View Full Version : distance of sound emitter

grav

2008-Nov-15, 01:36 PM

Let's say that we "see" in a similar way that bats do, except the sound is emitted by an object and travels to us through the air, instead of simply reflecting off of the object. If we are stationary with the object, the object is "seen" at a distance of d. So if we were moving toward the object, would the angles of observation change so that the object appears to be at a different distance? The distance of the object is not measured by the intensity of the sound, like an object producing light might be measured by its illumination, but only by the angles that the sound waves fall upon the observer.

01101001

2008-Nov-15, 01:43 PM

So if we were moving toward the object, would the angles of observation change so that the object appears to be at a different distance?

I haven't had my morning coffee yet.

If we are moving directly toward an object why would any angles be changing? The angle is defined by what rays?

grav

2008-Nov-15, 01:50 PM

I haven't had my morning coffee yet.

If we are moving directly toward an object why would any angles be changing? The angle is defined by what rays?I'm not sure. I'm not even sure if this question can be answered or not in the traditional sense. I need my morning coffee too. Good idea, thanks. :)

grav

2008-Nov-15, 02:13 PM

Okay, let's just try it for light first. If Alice and Bob are stationary to each other at some distance, and Carl is moving past Alice toward Bob, would Carl optically observe a different distance to Bob due to a change in the angles of the light with a relative speed?

grant hutchison

2008-Nov-15, 02:27 PM

Okay, let's just try it for light first. If Alice and Bob are stationary to each other at some distance, and Carl is moving past Alice toward Bob, would Carl optically observe a different distance to Bob due to a change in the angles of the light with a relative speed?Yes.

The full treatment of relativistic aberration gives a change in apparent distance as well as a change in apparent angle to the line of flight. The change in apparent distance would be visible if you attempted to triangulate a distant object, or if you judged its distance by its angular size.

Grant Hutchison

grav

2008-Nov-15, 04:35 PM

Yes.

The full treatment of relativistic aberration gives a change in apparent distance as well as a change in apparent angle to the line of flight. The change in apparent distance would be visible if you attempted to triangulate a distant object, or if you judged its distance by its angular size.

Grant HutchisonThe line of flight is straight toward the emitting object, so that doesn't change. As far as the apparent angular distance, would that become closer or further away than for an observer that is stationary with the object?

On a side note, I'm glad they fixed the forum pages, stuck on yesterday's posts. I had to go to my recent posts to access this thread. Kind of a strange way to start off my birthday, but it makes things interesting. :)

Jeff Root

2008-Nov-15, 04:45 PM

Is it necessary to bring in relativity to get a changing angle? If so, that

seems to imply that ordinary speeds would not show any effect, for either

light or sound.

-- Jeff, in Minneapolis

grav

2008-Nov-15, 04:46 PM

The line of flight is straight toward the emitting object, so that doesn't change. As far as the apparent angular distance, would that become closer or further away than for an observer that is stationary with the object?Oops. Never mind. I worked it out using the Relativistic aberration formula, but I rushed it and used a positive v/c instead of negative for approach and got impossible results. I'll try doing it the correct way this time. :) So for approach, it seems the object will appear futher away.

grav

2008-Nov-15, 04:50 PM

Is it necessary to bring in relativity to get a changing angle? If so, that

seems to imply that ordinary speeds would not show any effect, for either

light or sound.

-- Jeff, in MinneapolisI found relativistic effects of contraction between frames performing that classically as well, like for light travelling through a medium, even at different measured speeds of light in different frames, so I am exploring that, but I originally used the same optical distance for the stationary observer as for one that coincides at the same place but has some relative speed. So I will need to find out how the angle of observation changes classically as well, and compare effects.

grant hutchison

2008-Nov-15, 04:50 PM

Is it necessary to bring in relativity to get a changing angle?No, you get a similar effect with classical aberration (the sort of angular shift you get in the apparent direction of the rain when you run through it). Relativistic length contraction merely modifies the effect.

Grant Hutchison

grav

2008-Nov-15, 05:18 PM

Okay, the formula for Relativistic aberration is cos 0' = (cos 0 - v/c) / [1 - (v/c) (cos 0)], where v is negative upon approach, or cos 0' = (cos 0 + v/c) / [1 - (v/c) (cos 0)] if we make v/c positive upon approach. We want to find the distance observed with some relative speed according to the new angles. As you said Grant, we can triangulate upon the observed position by finding the distance from the center of the eye or camera lens that the light falls upon and the angle from that point to the distant object. We then find the new angles accordingly and find where they would intersect from the same points of reception to the observer. The new observed distance becomes d_obs = d (tan 0') / (tan 0). However, doing this using the Relativistic aberration formula, the new angles become somewhat distorted where they intersect, although not by very much for small angles of reception, as they would be over a small angles of reception from a large distance to the object anyway.

