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Einstein's Box

  1. Jan 16, 2004 #1
    In 1906 Einstein utilized the center-of-mass theorem to derive an expression for the inertia of energy and an expression for the center of mass vector. The paper consisted of two parts the first of which was a spring board for the second part. The first part was entitled "A special case" and the second part was entitled the "On the Principle of the Conservation of the Center of the motion of the center of gravity"

    I've created a new web page which describes Einstein's derivation. The page also describes one modification given by Antipa in the Am. J. Phys. I've made one slight modification of his arguement by using the principle of the conservation of (relativistic) mass.

    This derivation is commonly known referred to as "Einstein's Box"

    http://www.geocities.com/physics_world/sr/einsteins_box.htm

    Caution: Please don't confuse the center of mass "frame of reference" with the center of mass "vector." Einstein's work as well as Antipa's, mine and anyone else who describes this work of Einstein's pertains to the center of mass *vector*
     
  2. jcsd
  3. Jan 19, 2004 #2
    So if you take Einstein's box, and put two photons in it, going from opposite sides, would the result be different? Now boast that frame. Are the two photons still in opposition? What changes as the box goes relativistic speeds to the observer?
     
  4. Jan 19, 2004 #3
    Which result are you refering to? Einstein assumed that the box was a rigid body and the photon gave a kick to the box. The box had to move otherwise momentum would not be conserved. The mass of the box moving was then used with conservation of the center of mass and the result of the calculation was m = E/c^2/. If you have two photons kicking then not momentum is transfered to the box. However more modern versions of this experiment don't assume the box is rigind - e.g. French's example of Unhinging Einstein's Box etc.

    In your example you're using a situation which is a bit different. In the example I mentioned with your two photons it is assumed that the photons kick at the same time. But under such a boost that you suggest, presumably parallel to the direction paralle to the photons motion, simultaneity is not conserved.

    See
    http://www.geocities.com/physics_world/photon_in_box.htm
     
  5. Jan 30, 2004 #4
    Actually, this question also bothered me in a different way. I did not bother about how to derive the effect like Arcon did.

    I was bothered by that when two photons in exact wave length and different phase actaully will supercede each other and look like no photons or light waves at all.

    But since there is a phase difference, I would guess their original emited time need to be different. Will this resolve the issue?

    Or if, we carefully do it, maybe we can emit two EM waves at the same time but their phases will be completely in reverse and cancel out the EM waves and make no EM effects on a conduitnearby. Is this possible?
     
  6. Jan 30, 2004 #5

    russ_watters

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    Staff: Mentor

    Thats the basics of interference, Sammy. I don't think it can be made to work in the way you suggest, but it does work.
     
  7. Feb 8, 2004 #6
    Arcon,

    In reading your "Gravitational Force", I found you lower the force index in your EQ (1), so it now seems to be a 1-form. Also, comparing it with Sean's lecture note, the Chris.. connection is a plus item not a minus item. Is it a result of lowering the index?
     
  8. Feb 8, 2004 #7
    Yes. The reason is because this must hold in all limits, especially in the weak field limit. When you take the derivative of the Lagrangian with respect to xu the result is the generalized force and such a force is neccesarily a 1-form. In the weak field limit you treat the Lagrangian as a scalar field on a flat spacetime background. Taking the derivative with respect to xu produces a 4-force in that sense. That force is a 1-form. This also makes it consistent with how the generalized 4-force is defined in special relativity.


    I don't see what you mean. Perhaps you're confusing the sign in the covariant derivative with the sign in the Christoffel symbol?

    If V is a vector and p is a 1-form then

    [tex] V^{\alpha}_{;\beta} = V^{\alpha}_{,\beta} - V^{\mu}\Gamma^{\alpha}_{\mu \beta} [/tex]

    [tex] p_{\alpha;\beta} = p_{\alpha,\beta} - p_{\mu}\Gamma^{\mu}_{\alpha \beta} [/tex]
     
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