# Describing 3-component spin 1 with a 4-Vector

1. Sep 21, 2014

### geoduck

For a spin 1 field described by a 4-vector, the condition $$m \partial_\mu A^\mu=0$$ can be derived from the equations of motion. This condition reduces the degrees of freedom from 4 to 3 for a massive particle (when m is not equal to zero).

However, for the photon, $$\partial_\mu A^\mu=0$$ has to be imposed by hand since the above equation is trivially zero on the LHS for m=0.

So shouldn't the photon still have 3 degrees of freedom once you add the Lorentz gauge condition $$\partial_\mu A^\mu=0$$ ?

But I read that it's the gauge condition that reduces the degrees of freedom from 3 to 2. But it seems it should reduce it from 4 to 3, the same as imposing the Lorentz gauge condition on a massive vector field.

2. Sep 21, 2014

### George Jones

Staff Emeritus
Yes.

Very roughly, imposing the Lorenz condition (Ludwig Lorenz, not H.A. Lorentz) reduces the number of degrees of freedom by 1, and imposing a particular gauge that is compatible with the Lorentz condition further reduces it by 1.

3. Sep 21, 2014

### geoduck

I don't ever recall making a further choice after the Lorenz gauge.

Instead I read that the longitudinal mode decouples. So it's not eliminated with a further gauge condition, but it decouples.

So can you just ignore the longitudinal mode rather than imposing another gauge condition?

4. Sep 22, 2014

### vanhees71

The point is that for a free em. field the Lorenz gauge condition doesn't fix the gauge completely, i.e., the constraint
$$\partial_{\mu} A^{\mu}=0$$
$$A_{\mu} \rightarrow A_{\mu}'=A_{\mu} + \partial_{\mu} \chi$$
with
$$\Box \chi=0.$$
So you can impose a further gauge constraint. A convenient choice is to demand that
$$A_0=A^0=0,$$
the socalled "radiation gauge". It fulfills both the Lorenz and the Coulomb condition.

Of course, this is only compatible for free em. fields, because if charges and currents are present, in the Lorenz gauge the inhomogeneous Maxwell equations read (in Heaviside-Lorentz units with $c=1$ as usual in modern QED/QFT/HEP texts)
$$\Box A^{\mu}=j^{\mu}.$$
Then you cannot set $A^{0}=0$, because that would contradict the wave equation for $A^0$, if $$j^0 \neq 0$$.

For the free field, however, you can choose the radiation gauge, and then the gauge is completely fixed. You are left with two independent field-degrees of freedom, as it must be for massless particles with non-zero spin.

Last edited: Sep 22, 2014
5. Sep 22, 2014

### dextercioby

Just perform the Hamiltonian analysis and you'll see how 4 DOF turn to 2. It's simple: the Hamiltonian system is 1st class with 2 constraints:

$$\pi_{0} \mbox{weakly} ~ 0$$ and $$-\partial^{i}\pi_{i} \mbox{weakly} ~ 0$$

As known, the only way to cure 1st class constraints is to convert them to 2nd class wrt the Dirac bracket. This adds 2 more first class constraints

$$A_{0} \mbox{weakly} ~ 0$$ and $$-\partial^{i}A_{i} \mbox{weakly} ~ 0$$

So the free e-m field has 2 times less DOF than the reduced phase space = (8-4)/2 = 2 DOF.

6. Sep 22, 2014

### geoduck

For scattering, don't you assume all initial and final particles are free, hence you can use the Coulomb gauge (and additionally set $A^0 =0$). In the Coulomb gauge $\nabla \cdot \vec{A}=0$ which implies that the vector potential $\vec{A}$ is orthogonal to $\vec{k}$, and hence the external polarization vectors whose linear combination equals $\vec{A}$ must be orthogonal to $\vec{k}$, which is a property of E&M waves (polarization vectors orthogonal to wave-vector)?

Can you choose a gauge for your free particles that doesn't have the property that $\vec{A}$ is orthogonal to $\vec{k}$?

In the interacting case (internal lines) then your propagator has a sum over all 3 polarization vectors and this sum gives the $g^{\mu \nu}-\frac{k^\mu k^\nu}{k^2}$ numerator in the propagator. Is the reason you sum over 3 rather than 2 polarization vectors because the photon need not be on-shell for virtual particles, so the the virtual photon in general has mass hence you need to sum over 3 polarizations?