For instance, at v/c = .6, with the light emitted at a distance of d from the observer's current position, an original angle of one degree becomes .500009534 degrees observed and an original distance of two degrees becomes 1.000076183 degrees. So the observed distance with an original angle of one degree is 2.000114188 d and for two degrees it becomes 2.000457112 d. We find that with a very large actual distance for the object, so over very small angles of reception, the observed distance then works toward a limit of sqrt[(1 + v/c) / (1 - v/c)] d. So ignoring the distortion, that is the observed distance I will go by. It gives sqrt(3) times the observed distance as the actual distance with a relative speed for v/c of .5, twice as great for .6, 3 times as great for .8, etc.

grav

2008-Nov-15, 05:28 PM

What is the formula for classical aberration? I can't find it, but I have a link where I could determine eventually from the numbers.

grav

2008-Nov-15, 05:43 PM

What is the formula for classical aberration? I can't find it, but I have a link where I could determine eventually from the numbers.Okay, no. That link (http://www.fourmilab.ch/cship/aberration.html) only shows it for an original angle between the emitted rays and the line of travel of 90 degrees, like running through rain. The results they give for the observed angle with different relative speeds comes out to 90 - atan(v/c) degrees. So would that become just 0 - atan(v/c) depending upon the original angle, then?

grav

2008-Nov-15, 06:17 PM

Well, let's just start with the Relativistic aberration first. If Alice is stationary with Bob at a distance of d, and Carl coincides with Alice, travelling directly toward Bob, then Alice observes Bob at an optical distance of d and would say that he is currently at a distance of d also, since he remains stationary to her. Carl, on the other hand, with a relative speed of v to Bob, will optical observe Bob at a distance of d * sqrt[(1 + v/c) / (1 - v/c)]. So Carl would say that the light that he observes has taken a time of t = d / c to reach him, and that Bob has moved (d / c) v closer to Carl over that time. So Carl says that Bob's actual current position is d * sqrt[(1 + v/c) / (1 - v/c)] - d v / c. For v/c = .6, that gives a current distance of 1.4 d according to Carl, so Carl says that Alice and Bob are currently separated by that distance, 1.4 times greater than Alice would say. For v/c = .8, it would be 2.2 times greater. Is all of that right?

astromark

2008-Nov-15, 08:45 PM

Can you see what I can ? Do I know how to impart that information ?

Talking of sound as a tool for 'imaging' the world around us is foreign to us. So its a big ask. Some of the oceans dwellers do it as a matter of course. Ask a whale !... sounds a little off hand, but is true. Thats the world they live in. So how to describe this easly... Looking at the math equations above nearly frightened me away. So that does not cut the mustard for me. Even if it is right. As I will claim to understand the Doppler effect as sound tone is easy to demonstrate this trick. Is Aguilar momentum going to change my perspective of where that object is. Not if I am able to calculate that movement va the sound tone changes. If I have been unable to paint this image well, that would not surprise me. Try communicating with a ocean going giant submarine... The world they move through is sonar in nature but they are never blind. I have watched Dolphins play with a ball... darken the water to prohibit the vision. They still see the ball.

As for the maths... un-necessary. We learn to catch the ball without ever considering all the math curves. We have adapted to this environment we live in as has the whale. Those creatures living in the atmosphere of Jupiter can not use vision as we know it. ( Thats a curve ball ) Right out of left field., but I wonder where this is going ?

grav

2008-Nov-15, 09:02 PM

Can you see what I can ? Do I know how to impart that information ?

