Maths problem

Post by **pErvinalia** » Sun Feb 26, 2017 12:48 am

If the speed of sound was constant when observed from all inertial frames, then length contraction would have to occur. But the speed of sound isn't.

Post by **mistermack** » Sun Feb 26, 2017 5:28 am

pErvin wrote:If the speed of sound was constant when observed from all inertial frames, then length contraction would have to occur. But the speed of sound isn't.

No, because you are talking about the speed of sound, as measured by light clocks. Not sound clocks. You keep confusing the two.

That's the essence of the argument. In special relativity, we measure the speed of light, using clocks that slow with their velocity relative to the speed of light.

In the equivalent in sound, you are measuring the speed of sound, using clocks that slow with their velocity relative to the speed of sound.

I don't think you will ever get it.

Post by **pErvinalia** » Sun Feb 26, 2017 5:56 am

mistermack wrote:In the equivalent in sound, you are measuring the speed of sound, using clocks that slow with their velocity relative to the speed of sound.

No clock slows with it's velocity relative to the speed of sound. Well a sound clock will slow, but not due to any relativistic effects. It will slow because the sound has to travel further. There is no equivalence to the speed of light and relativity. Clocks slow in Einstein's relativity because of relativistic effects, not because the light has to travel a greater distance the faster the clock moves (like in the case of a sound clock).

I don't think you will ever get it.

It looks that way. I'm not sure what there is to get. ALL clocks slow when observed from another inertial frame as the relative speed between the two frames increases. I think you are confusing what time dilation actually means. It's not a concept that is dependant on clocks. Time itself literally slows from an observers position. It doesn't change depending on whether there is one type of clock or another measuring it. Every physical process slows down. But due to length contraction the speed of light stays the same, and therefore the physics in each frame remain equivalent. This isn't in anyway equivalent to the slowing of a sound clock.

Post by **pErvinalia** » Sun Feb 26, 2017 6:42 am

mistermack wrote: I originally just made the comment that you can do a similar exercise in sound and get an equivalent speed of sound that all observers will measure the same, IF their clocks operated by sound only.

Can you expand on that? I don't think this would be the case.

Post by **mistermack** » Sun Feb 26, 2017 12:05 pm

pErvin wrote: Can you expand on that? I don't think this would be the case.

No, because

hurts after a while.

Post by **pErvinalia** » Sun Feb 26, 2017 12:23 pm

Ok, then I'll dismiss it as the wibble it appears to be.

Post by **Brian Peacock** » Mon Feb 27, 2017 11:16 pm

I think this one explains my point a bit better...

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The sound clock on a train (SCOAT!) takes a few things for granted: the speed of the train and the speed of sound are constant; the train does no accelerate, just goes from stationary to a full and constant speed, and; the reflector in the model remains at a fixed distance from the clock regardless of the angle.

The speed of sound in a stable (ideal, non-dispersive) medium is likewise fixed. In a dispersive medium, like air, different gasses absorb different frequencies (pitches, Hz) at different rates, so the 'ping' in the model represents either sound in an ideal medium or a single frequency in a dispersive medium.

In this version I've negated software time-logging, as it was too reliant on precocious computer CPU resource allocation, instead favouring a simple distance/speed of sound calculation.

The 't' (time) values displayed when the clock reaches the edge of the animation canvas are not seconds but arbitrary, though fixed, units determined relative to the dimensions of the animation canvas itself, which in turn is determined by the size of the users browser's window. All calculations are performed to 12 decimal places and only rounded up to 3 decimal places for display purposes.

The speed of the train in the model is 0.0006 of the width of the animation (in pixels), and the speed of sound is 2.466 × train speed. These values are incremented at the refresh rate of the viewer's screen (c.60-120 fps). This ensures that viewers on different devices at different screen sizes get a roughly similar experience, and so that the 'ping' circles remain inside the boundary of the animation canvas.

As I was trying, and apparently failing, to show earlier, because the train and the speed of sound are constant the model is symmetrical around 90° to the direction of travel. This is obviously because the distance to the reflector is constant, regardless of angle, and because the distance the 'ping' travels to-and-from the reflector at 90°-n is equal to the distance at 90°+n. You can test this by comparing results at 45° and at 135°.

