Maths problem

[mistermack]

I struggle with this, but those of a mathematical bent might find it easy.

Imagine you have made a clock that works on sound. A sender makes a ping. A reflector reflects it back. And your sender sends another, as soon as it hears the previous echo.
So your clock goes ping ping ping when stationary.

When the clock is moving in the air, the sound has further to travel to the reflector, and further back. ( picturing a reflector at right angles to the motion ) So the pinging (ie the clock) HAS to go slower.
Point the clock at 180 degrees the other way. It's still at the same 90 degrees to the motion.
The clock will slow exactly the same amount.

The question is, will the clock slow the same amount WHATEVER angle you point it at?
Or will the pinging speed up or slow down, depending on where you point it?

My brain goes on strike when I try to work out stuff like that.
But it's relevant to special relativity. Does air have it's own version of special relativity, if you use a clock that relies on the speed of sound in air?

[Brian Peacock]

Changing the angle will effect the amplitude of the ping, relative movement will effect it's fequency, but as sound travels at a fixed rate in a stable medium only variation in distance will effect the timing.

[mistermack]

No but the point is that the distance covered by the ping increases, with the velocity of the clock.
So there is more time between each ping, slowing the clock.
As far as I can see, this happens no matter where you point the clock.
So, if the clock is moving at a steady rate, does the interval between pings stay the same, wherever you point the clock?

[Brian Peacock]

As the distance between emitter→reflector→receiver increases so the clock will slow, because sound travels at a constant in a stable medium.

[NineBerry]

The cake eh the clock is a lie.

[mistermack]

As the distance between emitter→reflector→receiver increases so the clock will slow, because sound travels at a constant in a stable medium.

Yes, I think that's a given.
But, at a constant speed, does the clock run at the same speed, no matter what angle it is pointing to?
In other words, does the sound wave cover the same distance, whatever angle you point the clock.
Will the clock ping at a different rate, if say, the clock is aligned at 20, 30, 50 degrees to the direction of travel?

It's actually not a simple thing to work out. I tried but my head hurt. I was hoping some sad mathematician could rule on it.

[Brian Peacock]

Ah, you're thinking about the relativistic Doppler effect which comes into play when emitter and receiver are moving relative to each other. As mentioned this changes the frequency of the ping, with wavelengths compressing as the distance reduces and lengthening as it increases. Still, what I've said still holds - the clock's time depends on the distance of the loop. Soundwaves are diffused by the friction of the medium, but on the presumption that the receivers are perfectly sensitive, that the medium is stable, and that there's no sound absorbing, occluding or reflective surfaces in the system, and that emitter and receiver are not moving faster than the speed of sound for the medium, the relative angle of movement will still time the clock according to the basic geometry of the total length of the loop: time = distance/speed.

[pErvinalia]

I struggle with this, but those of a mathematical bent might find it easy.

Imagine you have made a clock that works on sound. A sender makes a ping. A reflector reflects it back. And your sender sends another, as soon as it hears the previous echo.
So your clock goes ping ping ping when stationary.

When the clock is moving in the air, the sound has further to travel to the reflector, and further back. ( picturing a reflector at right angles to the motion ) So the pinging (ie the clock) HAS to go slower.
Point the clock at 180 degrees the other way. It's still at the same 90 degrees to the motion.
The clock will slow exactly the same amount.

The question is, will the clock slow the same amount WHATEVER angle you point it at?
Or will the pinging speed up or slow down, depending on where you point it?

My brain goes on strike when I try to work out stuff like that.
But it's relevant to special relativity. Does air have it's own version of special relativity, if you use a clock that relies on the speed of sound in air?

Sound, unlike light, isn't a constant speed in all frames of reference. Therefore, the angle the clock points relative to the direction of travel will impact on it's recording of time passing. Imagine the clock travelling at 0.9 the speed of sound. The speed of the sound wave, from the reference of the clock, will be 0.1 of the speed of sound in a stationary reference frame. So the sound will take way longer to reach the reflector.

[mistermack]

Ah, you're thinking about the relativistic Doppler effect which comes into play when emitter and receiver are moving relative to each other. As mentioned this changes the frequency of the ping, with wavelengths compressing as the distance reduces and lengthening as it increases. Still, what I've said still holds - the clock's time depends on the distance of the loop. Soundwaves are diffused by the friction of the medium, but on the presumption that the receivers are perfectly sensitive, that the medium is stable, and that there's no sound absorbing, occluding or reflective surfaces in the system, and that emitter and receiver are not moving faster than the speed of sound for the medium, the relative angle of movement will still time the clock according to the basic geometry of the total length of the loop: time = distance/speed.

No Brian, the emitter and receiver are fixed relative to each other. But because the clock is moving, the distance that the sound travels through the still air is increased.

Imagine a train, with a long arm sticking out from it, and you are running on the ground.
You run between a point on the train, and the reflector. When the train is stationary, you just have to run X, the length of the arm. As the train gets going, you have to run X + extra distance, because the reflector moves ahead, and then the emitter moves ahead as well on your return. So, if you move the hands of a clock each time you return, it will run slower, the faster the train travels, because you are having to run farther and farther.

