Question

A Vulcan spacecraft has a peed of 0.600c with respect to earth. The Vulcan determine 32 h to be the time interval between two events on earth. What value would they determine for this time interval if their ship had a speed of 0.940c with respect to earth

Answer 1

This is a little twisted! That's a pun.. I'll get there.. This is something that is important conceptually, because it demonstrates relativistic principle that are used for particles. Also, it might be fun to know if you were on that ship.
I say twisted... Since you know the time elapsed according to the Vulcan spacecraft, and the relative speed, you can find the time it took according to Earth. Since you'll then know the time it took according to Earth, you can find how much time it takes to any speed of observer.

Use the spacecraft time interval and speed
to
find the Earth time interval
to
find any-speed spacecraft time interval.



So, just to think about it - "moving clock ticks slower." That's a phrase I was taught. Spacecraft and Earth experience relative movement, and so they perceive each others activities to be slower than each thinks of the event happening at rest or non-relativistic speeds. Meaning, Vulcan person says Earth event took long, Earth person might disagree. The complement is also true, but not relevant to the problem, that if Earth person says Vulcan ship event took long, then Vulcan person might disagree. So, we expect Earth person elapsed time to be shorter.

First lets find the time elapse on Earth, using ship's speed and elapsed time.
Here's your equation:
Using T's to represent intervals and the gamma u to be the Lorentz transformation constant (I think that's what it's called):\[\Large T_{\text{at rest with event}}=\left( T_{\text{moving relative to event}}\right) \left( \gamma_u\right)\]We have the interval at rest with the event isolated already - that's an Earth person with the Earth event. That's what we want. So we substitute! Note that:\[\Large\gamma_u=\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}\text{, where }u=\text{ relative velocity}\]

Then you have that T for elapsed time relative with the event.

Next, you have to find out what the time interval is according to a ship moving relative to Earth at a different speed. So, solve for that T.\[\Large {T_{\text{at rest with event}}=\left( T_{\text{moving relative to event}}\right) \left( \gamma_u\right)\\\Downarrow\\T_{\text{moving relative to event}}=T_{\text{at rest with event}}\div \gamma_{u_2}}\\\text{where }u_2\text{ indicates we have a different relative velocity.}\]And\[\Large \gamma_{u_2}=\frac{1}{\sqrt{1-\frac{u_2^2}{c^2}}}\]Substitute again, and see what you get!

\[\text{Note: The time interval for at rest with event is usually }T\text{,}\\\text{while the time interval according to a frame}\\ \text{moving relative to the event is }T_0\text{.}\]
Source: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html for relativistic equations