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

Shady Crypt Observatory E=mc^2 derivation

  1. Feb 16, 2010 #1
    http://www.shadycrypt.com/

    After equation (27) it says:

    "Keep in mind that we are not talking about the kinetic energy of the entire apparatus – only the kinetic energy of the mass associated with the pair of light rays that were measured. Since the moving observer measures higher radiated energy, the associated component of that apparatus's Kinetic energy must have been reduced by the same amount (conservation of energy). The only conclusion that follows from equation (27) is that if the apparatus gives off energy [itex]E_0[/itex], its mass must decrease by a corresponding amount as a result."

    I think the first bit is saying that the total energy of the two light rays (travelling in opposite directions) in the (instantaneous comoving) rest frame of the spaceship is greater than their total energy in the rest frame of the apparatus.

    [tex]\gamma E_0 = E_0 + K.E. > E_0[/tex]

    I've just added subscript zeros, for clarity, to the notation used at the site. But what does it mean by "the associated component of that apparatus's Kinetic energy must have been reduced by the same amount (conservation of energy)"? The apparatus, i.e. the pair of light rays, has no kinetic energy in its own rest frame, as its entire energy is accounted for by [itex]E_0[/tex], the rest energy, so how can it be that K.E. is reduced from nothing (in the rest frame of the apparatus) to less than nothing (in the rest frame of the spaceship)? It must mean "reduced" in some other sense, given off, I suppose, even though the total energy is constant in either frame.

    I don't understand this point:

    "So the stationary observer measures [itex]E_0[/itex] energy for a given pair of light rays, and the moving observer measures [itex]\gamma E_0[/itex] energy for the exact same rays. But energy is energy. One can't measure different values of energy in different frames of reference or conservation of energy would be violated; unless there is something else going on related to energy in the different frames. And there is. In the stationary frame, the apparatus has no kinetic energy relative to the stationary observer. But the apparatus does have kinetic energy according to the observer in the moving frame."

    The total energy is greater in the spaceship's frame, and the K.E. is greater in the spaceship's frame, so how can it say that both frames must have the same value for either kind of energy. I thought conservation of energy meant that kinetic energy plus potential energy is constant in any frame you choose to measure them in, rather than that a particular value of kinetic energy plus rest mass must be unchanged by a change of frame; how can it be unchanged, given that it's a sum of kinetic energy and rest mass, and kinetic energy is frame-dependent while rest mass is frame-invariant? And how can we draw any conclusions about whether energy is conserved when no mention is made of potential energy?

    Or maybe conservation of energy means (as Taylor and Wheeler say in Spacetime Physics) that total relativistic energy, [itex]\gamma E_0 = E_0 + K.E.[/itex], is conserved. Still no mention of frame-imvariance, and indeed the gamma factor tells us that it's not frame-invariant. (I wonder how the concept of potential energy relates to these relativistic definitions.)
     
    Last edited: Feb 16, 2010
  2. jcsd
  3. Feb 16, 2010 #2

    JesseM

    User Avatar
    Science Advisor

    Light rays can't have their own rest frame in relativity, since all inertial frames move slower than light, so that can't be what they mean by "the apparatus". In the diagram they represent the apparatus by a light bulb connected to a battery giving off light rays in different directions, the light rays would still have kinetic energy in that object's rest frame (for a light ray the kinetic energy would just be c times the momentum p).

    However, I don't know what they mean when they say "But energy is energy. One can’t measure different values of energy in different frames of reference or conservation of energy would be violated; unless there is something else going on related to energy in the different frames." Conservation of energy does not imply the energy of a system must have the same value in different frames; energy is frame-dependent, just like momentum. It seems like either the argument is wrong, or it's at least badly-worded and I don't understand what they really mean here. (Could it be that they're alluding to the energy-momentum 4-vector, whose length, i.e. square root of the dot product of the vector with itself, is frame-invariant? I'd have to look over the argument in more detail to see if this interpretation could make sense, don't have time right now)
     
  4. Feb 16, 2010 #3
    I saw their proof yersterday right after you talked about this site in a thread where you had a problem in showing an equation they give. I think they are just trying to prolong the derivation of Einstein's equation while blending his own sort of mind-based proof procedure with arguments based upon the conservation laws of momentum and energy which sounds really terrific and I don't really know how they could trudge through that mess! But I assume you are trying to find another derivation of [tex]E=mc^2[/tex] being more realistic than Einstein's own, so http://www.karlscalculus.org/pdf/einstein.pdf" [Broken] is a very beautiful and simple one which is based on some elementary arguments and achievements of special relativity.

    AB
     
    Last edited by a moderator: May 4, 2017
  5. Feb 16, 2010 #4
    Although one light ray can't have a rest frame, unless I'm mistaken, a system of two light rays travelling in opposite directions can, since we can specify a frame in which the vector sum of their 3-momenta (or 3-velocities) is zero. The energy of one light ray would be c times the magnitude of its 3-momentum, because the (rest) mass is zero, but for two light rays travelling in opposite directions, the vector sum of their 3-momenta is zero, so the magnitude of their combined 3-momentum is zero, and kinetic energy is zero, and the total energy is the (rest) mass of the system times c squared. Is that right?

