# Is Energy Really Equivalent To Mass? (a Paradox)

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 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 correspond to the rest mass, in fact this frame is the “CoM frame”. 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!! hmm…. something looks wrong here, isn’t it?… waitwaitwaitwaitwaitwait…YES IT IS WRONG.

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

What’s happening?? If this is true, then sdl3$fh%^k@h, and then Einstein will be angry, and then $1/0$ and then $0/0$ and then… and then….and then…… Alright calm down bro… let’s look at this 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. Okay I need someone to help me. Wait a sec… Hey Spiderman! I kidnapped your girlfriend. If you don’t help me, I’ll make sure you won’t be able to see her again! Ok there you go Let’s assume that the Spiderman is massless. In the CoM frame he must hold the force which is equal to the force between the plates $F=\int (\frac{E}{2}) \sigma dA=\frac{\epsilon_{0} E^2 A}{2}$ And it is also the same 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 still stuffs in the moving frame still 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 others, 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 transfer it into the right plate to account for the attraction. 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 in other words: The “missing mass” might be hiding inside the Spiderman!!! Yeahhh there you are! Gotcha! 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 already 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 the energy?? Let’s come back to our home, to the CoM frame, things are clearer there. We know that the event one “force start acting on the left plate” and event two “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, the first event happens $\Delta t'=\frac {\gamma vL}{c^2}$ before the second event (rear clock ahead), Thus during this time the Spiderman has stolen an amount of energy from the left plate $\Delta U=\vec{F}.\vec{v} \Delta t'=\frac {\gamma \epsilon_{0} E^2 Ad}{2} \frac{v^2}{c^2}$ What a thief! After the second event occurs energy enters and leaves Spiderman at the same rate. So no energy will be stolen anymore. What else? Of course we just need to stick this $\Delta U$ to 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 $m'=\gamma m$ btw, thanks Spiderman. # The Water Jumper *Sniff, sniff..* You guys smell that? The smell of awesome in the air. Enjoy the video first… About four years ago, during an epic sunset, I was standing at a windy seashore throwing stones. The feeling when I watch 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 it much better than me, but beware when you start doing this thing it’s kind of hard to stop. Once you start, you won’t stop until the sun says byebye to you. I promised myself to think about the physics behind it once I get back to my room. But somehow it ran away from my mind, it takes about four years for me to remind myself about it. And 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 flat one. After trying many times I concluded that the only way to make 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 to give the stone a sufficiently fast spin. So I will suppose that these three factors, the horizontal velocity, the orientation and the spin are the main factors to give an awesome throw. My first naive suspect was the surface tension but I think I can get rid of it pretty quickly. First of all, it is extremely weak. Here’s a quick estimation. The mass of a good skipping rock is about $300 g$, so its weight is $W=3 N$. 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 roughly $F_{st}=7.2 \times 10^-3 N$ which is only sufficient to cause the stone feels itchy. And no matter how hard I tried, I could not find 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. Some of the kinetic energy is stored as 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? But it turns out that we don’t 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 which gives a lift force. It requires the stone to be in a correct orientation so that as the stone travels a distance on the surface of water it pushes some water downwards along the 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 maybe enough to fling the stone back to the air. Also most importantly the stone can still deflect water downward even when it has no vertical velocity. It is somewhat similar to how airplanes fly but this time water is deflected instead of air. 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, unfavorable for skipping. If the stone gets really lucky all the time, it may looks like this Okay 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 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. We can observe this as the splashes around the water, I am also going to ignore this effect mainly because it is too complicated. 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 the 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 also 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 be really wild if the shape is not alright. Again, to simplify things let’s assume that the stone is coin shaped with radius $R$ so that the immersed area depends only on $y$, the 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 leave the angle as a constant throughout the collision. Now the two equations of motion are complete, however they are nonlinear and unsolvable. 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(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. I think this result seems to be quite reasonable. It is also interesting to check out the minimum horizontal velocity we need to give in order to bounce at least once. My laptop says that we should at least throw the stone faster than $v_{x}=5.43 m/s$ in order for it to bounce once. 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. Edit: 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(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). I think this cannot occur if the velocity of water flow is always constant along the streamlines. I think the only way to make the streamline less deflected is that the water streamlines should get thinner as it flows around the stone. And in order to be thinner the water flow must accelerate in the direction of the flow(continuity). This can only works if water pressure in front of the stone is higher than behind the stone, I think intuitively it’s obvious that the front pressure is indeed higher, the surface of water in front of the stone is also higher. So from Bernoulli we can see how much the water flows can accelerate and how much it’s area would decrease, and thus how much the deflection will decrease. And from these we can obtain the correction terms. All of this can be calculated analytically, but think this involves a lot of messy works. While the effect might not be significant and interesting, so I decided not to work on the details. # New Technique!!!-Part 2 I was thinking about some stuffs while taking a piss… I imagined something in my head without really taking it seriously. Then suddenly BAM!! I realized that actually this thing might work, it turns out that I found another application of the! Holy sh*t, I felt like it was a piss which would determine the fate of the world. Here is the problem (actually I made it up after I found the solution), most electrodynamics textbooks simply state that the magnetic field outside a solenoid is negligibly small without giving clear reasoning. Then how small is negligibly small? We need to know exactly what are the cases where we can still safely neglect the external field of a solenoid! Otherwise when the time comes we may accidentally neglect some important terms! Okay, let’s welcome our guest star today: A long solenoid with length $L$, and cross sectional area $A$. Such that $L>>\sqrt{A}$. As a rescue team our mission is to calculate the magnetic field outside of a solenoid, at a point with a distance $r$ from the axis of the solenoid with $r<. And not close to any of the ends of the solenoid. Because, you know, things get insane there. Now here is the time when the hero comes in, let’s use special technique! We need to make a bada$\$  loop so that the flux through this loop due to the solenoid’s outer field depends only on $B_{out}(r)$ and $r$. Does such kind of loop really exist?? Indeed there is a way to do that, by combining two current loops. It is kinda hard to describe, so I’ll just show you how it looks like.

