Gauge conditions in interaction theory

[Haorong Wu]

TL;DR

If the interaction strength depends on the time, the gauge conditions may not be satisfied. Is this true, and how do we interpret this result?

Suppose we choose the Lorenz gauge conditions for an EM field, $\partial_\mu A^\mu=0$. The EOMs for the field are given by $\Box A^\mu=J^\mu$, with $\partial_\mu J^\mu=0$.

If the interaction time is ranged from $-\infty$ to $t$, $A^\mu$ satisfy the Lorenz gauge, because $A^\mu(x) = \int d^4x' D_{ret}(x - x')J^\mu(x')$ satisfy $\partial_\mu A^\mu(x) = \int d^4x' \, D_{ret}(x - x') [\partial'_\mu J^\mu(x')]=0$, where $D_{ret}(x - x')$ is the retarded Green's function.

But, in reality, the interaction time should not be infinite. Therefore, suppose $K^\mu=g(t)J^\mu$, where $g(t) = e^{-\epsilon |t|}$ is introduced to limit the interaction time to a finite range. In this case, $\partial_\mu K^\mu=J^0\partial_0 g(t)+g(t)\partial_\mu J^\mu=J^0\partial_0 g(t)\ne 0$. Then, the Lorenz gauge conditions are not satisfied.

Is this derivation valid? How do we interpret this result?

 

[Haorong Wu]

Hi, @DelajaSchuppers. Thank you for your answer. I am not familiar with the UniPhiEd theory.

I understand the continuity equation is broken in the finite time interaction. Then, do we lose the privilege to choose the gauge, even before the interaction starts? For example, if the interaction time is restricted to $[0,t]$, with $g(t')\ll 1$ for $t'\ll 0$ or $t' \gg t$, how do we choose the gauge at the earlier time $t'\ll 0$?

 

[Paul Colby]

The EOM you quote conserve charge, $\partial_\mu J^\mu =0$. You then violate charge conservation with your assumption, $K^\mu = g(t)J^\mu$. Doesn’t seem surprising things break.

 

[Haorong Wu]

The EOM you quote conserve charge, $\partial_\mu J^\mu =0$. You then violate charge conservation with your assumption, $K^\mu = g(t)J^\mu$. Doesn’t seem surprising things break.

Hi, Paul Colby. Then, after the interaction, what should I do next? Should I accept that the EM fields now violate the Lorenz gauge conditions, or should I throw away the temporal and the longitudinal components to restore the gauge conditions, or should I resort to the Ward identities to restore the gauge conditions (this may work for interaction time ranged from $-\infty$ to $t$, but may fail for finite interaction time)?

 

[PeterDonis]

after the interaction, what should I do next?

I think you're missing Paul Colby's point. The EM field equations only make sense in the first place if the current $J^\mu$ satisfies the conservation equation $\partial_\mu J^\mu = 0$. You can't just arbtrarily decide that the actual source is $K^\mu$ instead of $J^\mu$. You have to specify $J^\mu$ so that it satisfies the conservation equation and also describes whatever "sources" are present in your scenario--and satisfying the conservation equation means sources can't just magically appear or disappear. The specification for $J^\mu$ then determines what "interactions" there are. There is no "after the interaction" where you're allowed to have something different from $J^\mu$ as the source, or just allow $J^\mu$ to magically disappear.

Here's another way of putting this: you speak of "interaction strength" in your OP. What is this "interaction strength" and where does it appear in your equations? Nowhere that I can see. So what does it even mean for the "interaction strength" to depend on the time?

In other words, so far you haven't even specified a consistent scenario, so it's no surprise that you're getting nonsensical answers.

 

[PeterDonis]

Should I accept that the EM fields now violate the Lorenz gauge conditions, or should I throw away the temporal and the longitudinal components to restore the gauge conditions, or should I resort to the Ward identities to restore the gauge conditions

None of the above. You need to go back to the beginning and specify a consistent scenario.

 

[Paul Colby]

One is always free to change gauges. One is not free to violate charge conservation. Beyond that, I have no idea what you’re trying to accomplish.