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Visualisation of relativity

  1. Dec 30, 2004 #1
    Hi,
    This is a question to which I haven't found an answer anywhere, so I'm probably seeing this wrong but I can't seem to figure it out.
    I realize time is warped relative to speed, what I don't understand is the following:
    -Considering 2 people, A and B. Let us say that A is stationary and B is moving away from A at a speed very close to that of light (such as the famous twin paradox).
    -Now consider that A is emitting an information bearing, electromagnetic signal towards B.
    -The way I see it, while B is moving away from A, it is receiving a red shifted beam of information. B is "learning" a lot slower than if he were at rest.
    -Now let B turn around at some point and move towards A at the same speed as he was moving away. B is now receiving a blue shifted signal and is "learning" at very fast pace.
    -When B finally reaches A, he will have been subjected to all the information broadcasted by A, but at some point will have received it a lot faster, and at some point a lot slower.
    - If B's time is running a lot slower as it is proven to, and A has broadcasted one year of his life, does this mean that B has seen everything that has happened in fast forward all the time? Even though the signal was red shifted for half of his trip?
    I can't seem to put together the idea of an information exchange with time compression... I visualize electromagnetic waves as constant speed separated crests, which naturally seem closer together if you're traveling towards them, and further apart if you are traveling away.
    In the same way, if B is sending EM waves while coming closer to A, A will receive a compressed signal followed from very close by the arrival of B (lets say B is traveling at 99.999999% of c)...
    I know a drawing would help but if somebody understands what I'm trying to say could you please enlighten me? Thanx :)
     
    Last edited: Dec 30, 2004
  2. jcsd
  3. Dec 30, 2004 #2

    pervect

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    You might find this particular subsection of the sci.physics.faq helpful

    http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html

    But i'll give a thumbnail sketch of what the traveller sees. The doppler shift factor, k is a function of velocity. So let's say that the doppler shift factor is 10:1.

    On the outbound trip, B travels for some time period (let's say 1 year, to make it simple), at a doppler shift factor of 10, and sees the signals from .1 years of events on the planet.

    On the inbound trip, B travels 1 year at a doppler shfit factor of 10, and since it turns out that the doppler shift factor is the recirpocal of what he saw on the outgoing trip - so he sees 10 years of signal in his 1 year return trip.

    The net result is that B, on a 2 year trip, has seen 10.1 years elapse for the person who stayed behind.

    When we look at the relativistic equations, this checks. The doppler shift factor is [tex]\sqrt{\frac{c+v}{c-v}[/tex]

    ref:
    http://scienceworld.wolfram.com/physics/DopplerEffect.html

    So if your velocity is (99/101)*c, you will experience a doppler factor of 10 (substitute and check - also substitue v= -(99/101)*c into the formula above, and see that the doppler shift factor is indeed .1 on the outbound trip).

    The formula for relativistic time dilation is

    [tex] gamma = \frac{1}{\sqrt{1-(v/c)^2}} [/tex]

    If you substitute v=(99/101) * c in the above expression, you get a gamma of 5.05.

    This checks - B has travelled 2 years, and with a time dilation factor of 5.05, we expect that 2*5.05 = 10.1 years have elapsed on earth. Which is the same result we found earlier.
     
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