# Boosted Capacitor

I got bored, so I put on my “relativistic glasses.”

Cool!! Now everything looks relativistic to me…

Let’s take a look at a massless-plates capacitor with area $A$ and width $d$ such that $\sqrt{A}>>d$.

Now I charge the plates with $+Q$ and $-Q$, as a result there is an electric field $E=\frac{Q}{\epsilon_{0} A}$  between them. Since the capacitor is extremely thin, it’s very easy to calculate the field energy

$U_{E}=\int \frac{\epsilon_{0}}{2} E^2 dV=\frac{\epsilon_{0} E^2A d}{2}=m c^2$

The $mc^2$ at the end says that the energy of this system corresponds to the rest mass, i.e. this frame is the “CoM frame” or the frame of zero momentum. As you might have guessed I’ll move to a new frame after this.

Let’s see if I start running to the left with a relativistic velocity $v$, woahh now the system looks like this

The capacitor get contracted by a factor of $\gamma$. And as we know the electric field transform into itself if it points parallel to the velocity. Therefore the energy density stays unchanged… and since the volume is contracted by a factor of $\gamma$, so is the total energy

$U'_{E}=U_{E}/\gamma$

or

$m'=m/\gamma$

hey wait! Something looks wrong here, isn’t it?… waitwaitwaitwaitwaitwait…yes it is wrong

Shouldn’t the mass transform like $m'=\gamma m$ ???

Alright calm down bro… let’s look at it more carefully..

The picture above is bugging me, I imagine that if we left the system that way then the plates will move toward each other and BAM. So let’s put something between the plates to avoid the plates crashing into each others.

Let’s assume that the Spiderman is massless. In the CoM frame he must withstand the squeezing forces from the left and right plates

$F=\int (\frac{E}{2}) \sigma dA=\frac{\epsilon_{0} E^2 A}{2}$

And the same happens in the moving frame, since force is invariant under parallel Lorentz transformation. Well, I can assure myself that everything is fine in CoM frame, but stuffs in the moving frame look quite suspicious and also interesting (I’ll make this more precise in a moment). Who knows, maybe the Spiderman can somehow give contribution to the total mass.

In the moving frame, the Spiderman plays a role as the energy transmitter. I mean, first imagine that there is no Spiderman between the plates. Since the plates are attracting each other, the right plate will slow down and the left plate will speed up. Now if there is a Spiderman between them, their velocities will not change at all. So in other words, the Spiderman is taking energy from the left plate at a rate $\vec{F}.\vec{v}$ and transfering it to the right plate. But, remember that the energy can’t teleport from one plate to the other instantaneously. Some of it might not reach the right plate yet, still located between the plates. Thus the “missing mass” might be hiding inside the Spiderman!

However don’t be too happy, let’s see if we really calculate the energy located between the plates. Let’s assume that the spiderman has maximal rigidity, that is energy is transfered through him at the velocity of light $c$. Thus the time it takes for the energy to be transferred from the left plate to the right one is

$\Delta t=\frac{d'}{(c-v)}=\frac{d}{c} \sqrt{\frac{1+v/c}{1-v/c}}$

And the total energy taken from the left plate during this time is

$\Delta U= \vec{F}.\vec{v} \Delta t=\frac{\epsilon_{o}E^2 Adv}{2c}\sqrt{\frac{1+v/c}{1-v/c}}$

Notice that if we add this term to $U_{E}/\gamma$, we still won’t be able to get the $\gamma U_{E}$ that we wanted because of the annoying square root factor in $\Delta U$.

Hold on! Maybe even before the energy from the left plate reached the right plate, the right plate has started receiving energy from the Spiderman (Perhaps it can “borrow” some energy from the Spiderman’s elastic energy). So how do we know when does the right plate start receiving energy? Let’s come back to our home, to the CoM frame, things are clearer there. We know that  event 1: “force start acting on the left plate” and event 2: “force start acting on the right plate” must happen simultaneously in this frame due to symmetry. From the loss of simultaneity we obtain that these two events aren’t simultaneous in the moving frame, event 1 happens

$\Delta t'=\frac {\gamma vL}{c^2}$

before the event 2(rear clock ahead), thus during this time the Spiderman has stolen an amount of energy from the left plate

$\Delta U=\vec{F}.\vec{v} \Delta t'=\frac {\gamma \epsilon_{0} E^2 Ad}{2} \frac{v^2}{c^2}$

After the second event occurs energy enters and leaves Spiderman at the same rate. So no energy will be stolen anymore. So what’s left is to combine this $\Delta U$ with the $U_{E}/\gamma$. Are you ready? 3..2…1……

$U_{E}/\gamma+\Delta U=\frac{\epsilon_{0} E^2 A d}{2\gamma}+\frac {\gamma \epsilon_{0} E^2 Ad}{2} \frac{v^2}{c^2}=\frac{\gamma\epsilon_{0} E^2 A d}{2}=The\hspace{2mm}Damn\hspace{2mm}\gamma U_{E}$

or

$m'=\gamma m$