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Understanding the speed of light between 2 ships racing away from each other

  1. Dec 12, 2012 #1
    I'm not sure if this post belongs with general physics or relativity, so feel free to move it. I am trying to get a better grasp of special relativity and I think the following scenario will help.
    Say rocket X moves at -0.51c and rocket Y moves at +0.51c and they take off from point Z. When the rockets are 1 light day apart, rocket X turns on rear lights. I would think that an observer at point Z would see the light from rocket X in 12 hrs. What about Rocket Y? Would they see the light in 1 day? Would they see the light at all?
  2. jcsd
  3. Dec 12, 2012 #2


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    In coordinate system of Z, it's very simple. Light travels distance ct, while rocket travels 0.51ct, and light needs to travel an extra light-day to catch up. So in units of days, 0.51ct+c = ct. t = 1/0.49, or just over two days for light from X to catch up with Y. At this point, there isn't really any relativity involved.

    From perspective of X and Y, however, the time will be different from Z and from each other. That's where relativity kicks in.
  4. Dec 13, 2012 #3


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    I've made a spacetime diagram to illustrate K^2's answer:


    The thick red line is showing the trajectory of Rocket X while the black line is showing the trajectory of Rocket Y. They both start out from position Z in blue. The thin red line shows the trajectory of the light that Rocket X turns on when the rockets are one light-day apart.

    The dots represent one-day increments of time on a clock carried with each observer.

    Attached Files:

  5. Dec 13, 2012 #4
    Thanks K^2. I need a clarification on the above quote. If relativity were not involved, would not the observer at K see the light beam approaching at .49c? My thought is that with relativity, the light beam will travel at c (meaning relativity is involved). Where am I going wrong?

    ghwellsjr, thanks for the awesome graph. According to the graph, light from rocket X would take 1/2 a day to reach point Z and 2 days to reach rocket Y. But what is happening to the observer in rocket Y. How is time perceived in rocket X compared to rocket Y?
  6. Dec 13, 2012 #5
    ghwellsjr, somehow I missed the note about the dots on the graph. But now I see it would take 2.5 days for the light from X to reach Y. And it looks like an observer from Z would think that the light would take two days to travel from X to Y. Its clear that the dots should be placed at each grid tic for the blue line, but how did you know where to place the dots on the red and black line?
  7. Dec 13, 2012 #6


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    The dots are placed according to the time dilation factor which is represented by the Greek letter gamma, γ. In case you don't know how to calculate it, it is:

    γ = 1/√(1-β2) where β (beta) is the speed as a fraction of the speed of light. In your case, the rockets are traveling in the rest frame of Z at 0.51c so β=0.51 and we can calculate gamma as:

    γ = 1/√(1-β2) = 1/√(1-0.512) = 1/√(1-0.2601) = 1/√(0.7399) = 1/0.86 = 1.1626.

    So this means that the rocket dots are placed so that the spacing of their time coordinates is 1.1626 times the spacing of the time coordinates.

    Does that make sense?

    I'm going to upload two more diagrams, one in which the red rocket is at rest and one in which the black rocket is at rest and I'll have more discussion at that time including answering your other questions.
  8. Dec 13, 2012 #7


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    Light travels at c with respect to the coordinates of an Inertial Reference Frame (IRF). It does not travel at c with respect to observers that are moving in the IRF. But no matter what the IRF, no observer can see the speed at which light is propagating.
    That question does not depend on the IRF and is not a question about time dilation. Rather, we use Relativistic Doppler as an easy way to analyze how each rocket perceives time in the other rocket. First we have to calculate the relative speed between the two rockets using the Velocity Addition formula (see the wikipedia article for more info). The calculation is:

    (0.51+0.51)/(1+0.51*0.51) = 1.02/1.2601 = 0.809

    Now we use the Relativistic Doppler factor (see the wikipedia article):

    √((1+β)/(1-β)) = √((1+0.809)/(1-0.809)) = √(1.809/0.191) = √9.471 = 3.078

    This means that the black rocket Y will see his own clock advance at just over three times the rate that he sees the red rocket X's clock advance which we can easily see from the first graph in post #3.

    OK, now let's see those other two graphs. First the IRF in which the red rocket X is at rest:


    Now you can see that the red rocket X's clock is ticking in step with the coordinate time and is not time dilated but the black rocket Y's clock and Z's clock are time dilated. Note also that the distance between the rockets when X turns on his light is less than one light-day. However, the Relativistic Doppler factor is the same as it was before. Note how the light travels at c (it's defined to do that) and now takes longer, about 3.5 days to get from X to Y.

