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Simple way to understand mass-energy equivalence?

  1. Sep 28, 2014 #1
    In the Wikipedia entry on mass-energy equivalence*, there is a reference to an elementary derivation of mass-energy equivalence in an article by Rohrlich (AJP 1990), which apparently expands on the original thought experiment by Einstein (ie observer in middle of moving train, light pulses emitted at front and back of train). In the Wikipedia explanation, an object moves to the right in front of you, and emits light pulses forward and backward. In the frame of reference of the moving object, both light pulses have the same momentum.


    But in your resting frame, you see the object emitting more momentum to the right (blue-shifted light), and less momentum to the left (red-shifted light). Because the total momentum of the moving system must be conserved, and you do not see a change in velocity of the object when the light is emitted, you conclude that the object has lost mass, because you need to balance the “net momentum now moving to its right in the two light pulses it emitted”, despite its unchanging velocity.


    I had difficulty thinking I really understood this because the issue of Doppler shifts make me want to think of waves and cycles rather than single photons. So I couldn't convince myself I really understood the issue. This go me to thinking if the thought experiment described below couldn't perhaps lead to the same conclusion through analogous logic without having to accept that one pulse of light has less momentum than the other.


    The situation I imagine is identical to that of Wikipedia and Rohrlich, but has an object emitting two photons (or pulses of light) “up” and “down” rather than “forward” and “backward”, as it moves past you to the right.


    If this is the case, do you not, after the emission, see three objects with right momentum (one photon up and right, one photon down and right, the object to the right)? Some of the total right momentum of the system is now present in the (?partly) right momentum of the two photons or light pulses. The object does not appear to change velocity after emission, therefore it must have lost mass, because it has less right momentum after emission.


    Once you accept that its just a fact that light carries momentum, and accept that momentum is conserved in the system described, isn't this example sufficient to argue that the mass of the emitting object must change because it emitted energy? Is there a flaw here, perhaps resulting from the fact that the motions of the objects now aren't “purely” all in the left-right axis?

    * http://en.wikipedia.org/wiki/Mass–energy_equivalence#Einstein:_mass.E2.80.93energy_equivalence (section 9.5.2)
     
  2. jcsd
  3. Sep 28, 2014 #2

    PeterDonis

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    Don't get too hung up on the term "photon". It is often used in classical (i.e., non-quantum) relativity as a shorthand way of referring to a very short-duration light pulse (more technically, it refers to a short-duration light pulse in the geometric optics approximation, where we can assume that the wavelength of the light is so much shorter than any other length in the problem that it can be ignored and we can treat the light pulse as a "particle"). Such a pulse is certainly not a photon in the quantum sense.

    Yes.
     
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