For us to measure any movement, the "something" has to have a different position to some reference frame. now speed is defined by the amount of changed position( which we can tell by the reference frame) inside a certain amount of time. now assuming light moves at a certain speed, now has a reference frame implied by the very definition of speed. Following this, anything that has displacement over time can so on measure different speeds depending on the frame. However the "speed of light" does not follow these same rules in logic, so my question would be why call it speed when it clearly devies what speed is supposed to represent?
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2$\begingroup$ If you send a flash of light (using e.g. a flashlight or a laser) from a source to a camera sensor that is placed a certain distance away from the source, the camera sensor doesn't react until a few moments (that we can measure!) after the source was turned on. How do you interpret that? $\endgroup$– Marius Ladegård MeyerCommented Oct 2, 2023 at 12:24
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2$\begingroup$ "however, it doesnt move relative". I don't understand this statement. The fundamental postulate of special relativity is not "light does not move relative to anything". It is "in any inertial frame of reference, light in vacuum moves at speed $c$ relative to any observer." $\endgroup$– Marius Ladegård MeyerCommented Oct 2, 2023 at 12:36
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1$\begingroup$ Maybe you are confusing this with the fact that light does not propagate through any medium, like the old aether concept...? $\endgroup$– Marius Ladegård MeyerCommented Oct 2, 2023 at 12:38
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1$\begingroup$ "it shouldt travel any distance. because the definition of moving is always relative." This is a non-sequitur. From the fact that motion is relative we cannot conclude that it doesn't travel any distance. In fact, there is no frame where that is true. $\endgroup$– DaleCommented Oct 2, 2023 at 12:55
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2$\begingroup$ BTW, @michaeloppenheimer is asking for clarification of the question. To me it doesn't sound like an insult. It sounds like reframing the question or trying to understand what you mean by the terms you use. $\endgroup$– garypCommented Oct 2, 2023 at 13:31
2 Answers
You are quite wrong to say that light does not move relative to anything. If you and I stand together and flash a light, it moves relative to both of us at the speed c. If you walk at a metre per second in the direction of the light, then light moves at a speed c relative to me and at a speed c relative to you. The fact is that light moves at a speed c relative to everything, which is not the same as your claim that it moves relative to nothing.
The speed c is about a foot per nano-second. If you were to flash a light at a detector a thousand feet away, it would be detected after a microsecond. There is no conceptual difficulty in defining the speed relative to you- it is simply the distance travelled by the light in your frame divided by the time taken in your frame. If were to speed past you at 0.5c toward the detector just at the moment you flashed the light, the speed of light would be c relative to me too. In my frame the light would have travelled less than 1000 feet to reach the detector, and it would have taken less than a microsecond to do so- the distance and the time would both be reduced in my frame so that dividing the former by the latter would still be c.
Conversely, if I sped past you at 0.5c in the opposite direction just as you flashed the light, then in my frame the light would travel a longer distance to reach the detector and take a longer time, but again the ratio between the two would still give the same speed, c, in my frame.
So the speed of light is the same in every frame, and is simply the distance travelled in any given frame divided by the time taken in that frame.
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$\begingroup$ Thank You for your answer! $\endgroup$ Commented Oct 2, 2023 at 22:11
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$\begingroup$ However I still have a few questions; For us to measure any movement, the "something" has to have a different position to some reference frame. now speed is defined by the amount of changed position( which we can tell by the reference frame) inside a certain amount of time. now assuming light moves at a certain speed, now has a reference frame implied by the very definition of speed. Following this, anything that has displacement over time can so on measure different speeds depending on the frame. $\endgroup$ Commented Oct 2, 2023 at 22:36
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$\begingroup$ However the "speed of light" does not follow these same rules in logic, so my question would be why call it speed when it clearly devies what speed is supposed to represent? $\endgroup$ Commented Oct 2, 2023 at 22:43
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$\begingroup$ One of the counterintuitive things about special relativity is that there is no inertial reference frame in which light is at rest. Light moves in every inertial reference frame. You can measure the speed of light in any inertial reference frame. It is always moving at c. You can run faster and faster to try to catch up to light. You never get any closer to catching it. See A photon travels in a vacuum from A to B to C. From the point of view of the photon, are A, B, and C at the same location in space and time? $\endgroup$ Commented Oct 2, 2023 at 23:37
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$\begingroup$ But light does follow the same rules as all other speeds! The relativistic law of velocity addition means that speeds do vary between reference frames, but they vary less and less as they approach c, and c just happens to be the limiting case. $\endgroup$ Commented Oct 3, 2023 at 5:21
This question really shouldn't receive down votes (punctuation notwithstanding) since it is entirely logical for Galilean relativity. You fire a laser pulse with wavenumber $\vec k$, and then just boost along $c\hat k$, et voila: the laser is stationary.
Of course Galilean relativity is wrong, and there are no reference frames at $c$, and we need to use Lorentz transformations.
Back to the question: distance over time. Say you fire the laser at
$$ E_0 = (ct_0=0, x_0=0) $$
and then detect at the end of your length $L$ table:
$$ E_1 = (ct_1 = cL/c = L, x_1=L) $$
and measure the speed:
$$ v_S = \frac{x_1-x_0}{t_1-t_0} = \frac{L}{L/c} = c $$
($S$ is the lab frame).
Now boost to a frame moving along $x$ at $v$:
$$ E_0 = (ct_0'=0, x'_0 = 0) $$ $$ E_1 = (ct_1' = c\gamma[t_1-\frac{vx_1}{c^2}], x'_1 = \gamma[x_1-vt_1]) $$ so $$ v_{S'} = \frac{x_1-x_0}{t_1-t_0} = \frac{\gamma L}{\gamma L/c} = c $$
No problem. Everyone agrees, $c$ is the limit.
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$\begingroup$ Thank you for your explanation and comment on down votes, thanks for that !however how did you go from saying cL/c to just L/c,. for the calculating speed Vs $\endgroup$ Commented Oct 2, 2023 at 22:56
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$\begingroup$ and also using the Lorentz transformations and so using gamma, doesnt that assume the speed of light to be constant. Thank you $\endgroup$ Commented Oct 2, 2023 at 22:58
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$\begingroup$ 2) it assumes relativity, so yes. 1) I didn't do the math, I just went from memory, so check for mistakes when you work it out on your own. $\endgroup$– JEBCommented Oct 3, 2023 at 13:53