My name is Dennis. This might sound just straight up ignorant of basic concepts of physics.
I was having a conversation with my friend in regards to streaming a live event on earth to a spaceship that is traveling to a certain destination 10 light years away.
This live stream, lets just say it is a constant live stream and it is live for 10 years on earth, nonstop.
Without considering stuff like signals becoming weaker or what not, if it was possible to transmit the live stream signal to the spaceship at the speed of light, or slightly slower than speed of light but still faster than the spaceship, while the spaceship is traveling away from earth slightly slower than the speed of the live stream:
1st question: Would the person in the spaceship be able to stream the live video feed? So before taking off in the direction of the 10 light year destination, the person inside the spaceship turns on the stream and is watching the Live broadcast in the spaceship, maybe while waiting for take off. After he takes off and starts traveling at speeds close to speed of light, but not faster than the live stream would he be able to continue streaming this live steam video?
2nd: What would that video feed look like to the guy in the spaceship? How would this work in regards to time? Will the video be playing like in fast forward mode as soon as the spaceship starts traveling at the high speed?
3rd question: If the spaceship takes 10 years to reach the destination, the observer on earth will feel 10 years passing by while observing this spaceship reach the destination. But the person in the spaceship will not feel like 10 years passed, but less, right? If this is the case, wouldn't the live stream start fast forwarding?
I mistakenly posted this on worldbuilding beta… so I posted it again here. I hope it isn't against some rules…
Best Answer
The signal would be received redshifted by a factor of
$$f=\sqrt{\frac{c-v}{c+v}} \text{ , }\text{ } z=\frac{1}{f}-1$$
and it would run slower or faster by that factor, depending on if the ship moves away from or towards the receiver. So if your ship is receding away from you with $v=c/2$ then $1 \text{ sec}$ recorded on the ship would take $\sqrt{3}\text{ sec}$ to watch on earth. If the ship is moving towards your direction you receive the signal by that factor faster, then $1 \text{ sec}$ send from the ship would be received in $1/\sqrt{3}\text{ sec}$.
This is because you have to combine relativistic time dilation and classical doppler, which leads to this formula:
$$(1-\frac{v}{c}) \text{ }/ \text{ } \sqrt{1-\frac{v^2}{c^2}}=\sqrt{\frac{c-v}{c+v}}$$
where the first term is for the newtonian doppler, the second one for the time dilation (which tells you how fast the clocks on the ship are running without the extra distortion from the doppler effect) and the result is the relativistic doppler, which combines both.
You can also find it here in a slightly different notation.