# C as the speed of light and the speed of sound in Michelson-Morley.

Tags:
1. Jun 29, 2014

### GerryB

The symbol, c, represents both the speed of light and the speed of sound in most scientific reference texts. Can the speed of sound be substituted for the speed of light in the Michelson-Morley (MM) formula: T = [L / (c - v)] + [L / (c + v)]?

Consider an observer on a train of length, L. It is moving with a constant velocity, v, along a level straight section of track. Can this observer find the train’s velocity relative to the embankment (Earth), with a single clock? That is, if there is an inertial reference frame attached to the moving train, and another inertial reference frame, with another observer, attached to the Earth (which is considered at rest), can the caboose observer find the train’s relative velocity? Will this value be in the form of the simplest law of motion: v = d / t, or in a form that violates the classical principle of relativity? Alternatively speaking, will the two observers, one in motion and one at rest, measure two different values (via Galilean transformation) for the velocity, v of the train. Or, is it possible for them to measure the same value, v?

It is a windless day (air, medium at rest). The observer is in the caboose with a lantern to signal the engineer in the locomotive. When he sees the flash of light (effectively instantaneous at this distance), he blows the whistle. She starts her clock at the same moment she sends the light flash, and thusly she can measure the time, t, for the sound wave of the whistle to travel along the length, L. When the sound wave reaches her, she flashes the lantern once more. An observer at rest on the platform sees the first flash and the second flash, and with his clock measures the time between these flashes. Will each observer measure the same time, t, between the two flashes?

If the platform observer has a means to establish the distance between the flashes (by some landmarks running parallel to the track, for example). The length of the train is obtained from the specifications; the constant velocity of sound in still air is agreed upon amongst the observers; their clocks are mechanically similar. Will he then be able to determine the same velocity for the train? Will these two observers use the same formula: L = ct + vt? This formula describes the idea that as the sound wave (velocity, c) travels rearward, it meets the caboose (velocity, v) traveling forward during the sane time. Each begins at the endpoints of the distance, L. This formula can be rearranged to the MM form: L / (c + v) = t; v = [L / t] - c, to find the velocity of the train relative to the earth.

Last edited: Jun 29, 2014
2. Jun 29, 2014

### Simon Bridge

Welcome to PF;
If you used sound interference, I suppose you could use the same equation over short distances.
Not sure if you could use the exact formula because sound waves work a bit differently to light.
You will not get the null result that you get for light because the speed of sound is frame dependent and there is no intrinsic problem with objects (the train say) going faster than the speed of sound.

You should be able to work out the answers to your questions using normal Galilean relativity, since the relative speeds are very slow. If you want to explore the special relativity with sound pulses, you will need to postulate a very fast train to see the effects... recall that sound is usually non-relativistic.

3. Jun 30, 2014

### GerryB

I was just mainly concerned whether the substitution was valid. If it is valid, then sound could be substituted for light in ANY formula involving c. I was not thinking about doing an interference experiment, but using the formula of one arm of interferometer as a replacement for the echo formula.

As you say, "...relative speeds are very slow," but I think that is precisely what I want. I apologize that this might be a new theory, but I am mostly just exploring ideas as a mental exercise. I will try to more properly phrase my questions in the future.

Last edited: Jun 30, 2014
4. Jun 30, 2014

### Staff: Mentor

You could, in principle, make an acoustic interferometer. It would allow you to detect any anisotropy in the speed of the air flow through the interferometer.

The main difference between sound and light is that the speed of light is invariant whereas the speed of sound is not.

5. Jun 30, 2014

### GerryB

Yes DaleSam, I was struggling to set up the train as the one arm of the interferometer. Then, use a clock to time the sound wave traveling through still air (medium). I think that sound waves are invariant under certain conditions such as this. I was just looking for a second witness on the validity of the substitution.

6. Jun 30, 2014

### Staff: Mentor

The speed of sound waves are never invariant.

But a MM type interferometer does not detect invariance. It detects isotropy relative to the apparatus.

7. Jun 30, 2014

### GerryB

We may have to agree to disagree. The speed of the sound is independent of the speed of the source of the emitter of the the wave, like light. When the source and receiver are traveling in tandem, and their motion is disconnected from the medium, then I am supposing that the MM formula would apply. But this only works if the speed of sound can be substituted for the speed of light, all be it on a very local scale. I will try to narrow my question.

8. Jun 30, 2014

### Staff: Mentor

This is not a matter of opinion where two people can express different but equally valid opinions.

