# Invariance of EM Lagrangian

1. Apr 11, 2016

### ShayanJ

1. The problem statement, all variables and given/known data

Show that in order for the free Lagrangian to be invariant when $A^\mu$ is transformed by a transformation U, it has to transform as below:
$A'^{\mu}=\frac i g (\partial^\mu U) U^{-1}+U A^\mu U^{-1}$

2. Relevant equations

The wording of the problem is a bit awkward I think, but it seems it means the free EM Lagrangian!

$L_{free \ EM}=-\frac 1 4 F_{\mu \nu} F^{\mu \nu}$
$F_{\mu \nu}=\partial_\mu A_\nu - \partial_\nu A_\mu$

3. The attempt at a solution

I've never seen it like this before. It was always figuring out the transformation rule for $A^\mu$ somehow that it leaves another field's Lagrangian invariant!
The only method that comes into my mind is substituting $A'^\mu$ in $F'_{\mu \nu} F'^{\mu \nu}$ and expanding it. But it doesn't show any sign of convinient cancellations early enough and it seems I have to expand it all the way. But it would be a giant with 100 terms in it! So I think there has to be a better way for doing this. But I just can't figure it out!

Any small hint is appreciated.
Thanks

2. Apr 11, 2016

### Fred Wright

I think you are dealing with scalar O(n) gauge theory. The Lagrangian density can be expressed as L=1/2(∂μΦ)TμΦ- 1/2m2ΦTΦ where Φ is a vector of fields. Now, the Lagrangian is invariant under the transformation Φ' |→ UΦ whenever U is a constant matrix belonging to the nxn orthogonal group O(n). Also, (∂μΦ) |→ (∂μΦ)' = U(∂μΦ) and this expresses global symmetry. However, if you require local invariance the U will not pass through the derivative when U=U(x); obviously ∂μ(UΦ) ≠ U(∂μΦ). I suggest that you introduce the covariant derivative Dμ = ∂μ + igAμ, where g is a coupling constant and A(x) is the gauge field. The required result should immediately follow after a simple calculation.

Last edited: Apr 11, 2016
3. Apr 11, 2016

### ShayanJ

Thanks for the comment but I know that and the problem seems to require that only the EM Lagrangian gets involved.

4. Apr 11, 2016

### Fred Wright

I think that "free Lagrangian" in the problem statement refers to no interactions between the scalar fields. I think you are making a mistake in identifying the vector potential of classical EM with the gauge field of QFT.

5. Apr 11, 2016

### ShayanJ

The vector potential of classical EM is the gauge field of QFT if the gauge group is U(1)!
And without mentioning any specific Lagrangian, it could be a complex scalar field or a spinor field!
And actually it was a problem on a test. I asked the professor which Lagrangian he meant, is it the EM Lagrangian? He said its obvious! He's really bad at teaching!

6. Apr 12, 2016

### CAF123

Hi Shyan,
I am still learning this as well so take what I say with a pinch of salt :) From the structure of the transformation law for $A_{\mu}$, I would guess the lagrangian is pure Yang-Mills, i.e $$\mathcal L = -\frac{1}{4} F_{\mu \nu}^a F^{\mu \nu, a}$$ with gauge group SU(N). So the $F_{\mu \nu}$ has an additional piece in its structure.
Possibly?

7. Apr 12, 2016

### ShayanJ

Thanks for the comment but no. If you set $U=e^{i\alpha(x)}$ in the transformation law given in the first post, you see its exactly a U(1) gauge transformation.

I'm starting to think that the professor just confused me with the wording of the problem and his unclear answer to my question and I simply should have used a complex scalar field or a Dirac field.

It would be good if I had some confirmation from science advisors whether people actually check the transformation rule given in the first post using only the free EM Lagrangian or not!(@vanhees71 , @strangerep ,@samalkhaiat ,@Demystifier , and any other advisors...you have no comment?)