An angle of 90° represents the shortest possible distance to-and-from the reflector on a moving train, and an angle of 0|180° is the longest. The clock will run slightly slower at 0|180° and faster as it tends towards 90°. Again, the speed of sound is constant and the clock's timing is effected only by the angle of the reflector, it not being multiplied by the speed of the train - assuming the speed of the train does not exceed the speed of sound of course.

Neither this model, nor the thought experiment itself (as interesting as it is), has any bearing on the time dilation effects associated with speed-of-light travel. All it shows is that the timing of a SCOAT is dependent on the angle of the reflector relative to the direction of travel. Time does not slow when the reflector is directly fore and aft, just that the periodic interval between the clock and the reflector and back to the clock increases.

My earlier references to the Doppler effect were only brought in to highlight and contrast the regularity of on-board time with the fact that, were this a real-life experiment, the ping sound returning from the reflector at an angle of +90° would be at a slightly lower pitch to the ping sent out by the clock, and likewise at a slightly higher pitch returning from a reflector at -90° - due to the effects of motion on waves. This is to be considered an interesting observation, and has no bearing on the SCOAT's timing. I accept I could have been clearer on this point. I also apologise for any confusion caused by the editing of my bus-posted comments a few days ago.

Post by **pErvinalia** » Tue Feb 28, 2017 1:28 am

Again, if you're not adding and subtracting (it's not "multiplication", again) the train velocity to/from the speed of sound, then you are doing it wrong. Unless you are assuming the clock is a closed unit (which you haven't told us).

And your assessment of Doppler is still wrong. For a start, both the emitting and return ping from a +/-90deg clock would be travelling ahead of the emitters by the exact same amount, so in both cases (in a stationary reference frame) would be of a shorter wavelength and therefore higher frequency. But it's not a stationary frame. The receivers are moving away relative to the sound wave, by the same ratio that the frequency is increasing (due to the emitters moving into the sound wave). So the net effect would be no change in frequency at all when received.

Post by **Brian Peacock** » Tue Feb 28, 2017 2:57 am

We're talking about a two point system where the speed of the train and the speed of sound are constant. Given those constants, the clock's timing, its periodic interval, is a function of the distance from emitter -> reflector -> receiver/emitter. Basic geometry is all that's required once the emission point, reflection point, and return points have been identified. I'm not giving a full account of Doppler, just pointing out something noteworthy - the tendency towards an apparent shift in wave length (pitch) depending on the angle to the reciever, because the reciever is moving relative to the point of reflection. The point of frequency equilibrium is at reflector angle somewhere ahead of the emitter: x - 90deg. But if we're going to get that knitpicky we might as well factor in things like the ping's frequency range, angle dependant amplitude variance, reflector surface characteristics, and an account of the possibility of noise cancelling standing waves forming in specific relationships in a specific media. I'm not trying to build the bloody thing on a web page, I'm just showing the geometric relationship given two points, emitter/receiver and reflector, in a fixed relationship and two constants, the speed of the train and the speed of sound, in a simple system.

Post by **pErvinalia** » Tue Feb 28, 2017 4:14 am

The problem is that it is wrong. You are showing a singular ability to totally ignore every problem I point out with your model. For about the 5th time, it's not basic geometry if the clock is open to the (still) air. This is because the sound ping is moving through the still air at a velocity relative to the train/clock travel. That means that the path travelled by the sound ping will be a scalene triangle (when the x-axis/ground is drawn in) - other than at 90deg clock angle to train travel.

Regarding Doppler, all the niticking wibble you went on about is totally irrelevant to the flaw in your thinking. If the clock is open to the air, the wavelength of the sound ping will always be compressed in the direction of travel as long as the vector component of the velocity in the direction of train travel is greater than the train velocity. What this means is that for a stationary observer watching the train go past, they will experience the Doppler effect (higher frequency as the train is moving towards them, and lower as it is receding). BUT, this is the point you aren't getting, the clock isn't stationary. The clock is moving at the same speed as the train. Therefore, there will be no Doppler effect at the clock receiver because the wavelength will be expanded due to movement away from the sound wave in still air by the same amount that it is compressed at the emitter due to it's movement through the sound travelling medium (i.e. still air).