What I'm asking is, for a fixed speed of the train, will the distance I run change, depending on the angle of the arm to the train?

pErvin, the same question to you. I know that the clock will slow with the train moving, but will altering the angle of the arm, at a constant speed, change the speed of the clock?

[Brian Peacock]

Ah, you're thinking about the relativistic Doppler effect which comes into play when emitter and receiver are moving relative to each other. As mentioned this changes the frequency of the ping, with wavelengths compressing as the distance reduces and lengthening as it increases. Still, what I've said still holds - the clock's time depends on the distance of the loop. Soundwaves are diffused by the friction of the medium, but on the presumption that the receivers are perfectly sensitive, that the medium is stable, and that there's no sound absorbing, occluding or reflective surfaces in the system, and that emitter and receiver are not moving faster than the speed of sound for the medium, the relative angle of movement will still time the clock according to the basic geometry of the total length of the loop: time = distance/speed.

No Brian, the emitter and receiver are fixed relative to each other. But because the clock is moving, the distance that the sound travels through the still air is increased.

No MrM, the distance (therefore the time for sound to travel) from the emitter to the reflector is always fixed at the time the ping is sounded, but if the emitter is moving towards the reflector the return time will be shorter (i.e. distance from reflector to emitter) than if the emitter is moving away from the reflector. Presuming that the ping is travelling unencumbered at a constant, in a straight line, through a stable medium, the total distance (therefore the time) from emitter→reflector→emitter can be worked out with basic geometry.

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        <h1 id="page_title">Brian Peacock <span>Ping</span></h1>
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Copy and paste that into plain text editor like Notepad. Save it with a 'html' extension (eg: ping.html) and then open it in a browser. The timing component may wander depending on your computer's CPU but the distances will remain the same.

[pErvinalia]

Ah, you're thinking about the relativistic Doppler effect which comes into play when emitter and receiver are moving relative to each other. As mentioned this changes the frequency of the ping, with wavelengths compressing as the distance reduces and lengthening as it increases. Still, what I've said still holds - the clock's time depends on the distance of the loop. Soundwaves are diffused by the friction of the medium, but on the presumption that the receivers are perfectly sensitive, that the medium is stable, and that there's no sound absorbing, occluding or reflective surfaces in the system, and that emitter and receiver are not moving faster than the speed of sound for the medium, the relative angle of movement will still time the clock according to the basic geometry of the total length of the loop: time = distance/speed.

No Brian, the emitter and receiver are fixed relative to each other. But because the clock is moving, the distance that the sound travels through the still air is increased.

Imagine a train, with a long arm sticking out from it, and you are running on the ground.
You run between a point on the train, and the reflector. When the train is stationary, you just have to run X, the length of the arm. As the train gets going, you have to run X + extra distance, because the reflector moves ahead, and then the emitter moves ahead as well on your return. So, if you move the hands of a clock each time you return, it will run slower, the faster the train travels, because you are having to run farther and farther.

What I'm asking is, for a fixed speed of the train, will the distance I run change, depending on the angle of the arm to the train?

pErvin, the same question to you. I know that the clock will slow with the train moving, but will altering the angle of the arm, at a constant speed, change the speed of the clock?

Yes it will, as the distance needed to travel by the blip to record one second changes as the angle of the clock changes. I just did some calculations and I'll make some diagrams and bung them up in a little bit. But if we assume speed of sound as 300m/s and the train is travelling at 280m/s and the clock is perpendicular to the direction of train travel, after one second of the train travelling the clock will only have recorded 0.45 seconds. If the clock is at an angle of 45deg to train travel it will have recorded 0.59sec after one second of train travel. As the angle of the clock to the direction of travel reduces, so too does the distance the sound blip needs to travel, and therefore the slowing of the clock slows too (that is, it speeds up). My brain isn't awake enough yet to be able to work out accuracy of the clock when it is facing in the direction of travel. Maybe after some caffeine..

edit: actually, that figure of 0.59sec is wrong (although, qualitatively right; that is, the clock will be faster than when perpendicular due the shorter distance needed to cover). I'm not sure if I possess the math-fu to work out the exact value.

[mistermack]

Brian, that's a beautifully built demonstration, but I have to repeat, in the case that I am postulating, the emitter and the reflector CO-MOVE, they are all part of a one-piece clock.
I wish I'd stressed that more.
So the orange and green dots move in the same direction at the same speed.

What changes is just the direction that you point the clock.

[pErvinalia]

Not surprisingly my calculations were all wrong (and possibly my broad conclusions as well). I'm going to start on this one again..

[pErvinalia]

1. When the mirror is parallel to the direction of train travel, it will take 15.52 real seconds to register a second (using my parameters above). The sound wave will be travelling at 20m/s relative to the mirror/train. To get to the reflector 300 metres away it will take 15 seconds. On the return journey it is travelling at 580m/s second relative to the reflector/train. That 300m will be covered in 0.52 seconds.

[pErvinalia]

2. So as the clock moves away (i.e. is angled) from parallel, the vector component of the sound wave moving in the same direction as the train gets smaller and smaller. So on a very fast train like my example, eventually the clock would stop working as the vector component of the sound wave in the direction of travel would be less than zero. i.e. the sound would never make it to the reflector.

So, I'm going to switch over to a train travelling at more regular speeds now.