Once again this semester, I’m teaching a ridiculously fun course – Physics 312: Relativity and Cosmology; Einstein and Beyond. As I’ve mentioned before, this course is so enjoyable because one gets to expose undergraduates with not much physics background to some of the mind-bending results of relativity, and watch them struggle with it and, usually, finally come to understand it. Great stuff.
While I have a blast with the later parts of the course – general relativity and cosmology – I have a particular soft spot for something rather close to the beginning, in the special relativity portion – the most famous equation in physics – E=mc2.
So how does one go about motivating this equation for a class of students with only a little physics background, but who know some calculus? Well, perhaps the first thing to say is that, for the purposes of this course, it is much more important that they understand why an equation relating mass and energy is required, and how one might derive its form, than that they actually be able to do the detailed derivation themselves. So that is the tack that I take.
Early in the course, we review the idea that light is an electromagnetic wave. We do this by starting with Maxwell’s equations, which describe how moving and spatially varying electric and magnetic fields are related, and using them to show that, even in vacuum, if one tweaks the electric or magnetic field, then that disturbance propagates as a wave, with a given speed. We then see that the speed that arises is empirically equal to the speed of light, and hence we identify light itself with electromagnetic waves. This is a powerful idea, because students have a great deal of intuition about waves. In particular, they know that waves carry energy and momentum. So, at this point, students are pretty comfortable with the idea of light as a wave, and that light therefore carries energy and momentum.
Now, this is a great point for one of those staples of relativistic reasoning – the thought experiment. We start with one that doesn’t involve any of those worrisome relativity ideas. Think of a physicist, standing at one side of a large box, which itself is sitting on a perfectly frictionless surface (think of ice if you like). The physicist possesses a large cannon, which she is using to hurl heavy cannonballs across the box. What happens to the whole system?
Well, the box, physicist, cannon and cannonballs are a closed system, with no external forces acting on it. So one thing we know is that the center of mass of the system won’t move. Of course, that doesn’t mean that nothing will happen. As a cannonball is launched, it acquires a certain momentum, and conservation of momentum means that the box acquires the equal and opposite momentum, and sets off sliding backwards on the ice.
The next important event is that the cannonball collides with the opposite wall of the box, imparts it’s momentum to the box, and both cannonball and box come to a halt. At this point, the distribution of mass in the box is different from at the beginning (a cannonball has been transferred from one side to the other), and the position of the box has shifted. These two differences conspire in such a way that the center of mass of the system as a whole remains in the same place. All is right with the world.
Now let’s think about a second thought experiment, which is closely related to the first. All I want to make different is to replace the cannon by a powerful laser. Instead of a cannonball being propelled across the box, we’ll now think about the laser firing a pulse of light. Now, the light carries momentum, and so when the laser fires and the pulse sets off, the box will once again begin a backwards slide in order that momentum be conserved. Also once again, when the light reaches the other side and is absorbed by the opposite wall, the momentum will be transferred back to the box, which will then come to a halt. But now you see the problem. The distribution of mass in the box is the same as it was at the beginning, and no external forces have acted on the system, and yet because the box has slid backwards and no mass has been moved, the center of mass of the entire system has moved! All no longer seems right with the world.
This kind of thought experiment is what forces one to the conclusion that the idea of “center of mass” needs to be replaced by a more general concept – that of a “center of energy”. Obviously, this means that one must take into account how the distribution of energy in the system has changed, as well as the mass, when figuring out how a system should behave under no external forces. Another way to say this is that moving energy to one side of the box to the other is equivalent to moving some mass across the box – mass-energy equivalence!
This is the punch line, but one can do a little better. One can, of course, ask, when I’ve fired my laser pulse and had it absorbed on the other side, how far has the box moved? One can then ask, how much mass would I have to have moved from one side to the other in order that this movement of the box, combined with the mass movement, leave the center of mass unchanged. Equating the answer, m, to this question, with the energy, E, of the pulse, moving at the speed of light, c, yields: E=mc2.
February 23rd, 2007 at 3:02 pm
What happened to the physicist who was standing behind the cannon when she fired it? I would think our professor is now a red goo on the opposite wall from the box, where she’s been smashed by her cannon. Even if we assume she’s a gymnast too and did a double back flip out of the way, it seems like the action of the cannon is a consideration.
February 23rd, 2007 at 3:28 pm
The cannon is jammed up against the wall and she’s standing to the side.
February 23rd, 2007 at 3:47 pm
My understanding is that c is a constant that most have accepted to be equal to the speed of light.
February 23rd, 2007 at 3:48 pm
Hi Wayne. c is defined as the speed of light.
February 23rd, 2007 at 3:53 pm
Hm. If I lived closer to Syracuse, I’d be tempted to try to sneak into your lectures. Unfortunately, an hour commute is a bit much.
Wish someone over at RIT had taught this… or, if they did, that I had noticed it while I was there.
February 23rd, 2007 at 4:08 pm
quality. This is why I read cosmic variance.
February 23rd, 2007 at 4:17 pm
Thanks Mark; that is a great exposition. Can you show us the math for the last step? Why must m = E/c^2, and not 2E/c^2, or E/c, or E/c^3, etc.
February 23rd, 2007 at 5:14 pm
I had no idea taught at Syracuse. I live in upstate NY, starting at Corning Community College, transfering to Keuka College for my Bachleors. I was a hairs breath away from attending SYU instead, and now sorry I didn’t. I’d have loved to have taken every class you taught.
February 23rd, 2007 at 5:42 pm
Sweet. I can probably even remember this to amaze and amuse my friends. I love being a student, even if I’ve been out of college for 30 years…
thanks
February 23rd, 2007 at 5:57 pm
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February 23rd, 2007 at 6:22 pm
Joe:
Consider a photon classically as a cylindrical chunk of light of length d. It moves at speed c and contains a total energy E. We know that energy is force times distance for constant force, E=F*d, and force is the change in momentum per unit time, F=p/t=p/(d/c) for constant p. Thus, the momentum imparted on the far wall of the box is p=E/c.
If the box is of length L, the photon travels a time T=L/c, during which time the box moves with speed v=p/M, where M is the mass of the box. Therefore, the box shifts a distance backward x=p/M*L/c. For the center of mass to be constant, we require M*x=m*L, where m is the “mass” of the photon. Finally, we find m=p/c=E/c^2.
February 23rd, 2007 at 7:10 pm
Does the fact that the speed of light can be less than c affect this in anyway?
February 23rd, 2007 at 7:34 pm
Hi Mark,
With regard to Wayne’s (#3) comment, I also had the understanding that the speed of light and c were technically two different things– c being Einstein’s “speed limit of the universe,” and that the speed of light just happens to be equal to c because of light’s lack of mass. If neutrinos had turned out to be massless then wouldn’t “speed of neutrinos” have been just as good a description of c as “speed of light?” Since it appears that photons are massless, of course, it makes sense to define the _value_ of c as the current measured value of the speed of light.
On a more speculative note, wouldn’t this make a (hypothetical) discovery of Lorentz non-invariance more difficult to prove as such? One would have to rule out the photon having mass before you scrapped special relativity, and I have absolutely no idea how you’d do that. Real pair production without a recoil mass?…good luck measuring that with instruments really really far away…:)
February 23rd, 2007 at 7:53 pm
zwa:
No. “c” is the speed of any massless particle along an inertial trajectory. When light is slowed, as in glass, it’s because the medium is electrically responsive, so photons are being constantly absorbed and emitted — the light wave is an ensemble of particles, not a single one. I can still send information via neutrinos at nearly c through the same glass (i.e., causality), and the kinematical arguments above still hold.
February 23rd, 2007 at 9:24 pm
Wonderful!
Pure poetry… which is what I teach.
The best poets, artists, musicians, apply a form of reasoning with comparable discipline, rigor, and attention to detail–to very different subjects, with very different methods.
A beautiful explication of this most famous equation.
I will use this as an example of the quality of reasoning I would like to see my students apply to Wordsworth, Keats, Shelley.
February 23rd, 2007 at 10:37 pm
de-lurker (#13) and Wayne (#3) – I see what Wayne meant now – thanks! Yes, indeed, c is the speed of massless particles and, since light is massless, it travels at this speed. If, for example, there was a phase transition that spontaneously broke electromagnetism and gave a mass to the photon, there would still be a universal speed limit in local inertial frames that corresponds to the speed of massless particles. Light, however, would travel more slowly.
Hope this makes sense.
February 23rd, 2007 at 11:03 pm
Re #11, it would seem that a classical chunk of fluid of length d moving at velocity v would have momentum E/v.
February 24th, 2007 at 12:38 am
I must be really stupid. I can’t visualize or follow the thought experiment as described. (For example, what does it mean to hurl a cannonball “across” the box?)
February 24th, 2007 at 1:46 am
I must be really stupid. I can’t visualize or follow the thought experiment as described.
Imagine yourself at one end of the inside of a railroad car on a frictionless track. You’re firing cannonballs (or light rays) at the other end so that they slam into the far wall.
February 24th, 2007 at 4:23 am
Arun (#17):
That’s a good question. For classical, massive fluids, the momentum is proportional to the velocity, so you get a factor of 1/2 for the kinetic energy from applying the work-energy theorem. For light, the imparted momentum depends only on the energy density of the electromagnetic field, and all light moves at speed c in a vacuum.
February 24th, 2007 at 9:45 am
[...] Mark at Cosmic Variance explains the equation e = mc^2 using a thought experiment. The basic point of the thought experiment is that since light is a wave, it imparts momentum, so conservation of momentum eventually implies mass-energy equivalence. [...]
February 24th, 2007 at 9:56 am
This is speculative analogy, hopefully with some insight, since I do not yet have the skill to be rigorous.
F = m * v^2 [or a(cceleration) in lieu of v^2] is analogous to E = m * c^2
Newton modeled mechanics after ballistics.
Einstein demonstrated how mechanics can be like relativity ballistics.
Ballistics and mechanics are both pursuit-evasion games consistent with the description by Basar and Olsder in ‘Dynamic Noncooperative Game Theory.
David Hestenes, abstract: from ‘The Kinematic Origin of Complex Wave Functions”:
“A reformulation of the Dirac theory reveals that i*h-bar has a geometric meaning relating it to electron spin. This provides the basis for a coherent physical interpretation of the Dirac and Schroedinger theories wherein the complex phase factor exp(-i*psi/h-bar) in the wave function describes electron zitterbewegung, a localized circular motion ["helical trajectory", p6] generating the electron spin and magnetic moment. Zitterbewegung interactions also generate resonances which may explain quantization, diffraction, and the Pauli principle.”
http://modelingnts.la.asu.edu/pdf/Kinematic.pdf
I am reading a Cambridge Monographs on Mathematical Physics by Terry Gannon, ‘Moonshine beyond the Monster: The Bridge Connecting Algebra, Modular Forms and Physics’.
On page 236 “… Space-time is not the vector space R^4, but rather a nontrivial (curved) four-dimensional pseudo-Riemannian manifold. Gravity is the convergence or twisting of nearby geodesics; what we perceive as the elliptical revolution of the Earth about the Sun is merely the gentle entwining of the Earth’s geodesic with the Sun’s (Fig 4.3) …” [the Earth has a helical trajectory about the sun].
This appears to suggest a zitterbewegung relationship between the Earth and Sun with respect to gravity, In turn, this seems to suggest that gravity may be surprisingly similar in function to electromagnetism, although at a much different scale or gauge.
The most interesting curved cube, possibly related to gravity waves, may be the super-ellipsoid (e,n). As n -> oo, the structure becomes more of a cube.
http://mathworld.wolfram.com/Superellipsoid.html
As the catenoid becomes more important in curved space-time, the helicoid transformation becomes more important.
http://mathworld.wolfram.com/Catenoid.html
and
Derek K Wise ‘MacDowell-Mansouri gravity and Cartan geometry’
http://arxiv.org/PS_cache/gr-qc/pdf/0611/0611154.pdf
The possible catenoid [and therefore helicoid] relationship to gravity waves might be apparent when the catenary is considered:
“The curve a hanging flexible wire or chain assumes when supported at its ends and acted upon by a uniform gravitational force.”
http://mathworld.wolfram.com/Catenary.html
February 24th, 2007 at 11:02 am
So, what is the easiest demonstration that light carries momentum related to its energy by p = E/c?