    The line "we are not talking about the kinetic energy of the entire apparatus – only the kinetic energy of the mass associated with the pair of light rays that were measured" suggested to me that they were referring only to the energy of two light rays travelling in opposite directions along the x-axis. But maybe they meant the lightbulb thing too, or the energy of the bulb before it emits the light rays. I don't know.

    That's what I thought.

    Maybe... After equation (15) where they define the energy-momentum 4-vector, they deride derivations which "stop about here and claim something to the effect of, ...since P is called the energy-momentum 4-vector, E is therefore equal to mc2."
     
  6. Feb 16, 2010 #5
    Another thing I wondered was how it's possible to rely on the nonrelativistic identity in their eq. (21), where they say: "we know from the classical physics, by nearly the very definition of kinetic energy, K.E., that

    [tex]\vec{f} \cdot \vec{v} = \frac{\mathrm{d} }{\mathrm{d} t} \left ( \mathrm{K.E.} \right).[/tex]"

    I can see how this follows from the nonrelativistic definition of 3-force and 3-velocity, including

    [tex]\vec{f} = m \vec{a},[/tex]

    but earlier they showed that in relativity, this equation must be replaced by

    [tex]\vec{f} = m \left( \frac{\mathrm{d} \gamma }{\mathrm{d} t} \vec{v} + \gamma \vec{a} \right).[/tex]

    So if

    [tex]\vec{f} = m \vec{a}[/tex]

    doesn't hold in general, how do we know that this equality involving kinetic energy will hold in relativity.
     
  7. Feb 16, 2010 #6
    Thanks for the link, Altabeh. I'll check that out. At the very least it's giving me lots of algebra practice, and highlighted some pitfalls to watch out for (such as mistaking the derivative of speed for the magnitude of acceleration).
     
    Last edited by a moderator: May 4, 2017
  8. Feb 16, 2010 #7

    JesseM

    User Avatar
    Science Advisor

    Since all inertial frames agree both beams move at c, all frames which see the beams moving in opposite directions will agree the sum of their 3-velocities is zero. Momentum is more subtle since in quantum mechanics it depends on their frequency, so I think there would probably be a unique frame where their 3-momenta sum to zero.
    Only if they have equal frequencies.
    Hmm, I'm not absolutely sure how the kinetic energy of a composite system is defined, since kinetic energy isn't a vector...if you heat up a box of gas that's at rest in your frame, does the kinetic energy increase since the sum of the kinetic energies of all the molecules increases? I think the answer is no, that this extra energy is just folded into the "rest mass" of the composite system, which is not defined as the sum of the rest masses of all its parts, but rather as the sum of rest masses of part + kinetic energy of parts + potential energy of parts relative to one another (the potential energy part explains why a hydrogen molecule has a smaller rest mass than the sum of the rest masses of a proton and electron, since their potential is lowered when brought together into an atom...I think this type of intrinsic potential energy is just the negative of what's called the 'binding energy').
    I didn't catch that line, but that just adds to my confusion, since I'm not sure if they're talking about the sum of the kinetic energies of the two rays or if they're treating them as a composite system whose kinetic energy should be zero in the frame where their total momentum is zero.
     
  9. Feb 17, 2010 #8
    Shady Crypt aims to show that rest energy = mass times c squared, while Hahn uses the same symbols to mean that relativistic energy = relativistic mass times c squared. But I suppose once a connection between energy and mass is established, we can get from one to the other.
     
    Last edited by a moderator: May 4, 2017
  10. Feb 17, 2010 #9
    Yes of course and that is the reason I'm not interested in the Shady Crypt's twisted and sort of erroneous proof because the mass and energy equivalence from the perspective of classical physics has been proven beautifully by Einstein and many others over and over! If you are searching for better derivations which are way prettier worded than Shady's article, see the below papers I've provided for you:

    1- Steck, D. J. 1983. An elementary development of mass-energy equivalence.
    Am.J.Phys. 51: 461-462

    2- Peters, P. C. 1986. An alternate derivation of relativistic momentum. Am. J. Phys.
    54: 804–808

    3- Fegenbaum, M. , Mermin, N. 1988. E = mc2 . Am. J. Phys. 56:18–21

    4- Plakhotnik, T. 2006. Explicit derivation of the relativistic mass-energy relation for
    internal kinetic and potential energies of a composite system. Eur. J. Phys. 27:
    103–107, 2006.

    5- Sebastiano, S. , Massimo P. 2005. Deriving relativistic momentum and energy
    Eur.J.Phys. 26: 33-45.

    6- Rohrlich, F. 1990. An elementary derivation of E = mc2 . Am. J. Phys. 58:
    348–349

    These papers are all built on using the approaches of conservation laws in particles collisions and the postulates of Special Relativity! Since they have been published in the remarkable journals, so their language and highlighted calculations can help you much more than Shady Crypt!

    AB
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook