Maths problem

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Re: Maths problem

Post by Feck » Fri Feb 24, 2017 2:52 pm

mistermack wrote: I made a comment on the science forum about the invariance of the speed of light, .
Yes I tried that, the answers normally get quite confused and angry :zilla:
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Re: Maths problem

Post by Brian Peacock » Fri Feb 24, 2017 7:08 pm

Here you go MrM. Revised model.

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If I were being scrupulous about this I'd have generated the data first and then animated it, but who's got time for that? As it is it just logs animation events as the script runs the differential equations. Any variation in timing will be down to allocation of CPU resources. There's some slight variation in distance but that's down to pixel management plus the aforementioned CPU, but the trigonometry is simple and the results demonstrate the horizontal symmetry in the system: i.e. 90&#176;-n <=> 90&#176;+n etc.
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Re: Maths problem

Post by Scott1328 » Fri Feb 24, 2017 11:25 pm

It seems to me this question comes down to: why don't I hear a Doppler effect from the radio or other passengers in the car when we are moving down the highway at 90 kph but a pedestrian certainly does hear a Doppler effect from those same sources?

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Re: Maths problem

Post by pErvinalia » Sat Feb 25, 2017 1:21 am

Brian Peacock wrote:Here you go MrM. Revised model.
That's such a cool animation, B. :awesome:

But it's probably wrong. I suspect you haven't added the train velocity to the vector component of the speed of sound against the direction of travel. Both Janus and I have done that (edit: actually, I think both of us forgot to do it for the 90deg example). When you do that, the calculation for angles between 0 and 90 degrees becomes non-trivial. That is, the length travelled by the sound wave isn't the simple case of 2x the hypotenuse as it is at 90deg. That is, the full path travelled doesn't form an isosceles triangle. It forms a scalene triangle.

As far as I understand it, that would be the case if the clock was open to the air (which we assume is still). If the clock was enclosed, then the clock and the medium through which the sound travels would have no relative velocity difference, in which case sound would travel the distance between the two reflectors/emitters at the stable speed of sound in both directions.
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Re: Maths problem

Post by Brian Peacock » Sat Feb 25, 2017 2:17 am

Perhaps then I was wrong about the symmetry - I just assumed the variance was down to pixel management and clock cycles (though the numbers are calculated in 32-floats and only rounded to two decimal places for the final display) - I am working on an 8 year old netbook that's feeling the strain though. I presumed the discrepancies were down to processing. If only I had something with a bit more grunt and a dedicated graphics processor... :sigh: As I said - if I was being a bit more rigorous I would have generated the data first and then animated it, and I'm pretty sure a suitably sufficient dataset would exceed the character limit for a forum post. Ah, well..

Running the model a number of times at 90&#176; consistently produces a slightly scalar triangle - a knat's pube further to the reflector than back again. Granted, we're talking about the ping travelling through an ideal (non-dispersive) medium, and that I'm only dealing with the wave front in a frictionless system, but still, sound travels at a constant in a stable medium independent of frequency or amplitude - (in a dispersive medium different frequencies travel at different rates, but for our purposes the ping represents a fixed frequency I guess) - which the animation shows both at the point of emission and the point of reflection. In this scenario the speed of sound, that is, the progression of the wave front, isn't multiplied by the speed of the train. That's basically what I was trying to show. I might run the output through some relativistic equations later just to check.
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Re: Maths problem

Post by pErvinalia » Sat Feb 25, 2017 3:16 am

It will never be multiplied by the train velocity. It will be added/subtracted if the clock is open to the surrounding non-moving air.

Can you tease out the algorithm you used to write the animation and show us how it would compute a clock at say 45deg to direction of train travel? I'm not saying it is potentially wrong due to clock cycles and rounding. It is seemingly wrong in concept (if we are dealing with a clock open to the air). I guess it depends on how we assume the clock to be constructed. Since it was MM's thought experiment, perhaps he could tell us which method of clock construction/operation fits his query the best.
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Re: Maths problem

