Q: How can photons have energy and momentum, but no mass?

Posted on September 8, 2010 by The Physicist

Physicist: Classically (according to Newton) kinetic energy is given by E=\frac{1}{2}mv^2 and momentum is given by P=mv, where m is mass and v is velocity.  But if you plug in the mass and velocity for light you get E=\frac{1}{2}0c^2=0.  But that’s no good.  If light didn’t carry energy, it wouldn’t be able to heat stuff up.

The difficulty comes from the fact that Newton’s laws paint an incomplete (and ultimately incorrect) picture.  P=mv is very accurate for slow moving (compared to light) objects with mass, but it’s not true in general.  When relativity came along it was revealed that there’s a fundamental difference in the physics of the massive and the massless.  Relativity makes the (experimentally backed) assumptions that: #1) it doesn’t matter whether, or how fast, you’re moving (all physical laws stay the same) and #2) the speed of light is invariant (always the same to everyone).

Any object with mass travels slower than light and so may as well be stationary (#1).

Anything with zero mass always travels at the speed of light.  But since the speed-of-light is always the speed-of-light to everyone (#2) there’s no way for these objects to ever be stationary (unlike massive stuff).  Vive la différence des lois!  It’s not important here, but things (like light) that travel at the speed of light never experience the passage of time.  Isn’t that awesome?

The point is: light and ordinary matter are very different, and the laws that govern them are just as different.

Light and Matter: different

That being said, in 1905 Einstein managed to write a law that works whenever: E^2=P^2c^2+m^2c^4.  The same year (the same freaking year) he figured out that light is both a particle and a wave and that the energy of a photon isn’t governed by it’s mass or it’s velocity (like matter), but instead is governed entirely by f, it’s frequency: E=hf, where h is Planck’s constant.

For light m=0, so E=Pc (energy and momentum are proportional).  Notice that you can never have zero momentum, since something with zero mass and zero energy isn’t something, it’s nothing.  This is just another way of saying that light can never be stationary.

Also!  Say you have an object with mass m, that isn’t moving (P=0).  Then you get: E=mc2 (awesome)!

 

Unrelated tangent: It took a little while, but the laws governing the massive and the massless are even more inter-related than the ‘Stein originally thought.  He figured out that the energy of a photon is related to it’s frequency (E=hf), but why are photons so special?  Why do they get to have frequencies?  They’re not special.  Years later (1924) de Broglie drew the most natural line from Einstein’s various equations from light to matter.  mc^2=E=hf  So for a given amount of matter you can find it’s frequency.  Holy crap!  Everything has a frequency!

On the off chance that anyone out there got unduly excited about that last statement: the frequencies never go out of wack, you can’t tune them, more importantly they are utterly unimportant on the Human scale, or even the single-cell scale, and don’t ever buy a bracelet or anything else with “quantum” in the name.

No, no, no, no, no, no, no, no, no.

Posted in -- By the Physicist, Relativity | 227 Comments

Q: If you were on the inside of the Sun falling in, the matter closer to the surface doesn’t affect your acceleration, but the matter closer to the core does. Why is that?

Posted on September 2, 2010 by The Physicist

The original question was: Plait talks about the “physics of solid bodies” and why, specifically, if you were on the inside of the Sun falling in, the matter “behind” you- closer to the surface- doesn’t affect your acceleration at all, and all that matters is the matter “in front” of you- closer to the core. Why is that?

Physicist: The short, uninteresting answer is that the gravity from any layer above you cancels itself out. If you take any sample layer above you, and you happen to be closer to one side, then you’ll find that the side you’re closer to has more pull on you, but there’s less of it. Conversely, the far side has less pull, but there’s more of it. For a sphere (but not a ring) these forces cancel exactly. So as you fall in you can ignore all the layers above you.

Pick a layer. Anything inside will experience exactly the same amount of pull in every direction, and so, no pull at all.

Answer gravy: One of the greatest tools in the physicist’s tool kit is “Gaussian Surfaces“. They let you shortcut really difficult math problems using pictures and a little reasoning.  Even better, you come across smarter than perhaps you deserve, which is a big plus.

