- #1
Albertgauss
Gold Member
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Hi all, this is not a HW problem but one I made up myself, (so you won't find it in any txtbooks or solutions etc.). It also is much more advanced than for someone on the physics HW site, as I can see, so I thought it best to put here.
I'm having trouble working out simultaneity and Lorentz Contraction for the following case:
PROBLEM: Earth sees a rocket moving at β=+0.99999c (in the direction of the positive x-axis). At t=0, x=0, the coordinates of the rocket are t'=0, x'=0 (origins coincide). In the Earth frame, there is an obstacle at 2000 m at t=0. The obstacle does not move from this coordinate the whole time, as viewed from Earth. At t=t'=0, a light from the obstacle travels back towards the rocket at -c (the negative denotes velocity and c is light speed). Now, the rocket travels towards the obstacle, but the rocket won't know the obstacle is there until the rocket receives the light pulse from the obstacle.
!
MY QUESTION when the rocket receives the pulse from the obstacle, how much time will the rocket have to avoid the obstacle and how far will the obstacle be from the rocket?
!
If you skip down to SUMMARY I will have the main discussion there, also, my questions and problems with my own analyses. Below is how I got my numbers:
Parameters
β for rocket is = 0.99999 thus γ=223.6073568 (decimals matter, as I panfully learned)
Origin
Earth says rocket is at: x=0, t=0
Rocket says Rocket is at: x'=0, t'=0
Earth says obstacle is at: x=2000 m, t = 0
Use Lorentz on the above to get to the coords
x'=γ(x-βct); t'=γ(t-βx/c)]
to get the coords for the rocket below:
Rocket says obstacle is at: x' = 4.47214(10^5) m, t' = -.0014907 secs
Rocket Recieves Light from obstacle at
Earth says: [condition is, in Earth's frame]: βct = 2000-ct
x_rocket_rcvs_light = βct = 999.995 m
t_rocket_rcvs_light = 2000/[c(1+β)] = 3.33335(10^-6) secs.
Use Lorentz on the above to get to the coords
x'=γ(x-βct); t'=γ(t-βx/c)]
to get the coords for the rocket below:
rocket says
x'_rocket_rcvs_light ≈ 0
t'_rocket_rcvs_light = 1.4907(10^-8) secs
At the time the rocket receives the light from the obstacle, the obstacle is at:
Earth says obstacle is at (obstacle does not move in Earth's frame):
x_obstacle = 2000 m
t_obstacle = 3.33335(10^-6) secs
Use Lorentz on the above to get to the coords
x'=γ(x-βct); t'=γ(t-βx/c)]
to get the coords for the rocket below:
Rocket says obstacle is at:
x'_obstacle = 223608 m
t'_obstacle = -7.453(10^-4) secs
If the rocket were to continue and collide with the obstacle, this would happen at,
Earth says:
x_rocket_no_swerve = 2000m
t_rocket_no_swerve = 2000/(0.99999c) = 6.6667(10^-6) secs
Use Lorentz on the above to get to the coords
x'=γ(x-βct); t'=γ(t-βx/c)]
to get the coords for the rocket below:
x'_rocket_no_swerve = 0
t'_rocket_no_swerve = 2.236(10^-8) secs
Notice that, all the x's are all zero, as the rocket will always feel itself at the origin of its own frame. In the rocket's frame, the rocket will receive the light from the obstacle at the rocket's own origin , and if the rocket hits the obstacle, the rocket will do so at the origin of the rocket's frame. So this is a good check I have used the Lorentx Equations correctly, at least up to there.
SUMMARY:
Earth:
The Earth says the rocket starts at x=0,t=0, then the rocket receives the light from the obstacle at x=999.995 m, t=3.33335(10^-6)secs. The obstacle is at x=2000, t= 3.33335(10^-6)secs at this point. If the rocket plows into the obstacle, the Earth will say that this will happen at x=2000 m, t = 6.6667(10^-6) secs.
Rocket:
The rocket says the rocket starts at x'=0,t'=0, then the rocket receives the light from the obstacle at x'= 0 , t'= 1.49(10^-8) secs. The obstacle is at x = 223608 m, t= -7.453(10^-4) secs at this point. If the rocket plows into the obstacle, the rocket will say that this will happen at x'=0 m, t = 2.236(10^-8) secs.