February 24th, 2007 at 12:01 pm
The quantitative experimental confirmation of the photoelectric and Compton effects
February 24th, 2007 at 3:45 pm
This question may reveal that my scientific background is limited to poli sci
If photons are massless then why does gravity affect them as seen with gravitational lensing? Or is the light not affected, but rather that the space is bent and the light travels “straight” along bent space thereby turning its path?
February 24th, 2007 at 6:38 pm
Amazingly good! Thanks very much. If you take requests, could you write something similar about the difference between inertial and accelerating reference frames, and how general relativity is needed for the accelerating frame? Re#25, I think the answer has something to do with imagining firing a laser from one side of an accelerating elevator to the other; the elevator moves in the interim, and the light appears to bend to an observer within the elevator …don’t take my word for it, though.
February 24th, 2007 at 7:42 pm
Rob (#25), there are two answers to your question. The first (which you provide) is that space is curved, and although the light travels along what it takes to be a straight line, to a distant observer its path will appear to have been bent. Thus photons feel the gravity of other objects. In this picture, the photons don’t cause any curvature (and, thus, have no gravity) themselves. However, photons represent energy. And, as Mark explained above, energy is equivalent to mass. This means that photons do have a slight gravitational attraction. If you have enough photons in one place, then they have an effective mass which will cause space to bend around them. For example, in an extreme case, this can cause a self-bound object, where the gravity of the photons keep them together. This is called a geon.
February 24th, 2007 at 8:04 pm
Ike, I’m not going to fully answer your question, but: general relativity is not needed to describe accelerating frames! (I’m restraining myself from including additional exclamation points.) In special relativity, the thing about inertial frames is that they’re all created equal, while accelerating frames clearly are not — you can measure your acceleration! If you couldn’t, the world would be a very different place.
The difference between GR and SR is that GR includes gravity, and describes it as the curvature of spacetime. And that in GR inertial frames are purely local; they can’t be extended throughout spacetime. Just like you can’t extend Cartesian coordinates all over a sphere, but they do a good approximation in a very small region.
February 24th, 2007 at 8:30 pm
Thanks Sean,
I suppose I put that backwards; I’m no expert but I thought Einstein began general relativity by picturing an elevator accelerating through space and by noting that you couldn’t tell the difference between being in an accelerating elevator in free space and being in a gravitational field. Thus, would it be correct to say that general relativity requires accelerating reference frames?
What I was really wondering about the box thought experiment is if anything would change if the box was accelerating at right angles with respect to the firing direction (as you say, you could measure the acceleration within the box by firing the cannon and measuring the displacement from the target). In other words, what does an accelerating elevator in outer space have to do with general relativity?
February 24th, 2007 at 8:47 pm
http://www.astronomynotes.com/relativity/s3.htm
Should have looked it up…nice site.
February 25th, 2007 at 12:05 am
Mark,
Sorry but your take is too mechanical, too mathematical, too simplistic.
WE DON’T KNOW WHY E=MC^2. Because we don’t know why photons travel through space, has a maximum speed limit of c, and why this speed limit has anything to do with mass and energy. Hell, we don’t even KNOW why there’s mass and where it comes from (Higgs is unproven). And embarrassingly, we don’t know what energy is made of and why it has, and must have, the vast properties that we now see. Can anybody say e=mc^2 holds true at Plank’s time? And why this equation is not time dependent. One should be humble when talking about e=mc^2, because it’s like a fish trying to explain why the moon goes around the Earth.
February 25th, 2007 at 12:35 am
dark-matter, one should also be humble when trying to “correct” people who know what they’re talking about. For one thing, most of the mass of the objects around you? Nothing to do with the Higgs mechanism. It’s good old-fashioned chiral symmetry breaking, largely understood theoretically and experimentally 30 years ago.
February 25th, 2007 at 12:53 am
Like you, I love introducing undergraduates (and even higher secondary students) to the wonders of contemporary physics, which I do at the Sri Aurobindo International Centre of Education in Pondicherry, India. The less the students are spoiled by exposure to classical physics, the easier it is for them to get into contemporary physics.
