1. The problem statement, all variables and given/known data
A member of a colony on Jupiter is required to salute the UN flag at the same time as it is being done on Earth at noon in New York. If observers in all inertial frames(i.e. any observer traveling at any arbitrary velocity) are to agree that he has performed his duty, how long must he solute for(i.e. seeing you don't know how fast the observer is traveling, what time iterval ensures that the rising of the falg and the saluting are simultaneous for all possible speeds of the observer)? (Distance between the planets is approximately 8 x10^6 km, ignore any relative motion of the planets)


2. Relevant equations

Lorentz Transformation

3. The attempt at a solution

I need help getting started on this. I have no idea what to do at all(partly because of not understanding what the question wants) I'm not sure if I'm suppose to derive an equation, or there is a number answer to this.

I'm totally stuck on the thought process portion and cannot/don't know how to translate anything onto paper.
I know that the arbitrary observer can have velocity from the range of -c to c but I don't know if that means I should end up with two solutions which will give us the time interval.

So any advice on how to begin this would be greatly appreciated. Thanks

I'm off by a negative sign and I'm not sure why. But here it goes.

I'm going to make the simplifying assumption that the observers' origin is at the Earth at the time that the salute is supposed to be given. Let's synchronize all the observers' clocks with the Earth clock and let t = 0 be when the salute is supposed to be performed. The saluter has to salute somewhat before this time in order for the signal to be received at t = 0 on Earth, so he/she has to salute at t = -L/c where L is the Earth-Jupiter distance. Now an observer moving at speed v sees the salute at:
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So as v goes from 0 to c the time t' goes from -L/c to (ahem) minus infinity.

Given the result I strongly suspect our poor saluter is going to have to salute forever (and we can logic that out.) I just can't find a way to get rid of the pesky negative sign.

-Dan

__________________
"I must not fear. Fear is the mind killer. Fear is the little death that brings total obliteration. I will face my fear. I will permit it to pass over me and through me. And when it has gone I will turn the inner eye to see its path. Where the fear has gone there will be nothing. Only I will remain." - The Litany Against Fear, "Dune" by Frank Herbert

thank you for your help. I actually understand what I should be looking for now

I'd be interested in how you get the final answer, once you figure it out. That negative sign is driving me crazy!

-Dan

__________________
"I must not fear. Fear is the mind killer. Fear is the little death that brings total obliteration. I will face my fear. I will permit it to pass over me and through me. And when it has gone I will turn the inner eye to see its path. Where the fear has gone there will be nothing. Only I will remain." - The Litany Against Fear, "Dune" by Frank Herbert

yep so we use initial t' = gamma(t - vx/c^2)

what we're suppose to assume is that t' = 0 because all observers have to see the event simultaneously. At this point we know the range of speed an observer can have ranges from -c to c so we just plug that in and solve for t for the two different velocities

Ah. Working backward. I like it. Thank you for sharing.

-Dan

__________________
"I must not fear. Fear is the mind killer. Fear is the little death that brings total obliteration. I will face my fear. I will permit it to pass over me and through me. And when it has gone I will turn the inner eye to see its path. Where the fear has gone there will be nothing. Only I will remain." - The Litany Against Fear, "Dune" by Frank Herbert