Recent content by Lissajoux

I'm just looking at the spacetime diagram now. So the signal was sent at t=2.5h, where x=1.5lh. And the signal was received at t=4h, where x=0lh. Is that interpretation correct? And was the other parts of the last few posts correct i.e. the diagram then I assume so and also the speed calc.

.. the signal has to be received at x'=0 lh right? :smile: does the x=2.5lh apply to the location of the Earth instead? or is that just wrong?

I think this still applies? So the ST diagram would be: http://yfrog.com/4jrelativ2j So the speed of signal: v=\frac{\Delta d}{\Delta t}=\frac{1.5lh}{1.5h}=1l=c (not sure about the correct notation for the last bit of that equation) .. any of this any good?

Yes I think I do see that now. So the ship receives the signal at different times, but at the same location on the diagram as those on the ship don't think there moving. If that makes sense, maybe I'm not wording it in the best way. I get it though. Here's a new spacetime diagram...

http://yfrog.com/10relativ2j how's that? Then should there be a line from the Earth line to the shuttle line, from t=2.5h to the point that the signal is received. I've noted that this would mean that whilst the Earth and the ship continue to move further apart, the signal would move towards...

In regards to the diagram: 1. The world line for the ship extends vertically upwards from the origin. That's shown on there. 2. I took those values. But that's clearly wrong, that's in the Earth reference frame and we're in the spaceship's reference frame for this diagram. So should be...

So that second diagram was correct? And that is for the spaceship's reference frame isn't it? I should use this latest diagram now then and ignore the first I take it. Those answers were based on my first diagram, which was wrong. But for the signal: Start: t=2.5h, x=0lh End: t=4.0h, x=2.5lh...

I've done so many different calculations it's a bit hard to keep track of the right ones (the few there have been :wink:) At least I should be able to answer the questions with the help of the diagram. Time between emission and reception of signal: 4.0-2.5=1.5h Distance traveled by signal...

Yep :smile: http://yfrog.com/5orelativ1j (sorry it's not very good, was best I could do in paint :wink:)

Back in post 18 I believe we established that according to the ship, the signal left Earth when the ships clock read 2.5h and the ship at that time was 1.5lh away from the earth. Then in post 19, that according to the ship the signal was received when the ships clock read 4.0h and at that point...

Yeah think I've figured it out now :smile:

I've just looked through my notes again and thought I could use this to find the speed of the signal: u_{x}'=\frac{u_{x}-v}{1-\frac{u_{x}v}{c^{2}}}=\frac{c-0.6c}{1-\frac{0.6c^{2}}{c^{2}}}=\frac{(1-0.6)c}{1-0.6}=\frac{c}{1}=c where u_{x}' is the speed of the signal in the ship's reference frame...

Homework Statement An electron has a total energy of exactly 5000MeV just before a collision. I need to calculate: 1. The mass energy in MeV 2. The Lorentz factor \gamma 3. The speed as a fraction of c 4. The momentum in MeV/c^{2} Homework Equations Within the problem...

(Ignore my previous post if you read it, this is the better replacement) Right I've gone back through my calculations to try and make sense of it all and spot where I've gone wrong. So in the Earth reference frame, the signal reaches the ship at t=5.0~\textrm{h} and at that time the Earth and...

Hmm. Method: x_{ship}=0.6t and x_{shuttle}=0.8c(t-4) Hence: 0.6ct=0.8c(t-4) Now solve for t: 0.6ct=0.8ct-3.2c\implies 3.2c=0.2ct\implies t=\frac{3.2}{0.2}=16hours Therefore: t=16~\textrm{h} x=16-4=12~\textrm{lh} \beta=\frac{v}{c}={0.8}{c}[/itex]? (not sure on this though, as it stayed...