Trouble understanding frames in relativity

[approximatelysphere]

Homework Statement

I'm having trouble with this specific problem/though experiment: Suppose you have a satellite travelling at v=0.8c and at t=0 it aligns with the earth. At t=1 year in the earth's frame of reference, a signal from mission control on earth is sent to the satellite. How long does it take for the signal to travel from earth to the satellite, in the earth's frame and in the satellite's frame?

Relevant Equations

dt=dt'/(1-v^2/c^2)^(1/2)

L=(1-v^2/c^2)^(1/2)*L'

I have attached a copy of my work as a pdf file. Please excuse my non-rigorous usage of notation.

Thank you so much!

 

[BvU]

Hi,

Your diagrams are not very useful. You want to draw 2D Minkovski diagrams:

For the diagram in the earth's frame you have one line for the satellite position: $x = 0.8 ct$ and one for the light signal: $x = c(t-1)$. Where do they intersect ?

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[BvU]

Where do they intersect ?

Very good. At $x= 4, \ ct = 5$ light years ! (*)
In agreement with your PDF page 1.

Then, on page 2: the time dilation formula yields $\Delta t' = 2.4$ years. (never forget the units)

Here, however, your confusion has already started: Four years have passed in the earth's frame. Equivalent to 2.4 years in the satellite's frame-- but these intervals DO NOT START OR END AT THE SAME TIME in both frames!

The four years is the right answer to the first part of the problem. For the second part some work needs to be done.

Then:

L = 0.8ly is the “proper length” of
the separation between the earth and the satellite in the earth’s reference frame.

is still correct. It is also the “proper length” of the separation between the earth and the satellite in the satellite’s reference frame (of course )

Even the 0.48 ly is right, but it is not the distance the light has to travel in the satellite frame !!!!

At this point I would like to refer to a PF singularity's signature:​

​

"You did not take relativity of simultaneity into account." - The answer to 99% of all paradox threads in the relativity forum​

PF Insights, including The Birth of a Textbook​

Earth and satellite do NOT, however have the same times nor x-coordinates for the event 'departure of the signal from earth', as becomes clear from the Minkovski diagrams, and also when one does the calculations :

(*) to check things, we are going to use the Lorentz transform and see what the satellite frame coordinates for the events A and B are.
Notation: $x, \ ct$ unprimed is in earth frame,
$x', \ ct'$ primed is in satellite frame. Units are light years.

For the earth frame we have $\beta = v/c = 0.8,\ \gamma = 1/\sqrt{1-\beta^2} = 1/0.6$
For the satellite frame we have $\beta' = v'/c = -0.8$ !

For A, the arrival of the signal at the satellite: $$\begin{align*}
x'&= \gamma(x - \beta ct) &= 0 & \phantom{x} \\
ct'&= \gamma(ct - \beta x) &= 3 &\phantom{x} \\ \end{align*}
$$For B, the departure of the signal from earth: $$\begin{align*}
x'&= \gamma(x - \beta ct) &= -0.8/0.6 &=-4/3 \\
ct'&= \gamma(ct - \beta x) &= \phantom{-}1/0.6 &=\phantom{-}5/3 \\ \end{align*}$$
It is instructive to do the inverse transform too !

At last, we can work out the second part of the answer to our problem: arrival at $ct'= 3$ ly, departure at $ct'= 5/3$ ly, so the light has traveled for 4/3 year on the satellite's clock.

Note that

non-rigorous usage of notation

is an invitation for errors to sneak in.

And the diagrams:

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