Why Does E=mc2?

By Mark Trodden | February 23, 2007 1:28 pm

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.

  • Michael

    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.

  • http://blogs.discovermagazine.com/cosmicvariance/mark/ Mark

    The cannon is jammed up against the wall and she’s standing to the side.

  • Wayne

    My understanding is that c is a constant that most have accepted to be equal to the speed of light.

  • http://blogs.discovermagazine.com/cosmicvariance/mark/ Mark

    Hi Wayne. c is defined as the speed of light.

  • Benabik

    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.

  • Mr Knightley

    quality. This is why I read cosmic variance.

  • Joe

    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.

  • Thomas Francis Ryan III

    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.

  • Bruce

    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…

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  • Sourav


    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.

  • zwa

    Does the fact that the speed of light can be less than c affect this in anyway?

  • de-lurker

    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…:)

  • Sourav


    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.

  • Jacob Russell


    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.

  • http://blogs.discovermagazine.com/cosmicvariance/mark/ Mark

    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.

  • http://arunsmusings.blogspot.com Arun

    Re #11, it would seem that a classical chunk of fluid of length d moving at velocity v would have momentum E/v.

  • Jeff Chamberlain

    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?)

  • Analyzer

    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.

  • Sourav

    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.

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  • Doug

    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.”

    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.

    As the catenoid becomes more important in curved space-time, the helicoid transformation becomes more important.
    Derek K Wise ‘MacDowell-Mansouri gravity and Cartan geometry’

    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://arunsmusings.blogspot.com Arun

    So, what is the easiest demonstration that light carries momentum related to its energy by p = E/c?

  • Perl

    So, what is the easiest demonstration that light carries momentum related to its energy by p = E/c?

    The quantitative experimental confirmation of the photoelectric and Compton effects

  • Rob

    This question may reveal that my scientific background is limited to poli sci :)

    Yes, indeed, c is the speed of massless particles and, since light is massless, it travels at this speed.

    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?

  • Ike

    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.

  • http://blogs.discovermagazine.com/cosmicvariance/daniel daniel

    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.

  • http://blogs.discovermagazine.com/cosmicvariance/sean/ Sean

    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.

  • Ike

    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?

  • Ike

    Should have looked it up…nice site.

  • dark-matter

    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.

  • anon

    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.

  • http://thisquantumworld.com Ulrich Mohrhoff

    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.

  • http://quasar9.blogspot.com/ Quasar9

    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

  • http://blogs.discovermagazine.com/cosmicvariance/mark/ Mark

    Quasar9. In the first case, the box moves backwards, then forwards, then stops. In the second case, it keeps moving backwards and forwards.

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  • Roberto C

    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.

  • Z

    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.

  • franjesus

    I remember reading a very similar explanation on French’s Special Relativity.

  • Sourav


    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

  • Rob

    Thanks Ike # 26 and Daniel #27!

  • dark-matter

    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?

    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

  • http://blogs.discovermagazine.com/cosmicvariance/mark/ Mark

    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.

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  • http://arunsmusings.blogspot.com Arun

    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?

  • Sourav


    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.

  • http://arunsmusings.blogspot.com Arun


    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?

  • Sourav


    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 :) ).

  • raj

    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.

  • http://predelusional.blogspot.com/ Stephen Uitti

    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?

  • Keller Waud

    I Believe in the Strausenberg theory of space and time as it has far greater substance in its argument: http://www.Strausenbergsociety.org

  • holdsky28

    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.

  • http://blogs.discovermagazine.com/cosmicvariance/mark/ Mark

    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.

    Hi holdsky28. Are you referring to the argument I gave, or to comment #51?

  • http://countiblis.blogspot.com Count Iblis

    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},


    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.

  • Paul Valletta

    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?

  • Paul Valletta

    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?

  • http://quantumnonsense.blogspot.com/ Qubit

    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!

  • brett

    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?

  • http://countiblis.blogspot.com Count Iblis

    Brett, see here

  • http://quantumnonsense.blogspot.com/ Qubit

    “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?”

    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!

  • http://blogs.discovermagazine.com/cosmicvariance/mark/ Mark

    Qubit – I really don’t know what your last comment is supposed to mean. Want to clarify?

  • http://countiblis.blogspot.com Count Iblis

    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.

  • http://quantumnonsense.blogspot.com/ Qubit


    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?


  • http://quantumnonsense.blogspot.com/ Qubit

    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? :)


  • http://blogs.discovermagazine.com/cosmicvariance/mark/ Mark

    Ok Qubit – I’ve no idea what you’re getting at, or where the quotes came from. Sorry.

  • http://quantumnonsense.blogspot.com/ Qubit

    Oh Well! I Tried.

  • Pingback: Talking Right Past Each Other - Deep Grace Is Needed « Deep Grace of Theory()

  • http://www.deepgraceoftheory.wordpress.com Janet Leslie Blumberg

    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.

  • http://www.deepgraceoftheory.wordpress.com Janet Leslie Blumberg

    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!

  • jon

    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?

  • http://blogs.discovermagazine.com/cosmicvariance/mark/ Mark

    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.

  • Pingback: E=MC2: why? - Page 2 - Bad Astronomy and Universe Today Forum()


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Cosmic Variance

Random samplings from a universe of ideas.

About Mark Trodden

Mark Trodden holds the Fay R. and Eugene L. Langberg Endowed Chair in Physics and is co-director of the Center for Particle Cosmology at the University of Pennsylvania. He is a theoretical physicist working on particle physics and gravity— in particular on the roles they play in the evolution and structure of the universe. When asked for a short phrase to describe his research area, he says he is a particle cosmologist.


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