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I am new in General Relativity. I know that electromagnetic field (or, the electromagnetic energy tensor, $T^{ik}=1/4\pi[1/4F_{mn}F^{mn}g^{ik}-F^i_lF^{lk}]$) can affect gravitation. Now if we take a hollow sphere made up of conducting or magnetic material, it will shield the electromagnetic field. See the diagram below,

enter image description here

So there is no electromagnetic field inside the sphere (assume the shielding material is ideal and completely shielding the inner region of the sphere). Now will there (inside the sphere) be any gravitational effect due to the electromagnetic field applied outside the sphere? If yes then please quantify (if possible). Detailed calculations may be appreciated.

Edited to add: More precise question is to calculate the metric tensor inside and outside of the sphere satisfying boundary conditions.

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The picture you provide is a difficult problem. However there is a simple example consistent with the words you wrote. You can make a thin spherical shell and place an electric charge Q on it.

Outside the shell the metric will be the Reissner–Nordström metric:

$$ ds^2=\left(1-\frac{2M}{r}+\frac{Q^2}{r^2}\right)dt^2-\left(1-\frac{2M}{r}+\frac{Q^2}{r^2}\right)^{-1}dr^2-r^2\left(d\theta^2+\sin^2\theta d\phi^2\right). $$

Inside the shell ($r<R$) the metric will be the spherical Minkowski metric: $$ ds^2=\left(1-\frac{2M}{R}+\frac{Q^2}{R^2}\right)dt^2-dr^2-r^2\left(d\theta^2+\sin^2\theta d\phi^2\right). $$

So now you can see the effect of the charge inside the sphere. Which is locally nothing, everything is flat, it's just that clocks tick at the rate $d\tau=dt\sqrt{1-\frac{2M}{R}+\frac{Q^2}{R^2}}$ where $dt$ is the rate that clocks tick far far away. As long as you stay inside you don't notice that your clocks tick at a different rate than clocks far far away.

Note that you do want to have $R$ be large enough that your shell doesn't form a black hole.

Can I notice the different rates of clocks if I look from outside? Yes, you notice the different rates of clocks if people inside look out or if people outside look in.

Do all the clocks inside the sphere tick at same rate? The clocks inside tick the same way as if in an empty universe, so they tick differently if moving, just like in an empty universe.

Will there be any effect like 'acceleration due to gravity' inside the sphere? There is no acceleration due to gravity inside the sphere.

What if the charge on the sphere isn't static or there is a current? If the charge is moving then you either have it go in and out and nothing much changes or else you break the spherical symmetry.

What about a magnetic field? A magnetic field breaks the spherical symmetry so makes even writing down answers much more complicated, and solving them even harder still.

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  • $\begingroup$ @TimaeusWhat is about magnetic field? Can I notice the different rates of clocks if I look from outside? Do all the clocks inside the sphere tick at same rate? Will there be any effect like 'acceleration due to gravity' inside the sphere? What if the charge on the sphere isn't static or there is a current? $\endgroup$
    – Aniruddha
    May 4 '15 at 23:38
  • $\begingroup$ @Aniruddha Edited to refer to those questions as well. $\endgroup$
    – Timaeus
    May 5 '15 at 0:40
  • $\begingroup$ if we make the magnetic field spherically symmetric by assuming $B_{r,\theta,\phi}=A(r)r$ (outside the sphere), $B_{r,\theta,\phi}=0$ (inside the sphere). then, can it be solved? $\endgroup$
    – Aniruddha
    May 5 '15 at 3:19
  • $\begingroup$ @TimaeusOne more thing, can you tell me by physical understanding, not by mathematical calculation, that if there outside the sphere we apply a temporally variable electromagnetic field ($E=E(t),B=B(t)$), then will there inside the sphere be any thing like 'acceleration due to gravity' or tidal forces or any measurable gravitational effect (no matter how small) which will vary with time? $\endgroup$
    – Aniruddha
    May 5 '15 at 3:27
  • $\begingroup$ @Aniruddha I don't understand. If $\vec B=B(r)\hat r$ in flat space then it can't satisfy $\vec \nabla \cdot \vec B = 0$ unless $B(r)=k/r^2$ and all those field lines converge towards the center. If you try to transition to zero B field inside without violating $\vec \nabla \cdot \vec B = 0$ all the field lines are coming out, so there is no place for them to come from and they can't start from nowhere. $\endgroup$
    – Timaeus
    May 5 '15 at 3:30

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