Radiative and Longitudinal degrees of freedom of EM field

Question

I'm having some issues trying to understand some of the affirmations my professor made in the last class. After working on the Maxwell's equation of the non-static regime for a quite a while, we arrived at the following expressions:

$$\nabla^2 \vec{A}_T - \frac{1}{c^2} \frac{\partial^2 \vec{A}_T}{\partial t^2} = -\frac{4\pi}{c} \vec{j}_T$$

$$\nabla^2 \phi + \frac{1}{c} \vec{\nabla} \cdot \frac{\partial \vec{A}_L}{\partial t} = -4\pi\rho$$ $$\frac{1}{c} \frac{\partial^2 \vec{A}_L}{\partial t^2} + \vec{\nabla} \frac{\partial \phi}{\partial t} = 4\pi \vec{j}_L$$

where the subscripts $T$ and $L$ stand for transverse or longitudinal as in the Helmholtz decomposition.

At this point, he said that this equations could lead us to think that there are two radiative (transverse) degrees of freedom and two longitudinal.

He then continued by saying that we can use the continuity equation to show that the last two equations are actually equivalent, and so there is only one longitudinal degree of freedom. I've showed that:

$$\frac{\partial}{\partial t} \left(\nabla^2\phi+ \frac{1}{c} \vec{\nabla}\cdot \frac{\partial \vec{A}_L}{\partial t} \right) = \vec{\nabla} \cdot \left(\vec{\nabla} \frac{\partial \phi}{\partial t}+\frac{1}{c}\frac{\partial^2 \vec{A}_L}{\partial t^2}\right)$$

where the LHS comes from the second equation and the RHS comes from third equation. What I don't get is: the 2nd and the 3rd are equivalent, but still they depend on $\phi$ and $\vec{A}_L$, which for me should provide more two degrees of freedom, one for each. Is this longitudinal degree of freedom the combination of both of them?

I have done all the pertinent passages by myself, the only thing in my way is his interpretation. I appreciate any guiding thoughts on this matter.

Answer 1

The time derivative of the second equation is identical to the divergence of the third one.

Edit. So the two field components are dependent and are one degree of freedom.

In fact the charge-current has four components but only 3 degrees of freedom. This is the deeper reason why you need exactly three field DoFs in gauge theory.

Answer 2

Okay, reading the answer of @my2cts I've come up with a reasoning that allows me to sleep at night. The 2nd and 3rd equation are indeed the same equation, say $$\nabla^2 \phi + \frac{1}{c} \vec{\nabla} \cdot \frac{\partial \vec{A}_L}{\partial t} = -4\pi\rho$$

which allow us to write one field in terms of the other, for example

$$\nabla^2 \phi = - \frac{1}{c} \vec{\nabla} \cdot \frac{\partial \vec{A}_L}{\partial t}-4\pi\rho$$

and so the two are not independent, and we have only one longitudinal degree of freedom.