The flux through this loop due to the solenoid’s outer field is

$\phi_{RS}=B_{OUT}(r).2\pi r\delta r$

Note that there is a component of $B_{OUT}$ which is parallel to the loop, but don’t worry about that. Since we already chosen a point where $r<, we can only have field lines with small slopes. Which means that $B_{parallel with loop}$ compared to $B_{OUT}$ is like an ant compared to an elephant.

On the other hand the awesome loop also produces it’s own magnetic field , we can neglect the field due to the “connecting wires” because they are so tiny. So basically this field is just the difference of magnetic field due to circular loops of radius $r$ and $r+\delta r$. The magnetic field due to one ring only is

$B_{ring}=\frac {\mu_0 r^2I}{2}\frac{1}{(x^2+r^2)^{3/2}}$

The difference of magnetic flux through the whole solenoid due to two such rings with slightly different radius is

$\phi_{SR}=\delta ({\int_{-x}^{L-x}\frac{\mu_0 I_Rr^2}{2(x^2+r^2)^{3/2}}Andx})$

Kick out all the constant factors!

$\phi_{SR}=\frac{\mu_0I_RAnr^2}{2}\delta(\int_{-x}^{L-x}\frac{1}{(x^2+r^2)^{3/2}}dx)$

Integrate!

$\phi_{SR}=\frac{\mu_0I_RAn}{2}\delta(\frac{L-x}{\sqrt{(L-x)^2+r^2}}+\frac{x}{\sqrt{x^2+r^2}})$

The square root of blah blah blah makes me sick, anyone will get tired of seeing this. So let’s take the first order approximation using $L>>r$. But actually we HAVE to do this, because we have used the approximation once when calculating the field due to the ring, we should be consistent. Here we go

$\phi_{SR}=\frac{\mu_0I_RAn}{2}\delta(2-\frac{1}{2}[\frac{r^2}{(L-x)^2}+\frac{r^2}{x^2}])$

Taking the differential

$\phi_{SR}=-\frac{\mu_0I_RAn}{2}(\frac{1}{(L-x)^2}+\frac{1}{x^2})r\delta r$

Then using the mutual inductance symmetry relation

$\frac{\phi_{RS}}{I_S}=\frac {\phi_{SR}}{I_R}$

Finally we get

$B_{OUT}(r)=-\frac{\mu_0I_SAn}{4\pi}(\frac{1}{(L-x)^2}+\frac{1}{x^2})$

This only holds at the points which aren’t near the ends of the solenoid, so $x$ is about the same order as $L$. In other words $B_{OUT}$ is in the order of $B_{IN}\frac{A}{L^2}$. We can check that if $L>>\sqrt{A}$ then $B_{OUT}(r)$ is vanishingly small.

That was seriously cool, I thought we would never be able to calculate solenoid’s outer magnetic field analytically. My computer is about to explode because of the awesomeness.

# New Technique!!!

That moment strikes again, I have been like this: http://abstrusegoose.com/426 since yesterday.

One day, a little boy asked you what is the value of magnetic field due to a circular current loop at a point along the symmetrical axis of the loop?

You got tired as soon as you hear that question “Urghhh not this again”

Well a simple Biot-Savart’s would easily do the job. Or perhaps if you want to show him a fancier method you can calculate $\vec{A}$ and then just curl the sh*t. $\vec{B}=\nabla\times( sh*t)$

What a boring problem…

BUT

I can turn this problem into an interesting one! Because I found another method to calculate the $B$ using the concept of….. dum.. dum.. dum….duuum…. MUTUAL INDUCTANCE!

See? Now the question is changed into: ”How could mutual inductance has anything to do with this problem?”

Any idea?

Okay imagine a circular current loop.. Hmm well let’s make it even more interesting, imagine an ARBITRARY closed current loop. Remember that the goal is to calculate the magnetic field $B_P$ due to this loop at a point, say point P. Now let’s turn this problem into a mutual inductance problem.

Here is the key, put a really really small closed current loop with magnetic dipole moment $\vec{m_S}=I_s\vec{A_S}$ at point P. Our plan is to calculate the magnetic flux $\phi_{SA}$ through this small loop due to the arbitrary loop and then divide this flux with $A_S$. It works because the loop small loop is so tiny that we can simply write $\phi_{SA}=\vec{B_P} .\vec{A_S}$. Here is how it looks like:

The coolest thing about mutual inductance is that it is always symmetrical, in this case $M_{AS}=M_{SA}$. If you want a proof go take a look at Feynman Lectures on Physics Vol.2 pg.17-10. So we have

$\frac{\phi_{SA}}{I_A}=\frac{\phi_{AS}}{I_S}$

Which means that we can calculate $\phi_{SA}$ by calculating $\phi_{AS}$. We know that the magnetic field due to a very tiny loop is similar to that of a magnetic dipole $\vec{m}$ . Or we can picture it as two magnetic charges $\pm q_m$ separated by an infinitely small distance $d$ so that  $\vec{m}=q_m\vec{d}$.

Let’s calculate first the magnetic flux through the arbitrary loop due to  $+q_m$ alone. Thanks to magnetic Gauss law we can choose any surface as long as it is bounded by the arbitrary loop. Since the field is radial, it is best to choose a part of spherical shell of radius $r$ as the flux surface. But in general we can’t find part of spherical shell that is completely bounded by the arbitrary loop, thus we need  additional “radial surface” to make it fit to the arbitrary loop. However it’s not a problem at all since the flux through this “radial surface” is of course zero, because the field is tangential to it. So the flux is simply

$\phi_{+q_m}=\frac{\mu_0}{4\pi}\frac{q_m}{r^2}A_+$

Where $A_+$ is the area of spherical shell portion. Then by the definition of solid angle $\Omega$, we have $\frac{A_+}{r^2}=\Omega_+$. Therefore

$\phi_{+q_m}=\frac{\mu_0}{4\pi}{q_m}\Omega_+$

Don’t forget about $-q_m$ buddy, we can calculate its contribution the flux using the same method. Let’s put them together

$\phi_{AS}=\frac{\mu_0}{4\pi}{q_m}[\Omega_+-\Omega_-]$

$\phi_{AS}=\frac{\mu_0}{4\pi}{q_m}[\Omega(\vec{r})-\Omega(\vec{r}+\vec{d})]$

Recall the definition of derivative, the $(\Omega(\vec{r})-\Omega(\vec{r}+\vec{d}))$ on the left hand side can be replaced with $-\frac{d\Omega}{dr}d$. Note that $dr$ is a small displacement element in the direction of $\vec{d}$. Thus we get