QUESTIONS
1. I'm having some trouble understanding this. I know that what is simultan in one frame is not in another. Here, when the rocket receives the light pulse, the Earth says the rocket is at 999.995 m and the obstacle is at 2000m, and the time these two things happen occurs, for the Earth at the same time of 3.33335(10^-6) secs. But, in the rocket frame, the rocket will receive the light signal first, and then some time later, the obstacle will be at the pos, time Earth sees for the obstacle at 2000 m, 3.33335(10^-6) secs. But, when the rocket recives the light signal, he will know the obstacle is there. Notice that for the second coordinate set, the rocket receives the signal at 745 μsecs before the origins even line up (that's from the negative from the time transform) In the rocket's frame, once the rocket receives the light signal, far does the rocket say the obstacle is and how much time does the rocket think the rocket needs to get to the obstacle? How do I use the Lorentz Transforms correctly for this? .
2. Now, the distance from the Earth to the obstacle is 2000m, but shouldn't this length be shorter according to the rocket's frame? This 2000m is a proper length, since it does not move in the Earth's frame. The rocket should see the 2000 m shrink. But, if you notice, when the rocket receives the light pulse from the obstacle, the rocket thinks the obstacle is at 223608m !
I tried to use the famous Lorentz Length Contraction Formula of L=Lo/γ (which assumes the origins line up at x=x'=0, t=t'=0) for 2000m, and the Length Contraction Formula gives me a length of 8 m or so for the rocket's frame. If I do an inverse lorentz transformation, then I get the 2000 m coordinate of the obstacle contracted down to 8 m as viewed by the rocket, in agreement with the Lorentz Contraction Formula. But why should I use an inverse lorentz transformation? Why didn't the standard lorentz transformation give a length contraction on the coordinate? Why did the coordinate of the obstacle appear farther away in the frame of the rocket using the standard Lorentz Transformaion?
3. Also, the rocket says it received the light pulse 7.453(10^-4) secs BEFORE the origins were ever aligned. Haven't I eliminated this problem by trying to force the origins to be x'=0, t'=0?
I'm having trouble working out simultaneity and Lorentz Contraction for the following case:
PROBLEM: Earth sees a rocket moving at β=+0.99999c (in the direction of the positive x-axis). At t=0, x=0, the coordinates of the rocket are t'=0, x'=0 (origins coincide). In the Earth frame, there is an obstacle at 2000 m at t=0. The obstacle does not move from this coordinate the whole time, as viewed from Earth. At t=t'=0, a light from the obstacle travels back towards the rocket at -c (the negative denotes velocity and c is light speed). Now, the rocket travels towards the obstacle, but the rocket won't know the obstacle is there until the rocket receives the light pulse from the obstacle.
!
MY QUESTION when the rocket receives the pulse from the obstacle, how much time will the rocket have to avoid the obstacle and how far will the obstacle be from the rocket?
!
If you skip down to SUMMARY I will have the main discussion there, also, my questions and problems with my own analyses. Below is how I got my numbers:
Parameters
β for rocket is = 0.99999 thus γ=223.6073568 (decimals matter, as I panfully learned)
Origin
Earth says rocket is at: x=0, t=0
Rocket says Rocket is at: x'=0, t'=0
Earth says obstacle is at: x=2000 m, t = 0
Use Lorentz on the above to get to the coords
x'=γ(x-βct); t'=γ(t-βx/c)]
to get the coords for the rocket below:
Rocket says obstacle is at: x' = 4.47214(10^5) m, t' = -.0014907 secs
Rocket Recieves Light from obstacle at
Earth says: [condition is, in Earth's frame]: βct = 2000-ct
x_rocket_rcvs_light = βct = 999.995 m
t_rocket_rcvs_light = 2000/[c(1+β)] = 3.33335(10^-6) secs.