A Gaussian surface is nothing more than an invisible bubble that you draw in space. The “inverse square law” of gravity can actually be rewritten as “the total amount of gravity pointing into the bubble is proportional to the amount of matter inside the bubble”. The arrangement of matter (both inside and outside the bubble) certainly changes how gravity points into (or out) of the bubble, but the total amount of gravity pointing through depends only on the amount of matter inside.

(upper left) when the mass is symmetrical and in the middle, then the gravity is exactly the same everywhere on the surface. (upper right) if the matter is off to the side, then gravity will be stronger, or weaker, or point in different directions at different points on the surface, but the total stays the same. (bottom) matter outside of the Gaussian surface can affect how gravity pokes through, but it can't affect the total.

Now say that your bubble is an exact fit around a sphere of matter. Everything is perfectly symmetric, so there’s no reason for gravity to be any stronger or weaker anywhere and, given the amount of mass inside the sphere, you can figure out how strong the gravity is. Now say you add more matter, but uniformly, on top of your original sphere.

In both situations the total amount of matter inside the bubble is the same, and everything is nice and symmetrical, so the gravity along the surface of the bubble is the same.

The matter inside the sphere has remained the same, so the pull at the surface of that sphere remains the same. As a result, so long as the matter above is at least fairly symmetrical (which is the case for any planet or star you can think of), you can ignore the layers above the surface of the bubble.

Specifically, as you fall farther and farther into the Sun (or Earth, or whatever else is round) you can figure out how much gravity you’re feeling by using a Gaussian surface, for which you only need the matter below you. The layers above will exert no pull on you, and you will exert no net pull on them (for every action/force there is an equal and opposite reaction/force).

 

This part has nothing to do with the question: You can use Gaussian surfaces to prove some surprising things. Specifically: Dyson spheres work, and black holes have no more gravity than the stars they came from.

From the last argument (above) you know that the layers above you have no net gravitational effect on you. But what if you fall a little way into a planet, and suddenly find that the inside of it is completely hollow? One you’re inside all the layers are layers above you. So there’s no gravity at all inside of a large hollow sphere (at least, none caused by the sphere). If you built a really huge sphere around a star you’d have a “Dyson’s sphere”. The sphere doesn’t pull the star, and the star doesn’t pull the sphere. It’s stable no matter where the star is inside the ball. So long as no one shoves anything, everything will just float neutrally right where it is.

(left) the set up for a Dyson Sphere. The perfectly spherical shell has no gravitational effect on anything inside the sphere, including the star, and vice versa. (center) Miles Dyson, inventor of the Dyson sphere, and future inventor of Skynet. (right) an artist's interpretation of a Dyson sphere.

Now, put a Gaussian surface around a star. There’s a certain amount of matter in the star, and that tells you how much gravity is pointing through the surface. If the star shrinks, who cares? Same amount of mass = same amount of gravity.

The gravity through the outer Gaussian surface stays the same, since both contain the same amount of matter. The gravity through the inner Gaussian surface increases dramatically after the star collapses, because it contains all of the star's mass, instead of just a small part of it.

But if you draw a small Gaussian surface around the core of the star you’ll find that the gravity along the surface is small, because there is (relatively) little mass inside of it. If for some reason you found yourself in the center of the Sun, you’d be floating in zero G’s. Point of fact; you’d also be on fire.

Now when the Star collapses, all of the matter is drawn into a tiny region. Both spheres (see diagram on the right) now contain all of the star’s matter, and thus the same total amount of gravity pokes through them. The only difference is that the inner sphere is smaller, so the gravity has to be more intense to get the same total as the outer sphere.
Black holes do have very intense gravity, but only in the region where the star used to be.

Posted in -- By the Physicist, Astronomy, Math, Physics | 14 Comments

Q: How do surge protectors work?

Posted on August 29, 2010 by The Physicist

Physicist: To control power in a house or an outlet you’d generally use a fuse.  But fuses are slow, they need time to heat up.  A surge (or the faster “spike”) happen too fast, so reacting to a surge is no good.  Dealing with a surge properly has to be built into the nature of the machine.  There are two (general) ways to do this: “choking” and “shunting”.

Shunting is what a sink does when it over-flows.  You never worry about water getting to the ceiling above your sink, because the moment the water gets as high as the edge the sink starts overflowing and stops filling up.  The ceiling is protected by the innate nature of the sink (which doesn’t need to react, it just “does”).

If the water in the sink gets too high it simply falls over the edge. Similarly, if the voltage in the load line gets too high, the resistance of the varistor (variable resistor) drops from very, very high to nearly zero. This allows a connection to ground, and electricity can then flow out of the circut, instead of through it.

The zero-finesse surge protector is just a varistor placed between the power line and the ground line.  The ground line (in this metaphor) is the floor around the sink, where all the overflowing water gets dumped.

While there is some fancy quantum mechanics tied up in varistors (I’m talkin’ valence and conduction bands here), what really gets my physics juices jumping is waves and frequencies.

The other type of surge protector is essentially a “band pass filter” centered around 60 hz (the frequency of the current in the electrical grid).  You can think of this as the “radio” of the surge protector tuning into the “station” of the wall outlet (and tuning out everything else).

Now, I hope this doesn’t come as a shock to anyone, but most surges (e.g., lightning strikes) are fast.  It turns out that because of the Uncertainty Principle, things that happen really fast are necessarily spread out over a lot of frequencies (small time uncertainty means large frequency uncertainty).

The frequency spectrum for a wall socket (top) and a lightning bolt (bottom). Keep in mind that this is not a graph in time, but in frequency. If you played a single note on a piano you'd get something like the top graph, and if you banged on all the keys at once you'd get something like the bottom graph.

So while a power surge may have a lot of energy overall, the amount of energy right around 60 Hz, where it can get into the circuit, should be fairly small.  You can build chokes to limit the frequencies that get past the surge protector by using carefully tuned LRC circuits, but generally (since most of the energy is in frequencies much higher than 60 Hz) you can just build a “low pass filter”.

Which is just fancy talk for “an inductor”.

Which is just fancy talk for “a coil of wire“.

Posted in -- By the Physicist, Engineering, Physics | 2 Comments

Relativity and Quantum Mechanics: the elevator pitch

Posted on August 24, 2010 by The Physicist

Physicist: A woman on the subway, about two stations away from her stop, asked us “what are relativity and quantum mechanics?”
So, this is a two-stop elevator pitch for the two most pivotal sciences since slicedbreadology.

Elevators: Wonderful, mechanical rooms, quietly skirting the ever-thinning line between broom closet and robot.

Relativity: Speed is just distance over time (as in “miles per hour”). Normally when you change you’re own speed, the speeds of everything else changes (for your point of view). For example, if you’re walking slowly down the street everyone else will be moving quickly (and, for the sake of this example, in the same direction), but if you pick up the pace and walk normally, then everyone else will barely be moving at all.

But the speed of light is different. No matter how you move, it will always stay the same. Since that particular speed refuses to change, distance and time have to change instead.  Relativity is the study of how distance and time change with speed, and the consequences that follow from those changes.

Quantum Mechanics: When you look at very, very small objects, like individual particles, you begin to find that they don’t behave right. If particles were like ordinary objects, but smaller like tiny billiard balls, then you’d expect them to act like ordinary (but tiny) objects. Instead, they “ooze” from place to place, move through impassable barriers, exist in several places at the same time, and interfere with each other. It’s impossible even to say exactly where they are.
All of these are impossible (or at least very unlikely) behaviors for solid “particle-ish” objects. But all of these behaviors are explained, and even expected, if all of matter is actually some kind of wave.
So, quantum mechanics is the (more-accurate-than-every-other-science) study of the universe from the perspective that everything, at the lowest levels, is made up of some kind of waves.

Posted in -- By the Physicist, Quantum Theory, Relativity | 3 Comments

Q: Why are orbits elliptical? Why is the Sun in one focus, and what’s in the other?