Yes. This is true, but it is also not what invariance means. Invariance is when different inertial frames agree on a quantity.

There can logically only be one invariant speed, and experiments show that it is the speed of light, not sound.

9. Jun 30, 2014

### Staff: Mentor

The speed of sound waves is constant with respect to the air through which they are moving. Therefore, that speed cannot be invariant, because the speed of the air is not invariant.

10. Jul 1, 2014

### Simon Bridge

No. That is not correct.
Light and sound do not always follow the same equations.

11. Jul 1, 2014

### GerryB

Ah Simon, this is what I am looking for, to which equations can both light and sound be applied? At the slow speeds of sound waves, Special Theory relativistic effects should be negligible.

12. Jul 1, 2014

### GerryB

The earth drags a several miles thick layer of atmosphere along with it as it hurtles through the galaxy. So, the air / medium moves with the earth, fortunately for us.

Let us assume that air near the surface of the earth, in the vicinity of the train, behaves uniformly in all directions when sound waves pass through it, isotropy. I am not worried about the behavior of the medium at this point, except that the air molecules do not share in the motion of the train.

On a windless day, through an open window, the train observer in the moving reference frame will hear the same sound wave moving at the same speed as the platform observer in another reference frame at rest.

By Doppler, the troughs and crests of the wave will stretch out near the engine whistle, but compress near the caboose observer by an equal amount due to the tandem motion of the train. They thus cancel out, so that there is no change in frequency for the train observer. For the platform observer there is a change in frequency, but each observer will not notice any change in the wave speed in the same still air, by definition. The sound wave will have no velocity added to it by the moving source. They will agree on the speed of the sound wave, though their reference frames are in motion relative to one another .

13. Jul 1, 2014

### Staff: Mentor

He will not, because on a windless day the train observer will not experience still air outside the window; there's a strong wind blowing towards the rear of the train. The platform observer doesn't experience that wind (because it's a "windless day", right?).

Suppose the length of the train is $L$, the speed of the train relative to the air (and thus the the platform) is $v$,and the time between emission of the sound at the front of the train and detection of the sound at the rear of the train is $T$. The train observer will calculate the speed of sound to be $L/T$, just by using the simple formula that speed equals distance traveled divided by time. The platform observer, using the exact same formula, finds that the speed of sound is $(L-vT)/T$.

Note that this has nothing to do with the motion of the source; it's irrelevant whether the sound is produced by something on the front end of the train or something on the ground that makes noise when the front of the train passes it.

14. Jul 1, 2014

### Staff: Mentor

None of this is correct. I have no idea why you would believe any of it.

15. Jul 1, 2014

### bahamagreen

Nugatory,
Did you get the train and platform observers' equations switched?

Doesn't the train observer's measure of the sound path length need to be L-vT because the train is moving, shortening the distance between the origin of the sound in the still air and the moving observer's interception of that sound... he's closing the gap while the sound is in transit, so the length he wants to time is not the length of the train, but the length of sound travel through still air from origin to interception?

For the platform observer, won't he use the simple L/T calculation because the origin of the sound in the still air remains fixed, as does himself the platform observer... no gap is being closed between the origin of the sound and the intercrption of it, the distance between the origin of the sound and the platform observer remains constant during the transit of the sound; it does not matter what the motion of the train might be after it emits the sound?

Doppler effects will make the sound have different measured frequencies, but the sound speed itself should measure the same value for each observer, right?

DaleSpam,
I'm shocked! All of that paragraph seems to be correct to me... I have no idea why you would not believe any of it... I must have some grave misunderstanding; could you elaborate on what's incorrect?

16. Jul 1, 2014

### Staff: Mentor

No.

The train observer records sound being emitted at the front of the train at time zero and detected at the back end of the train at time $T$. As far as he's concerned he's at rest while the air and platform are moving backwards, so for him the distance between those two points is $L$ and the speed of travel is $L/T$.

The platform observer sees the sound emitted at time zero. Call the position of the front of the train at that time position zero, and then the back of the train is at position $-L$ at time zero. But when the sound reaches the back of the train at time $T$, the back of the train has moved to position $-L+vT$, so the distance between the point of emission and point of detection is $L-vT$.

It's worth noting that $L-vT$ is the speed of sound in still air in this thought experiment; the still-air frame is the one in which is $v$ is zero.

17. Jul 1, 2014

### bahamagreen

I'm providing a long response to help identify the problem or misunderstanding... I'm continuing to think the calculations are switched...?

