Image that a photon is emitted from a sphere, moving orthogonally away from the sphere's surface. As the weight of this sphere increased, how would the increasing gravitational force affect the motion of the photon? Imagine that the mass of the sphere was sufficiently high to form a black hole - but the photon is emitted outside the black hole's event horizon - what happens to the photon then?
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2$\begingroup$ It gets redshifted, therefore it loses momentum. Other than that, it just moves forward in a straight line. $\endgroup$– WoodCommented Oct 18, 2016 at 9:08
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$\begingroup$ So as the force approach infinity, would the only thing that occured be the photon losing energy? $\endgroup$– HonwangCommented Oct 18, 2016 at 9:14
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$\begingroup$ Additionally, where does this energy go? $\endgroup$– HonwangCommented Oct 18, 2016 at 9:14
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2$\begingroup$ I think it just goes to gravitational potential energy, just like any massive particle. $\endgroup$– WoodCommented Oct 18, 2016 at 9:33
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Photon momentum does not correspond to its velocity, but its frequency, so if you applied a force opposite to its momentum, it would not slow down, but redshift.
As photons carry no charge, the only force applicable is gravity, and its general relativistic description is more kinematical than dynamical (the photon gets redshifted because the relative orientation of the velocity vectors of photon and observer changed while moving along a straight line through curved spacetime - no forces anywhere in sight).
As to where the energy goes, the classical answer is into the gravitational potential. The relativistic answer is a bit more complicated as it's not possible to associate an energy density with the gravitational field. Morally speaking, it's arguably nevertheless the right idea.
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$\begingroup$ Thanks - as the gravitational force got sufficiently high, I'm guessing the photon would continue to redshift until it has no appreciable frequency - so it would have no energy. Is this a safe assumption to make? $\endgroup$– HonwangCommented Oct 18, 2016 at 12:16
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$\begingroup$ @Honwang: more or less; the gravitational frequency shift when increasing the distance from the center of mass from $r_0\to r$ is given by $1+z = \sqrt{(1-r_s/r)/(1-r_s/r_0)}$, which goes to infinity at the Schwarzschild radius $\endgroup$ Commented Oct 18, 2016 at 16:38