Posted on August 18, 2010 by The Physicist

Physicist: This question always bothered me too.  The short answer is: it falls out of the math.  Specifically, the math of first year physics and second year calculus.  The fact that the Sun is in one focus is just one of those things.  It’s nothing special.  Even less special is the other focus, which contains nothing at all.

Ellipses and their foci have a lot of useful properties. It so happens that an orbiting object traces out an ellipse, with the thing it orbits around at one of the focuses. Coincidence? Yes.

I can’t find a good intuitive reason why orbits are elliptical.  In fact, I can’t even find a mathematical derivation.  So, because it should be found somewhere, I’ll leave the derivation floating in the answer gravy.

Answer gravy: The force of gravity is usually written as ma=F=-\frac{GMm}{R^2}.  You can rewrite this using vector notation as m\ddot{\vec{x}}=-\frac{GMm\vec{x}}{|\vec{x}|^3}, where the dot on top is a time derivative.  To keep the notation both standard and confusing, \vec{x}=(x,y).

\begin{array}{ll}m\ddot{\vec{x}}=-\frac{GMm\vec{x}}{|\vec{x}|^3}\\\Rightarrow\ddot{\vec{x}}=-\frac{GM}{|\vec{x}|^3}\vec{x}\\\Rightarrow\dot{\vec{x}}\cdot\ddot{\vec{x}}=-\frac{GM}{|\vec{x}|^3}\vec{x}\cdot\dot{\vec{x}}\\\Rightarrow\dot{\vec{x}}\cdot\ddot{\vec{x}}=-\frac{GM}{(\vec{x}\cdot\vec{x})^{3/2}}\vec{x}\cdot\dot{\vec{x}}&\left\{\vec{x}\cdot\vec{x}=|\vec{x}|^2\right.\\\Rightarrow\frac{d}{dt}\left[\frac{1}{2} \dot{\vec{x}}\cdot\dot{\vec{x}}\right]=\frac{d}{dt} \left[\frac{GM}{(\vec{x}\cdot\vec{x})^{1/2}} \right]&\left\{\frac{d}{dt}\left(\vec{x}\cdot\vec{x}\right)=2\vec{x}\cdot\dot{\vec{x}}\right.\\\Rightarrow \frac{d}{dt}\left[|\dot{\vec{x}}|^2 \right]=\frac{d}{dt}\left[\frac{2GM}{|\vec{x}|}\right]\\\Rightarrow |\dot{\vec{x}}|^2=\frac{2GM}{|\vec{x}|}+c\end{array}

c is an “integration constant“, it can be any number.  Jumping over to polar coordinates \left(\begin{array}{l}x=R\cos{(\theta)}\\y=R\sin{(\theta)}\end{array}\right) you can rewrite the usual velocity in terms of how fast you’re moving toward or away from the Sun (\dot{R}) and how fast you’re going around (\dot{\theta}).

\begin{array}{ll}\Rightarrow\dot{R}^2+R^2\dot{\theta}^2=\frac{2GM}{R}+c&\left\{\begin{array}{ll}|\vec{x}|=R\\|\dot{\vec{x}}|^2=\dot{R}^2+R^2\dot{\theta}^2\end{array}\right.\\\Rightarrow\left(\frac{dR}{d\theta}\dot{\theta} \right)^2+R^2\dot{\theta}^2=\frac{2GM}{R}+c&\left\{\frac{dR}{dt}=\frac{dR}{d\theta}\frac{d\theta}{dt}\right.\\\Rightarrow\left(\left(\frac{dR}{d\theta}\right)^2+R^2\right)\dot{\theta}^2=\frac{2GM}{R}+c\\\Rightarrow\left(\left(\frac{dR}{d\theta}\right)^2+R^2\right)\frac{L^2}{R^4}=\frac{2GM}{R}+c&\left\{R^2\dot{\theta}=L\right.\\\Rightarrow\left(\frac{1}{R^2}\frac{dR}{d\theta}\right)^2+\frac{1}{R^2}=\frac{2GM}{L^2R}+\frac{c}{L^2}\\\Rightarrow\left(\frac{1}{R^2}\frac{dR}{d\theta}\right)^2+\frac{1}{R^2}=2\alpha\frac{1}{R} +C&\left\{\begin{array}{l}C=\frac{c}{L^2}\\\alpha=\frac{GM}{L^2}\end{array}\right.\end{array}