7. Sep 22, 2014

### samalkhaiat

Consider the Maxwell’s equations
$$\partial^{ 2 } A^{ \mu } = \partial^{ \mu } \partial_{ \nu } A^{ \nu } , \ \ (1a)$$
and the gauge transformation
$$A^{ \mu } \rightarrow A^{ \mu } + \partial^{ \mu } f ( x ) . \ \ \ (2a)$$
Now if you insert the solution
$$A^{ \mu } ( x ) = \epsilon^{ \mu } ( p ) \ e^{ i p \cdot x } ,$$
in equations (1a) and (2a), you get
$$p^{ 2 } \ \epsilon^{ \mu } = ( p \cdot \epsilon ) \ p^{ \mu } , \ \ \ \ (1b)$$
and
$$\epsilon^{ \mu } ( p ) \rightarrow \epsilon^{ \mu } ( p ) + i f ( p ) \ p^{ \mu } , \ \ \ (2b)$$
where
$$f ( p ) \equiv f ( x ) \ e^{ - i p \cdot x } .$$
For $p^{ 2 } \neq 0$, (1b) implies that the polarization vector is proportional to $p$,
$$\epsilon^{ \mu } = \frac{ p \cdot \epsilon }{ p^{ 2 } } \ p^{ \mu } .$$
However, this “massive” mode is unphysical because it can be gauged away by the choice
$$f ( p ) = i \frac{ p \cdot \epsilon }{ p^{ 2 } } .$$
For massless mode, equation (1b) implies the Lorentz condition $p \cdot \epsilon = 0$. So, in the frame $p^{ \mu } = ( \omega , 0 , 0 , \omega )$, the polarization vector has only 3 degrees of freedom $\epsilon^{ \mu } = ( \epsilon^{ 0 } , \epsilon^{ 1 } , \epsilon^{ 2 } , \epsilon^{ 0 } )$.
Again, the $\epsilon^{ 0 }$ component is not physical for we can get red of it by the choice
$$f ( p ) = i \frac{ \epsilon^{ 0 } }{ \omega } .$$
So, for EM wave moving in the z-direction, the physical polarization vector is confined in the xy-plane and has only 2 degrees of freedom
Sam

8. Sep 22, 2014

### vanhees71

Yes, that's precisely the point! In radiation gauge, the only plane-wave modes are the two transverse solutions, and that's why the free em. field has only two polarization states and not three.

9. Sep 22, 2014

### geoduck

It seems from your derivations that the Lorenz gauge $\partial_\mu A^\mu=0[ /itex] is forced. It's not a gauge choice. In all gauge choices [itex]\partial_\mu A^\mu=0[ /itex] will be true. The only gauge-fixing you do is to get rid of the time-component. This is really your only choice since you can't cause [itex]\partial_\mu A^\mu \neq 0[ /itex]. Does this sound right? In samalkhaiat's response, he sets [itex]p^2=0$ for a massless photon, forcing the 4-divergence of 4-potential equal to zero. If the 4-divergence of the potential is not zero, then the photon is not massless?

Last edited: Sep 22, 2014
10. Sep 22, 2014

### samalkhaiat

No. A solution $A^{ \mu }$ (to Maxwell’s equations), with $\partial_{ \mu } A^{ \mu } \neq 0$, does not imply $p^{ 2 } \neq 0$. That part of my post only shows that Maxwell’s equations admit unphysical non-zero mass solutions.
If you have a specific solution $B^{ \mu }$ satisfying $\partial_{ \mu } B^{ \mu } \neq 0$, you can always choose the gauge function to find an $A^{ \mu }$ such that the Lorenz condition, $\partial_{ \mu } A^{ \mu } = 0$, is satisfied:
Since $B^{ \mu }$ is given, we can calculate $\partial_{ \mu } B^{ \mu }$ and therefore solve the equation $\partial^{ 2 } f = - \partial \cdot B$. Then, you can use the solution $f ( x )$ to find $A^{ \mu } = B^{ \mu } + \partial^{ \mu } f$ which satisfies the Lorenz condition.
Observe that the new solution is not unique: (from the theory of differential equations) you may always add a solution to the homogeneous equation $\partial^{ 2 } f = 0$. Thus, you are still allowed to make gauge transformation without spoiling the Lorenz condition, provided that $\partial^{ 2 } f = 0$. In other words, Lorenz condition does not specify the gauge fields uniquely.
The conclusion of my previous post is the following: Maxwell equations allow solution of the form $A_{ \mu } = \epsilon_{ \mu } \exp ( i p \cdot x )$ with $p^{ 2 } = 0$ and $p \cdot \epsilon = 0$. The gauge freedom then used to show that $\epsilon_{ \mu }$ has only 2 physical degrees of freedom.

11. Sep 23, 2014

### vanhees71

No! You can choose different gauges than Lorenz gauge, which do not fulfill $\partial_{\mu} A^{\mu}=0$. Take any four-potential $A_{\mu}$ fulfilling the Lorenz condition. Then I'm free to choose another four-potential
$$A_{\mu}'=A_{\mu} + \partial_{\mu} \chi,$$
with an arbitrary scalar field $\chi$. If $\Box \chi \neq 0$, $A_{\mu}'$ does not fulfill the Lorenz gauge condition, but of course is equivalent in the sense that it describes the same physics.

You can also choose, in the general case, temporal gauge $A_0=A^0=0$ or axial gauge $A_3=-A^3=0$. Then you cannot in addition demand that also the Lorenz gauge condition is fulfilled. An exception is the free field, where you can add a 2nd condition (and you must if you want to make the solutions unique). A convenient choice is the radiation gauge, where $\partial_{\mu} A^{\mu}=0$ and at the same time $A_0=0$.