8. Apr 13, 2016

### samalkhaiat

Shyan, I rarely visit the section of the forums. So, please notify me by PM in the future.
Okay, I think your instructor has messed the problem up very badly. As it stands, the wording of the problem makes no sense at all. I believe that he had the following problem in mined, but failed to formulate it properly.
The problem: Let $\mathcal{L}(\Phi , \partial \Phi)$ be a free generic matter field theory. Assume that $\mathcal{L}(\Phi , \partial \Phi) = \mathcal{L}(\Phi^{'} , \partial \Phi^{'})$ is invariant under the following (global) transformations $$\Phi^{'} = U \Phi , \ \ \ \partial_{\mu}\Phi^{'} = U \partial_{\mu}\Phi ,$$ where $U$ is a constant (matrix).
Show that demanding invariance under local transformation (achieved by letting $U \to U(x)$) forces you to introduce a (matrix-valued) vector field $A_{\mu}$, which couples to the matter field $\Phi$ and transforms according to $$A^{'}_{\mu} = i (\partial_{\mu}U) U^{-1} + U A_{\mu} U^{-1} . \ \ \ \ \ \ (1)$$
Can you do this problem?
Remark: (i) If $A_{\mu}$ is a number (not matrix), i.e., if the gauge group is $U(1)$, then the second term in the transformation law is $$U A_{\mu} U^{-1} = A_{\mu} U U^{-1} = A_{\mu} .$$ In this case, you can easily show that the field tensor $F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$ is invariant under the $U(1)$ gauge transformation $$A^{'}_{\mu} = i (\partial_{\mu}U) U^{-1} + A_{\mu} . \ \ \ \ \ \ (2)$$ This means that you need to add the gauge invariant Maxwell’s Lagrangian $- \frac{1}{4}F^{2}$, to the gauge invariant matter field Lagrangian that you obtained by introducing the gauge field $A_{\mu}$. This leads you to the QED Lagrangian: $$\mathcal{L}_{QED} = \mathcal{L}( \Phi , D_{\mu}\Phi ) - \frac{1}{4} F_{\mu \nu} F^{\mu \nu} .$$
(ii) If $A_{\mu}$ is a matrix-valued vector field, i.e., if $U(x)$ belongs to a non-abelian gauge group, then the matrix-valued field tensor is given by
$$[D_{\mu} , D_{\nu}] \Phi \equiv i F_{\mu \nu}\Phi , \ \ \ \ \ (3)$$
$$F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu} + [A_{\mu} , A_{\nu}].$$
In this case, you can show (using 3) that the field tensor transforms according to
$$F^{'}_{\mu \nu} = U F_{\mu \nu} U^{-1}.$$
Thus, to construct a gauge-invariant Lagrangian for the gauge fields you need to trace over the group index: $$\mathcal{L}_{YM} = - \frac{1}{4} \mbox{Tr} ( F_{\mu \nu} F^{\mu \nu}) .$$ In this case, the resulting gauge-invariant interacting theory is described by the following total Lagrangian
$$\hat{\mathcal{L}} = \mathcal{L}( \Phi , D_{\mu}\Phi ) - \frac{1}{4} \mbox{Tr} ( F_{\mu \nu} F^{\mu \nu}) .$$

9. Apr 13, 2016

### ShayanJ

Thanks Sam, that's very kind of you.
I think...but I'm not getting the right result!

At first I note that for the local transformation, we'll have
$\mathcal L(U\Phi,\partial (U \Phi))=\mathcal L(U\Phi,(\partial U)\Phi+U\partial \Phi)$,
this can't be equal to $\mathcal L(\Phi,\partial \Phi)$ for a general U.

In order to achieve invariance, I consider the Lagrangian
$\mathcal L(\Phi,\partial \Phi+A \Phi)=L(\Phi,\partial \Phi)+\frac{\partial \mathcal L}{\partial(\partial_\mu \Phi)} A_\mu \Phi$
instead. (Actually expanding the Lagrangian and retaining only the first term here doesn't seem quite right to me because it assumes that $A_\mu$ are small!)

Then I apply a local transformation U to the fields to get:
$\mathcal L(U\Phi,(\partial U)\Phi+U \partial \Phi+A' U \Phi)=\mathcal L(U\Phi,U\partial \Phi)+\frac{\partial \mathcal L}{\partial(\partial_\mu \Phi)} \left[ (\partial_\mu U)+A'_\mu U \right] \Phi$
So for the new Lagrangian to be invariant under a local transformation U, we should have:
$A_\mu=(\partial_\mu U)+A'_\mu U$
But this is not correct! I can't find out what went wrong!