$\phi_{AS}=-\frac{\mu_0}{4\pi}{q_m}d\frac{d\Omega}{dr}=-\frac{\mu_0}{4\pi}m\frac{d\Omega}{dr}=-\frac{\mu_0}{4\pi}I_SA_S\frac{d\Omega}{dr}$

Then using the mutual inductance symmetry relation we have

$\phi_{SA}=\frac {I_{A}}{I_{S}}\phi_{SA}=-\frac{\mu_0}{4\pi}I_AA_S\frac{d\Omega}{dr}$

Looks like we get the magnitude of magnetic field at point P

$B_P=-\frac{\phi_{SA}}{A_S}=-\frac{\mu_0I_A}{4\pi}\frac{d\Omega}{dr}$

The direction of $\vec{B_P}$ that we have just calculated is the same as that of $\vec{d}$ which we have arbitrarily chosen since I had secretly used $\vec{B_P}.\vec{A_S}=B_PA_S$ without telling you. So

$\vec{B_P}=-\frac{\mu_0I_A}{4\pi}\frac{d\Omega}{dr}\hat{d}$

“Wait! How could..” Yes you are right this is in fact not the total magnitude of $\vec{B_P}$. What we have just calculated is merely the projection of $\vec{B_P}$ into the direction of $\vec{d}$ that we chose. So how do we find the real value of $\vec{B_P}$? It’s not quite a big deal, we can simply change the orientation of $\vec{d}$ until we get the maximum value of $\vec{B_P}$, and that is the actual value of $\vec{B_P}$. Notice that only $\frac{d\Omega}{dr}$ that depends on the direction of $\vec{d}$ that we chose. So we have to find the maximum value of this little $\frac{d\Omega}{dr}$. If we understand carefully the meaning of gradient of a function $\nabla f$, it actually represents the maximum vectorial derivative of a function. Therefore finally we arrived at the final result

$\vec{B_P}=-\frac{\mu_0I_A}{4\pi}\nabla\Omega$

Isn’t it cute? Excellenteee right!

I hope that this new handy skill can be used creatively to tackle some of the most awesome physics problems in the world!! Can’t wait to apply this mutual inductance method on anything!

Now I think it is the time for this song

# From Water Jet to Little Droplets

A: Hey B, you accidentally left the water faucet not fully closed!

B: Really??? Cause I don’t hear anything even though the faucet is just behind this door.

A: Well I didn’t hear it, I saw it!

B can’t say a word after that..

So is there anything wrong with B’s hearing? Perhaps no, we can’t blame her/him; sometimes a small water stream can make a lot of noise, while some other times it is really quiet. The quiet stream is identical with a cylindrical column of water giving a continuous disturbance to the sink. However it may happen that the column of water reaches a critical length and it breaks up into tiny droplets and falls in disjoined cluster. If the water jet breaks up into droplets before hitting the sink, the disturbance will be more concentrated, hence it will give rise to louder and more annoying sounds. Let’s call the point where the water jet breaks up into droplets point x. I observed this point x is also moving up and down in somewhat jerky manner or probably the motion is actually periodic but it is too fast that our eyes won’t be able to catch up.

Okay let’s play with it experimentally! Take a plastic cup and make a small hole at its bottom, I used a hot nail to melt a hole on it. Then put some water on it, and observe the breaking point. It is hard to observe where exactly the point x is, so I illuminated the water with a flashlight. Since the water column won’t scatter many photons, this way we can clearly distinguish it with water droplets which can scatter many photons. This is how the set up looks like

But just a moment, there is an unwanted effect, which is the water vortex. If the water is spinning before exiting the hole, the resulting jet won’t be cylindrical and it can breaks into droplets much easily and it will be crazy and complicated. Here is one simple way to minimize this effect:

Yupee it works! Ok good, now the experiment can be more peaceful. The length of the water jet depends of the height of water in the container at that instance. Thus I measured the length of water jet as function of the height of water in the container and I found that they are proportional. The higher the height of water column, the longer the jet length will be. And further observation shows that the length of the jet is longer for a larger orifice diameter. Then I think for a while, go to pee, realized that the breaking up occurs independent on the direction of gravity, and decided to stop the measurement. Later I confirmed it using holed water bottle to make a parabolic water jet, and it really turned out to be independent of the direction of gravity.

