Trouble understanding frames in relativity

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SUMMARY

The discussion focuses on understanding frames in relativity, specifically through the use of 2D Minkowski diagrams. Key calculations include the intersection of satellite and light signal positions at ##x = 4, \ ct = 5## light years, and the time dilation formula yielding ##\Delta t' = 2.4## years in the satellite's frame compared to four years in the Earth's frame. The participants emphasize the importance of recognizing the relativity of simultaneity, which is crucial for accurately interpreting events in different reference frames. The discussion concludes with a recommendation to utilize Lorentz transformations for deeper insights into the relationship between the Earth and satellite frames.

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  • Understanding of 2D Minkowski diagrams
  • Familiarity with time dilation and Lorentz transformations
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  • Ability to interpret spacetime coordinates
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approximatelysphere
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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!
 

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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 ?

##\ ##
 
BvU said:
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 :wink:)

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​

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
approximatelysphere said:
non-rigorous usage of notation
is an invitation for errors to sneak in.

And the diagrams:

1727789664983.png


1727789626361.png



##\ ##
 
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Hi, thank you so much for your detailed answer. Please excuse me for not replying for so long. Life has been very very chaotic.

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

Could you please explain what this o.48 ly actually is a measure of?
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!
By this, do you mean that 4 years passed is indeed 2.4 years in the satellite's frame, but since the event of the departure of light does not happen simultaneously in both frames, 2.4 is not necessarily the number of years needed for light to travel from the earth to the satellite in the satellite's frame?

BvU said:
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 :
Aha! So my main mistake is assuming that light departs from the earth in the earth's frame at the same moment as in the satellite's frame?
 
approximatelysphere said:
Aha! So my main mistake is assuming that light departs from the earth in the earth's frame at the same moment time coordinate as in the satellite's frame?
I've made a change to what you wrote. To understand reference frames, you should start thinking in terms of coordinates in spacetime.
 
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