Is the speed of light in vacuum always the same value?
Solution 1
As far as we can tell, the local speed of light in vacuum is indeed constant.
Photons don't slow down or speed up as they fall into or rise out of a gravity well. However, just as a massive object's kinetic energy changes as the object falls into or rises out of a gravity well, photons also gain or lose energy. In the case of photons, this energy change manifests itself as a change in frequency (or wavelength) rather than a change in velocity.
Solution 2
No, in perfect vacuum, photons do not slow down. Although, gravity of massive objects like stars or planets can bend the trajectory of photon (the Theory of General Relativity) like a lense.
If you are referring to the fact that Black Hole is black because no photons can escape its massive gravitational force and you thought it is because the gravity of the black hole is slowing photons down, which is not the truth.
An analogy:
The stronger the gravitational force of a body, the faster space around that body is falling into the body itself. For a black hole, the closer any photon get to it, the faster the space around the black hole falling. Eventually, there is a point where the rate of the falling is equal to the speed of light. That point is called the EventHorizon. That is why no photon can escape, and the black hole is, well, black.
Solution 3
If you with your rods and clocks are in free fall (ie: your metric is the Minkowski diag(1,1,1,1) ) in a vacuum and the light ray passes near you, you will always measure the standard speed c= 2.99792458 E+8 m/sec.
However, the speed of light is observed to be different if the observer and his rods and clocks are in a different gravitational environment than the light. One experimental measurement of this is the Shapiro delay (http://en.wikipedia.org/wiki/Shapiro_delay). The time for a radar pulse to travel from earth to Venus and back is longer when the sun is near the path compared to when the sun is not near the path (and therefore the observed speed of light is slower). The increased time is not just due to the increased length of the slightly bent path. In fact the bending of the light can be calculated from the part of the wave front away from the sun going a little faster than the part of the front near the sun … just like is done for the bending of light at the interface between materials with different indices of refraction in optics. If we were in free fall near the path of the radar photon (instead of at the earth’s position in the sun’s gravitational potential), we would measure the standard speed c= 2.99792458 E+8 m/sec using the same standard rod and clock we had carried from earth after doing the previous measurement.
As a result of special relativity, we are used to hearing “the speed of light is a constant in all reference frames”. The statement is actually “the speed of light is constant in all frames related by a Lorentz transformation”. Lorentz transformations include spatial rotations and boosts which leave invariant the (1,1,1,1) diagonal metric. However, this metric is not left invariant by the transformation that moves to a different gravitational potential. If the metric becomes the Schwarzschild metric, $$ ds^2=(12GM/r)(cdt)^2+(12GM/r)^{1}dr^2+r^2(d\Theta ^2 + sin^2(\Theta) d\Phi^2) $$ the free falling observer, infinitely far from the central mass M, observes the speed of light (defined by ds=0) in the direction $dx=rd\Theta$ to be $$ dx/dt = c * (12GM/r)^{1/2} $$
Solution 4
The answer to your question depends on fine definitions.
Locally the speed of light is always the same; more precisely, the universal, Lorentz invariant speed $c$ (which is also the maximum speed of a causeeffect relationship and experimentally observed to be the same as the speed of light) is constant. This means that any measurement of light speed in any laboratory with small enough time and spatial extent[1] will always yield the same value $c$;
Globally the speed of light can vary. A light clock's period can indeed be observed from afar to vary by an observer at a different point in a gravitational field: witness  for example  the varying $g_{0\,0}$ term in the Schwarzschild or Rindler metrics.
However, the failure of light to escape a black hole is not well thought of in terms of Newtonian notions, although if you calculate the Schwarzshild radius with Newtonian physics you will indeed get the right answer! The energy of the light decreases as described by Newtonian physics, but this manifests itself as a redshift rather than a local "slowing" as discussed in David Hammen's answer. In Newtonian mechanics, the analogous concept to a black hole is called a Dark Star and, in this paradigm, you can escape a Dark Star by climbing a rope dangled from a passing spaceship. You cannot do the same thing from a Schwarzschild black hole without going backwards in time.
[1] Here we mean in the Weierstrassstyle limit sense: as we make our laboratory smaller and smaller so that the spacetime manifold appears more and more like a small chunk of flat Minkowskian spacetime, the limiting result in this thought experiment is always $c$.
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axelonet
Updated on August 01, 2022Comments

axelonet 5 months
The escape velocity of different planets and stars vary. If they vary, the velocities of bodies escaping from the respective stars or planets should also vary.
Like, if I want a ball to reach 10 meter height, I should throw with different velocities on earth compared to some other planet. As light (say, a photon) crosses many stars and planets, doesn't its speed vary compared to another photon?

Qmechanic over 7 yearsRelated: physics.stackexchange.com/q/2230/2451 and links therein.