    Can you do the calculations to support the spacing of the dots for the time dilations in this IRF?

    Finally, the graph for the IRF in which the black rocket Y is at rest:


    Once again, the time dilations are different based on the speed of each observer's clock in the IRF and the time for the light to travel from X to Y is about 1 day and the distance between the rockets when X turns on his light is now more than one light-day but the same Doppler Factor applies.

    Does this all make perfect sense to you?

    Attached Files:

    Last edited: Dec 14, 2012
  9. Dec 13, 2012 #8


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    If it were not for relativity, you would conclude that a baseball thrown backwards with velocity c from a ship going .51c would appear to travel at .49c. But electromagnetic wave is governed by Maxwell's Equations, which give you speed of c even if the source is traveling away at .51c. So in classical mechanics + classical electrodynamics, you still get the correct answer, and you would still conclude that light takes just over two days to reach the other ship.

    So long as you only consider frame of reference of Z, there is nothing wrong with this proposition even without considering relativity. It's only when you say, "Wait a minute. But shouldn't the person on rocket X also see light departing at c," that it becomes apparent that relativity is necessary. In other words, when you start considering two different frames of reference you have to do so with relativity in mind. And ghwellsjr's diagrams for the other two frames of reference demonstrate that clearly.

    From perspective of Y, and counting from when the two rockets left. From perspective of Z, it still takes just a touch over 2 days. The grids on each graph show coordinates for the observer that's stationary in that coordinate system. In other words, whose world-line is a vertical line.
  10. Dec 14, 2012 #9
    Yes - excellent.

    Can you clarify this? Does this mean that the observers will see the light traveling at c but only because their clocks are different?

    I believe it will still be 1.16 for the blue line, but I'm not sure how to calculate beta for the black line.
  11. Dec 14, 2012 #10
    Thanks for your response. In my post 5, I started reading the graph as if time were on the x-axis instead of the y-axis. I do see that from the perspective of Z, it takes just over 2 days.
  12. Dec 15, 2012 #11


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    No one can ever see light traveling. Have you thought about this? We can see other things traveling because we illuminate them with light and because the light always travels faster than those other things. But how would you see light traveling? The best thing we can do is put stationary things in its way so that the light bounces off of them but now we always have this problem that we are really looking at a round trip for the light and so we have to make some assumptions about how the light is traveling in order to answer the question and the assumptions create the answer so we will be fooling ourselves if we think we can actually answer the question.

    But Einstein's assumption is that light travels at c in every IRF and so that is how I drew the graphs and both graphs provide a description that is consistent with all the observations and measurements, even though they both give a totally different picture. Just remember that the differences are caused by our arbitrary assignment of the speed of light being c in each IRF.

    So it's not just that their clocks are different that permits them to get the same value for c when they measure the round-trip speed of light, it's also because their rulers contract along the direction of motion in the IRF.
    It's really very simple. You just look at how far black has progressed at some point and how long it took him to get there and divide to get beta. If you look at the black line on the first graph, you will see that it passes very close to the grid marks corresponding to 4 light-days and 5 days for a speed of about 0.8c. Actually, I calculated the speed using the Velocity Addition formula and got a value of 0.809c and so that's how I drew the graph.

    Does that make sense?
  13. Dec 15, 2012 #12
    I thought it was clear that I was referring to "measuring" the light speed. But I understand the confusion due to the poor word choice on my part. I know it's off topic, but you might be interested in the following videos on you tube. The first is from MIT, which uses an imaging system that can collect 1 trillion frames per second. You can literally see the light move and bounce around a 2 liter coke bottle.
    The second video is about Danish physicist, Lene Vestergaard Hau out of Harvard University. I'm sure you are familiar with this one but if not… she uses a Bose Einstein condensate to slow light down to speeds of around 15 mph.

    Yes. You previously defined Beta as "the speed as a fraction of the speed of light". I did not make the connection of your definition of Beta with the slope of the line on the graph. When you spelled it out, it is very obvious.
    With a much better understanding, I can see that the spacing for the blue line is 1.16 and the spacing for the black line is 1.67. You have been an incredible help!!
    Last edited by a moderator: Sep 25, 2014
  14. Dec 15, 2012 #13


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    You are not actually seeing the light move, however. You are seeing the light that got scattered from these points in space some time ago, with light scattered that way reaching camera now. This is a very important distinction in relativity, because time delay will depend on speed of light and distance, so you will not see exactly how light was moving. Furthermore, you never observe the same quanta of light moving along. The portion of light you saw at one point did not make it to the next point.

    You might be aware of all of this, but I just want these points to be clear.
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