The first plausible explanation that came to my mind is:

Suppose we drop one ball per second from the roof of a building, initially the distance between two consecutive balls is $g(1)^2/2$. We know that at any time, a ball is always faster than the ball above it, and is always slower than the other ball below it, it means that the distance between two consecutive balls will increase as the balls go down. Similarly this effect may also happens in water stream, as it goes down, it tends to move apart.
The independent of gravity direction argument obviously crushes this explanation.. At least this effect might not be dominant.

Let’s start over and contemplate at this problem more carefully. First it would be awesome to get rid of viscosity, the problem will be much simpler without it. Let’s compare it with surface tension.

$\frac{F_{viscosity}}{F_{surface tension}}=\frac{\eta v}{\gamma}=0.0172$

Where $\eta$ is the viscosity of water, $\gamma$ is the surface tension of water, and $v$ is the characteristic velocity between adjacent layers of water, I assumed it to be equals to the jet velocity which is about $v=1.25 m/s$(I am a bit too generous here). At $20^0C$, $\eta=1.002\times10^{-3} P_a s$, and $\gamma=7.28\times 10^{-2} N/m$.

So the viscosity is no match for surface tension. The surface tension has no alibi! Let’s get him!

A liquid desires to be in a minimal energy state, any movement that will lead to reduction of surface area is favorable by surface tension.  On a level of quasi-static motion it would thus be desirable to collect all fluid into one sphere, corresponding to the smallest surface area. Evidently it does not happen. The surface tension has to work against inertia, which opposes fluid motion over long distances.

The cross section of water jet with circular shape has the lowest energy, in other words the cross section of the jet tends to be circular. In general, the faucet’s hole is not circular, thus it will induce a cross sectional oscillation. Initially the cross section of the water jet is not circular and it tends to be circular. When it reach the circular form, the water still has some kinetic energy (inertia), therefore it overshoots and the cross section continues to evolve further until all the kinetic energy goes into the surface tension potential energy after that it tends to be circular again, and so on. I suspect that this disturbance is amplified as it is transmitted down the jet until it breaks up into drops at a point. Then I take a fork and using its tip I give a small perturbation on the jet, I could see the perturbation propagates upwards and downwards like a wave.  The perturbation that goes upward decays and eventually vanished, but the perturbation that goes downward grows in size, and it moves the point x higher or the water jet shorter. Then I did another experiment, If I put the disturbance higher the point x moves down, if I put the disturbance lower the point x moves up. Which proves my hypothesis that the disturbance amplifies as it goes down. The longer the disturbance travels down the jet the stronger it will be. Thus the measurement of the length of water jet as function of the height of water in the container is pointless because it depends on how uncircular the orifice is. It just proves that for higher water height in the container, the velocity of jet is higher thus it allows the jet travels more distance before the disturbance goes sufficiently large to breaks into drops. And it also proves that for smaller orifices radius, the uncircularness becomes more effective.

The last thing that I tried: Dimensional analysis. I wonder that the skill of using dimensional analysis might be useful for me in the future. The argument that sounds like “there’s only one way to construct <insert a dimension here> by combining the relevant parameters alone” might be useful, it may give us some physical insights, the meaning of that unique dimensional combinations. Some physics problems can be solved easily using this method, e.g. a hoop of rope problem; to find the acceleration of the end point of the rope if the rope is allowed to slide down through a hole. Now back to the case, The only time scale that can be formed from fluid parameters alone is

$t=\sqrt{\frac{r^3 \rho}{\gamma}}$

But what does it means? I don’t know, it looks like some inertial terms divided by the surface tension term. The get some idea, let’s try to plug in the numerical values

$r=2mm$

$\rho=1 kg/m^3$

$\gamma=7.28\times 10^{-2} N/m$

We get

$t=0.331 ms$

This is sooo small, it is not possible that this is the time it takes from the orifice to breaking point. I think it might be the time scale of the point x’ periodic motion, probably it is the time it takes to form a single droplet.

