy associated with the scalar field.

## 2. What is the physical significance of the Lagrangian for the electromagnetic field coupled to a scalar field?

The Lagrangian for the electromagnetic field coupled to a scalar field is a fundamental tool in theoretical physics that helps us understand the behavior of these two fields. It allows us to calculate the equations of motion and study the interactions between the electromagnetic and scalar fields.

## 3. How does the Lagrangian for the electromagnetic field coupled to a scalar field relate to the Standard Model of particle physics?

The Lagrangian for the electromagnetic field coupled to a scalar field is a crucial component of the Standard Model of particle physics. It describes the interactions between the electromagnetic and scalar fields, which are essential for understanding the behavior of fundamental particles such as electrons, protons, and quarks.

## 4. What is the role of the scalar field in the Lagrangian for the electromagnetic field coupled to a scalar field?

The scalar field in the Lagrangian for the electromagnetic field coupled to a scalar field represents a type of particle called the Higgs boson. This field is responsible for giving mass to other particles in the Standard Model and plays a crucial role in the mechanism of electroweak symmetry breaking.

## 5. How is the Lagrangian for the electromagnetic field coupled to a scalar field used in practical applications?

The Lagrangian for the electromagnetic field coupled to a scalar field is used in various practical applications, such as in the development of quantum field theories and in particle physics experiments. It is also used in cosmology to study the behavior of the early universe and in condensed matter physics to understand the behavior of materials at a microscopic level.

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