Mathematical Obscurities in Special Relativity

As I mentioned, I worked with magnetic and optical sound tracks, so that is how I know this about the Doppler effect.

The air controls the speed of the whistle sound. If the train is stationary, the sound wave is not stretched out, and it is moving at 1,100. Consider that to be like a standard record being recorded at 33 rpm. The 33 rpm is like the speed of the sound in air.

Ok, now add more speed to the recording, by, let’s say, moving the needle by 1 rpm. So, the record is moving at 33 rpm and the needle is moving in the opposite direction at 1 rpm. That causes the sound wave to be “stretched out” on the record.

Play back the record at 33 rpm and you will hear a lower tone because the wave (the wiggles on the groove of the record) is stretched out a little by the motion of the needle. But if you play back the record at 34 rpm, even though the wave is stretched out, you will hear it at the original normal tone that was being emitted by the musical instrument while the sound was being recorded.

The same with the train. The whistle moving through the air causes the normal tone sound wave to be “stretched out” in the air. If the observer were stationary with the air, such as by standing on the track at the rear of the train, he will hear a lower tone because he is encountering the stretched out wave at V, 1,100. But regarding the passenger on the rear of the train, the train is pulling him along at v, and he will encounter the stretched out wave at V + v, and he will hear a normal tone, while the guy who is stationary on the track will hear the lower tone which he receives at V.

The guy on the front of the train, assuming the whistle is in the middle of the train, will hear a compressed or “bunched up” wave at the normal tone, because he is encountering the short wave at V – v, that is the speed of sound minus the speed of the train. So, he perceives the compressed wave over a longer period of time, and he perceives it as not being compressed. While a guy standing on the track in front of the train will encounter the compressed wave at the speed of V, and he will hear a higher pitch.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by DAVID</i>
<br /><blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by 1234567890</i>
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This sounds suspiciously correct, but it's wrong. Since you stipulated that the soundwaves are traveling through the air outside
even in the moving train, they are being detected at c+v and c-v by the rear and front detectors in the moving train respectively. There are two effects the one you refer to is cancelled, a second one I was talking about is not. It becomes obvious if you set v = c, in which case the front detector never detects any soundwaves at all- now THAT's a REAL shift.


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I apologize. In one of my earlier posts I used “c” for the speed of sound. I didn’t not mean to imply it was the speed of light.

I should have used V for the speed of the sound and v for the speed of the train.

So, the rear moving observer on the moving train encounters the stretched out sound wave moving toward him at the additive speed of V + v rather than just V. So, he encounters the stretched out wave at 1,188 fps, instead of the normal speed of sound, which is about 1,100 fps.

This is the same as recording a normal sound at about 34 rpm on a phonograph record. When playing it back at 33 rpm, you will hear a slightly stretched out sound wave that has been physically recorded onto the record as being a little longer than normal. But if you adjust your playback speed up to 34 rpm, you will hear the stretched out wave as a normal tone, because even though it is stretched out, it is passing your needle at a higher speed.

This is a law of nature. This is not a “theory”. I apologize for using the letter c earlier for the speed of sound.





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No, I wasn't quibbling about your usage of the letter c in lieu of v. Let's work out the math and see if what you say is correct:

Let's say the velocity of sound in air is 543 meters per second. There is a sound source in the middle of the outside of the train and two sound recorders mounted on the outside of the front and rear of the train. The distance from the source to the rear is 543 meters as is the distance from the source to the front detector.

Now we perform an experiment in a stationary train. Assume the sound source is generating pulses at 1 beat per second. The detectors at both ends of the train are tape recorders with 9 seconds of tape. Both tapes will record 9 beats before it runs out of tape.

Let's assume now the train is moving at 543 meters per second with respect to the air outside. Now the first beat will be recorded by the rear of the train in 0.5 seconds instead of 1 second in the stationary train. 0.5 seconds later, the source emits the second beat, which also takes 0.5 seconds to get to the recorder, for a total time between beats of 1 seconds, the same as in the stationary train. But the tape finishes recording the last beat at 8.5 seconds instead of the 9 seconds of tape used in the stationary train.

