Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

E = MC2 - how was it initially proven?

  1. May 23, 2005 #1
    Hi Guys,

    I got into a silly argument re: e = mc2 this weekend. A friend of mine argued that the furmula was still just a theory - and although I realize almost all great science does start as a theory this particular one was proven a long time ago.

    The problem was that other than using practical examples like the atom bomb I could not remember how Einstein initially proved the theory - can someone refresh my memory please.

    Thanks
    Warrick
     
  2. jcsd
  3. May 23, 2005 #2
    well I always though that the A-bomb was the fist real life application of e=mc^2, I mean sure you can throw around some fancy mathematics, but until you actaully demonstrate something its still a thoery. So I am not sure Eistine actually "proved" anything, but maybe im wrong.

    Any ideas?
     
  4. May 23, 2005 #3
    Yeah, it's called the special THEORY of relativity, so I'm guessing no proof is likely. The earliest practical usage that I'm aware of was when Marie Curie was trying to account for the loss of mass of radioactive materials. E=mc^2 accounted for it perfectly, so the story goes.
     
  5. May 23, 2005 #4
    Quite an open and undefined question. The formula follows from the two assumtions that the speed of light is constant in any frame, and that the laws of physics hold in any inertial (ie unaccelerating) frame.

    I'ts also a bit bastardised, as the 'true' formula is [itex]E^2=m^2c^4+m^2 p^2[/itex].

    As has been said, the atom bomb was a pretty good 'proof' of the at-rest equation (ie p = 0). The 'true' formula is used all over for relativistic kinematics in areas such as particle physics, solid state physics etc.
     
  6. May 23, 2005 #5

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    YIKES!

    What made you think that the word "theory" automatically means something that is derived without "proofs"? Please do not confuse the usage of that word in science as that used in the pedestrian sense.

    There are theories that begin from First Principles derivation. In physics, that is as solid and fundamental as anything. This is not to be confused with phenomenological theories that are meant to be an ad hoc description till something more fundamental comes along.

    Eintein's infamous equation did not appear out of nowhere. It was DERIVED after one adopts ALL the fundamental postulates of Special Relativity and reformulates classical mechanics. Now those postulates are not derived but rather based on an initial ansatz. The consequences of such postulates are then tested and seen to conform with experimental verifications. That's why we accept those postulates to be valid, and we continue to test them.

    Zz.
     
  7. May 23, 2005 #6

    arivero

    User Avatar
    Gold Member

    In some way most proofs of electromagnetism are proofs of special relativity, because electromagnetic theory is not consistent with Galilean invariance; it needs Lorentz transformations. So E=mc^2 was implied by Lorentz, and Einsten "just shows" how to derive it from axioms.

    Of course the theory of nuclear stability (a.k.a. "atomic bomb", but also plainly alpha and beta radiation) is an early consequence of the relationship between mass and energy, but it is more of a historical coincidence, with the Curies around showing all kind of rays just at the same age that relativity was developed.

    Now, if I am to think which is the more spectacular consequence, I will say "pair creation", the fact that above a certain energy level, photons are able to create pairs of elementary massive particles, say electron and positron. The make nice "V for Vendetta" plots in cloud chambers.
     
    Last edited: May 23, 2005
  8. May 23, 2005 #7

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

  9. May 23, 2005 #8
    Mmm... I was thinking of 'proof' more in the mathematical sense. So as long as a 'theory' predicts experimental outcomes, it is deemed 'proven'? I did not know that. I read in some pop-sci book years and years ago the progression of truth from conjecture to law, and it said that if a 'theory' (which can predict experimental outcomes in a field) can be proven to be universal (i.e. apply to all fields) then it is a law. Hence my impression that theories are not proven. But I obviously got the wrong end of the stick (although I did say it was only a guess).
     
  10. May 23, 2005 #9

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    If u consider the lagrangian action for a free particle

    [tex] S^{L}_{0}=\alpha\int_{(1)}^{(2)} ds [/tex]

    in the noncovariant formalism and u apply Noether's theorem for time translations,u'll get the conserved quantity called "energy".

    If u compute the Hamiltonian,u'll run into

    [tex] H\left(\vec{r},\vec{p},t\right)\longrightarrow H\left(\vec{p}\right)=\sqrt{\vec{p}^{2}c^{2}+m^{2}c^{4}} [/tex]

    Daniel.
     
  11. May 23, 2005 #10

    jtbell

    User Avatar

    Staff: Mentor

    No physicist worth his salt should ever deem a physical theory (in its entirety) strictly "proven" in the mathematical sense. The most that one can say is that a theory has been verified or supported by experimental results to date. When a physicist says "proven", he probably means something like, "there's enough evidence supporting it that I'd probably keel over with a heart attack if something came along to disprove it."

    And remember that even in the strict mathematical sense, "proven" is relative to a given set of axioms which must be accepted without proof. Remember Lobachevsky? :wink:
     
  12. May 23, 2005 #11

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    The explanation from Einstein's "m = L/c^2" paper in 1905 is also quite understandable. The original translation is here:
    http://www.fourmilab.ch/etexts/einstein/E_mc2/www/

    Essentially it is this: if light carries momentum, and if momentum is always conserved, light must increase the mass of the object that aborbs it and decrease the mass of the object that emits it.

