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THE NON-APPLICABILITY OF ADDITION OF VELOCITIES TO PROPAGATING SOUND WAVES


Geryllax Vu

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On a windless day a train of length travels along a level straight section of track at the constant velocity v.  An observer in the caboose has a clock and a light source with which she will send a signal to the engineer at the front of the train.  Upon seeing this signal, he will blow the whistle which will send out a sound wave that has the constant velocity c  through the still air/medium.

 

At this short distance a light signal is effectively instantaneous so upon sending the light signal she also starts the clock that she has.  When the sound wave reaches the caboose observer’s ear she will stop her clock.  She should then measure approximately the Newtonian universal time interval t  between the departure and arrival events of the sound wave in the train reference frame.

 

The caboose and the engine are at a fixed distance apart.  They form a tandem which is moving through the still air at the single velocity v  each endpoint maintaining their distance of separation. The sound wave and the caboose begin their journeys at the endpoints of L and will meet at some location in space between the original locations of the endpoints along their adjoining line.  The sound wave travels the distance ct  rearward towards the caboose and the caboose travels the distance vt  forward towards the sound wave during the same interval of time t  (distance = speed × time) and adding these two distances should equal L.  Thus, all the variable values are available from within the train reference frame:

 

♦ L = ct + vt ; t = L / (c + v )

 

This formula (similar to the Michelson-Morley experiment) could be used by both the train observer (in the train reference frame) and an observer that she need not communicate with at rest on the nearby platform (in the platform reference frame).  The train observer might assume the train to be in motion and would thus measure with her clock an interval of time that would indicate that the sound wave has travelled at the unchanged velocity c  for a lesser or greater distance than when the train is at rest.  This is a result of the consideration that the air molecules pass easily through the porous conceptual walls of any reference frame that is not an enclosed compartment.  Alternatively, the train moves through a cloud of stationary air molecules that are not carried along by the train reference frame.  There will be no addition to or subtraction from the velocity of the sound wave only a change in the distance the sound wave travels.  The train observer will then not have to apply the principle of addition of velocities from the Galilean or Lorentz transformation between the two reference frames that are in relative motion.

 

Let, L = 1000 meters; c = 343 meters/second; assume v = 30 meters/second:

 

♦not, t = L / c, (train, air, and platform at relative rest) = [1000 m] / [343 m/s] = 2.92 s

 

♦but, t = L / (c + v ), (train in motion through air) = [1000 m] / ([343 m/s] + [30 m/s]) = 2.68 s

 

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If the space inside the train is wide and uncluttered (a passenger train with large, open doors between each carriage) - and also sealed from the outside, carrying the air inside with it, then you can assume that the speed of sound through the air inside the train will also be c. So you would expect the sound of the whistle, through the inside, to arrive in L/c (=2.92 s). So you can expect two versions of the tweet to arrive at the rear of the train. Tweet..tweet, separated by 0.24s.

The frequency of the sound heard by the platform observer would be lower than that of the observer in the calaboose. You could say that the Doppler Effect is applied twice on the way from the whistle and the observer at the back, one, a decrease in frequency and the other an increase back to the original.

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On 10/7/2016 at 11:53, sophiecentaur said:

If the space inside the train is wide and uncluttered (a passenger train with large, open doors between each carriage) - and also sealed from the outside, carrying the air inside with it, then you can assume that the speed of sound through the air inside the train will also be c. So you would expect the sound of the whistle, through the inside, to arrive in L/c (=2.92 s). So you can expect two versions of the tweet to arrive at the rear of the train. Tweet..tweet, separated by 0.24s.

The frequency of the sound heard by the platform observer would be lower than that of the observer in the calaboose. You could say that the Doppler Effect is applied twice on the way from the whistle and the observer at the back, one, a decrease in frequency and the other an increase back to the original.

I have been struggling to find a clear and concise way to describe the point of the experiment and you have grasped it.  I believe that as I have presented it your assumption is correct that an enclosed compartment  is measurably different from an open reference frame.  I agree that it would be possible to hear two tweets as you say because the air molecules in the moving compartment  may slow down or speed up the sound wave depending on the sound wave's direction of travel.  But a sound wave traveling through still air will always be c.

I agree with your observation that there will be no Doppler effect for the train observer because the source and receiver are moving in the same direction at the same speed so the waves spread out at the source but squeeze together at the receiver thus cancelling each other out and leaving the original frequency.  I plan on using this difference for the two observers in the future.

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