Experiments on the constancy of the speed of light with a moving detector

Question

Are there experiments that verify the constancy of the speed of light independently of the speed of the detector, where the moving detector’s frame of reference is not the earth frame of reference as in Michelson–Morley experiment? For example, the light source is attached to earth, and the detector is moving with respect to earth.

Answer 1

As far as I know there is not an experiment explicitly designed to detect the speed of light with a detector moving with respect to the earth. However, there are two types of devices that I know of whose design and operation is based on the constant speed of light for objects moving relative to the earth.

The first is GPS. The GPS system relies very heavily on the 2nd postulate. All of the elements of the system are moving in the inertial reference frame that is used. Furthermore, GPS receivers in vehicles are moving relative to the non-inertial reference frame of the surface of the Earth. So this involves both the invariance of light speed and moving detectors.

The second is a free electron laser. In this highly relativistic electrons produce stimulated emission of radiation. As such the lasing electrons are both emitters and detectors. This device is a recognized experiment that depends on the invariance of c where the detectors are highly relativistic.

Answer 2

What experiment would you accept to measure that lightspeed doesn't change?

For example, Michaelson-Morley did not show that light always has the same speed. If you look at how interference patters are made, it's strictly from geometry. No matter how fast light moves, if it has the same wavelength it will make the same interference patterns.

What they showed was that in their case the light did not change speeds when it was reflected into different directions. It went the same distance in different directions and the interference pattern did not change.

How did they know it was the same distance? They measured it with an inteferometer and it came out the same.

If the ether was moving, then light would move at different speeds in different directions, and if you compared the speeds when the ether moved in a different direction than when you did the interferometer calibrations, you'd get a different result.

But light that moved at a different speed all the time, but kept the same wavelength, would look the same. Of course, for that to happen the frequency would have to change. And how can you change the frequency of light while keeping the wavelength the same? The source and the observer would have to have clocks that ran at different speeds, and of course they knew that was impossible!

One way to measure lightspeed is to measure the wavelength and the frequency at the same time. That used to be inconvenient. If the frequency was slow enough to measure, the wavelength would be inconveniently long, and vice versa. That's why they came up with other methods.

I remember reading about somebody who figured out how to inscribe a whole MM experiment onto a single quartz chip, and then they rotated and revolved and accelerated their chip various ways to demonstrate that there was no difference. I didn't find that now in a quick search, but here's something.

Answer 3

The Michelson-Morley experiment is a 'double' Doppler effect.

One of the arms of the interferometer which moves along the axis of $(x)$ is composed of two mirrors separated by a distance L, the light which separates on the first mirror arrives at the second in a time $T'_{1,2}=L'_{1,2}/c$, According to the Lorentz transformations:

$$ \Delta x'=L'_{1}=\gamma(\Delta x-v\Delta t )$$ $$ L'_{1}=\gamma(1-\beta )L=\sqrt{1-\beta^{2}}\,\frac{L}{1+\beta} \;\;\;\;(1)$$

The same thing for the return of the reflected light: $$ L'_{2}=\gamma(1+\beta )L=\sqrt{1-\beta^{2}}\,\frac{L}{1-\beta} \;\;\;\; (2)$$

One recognizes the hypothesis of Lorentz:

$$ L'_{1}+L'_{2}=\sqrt{1-\beta^{2}}\,\left(\frac{1}{1+\beta}+\frac{1}{1-\beta} \right )L $$

$$ L'_{1}+L'_{2}= \;\frac{2L\,\sqrt{1-\beta^{2}}}{1-\beta^{2}} \;\;\;\; (3)$$

The multiplication of (1) and (2) gives : $$L'_{1}.L'_{2}=L^{2}$$ which is demonstrated here for the 'double' Doppler effect.

To get (3) we have made the sum of (1) and (2), if there is something that does not work, it is that the Doppler effect does not hold the road too.

Answer 4

Yes, I performed an experiment in which the light source- an Eveready R3000 torch- was firmly clamped to a stainless steel upright embedded 1432mm into the ground (ie the Earth frame of reference), while the detector was Gorilla-Taped to a bracket on the front of my Mercedes CL420. I measured the elapsed time for a pulse of light emitted by the torch to reach the detector 600 times at each of three closing speeds measured by the speedometer of the Merc, namely 20mph, 40 mph and 60mph. The results were exactly in accordance with the predictions of special relativity to within nine standard deviations, and were a glowing tribute to the prescience of A. Einstein.