Talking of sound as a tool for 'imaging' the world around us is foreign to us. So its a big ask. Yes, thanks Astromark, I agree. :) I realized that as soon as I posted it. I've since changed it back to the aberration of light. I'm having a hard enough time dealing with that as it is.

grav

2008-Nov-17, 02:05 AM

Well, I've finally worked out the optical distance classically. It took a while but seems pretty simple now that it's done. It comes down to simple addition of vectors. If the light travels from an emitter at c, with an angle of 0 between the path of the light and an observer's line of travel, then it reaches the observer at a speed of (sin 0) c perpendicular to his line of motion and (cos 0) c along his line of motion. To find the new angle of observation, we simply add the relative speed along the line of travel to the speed of the light along that path, giving (cos 0) c + v. Now, the vectors for the distances along each path also add in exactly the same way along the same vectors. So if (cos 0) c is the same as the old path in the line of travel and (cos 0) c + v is the new one, then the ratio between those is just [(cos 0) c + v] / [(cos 0) c] = 1 + (v/c) / (cos 0), and since we only want the distance in the forward direction only anyway, with the observer travelling straight toward the emitter, then (cos 0) = 1, and the ratio of the optical distance to the actual distance is simply 1 + v/c. So if Carl coincides with Alice while travelling directly toward Bob, and Alice observes a distance of d between Alice and Bob while stationary to Bob, then Carl will observe a distance of (1 + v/c) d. So in the time that the light has taken to reach Carl, Carl would say Bob has since travelled a distance of d v /c directly toward Carl, whereas Bob's current distance is then (1 + v/c) d - d v / c = d, the same as Alice would say. So it appears that classically Carl and Alice would agree upon the distance between Alice and Bob.

I could have just left it at that, but I wanted to be absolutely sure, so I also worked out the formula for the new angle in order to compare it to the numbers in the link from a few posts back. The distance along the cross-wise path of the light would remain the same with the old and new angles, and so would the speed along that vector, so (sin 0) c = (sin 0') c', where c' is some new speed for the light along the new resultant vector classically, and (cos 0') c' = (cos 0) c + v also. Finding for and cancelling out the c' in each case, we get

c' = (sin 0) c / (sin 0') = [(cos 0) c + v] / (cos 0), so

(sin 0) / (cos 0) = [(sin 0') / (cos 0')] [1 + (v/c) / (cos 0)]

(tan 0) = (tan 0') [1 + (v/c) / (cos 0)]

(tan 0') = (tan 0) / [1 + (v/c) / (cos 0)]

Using that, one can find the same new angles as were found in the link, but since the rain is falling at 90 degrees, tan 90 becomes infinity, so is a little difficult to work with, but one can come very close to the correct new angle by using some angle just under 90 degrees originally. It would be better if we used angles that can be dealt with better, though, like the standard cosine, so let's translate it to that and we can compare it to the relativistic formula as well. That becomes

(tan 0') [1 + (v/c) / (cos 0)] = (tan 0)

(tan 0') [(cos 0) + (v/c)] = (tan 0) (cos 0)

[(sin 0') / (cos0')] [(cos 0) + (v/c)] = (sin 0)

(sin 0') [(cos 0) + (v/c)] = (sin 0) (cos 0')

sqrt[1 - (cos 0')^2] [(cos 0) + (v/c)] = sqrt[1 - (cos 0)^2] (cos 0')

[1 - (cos 0')^2] [(cos 0 + (v/c)]^2 = [1 - (cos 0)^2] (cos 0')^2

[1 - (cos 0)^2] [(cos 0)^2 + 2 (v/c) (cos 0) + (v/c)^2] = [1 - (cos 0)^2] (cos 0')^2

(cos 0)^2 + 2 (v/c) (cos 0) + (v/c)^2 - (cos 0')^2 (cos 0)^2 -2 (v/c) (cos 0) (cos 0')^2 - (v/c)^2 (cos 0)^2 = (cos 0')^2 - (cos 0)^2 (cos 0')^2

(cos 0)^2 + 2 (v/c) (cos 0) + (v/c)^2 - 2 (v/c) (cos 0) (cos 0')^2 - (v/c)^2 (cos 0)^2 = (cos 0')^2

(cos 0')^2 [1 + 2 (v/c) (cos 0) + (v/c)^2] = (cos 0)^2 + 2 (v/c) (cos 0) + (v/c)^2

(cos 0') = [(cos 0) + (v/c)] / sqrt[1 + 2 (v/c) (cos 0) + (v/c)^2]

So that is the new classical angle using cosines with 'v' positive for the observer moving toward the emitter. The standard convention, however, seems to be a positive 'v' when the observer is moving away, so in that case, the formula would then become (cos 0') = [(cos 0) - (v/c)] / sqrt[1 - 2 (v/c) (cos 0) + (v/c)^2].

