Tuesday, June 19, 2018

electromagnetism - Determining the Lorenz gauge condition



I'm having a bit of trouble understanding how the gauge condition is found.


Consider the potentials $V$ and $\vec A$ and $V'$ and $\vec A'$ so that
$$\vec E = -\vec\nabla V -\frac{\partial\vec A}{\partial t} = -\vec\nabla V' -\frac{\partial\vec A'}{\partial t}$$ $$\vec B = \vec\nabla \times \vec A=\vec\nabla \times \vec A'$$


The 2 sets of potentials are then related like this:


$$\vec A' = \vec A+\vec\nabla\theta$$ $$V' = V-\frac{\partial\theta}{\partial t}$$ For a function $\theta$.


The equations for the potentials are the following:


$$\vec\nabla^2\vec A-\varepsilon\mu\frac{\partial^2 \vec A}{\partial t^2}=-\mu\vec J+\vec \nabla(\vec \nabla.\vec A+\varepsilon\mu\frac{\partial V}{\partial t})$$


$$\vec\nabla^2V -\varepsilon\mu\frac{\partial^2 V}{\partial t^2}=-\frac{\rho}{\varepsilon} - \frac{\partial}{\partial t}(\vec\nabla . \vec A+\varepsilon\mu\frac{\partial V}{\partial t})$$


They can be decoupled by choosing the right potentials so that $(\vec\nabla . \vec A+\varepsilon\mu\frac{\partial V}{\partial t})=0$


The above part is the theory in general.



Now to find the correct potentials where this is the case, let's say that this last part is true for potentials $\vec A'$ and $V'$, i.e.


$$\vec\nabla . \vec A' + \varepsilon\mu\frac{\partial V'}{\partial t}=0$$


Filling in the relations between the 2 sets of potentials from earlier gives:


$$\vec\nabla . \vec A + \nabla^2\theta + \varepsilon\mu\frac{\partial V}{\partial t} -\varepsilon\mu\frac{\partial^2\theta}{\partial t^2}=0$$


Yet my book tells me that this means that


$$\nabla^2\theta - \varepsilon\mu\frac{\partial^2\theta}{\partial t^2}=0$$


I don't see why this is the case.



Answer



You've just run into the fact that the Lorenz gauge is only a partial gauge choice; it does not uniquely specify the vector and scalar potentials. The condition given on $\theta$ ensures that, if your original $A, V$ obeyed the Lorenz condition, then the new potentials $A', V'$ will also obey the Lorenz condition.


Unique solutions for $A, V$ arise only with additional conditions imposed (for instance, vanishing sufficiently fast at infinity).



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