Massive Vector Field: Questions & Answers

[Aleolomorfo]

Hello everybody.
The Lagrangian for a massive vector field is:
$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{m^2}{2}A_\mu A^\mu$$
The equation of motion is $\partial_\mu F^{\mu\nu}+m^2A^\nu = 0$
Expanding the EOM with the definition of $F^{\mu\nu}$ the Klein-Gordon equation for $A_\mu$ is obtained: $\Box A^\nu + m^2A^\nu = 0$.
Then taking the derivate of the EOM with respect to $\partial_\nu$, the Lorenz gauge condition is obtained $\partial_\nu A^\nu = 0$

I have two question.
The first one is related to the gauge condition. What does the fact that the theory still has the Lorenz condition (without the need to impose it) means? Because the Lagrangian is not gauge-invariant, but it satisfies a gauge condition.
The second is about the degrees of freedom. A massive vector field has three degrees of freedom. However, $A_\nu$ is a 4-component object which satisfy the equation of motion and the Lorenz condition, but these are two conditions, so 4-2=2 degrees of freedom. Where does my reasoning fail?

 

[Orodruin]

Just from the equation of motion you have 4 degrees of freedom that each satisfy the KG equation. The Lorenz condition removes one of those.

 

[Mudit_Jain]

First of all, your reasoning in the first paragraph is backwards. You begin with the EOM
$$\partial_{\nu}F^{\mu\nu} = -m^2A^{\mu}$$
which when expanded gives
$$\partial^{\mu}\partial\cdot A - \Box A^{\mu} = -m^2A^{\mu}.$$
Taking divergence of this gives
$$\Box\partial\cdot A - \Box\partial\cdot A + m^2\partial\cdot A=0 \implies \partial\cdot A = 0$$
and you realize that the vector field must satisfy this Lorentz condition. Then you can go back a step, make use of this Lorenz condition, and obtain your Klein-Gordan equation $$\Box A^{\mu} = m^2A^{\mu}.$$
Now to your question 1: " What does the fact that the theory still has the Lorenz condition (without the need to impose it) means? Because the Lagrangian is not gauge-invariant, but it satisfies a gauge condition."
--You must not think of the Lorenz condition as a gauge condition, since as you yourself pointed out, we're not dealing with a gauge theory here. The Lorenz condition simply reminds us that in general if I write the equation of motion as $\partial_{\nu}F^{\mu\nu} = J^{\mu}$ where $J^{\mu}$ is some current, it has to necessarily be conserved, meaning $\partial_{\mu}J^{\mu} = 0$ (because $F^{\mu\nu}$ is anti-symmetric). In this case of the free massive vector field, $J^{\mu} = m^2A^{\mu}$ which must now be conserved, and hence the so called Lorenz condition.

To your question 2: "A massive vector field has three degrees of freedom. However, $A^{\mu}$ is a 4-component object which satisfy the equation of motion and the Lorenz condition, but these are two conditions, so 4-2=2 degrees of freedom. Where does my reasoning fail?"
-- You're not quite right in counting the Lorentz condition as a technically true constraint, but I sympathize over this with you if we're talking intuitively (more on this later). But you definitely cannot count the equation of motion as a constraint. It just goes against the very definition of an "equation of motion". Equation of motion specifies how your degrees of freedom propagate in time, given the appropriate amount of initial conditions. Usually in relativistic field theories, we deal with 'second order' degrees of freedom, which is to say that for each degree of freedom, we need two sets of initial data to fully know how they are going to evolve in time. This is because we are dealing with PDEs that are second order in time (e.g. the sets of data could be (1) the initial field configuration over the whole of space, and (2) the initial conjugate momenta field over the whole of space). In this case for a massive vector field, you have three degrees of freedom and therefore need 6 sets of data: e.g. the three initial field configurations over the whole of space, and the three initial conjugate field configurations over the whole of space. Now I mentioned that the Lorenz condition is technically not quite a true constraint in some sense: From this Lorenz condition we have $\dot{A}^{0} = -\partial_{i}A^{i}$ which is not quite constraining $A^{0}$ but rather only $\dot{A}^{0}$. Now you may first think that you can integrate $\dot{A}^{0}$ w.r.t. time and obtain $A^{0}$, but you will quickly notice that I can add any time independent spatial function $f({\bf x})$ (viz a viz the 'integration constant'). This is precisely what I meant by the Lorenz condition not being a true constraint in that it does not reproduce $A^{0}$ for you uniquely. But then how do I reconcile the fact that I only have 3 degrees of freedom? Simple: Just expand the kinetic term $F^{\mu\nu}F_{\mu\nu}$ in terms of the vector field and its derivatives. You obtain
$$-\dfrac{1}{4}F^{\mu\nu}F_{\mu\nu} = \dfrac{1}{2}\left(\dot{\vec{A}} + \nabla A^{0}\right)^2 - \dfrac{1}{2}\left(\nabla\times\vec{A}\right)^2.$$
And now notice that the $A^{0}$ does not come with any terms that are quadratic in its time derivative. Therefore we finally realize that it is not a degree of freedom. You can solve for $A^{0}$ by varying the action (including the mass term) w.r.t. it and plug it back into the action to obtain the more direct and elucid action for the actual three degrees of freedom.

 

[vanhees71]

Another elegant method to treat the massive vector field is to treat it as an Abelian gauge theory using the Stueckelberg formalism. One should note that massive gauge bosons without a Higgs mechanism is possible for Abelian U(1) (but not for non-Abelian) gauge theories. The funny thing then is that in the quantized version you have to introduce a "Stueckelberg ghost" in addition to the usual "Faddeev Popov ghosts", which however are all non-interacting because the gauge group is Abelian.