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First of all, I know that light does indeed travel slower in a medium like air or water, but that's because the photons are bouncing off of the medium's particles and in different directions so the light beam as a whole is slowed down (or at least that's the classical explanation I've read somewhere).

But my question is about light traveling freely in the vacuum.
What if an astronaut during a spacewalk switches on a flashlight towards the opposite direction he is moving. Will that light travel slower than normal?

If not, is there another way to make light move slower?
For example, can you make a beam of light move so slow that you can visually perceive it being affected by gravity?

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To address the specific question about an astronaut and the flashlight: special relativity guarantees the speed of light will always be $c$ in this case. –  Javier Badia Sep 10 '13 at 22:10
    
related: physics.stackexchange.com/q/76492 –  Ben Crowell Sep 10 '13 at 22:23
    
also related: physics.stackexchange.com/q/76240/28528 –  Owens Sep 10 '13 at 23:24
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BTW I dont believe it's accurate to say that light slows down because it's bouncing around from particle to particle in different directions. I commented as much in the previous discussion. If this was correct, photons would emerge from the medium in all random directions and at varying apparent speeds, which is not how it works. –  Owens Sep 10 '13 at 23:28
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5 Answers

up vote 3 down vote accepted

Not only is the constant nature of the speed of light guaranteed by theory, it is also shown experimentally. In fact, as you may know, it was the experimental discovery that the speed of light is constant irrespective of the (inertial) frame of reference which formed the inspiration for the development of special relativity by Albert Einstein.

Mathematically, purely from the relativistic formula for velocity-addition it can be seen that the light would still travel at a speed $c$ in a vacuum. Indeed, say the astronaut has a speed of $-v$ with respect to the ground. He observes the light leaving his flashlight with a speed of $u = +c$ (with respect to him).$^1$ The relativistic theory then tells us the light is travelling at a speed $s$ with respect to the ground, given by

$$\begin{align} s &= \frac{(-v)+(+c)}{1+\frac{(-v)(+c)}{c^2}} \\ \\ &= \frac{c-v}{1-\frac{v}{c}} \\ \\ &= \frac{c-v}{\frac{c-v}{c}} \\ \\ &= \frac{c-v}{c-v}c \\ \\ s &= c. \end{align}$$

You can walk at whatever speed you like, you'll never get a different result. Except if you insert $-c$ instead of $-v$, then the answer is undefined - however, you do still get $c$ if you calculate it using limits.


$^1$ Why can we say this? Well, remember that in the reference frame of the astronaut, he himself is not moving at all and he would expect the light to leave his flashlight at a speed of $c$ (w.r.t. him). This becomes clearer when you replace the flashlight with a small cannon and the light with a little ball.

Suppose you can set the exit speed for this ball out of the cannon. Say you set it at $v$. When you walk around carrying this small cannon, you expect the ball to still leave the cannon at a speed $v$ with respect to you when you fire it. Classically, if you’re walking at a speed $u$ with respect to the ground, the ball will leave the cannon at a speed $u+v$ with respect to the ground.

Relativistically, the only thing that is invalid in all the above is this simple summation of speeds. You need the formula I used in the main text of this post, which reduces to the simple summation if both $u$ and $v$ are much smaller than $c$. So it’s perfectly alright to say that the light leaves the astronaut’s flashlight at a speed $c$ w.r.t. the astronaut, even without knowing about the constancy of the speed of light in a vacuum.

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I'll accept your answer because you mention the fact that this is supported on empirical observations and not just somebody's theory. However, you say "He observes the light leaving his flashlight with a speed of u=+c" like it was an obvious fact, but that's exactly my question: why light's speed is not reduced like a tennis ball would. I guess the answer is... just because it's been seen experimentally that light behaves that way. –  GetFree Sep 14 '13 at 0:27
    
@GetFree That’s the idea special relativity is built on: the – indeed experimental – fact that the speed of light in a vacuum is constant and equal to $c$ in all inertial frames. But we can say that the astronaut observes the light coming out of his flashlight at a speed $c$ (with respect to him) without needing that constancy. I’ve edited my answer to hopefully make that clear. –  Wouter Sep 14 '13 at 11:42
    
But if you do know about the constantness of light's speed, then you would think... if we flash a light backwards, and we are moving the other way at speed v, then w.r.t us the light should be moving at speed c + v. But evidence says otherwise. It's the perception of the speed what's constant, not the speed per se. –  GetFree Sep 15 '13 at 7:32
    
@GetFree No, remember again that w.r.t. yourself, you are never moving. So $v$ in your own reference frame is always $0$. Therefore the light will - in your own reference frame - move away from you at speed $c$ regardless of the constancy of the speed of light. The strange thing is that it also moves at speed $c$ in the reference frame of the ground, while you are moving at speed $v \neq 0$ w.r.t. to that reference frame (where $v$ can be positive or negative). Classically, you would expect it to move at speed $c+v$ w.r.t. this reference frame (but not w.r.t. your own!). –  Wouter Sep 15 '13 at 7:56
    
I'm being misunderstood. What I'm saying is that if you know about the constantness of light's speed and if you wrongly assume that that constantness is w.r.t a fixed motionless point (like I initially did) then you would think that light should be measure at different speeds depending if you are moving of not. So my conclusion is that it's not the speed itself what's constant but the perception of it. –  GetFree Sep 16 '13 at 8:05
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There would be frequency shift (color-shift) in the light of the torch relative to a stationary observer due to the Doppler effect. However, each instantaneous "piece" of light would still emanate from the torch at the same velocity (C, the speed of light) and travel from its instantaneous point of emanation at that constant maximum speed.

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light is electromagnetic wave in essence. $c_0^2=\frac1{\varepsilon_0\mu_0}$ where $\varepsilon_0$ is the dielectric constant and magnetic dielectric constant in vacuum. so, it doesn't matter whether the astronaut turns on the flashlight towards any direction. the speed is determined by the natural property of medium or vacuum the light travels in. So, the answer to your question is "No".

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The velocity of light is locally invarient, that is if you measure it at your position you will always get the value $c$. The other answers explain why in your example of the astronaut with the torch we would still measure the speed of light to be $c$.

However in General Relativity the speed of light is not globally invarient. For example if your astronaut is hovering just above the event horizon of a black hole and shines his torch outwards we would see the light initially travelling at less than $c$ and accelerating towards $c$ as it gets farther from the black hole. At the event horizon itself we would see the speed of the light slow to zero. See this Wikipedia article for some useful links, though the article itself doesn't say much more than I've said here.

You need to be a bit cautious attaching physical relevance to this, because any velocities we measure is not invarient quantities in GR. The proper velocity of light is an invariant and remains $c$ even at the black hole event horizon. Still this is an example of how the speed of light can be changed and the light can be affected by gravity.

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What if an astronaut during a spacewalk switches on a flashlight towards the opposite direction he is moving. Will that light travel slower than normal?

In Newtonian mechanics, object A is moving at velocity $u$ relative to B, and B is moving at $v$ relative to C, then A's velocity relative to C is simply $u+v$. But in special relativity, the equation is $(u+v)/(1+uv)$ (assuming that $u$ and $v$ are both relative to the speed of light, i.e., the units are such that $c=1$). If $u=1$, then the result of the combination of velocities is still 1.

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