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Where does binding energy, or internal kinetic energy exist?

  1. Mar 22, 2012 #1
    If I have 2 electricly charged bodies orbiting one another, the binding energy must contribute to the gravitation of the system. Their reletive KE must also. So the total gravitational mass of the system is greater then the sum of the indivual objects. If I used a laser and gravitational lensing to map mass distribution of the system where would I find the extra mass?

    Next question: Does gravitational binding energy work the same way? How large is the gravitational binding enery and the internal kinetic inergy of a galaxy in comparison to the missing mass-energy of dark matter?
     
  2. jcsd
  3. Mar 22, 2012 #2

    Mentz114

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    There does not have to be any extra mass. The energy and momentum will also gravitate. So will the electric field, in its own right. The energy-momentum tensor of static charged matter includes terms for the mass and the electric field.
     
  4. Mar 22, 2012 #3
    That's right but the sign of binding energy depends on the interaction. For example, binding energy of helium nucleus is negative (you gain energy by fusing hydrogen into helium) while binding energy of uranium is positive (you gain energy by fission instead). With gravity, binding energy is always negative, so a bound system has less energy than an unbound system. This is because gravity always sucks.

    As a side note, if the binding energy would be so massive that it could explain dark matter, gravity would be entirely unstable, because the binding energy would feed to itself, basically making all massive objects have infinite energy.
     
  5. Mar 22, 2012 #4

    pervect

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    There's a consistent definition of energy for the electrically bound system - you'll find that the total energy is the kinetic energy plus the energy stored in the electromagnetic fields - in geometric units said field energy is proportional to E^2 - B^2. In the case where B=0, I believe this expresssion become 1/2 epsilon E^2 in non-geometric units (from wiki).

    If you have point charges you have the usual problem with the infinities of the electromagnetic field energy - I'm not quite sure how this is dealt with, except by avoiding point particles.

    You can't, in general, localize the gravitational binding energy the same way you did the electromagnetic energy. If you are in an asymptotically flat space-time you'll be able to define a conserved energy which should be the ADM and/or Bondi energy of the system (and as far as I know the two energies should be equal if there are no gravitational waves). But there won't be a single, consistent way to localize this in terms of an energy density.

    Being in an asymptotically flat space-time is more-or-less necessary to carry out your experiment anyway, or at least very highly convenient. Asymptotically flat space-time means that there isn't any effect from the rest of the universe. If you restrict yourself to a small enough area, you can probably use the asymptotically flat calculations as a reasonable approximation to what you actually measure.

    A coarse scan will show, from the outside, a gravitational field that looks like that of a single rotating mass with the appropriate energy - I believe the system will have a metric that looks pretty much like a Kerr metric.

    A finer scan will reveal the details with two orbiting particles.

    You'll be able to predict light deflections - but there won't be a unique intepretation of them as an energy density that transforms in a covariant manner, alas. For instance, pseudotensor methods will give results that are both non-covariant and non-unique.
     
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