# 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

$m'=\gamma m$

# The Water Jumper

*Sniff, sniff..* You guys smell that? The smell of awesome in the air.

About four years ago, during an epic sunset, I was standing at a windy seashore throwing stones. The feeling when I watched the stones skipping was like HOLY! Holy! Hooolholy! holy cow! It could be longer. It’s just because my record was only five skips.  I’m sure you could do much better than me, but beware when you have started doing it, it’s kind of hard to stop. You won’t stop until the sun says byebye to you.

I promised myself to think about the physics behind it once I got back to my room. But somehow it ran away from my mind, it takes about four years for it to come back. Unfortunately I haven’t got a chance to skip stones again since then. However I still remember some important details. From my experiences, I intuitively think that a good stone for stone skipping is a flatter one. After trying many times, I concluded that the only way to make a stone skips is to give the stone enough horizontal velocity with a correct orientation, and also a quick flick at the end of the throwing process is need, to give the stone a sufficiently fast spin. So I suppose that these three factors, the horizontal velocity, the orientation and the spin are the main factors to have an awesome throw.

What makes the stone bounces?
My first naive suspect was the surface tension but now I think I can get rid of it pretty quickly. First of all, surface tension is extremely weak. Here’s a quick estimation. The mass of a good skipping rock is about $300 g$, so its weight is roughly $W=3 N$. The surface tension of water at $25 ^\circ C$ is $\gamma=7.2\times 10^-2 N/m$. Thus if the planar dimension of the stone is about $L=10 cm$, then the order of the surface tension force is about $F_{st}=7.2 \times 10^-3 N$ which is only sufficient to cause the stone feels itchy. And no matter how hard I tried, I could not think of any possibility of surface tension force having dependence with the three factors above, so I guess we can safely forget about surface tension here. I also remember the motion of the water where I throw the stones at was very unfavorable for surface tension. Perhaps we can test this putting some detergent into a pool of water, to see if the stone can still skip on it.

Okay so what else could possibly cause the bouncing effect? In order to bounce the stone we need a force that can flip the vertical velocity of the stone. It means, we need a force that continues to exist even when there is no vertical velocity. For a typical object bouncing on the floor, the one that makes sure the flipping to happen is the elasticity of the object. During a collision, some of the kinetic energy is converted to potential energy ,which later will be used to renew the kinetic energy again. Okay, but what kind of elasticity can we expect from a collision of a solid with a fluid?

It turns out that we don’t really need one. Maybe it is just a simple Newton’s third law. As the stone hits the water surface, it displaced an amount of mass downward, as it travels a distance on the surface of water, which gives a lift force. It is somewhat similar to how airplanes fly but this time water is deflected instead of air. However, we require the stone to be in a correct orientation to make sure that the water is pushed in a right way. As we can see, the higher horizontal velocity of the stone , the more mass it can deflects in a time interval. Which means that if the stone  has a sufficiently high velocity, the vertical momentum of the water pushed may be enough to fling the stone back to the air. Also most importantly, a stone can still deflect water downward even when it has no vertical velocity.

But wait, how about the spin? We know that it won’t work without the spin, so it must contributes somewhere. I think most likely the spin plays a role as stabilizer, maintaining the correct orientation favorable for the skipping. Otherwise if there is no spin, the stone may easily change its orientation and hit the water with bad orientations that are unfavorable for skipping. If the stone gets really lucky all the time, it may look like this

Okay, now let’s analyze the collision process quantitatively. To get some idea about the flow of water around the rock, let’s estimate the Reynold’s number. It is not hard to find the viscosity of water at  $20^\circ C$, it is about $\mu=1.002\times 10^-3 Pa.s$. I suppose typical velocity of the stone would be about several meter per second, so let’s just take $v=1 m/s$. And then the planar dimensions of the stone are assumed to be about $L=10 cm$. Finally the density of water is of course $\rho=1000 kg/m^3$. Using these numerical values we can calculate the Reynolds numer

$Re=\frac{\rho v L}{\mu}=9.98 \times 10^4\approx 10^5$

Which says that in this regime there would be a big gap behind the stone as it moves on the water surface. Roughly it should looks like this

After looking at the video in the begining, it seems reasonable to assume that the velocity of the rock to be horizontal, that is $v\approx v_{x}>>v_{y}$. I guess I can also safely ignore the adhesion force between the rock and the surface of water which cause a portion of water to move with the velocity of the stone surface in contact, this is somewhat similar to friction. Also note that there could be some water being deflected upwards following the surface of the stone , take a look at the red arrow above. Some water can be deflected in a way depicted by the red arrow. We can observe this effect as splashes around the stone. I am also going to ignore this effect mainly because it is too complicated and it does not add a new physics. We’ll see that we can take into account this effect to some degree by introducing a constant. Or perhaps I could say I will only consider the case where the these effects are small. So the only forces that we will consider here is the force required to deflect an amount of water coming from the front of the stone and the gravitational force.