Well, that was fun. Yeah, right. Are you still there, Astromark? If so, don't worry. I doubt I'll be posting like that again any time soon. Not too awful soon, anyway. :)

grav

2008-Nov-17, 02:42 AM

It looks like the only difference in the formulas for the classical version and relativistic version of the cosine of the new angles is that the denominator of the classical version is sqrt[1 - 2 (v/c) (cos 0) + (v/c)^2] while the denominator of the relativistic version is sqrt[1 - 2 (v/c) (cos 0) + (cos 0)^2 (v/c)^2] = 1 - (cos 0) (v/c).

grav

2008-Nov-17, 03:30 AM

Well, that was fun. Yeah, right. Are you still there, Astromark? If so, don't worry. I doubt I'll be posting like that again any time soon. Not too awful soon, anyway. :)Whoops. Sorry, Astromark. Here we go again. Close your eyes. :)

Now I want to do the exact same thing as I did in the next to last post to find the classical formula for aberration, but in reverse, and starting with the relativistic formula to find what the starting point for that would be. So using the positive convention for 'v' upon approach, we have

(cos 0') = [(cos 0 + (v/c)] / [1 + (v/c) (cos 0)]

(cos 0') = [(cos 0 + (v/c)] / sqrt[1 + 2 (v/c) (cos 0) + (cos 0)^2 (v/c)^2]

(cos 0')^2 [1 + 2 (v/c) (cos 0) + (cos 0)^2 (v/c)^2] = (cos 0)^2 +2 (v/c) (cos 0) + (v/c)^2

[1 - (cos 0')^2] [(cos 0^2 + 2 (v/c) (cos 0) + (v/c)^2] = [1 - (cos 0)^2] (cos 0')^2 + (v/c)^2 (cos 0)^2 (cos 0')^2 - (v/c)^2 (cos 0')^2

[1 - (cos 0)^2] [(cos 0)^2 + 2 (v/c) (cos 0) + (v/c)^2] = [1 - (cos 0)^2] [1 - (v/c)^2] (cos 0)^2

(sin 0') [(cos 0) + (v/c)] = (sin 0) sqrt[1 - (v/c)^2] (cos 0)

(tan 0') = (tan 0) sqrt[1 - (v/c)^2] / [1 + (v/c) / (cos 0)]

With the classical version, it began with (tan 0') = (tan 0) / [1 + (v/c) / (cos 0)], so the relativistic version is just the exact same thing with the tangent of the angles except that it includes a multiplication of the Lorentz contraction. That would give us an optical distance that is greater than the actual distance by [1 + (v/c) / (cos 0)] / sqrt[1 - (v/c)^2]. If the observer is travelling directly toward the emitter as Carl is toward Bob, then (cos 0) = 1 and we get an optical distance for Carl that is greater than what Alice sees by [1 + v/c] / sqrt[1 - (v/c)^2] = sqrt[(1 + v/c) / (1 - v/c)], the same as we found earlier in the thread.

astromark

2008-Nov-17, 05:18 AM

Aaugh !............. slam.

grav

2008-Nov-17, 05:31 AM

So if Carl coincides with Alice while travelling directly toward Bob, and Alice observes a distance of d between Alice and Bob while stationary to Bob, then Carl will observe a distance of (1 + v/c) d. So in the time that the light has taken to reach Carl, Carl would say Bob has since travelled a distance of d v /c directly toward Carl, whereas Bob's current distance is then (1 + v/c) d - d v / c = d, the same as Alice would say. So it appears that classically Carl and Alice would agree upon the distance between Alice and Bob.I made a mistake there. If Carl observes a distance of (1 + v/c) d between himself and Bob while also observing Bob travelling toward him at v, then he should conclude that it took the light a time of d (1 + v/c) / c to reach him, and in that time, Bob has travelled closer to Carl a distance of d (1 + v/c) v / c, so Bob's current distance would become d (1 + v/c) - d (1 + v/c) v / c = d (1 + v/c) (1 - v/c) = d [1 - (v/c)^2].