"As far as he's concerned he's at rest while the air and platform are moving backwards..."

But, if what he wants to measure the air speed of sound through the air (as in with respect to the air, meaning how fast the sound moves through the air from the "at rest" perspective of the air itself), then the train observer does not want to measure the length of the train and use that as L, he wants to measure the still air distance through which the sound travels between origin and interception. That still air distance is shorter than the length of the train moving through the air. This is the very same argument to present for the platform observer's calculation.

Either that, or he needs to take the apparent wind speed velocity into account and adjust the time measure longer...

He wants to measure the speed of sound through air itself, through air at rest... it seems he must take either the closing travel length gap or the additional wind speed into account to measure speed of sound. If he does not do the adjustment, his measurement is confounded and is not m/s wrt air at rest.

You are not suggesting that "air speed of sound" measurements mean anything taken from moving windy conditions without compensating those factors? Air speed of sound always means speed of sound with respect to still air, the air itself, right?

If the train were not moving and there was a steady wind, the observer would adjust for the wind speed when calculating and reporting "the airspeed of sound"... on the moving train, or on a stationary train on a windy day, if he just uses L/T I don't think you can say that he is measuring sound speed. He is measuring elapsed time and the wrong distance, but not the air speed of sound.

I'm not following your explanation of the platform observer's measurement... why is he looking for the sound to reach the back of the train? He is measuring from the emission at the front of the train along the direct path to his own location... that distance is fixed, so he takes that distance "L" (not the train length), and divides by his own measure of "T". The motion of the train, the length of the train, and anything happening at the back of the train are all irrelevant to him. Neither the point of emission of the sound, nor the point where he intercepts the sound on the platform are moving, no gap is closing, and no wind is blowing... it is a direct measure of the air speed of sound between the place where it starts at the front of the train (even the train itself is irrelevant, could have just been a horn set stationary on the tracks) and his stationary position on the platform.

The platform observer sees the sound emitted at time zero. Call the position of the front of the train at that time position zero, and then the back of the train is at position −L at time zero. But when the sound reaches the back of
the train at time T, the back of the train has moved to position
−L+vT, so the distance between the point of
emission and point of detection is L−vT.

actually applies to the train observer... that is why I think you might have the calculations switched between the observers... this is not a Relativity problem... the frame of reference for both observers is the frame within which the still air is at rest, which corresponds to that of the ground, and the platform, and with respect to which both observers will base their calculations of the air speed of sound through that frame.

18. Jul 2, 2014

### Simon Bridge

Well for instance p=hc/λ does not apply to the momentum of air particles in a sound wave.

Just use your good sense -
- the speed of sound is not invarient so anything that relies on invarience will not work.
(...the doppler shift equations are different for light for eg.)

- light is a transverse wave with magnetic and electric vectors, by comparison sound waves are logitudinal in only one vector (displacement).

- and so on.

Light and sound are inherently different phenomenon - you cannot just blindly substitute equations just because some of the letters are the same.

19. Jul 2, 2014

### Staff: Mentor

Yes, of course if he wants the speed of sound relative to the air it is propagating through, then he has to allows for his speed relative to the air. But we should carefully state when we are switching frames - the train observer is MEASURING the speed of sound in his frame in which the air is moving, and then applying the Galilean transforms to CALCULATE the speed of sound in a frame in which the air is at rest.

Your immediate reaction may be to say that I'm quibbling because it's obvious that the "real" speed of sound is the speed of sound in still air, and therefore that the train frame results are bogus and distorted by the air movement. But consider that:
- The $T$ and $L$ values are the only measurement results directly available to the train observer. He has to work with uncorrected speed measurements because that's all he has in his observational data.
- Suppose neither the platform nor the train observers knew up front that air existed and that sound waves propagated through it. Just by comparing their observations of distance traveled and elapsed time in their respective frames, they would be able to advance the hypothesis that there is air, that sound travels at a constant velocity relative to air, that the platform observer is at rest relative to the air, and the train observer is not.
- Consider a different observation in the same thought experiment setup: We look at the damage done by a windblown object striking the observer at the back of a train. It's going to be very hard to persuade the train observer that the quantity $L/T$ is not a physically meaningful and completely real speed; conversely the platform observer will begin analyzing the collision by applying the Galilean transforms to get to the moving-air frame.

20. Jul 2, 2014

### Staff: Mentor

Sure.