L is the angular momentum of the planet in question, and it’s constant.  It may seem silly but, with the advantage of foresight, it’s better to solve this problem in terms of 1/R instead of R.

\begin{array}{ll}\Rightarrow \left(-\frac{dS}{d\theta}\right)^2+S^2=2\alpha S +C&\left\{\begin{array}{l}S=\frac{1}{R}\\\frac{dS}{d\theta}=-\frac{1}{R^2}\frac{dR}{d\theta}\end{array}\right.\\\Rightarrow-\frac{dS}{d\theta}=\sqrt{-S^2+2\alpha S+C}\\\Rightarrow d\theta=\frac{-dS}{\sqrt{-S^2+2\alpha S+C}}\\\Rightarrow\int d\theta=-\int\frac{dS}{\sqrt{-S^2+2\alpha S+C}}\\\Rightarrow\theta+D=-\int\frac{dS}{\sqrt{C+\alpha^2-(S-\alpha)^2}}\\=\int\frac{\sqrt{C+\alpha^2}\sin{(u)}du}{\sqrt{C+\alpha^2-(C+\alpha^2)\cos^2{(u)}}}&\left\{\begin{array}{l}S-\alpha=\sqrt{C+\alpha^2}\cos{(u)}\\dS=-\sqrt{C+\alpha^2}\sin{(u)}du\end{array}\right.\\=\int\frac{\sin{(u)}du}{\sqrt{1-\cos^2{(u)}}}\\=\int\frac{\sin{(u)}du}{\sqrt{\sin^2{(u)}}}\\=\int du\\\Rightarrow\theta+D=u\\\Rightarrow\cos{(\theta+D)}=\cos{(u)}\\\Rightarrow\sqrt{C+\alpha^2}\cos{(\theta+D)}=\sqrt{C+\alpha^2}\cos{(u)}\\\Rightarrow\sqrt{C+\alpha^2}\cos{(\theta+D)}=S-\alpha\\\Rightarrow\frac{1}{R}=S=\alpha+\sqrt{C+\alpha^2}\cos{(\theta+D)}\\\Rightarrow R=\frac{1}{\alpha+\sqrt{C+\alpha^2}\cos{(\theta+D)}}\\\Rightarrow R=\frac{P}{1+\epsilon\cos{(\theta+D)}}&\left\{\begin{array}{l}P=\frac{1}{\alpha}=\frac{L^2}{GM}\\\epsilon=\sqrt{\frac{C}{\alpha^2}+1}\end{array}\right.\end{array}

The choice of P and ε may seem arbitrary (and it is), but it has some historical relevance.  P is called the “semi-latus recturn” and it basically describes the size of the orbit.  ε is called the “eccentricity”, and it describes how lopsided the orbit is.  ε=0 means the orbit is a circle, 0<ε<1 means the orbit is elliptical, and 1≤ε means that the orbit is open (not actually orbiting).  For reference, the Earth’s eccentricity is ε=0.01671123 and Halley’s comet’s is ε=0.967.

D just describes what direction the far side of the ellipse points in, so it’s not actually important to the overall shape.

