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Confusing myself with Lorentz transformation equations.

(This is my own question so any information left out is my mistake). Consider a spaceship passing a point on Earth with position (x,y,z) = (0,0,0) at time t = 0, and at a speed v = 0.5c. Take t' = 0 at this moment in time (the time inside the space ship). How long would it take the ship to reach a point one light year away from (x,y,z) = (0,0,0) as measured by the observer on Earth and by an observer in the space ship?

Clearly the observer measures it taking two years.

Now for the space ship. I'm confusing because there are two equations, but surely only one is right;

[br][br]t = t_{0} \gamma

[br][br]t_{0} = \gamma (t - vx/c^2)

Where

t_0

would be the time taken as measured in the space ship's frame.

t

being the time taken as measured in the Earth's reference frame. However, re-arranging shows;

[br][br]t_{0} = \frac {t}{\gamma}

[br][br]t_{0} = \gamma (t - vx/c^2)

x

would be 1 light year. But surely only one of those is right? Which is it and for which situation do I use either?

Reply 1

SDA

I haven't got an answer for you fully yet but what I can say is that:

t_0=t / \gamma

applies to time intervals so should strictly be written

\Delta t_0 = \Delta t / \gamma

whereas the other equation applies to the time coordinate not time interval. I don't know if this helps or if you already knew that but I will keep thinking and report back when I have an idea.

Reply 2

3nTr0pY

A word of advice:

You have to be very, very clear about frames of reference in special relativity problems. Otherwise you'll get into a hopeless mess. Usually we denote the frame of the stationary observer as S, with time t, and the frame of the spaceship as S', with time t'.

The way I like to explain this is through spacetime intervals. In SR,

ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2

is always conserved irrespective of the frame of reference. We thus get:

c^2dt^2 - dx^2 - dy^2 - dz^2 = c^2dt'^2

This is because in S' the spaceship isn't moving, making dx' = dy' = dz' = 0

Dividing both sides by c^2 and rearranging gives:

dt = \gamma dt'

Which is the correct expression. Replacing dt with delta t, and using delta t = 2 years allows delta t' to be found. It it is less than delta t as expected by the twin 'paradox'.

(edited 11 years ago)

Reply 3

Anti Elephant Mine

I wrote this when I was very tired so apologies for not being entirely specific. My main confusion comes from the use of the two equations.

At t = 0 (in the reference frame of the observer on Earth) the spaceship passes a point (0,0,0) on Earth. At this moment we take the time of the reference frame of the observer on the Space Ship to be t' = 0.

So surely we can use either equation. The observer on Earth measures the ship taking two years to go a point one light year away from (0,0,0).

In the reference frame of the ship, the time at which it passes this point can be given by; (since it started at t' = 0, so we can omit the necessity for intervals).

t_0 = t/ \gamma

Similarly

[br][br]t_{0} = \gamma (t - vx/c^2)

is another equation for Lorentz contraction for 'events'. If we define the 'event' to be the ship passing the point one light year away, then how can this be any different from the previous equation? This gives the time of the event in the reference from of the space ship, so placing

v = 0.5c

x = 1 light year

t = 2 years

gives a completely different answer surely?

(edited 11 years ago)

Reply 4

3nTr0pY

Using spacetime intervals is a really nice way to understand what's going on, but I guess you may not have been introduced to them before.

So, talking only about Lorentz transforms, I've (finally) worked out exactly what you've done wrong. Turns out it was the maths:

This gives the time of the event in the reference from of the space ship, so placing

v = 0.5c

x = 1 light year

t = 2 years

gives a completely different answer surely?

It actually gives exactly the same answer, if you substitute in properly. In fact, it's easier to see why by keeping it in symbols. We have:

[br][br]t_{0} = \gamma (t - vx/c^2)

Using x = vt we have:

[br][br]t_{0} = \gamma (1 - v^2/c^2)t

and so:

[br][br]t_{0} = t/\gamma

The whole thing could be done a lot more simply using the inverse Lorentz transform though. This is:

[br] t = \gamma (t_{0} + vx_{0}/c^2)

We know that

[br][br]x_{0} = 0

And so we get the desired answer straight away.

Reply 5

Anti Elephant Mine

Thanks! I guess I typed it wrong in my calculator... making me believe that the different equations provided different answers. I did, however, understand your first post despite not having done it before.