# Stuck on SR

1. Feb 8, 2008

### barnflakes

Can somebody give me some insight as to how I solve the following questions:

1) A rocket travels from the earth to the moon(distance measured from earth 380,000km) at a speed of v = 0.8c. (i) How long does the trip take according to an observer on earth? (ii) How long does the trip take according to the astronaut (iii) What is the earth-moon distance measured by an astronaut on the rocket?

So for part (i), I'm thinking along these lines:

The earth is reference frame S, the rocket S'.

Therefore, the relative velocitys between the reference frames is 0.8c?

The distance to the moon(the distance the rocket travels) according to the man on earth is 384000x10^3km which is x. This is where I get stuck.

t = (t' + (vx/c^2))x (1-u^2/c^2)^-0.5

I seem to have too many unknowns?

2. Feb 8, 2008

### barnflakes

I've just had a thought, is it a simple case of x = ut for part (i)?

3. Feb 8, 2008

### bernhard.rothenstein

In the experiment you dscribe the astronaut measures a proper time interval t'-0 whereas the observers on earth measures a coordinate time interval t-0 related by the time dilation formula.
The way in which you use the Lorentz transformation is not correct. In its right side you should out x' and not x and to consider x'=0.

4. Feb 8, 2008

### JesseM

In the equation you're using, x and t are coordinates, not intervals. If you want to use that equation, you should pick the coordinate of the Earth and moon in the Earth's frame (say, x=0 km for Earth and x=380,000 km for the moon) and then figure out the time coordinates for the rocket leaving Earth and the rocket arriving at the moon (for example, if we set our clock so that at t=0 the rocket leaves earth, and it's moving towards the moon at 0.8c in this frame, at what time coordinate will it arrive at the moon in this frame)? Then for each of these two events, you can find the corresponding time coordinate t' of the event in the rocket's frame using the equation:

t' = (t - vx/c^2) * (1 - v^2/c^2)^-0.5

with v=0.8c.

A simpler way is just to use the time dilation equation $$\Delta t = \Delta \tau / \sqrt{1 - v^2/c^2}$$, where $$\Delta \tau$$ is the time elapsed on a clock between two events on its worldline (which is the time between these events in the clock's own rest frame), and $$\Delta t$$ is the time between those same events in the frame of an observer who sees the clock moving at speed v. So, if you already know the time in your frame for a moving clock to travel between two points, you can flip this equation around to find the time elapsed on that clock between those points. Similarly, for part (ii) you could make use of the length contraction equation $$\Delta x = \Delta X * \sqrt{1 - v^2/c^2}$$, where $$\Delta X$$ is the distance between two objects (or between two ends of a single object) in their own rest frame, and $$\Delta x$$ is the distance between those objects in a frame where they are moving at speed v along the axis between them. Or, as before, you could just use the full Lorentz transformation equations to deal with part (ii).

For more on the difference between the time dilation equation and the Lorentz transformation equation which you were using, see this thread.

Last edited: Feb 8, 2008
5. Feb 9, 2008

### barnflakes

If I'm honest, I'm still really confused.

I understand what you say about the Lorentz contractions being based on coordinates, that now makes sense, however I can't see how to choose the correct coordinates for this situation.

For the initial state of the earth and rocket reference frame, set the coordinates of the earth to be 0 and the moon to be 384000km:

If I say at t1 = t1' = 0, x = 0 and x' = 0

for earth at t2:

at t = t2, x = 384000km, v = 0.8c

for rocket at t' = t2', x' = x', v = 0.8c ??

I'm lost :(

I still can't see how I can use the time dilation equation because I don't know the time difference of the two events happening on the world line?

6. Feb 9, 2008

### JesseM

But you don't have to use t2, you can figure out the actual value of the time in the S frame. After all, you know the distance from Earth to moon in the S frame, you know the speed the rocket is moving in the S frame, and speed is just (change in position)/(change in time)
But that's what the Lorentz transformation is for, if you know the coordinates x and t of an event in the S frame, it will tell you the corresponding coordinates x' and t' of the same event in the S' frame:

x' = (x - vt)/sqrt(1 - v^2/c^2)
t' = (t - vx/c^2)/sqrt(1 - v^2/c^2)

Again, it's easy to figure out the time in the S frame, since you know the distance and you know the speed, and speed = (change in position)/(change in time).

7. Feb 9, 2008

### barnflakes

Thank you JesseM.

This is what I've got so far then

t according to observer in S(on earth) : 384000km/0.8c = 1.6s.

t according to man on rocket: change in t' = change in t/(1-v^2/c^2)^-0.5 = 0.96s

Assuming that is correct, distance as measured by man on rocket = 384000km x (1-(0.8c)^2/c^2)^0.5 = 2.304 x 10^8m.

Does this agree with what you expect?

For the time dilation bit - moving clocks tick slower according to an observer in a reference frame other than the one with proper time? So that is correct?

8. Feb 9, 2008

### JesseM

Yup, those are the results I get too.
Right. And likewise, if you have two objects at rest with respect to one another (or two ends of a single object), the distance between them in a frame where they're in motion will be smaller than the distance between them in their rest frame (unless their direction of motion in this frame is perpendicular to the axis between them).

Last edited: Feb 10, 2008
9. Feb 10, 2008

### barnflakes

Thank you JesseM, you have been very helpful!

One other question if I may.

Given:

x' = y(u/c)(x-ut) (1)
y' = y (2)
z' = z (3)
t' = y(u/c)(t-ux/c^2) (4)

where y(u/c) = (1-u^2/c^2)^-0.5

I have to show

x = y(u/c)(x' + ut')
y = y'
z = z'
t = y(u/c)(t' + ux'/c^2)

Now, I tried to simply rearrange (1) and put it into (4) and then rearrange but it got very messy and I didn't get the right answer, what is the correct substitution to make?

10. Feb 10, 2008

### barnflakes

no worries, I've managed to do it now.