Use Lorentz on the above to get to the coords
x'=γ(x-βct); t'=γ(t-βx/c)]
to get the coords for the rocket below:
rocket says
x'_rocket_rcvs_light ≈ 0
t'_rocket_rcvs_light = 1.4907(10^-8) secs
At the time the rocket receives the light from the obstacle, the obstacle is at:
Earth says obstacle is at (obstacle does not move in Earth's frame):
x_obstacle = 2000 m
t_obstacle = 3.33335(10^-6) secs
Use Lorentz on the above to get to the coords
x'=γ(x-βct); t'=γ(t-βx/c)]
to get the coords for the rocket below:
Rocket says obstacle is at:
x'_obstacle = 223608 m
t'_obstacle = -7.453(10^-4) secs
If the rocket were to continue and collide with the obstacle, this would happen at,
Earth says:
x_rocket_no_swerve = 2000m
t_rocket_no_swerve = 2000/(0.99999c) = 6.6667(10^-6) secs
Use Lorentz on the above to get to the coords
x'=γ(x-βct); t'=γ(t-βx/c)]
to get the coords for the rocket below:
x'_rocket_no_swerve = 0
t'_rocket_no_swerve = 2.236(10^-8) secs
Notice that, all the x's are all zero, as the rocket will always feel itself at the origin of its own frame. In the rocket's frame, the rocket will receive the light from the obstacle at the rocket's own origin , and if the rocket hits the obstacle, the rocket will do so at the origin of the rocket's frame. So this is a good check I have used the Lorentx Equations correctly, at least up to there.
SUMMARY:
Earth:
The Earth says the rocket starts at x=0,t=0, then the rocket receives the light from the obstacle at x=999.995 m, t=3.33335(10^-6)secs. The obstacle is at x=2000, t= 3.33335(10^-6)secs at this point. If the rocket plows into the obstacle, the Earth will say that this will happen at x=2000 m, t = 6.6667(10^-6) secs.
Rocket:
The rocket says the rocket starts at x'=0,t'=0, then the rocket receives the light from the obstacle at x'= 0 , t'= 1.49(10^-8) secs. The obstacle is at x = 223608 m, t= -7.453(10^-4) secs at this point. If the rocket plows into the obstacle, the rocket will say that this will happen at x'=0 m, t = 2.236(10^-8) secs.
QUESTIONS
1. I'm having some trouble understanding this. I know that what is simultan in one frame is not in another. Here, when the rocket receives the light pulse, the Earth says the rocket is at 999.995 m and the obstacle is at 2000m, and the time these two things happen occurs, for the Earth at the same time of 3.33335(10^-6) secs. But, in the rocket frame, the rocket will receive the light signal first, and then some time later, the obstacle will be at the pos, time Earth sees for the obstacle at 2000 m, 3.33335(10^-6) secs. But, when the rocket recives the light signal, he will know the obstacle is there. Notice that for the second coordinate set, the rocket receives the signal at 745 μsecs before the origins even line up (that's from the negative from the time transform) In the rocket's frame, once the rocket receives the light signal, far does the rocket say the obstacle is and how much time does the rocket think the rocket needs to get to the obstacle? How do I use the Lorentz Transforms correctly for this? .
2. Now, the distance from the Earth to the obstacle is 2000m, but shouldn't this length be shorter according to the rocket's frame? This 2000m is a proper length, since it does not move in the Earth's frame. The rocket should see the 2000 m shrink. But, if you notice, when the rocket receives the light pulse from the obstacle, the rocket thinks the obstacle is at 223608m !
I tried to use the famous Lorentz Length Contraction Formula of L=Lo/γ (which assumes the origins line up at x=x'=0, t=t'=0) for 2000m, and the Length Contraction Formula gives me a length of 8 m or so for the rocket's frame. If I do an inverse lorentz transformation, then I get the 2000 m coordinate of the obstacle contracted down to 8 m as viewed by the rocket, in agreement with the Lorentz Contraction Formula. But why should I use an inverse lorentz transformation? Why didn't the standard lorentz transformation give a length contraction on the coordinate? Why did the coordinate of the obstacle appear farther away in the frame of the rocket using the standard Lorentz Transformaion?
3. Also, the rocket says it received the light pulse 7.453(10^-4) secs BEFORE the origins were ever aligned. Haven't I eliminated this problem by trying to force the origins to be x'=0, t'=0?