# Isn't the “constant speed of light” postulate valid also for sound waves?

The second postulate of Special Relativity says:

As measured in any inertial frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body.

But isn't it also valid for sound waves? I mean, a jet flying above the speed of sound will not generate waves that travel at V+U (where V is the jet's velocity and U the speed of sound), but waves that travel at U. Provided that Maxwell's laws show that light is a wave, what's the point of this postulate?

• The difference is that sound travels through a medium, whereas there's no luminiferous aether. – lemon Jul 5 '16 at 8:15
• The point you are trying to make is not clear. Yes, the speed of sound is constant relative to the medium, not the source nor the observer. So what? Are you asking a question about sound or about the postulates of SR? – sammy gerbil Jul 5 '16 at 8:17
• @lemon So, the key point of the postulate is not that the speed of light is independent of the frame of reference (which could be assumed as "normal" provided that light is a wave) but that it actually propagates at constant speed without the need of a medium. – Claudix Jul 5 '16 at 8:19
• Just to set the physical record straight: Maxwell's equations don't say anything about light. One has to come up with experimental validation that light behaves according to Maxwell's equations, which isn't completely trivial, since one can't actually measure E and B-field components on light directly. Sound waves, on the other hand, do not behave according to these equations. They behave according to the equations of thermodynamics and fluid mechanics. – CuriousOne Jul 5 '16 at 8:27
• @Claudix, i have asked the same basic question about the 2nd postulate. . and that was a repeat of another question. i am still unsatisfied with the answers. seems to me that if all of the physics is the same for the two inertial observers (that are in motion relative to each other), that means that both have to experience the same $\mu_0$ and the same $\epsilon_0$. – robert bristow-johnson Jul 6 '16 at 1:06

Light doesn't travel at $c+V$ (where $V$ is the speed of the source), it travels at $c$.

What's the difference? It means that if you're flying towards someone at a speed $V$ and you shine a light at them, you measure the light to travel away from you at a speed $c$, but the other person measures it to fly past them also at a speed $c$ (i.e. not $V+c$).

In the case of sound, the source and observer may disagree on the relative speed. The source will measure the sound to propagate at a speed $U-V$, whereas the observer will measure it to propagate at $U$.

• I see... So actually sound waves do not travel at the same speed when measured from the source and an observer? – Claudix Jul 5 '16 at 8:27
• @Claudix That's correct. – lemon Jul 5 '16 at 8:32
• Imagine two people conversing within a super-sonic Concorde. All of the air within inside the plane will be traveling at supersonic speed relative to air outside the plane, but will be essentially stationary relative to the passengers. Consequently, the propagation of sound within the plane will be unaffected by the speed at which it is traveling. – supercat Jul 5 '16 at 18:48

The point is that the considered postulate states that the speed of light in the vacuum is $c$ with respect to each and every 'inertial' observer. It is independent of not only the source that is emitting the considered light quanta but also of the observer who is observing it as long as it is an inertial observer.

It is true that for a given observer, in a given medium, the speed of sound is independent of the source that emits the sound. But as 'lemon' has described, the speed of the sound does depend on upon the observer even if it is independent of its source - but the speed of light is independent of the both of them.

The assumption that the speed of light is independent of the speed of the light source is false but sounds reasonable, the reasonableness due to the fact that the assumption is valid for all waves other than light. However, when combined with the principle of relativity, this assumption entails that the speed of light is independent of the speed of the observer as well, a conclusion which is almost obviously absurd. When the initially stationary observer starts moving towards the light source with speed v, the frequency he measures shifts from f=c/λ to f'=(c+v)/λ, which can only mean that the speed of the light relative to the observer has shifted from c to c'=c+v.

• EM radiation does not undergo doppler shifts with respect to observers and sources in relative motion. Michelson and Morley's experiment in 1887 demonstrated this pretty conclusively. You say that light waves undergo such a frequency shift -- have you actually seen this? I bet not... – Zorawar Jul 5 '16 at 16:11
• "EM radiation does not undergo doppler shifts with respect to observers" – Pentcho Valev Jul 5 '16 at 16:38
• "EM radiation does not undergo doppler shifts with respect to observers". Not true. See this: physics.bu.edu/~redner/211-sp06/class19/class19_doppler.html "Let's say you, the observer, now move toward the source with velocity vO. You encounter more waves per unit time than you did before. Relative to you, the waves travel at a higher speed: v'=v+vO. The frequency of the waves you detect is higher, and is given by: f'=v'/λ=(v+vO)/λ." – Pentcho Valev Jul 5 '16 at 16:48
• The speed of light does not change - it is the wavelength that changes. This is a critical distinction. – J... Jul 5 '16 at 18:49
• "The speed of light does not change - it is the wavelength that changes." The observer starts moving towards the source, and the wavelength of the light he is going to meet miraculously changes so that the speed of the light relative to the observer can remain unchanged? Isn't that too absurd? See the quotation above: "Let's say you, the observer, now move toward the source with velocity vO. You encounter more waves per unit time than you did before. Relative to you, the waves travel at a higher speed: v'=v+vO." – Pentcho Valev Jul 5 '16 at 19:38