Look at this, it is really so fast

# Physics of Walking While Holding a Plastic Bag

When we are walking in a straight line, while holding a plastic bag on our hand, with an acceleration much smaller than $g$. There are two possible cases that may happen….

Sometimes the plastic bag would swing so peacefully that it makes almost no difference whether it is there or not….

But sometimes it may swing with a large amplitude, annoying enough to force us to make a conscious attempt to stop it.

Then how should we walk in order to avoid the latter case?

Let’s take a look at the plastic bag:

Yes, that is how it looks like in the physics world!

We can neglect the mass of the plastic bag relative to the total mass of the stuffs inside the bag. Due to centrifugal force, the stuffs inside will distribute themselves as far as possible from the pivot(finger holding the bag), so it is quite likely that they will end up having moment of inertia close to $ml^2$ relative to the pivot. And since the plastic bag has a large cross sectional area, we cannot underestimate the power of air drag. Let’s say it is proportional to the velocity of the bag, so we can write the torque due to the drag force as $\tau_{drag}=-k l^2 \frac{d\theta}{dt}$. Further I will assume that the mass of the plastic and the stuffs inside is small enough that we can neglect the transient effect. Yet another assumption, we may say that while walking human’s center of mass translates in approximately sinusoidal motion so that we can write the acceleration of center of mass as $a=a_0sin\omega t$. Since we are working in the hand’s frame, actually the drag force will change the equilibrium position so that it is not vertical anymore, but we will assume that this effect is small.

Okay now let’s write down the equation of motion, here is the torque about pivot equation:

$ml^2\frac{d^2\theta}{dt^2}+kl^2\frac{d\theta}{dt}+mgl\theta=ma_0 lsin\omega t$

Forget about the homogeneous solution, the steady state solution for this differential equation is:

The amplitude of oscillation

$\theta_0= \frac{a_0}{l\sqrt{(\gamma^2\omega^2+((g/l)^2-\omega^2)^2}}$

where $\gamma=k/2m$

So the high amplitude motion will occurs at resonance, when the $\omega$ of walking pace close to $\sqrt{\frac{g}{l}}$

But it is not that simple.. I measured my own standard walking pace using my ipod, the average time it takes for me to do ten steps is $4.91 seconds$, hence the average time between steps is $0.491 seconds$ or the frequency of pace is $2.04 Hz$. And the natural frequency of oscillation of a random plastic bag filled with snacks and drinks can be calculated using $f=\frac{1}{2\pi}\sqrt{\frac{g}{l}}$ , and the result is $0.760 Hz$ for $l=43cm$. Thus resonance won’t occur at around the standard walking pace.

But we may pass through the resonance frequency during the process of building up speed. During the process, the frequency pace is increased from zero to $2.04 Hz$, at some point it will hit $0.760 Hz$. Now the thing is during the process how long will it stay inside the “near resonance region” which is roughly $\omega_r -\gamma<\omega<\omega_r+\gamma$? Where the amplitude of the oscillation is at least half the maximum amplitude. The longer it stay in that region the more effective the resonance will be. We know that the characteristic time required for the oscillations to die out is $1/\gamma$. Therefore our walking pace must pass through this “near resonance region” in time more than $1/\gamma$ to get significant excitation of the resonance. The value of gamma can be calculated by comparing the natural frequency of oscillation with the experimental frequency of oscillation which turns out to be $f_1=0.735Hz$. We get $\gamma=1.45 Hz$

The average walking velocity is related to the pace frequency as follows

$v=d\frac{\omega}{2\pi}$

Where $d$ is the pace length. It can be measured by wetting my sandals before walking on the dry road, and then I came back and measure the distance between two  footsteps using a ruler.

The average distance turns out to be $d=110cm$ . Assuming constant pace length, the change in velocity corresponds to the change in $\omega$ as follows:

$\delta v=\frac{d}{2\pi}\delta\omega$

Thus the time required to pass through the resonance width $\delta\omega=2\gamma$ is

$\delta t=\frac{d\gamma}{a\pi}$

Which must be larger than the damping time $1/\gamma$ in order to excite an effective resonance. At the end we get

$a<\frac{d\gamma^2}{\pi}=0.736 m/s^2$

Therefore to avoid an annoying high amplitude swinging plastic bag, we must accelerate faster than $0.736 m/s^2$, at least for this plastic bag

# Childhood Question Finally Solved?

When you look at a light source you will notice something awesome, don’t you?

