Watching a live stream in spaceship traveling away from earth close to speed of light

Question

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...

Answer 1

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.

Answer 2

I have made some videos of the twin paradox from the traveller's point of view (i.e. first person) here. I'll post the channel notes here to add a bit of substance, I would appreciate feedback and corrections.

The twin "paradox" (in quotes because it is NOT a real paradox!) is a valuable learning tool for Special Relativity: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_intro.html

This is a series of visualizations of the journey from the point of view of the travelling twin, who flies 20 light years away from his home station then returns. The journey consists of four parts, joined together. The first quarter is an acceleration away from the station. During the second and third quarters the ship accelerates towards the station (so that at the half way point the ship is stationary 20 light years away). In the fourth quarter the ship accelerates away from the station in order to come to rest there.

The total coordinate travel time (shown as a red dot in the top left HUD clock) is 43.711/58.918 years for acceleration at earth/moon gravity levels, whilst proper time (green dot) is 12.101/38.694 years. The yellow dot represents the time the traveller would see on the station clock face through a very powerful telescope! The octahedral stations (spaced one light year apart and one light year to the left of the flight path) are all synchronized to coordinate time and rotate once over the course of the whole journey. There is a 2x2 light year wall one light year beyond the far end of the journey, and large rectangular frames every 5 light years. Where a floor is shown it is 1 light year per stripe, and there are small 1 ly milestone spheres along the way, with a larger one every five light years.

The flights are rendered without relativistic effects and then for two values of acceleration (currently earth gravity and moon gravity). The distortion artifacts are due to aberration of light, the Doppler effect and the headlight effect. The earth gravity videos exhibit some nice penrose-terrel "rotation" effects. Magenta markers in a circle show where you really are in the scene (Doppler shift = gamma for this circle), and grey markers show the circle where the doppler shift is 1 (gamma is in theory directly observable on this circle).

These videos were made with POV-Ray (http://www.povray.org/) and avconv/ffmpeg (https://www.ffmpeg.org/) video encoding software.