The measurement units we ordinarily use are conventions. Theoretical physicists use natural units, in which such universal constants as c are dimensionless and equal to unity. So if conventions are left aside, the formula says E=m. In other words, E and m are absolutely the same thing, only (conventionally) measured in different units. E=mc2 therefore is no more than a rather unexciting conversion formula, much like the one we use to convert miles into kilometers or Fahrenheit into centigrades. This of course does not touch on the deep and fascinating question of what this thing E=m really is.
February 25th, 2007 at 6:01 am
Hi Mark,
What if this imaginary box resting on frictionless ice has a ‘perfect’ mirror on its inner wall, reflecting the photon back to the laser gun
-
and what (or watt) if this imaginary box sitting on frictionless ice has ‘perfect’ mirrors on its inner walls, reflecting the photon back and forth
February 25th, 2007 at 8:27 am
Quasar9. In the first case, the box moves backwards, then forwards, then stops. In the second case, it keeps moving backwards and forwards.
February 25th, 2007 at 5:34 pm
왜 E= M c^2 ì¼ê¹Œ?
ì•„ì¸ìŠˆíƒ€ì¸ì˜ ìœ ëª…í•œ ì´ ê³µì‹ì€ ê¸°ì–µë ¥ì„ ë†’ì—¬ì£¼ê³ , ì§‘ì¤‘ë ¥ì„ í‚¤ì›Œì¤€ë‹¤ê³ ì£¼ìž¥í•˜ëŠ” 수ìƒí•œ 학습ë„ì›€ìž¥ì¹˜ì˜ ì´ë¦„ì— ì”°ì¼ ë§Œí¼ ëŒ€ì¤‘ì—게 ë„리 ì•Œë ¤ì ¸ 있다.
ê³ ë”±í•™êµì—ì„œ 물리 êµê³¼ì„œì—ë„ …
February 26th, 2007 at 3:54 am
Hi Mark,
Does the box really move? How can the far wall be told to move, before the laser light hit it? And wen it does, it is time to stop… Well, anyway I like the this thought experiment a lot, and I’ll keep it in my toolbox.
February 26th, 2007 at 4:28 am
Arun, #23,
The statement that light “carries” a momentum p = E/c is wrong, because in physics you can only say what momentum is delivered to an object by a photon.
A photon can deliver either p = E/c (if the photon is absorbed) or p = 2E/c (if the photon is reflected by a rigid surface).
It is like a molecule hitting a wall. If it sticks to the wall (i.e., like a water vapor molecule hitting a cold surface), you get momentum p = mv delivered. If it is reflected, you get effectively double the momentum being delivered because instead of just stopping the molecule, the wall reverses its momentum, which gives an additional momentum to the wall.
E.g., when a weak shock wave hits a wall at normal incidence and reflects, the reflected overpressure on the wall is twice the incident overpressure (in a sttron shock wave, the reflection factor is even higher than two because other factors come into play).
The momentum absorbed by a wall stopping a photon, p = E/c, if you postulate an “equivalent mass, m” in p = mc, would give mc = E/c or E = mc^2. The photon doesn’t have rest mass, but this formula shows what mass it would have to have to deliver the same momentum for a given amount of energy.
But with the reflected momentum, p = 2E/c, the equivalent energy for a unit mass is different: mc = 2E/c, hence E = (1/2)mc^2.
February 26th, 2007 at 4:41 am
I remember reading a very similar explanation on French’s Special Relativity.
February 26th, 2007 at 1:54 pm
Z,
If a photon bounces perfectly elastically off the far wall, its change in momentum is 2E/c (*), so it most certainly “carries” a momentum of E/c (remember, momentum is a vector quantity so direction matters). And, if it bounces off the far wall and is then reabsorbed inelastically by the back wall (Mark’s comment #35), you get the same result by the method I use in comment #11.
(*) Ignoring a small Doppler shift
February 26th, 2007 at 10:46 pm
Thanks Ike # 26 and Daniel #27!
February 26th, 2007 at 11:42 pm
anon:
Chiral symmetry breaking, while applicable to chiral (ordinary) matter, cannot be the last word on mass. All chiral matter can absorb/emit photons. But there is dark matter. Are you suggesting dark matter is chiral?
Mark:
It appears you have company in anon making grandiose statement of deep knowledge.
As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality.
- Albert Einstein
February 26th, 2007 at 11:59 pm
If you think this simple attempt to explain a well-known equation to people is grandiose, you should see the outrageously overdone statements I make just before deleting silly comments.
February 27th, 2007 at 5:24 am
[...] I found it very interesting, and highly recommend it. That blog, which might have the best name and subtitle of any blog, often has good intellectual candy on it. Anyway, the most interesting concept that the article introduced to me was the Overton window. [...]
February 27th, 2007 at 7:34 am
So, E = p c for light is either arrived at from experiment or mathematically derived from special relativity – is there no thought experiment which shows this?
February 27th, 2007 at 8:15 pm
Arun,
A thought experiment presumes some rules to follow, no? A theory is precisely what formalizes these rules, so I think you’re asking too much.
The fundamental symmetries that yield a theory are observed facts. The electromagnetic gauge symmetry ensures that photons are massless, and therefore by the Lorentz symmetry of relativity they have E=pc. (Newtonian mechanics follows as a small-velocity approximation of relativistic mechanics.)
I suppose if one develops and confirms a more fundamental theory then we might find these facts as a consequence after a moment’s thought.
February 27th, 2007 at 10:10 pm
Sourav,
E=mc² follows from the photon following p=E/c and the center of mass remaining unmoved in the thought experiment described by Mark. Or else one can derive it from theory. But it is natural to ask is there a thought experiment that establishes p=E/c, or is one forced to derive it from theory?
February 28th, 2007 at 4:05 am
Arun,
I apologize — I now understand that by “arrives at” you mean to derive/illustrate, rather than to establish scientifically.
I can’t think of a good thought experiment off the top of my head. An analogy for my first paragraph in comment #11 would be a shotgun blast. Your shotgun shells can have different “loads” — varying numbers of pellets — but the muzzle velocity will always be roughly the same. Then, the energy will be proportional to the momentum times the muzzle velocity. The size of the load is like the frequency of a photon: the target gets more “hits” in the same amount of time (though this is getting rather loose with the wave-particle duality
).
February 28th, 2007 at 7:50 am
This is interesting. I followed the equations in my SR book a long time ago, but this is an interesting perspective.
One other thing that I found interesting–and I don’t remember where I read it–is that the sum of the massses of an electron and proton, when measured separately, is actually greater than the mass of a hydrogen atom. The explanation being the addition of the energy that is required to separate the atom into its component parts.
February 28th, 2007 at 3:05 pm
Apparently, it is possible, though tedious, to derive e=mc^2 without calculus. What you do is describe the difference method as you go… and this has been done for high school students.
The idea that c is dimensionless eludes me. As an engineer, energy is a mass time distance squared over time squared. Mass has units of, well, mass, and c is a speed, so has distance over time. So, e = mc^2 makes sense from a dimensional analysis. Engineers use dimensional analysis as a cheap cross check that a formual is correct all the time (or at least, that the formula is not obviously wrong). It is somewhat reassuring that the dimensional analysis holds for relativity.
So, a neutron walks into a bar and orders a beer. The bartender serves it a beer. The netron reaches for it’s wallet to pay, but the bartender says, “No, no. For you – no charge.”
So, a neutrino walks into a bar and orders a beer. The bartender hands the neutrino a beer, which promptly falls through it’s hands onto the floor. What did you expect?
March 3rd, 2007 at 1:30 pm
I Believe in the Strausenberg theory of space and time as it has far greater substance in its argument: http://www.Strausenbergsociety.org
March 9th, 2007 at 10:26 pm
But still,it can’t explain why E=mc^2 ,maybe it can be write E=(mc^2)/2.or other form for caculates the energy of motion.
March 10th, 2007 at 7:43 am
Hi holdsky28. Are you referring to the argument I gave, or to comment #51?
March 11th, 2007 at 8:39 am
Ulrich #33 makes a good point. Energy and mass are the same thing. To address the point made about dimensions by Stephen #50, consider deriving nonrelativistic physics from relativistic physics. Actually, i.m.o. this is the only correct way to derive
E = m c^2
You start with the usual equations of special relativity, you put c = 1 everywhere. Mass is the same as energy, so there would be no point in introducing a quantity M if you are not going to use it (just like Galileo never wrote the equation t’ = t)
The energy and momentum are given by
E(v) = gamma(v) E(0)
P(v) = v E(v) = gamma(v) v E(0)
Here v is the dimensionless velocity (time and distances are measured in the same units), gamma(v) = 1/sqrt[1-v^2] and E(0) is the rest energy.