Since, in Newtonian mechanics, force is independent of frame reference, it is a good idea to move to the rock’s frame to simplify things. In this frame, the water is moving with velocity $v$ towards the rock before being deflected downwards at an angle $\alpha$. The amount of water being deflected depends on the cross sectional area of the stone $A(y)$ below the surface of  water as shown above. During the collision, the equations of motion are

$\frac{dp_{y}}{dt}=m \frac{d^2 y}{dt^2}=C_{1}\rho A(y) (\frac{dx}{dt})^2 sin\alpha-mg$

$\frac{dp_{x}}{dt}=m\frac{d^2 x}{dt^2}=-C _{2}\rho A(y) (\frac{dx}{dt})^2 (1-cos\alpha)$

I put the factor $C_{1}$  and $C{2}$ to account for a portion of water outside the cross-sectional area that is also get pushed downward by the neighbors and the portion of water that is pushed upwards as a splash. For  $C_{1}$ the two effects are competing, and we don’t know who win the fight. But we can guess that the value won’t differ too much from $1$, thus for simplicity let’s just choose $C_{1}=1$. On the other hand the two effects work together in the case of  $C_{2}$, so $C_{2}$ must be bigger than one. Let’s say  $C_{2}=1.5$. Now the $A(y)$ depends on the shape of the stone, and it can vary really wildly if the stone is amoeba-shaped. For the purpose of writing a nice formula, let’s assume that the stone is coin shaped with radius $R$ so that the immersed area depends only on $y$, i.e. vertical position of the lower edge of the stone. We can obtain $A(y)$ by integration

$A(y)=R^2[-(1-\frac{y}{R})\sqrt{1-(1-\frac{y}{R})^2}+\frac{\pi}{2}-arcsin(1-\frac{y}{R})]sin\alpha$

Finally we have to assume that the the spin is good enough at stabilizing the angle $\alpha$ that we can think of the angle as a constant throughout the collision. Now the two equations of motion are complete, the only problem is that they are nonlinear.
Thus we have no choice but to deal with it numerically. Using the following numerical values:

$m=0.5 kg$$C_{1}=1$$C_{2}=1.5$$R=10 cm$$\alpha_{0}=10^\circ$$\rho=1000 kg/m^3$$g=10 m/s^2$

And the following initial conditions(at the moment the stone leaves the thrower’s hand):

$y_{0}=0.5 m$

$x_{0}=0$

$vx_{0}=10 m/s$

$vy_{0}=0$

Python simulation shows the following beautiful trajectory of the stone’s CoM before sinking:

With these initial conditions, the stone bounces 14 times before eventually sinking, seems to be quite reasonable.

It is also interesting to check out the minimum horizontal velocity we need to give in order to make the stone bounces at least once. My laptop says that we should at least throw the stone faster than $v_{x}=5.43 m/s$. Seems easy right? Yeah it’s not the hard part, actually the hard part is to make the right orientation and to give enough spin to the stone to maintain the orientation.

I think this is good enough for now, I will refine my model if I get a chance to play at a seashore again.

Here are some thoughts to estimate $C_{1}$ and $C_{2}$:

Okay soo we want to see, if the main portion of water get pushed aside how much extra water will also get pushed together with it. Close to the main stream area (inside the cross section of the stone), the streamlines would be deflected strongly. As we move farther from the surface the streamline becomes more and more straight(less deflected).
Water streamlines are forced to get thinner as it flows around the stone, and the thinning gets less and less as we move farther away from the stone. The thinning requires the water flow to accelerate in the direction of the flow(continuity). This can only work if water pressure in front of the stone is higher than water pressure behind the stone. Intuitively, it’s obvious that the front pressure is indeed higher. Then, given how much a streamline’s cross sectional area decreases, from Bernoulli, we can calculate how much it will accelerate  and thus how much the deflection will decrease. And from these we can obtain the correction terms. All of this can be calculated analytically using integration, it involves a lot of messy works. While the effect might not be significant and interesting, so I decided not to work on the details.