It turns out that this last equation relating R and θ is all you need to define an ellipse, such that the center of the system, (0,0), is at one of the foci.  Here’s a proof:

An ellipse with a focus at (0,0) can be written \frac{(x+F)^2}{A^2}+\frac{y^2}{B^2}=1 where F is the distance from the center of the ellipse to the focus and F^2=A^2-B^2.

\begin{array}{ll}R=\frac{P}{1+\epsilon\cos{(\theta)}}\\\Rightarrow R+\epsilon R\cos{(\theta)}=P\\\Rightarrow \sqrt{x^2+y^2}+\epsilon x=P\quad\quad\quad\left\{\begin{array}{l}x=R\cos{(\theta)}\\y=R\sin{(\theta)}\end{array}\right.\\\Rightarrow \sqrt{x^2+y^2}=P-\epsilon x\\\Rightarrow x^2+y^2=P^2-2P\epsilon x+\epsilon^2x^2\\\Rightarrow (1-\epsilon^2)x^2+2P\epsilon x+y^2=P^2\\\Rightarrow x^2+2\frac{P\epsilon}{1-\epsilon^2} x+\frac{y^2}{1-\epsilon^2}=\frac{P^2}{1-\epsilon^2}\\\Rightarrow x^2+2\frac{P\epsilon}{1-\epsilon^2} x+\left(\frac{P\epsilon}{1-\epsilon^2}\right)^2+\frac{y^2}{1-\epsilon^2}=\frac{P^2}{1-\epsilon^2}+\left(\frac{P\epsilon}{1-\epsilon^2}\right)^2\\\Rightarrow \left(x+\frac{P\epsilon}{1-\epsilon^2}\right)^2+\frac{y^2}{1-\epsilon^2}=\frac{P^2(1-\epsilon^2)}{(1-\epsilon^2)^2}+\frac{P^2\epsilon^2}{(1-\epsilon^2)^2}\\\Rightarrow \left(x+\frac{P\epsilon}{1-\epsilon^2}\right)^2+\frac{y^2}{1-\epsilon^2}=\frac{P^2}{(1-\epsilon^2)^2}\\\Rightarrow \frac{\left(x+\frac{P\epsilon}{1-\epsilon^2}\right)^2}{\left(\frac{P^2}{(1-\epsilon^2)^2}\right)}+\frac{y^2}{\left(\frac{P^2}{1-\epsilon^2}\right)}=1\end{array}

Put it all together, and you’ll find that this is definitely an ellipse with a focus at the point (0,0), the location being orbited around (like the Sun for instance).

Posted in -- By the Physicist, Astronomy, Physics | 55 Comments

Q: What would happen if everyone in the world jumped at the same time?

Posted on August 15, 2010 by The Physicist

Physicist: Sounds like a party!

It would create a spherical wave that would descend through the Earth, focus at the core, and then expand again hitting the entire surface again, at more or less the same time, a little under 20 minutes later. In some areas of the Earth there’s a chance that someone might feel or hear a gentle bump. But probably not. Also, don’t worry about the core. It’s already suffering from much worse than a bunch of people jumping around.

A more interesting question might be “can you weaponize hopping?”

Good clean fun? Or the most deadly stealth weapon since HAARP?

There is some precedent for shock waves alone doing damage.  Both Mercury and Saturn’s moon Mimas have suffered massive impacts (the Caloris and Herschel impacts respectively) that created shock waves that moved through/around them and focused on the far side, causing geological scale damage (Allegedly.  It’s hard to say for sure what caused the damage).

Depending on how high the people of the world are willing to jump (without injury), we can generate energy on the order of approximately 1-2 kilotons of TNT. For comparison, the Little Boy nuclear bomb detonated over Hiroshima had an estimated yield of around 15 kilotons.

Now say that the entire world decides that Paris has created more than enough high art, and needs to be dealt with.  By carefully timing when everyone jumps off of their kitchen tables, so that the waves thus created by everyone arrive at the Eiffel tower at the same moment, a fair amount of damage could be done (maybe).  Due to fluctuations in density and material throughout the planet waves have a tendency to get scattered.  As a result, the best you can do is a “fuzzy focus”, like trying to burn something with a smoked-glass magnifying lens.  So here’s a guess:

I’d bet that if everyone on the Earth jumped in the right sequence then you could (mostly) focus the waves at some point on the surface and create an earthquake of no more than 5.0 on the Richter scale.

Here’s another example of “phased array” technology, the terribly named: “hypersonic sound“.

Posted in -- By the Physicist, Physics | 8 Comments