Post by **pErvinalia** » Tue Feb 28, 2017 4:47 am

If the clock was a closed unit then it would work as follows:

The sound would travel between the emitter and receiver at the speed of sound relative to the clock/train. That is, it doesn't matter how fast the train is going, the sound ping will always move at approx 330m/s. Compare this to an open clock where the sound ping will move at the speed of sound relative to the surrounding still air (i.e. not the clock/train). So in the case of an enclosed clock, all paths travelled by the sound ping would be a case of simple geometry as your model shows. However, your assessment of Doppler and frequency changes would still be wrong. As the medium of travel and source and receiver are all moving at the same speed, there would be no sound wavelength compression or expansion. Therefore there would be no frequency change within the clock.

Post by **Brian Peacock** » Tue Feb 28, 2017 9:53 am

Why didn't you just say that before.

So the ping coming back from a reflector directly in front of the clock returns at a greater wavelength, but then is reciprocally compressed by the clock as it moves into the on-coming wave. OK. I was wrong. But, the speed of sound is still a constant in an ideal gas - a perfect reflector will not return the ping to the clock slower from directly in front than directly behind.

Post by **pErvinalia** » Tue Feb 28, 2017 10:10 am

Well I did say it about 5 times...

I'm not sure what you mean about a ping returning slower. If the clock is open to the surrounding air, the sound wave will move with the speed of sound relative to the air body. Which is obviously slower than the speed the clock is travelling, by an amount equal to the speed of the train. So you have to subtract that speed from the speed of sound to determine the speed the sound wave will hit the receiver (if we are talking about a clock aligned with and pointing in the direction of train travel). And then on the return journey of the sound wave you have to add that to the speed of sound. If the clock is closed such that the air body and the emitter/receiver are all moving at the same speed, then you don't need to worry about the train speed.

Post by **pErvinalia** » Tue Feb 28, 2017 10:18 am

I've been scribbling away this arvo trying to come up with an equation to solve the case of an open clock where the paths of the sound wave aren't symmetrical (i.e. form a scalene triangle) and for any alignment angle of the clock. I came up with a scary equation that turns out to be a quartic equation, something which is a nightmare to solve (with multiple solution possible as well). I can't solve it, and stack exchange isn't producing the goods, but someone put me onto this awesome site - Wolframalpha.com - where you can feed it your equation and it outputs the solution. As this is a quartic equation, it has 4 solutions (although, only 2 are unique in this case), but you can plug in values for the variables to see the answer. Which is thankful if you take a look at the equation solutions it pumped out.

(I should point out, that it is giving impossible values for t for a given train speed (v). Which means my equation is wrong somehow. Back to the scribbling).

Post by **Brian Peacock** » Tue Feb 28, 2017 10:26 am

pErvin wrote:Well I did say it about 5 times...

I'm not sure what you mean about a ping returning slower.

I'm saying that the speed of sound is constant; that it doesn't speed up when emitted in the direction of travel or slow down against it.

If the clock is open to the surrounding air, the sound wave will move with the speed of sound relative to the air body.

I've referred to an stable (ideal, non-dispersive) medium. The air is not moving, and for the purposes of the thought experiment, and the animation, there's no turbulence or friction etc.

Which is obviously slower than the speed the clock is travelling, by an amount equal to the speed of the train. So you have to subtract that speed from the speed of sound to determine the speed the sound wave will hit the receiver (if we are talking about a clock aligned with and pointing in the direction of train travel). And then on the return journey of the sound wave you have to add that to the speed of sound. If the clock is closed such that the air body and the emitter/receiver are all moving at the same speed, then you don't need to worry about the train speed.

If you say so. But the speed of sound in a stable medium is constant, as is its rate of decay, whether that medium is in a sealed carriage inside the moving train or in whatever surrounds it. If we're to assume that that the density of the medium is invariable then surely we can say that the whole experiment is an enclosed system - an experiment in an enclosed thought-box.

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