The speed of the sound wave in one reference frame is related to the speed in another frame by the relativistic velocity addition formula. The two speeds are only equal if they are the speed of light, which is the one and only invariant speed.

If the whistle is in the very front of the train then by Doppler the wavelength near the engine observer is the same as the wavelength near the caboose observer, and both are longer than the wavelength would be if the train were at rest. If the whistle is between the engine and caboose observers then the wavelength near the engine observer is shorter and the wavelength near the caboose observer is longer than the at-rest wavelength.

In either case, there is no compression of wavelength near the caboose observer.

Note, the usual Doppler formula assumes that the medium is at rest, so it only applies directly to the ground frame.

The air isn't still in the train frame. It is moving. The train observers will definitely notice a change in the air speed. More importantly for an interferometer, the speed will be anisotropic, indicating the direction of the airflow. That is, after all, the whole point of the measurement.

The sound wave will indeed have no velocity added to it by the moving source, but it will have velocity added to it by the moving medium. This applies to sound and also to light moving in something other than vacuum, as was demonstrated in 1851: http://en.wikipedia.org/wiki/Fizeau_experiment

21. Jul 2, 2014

### bahamagreen

DaleSpam: "The speed of the sound wave in one reference frame is related to the speed in another frame by the relativistic velocity addition formula. The two speeds are only equal if they are the speed of light, which is the one and only invariant speed."

This is not a relativity situation...
With the train speed about 30mph the observers' sound speed measures differ due to evoking relativistic velocity addition on order of 10^(-15)... well inside sub-molecular measurement error noise level.

DaleSpam: "If the whistle is in the very front of the train then by Doppler the wavelength near the engine observer is the same as the wavelength near the caboose observer, and both are longer than the wavelength would be if the train were at rest."

But since both observers are moving through that still air toward the sound source, their interception of a wavelength will occur over a shortened period, so the apparent frequency will be increased to correspond to that mechanical frequency local to the whistle (same pitch as if the train was at a dead stop in still air)..

DaleSpam: "If the whistle is between the engine and caboose observers then the wavelength near the engine observer is shorter and the wavelength near the caboose observer is longer than the at-rest wavelength.

In either case, there is no compression of wavelength near the caboose observer.
"

But the forward observer is moving away from the waves, the rear observer toward the waves... the forward observer's motion acts to increase the time duration of the wavelengths' passage, the rear observer's motion acts to shorten the time duration of wavelength passage - this later is the compression to which GerryB refers... the wavelength's still air period is being shortened by the rear observer's movement through the air, just as if he was approaching a fixed sound source; intercepting more wavelength per unit time, shortening the measured wavelength, compressing the wave-train to higher frequency.

"The air isn't still in the train frame. It is moving."

"The sound wave will indeed have no velocity added to it by the moving source, but it will have velocity added to it by the moving medium."

This is not a relativity situation; the air is still, the train is moving, the sound source is moving, the medium is not moving.

22. Jul 2, 2014

### Staff: Mentor

It certainly is a relativity situation since we are trying to determine what results are like in different frames.

Clearly 30 mph is low enough that you are justified in using Galilean relativity instead of Special Relativity, but either way it is relativity and my answer stands. The speed of sound is not the same in the two frames in Galilean relativity either.

Yes, this is the correct explanation of the received frequency. Not that the wavelength is shortened, but that they pass over the longer wavelength in a shorter time.

Again, you are interested in two frames so it is a relativity situation, but you are perfectly fine using Galilan relativity for this. It doesn't change any of my comments. The air is still in the station and train is moving. The train is still in the train and the air is moving, the speed of sound as measured by an interferometer is anisotropic.

23. Jul 2, 2014

### bahamagreen

You are technically correct - the best kind of correct.

24. Jul 3, 2014

### Simon Bridge

The best kind of correct is actually the kind that makes you attractive and rich ;)

25. Jul 3, 2014

### GerryB

Well Simon, we seem to be moving from physics into philosophy. Are you saying that the train observer and the platform observer measure different speeds for the sound wave from the train whistle?

Well Nugatory, creating wind on a "windless day" seems to be more of a philosophical problem than a physics problem. A passenger will experience a "breeze" through an open window on a moving train. The wind has been created by the train moving through the air molecules, not the air moving past train. Even by the classical Doppler equations, these are not equivalent scenarios. The air and platform moving to meet a resting train is not philosophically equivalent to the train moving towards the stationary platform.

Well argued points bahamagreen.

Last edited: Jul 3, 2014