If Carl and Alice are to agree upon the current distance of Bob, even just classically, where Alice would say it is d, and Carl optically observes a distance to Bob of d x, where x is some variable, then Carl would say that the light has taken a time of d x / c to reach him, and that Bob's current distance is d x - (d x / c) v, whereas d x - (d x / c) v = d in order to match what Alice measures, then x - v x / c = 1, so x = 1 / (1 - v/c). Therefore, Carl must optically observe Bob at a distance of d / (1 - v/c), not d (1 + v/c) as found, in order to agree with Alice's measurement. That's interesting.

grav

2008-Nov-17, 06:26 AM

With the classical version, it began with (tan 0') = (tan 0) / [1 + (v/c) / (cos 0)], so the relativistic version is just the exact same thing with the tangent of the angles except that it includes a multiplication of the Lorentz contraction. That would give us an optical distance that is greater than the actual distance by [1 + (v/c) / (cos 0)] / sqrt[1 - (v/c)^2]. If the observer is travelling directly toward the emitter as Carl is toward Bob, then (cos 0) = 1 and we get an optical distance for Carl that is greater than what Alice sees by [1 + v/c] / sqrt[1 - (v/c)^2] = sqrt[(1 + v/c) / (1 - v/c)], the same as we found earlier in the thread.Yep, I thought there was something strange about this, since the Lorentz factor is divided by the classical optical distance instead of multiplied. But if the classical version should be 1 / (1 - v/c) instead of (1 + v/c), then the relativistic version is sqrt[(1 + v/c) / (1 - v/c)] = sqrt[1 - (v/c)^2] / (1 - v/c), whereas the Lorentz contraction is multiplied as it should be instead of divided. I thought I had worked the classical version out naturally and it matched the numbers in the link, so they must've found the same formula for the classical version, but there must be something else that's missing. I'll keep working on it.

As far as the relativistic version goes, it should also give a Lorentz contracted length for an object travelling with some relative speed, so if Bob lies down longwise, Alice will see his feet at d1 and his head at d2, with a length of d2 - d1. Carl will see his feet at d1 x and his head at d2 x, where x is some variable for the distances with some relative speed. So Carl will measure Bob's current positions as d1 x - d1 x v / c for his feet and d2 x - d2 x v / c for his head, and the difference between them should be a Lorentz contracted distance of what Alice measures, so (d2 x - d2 x v / c) - (d1 x - d1 x v / c) = sqrt[1 - (v/c)^2] (d2 - d1), (d2 - d1) x (1 - v/c) = sqrt[1 - (v/c)^2] (d2 - d1), x = sqrt[1 - (v/c)^2] / (1 - v/c), x = sqrt[(1 + v/c) / (1 - v/c)], which is exactly what we found for that. So everything seems consistent with the relativistic version so far.

grav

2008-Nov-17, 03:49 PM

Oh, okay. Apparently I actually made two mistakes for the same thing, when subtracting the distance Carl would say Bob has travelled during the time the light takes to reach him. Simple addition of vectors says that classically, Carl optically observes Bob at d [1 + (v/c) / (cos 0)], where d is the distance Alice observes and measures for Bob, being stationary to her. Now, the thing here is that if Alice measures the speed of the light travelling to her as c classically in the same frame, then Bob would measure it as c + v due to his relative speed. So Carl would say the light takes a time of t = d [1 + (v/c) / (cos0)] / (c + v) to reach him and during that time, Bob has travelled a distance of d [1 + (v/c) / (cos 0)] v / (c + v), placing Bob's current position at d [1 + (v/c) / (cos 0)] - d [1 - (v/c) / (cos 0)] v / (c + v), where (cos 0) = 1, so d (1 + v/c) - d (1 + v/c) v / (c + v) = d (1 + v/c) [1 - v / (c + v)] = d [(c + v) / c] [ ((c + v) - v) / (c + v)] = d [(c + v) / c] [ c / (c + v)] = d, the same as Alice measures, and the formula for the new angle classically that was found before is still correct.

Also, if Alice and Carl are to measure the same length for Bob, with his feet at d1 and his head at d2, then [d2 x - d2 x v / (c + v)] - [d1 x - d1 x v / (c + v)] that Carl measures must equal d2 - d1 that Alice measures, where x is some variable for the optical distance. So we get (d2 - d1) x [1 - v / (c + v)] = d2 - d1, x = 1 / [1 - v / (c + v)] = (c + v) / [(c + v) - v] = 1 + v/c. So that works out too now.

Apparently, then, if Carl is to measure the Lorentz contracted distance that Alice sees with the relativistic version, then first we must transform the greater optical distance of 1 + v/c that Carl sees classically with the light travelling toward him at c + v due to his relative speed to that of 1 / (1 - v/c) as we found earlier if the light travels to him at just c, for c in all frames. Then we can just apply the Lorentz contraction to that and get sqrt[(1 + v/c) / (1 - v/c)] for the relativistic version.

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