Post by pErvinalia » Sat Feb 25, 2017 5:15 am

Scott1328 wrote:It seems to me this question comes down to: why don't I hear a Doppler effect from the radio or other passengers in the car when we are moving down the highway at 90 kph but a pedestrian certainly does hear a Doppler effect from those same sources?
Been trying to get my head around exactly what happens to sound waves in moving objects (and I guess stationary objects in a wind?). I suspect you would hear the Doppler effect if a couple of conditions where met. Let's say instead of a car you are on an open train bed that is travelling down the tracks at some speed. In the centre of the train bed is a horn mechanism that is sounding continuously. If you are at the front of the train bed and then move to the back of the train bed, I imagine you would hear the Doppler effect. Likewise you would hear the Doppler effect in your car example if it was possible to stand on planks that extended out of the front and back of the car. But you'd only notice it if you walked from one plank to the other.
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Re: Maths problem

Post by pErvinalia » Sat Feb 25, 2017 6:13 am

Actually, I don't know if that is right. My head hurts. Now I'm thinking the waves wouldn't be compressed or expanded as perceived by the person on the train while ever the person is moving at the same relative speed as the train.
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Re: Maths problem

Post by Brian Peacock » Sat Feb 25, 2017 10:00 am

If you were on the train with the horn atop the middle car you'd hear a constant tone at a slightly higher frequency if you were at the front and a slightly lower frequency if you were at the back - but there's be no Doppler effect because your distance from the sound source would be fixed. Sound travels at a constant in a stable medium.
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Re: Maths problem

Post by pErvinalia » Sat Feb 25, 2017 10:30 am

But you'd be moving away from the sound waves if in front of the horn. So the frequency would decrease, wouldn't it?
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Re: Maths problem

Post by Brian Peacock » Sat Feb 25, 2017 11:00 am

Look at the animation. Note how the sound front - the orange expanding circle - seems to bulge behind the moving clock? Now imagine a ping going out every second as the clock moves along with the train, producing multiple expanding circles. Those circles now represent the periodic cycle of a wave - closer together in front of the clock, further apart at the back, and equidistant at the side. Closer bands represent higher frequencies = higher pitches. Now imagine you're sat on the reflector: whatever the angle from the clock the periodic frequency - the interval between the circles arriving - is constant, therefore the pitch, tone or frequency is constant = no Doppler. Just higher frequencies (higher pitches) at the front of the train, lower frequencies at the back.

Keep those multiple circles in mind but imagine you were stood alongside the track with the train coming towards you. The edges of the first circles to arrive would be close together - at a high frequency - with the interval between their arrivals increasing as the train pulled alongside you, and at ever lengthening intervals as the train moved away from you = Doppler.
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Re: Maths problem

Post by pErvinalia » Sat Feb 25, 2017 11:18 am

That's not what Doppler is. The frequency doesn't change with distance from source. It only changes in relation to being either in front or behind the source.
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Re: Maths problem

Post by mistermack » Sat Feb 25, 2017 11:23 am

The whole thing is actually fiendishly complicated, the more you go into it.

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.
Going by what we've looked at so far, it's obvious that a sound clock undergoes time dilation with motion, and stops altogether when it reaches the speed of sound.
So the similarity is there to special relativity.
Then Janus claimed that the essential difference was that a clock working on sound would run faster or slower, depending on where you pointed it, whereas a light clock would always tick at the same rate.

Then he said that the light clock would ALSO run faster and slower, for different angles, if it wasn't for length contraction. So the only difference seems to be length contraction.

But I just looked up length contraction, and according to wikipedia, it arises out of time dilation. So there's no reason, if you are using sound clocks, why you shouldn't be experiencing length contraction in the same way.
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Re: Maths problem

Post by pErvinalia » Sat Feb 25, 2017 11:26 am

Time dilation has nothing to do with clock type. I explained this already. Time dilation occurs between relatively moving inertial frames. At non-relativistic speeds the sound clock reads slower simply because of the greater distance the sound has to travel on a moving train.
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Re: Maths problem

Post by mistermack » Sat Feb 25, 2017 11:55 am

pErvin wrote:Time dilation has nothing to do with clock type. I explained this already. Time dilation occurs between relatively moving inertial frames. At non-relativistic speeds the sound clock reads slower simply because of the greater distance the sound has to travel on a moving train.
You're talking about time as measured by the clocks that you are familiar with, that's all.
If your time is measured by sound clocks only, then your time will dilate with motion of the clock in the medium.

Nobody's claiming that your normal everyday time is affected by the clock type. That's a red herring.
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