# Can sound waves be used as simple explanation of relativity effects in STR?

There are so many similarities (Doppler Effect, independence of wave velocity from source speed etc..). Try moving in your car with music and ask you friend outside record it while you moving towards him (playback speed faster + “blue shift”) and away from him (slow playback speed + “red shift”).

• You also need the wave velocity to be independent of the velocity of the receiver. That's where this model breaks down. Someone running towards a speaker sees the wave coming at them faster than someone running away from a speaker. This isn't true in relativity. – Jahan Claes Jul 12 '16 at 15:35
• I've also investigated that the speed of signal and speed of magnetic disturbance (that is c) are completely different phenomenons. The speed of sound is closer to signal speed (that could be even 0 or negative for light) I think... – m8labs Jul 12 '16 at 15:45

No. In "STR" the speed of light is the same value no matter how the measurer is moving relative to anything else. For sound waves, the speed of the wave is measured relative to the medium through which the wave travels.

Consider a sound source and receiver at rest relative to each other. The source and receiver have coordinated watches and have agreed on when the source will emit a sound pulse. That way the receiver can measure the time interval between emission and reception and calculate the speed of the sound. If the air is still (no wind), the speed will be one value. If there is wind, the measured speed will be different. This is not true of light.

Also, the mathematical form for the SR Doppler shift is different from the Doppler shift for a sound wave, due to the independence of speed of light.

• What about perfect weather always with no winds ever, and perfect aerodynamic of the car without a windshield (we don't drag medium with the car in this case)? – m8labs Jul 12 '16 at 12:18
• $\Delta x / \Delta t$ is constant for light no matter how fast the source or receiver is moving. For sound, you would have to adjust the calculation for the relative speed of the source and receiver. – Bill N Jul 12 '16 at 13:25
• Can we treat Δx as distance between two crests? In this case our traveling between two light wave crests in the car will be pretty the same as traveling between two sound wave crests - that's why you measure speed of light still the same as it measured by stationary observer, and that's why you hear sound without any distortions while you moving together with the source. – m8labs Jul 12 '16 at 13:32
• A pulse of light will always be measured to travel at c, regardless of motion and regardless of measurement reference frame. A pulse of sound may be measured to travel at different speeds depending on the reference frame. – Bill N Jul 12 '16 at 13:47
• I'm starting to understand this after investigating that the speed of signal and speed of magnetic disturbance (that is c) are completely different phenomenons. The speed of sound is closer to signal speed (that could be even 0 or negative for light) I think... – m8labs Jul 12 '16 at 15:49

No, they can't. In fact sound waves would lead you to Galilean relativity, and in all the cases where it differs from Einsteinian relativity they would lead you to the wrong result. This is a catastrophically bad model for Einsteinian relativity.

As a simple example of where things go most horribly wrong consider a supersonic aircraft, and that the velocity relative to the ground of such an aircraft is the sum of the velocity of it relative to the air and the velocity of the wind.

• What about absence of the wind ever? Always perfect conditions. – m8labs Jul 12 '16 at 14:14
• @MaratAl No, nothing can rescue it. It's the exact opposite of a good model, sorry. – tfb Jul 12 '16 at 15:29
• I've also investigated that the speed of signal and speed of magnetic disturbance (that is c) are completely different phenomenons. The speed of sound is closer to signal speed (that could be even 0 or negative for light) I think... – m8labs Jul 12 '16 at 15:46