Time dilation between two frames

[penroseandpaper]

Homework Statement

A rocket passes by a replenishment ship at a relative constant velocity of 3/4 c (where c is the speed of light).

a) Using standard configuration for the frames of reference, and the fact that the replenishment ship has a length of L' = 100m, calculate the time it takes for the rocket to pass the ship - in the ship's frame of reference.

b) With the rocket passing the start of the ship at (0,0,0,0) and the end of the ship at (0,0,0,T), use Lorentz transformation to turn them into the ship's frame of reference.

c) Calculate T from the rocket's frame of reference.

Relevant Equations

t' = x/v

x'=Lorentz factor × (x-Vt)
t'= Lorentz factor ×(t-((Vx)/c^2))

[Wow - special relativity is amazing but boy is there a lot to get your head around! We're only being introduced to it as a taster of college courses but the few lessons we've had have left me with more questions than answers; most fascinating topic yet, I think!]
Unfortunately I've tackled a question online that doesn't come with a solution and I'm looking for someone to see where I've inevitably gone wrong.

a) First, I believe the question wants us to find t', which is termed the proper time for the ship's frame of reference.

Also, it is this frame of reference (the ship's) that is at rest in this scenario. And t' is always larger than t.

Therefore, is t' simply the calculation of distance / velocity?
I.e. t' = 100 ÷ 0.75c?

b) Lorentz transformations replace Galilean transformations, and are valid at all speeds up to c (Galilean transformations are only valid for low velocities).

Relative velocity remains at 3/4c, so the Lorentz factor can be calculated as 1/(√1-0.5625) = 1.5118...

So, x' = 1.5118 × (x-vt) = 1.5118 × (0-0.75 c × 0) = 1.5118 and 1.5118 × (0-0.75 c × T)= 1.5118× -224850000 = -339928230 T *But this being a negative doesn't look right to me, even though the X coordinate of both events (start and end of passing the ship) are given as 0.

Then the times are given by 1.5118 × (t-(Vx)/(c^2)), so 1.5118 ×(0-0) and 1.5188 × (T-0) , if I'm right in interpreting X in the brackets for the X coordinate given for the rocket.

However, I'm sure I've gone awry above.

C) I assume this will involve manipulating the expressions found above. However, I haven't attempted this as I believe the above to be wrong.

Thank you in advance for your insight

 

[phinds]

I haven't looked at your writeup really but I think you might be overthinking this. You know the length of the rocket in its own reference frame and you know its speed relative to the replenishment ship so just compute its length in the reference frame of the ship and you're basically done. All you have to do then is add that to the length of the ship and compute how long it takes a point object moving at 3/4 c to travel that distance.

If I have that wrong, one of our more knowledgeable members will correct me.

 

[PeroK]

I don't see that the question is clear about the relationship between the frames. You might imagine that the ship would be identified as the unprimed frame; and the rocket, as the primed frame, moving at $v$ to the right (along the positive x-axis).

But, part b) seems to suggest the the ship has the primed frame and is moving to the right relative to the rocket.

First, you should state which of these you are using.

Second, you have two events: the rocket passing the front of the ship and the rear of the ship. I suggest you label these events ($A$ and $B$ perhaps).

It's going to be a lot easier to read what you've done if you retain $\gamma$ and $v$ and $c$ until the final calculations. Something like this:

So, x' = 1.5118 × (x-vt) = 1.5118 × (0-0.75 c × 0) = 1.5118 and 1.5118 × (0-0.75 c × T)= 1.5118× -224850000 = -339928230 T *But this being a negative doesn't look right to me, even though the X coordinate of both events (start and end of passing the ship) are given as 0.

Is pretty hard to decipher.

The answer to question a) is correct. In the ship frame, you have plain old $\Delta t = L/v$.

For part b) you might get a minus sign for the time of event. It appears to me that $x$ is given as a variable for the position (in the rocket frame) of the nose of the rocket I imagine. If so, all your answers are going to be functions of $x$.

 

[penroseandpaper]

I haven't looked at your writeup really but I think you might be overthinking this. You know the length of the rocket in its own reference frame and you know its speed relative to the replenishment ship so just compute its length in the reference frame of the ship and you're basically done. All you have to do then is add that to the length of the ship and compute how long it takes a point object moving at 3/4 c to travel that distance.

If I have that wrong, one of our more knowledgeable members will correct me.

It states that the ship has a length of 100m in its frame of reference. And the relative velocity is 3/4c. But it doesn't state the rocket's length, and wants the time t' in the ship's frame of reference.

 

[PeroK]

It states that the ship has a length of 100m in its frame of reference. And the relative velocity is 3/4c. But it doesn't state the rocket's length, and wants the time t' in the ship's frame of reference.

That's another ambiguity in the question. I guess the events relate to the nose of the rocket passing the ship.