Time difference - camera's

1. Mar 30, 2005

Kilroy-was-here

Ok, first off, this is not a homework assignment, just something I've been wondering about for a while but can't seem to find the answer to. This is also my first post on these forums so excuse me if this is the wrong forum :)

Say, there's a situation like this: spaceship A and spaceship B both have a persistent video-connection, meaning that they can view each other's "bridges" (Star Trek, hurray!) constantly. This persistent connection can span a distance several light years (not feasible, but important for the story). Now, spaceship A is moving with a constant speed of, say, 0.25 c towards planet X. Spaceship B is also moving with a constant speed towards planet X, but spaceship B's speed (or velocity, whichever you want) is 0.5 c.

In this scenario, I think I am correct in saying spaceship B's clock is ticking slower than spaceship A's clock, from spaceship A's perspective. Now, the tricky part:

What will spaceship A see when they look into the "bridge" of spaceship B? And vice versa? Will the one see the people in the other "bridge" move in slow motion, or fast forward? Or will there perhaps be "gaps" in the transmission?

Thanks,

Peter

2. Mar 30, 2005

JesseM

If spaceship A and B are moving towards each other, then even though B's clock is "really" running slow in A's frame, if A is watching B live via a transmission moving at the speed of light, A will see B's clocks running fast due to the doppler effect. Suppose B is moving towards A at 0.6c, which means B's clocks should be slowed down by a factor of 1.25 in A's frame. Let's say that in 2005 B sends a signal from 20 light years away, and then in 2015 B sends a signal from 14 light years away. Since it took 10 years between these signals in A's frame, it must have taken 10/1.25 = 8 years according to clocks on B's ship. But then you have to take into account that the first signal will take 20 years to reach A, while the second will take only 14--this means A will receive the first signal in 2025, and the second signal in 2029. So only 4 years pass between the time A receives the signals, but 8 years passed onboard the ship between the time they were transmitted, which means that if A had been watching a continuous video feed between 2025 and 2029, A would see B sped up by a factor of 2, even though A could calculate that these signals must have been sent over a period of 10 years in his own frame, by taking into account the transmission time.

You can also calculate this more directly using the formula for the Relativistic Doppler Effect, $$\nu_{observed} = \nu_{source} \sqrt{(1 + V/c)/(1 - V/c)}$$. For example, if the frequency between two pulses in B's frame ($$\nu_{source}$$) is 1/(8 years), and V = 0.6c, then $$\nu_{observed}$$ is 1/(4 years), meaning A will see the pulses 4 years apart as I showed above.

Last edited: Mar 30, 2005
3. Mar 30, 2005

Staff: Mentor

Here's a blow-by-blow analysis of a similar situation: two twins watch each others calendars through telescopes as one travels out and back. I posted this in sci.physics.relativity about a year and a half ago, but I've still got the original file on my disk.

-----

The scenario: You stay behind on Earth while your twin goes on a space
journey. From your point of view he travels away from the Earth at a
speed of 0.8c for 5 years, covering a distance of 4 light-years in the
process, then immediately turns around and returns, again at a speed of
0.8c, taking another 5 years for the return trip. The total trip duration
is 10 years by your reckoning.

Relativity predicts that your twin experiences less elapsed time
because of time dilation:

(5 years) * sqrt (1 - 0.8^2) = 3 years

for each leg of the trip, and from his point of view the round trip lasts
only 6 years.

The relativistic time dilation equation predicts that each twin's clocks
"run slower" in the other twin's reference frame. So why can't your twin
conclude that the trip must be shorter for you, than it is for him? The
answer lies in the fact that your experiences are not symmetrical. Your
twin is at rest in two different inertial reference frames, one during the
outbound trip and another one during the inbound trip. You remain at rest
in a single inertial reference frame during the entire journey. Your twin
has to fire his spaceship's engines at the turnaround point. You do
nothing.

By examining what both of you actually *see* by watching each other's
clocks/calendars through telescopes, we can show that both of you must
come to the same conclusion: the trip lasts 10 years for you, and 6 years
for him.

First let's look at what's happening from your point of view.

To be specific, assume that your twin starts out on his journey at the
very beginning of the year 2004. You watch his clock (calendar?) through
your telescope as he recedes. The rate at which you receive the images of
each of his "new years" depends not only on his time dilation, but also on
the fact that he is moving away from you, so we have to use the
relativistic Doppler-effect equation. You see his calendar turn over to a
new year at intervals of

(1 year) * sqrt ((1 + 0.8) / (1 - 0.8)) = 3 years.

Therefore,

At the beginning of 2007, you see his calendar turn over to 2005.

At the beginning of 2010, you see his calendar turn over to 2006.

At the beginning of 2013, you see his calendar turn over to 2007.

This is the point at which you see him turn around and begin his return
trip. In your reference frame, he actually began the return trip at the
beginning of 2009 (2004 + 5), but you don't see it until 4 years
later because it happens 4 light-years away from you.

During his return trip, we have to switch the signs in the Doppler-shift
equation because now he's approaching, not receding. You now see his
calendar turn over to a new year at intervals of

(1 year) * sqrt ((1 - 0.8) / (1 + 0.8)) = 1/3 year.

Therefore,

At the beginning of May 2013, you see his calendar turn over to 2008.

At the beginning of September 2013, you see his calendar turn over to
2009.

At the beginning of January 2014, you see his calendar turn over to 2010,
and he arrives home. Ten years have elapsed on your calendar, and 6 years
have elapsed on his.

What does this look like from your twin's point of view, as he watches

As he is traveling away, he sees your calendar turn over a new year at
intervals of three years, just like you do his. Therefore,

At the beginning of 2007 (2004 + 3), he sees your calendar turn over to
2005.

But three years is the length of the outbound trip, according to him,
because of time dliation. So at this point he turns around and begins his
return trip. Now he sees your calendar turn over a new year at intervals
of 1/3 year, just like you do his. Therefore,

At the beginning of May 2007, he sees your calendar turn over to 2006.

At the beginning of September 2007, he sees your calendar turn over to
2007.

At the beginning of January 2008, he sees your calendar turn over to 2008.

At the beginning of May 2008, he sees your calendar turn over to 2009.

At the beginning of September 2008, he sees your calendar turn over to
2010.

At the beginning of January 2009, he sees your calendar turn over to 2011.

At the beginning of May 2009, he sees your calendar turn over to 2012.

At the beginning of September 2009, he sees your calendar turn over to
2013.

At the beginning of January 2010, he sees your calendar turn over to 2014.

But now three years have elapsed (on his calendar) since he turned around,
so he has now returned home. Six years have elapsed on his calendar, and
10 years have elapsed on yours, in agreement with what you observe.

4. Mar 30, 2005

Kilroy-was-here

jtbell,

thanks for posting that! It cleared up the problems I was having with imagining my scenario. It's really interesting to see how all these things happen, but for your feeling they can't happen. :tongue: