(Introductory SR) Light sent from a spaceship received by Earth

[malawi_glenn]

Homework Statement

A spaceship leaves Earth at a speed ##v= 0.5\cdot c##. An astronaut on the spaceship needs to send two light signals back to Earth. The first signal is sent so that it will be received by people on Earth exactly 1 year after the departure of the spaceship (according to the Earth frame). The second signal is then sent 1 year after the first signal was sent (according to the spaceship frame).

(a) According to the spaceship frame, when should the astronaut send the first signal after his departure?

(b) According to the Earth frame, what is the time interval between receiving the first and second signals on Earth?

Relevant Equations

The Lorentz-transformations:
##S'## moves in ##S## with constant speed ##v## in the ##x##-direction.
##x = \gamma (x' + vt') ## and ## t = \gamma (t' + x'v/c^2)##
##x'= \gamma (x - vt) ## and ## t' = \gamma (t - xv/c^2)##
Gamma-factor ## \gamma = 1/\sqrt{1-v^2/c^2}##

Doppler formula (longitudinal) ##T## is the observed (measured) period in ##S## and ##T'## is the emitted period as measured in ##S'##
## T = T' \cdot \sqrt{ \dfrac{c+v}{c-v}} ##

I am taking a summer course on special relativity and I stumbled across this problem and solution which I tought look neat. However, I think the solution provided for a) there is wrong. I will here present two of my solutions for a) and one solution for b) and ask if you think mine are okay? :)

For the first solution I used space-time diagram, and for the second solution pure Lorentz-transformation.

First solution for a)
##S## frame: spaceship frame. Spaceship is located at ## x = 0## in ##S##
##S'## frame: Earth, moves at ##v = 0.5c## in ##S##. Earth is located at ##x' = 0## in ##S'##.

##ct## axis vertical.
##x## axis horizontal.
##S## and ##S'## origins conincide at the start of the spaceships journey, i.e. ## x = x' = 0## at ##ct = ct' = 0##.

The worldline of the Earth in ##S'## frame is ##(ct',x') = (ct',0)##
Using the Lorentz-transformation
##t = \gamma\cdot (t' + x'\cdot v/c^2)##
##x = \gamma\cdot (x' + v\cdot t')##
we obtain for the ##ct, x## coordiantes for the worldline of the Earth in ##S## frame to be
##(ct, x) = (\gamma \cdot ct' , \gamma \cdot (v/c) \cdot ct')##

The first light signal is received at Earth ##(ct',x') = (ct',0) = (1 \text{light-year}, 0) = (c \cdot 1\text{year} , 0)##
we obtain for the ##ct, x## coordiantes of this event to be
##(ct, x) = (\gamma \cdot ct' , \gamma \cdot (v/c) \cdot ct') = (\gamma \cdot c \cdot 1\text{year}, \gamma \cdot 0.5 \cdot c \cdot1\text{year} )##

The first light signal worldline in ##S## is given by the equation
##ct = c \cdot \alpha + x##
where ##\alpha ## is the time in ##S##-frame when the first light signal should be sent in order to reach Earth at ## t' = 1 \text{year} ##.

The light signal worldline intersect the Earth worldline in ##S## at
## \gamma \cdot c t' = c \cdot \alpha + \gamma \cdot (v/c) \cdot t' ##

## \gamma \cdot c \cdot 1\text{year} = c \cdot \alpha + \gamma \cdot 0.5 \cdot c \cdot 1 \text{year} ##
which has the solution
## c \cdot \alpha = \gamma \cdot (1-0.5) \cdot c \cdot 1\text{year} ##
## \alpha = \gamma \cdot (1-0.5) \cdot 1\text{year} ##

with ##v = 0.5c##, ##\gamma = 1.1547## we obtain
##\alpha = 1.1547\cdot(1-0.5) \cdot 1\text{year} = ## 0.577 years.

The solution which I linked to states 0.447 years.
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Second solution for a)

Let ##S## frame be the Earth frame and ##S'## be the spaceships frame. ##S'## moves in ##S## with ##v = 0.5c##. Earth is located at ##x= 0## and spaceship is located at ##x' = 0##.

Let ##t_1## be the time in ##S##-frame when the spaceship emitts its first light signal.
At ##t_1##, the spaceship is located at ##x_1 = v \cdot t_1 = 0.5c\cdot t_1##
Because Earth is receiving the light signal at ##t=1 \text{year}## we must have ## t_1 + \dfrac{x_1}{c} = 1 \text{year}## since ##\dfrac{x_1}{c} ## is the amount of time the light signal must spend to reach Earth from the spaceship according to Earths frame (##S##-frame). But ##\dfrac{x_1}{c} = 0.5c\cdot t_1 ## so we have ## t_1 +0.5c\cdot t_1 = 1 \text{year}##, which has the solution ##t_1 = \dfrac{2}{3} \text{year}##.

That is, according to Earth-frame, the (first) light signal must be emitted by the spaceship at 2/3 years after the take-off. The spaceship was located at ## x_1 = 0.5c\cdot t_1 = 0.5c \cdot \dfrac{2}{3} \text{year} = \dfrac{1}{3} c \cdot 1\text{year} = ## 1/3 lightyear.

Using the Lorentz-transform ## t' = \gamma (t - xv/c^2)## we obtain ##t_1'##, the time according to the spaceship frame (##S'##) when the first light signal should be sent.
## t_1' = \gamma (t_1 - x_1 \cdot v/c^2) = \gamma ( \frac{2}{3}\cdot c \cdot 1\text{year} - \frac{1}{3} c \cdot 1\text{year} \cdot 0.5c/c^2 ) = 0.577 \text{year} ##
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Solution for b)
Let ##S## frame be the Earth frame and ##S'## be the spaceships frame. ##S'## moves in ##S## with ##v = 0.5c##. Earth is located at ##x= 0## and spaceship is located at ##x' = 0##.

I use the doppler formula for period, since we can think of the light signal interval on the spaceship is periodic signal with ##T' = 1 \text{year}##.

## T = T' \cdot \sqrt{ \dfrac{c+v}{c-v}} = 1 \text{year} \cdot \sqrt{ \dfrac{c+0.5c}{c-0.5c}} = ## 1.734 years

The solution which I linked to states 2.236 years.

Thank you all for reading this far, hope to hear from you reagarding who is correct on this nice problem :)

 

[PeroK]

I didn't check your solution, but I agree with you.

I did the kinematics in the rocket frame.

 

[kuruman]

Doppler formula (longitudinal) ##T## is the observed (measured) period in ##S## and ##T'## is the emitted period as measured in ##S'##
## T = T' \cdot \sqrt{ \dfrac{c+v}{c-v}} ##

What does the Doppler shift formula have to do with this?

 

[PeroK]

What does the Doppler shift formula have to do with this?

Using the Doppler shift is definitely the neatest solution for part b).

 

[kuruman]

Using the Doppler shift is definitely the neatest solution for part b).

Yes, I see it now.

 

[PeroK]

And the neatest solution for part a) is also to use the Doppler shift. Consider an initial signal send at time ##t = t' = 0## and the second is received at time ##T_E = 1##, so must have been sent at
$$T_S = T_E\sqrt{\frac{c-v}{c+v}} =\frac{T_E}{\sqrt 3}$$

 

[malawi_glenn]

And the neatest solution for part a) is also to use the Doppler shift. Consider an initial signal send at time ##t = t' = 0## and the second is received at time ##T_E = 1##, so must have been sent at
$$T_S = T_E\sqrt{\frac{c-v}{c+v}} =\frac{T_E}{\sqrt 3}$$

ah yes that is pretty neat! Never considered using doppler formula for a) by assuming spaceship sent out a signal immediately after start :)