Now consider phenomena at extremely low velocities. Just like if you want to see a plant growing you can take pictures every day and make a video such that 1 second corresponds to one day, in this case we should rescale time relative to space and formulate things in terms of the new variables. We define:
t_{slow} = t/c
where c is a (dimensionless) rescaling parameter. If we make c larger and larger, then for t_{slow} to stay of order unity, t must also become larger and larger. So, the large c limit is appropriate for studying phenomena that move very slowly.
If we define:
v_{slow} = dx/dt_{slow},
then:
v_{slow} = c v
We now consider what happens for c–> infinity while v_{slow} is kept constant, i.e. we run the video faster and faster but keep concentrating on things that appear to move at the same rate on the monitor.
The kinetic energy can be written as:
E_{kin}= E(v)-E(0)=
E(v_{slow}/c) – E(0) =
1/2 E(0) v_{slow}^2/c^2 + 3/8 E(0) v_{slow}^4/c^4 +…
And the momentum is:
P = gamma(v_{slow}/c) v_{slow}/c E(0) =
E(0) v_{slow}/c + …
Now, as we make clarger and larger, the things we see on the monitor may have E(0)’s that change too, so we should write:
E(0) = E(0,c)
Also, the energy and momentum should be rescaled to make sure we end up with equations that have finite coefficients in the limit c –> infinity. In this limit the equaton for the energy is:
E_{kin} = 1/2 E(0,c) v_{slow}^2/c^2
If we put:
E_{slow} = E c^p
where p is some exponent we’ll determine later, then:
E_{slow,kin} = 1/2 E_{slow}(0,c) v_{slow}^2/c^2
We have now expressed everything in “slow” variables and should now demand that in the c–> infinity limit, the equation does not become degenerate. This means that we must demand that
E_{slow}(0,c) = m c^2
for some constant m.
The momentum is given by:
P = E(0,c) v_{slow}/c =
mc^(-p+1) v_{slow}
And we see that we have to put:
P = P_{slow} c^(-p+1)
To find the exponent p, you can express E in terms of P, rewrite it in terms of the slow variables and demand that the equation does not become degenerate in the c –> infinity limit. You find that p has to be put equal to zero.
March 11th, 2007 at 10:52 pm
How about the outcome, in the thought experiment you suggest, with no space between the cannon/laser and opposite box wall?
A constraining box?..that has no transitional space available (except the length of the barrel of the cannon)..what would the outcome be?
Would an explosion occur? this of course nullifies the closed system concept.
The shortness of the available transitional space, dictates the outcome, think of a larger available space (a box that has a large distance from side to side)..then no evidence of the exchange of momentum would occur, not for a long time at least, for if the cannonball fails to interact with a boxside, then the result would be of a closed system box, travelling across the ice in ONE certain direction?
The term left-handedness come to mind?
March 11th, 2007 at 10:58 pm
Forgot to ask, if there is a thought experiment you can come up with, that shows within E=MC^2, the exchange rate of Mass to Energy, does ‘NOT’ equal Energy to Mass?
Creating Mass from Energy, cannot surely be equivilent to Creating Energy from Mass, there has to be subtle differences?
March 12th, 2007 at 5:48 pm
The mass of the box must effect the surface, otherwise how does the box “sit” on the surface? If the box itself effects the surface by its position. Then the box cannot be classed as a closed system. If there is no friction then what is the point of the surface?
This is an impossible thought experiment, because thinking about the box, has the same effect on the box as gravity; its an external force that penetrates the closed system of the box and the outside of the box, it makes the cannon ball heavy and light mass less, and the surface on the outside frictionless. How is it classed as a closed system?
I’ve heard another one about entropy and an ice cube melting on the table. How does the ice cube get there in the first place? Closed System? no way!
April 18th, 2007 at 1:32 am
Qubit, if the law of entropy holds true,how does one explain that the universe appears ordered or that complex entities such as ourselves even exist? Is the universe a closed system?