The bright point of the light will appear to be surrounded by a spray of radial light rays!

This thing is so cool that one may not be able to sleep because he or she can’t stop playing with it.

But what are those rays???

It is one of the biggest questions that I have since I was a child…

You may say, “Maybe it is due to scattering by the air”

OK now look at the light source again, and rotate you head, what do the rays do?

They rotate with you!

Then rotate the light source if you can.

Notice that the rays do not rotate!

So it is something to do with our eyes? Perhaps something related to our eyelids, eye liquid, brain perception or such things?

Not sure yet? Ok now choose an opaque object around you, it could be your hand, your Gundam toy, a book or etc..

Hold that thing between your eyes and the light source.

Try to block the rays without blocking the light source.

Notice that even if you hold the thing very close to your eyes the rays are always magically appears between the

the thing and your eyes. And also the rays will vanish if you cannot see the light bulb.

So it is a “biological” effect? But my camera does not agree, she says that she can capture that effect too!

What is Going On?

Wait a second… Look at that innocent little bottle of chili sauce over there!

Did you see the “white line” on the bottle?

Now why? Each “blob” on the bottle will produce one copy of mini light bulb, as a result we got a line of mini light bulbs. If the blobs are much smaller, our eyes won’t be able to distinguish the gaps between the mini light bulbs! I think water ripples also do the same thing:

Now look at this:

Even more epic one:

Thanks to the chili sauce bottle.. Now I have a special power to control light!! Maybe I am the chosen one!

I must protect the world from the dark forces!

Okay okay enough bullshitting.. What did I do??

When I rub the camera using my finger horizontally the light line becomes vertical however when I rub it vertically the line becomes horizontal. The light line is always perpendicular to the direction in which I rub the camera. And if I twist my finger while touching the camera, I got lines in every direction or you can say there is no line at all.

I think when I rub the camera in a certain directions, I will leave trace amounts of oil that I get on my fingers maybe from rubbing my forehead or nose. The trace is in a form of series of parallel lines, as in the chili sauce bottle case and water ripples case it will reflect light perpendicular to itself .

But most photos of such light rays that I saw have very neat six dominant lines. like this one

I think it is a completely different phenomena. Maybe it is because the camera’s light receptor is not isotropic, usually it has hexagonal symmetry.

Let’s get back to our eyes. I would assume that there are scratches on our eye lens and they are more or less randomly oriented. This scratches will act as reflectors as in above cases. The problem is if the scratches are randomly placed why are the rays radial? Only a few selected scratches glint light towards our retina.  The basic principle is a scratch will reflect light perpendicular to itself, thus only scratches perpendicular to our line of sight or “tangential” to the light source will reflect light to our eyes. Therefore it would be very unlikely to have a non radial line, the scratches must be very specially ordered. So the scratches produce lines of light radiating in all directions around the point source. We can also play with the lines by goggling and narrowing our eyes. That way we can change the shape of our eye lens, it’s like stretching the scratches vertically and thus the lines. When we goggle the lines are more “vertical”, when we narrow our eyes the lines are more “horizontal”.

There is another optical effect that we can observe. At first you can look at a light bulb without squinting. Then by squinting the eyes you can see the rays being extended. And when we look above or below the light bulb the the rays are extended further upward or downward. I think it is because when we move our eyelids some eye liquid is moved and covered the cornea , it will change the liquid’s thickness profile at some part. To prove this hypothesis, I tried to yawn several times to produce some tear before looking at the light source. And what did I get? I can see similar long line even without squinting!

Yet another interesting stuff that I discovered while playing with light:

If we look at a bright light source for a long enough  time and then we look the other way, the part where the bright light was will become dark. If we put the source at the center of our vision, it would be awesome. After the light source is removed, the dark spot will show us where we are looking at. For example I tried to read something while the dark spot still there, I found that upon reading my eyes are focused at left edge, middle, and right edge of the text repeatedly. Moreover you can also know where your eyes are focusing while staring at optical illusion picture!