Lorentz Transformation - Speeds relative to different observers

In summary: You should use the velocity of spaceship 1 relative to Earth since that's the frame of reference you're using.
  • #1
R3ap3r42
30
3
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.
 
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  • #2
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
 
  • #3
R3ap3r42 said:
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.
 
  • #4
R3ap3r42 said:
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)
 
  • #5
mattr808 said:
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.
 
  • #6
PeroK said:
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"?
 
  • #7
mattr808 said:
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?
 
  • #8
mattr808 said:
Sorry I'm not sure I follow, what do you mean by the "full Lorentz transformation"?
I mean the Lorentz Transformation.
 

Related to Lorentz Transformation - Speeds relative to different observers

What is the Lorentz Transformation?

The Lorentz Transformation is a mathematical formula used in special relativity to describe how the measurements of time and space are affected by the relative motion of two observers.

What is the significance of the Lorentz Transformation?

The Lorentz Transformation is significant because it allows us to understand how the laws of physics are the same for all observers, regardless of their relative motion. It also explains the phenomenon of time dilation and length contraction.

How do different observers measure speeds?

Different observers measure speeds differently due to the effects of time dilation and length contraction. For example, an observer moving at a high velocity relative to an object will measure a shorter distance and longer time for the object to pass by, resulting in a different speed measurement than an observer at rest relative to the object.

Can the Lorentz Transformation be applied to any speed?

Yes, the Lorentz Transformation can be applied to any speed, although it becomes more noticeable and relevant at speeds close to the speed of light. At everyday speeds, the differences in measurements between observers are negligible.

How does the Lorentz Transformation affect our understanding of time and space?

The Lorentz Transformation challenges our traditional understanding of time and space as absolute concepts. It shows that they are relative and can be affected by the observer's frame of reference. This has significant implications for our understanding of the universe and the laws of physics.

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