    Two equal masses, m1 and m2 (=m) at co-ordinates -d,0 and d,0 (origin at centre of mass). Since light has energy E = hf and momentum E/c (= hf/c), when a photon leaves m1, m1 recoils with momentum E/c. When the photon is stopped by m2, m2 takes on momentum E/c = mv (so v=E/cm). In time t=2d/c, m1 moves distance s=vt = E2d/mc^2. At time t after m1 begins moving, m2 receives momentum E/c.

    Now you can see where there is a problem. Unless some mass is transferred, the centre of mass has moved!! Newton's third law takes care of this where the masses are not separated by a distance as the changes in motion occur at the same instant. But when the momentum change is provided by light, there will be a shift in the centre of mass unless light transfers mass.

    How much mass does it have to transfer? Work it out: In order to conserve the centre of mass at time t, m1(-d) + m2d = 0 = m1'(-d1') + m2'd. Since d1'=d+s = d+E2d/mc^2, we have:

    [tex](m + \Delta m)(d) = (m-\Delta m)(d + E2d/mc^2)[/tex]

    [tex]md + \Delta md = md - \Delta md + mE2d/mc^2 - \Delta mE2d/mc^2)[/tex] or:

    [tex]2\Delta md + \Delta mE2d/mc^2 = 2Ed/c^2[/tex]

    Ignoring the negligible 1/mc^2 term (this actually disappears if you take into account the [itex]\gamma[/itex] factor but we can avoid the math because we can see that [itex]m>>\Delta m[/itex]):

    [tex]\Delta m = E/c^2[/tex] or:

    [tex]E = \Delta mc^2[/tex]

    AM
     
  13. May 23, 2005 #12

    russ_watters

    User Avatar

    Staff: Mentor

    Well, think about that for a sec: how can a theory be proven to be universal? The only way to know if a theory applies in all cases would be to test it in all cases. That's an infinite number of tests and physically impossible.

    So no, everything we know of science is "only" theory. Everything is provisional - conditional on the outcome of the next experiment.
     
  14. May 24, 2005 #13
    Hey, I didn't write it! But I'd guess if equivilent mathematical models can be derived in several different fields, or you can show how a certain theory can be derived mathematically from different axioms in another field, then you'd be onto something. Was this not the case (I don't know - I'm asking) with the equation for entropy being the same as the equation for the number of configurations of position & velocity of molecules in a gas? Or was the latter just chosen as a measure of the former?
     
  15. May 24, 2005 #14

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Nope.Both Boltzmann,Gibbs & von Neumann's formulae don't have direct connection to the "configurations & velociy of molecules in a gas".

    :wink:

    Daniel.
     
  16. May 24, 2005 #15
    Eh? That's a hard sentence to read. Boltzmann's equation for entropy of an equilibrium state of a system is a measure of the number of available configurations of position and velocity of molecules associated to that equilibrium state: S = k log W.

    But you're winking and that means I'm missing a joke. What is it? :uhh:
     
  17. May 24, 2005 #16

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Good,it was a joke about Boltzmann (i think you meant momenta :wink:);i was hoping to get u to search for the 3 formulae and tell me which applies and which is "more fundamental" (i know it's pleonastic).

    Daniel.
     
  18. May 24, 2005 #17
    Ah. No I'm far too lazy to do that.
    ?
     
  19. Jun 3, 2005 #18
    i do not have a vast amount of wisdom
    E=mc^2

    The coefficent varies differently with each element. As long as the correct coefficent was found to fit the equation and the element, the formula would be correct.

    :biggrin:
     
  20. Jun 3, 2005 #19

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    So, there are two notions of proof, and they're different. I assume that you mean prove in the sense of 'test' or demonstrate. In the sense of, for example, "prove yourself."

    The initial famous experiment (and sadly I can't remember the particulars.) that indicated that [itex]E=mc^2[/itex] was indeed correct involved radioactive materials that shed heat. There is, in fact, a measurable decrease in mass for radioactive materials that only shed heat (although the decrease in mass is very small in most circumstances). I would guess that it took place in the 1950's in Chicago as part of the research there.

    With modern science there are demonstrations of this property to be found in many different places.

    For example, it's possible to show that the mass of two hydrogen atoms and one oxygen atom adds up to more than the mass of two water atoms - in an amount that corresponds to the binding energy.

    Another good example is anti-matter - matter annihillation. The mass of the positron and electron can both be measured very precisely, and, the energy they produce when they annihillate each other corresponds with Einstein's theory.
     
  21. Jun 4, 2005 #20
    The proof can be done theoretically from the relativistic definition of force. It is shown here https://www.physicsforums.com/showthread.php?t=63380&highlight=Relativistic+Energy

    Kudos to physicsguru though, must have taken lots of effort to write that proof out.

    But as to how it was intially proven, that i'm not too sure how Einstein did it. But if you want to, you can check out transcripts of his original manuscript on SR. Great stuff.

    : )
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: E = MC2 - how was it initially proven?
  1. On E=mc2 (Replies: 1)

  2. In E=mc2, Why C? (Replies: 57)

  3. Energy in E=mc2 (Replies: 48)

  4. Concerning E=mc2 (Replies: 24)

  5. Questions about E=mc^2 (Replies: 25)

Loading...