10. Apr 14, 2016

### samalkhaiat

No, you can not expand the Lagrangian. Expansion requires infinitesimal transformation, but you are dealing with finite transformation $U(x)$.
Note that, by hypothesis, $\mathcal{L}(\Phi , \partial \Phi)$ is invariant under global (constant $U$) transformation, because both arguments of $\mathcal{L}$ transform in the same way (i.e., covariantly): $\Phi \to U\Phi$ and $\partial \Phi \to U \partial \Phi$. Now, keep this note in mined because we will need it bellow.
When we make the transformation local by letting $U \to U(x)$, we find $\Phi \to U(x) \Phi$, but covariance is lost because $$\partial_{\mu}\Phi^{'} = U \partial_{\mu}\Phi + (\partial_{\mu} U) \Phi . \ \ \ \ \ (1)$$ This means that the original Lagrangian is no longer invariant under the local $U(x)$ transformation. So, to achieve invariance, our Lagrangian must be in the form $\mathcal{L}(\Phi , \Psi_{\mu})$, with both arguments $\Phi$ and $\Psi_{\mu}$ having the same transformation law under $U(x)$. Thus, we only need to find the object $\Psi_{\mu}$ which transforms according to $$\Psi^{'}_{\mu} = U(x) \Psi_{\mu} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$
Once an expression for $\Psi_{\mu}$ is found, equation (2) will, therefore, insure the invariance of $\mathcal{L}(\Phi , \Psi_{\mu})$ under the local transformation $U(x)$.
Now, (i) Since we need to recover the original theory in the limit $\Psi_{\mu} \to \partial_{\mu}\Phi$, the object $\Psi_{\mu}$ must depend linearly on $\partial_{\mu}\Phi$. (ii) Since $\Psi_{\mu}$ and $\partial_{\mu}\Phi$ both carry Lorentz index, the new ingredient in the theory must also carry a Lorentz index. However, taking a Lorentz vector $V_{\mu}$ and setting $\Psi_{\mu} = \partial_{\mu}\Phi + V_{\mu}$ is not correct, because $\partial_{\mu}\Phi$ need not be a Lorentz vector. Indeed, $\Phi$ is an arbitrary matter field, it could itself be a vector field. So, the simplest choice which keeps Lorentz happy is achieved by taking
$$\Psi_{\mu} = \partial_{\mu}\Phi + i A_{\mu}\Phi \equiv D_{\mu}\Phi . \ \ \ \ (3)$$ Thus, the field $A_{\mu}$ is the only new ingredient in the theory. Now, from the local invariance condition eq(2), we can easily deduce the transformation law for the new field, just substitute (3) in (2)
$$\left( \partial_{\mu} + i A^{'}_{\mu} \right) U \Phi = U \left( \partial_{\mu}\Phi + i A_{\mu}\Phi \right) .$$
Okay, I leave you to expand the left-hand-side and obtain the required local transformation law for $A_{\mu}$. Notice that in the new theory, the field $A_{\mu}$ is coupled to the matter field $\Phi$. This means that imposing local invariance on a free globally invariant matter field theory leads to interacting field theory.
If you like to expand your knowledge and see how this is done in details, see the PDF below.

#### Attached Files:

• ###### Ch5 The Gauge Principle.pdf
File size:
343.6 KB
Views:
60
Last edited: Apr 14, 2016
11. Apr 15, 2016

### ShayanJ

Thanks for the explanations Sam.

Your PDF is nice, but I just have a question, In section 2, you try to prove $R_{\mu\nu}=R_{\nu \mu}$. This is trivially true for the spatial components. But about $R_{j0}=R_{0j}$, $R_{j0}$ is zero, but $R_{0j}=\int d^3 x \partial_j F(x)$ which you don't prove to be zero! But I really doubt that the integral over all space of the derivative of any given function is zero!

The name of the PDF file contains "Ch5" which suggests its the chapter 5 of something. A series of lecture notes of yours maybe?

12. Apr 15, 2016

### samalkhaiat

I did. If you go down two lines, you see the boundary condition on $F(x)$.
It is not any function.
The question was the following: What boundary condition the function $F(x)$ must satisfy in order for $\int d \sigma_{\mu} \ \partial_{\nu} F(x) = \int d \sigma_{\nu} \ \partial_{\mu} F(x)$ to be identically true?
Now, $$R_{0j} - R_{j0}= \int_{V} d^{3}x \ \partial_{j}F(x) .$$
By the divergence theorem
$$\int_{V} d^{3}x \ \partial_{j}F(x) = \int_{S^{2}} dS \ \hat{n}_{j} \ F(x) ,$$
we have the following behaviour
$$R_{0j} - R_{j0} \sim 4\pi | \vec{x}|^{2} F(x) |_{|\vec{x}| \to \infty} .$$
This, of course vanishes when $F(x)$ dies off faster that $\frac{1}{| \vec{x}|^{2}}$.

Actually, except for minor improvement, I made the material in that PDF when I was in grad school (22 years ago!). But, yes they are part of huge set of notes I delivered at various places on symmetries in classical and quantum field theories. The plan was (and still is) to organize them into a textbook, but unfortunately I haven’t got time to typeset them all yet.