As a photon has no mass and must always have velocity c, if I were to shine a laser straight up (so Earth's gravity would be pulling straight back on it), what would the effect be on the photon? It wouldn't slow it down nor divert it, correct? My understanding is that it would reduce the frequency of the photon (as it's kinetic energy must be reduced, just as a classical object would lose kinetic energy). If it's the case that only gravitational redshift would occur given this trajectory (and please correct me if I am wrong there), I have two similar questions:

Would not light leaving a galaxy therefore be affected by a gravitational redshift? Is that included when physicists perform calculations regarding the expansion of galaxies away from us (and how accurate could these calculations be, given general estimates of mass distributions, etc., particularly given dark matter's gravitational effects)? If not, could it be that what we now think is a separation of these galaxies is somewhat, primarily, or even completely just light being affected by gravity?

Also, would not light then be able to escape a black hole provided it entered in precisely perpendicular to the event horizon and the black hole was not moving at all orthogonally to the photon's trajectory? (Or, perhaps more plausibly, if a photon is emitted from inside the black hole with a relative velocity of c towards the event horizon.) And then just come out the other side severely redshifted (to a frequency of almost 0 Hz)? I'm familar with the GR equations for gravitational redshift, but it also does not work inside the Schwarzschild radius (as the denominator becomes a square root of a negative number).

Apologies if this is just confused ramblings of someone who knows just enough to be dangerous.

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    $\begingroup$ "would not light then be able to escape a black hole provided it entered in precisely perpendicular to the event horizon". --- Matthew, please think about this a bit. If light could escape, would it be a black hole? From Wikipedia: The defining feature of a black hole is the appearance of an event horizon—a boundary in spacetime through which matter and light can only pass inward towards the mass of the black hole. Nothing, not even light, can escape from inside the event horizon. $\endgroup$ Nov 20, 2014 at 1:36
  • $\begingroup$ That is the common definition, but that doesn't mean the math wouldn't indicate something different. Light moves relative to the black hole at c - if it is not straight away from the center, it will curve into orbit around it, still at c. But if it is moving straight away from the center, it would seem the gravity - at least per GR - would result in redshifting only. At least, if my math is right. $\endgroup$ Nov 20, 2014 at 2:44

2 Answers 2


For the first question: Sure, light emitted by a galaxy is affected by the gravitational redshift, but the effect is small and independent of the distance of the galaxy from us. (See also the question "Why is “gravitational” red-shift neglected in galaxy and galaxy cluster scales?".)

For the second question: Once inside the black hole, you can't emit the photon towards the horizon, because every valid direction either a photon or a massive particle could travel is towards the center of the black hole. In a sense, trying to avoid the singularity once inside the horizon is like trying to avoid tomorrow when outside.

  • $\begingroup$ You say, "[o]nce inside the black hole, you can't emit the photon towards the horizon", but I'm not saying I want to emit the photon towards the horizion (where I agree it would curve back to the center of the black hole), but vertically straight up from the core. A massive particle would, of course, engage in projectile motion - heading up for virtually no time before being pulled in via the black hole's gravity. But a photon does not react this way - it does not matter what the escape velocity is, because it is always going to move at c, no matter what, in every reference frame. $\endgroup$ Jan 30, 2018 at 7:19
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    $\begingroup$ There's an important distinction here: if you emit a photon inside the black hole horizon, it won't go away from the center and then bend and return to the center. In fact, every direction the photon can go is towards the center. But there is an interesting question here: what would happen if a particle were to emit a photon precisely when crossing the horizon, precisely away from the center? Theoretically, the photon would remain stationary on the horizon. See this question. $\endgroup$ Jan 31, 2018 at 6:21
  • $\begingroup$ I’m not sure I follow with the idea that “every direction the photon can go is towards the center”. Such a spacetime geometry would seem to imply also that nothing can “fall in” to the black hole, either, because there’s no curvature that would accelerate a particle that way. To me, it seems more like the event horizon is a cliff, but I also know from special relativity that no matter my reference frame, the light must travel at the speed of light. Putting those two intuitions together, it appears logically that the photon must be capable of escaping the black hole. $\endgroup$ Feb 1, 2018 at 7:14
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    $\begingroup$ It is indeed surprising that inside the black hole horizon every direction a particle can go is towards the center. On the other hand, it is not that surprising that outside the black hole every direction you could go is towards the future. (In general relativity time is just another coordinate - it just so happens that outside the black hole the time-coordinate of any particle always increases. Inside the black hole time loses its usual meaning, but the distance from the center takes a similar meaning as time has outside. Hence, avoiding the center is like avoiding tomorrow.) $\endgroup$ Feb 1, 2018 at 20:44
  • $\begingroup$ You’ll have to help me envision this. To simplify, if I take a 2D grid and warp it, no matter how I do so, every path between two points is reversible. The same is true with a 3D grid. I don’t see a geometric solution that produces this counter-intuitive concept that something could go from A to B, but not from B to A. The “gravity well so deep, you can’t get out at any possible speed” makes sense, but not the idea that all routes head towards the center. Even with time as another coordinate, I’m not seeing this contortion. Are you saying time stops? What’s the actual math? $\endgroup$ Feb 1, 2018 at 21:00

The question presupposes that photons would be emitted from the hard core of the black hole - that is, that they would fly into the air and then fall back in.

The black hole is not black on the inside when looking out. On the contrary, on the inside of the black hole, the sky would be bright.

Any laser pointed into the sky would transfer less energy to the sky than it received from it, therefore the net transfer of energy between the sky and the laser (from a laser of ordinary power) would still be into the laser, not out of it.

Incidentally, this does seem to suggest that we could communicate into black holes if there was appropriate equipment on both sides - by measuring the variance in the amount of energy going into a certain point (since a laser pointed out from inside the black hole would attenuate the absorbtion of energy from outside), though the measurement would have to take place over extraordinary amounts of Earth-time relative to time in the black hole.

The effective refractive index of the internal sky of the black hole would also require a laser there of extraordinary fine focus and alignment.

  • $\begingroup$ The answer is completely incorrect, because its assumption is wrong - if you aim a laser anywhere inside a black hole, that direction is towards the center (and not towards the horizon or anywhere else.) It's just as impossible to aim a laser from inside the black hole towards outside, as aiming it towards yesterday. $\endgroup$ Apr 18, 2020 at 15:24
  • $\begingroup$ @MikeRosoft, if all directions lead to the centre, then how does anything get into the black hole, and where would those things coming in appear to be coming from? Bear in mind that my contention was not that the laser would shine out of the black hole if aimed to the horizon - it was that the net effect would be to exert a modicum of resistance against what was inexorably being drawn in (and this whole scenario must be taken to be entirely hypothetical). $\endgroup$
    – Steve
    Apr 18, 2020 at 15:43
  • $\begingroup$ See the discussion at the other answer. Just like every direction when you are outside a black hole leads towards the future, so when you are inside every direction leads towards the center. And just like when you shoot a laser, you can't hit the target yesterday, so when you do the same inside the black hole you can't hit the target further from the center of the black hole. $\endgroup$ Apr 18, 2020 at 15:49
  • $\begingroup$ @MikeRosoft, it didn't involve hitting a target further from the centre. $\endgroup$
    – Steve
    Apr 18, 2020 at 18:47

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