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Thought Experiment

  1. Feb 25, 2010 #1
    There is an empty container on some weighing scales.
    It is filled with electrons & positrons, the lid is closed and a reading of the weight is taken.

    Now they go boom and are turned into gamma rays...

    Lets say the box does not let the rays escape, instead are reflected back inside the container.

    What happens to the scales, do they read less than before, more than before, or the same?


    (Sorry if this is the wrong board)
     
  2. jcsd
  3. Feb 25, 2010 #2

    tiny-tim

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    Hi Raekwon! :smile:

    Mass is energy, energy is mass …

    if nothing escapes from the container, and if no energy was supplied to cause the "boom", then the mass, and therefore the weight, will be exactly the same.
     
  4. Feb 25, 2010 #3
    Exactly my thoughts, but someone said because of E=mc2 that it would weigh less, which got me doubting myself, I guess they must have thought that light is massless and disregarded the conservation of mass. Thanks.
     
  5. Feb 25, 2010 #4

    tiny-tim

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    (please post on the forum, rather than using a PM)

    Yes, energy is mass, so stretching something, or compressing it, away from its equilibrium state, will increase its mass.
     
  6. Feb 25, 2010 #5
    Positrons, like electrons, have mass. But do positrons have the same "weight" as electrons? It is not obvious that the gravitational force is the same for positrons and electrons. Does matter gravitationally attract or repel antimatter? There are known subtle differences between particles and antiparticles, notably CP asymmetry, so why not gravitational force also? There are proposals for testing the gravitational force on anti-hydrogen in the low energy antiproton ring at CERN. See the proposed AEGIS experiment listed in

    http://en.wikipedia.org/wiki/Antiproton_Decelerator

    See discussion at

    http://en.wikipedia.org/wiki/Gravitational_interaction_of_antimatter

    If antimatter were known to be attracted to matter, then the AEGIS experiment would be unnecessary.

    Bob S
     
  7. Feb 25, 2010 #6

    tiny-tim

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    It depends what you mean by "unnecessary" …

    as the http://en.wikipedia.org/wiki/Gravitational_interaction_of_antimatter" [Broken] you quote says …
    :wink:
     
    Last edited by a moderator: May 4, 2017
  8. Feb 25, 2010 #7
    This is what never understood, if energy is mass, what is the photon?

    Why is the photon massless?
     
  9. Feb 25, 2010 #8

    tiny-tim

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    The photon has plenty of mass … energy is mass, and photons have energy.

    The photon has zero rest-mass (it needs it so that it can travel at the speed of light).
     
  10. Feb 25, 2010 #9
    So when it's resting (never) it has zero mass, yet when it's moving at the speed of light it has mass - how much?

    If it has mass does it have weight?

    Sorry for the dumb questions but google doesn't help, I found this and i'm more confused:
    http://www2.corepower.com:8080/~relfaq/light_mass.html [Broken]
     
    Last edited by a moderator: May 4, 2017
  11. Feb 25, 2010 #10

    tiny-tim

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    Its mass equals its energy.

    If it's contained inside a box (eg a fibre optic cable), then the box will have more mass, and will weigh more.
     
  12. Feb 25, 2010 #11
    The actual answer to the author's question is that the box would weigh a bit less after the recombination. How much less? Answer: Box Weight minus 1/4-total positron/electron pair weight.
    The positron/electron pairs, being massive, will have weight. Gamma rays, being massless yet possessing momentum will have no weight. Thus the momentum from the confined gamma rays will act on all sides of the box as to provide a downwards thrust equal to 1/4 the weight of the original positron/electron pairs.
     
  13. Feb 25, 2010 #12

    Vanadium 50

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    Glen, that turns out not to be the case. The box weighs the same. Gamma rays do have weight, although the simplest way to show this is simply to show E2 - p2 is the same before and after.
     
  14. Feb 26, 2010 #13
    What if we have:
    p2 = E2/c2 - m2c2
    (E2 = p2c2 +m2c4)

    If we apply energy to it instead of mass, we get:
    p = E/c

    Therefore:
    p2 = p2 - m2c2

    Therefore:
    m2c2 = 0

    We know c is more than 0, so m is 0

    So, E = mc2/sqrt(1 - v2/c2) will be the only way a particle can have a non-zero mass.
     
  15. Feb 26, 2010 #14

    Dale

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    Hi Raekwon,

    The easiest way to keep this clear in your mind is to distinguish between the mass of an individual particle and the mass of a system. Energy and momentum can be placed together into one single geometrical object, called a http://en.wikipedia.org/wiki/Four-vector" [Broken]. If you do this, then mass is simply the length (Minkowski norm: s² = c²t²-x²-y²-z²) of the energy-momentum four-vector. This is important because, in any interaction, the four-momentum is conserved which, in turn, means that mass is conserved.

    Now, just like the length of the sum of two vectors is, in general, different than the sum of the lengths of the two vectors (http://en.wikipedia.org/wiki/Triangle_inequality" [Broken]), similarly with four-vectors. So the mass of a system (the length of the sum of the particle's four-momenta) is in general different than the sum of the masses of the individual particles (the sum of the lengths of the particle's four-momenta).

    So, one photon is massless, but a system composed of two or more photons may have mass.
     
    Last edited by a moderator: May 4, 2017
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