I When does the second uniqueness theorem apply?

[Ahmed1029]

For the second uniqueness theorem of electrostatics to apply, does the outer boundary enclosing all the conductors have to be at a constant potential?

 

[vanhees71]

No, why?

 

[Ahmed1029]

No, why?

Because in the proof he assumes that V3 (the difference between the two supposed potentials) is constant at all surfaces and at the boundary, but that would mean that every supposed potential is constant at the outer boundary. What am I getting wrong?

 

[Delta2]

No the difference being constant doesn't necessarily imply that each potential is constant.

Take for example the functions $$\phi_1(x,y,z)=\frac {e^{-x-y-z}}{x^2+y^2+z^2}$$, $$\phi_2(x,y,z)=\frac {e^{-x-y-z}}{x^2+y^2+z^2}+1$$their difference is constant and equal to 1 (or -1) but the functions are not constant.

It just implies that their gradient is the same (under some other conditions too).

 

[Ahmed1029]

No the difference being constant doesn't necessarily imply that each potential is constant.

Take for example the functions $$\phi_1(x,y,z)=\frac {e^{-x-y-z}}{x^2+y^2+z^2}$$, $$\phi_2(x,y,z)=\frac {e^{-x-y-z}}{x^2+y^2+z^2}+1$$their difference is constant and equal to 1 (or -1) but the functions are not constant.

It just implies that their gradient is the same (under some other conditions too).

oh so that's it! Thanks!

 

[Ahmed1029]

No the difference being constant doesn't necessarily imply that each potential is constant.

Take for example the functions $$\phi_1(x,y,z)=\frac {e^{-x-y-z}}{x^2+y^2+z^2}$$, $$\phi_2(x,y,z)=\frac {e^{-x-y-z}}{x^2+y^2+z^2}+1$$their difference is constant and equal to 1 (or -1) but the functions are not constant.

It just implies that their gradient is the same (under some other conditions too).

Sorry I have another question. Suppose now I have two supposed potential functions whose difference is not constant at the outer boundary, how does the theorem play out?

 

[Delta2]

Sorry I have another question. Suppose now I have two supposed potential functions whose difference is not constant at the outer boundary, how does the theorem play out?

Not sure here but I think we need the condition that their gradients should be equal at the boundary (but again not necessarily constant gradients). Otherwise there is no uniqueness.

 

[Ahmed1029]

Not sure here but I think we need the condition that their gradients should be equal at the boundary (but again not necessarily constant gradients). Otherwise there is no uniqueness.

If the boundary is my domain and the gradient of each function is the same, the functions only differ by a constant and there is no issue here. But why do the gradients have to be the same?

 

[Delta2]

Hmmm, maybe we should wait for @vanhees71 to reply here, he is THE expert but in the meantime my view is that we require this condition cause that's how the proof of the theorem goes, without that condition or without the condition that the potentials are equal at the boundary, the proof can't continue.

 

[vanhees71]

In short the uniqueness of the solutions of Maxwell's equations just uses the positive definiteness of the energy density
$$u=\frac{\epsilon_0}{2} \vec{E^2} + \frac{1}{2 \mu_0} \vec{B}^2.$$
Then assume that the initial conditions of two solutions $(\vec{E}_1,\vec{B}_1)$ and $(\vec{B}_1,\vec{B}_2)$ are the same as well as boundary conditions on some closed boundary $\partial V$ of a volume $V$ (i.e., that the tangential components of $\vec{E}$ and $\vec{B}$ along $\partial V$ are the same for both solutions). Then $(\vec{E},\vec{B})=(\vec{E}_1-\vec{E}_2,\vec{B}_1-\vec{B}_2)$ fulfill the Maxwell equations with $\rho=0$ and $\vec{j}=0$ and thus Poynting's theorem tells us that
$$\partial_t u + \frac{1}{\mu}_0 \vec{\nabla} \cdot (\vec{E} \times \vec{B})=0.$$
Integrating this over $V$ and using Gauss's theorem we get
$$\dot{\mathcal{E}}_V=0,$$
i.e., the total energy of the field $(\vec{E},\vec{B})$ in the volume is conserved, and thus, since at the initial time $(\vec{E},\vec{B})=(0,0)$ this must hold true at any time, since $u$ can nowhere be $>0$, because otherwise the volume integral over $u$ couldn't vanish.

Of course the solutions for the potentials need not be the same, because they are only defined up to a gauge transformation.

 

[Delta2]

Interesting, we were talking about uniqueness for electrostatics and Poisson's equation but @vanhees71 proves it for all 4 Maxwell's equations.

 

[Delta2]

Of course the solutions for the potentials need not be the same, because they are only defined up to a gauge transformation.

in Electrodynamics the scalar potentials can differ by any function $\chi(x,y,z,t)$, but in electrostatics they can differ only by a constant at most right?

 

[vanhees71]

Right, because in electrostatics you only have $\vec{E}=-\vec{\nabla} \phi$, and from $\vec{\nabla}(\phi-\phi')-0$ you get $\phi-\phi'=\text{const}$, i.e., if $\vec{E}=-\vec{\nabla} \phi=-\vec{\nabla} \phi'$, $\phi$ and $\phi'$ can only differ by a constant.