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E=mc^2 in gravitational fields

  1. Dec 2, 2005 #1
    hi, i have recently been studdying reletivity and the whole e=mc^2, mass energy exchange stuff... and i have a major question (in my mind) that i can't get answered, sincei don't have a teacher, i thought this would be the best palce to come...

    What happens when you start accelerating an electron and dump in in a sun's gravitational field. it gains gravitational potential energy, so what happens now? does alll that gravitational potential energy add to the energy you put into the electron to accelerate it to this speed? how does the mass of the electron change? does it get bigger the instant it enters the field and gravity starts acting on it? or does it gain mass as it starts to fall and pick up kinetic energy?


    please clear this up for me, it's really bothering me, i am able to see that the mass of an electrong doesn't instantly increase when you put it on a spring, but that's spring potential. (i think i just answered my own question). but please give me feed back.
     
  2. jcsd
  3. Dec 3, 2005 #2

    Essentially, potential energy does not add to the energy in the electron. The energy possessed by the electron is only its rest energy (its mass) and its kinetic energy. The potential energy is a quality of the system, not of the electron. A better formula to use is:

    [tex] E^2 = (pc)^2 + (mc^2)^2 [/tex]

    where p is momentum, E is energy possessed by the particle, c is the speed of light, and m is the proper mass (this also works for massless particles that have momentum, like photons. Its really a much better version of the formula). That is only kinetic and rest energy.
     
    Last edited: Dec 3, 2005
  4. Dec 3, 2005 #3

    JesseM

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    Note: you forgot the exponent on the second c, it should be [tex]E^2 = (pc)^2 + (mc^2)^2[/tex], or just [tex]E^2 = p^2 c^2 + m^2 c^4[/tex].
     
  5. Dec 3, 2005 #4

    Oops, you're right. Thanks for pointing out my typo.
     
  6. Dec 3, 2005 #5

    samalkhaiat

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  7. Dec 3, 2005 #6

    samalkhaiat

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  8. Dec 3, 2005 #7
     
  9. Dec 3, 2005 #8
     
  10. Dec 3, 2005 #9

    Physics Monkey

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    The dispersion relation can be modified in the presence of external fields. For example, in the presence of an electromagnetic field, the Hamiltonian of a charged particle (charge q) can be written as
    [tex]
    H = \left( \left(\vec{p} - q \vec{A}\right)^2 + m^2 \right)^{1/2} + q \phi.
    [/tex]
    which you can check has the correct non-relativistic limit of
    [tex]
    H = \frac{\left(\vec{p} - q \vec{A}\right)^2}{2m} + m + q \phi,
    [/tex]
    and reduces to the usual energy relation in the abscence of electromagnetic fields.

    For gravity, one can define the usual action as
    [tex]
    S = - m \int d\lambda \left(-g_{\mu \nu} \frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}\right)^{1/2},
    [/tex]
    which you can approximate in the Newtonian limit as something like
    [tex]
    S = -m\int dt \left( 1 - v^2 + 2 \phi \right)^{1/2},
    [/tex]
    where [tex] \phi [/tex] is the Newtonian potential. If you expand the square root you find the usual action of Newtonian classical mechanics. You can then obtain the Hamiltonian as
    [tex]
    H = \frac{p^2}{2m} + m\phi + m
    [/tex]
    which is the usual result with the rest mass included. Of course, energy and momentum have their own special subtleties in full GR.

    To treat other interactions one would need to know how these interactions couple to the particle. In practice, electrodynamics and gravity are the primary interactions of interest in relativity (outside of quantum field theory). Note that the mass of the particle is the same in each case. In the modern parlance we assign only the concept of invariant mass to a particle. Also, I have not considered the effect of the particle on the fields, etc. so this is a first order treatment and not a completely general description of the system.
     
    Last edited: Dec 3, 2005
  11. Dec 3, 2005 #10

    pervect

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    In SR, adopting the useful convention that c=1, we write that E^2 - p^2 = m^2, as many posters have pointed out.

    What hasn't been pointed out, yet, is that in GR this expression is modified. Given a coordinate system, there is a purely local defintion of energy and momentum, which yields an invariant mass. But the formula for this mass depends on the coordinate system one uses. The coordinate system used determines the metric, and the metric determines the formula for mass in terms of the energy-momentum 4-vector.

    In the case of a diagonal metric, for instance, we can say that

    g_00 E^2 - g_11 p1^2 - g_22 p2^2 - g_33 p3^2 = m^2

    where g_00, g_11, g_22, g_33 are the metric coefficients. E, p1, p2, and p3 here are the covariant components of the local energy-momentum 4-vector of the particle with respect to a specific coordinate system.

    There is a more elegant way to write this expression using tensor notation, which is probably beyond the scope of this short post to explain, but I will present the result anyway.

    [tex]
    m^2 = P^i P_i
    [/tex]

    Under the proper circumstances, GR also has a notion of global mass, but this is a tricker notion. I would suggest that for a non-technical introduction, the best source is the sci.physics.faq at

    http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

    However, it might be best not to worry too much about the global defintion of mass until one has studied GR in depth.

    If one knows how to compute the invariant mass of an energy-momentum 4-vector, and realizes that this is a local quantity, and not a global one, one is off to a reasonable start.

    It is also useful to note that for simple problems involving static space-times, GR has an "energy" that is different from the covariant energy in the energy-momentum 4-vector, which is called by various names, most commonly the "energy-at-infinity", or "effective potential", sometimes it is pointed out that this is the contravariant component of the energy-momentum 4-vector.

    It may be confusing to have different notions of energy flying around, but unfortunately that's the way it is.

    It also makes questions like your (you=cap.r) original quesiton hard to answer - since GR has several notions of energy, your question becomes unfortunately ambiguous in a way that's difficult to sort out.

    For an example of how the notion of "effective potential" works and its application for calculating the orbits of small objects around black holes, see for instance

    http://www.fourmilab.ch/gravitation/orbits/

    This concept is very useful, and easy to apply, but note that it only works in static space-times.
     
  12. Dec 4, 2005 #11

    pervect

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    Here's a short example which may (hopefuly) help illustrate a few of the different notions of energy in GR and their applications.

    Suppose you drop a small body straight into a black hole.

    The mass of the black hole increases not by the invariant mass of the object, but by the ("energy at infinity")/c^2.

    Meanwhile, if we assume that the object is falling straight into the hole (only radial motion) and a Schwarzschild metric, the quantity g_00 E^2 - g_11 p^2 will be a constant, equal to the square of the invariant mass of the falling object (this is a different concept of mass than the energy-at-infinity!).

    Note that g_00 is ill-behaved at the event horizon. The invariant mass of the falling object remains positive and constant, but you wouldn't guess that from the formula given that g_00 ->0. This is an artifact of the bad behavior of the Schwarzschild coordinates.
     
  13. Dec 5, 2005 #12
     
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