For a relativistic spaceship, what is the most recent news the ship gets from Earth?

[Albertgauss]

TL;DR Summary

What is the most recent news on Earth a relativistic ship will receive when it lands far away?

Looking to see if I am in the ballpark correct about this. Let's say a spaceship flies at 99% the speed of light (B=0.99) to a planet 12 light years away. The spaceship is going to land on the planet once it arrives and stay put ever after. Ignore acceleration of start/stop. When it lands on the planet, what is the spaceship's most recent Earth news? Is my thinking correct below?

The math looked very difficult so I modified the spacetime diagram below for a ballpark answer (which is about all I want here).

The website is: https://www.physi.uni-heidelberg.de/~glassel/p3/twin.html

A few modifications. The blue graphics you see in the original post are for a ship of B=0.6. I could not filter that out quickly so ignore everything in blue. The purple axes are for the ship I am interested in, B=0.99. The ship lands on a planet 12 light years from Earth and is now at rest with respect to Earth's frame. It looks from my sketches that the ship will only know about the first year of things going on in Earth. Even though 13 years has passed on Earth, the ship only has news about what happened on Earth for that first year. Then, after the fourteenth year of life proceeds on Earth, only the year 2 information about what happened on Earth arrives to the planet. Is this basically correct?

 

[Orodruin]

Schematichally, yes, but your image has a major flaw:

Since the planet is 12 ly away, the light from Earth year 1 arrives at the time of Earth year 3. The black line, green line, and red dotted line should all meet at the same point.

Furthermore, the ship is represented by the t’ axis (x’ = 0). It arrives when the t’ axis crosses the planet line.

I don’t know why you think the math is hard here though. It is pretty straightforward. Your axes do not look tilted enough to correspond to beta = 0.99.

 

[Ibix]

Your diagram looks correct, except the inaccuracy @Orodruin noted (I think his "Earth year 3" is meant to be "Earth year 13", though).

The maths is fairly straightforward here since you don't even need relativity - just an intercept calculation. If the ship sets off at Earth time 0 and travels a distance $d$ to the planet at speed $v$, what time will it arrive? If light sets off at time $T$ and travels the same distance at speed $c$ what time will it arrive? If both those arrivals are simultaneous then you can solve for $T$.

 

[Ibix]

Your axes do not look tilted enough to correspond to beta = 0.99.

I would agree. The faster axis ($ct'$) appears to correspond to about 0.9c, since it takes 13-point-something years to cross 12 light years.

 

[Albertgauss]

Yes, looks like you're both correct. Below is a better image. Let's see if I got everything.

Since the planet is 12 ly away, the light from Earth year 1 arrives at the time of Earth year 3. The black line, green line, and red dotted line should all meet at the same point.

I think I got that drawn correctly now.

Your axes do not look tilted enough to correspond to beta = 0.99.

That's the trouble with trying not to cram everything together for a readable graphic. I wanted to keep things as clear as I could. But, yes, I agree, the ship axes would be very, very narrow, as narrow as I could make in powerpoint.

I don’t know why you think the math is hard here though. It is pretty straightforward.

The maths is fairly straightforward here, though, since you don't even need relativity - just a straightforward intercept calculation.

Yes, it turns out that you are both correct. If the ship travels at 0.99c, then it will arrive (as seen by Earth) at 12.12 years. It was just t=d/v=(12 c yr)/(0.99 c). So when the ship lands on the planet, the most recent news it has of Earth is up to 0.12 years or about 1.44 months. In plainer language, when the ship lands on the planet, it knows everything that happened on Earth up to 1.44 months after it left Earth. Is this correct?

Also, if Earth news leaves Eart at year 1, yes, then that Earth-year-1 news does arrive at the planet at year 13 as seen by Earth.

The last thing I wanted to do is figure out what time on the ship the Earth news at 1.44 months arrived? I use the simplest Lorentz transformation of

[t][']=γ(t-vx/[c][/2]) = √(1-[0.99][/2])*(12.12 yrs-0.99c*12c*yr/[c][/2])=0.03 years or 12. 4 days.

This assumes the origins line up and that the second event is the ship's landing on the planet, which occurs at Earth spacetime coordinates (x=12 c yr) and (t = 12.12 yrs)

One more calc, the ship says that the distance to the planet 12 light years away (Earth frame) would be:

[x][']=γ(x-vt) = √(1-[0.99][/2])*(12 c yrs-0.99c*12.12*yr)=1.69([10][/-4]) c yrs ~ 10.7 Astro Units or, in plainer language, in the ship's frame it thinks the distance to the planet is about the size of the solar system as far as Saturn.

How am I doing so far? Does everything look correct?

 

[PAllen]

Everything of earth seen on the planet is from 12 years earlier per the planet. So when the rocket first arrives on the planet, what they would see on a clock on earth would simply be 12/.99 - 12 years later than the earth clock time the rocket saw as it passed right by the earth. The time dilation factor for .99c is about 7.0888. So as the ship’s clock advances 1 year 253 days ( 12 / .99/ 7.0888 ) during the trip, it sees the earth clock advance advance 44.24 days. This latter is also the trip travel time per the ship divided by the relativistic Doppler factor for .99c.

 

[JimWhoKnew]

One more calc, the ship says that the distance to the planet 12 light years away (Earth frame) would be:

[x][']=γ(x-vt) = √(1-[0.99][/2])*(12 c yrs-0.99c*12.12*yr)=1.69([10][/-4]) c yrs ~ 10.7 Astro Units or, in plainer language, in the ship's frame it thinks the distance to the planet is about the size of the solar system as far as Saturn.

How am I doing so far? Does everything look correct?

In the moving spaceship's frame, both events (departure and arrival) occur at the same place $x'=0$ , ($~x-vt=0~$ trivially, when evaluated before substitution of rounded results). Both planets are moving in this frame. To get the constant distance between them in this frame, use the Lorentz contraction formula.