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Blueshift + Classical Physics = Golden Ratio

  1. Jan 30, 2008 #1
    I was doing calculations to see how far classical physics would take us in terms of the speed of an object never exceeding the speed of light in a reference frame. Here was the scenario I set up:

    http://la.gg/upl/light.jpg [Broken]

    So, if we want to find the time it takes the light to get from (A) to (B) and back to (A), we use the sum of the time it takes the light emitted from (A) at c to get to (B) and the time it takes the light reflected off (B) to get back to (A). This would equal (L/c) + ( L/(c+2V) ). Through algebra, this also equals 2L(c+V) / (c^2+2Vc).

    If the observers on (A) were to think that light always travels at c relative to themselves, they might use the time of this round trip to extrapolate the distance from (B) to themselves. If the light returns in the amount of time referenced above, the observers think it must have been moving at c for that amount of time if all it did was bounce off (B) and come back. Because the actual amount of time the light takes to return equals 2L(c+V) / (c^2+2Vc), the assumed distance the light had traveled would be 2Lc(c+V) / (c^2+2Vc) Since the assumption is that light is always traveling at c relative to (A), the observers would assume that the light took the same amount of time on the way to (B) as on the way back. This would yield an assumed distance of Lc(c+V) / (c^2+2Vc) from (A) to (B). Reducing this gets you L(c+V) / (c+2V). It is worth it to note that using twice this distance (for the trip there and back) and dividing it by the time for the trip will always result in the constant speed c for the observers calculations; this makes sense considering the assumption used by the observers (light always travels at c relative to themselves).

    We know the actual speed V of (B) relative to (A), so we can determine how far (B) actually travels in a given time period. Since speed=distance/time, it's also true that distance=speed*time (we already used this to determine the assumed distance of (B) from (A) in the last paragraph). Let's say that instead of sending out only a single light pulse, the observers on (A) also sent out another light pulse 1/n years after the first. We'll assume n is sufficiently large so that (B) will not have traveled far enough in that time period to hit (A). Planet (B) is traveling at V towards (A), so in a time period of 1/n, it will actually travel V*(1/n) closer to (A).

    The observers on (A), however, will be using the time of return of light pulses to determine the distance traveled by (B) in the time period 1/n. We already determined that the observers would come up with a value of L(c+V) / (c+2V) for their assumed distance away, and it's clear that the difference of this value from when (B) is at L away from (A) to when (B) is at it's new location will give the assumed distance traveled. We know that when the first light pulse hits (B), (B) is actually a distance of L away from (A), and after a time period of 1/n, (B) would have actually traveled V/n towards (A). This means the new location of (B) after a time period of 1/n would be L-(V/n). The distance traveled by (B) assumed by the observers on (A) would then be ((L)(c+V)/(c+2V)) - ((L-(V/n))(c+V)/(c+2V)) which equals (L - (L-(V/n)))((c+V)/(c+2V)) which equals (V/n)((c+V)/(c+2V)).

    Now the observers on (A) have the assumed distance (B) traveled in the period 1/n, and the duration of 1/n. From this, they can come up with an assumed speed for (B). Using speed=distance/time, they would assume the speed of (B) is ((V/n)((c+V)/(c+2V)))/(1/n) which equals (V((c+V)/(c+2V)) which equals (Vc+V^2)/(c+2V)). If you plug some numbers into this, you will find that observers on (A) will measure the speed of (B) to be lower than c, even if (B) is traveling at c toward (A).

    However, there is a limit to how fast (B) can actually be traveling before observers (A) will paradoxically measure the speed of (B) to be faster than c. To find this magic number, we must find the solution in terms of V to (Vc+V^2)/(c+2V))=c, which will give the speed (B) must actually be traveling at towards (A) for observers on (A) to assume its speed is c. Here is how it's done:
    V=c(1+ sqrt(5))/2)
    Ignoring the negative square root (as that would mean (B) would be moving away from (A)), we find that V is equal to approximately 1.61803399...

    When I saw this, I was astounded. I knew I recognized this number from somewhere. I punched it into Google, and sure enough, it was the Golden Ratio: http://en.wikipedia.org/wiki/Golden_ratio

    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 31, 2008 #2

    Doc Al

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    Staff: Mentor

    Your scenario is a bit ambiguous. What is L? Shall we assume that L is the distance between A and B (according to A) at the moment (according to A) that A emits the light signal? If so, then the time it takes for the light to reach B would be L/(c + V). And that would also be the time for the return trip as well. (All measurements according to A.)
    Last edited by a moderator: May 3, 2017
  4. Jan 31, 2008 #3
    L is the distance from (A) to (B) that observers on (A) would measure if only classical physics applied and both (A) and (B) were stationary. L is not the distance of (B) away from (A) at the time of the emission of the light signal, but the distance of (B) away from (A) at the time of the light striking (B).

    This makes the most sense because (A) does not know previously that (B) even exists (they are just sending out light pulses to map the things around them).

    I tried to make this clear by saying "Light emmited at c from (A) towards (B) hits a mirror on (B) when that mirror is a distance L away from the observer on (A)".
    Last edited: Jan 31, 2008
  5. Jan 31, 2008 #4

    Doc Al

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    Staff: Mentor

    OK, your first paragraph is clear enough. But doesn't your next sentence contradict that setup?

    All of a sudden relativity is back?

    I'd better go get some coffee.
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