
#1
Dec1212, 07:28 PM

P: 80

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
Dec1212, 07:52 PM

Sci Advisor
P: 2,470

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



#3
Dec1312, 02:56 AM

PF Gold
P: 4,518

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 lightday apart. The dots represent oneday increments of time on a clock carried with each observer. 



#4
Dec1312, 07:28 PM

P: 80

understanding the speed of light between 2 ships racing away from each otherghwellsjr, 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? 



#5
Dec1312, 08:17 PM

P: 80

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?




#6
Dec1312, 08:38 PM

PF Gold
P: 4,518

γ = 1/√(1β^{2}) 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/√(1β^{2}) = 1/√(10.51^{2}) = 1/√(10.2601) = 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. 



#7
Dec1312, 09:30 PM

PF Gold
P: 4,518

(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)/(1β)) = √((1+0.809)/(10.809)) = √(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 lightday. 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 lightday but the same Doppler Factor applies. Does this all make perfect sense to you? 



#8
Dec1312, 10:00 PM

Sci Advisor
P: 2,470

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. 



#9
Dec1412, 09:16 PM

P: 80





#10
Dec1412, 09:22 PM

P: 80





#11
Dec1512, 09:52 AM

PF Gold
P: 4,518

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 roundtrip speed of light, it's also because their rulers contract along the direction of motion in the IRF. Does that make sense? 



#12
Dec1512, 06:42 PM

P: 80

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. http://www.youtube.com/watch?v=yXHWJ4iUlZs http://www.youtube.com/watch?v=EK6HxdUQm5s 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!! 



#13
Dec1512, 07:27 PM

Sci Advisor
P: 2,470

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


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