Lorentz Transformation - Speeds relative to different observers

[R3ap3r42]

Homework Statement

Lorentz Transformation - Relative speeds to different observers

Relevant Equations

Lorentz Transformation, Time Dilation, Length Contraction

Two spaceships are heading towards each other on a collision course. The following facts are all as measured by an observer on Earth: spaceship 1 has speed 0.74c, spaceship 2 has speed 0.62c, spaceship 1 is 60 m in length. Event 1 is a measurement of the position of spaceship 1 and Event 2 is a measurement of the position of spaceship 2. These two events take place at the same time and the distance between them is 2.9 × 1012 m, again as measured by an observer on Earth.

a) Draw a diagram to illustrate the relative motion of the relevant inertial frames.
b) Determine the length of spaceship 1 as measured in its own rest frame.
c) Determine the length of spaceship 1 as measured by an observer at rest in spaceship 2.
d) According to an observer on Earth, how long before the spaceships collide?
e) According to an observer on spaceship 1, how long after Event 1 do the spaceships collide?
f) What is the time interval between Event 1 and Event 2 according to an observer on spaceship 1?I have done all but f)
I don't understand how to get to the expected value of 10635 secs.

 

[R3ap3r42]

Ah, I think I got it. We can use reverse Lorentz transsform

$$ t = \gamma (t' + \frac {vx'} {c^2}) $$

where the prime is the Earth so t' = 0 and x' = 2.9*10^12

 

[PeroK]

Ah, I think I got it. We can use reverse Lorentz transsform

$$ t = \gamma (t' + \frac {vx'} {c^2}) $$

where the prime is the Earth so t' = 0 and x' = 2.9*10^12

The problem defines three events, which need to be be given coordinates in the Earth frame. I would take the Earth frame to be the unprimed frame and the frame of spaceship 1 to be the primed frame. In any case, we can take event 1 as the common origin, event 2 is the position of spaceship 2 at time $t = 0$ (Earth frame) and event 3 is the collision.

You can then use the (usual) Lorentz Transformation to find the time of all three events in the frame of spaceship 1.

 

[mattr808]

Homework Statement:: Lorentz Transformation - Relative speeds to different observers
Relevant Equations:: Lorentz Transformation, Time Dilation, Length Contraction

Two spaceships are heading towards each other on a collision course. The following facts are all as measured by an observer on Earth: spaceship 1 has speed 0.74c, spaceship 2 has speed 0.62c, spaceship 1 is 60 m in length. Event 1 is a measurement of the position of spaceship 1 and Event 2 is a measurement of the position of spaceship 2. These two events take place at the same time and the distance between them is 2.9 × 1012 m, again as measured by an observer on Earth.

a) Draw a diagram to illustrate the relative motion of the relevant inertial frames.
b) Determine the length of spaceship 1 as measured in its own rest frame.
c) Determine the length of spaceship 1 as measured by an observer at rest in spaceship 2.
d) According to an observer on Earth, how long before the spaceships collide?
e) According to an observer on spaceship 1, how long after Event 1 do the spaceships collide?
f) What is the time interval between Event 1 and Event 2 according to an observer on spaceship 1?I have done all but f)
I don't understand how to get to the expected value of 10635 secs.

Would anyone mind explaining how to do part e? I thought you would use the velocity transformation to get velocity that spaceship 1 sees 2 coming towards it then use earth as the primed frame with t'=7108 (part d answer) and x'=2.9x10^12 and use t=gamma(t'+vx'/c^2) but I'm getting the wrong answer. (Correct answer is 4780s)

 

[PeroK]

Would anyone mind explaining how to do part e? I thought you would use the velocity transformation to get velocity that spaceship 1 sees 2 coming towards it then use earth as the primed frame with t'=7108 (part d answer) and x'=2.9x10^12 and use t=gamma(t'+vx'/c^2) but I'm getting the wrong answer. (Correct answer is 4780s)

You're probably forgetting that Earth and the spaceships have different simultaneity conventions. Relativity isn't just time dilation. You need the full Lorentz transformation.

 

[mattr808]

You're probably forgetting that Earth and the spaceships have different simultaneity conventions. Relativity isn't just time dilation. You need the full Lorentz transformation.

Sorry I'm not sure I follow, what do you mean by the "full Lorentz transformation"?

 

[vela]

Would anyone mind explaining how to do part e? I thought you would use the velocity transformation to get velocity that spaceship 1 sees 2 coming towards it then use earth as the primed frame with t'=7108 (part d answer) and x'=2.9x10^12 and use t=gamma(t'+vx'/c^2) but I'm getting the wrong answer. (Correct answer is 4780s)

Why are you using the velocity of spaceship 2 relative to spaceship 1 when you're transforming from the Earth frame to spaceship 1's frame?

 

[PeroK]

Sorry I'm not sure I follow, what do you mean by the "full Lorentz transformation"?

I mean the Lorentz Transformation.