What of the recorder in the front of the train? Well since the train is moving at 543 meters per second and sound only travelled at 543 meters per second in our example, it will record 0 beats in 9 seconds of tape. So we see that due to the velocity of the train,
the 9 beats that takes up 9 seconds in the stationary frame are reduced to 8.5 seconds in the rear of the moving train and are not detected at all by the front recorder.

That's the reason Einstein's two postulates contradict each other,
a fact he was well aware of as he spent half the 1916 book trying to resolve the "incompatibility" as he called it. C cannot be source independent if physics is the same inside all inertial frames, or conversely if c is source independent, physics is not the same inside different inertial frames.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by 1234567890</i>
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Let's assume now the train is moving at 543 meters per second with respect to the air outside. Now the first beat will be recorded by the rear of the train in 0.5 seconds instead of 1 second in the stationary train. 0.5 seconds later, the source emits the second beat, which also takes 0.5 seconds to get to the recorder, for a total time between beats of 1 seconds, the same as in the stationary train. But the tape finishes recording the last beat at 8.5 seconds instead of the 9 seconds of tape used in the stationary train.

What of the recorder in the front of the train? Well since the train is moving at 543 meters per second and sound only travelled at 543 meters per second in our example, it will record 0 beats in 9 seconds of tape. So we see that due to the velocity of the train,
the 9 beats that takes up 9 seconds in the stationary frame are reduced to 8.5 seconds in the rear of the moving train and are not detected at all by the front recorder.

That's the reason Einstein's two postulates contradict each other,
a fact he was well aware of as he spent half the 1916 book trying to resolve the "incompatibility" as he called it. C cannot be source independent if physics is the same inside all inertial frames, or conversely if c is source independent, physics is not the same inside different inertial frames.
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The source’s motion though the air does not alter the pulse rate of the source. Let’s say a guy at the middle of the moving train fires a series of shots one second apart while the train is first MOVING, then he does the same thing later while the train is STATIONARY. One second is still one second whether the train is moving at the speed of sound or not.

When do you start the tapes? If you start the rear recorder when the first shot is fired (using light signals, while the train is moving), the rear recorder will record only 8 shots, if the tape is only 9 seconds long.

The recorder at the end of the train will receive the first shot .5 seconds after it is fired (.5 seconds into the 9 second tape), which is a result of the V + v.

The recorder at the front of the train will never receive the sounds of the shots, since the train is moving forward at the speed of sound and so is the front recorder, so the sound will never catch up with the front observer.

The rear recorder will record the first shot .5 seconds after it is fired, and it will record the 8th shot at a time of 8.5 seconds on the recording. There is still .5 seconds to go before the 9th shot is fired. The tape runs out just as the last shot is being fired, so the last shot is not recorded on the tape.

If you start the tape as the first shot is being fired (using light signals), at the end of the experiment the tape has a recording of 8 shots on it. The first shot is recorded .5 sec into the tape, and the 8th shot is recorded .5 seconds before the end of the tape.

If we perform the experiment with a stationary train and recorder and start the rear recorder as soon as the first shot is fired, it will record the first shot 1 second into the tape, and it will record the 8th shot at the very end of the tape. It will not record the 9th shot.

If you want all the shots recorded, and if you start the tape when the first shot is fired, you will need about 10 seconds of tape for a stationary train and recorder. Actually, a couple of more seconds of tape would help, since a shot sound has some duration, and you want to be sure to record the reverb and echoes of the last shot.

Never go to a recording session without enough tape. Keep some spare tapes in your pockets.

No, 9 shots are recorded in the stationary train in 9 seconds while 9 shots are recorded
in 8.5 seconds in the rear of the moving train. (0-1) = shot 1, (1-2)= shot 2, (2-3) = shot 3, (3-4) = shot 4, (4-5) = shot 5, (5-6) = shot 6, (6-7) = shot 7 , (7- = shot 8 and (8-9) = shot 9 where the intervals in the parentheses represents the time of emission and time of detection , respectively.

In the moving train, the whole recording is shifted as thus in the rear of the train : (0-0.5) = shot 1, (1.0-1.5) = shot 2 , (2.0 -2.5) = shot 3, (3.0-3.5) = shot 4, (4.0-4.5) = shot 5, (5.0-5.5) = shot 6, (6.0-6.5) = shot 7, (7.0-7.5) = shot 8 and (8.0-8.5) = shot 9. Obviously the shots are shifted in such a way that all 9 shots are heard at an earlier time than when in the stationary train (0.5, 1.5, 2.5, 3.5, 4.5, et al).