Answer 5

Even a detector is moving with some speed to earth, it is adding some constant to earth's frame. Question is about constancy of speed of light in relative motion. Michelson-Morley's experiment was one such attempt, but failed. It failed to detect any fringe shift caused by relative speed of earth with aether. But it was interpreted as, there is no aether and speed of light is invariant in relative motion.

While I think that there is no way to determine motion of an inertial frame unless it's compared with other frame having absolute motion. An inertial frame is absolute with in itself. This is also postulate of theory of relativity, there is no preferred inertial frame. So how could an experiment on earth detect its own motion. So outcome of this result is not that there is no aether but there is no relative motion between earth and aether detected.

Reasons are numerous for failure or negative outcome of the experiment. First, the speed of light is not relative to aether. Second, there might be aether drag. Third, that light is not invariant in relative motion. Fourth, there is no motion of earth at all. We again revisit the experiment with classical relativity that is addition of velocities and see what it says. Also, if speed of light is constant then there should fringe shift, because there was relative speed between light and interferometer's arms, if there is motion of earth.

Let length of both arms were equal as given in numerous explanation of said experiment and was, $l$. Speed of interferometer was $v$ and speed of light independent of source is $c$. Then time taken by light in perpendicular arm trip is given by twice the time taken by light to travel distance of arm length which was elongated due to motion.

Length of elongated light path is, $l'=l/\cos{\theta}$, where $\theta$ is angle between interferometer's arm and light path. Similarly speed of light on light path is, $c'=c/\cos{\theta}$, where $\cos\theta=c/\sqrt{c^2+v^2}$

So total time taken for round trip is,

$t_\bot=\dfrac{2l/\cos\theta}{c/\cos\theta}=\dfrac{2l}{c}\tag 1$ It is clear that speed in perpendicular direction remain uneffected of relative motion, it doesn't mean speed remain same.

Now time taken by light for round trip in parallel direction is consists of two parts. One is time taken by light along the direction of interferometer's motion, $t_1$, and other is time taken by light against the motion, $t_2$. For $t_1$, speed of light is $c+v$ instead of $c$ and arm of interferometer moves $v\ t$ length during light travel distance $l$, so, $l+v\ t_1=(c+v)\ t_1 \implies t_1=\dfrac{l}{c}\tag 2$ Similarly for $t_2$, speed of light is $c-v$ and interferometer's arm travel distance $v\ t$ against the light's motion, so $l-v\ t_2=(c-v)\ t_2\implies t_2=\dfrac{l}{c}\tag 3$ Therefore total time taken in parallel direction is, $t_\parallel = t_1+t_2=\dfrac{2l}{c}\tag 4$

For there is to be fringe shift, there must be path difference between two lights to interfere, and for that there should be time difference $\Delta t$ for given arrangement. $\Delta t=\left\| t_\bot - t_\parallel \right\| = 0\quad \text{from (1) & (4)}\tag*{}$

So it is shown that there was no phase difference between two lights interfering in interferometer if treated with classical realtivity in which light not remain invariant to relative motion. Thus, this experiment was suffice to prove that there is no constancy of speed of light. If speed of light remain invariant then there should be fringe shift due to relative motion between interferometer and light, which was not detected. This proves that no further ecperiment needed to prove effect of relative motion on speed of light if treated with right approach.

Annexure

This is to view that what would be actual results if light is treated as constant as per lorentz's transformation in Michelson-Morley's experiment. As said above that if speed of light remain constant in relative motion, then there was fringe shift observed but not. All the calculation from aether's frame, as that can only make possible relative speed of earth with aether.

Now time taken by light for round trip in perpendicular arm of inteferometer to motion is given by, $t'_\bot=\dfrac{2l}{c}, \text{but}\ t'_\bot=\gamma t_\bot\implies t_\bot=\dfrac{2l}{\gamma c}\tag 5$ where $\gamma=1/\sqrt{1-\frac{v^2}{c^2}}$ is lorentz's factor, and $t'_\bot$ is time measured for round trip in proper frame.

Similarly, time taken in parallel arm of motion has encounter length contraction and time dilation both as round trip, thus $t_\parallel=\dfrac{2l'}{\gamma c}, \text{but}\ l'=l/\gamma\implies t_\parallel=\dfrac{2l}{\gamma^2c}\tag 6$

For there to be fringe shift, there must be time difference $\Delta t$ between time taken by light in perpendicular and parallel arms of interferometer, (5) & (6). So $\Delta t=\left\| t_\bot-t_\parallel \right\|=(\gamma-1)\dfrac{2l}{\gamma^2 c}\tag*{}$ This shows that there must be fringe shift if there is constancy of speed of light in relative motion as per Michelson-Morley's arrangement, if earth is moving.