April 18th, 2007 at 11:46 am
Brett, see here
April 18th, 2007 at 5:35 pm
Brett
By knowledge of something in advance of its occurrence; Precognition.
Do smart people really tell the truth?
That all I am saying at the moment, am full of cold!
April 18th, 2007 at 5:47 pm
Qubit – I really don’t know what your last comment is supposed to mean. Want to clarify?
April 18th, 2007 at 9:01 pm
Qubit, do you mean Laplace’s demon?
Laplace’s demon can be an effective Maxwell’s demon and lower the entropy of a system. Ordinary Maxwell’s demons always fail to be effective, because with incomplete information about the system they must make measurements. They must store and process information, but with finite memories they can’t ecape having to reset their memory which is an irreversible process leading to entropy increase.
April 19th, 2007 at 3:19 pm
Mark
Looking (so to speak) at a thought experiment is mostly done in what we regarded as the past, just like we have experienced it as a memory, but instead of experiencing it, we have created it in the way that we would have experienced it.
If the future is undecided then all possible outcomes have to be taking into consideration, if you have the knowledge of it occurrence in advance, then that question can be answered, without experiencing it in the past.
The 2nd law of thermodynamics can be over come by using tricks! But trick are exactly what they appear to be.
“I never think of the future – it comes soon enough.” You don’t come up with that sentence and E=MC2 without telling everybody what really E=MC2 is, unless your lying. Einstein thought of the future more than most!
“If A is a success in life, then A equals x plus y plus z. Work is x; y is play; and z is keeping your mouth shut.”
Who am I to tell you what Z is?
Qubit
April 19th, 2007 at 4:17 pm
Count Iblis and everyone maybe the demon can split its self into two. Have you read Rumpelstiltskin? The 1857 version. Mind you didn’t that story come from one of the stories of Isis getting the magic name from Ra, in order to resurrect Osiris, these are all Quantum physics. Looks like there is some serious cheating going on in this universe! And am sure that story as changed a little since last time I read it.
I’ve put my cards on the table (well on my blog), Mankind show me yours?
Qubit
April 19th, 2007 at 4:24 pm
Ok Qubit – I’ve no idea what you’re getting at, or where the quotes came from. Sorry.
April 20th, 2007 at 1:54 pm
Oh Well! I Tried.
April 22nd, 2007 at 12:17 am
[...]               And I can do this – because I’ve worked all my life to figure out how to do it. I can do this because my branches of philosophy are literary theory and epistemology (”the formal study of how humans come to know, and how they know that they know, when they think they do”). And for a long, long time I have wanted to explain these things, just the same way I now see gifted physicists on their popular weblogs explaining the complexities of all the various branches and sub-branches of physics to the uninitiated and partially initiated. (Check out Mark on “Why E=Mc^2″ at http://www.cosmicvariance.com) [...]
April 22nd, 2007 at 12:34 am
Mark,
I wish I could have had your course! As a humanities prof, I’ve worked with colleagues in physics to “finally come to understand it (relativity)” for about 15 years. They give me books and I read them. I follow the arguments. But “understand” it? Do you have other course materials on relativity to recommend? Or other people’s stuff? I’ve read Einstein’s own book on relativity, too. I get how different observers are measuring differently and all that. But I can’t visualize or comprehend how “light” is radiating through space-time, while everyone is measuring it from their own referential viewpoints. Am I supposed to be able to? I guess I’m confessing I don’t really understand how everything else obeys the addition of velocities relative to the observer except light. Do your undergrads really get there? I am fascinated by relativity and never stop wishing I could “get” it.
April 22nd, 2007 at 12:41 am
Hang on, I just clicked on your course materials. I’ll study them and get back to you if I’m still at sea. Thanks!
April 24th, 2007 at 9:13 am
sorry if i sound stupid, but since the light carries no mass how does it induce a force on the box such that it moves?
April 24th, 2007 at 11:51 am
Hi jon. Light carries momentum, that it imparts to the box. A particle need not have mass to do this. In fact, if you think of light as a wave, it is pretty easy to think of the wave as carrying momentum as it travels along.
July 7th, 2009 at 8:06 am
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