And of course the front of the train doesn't detect any shots at all.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by 1234567890</i>
<br />No, 9 shots are recorded in the stationary train in 9 seconds while 9 shots are recorded
in 8.5 seconds in the rear of the moving train. (0-1) = shot 1, (1-2)= shot 2, (2-3) = shot 3, (3-4) = shot 4, (4-5) = shot 5, (5-6) = shot 6, (6-7) = shot 7 , (7- = shot 8 and (8-9) = shot 9 where the intervals in the parentheses represents the time of emission and time of detection , respectively.

In the moving train, the whole recording is shifted as thus in the rear of the train : (0-0.5) = shot 1, (1.0-1.5) = shot 2 , (2.0 -2.5) = shot 3, (3.0-3.5) = shot 4, (4.0-4.5) = shot 5, (5.0-5.5) = shot 6, (6.0-6.5) = shot 7, (7.0-7.5) = shot 8 and (8.0-8.5) = shot 9. Obviously the shots are shifted in such a way that all 9 shots are heard at an earlier time than when in the stationary train (0.5, 1.5, 2.5, 3.5, 4.5, et al).

And of course the front of the train doesn't detect any shots at all.
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Ok, you are partly right, and I am partly wrong.

If the recorder starts on the moving train when the first shot is fired, 9 seconds of tape will record 9 shots, with the first shot recorded at .5 sec and the last shot recorded .5 seconds before the tape ends, like this:

_|__|__|__|__|__|__|__|__|_

On the stationary train, since it takes 1 second for the sound to reach the recorder, you’ll have to start the recorder .5 seconds after the first shot is fired to get the same recording timing.

On the stationary train, if you start the recorder when the first shot is fired, you’ll miss recording the 9th shot. The first shot will be recorded 1 sec into the tape, and the tape will end just as the sound of the 9th shot arrives at the recorder, like this:

__|__|__|__|__|__|__|__|__

If we consider the shots to consist only of single up and down sine wave spikes, lasting .1 seconds total, the recorded up and down spikes will look the same size and shape on an oscilloscope at the recorder, whether the train is moving or not.

The shot sound will last the same amount of time on the tapes, whether the train/recorder are moving or not.

In the air, the moving train spikes will be stretched out for .2 seconds in the air, and this is why a guy on the track at the end of the train will hear a lower tone. But remember that the moving guy (the moving recorder) at the rear of the train, and the stretched out sound wave, will encounter each other at twice the speed of sound, so the recorder will record .1 seconds of the up and down sine wave and the tone of the sound will sound normal.

In some of my earlier comments I pointed out that we don't know what time and space are. Here it becomes obvious that we don't know what velocity means as well. Einstein did a real good job of confusing this problem as well.

The issue boils down to understanding what is meant by the velocity of light being a constant. The resulting physical interpretation provided by Einstein completely confuses the problem and obscures what the correct conclusions ought to be.

The solution is to question the assumption that the velocity is a constant. This is based on a confusion in physics. We start by going back to Copernicus. He said Aristotle was wrong, the earth moves, not the sun. This began a long argument. Finally the proponents of the view that the earth moves won the argument. But when Michaelson made the first direct measurement of the motion of the earth, he discovered that the earth is stationary relative to the ether; or in other words the electromagnetic reference frame. This was a very unwelcome result. The direct experiment shows that the earth does not revolve around the sun. But, there is more, the experiment shows that the earth rotates upon its axis. The direct measurement shows that the earth rotates but does not revolve around the sun. The theory of relativity is an attempt to resolve the contradiction that this experiment raises.

Einstein's solution was to assert that the velocity of light is a constant independent of the observer's state of motion. Hence, any attempt to directly measure the earth's motion must fail. Hence the descrepancy is resolved. But this solution creates a host of other problems which have not been resolved. Now we don't know what time and space and velocity are.

The resolution of the problem by asserting that the velocity of light is constant solves the problem for the revolution of the earth around the sun, but fails to explain why the rotation of the earth on its axis is detected. If the light velocity is assumed constant, then we should not be able to detect either of these motions. But we actually detect one and not the other. This contradiction is not resolved by SRT. So until this problem is resolved, we should not believe physicists.