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Homework Help: How far is the relativistic star?

  1. Dec 1, 2011 #1
    1. The problem statement, all variables and given/known data

    Using natural units with c=1. We have our Sun and a star at rest relative to it one light year away. A space ship travels from our Sun to the star with v=1/2. During the journey how much time passes on the space ship's clock? How much distance does the captain of the space ship think has been traveled during the journey?

    2. Relevant equations

    -1 < v < 1

    t' = gamma(t - vx)
    x' = gamma(x - vt)
    y' = y
    z' = z

    gamma = 1/sqrt(1-v^2)

    3. The attempt at a solution

    The space ship and both Sun start with t=0, x=0. (0,0)
    In the Sun's frame the star is at (0,1), and the space ship will reach the star at (2,1).
    To get the space ship's view of the amount of time passed, we perform the Lorentz transform on (2,1).


    So by the ship's clock the space ship has reached the star in sqrt(4/3)*3/2=sqrt(4/3)*sqrt(9/4)=sqrt(3). That seems reasonable.

    How about the distance the ship's captain believes have been traveled. At t=t'=0 the star is at t=0 x=1. So if we perform the Lorentz transform on this we should get the distance perceived in the moving frame.

    t' = gamma(0 - 1/2)
    x' = gamma(1)
    y' = 0
    z' = 0

    so x'=sqrt(4/3) which is greater than one. That seems unphysical to me. But the captain will calculate his velocity as gamma/(gamma3/2) = 2/3 which seems reasonable.

    As a check I redid the problem with v=0.9 and the captain calculates his velocity as 4.73 which seems unphysical. What am I doing wrong?
  2. jcsd
  3. Dec 2, 2011 #2

    Simon Bridge

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    Star is 1ly away in Star frame - ship travels at 0.5c so it takes two years to get there.

    [itex]\gamma = 1/\sqrt{1-0.5^2} = 2/\sqrt{3}[/itex] which you got well done.

    Everybody agrees the ships clock reads [itex]\Delta t = 2/\gamma = \sqrt{3}yr[/itex] (well done) when it reaches the star, even though 2yrs have passed in the star frame. But in the ship POV this is fine because the star was only [itex]d=v\Delta t[/itex] away - so of course it took less time to draw close.

    You see - moving lengths are contracted ... and as far as the captain is concerned, the stars are the ones moving.
  4. Dec 2, 2011 #3

    D H

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    How you used the above is what lead to your error.

    The above is your error. What you are doing here is calculating the x coordinate of the spaceship in the spaceship frame. Think about it for a second: What is the spaceship's position with respect to itself?

    What you need to do is to
    • Calculate the target star's position in spaceship frame coordinates at start of the journey (t=t'=0). This yields the distance the spaceship will travel in spaceship coordinates, or
    • Calculate the sun's position in spaceship frame coordinates at the time the spaceship reaches the target star. This yields the distance the spaceship has traveled in spaceship coordinates.
  5. Dec 2, 2011 #4
    Not quite.

    The moving clock should be running slow.

    It certainly takes 2 years from the perspective of the Sun.

    Therefore the captain (who's moving) should see it take more than 2 years.

    The formula is [itex] \Delta t = \frac{t}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma t[/itex]

    which in this case means the captain sees it take [itex]\Delta t = \frac{4}{\sqrt{3}}[/itex] years.
  6. Dec 2, 2011 #5

    D H

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    The √3 years is correct. If the captain's spaceship is fast enough (0.5 c isn't even close to fast enough), the captain could make it to the Andromeda galaxy within her lifetime even though the galaxy is 2.5 million light years away -- to us, that is.
  7. Dec 2, 2011 #6
    I've seriously confused myself haha!

    Moving clocks run slow. Does that mean that the captain sees it take 2 years because he isn't moving with respect to the spaceship clock (i.e. both him and the spaceship clock are on the spaceship and so have no movement relative to one another).

    However, from the perspective of someone fixed at our Sun, they see the spaceship clock start to run slow because it is moving i.e. they will observe time dilation.

    The formula for time dilation is t' = gamma * t and gamma is definitely root(3)/2 so why isn't it 4/root(3)?

  8. Dec 3, 2011 #7

    Simon Bridge

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    I used to be confused by that "moving clocks run slow" saying all the time... and finally got it by remembering which twin gets to be younger.

    A clock is slow in the sense that it's hands are moving slower ... so, after a while, the time it shows lags behind. Slow right?

    It helps if you draw the space-time diagrams.
    For some reason nobody likes to draw the wee pictures these days.
    Everyone wants to plug numbers into equations.
  9. Dec 3, 2011 #8

    D H

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    No, it means the captain sees that that the journey takes less than 2 years.
    You are forgetting that from the perspective of someone on the moving spaceship, it is the Earth clocks that are running slow. This is one of those seemingly paradoxical aspects of special relativity: A says B's clock is running slow while B says that it is A's clock that is running slow.

    So, how to solve this particular problem? There a lot of ways to look at it, and all will give the same answer: The captain's clock indicate that the trip takes √3 years. One way is via the Lorentz transform, as was done in the original post. Another way is to divide the length-contracted distance by the velocity. From the captain's POV, the target star is initially at a distance of (√3)/2 light years (not one light year) and is moving toward the stationary spaceship at 0.5 c. d/v=(0.5√3 light years)/(0.5 light years/year) = √3 years.
  10. Dec 3, 2011 #9
    So surely the captain doesn't see the ship clock run slow since he is in the ship frame and so he sees the clock on Earth run slow.

    Meanwhile, the clock on Earth takes two years and it is the observer on Earth who thinks that the captains clock is running slow and taking \root(3) years.


    Disclaimer: It's been a long time since I did this and I seem to have forgotten it!
  11. Dec 3, 2011 #10

    D H

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    The captain indeed does not see her clock as running slow. To her, her clock is working just fine. (Glance at the clock and then count: One thousand one, one thousand two, one thousand three, ... one thousand sixty. Glance at the clock again: One minute!)

    She also sees the target star as coming toward her at 0.5c, easily verified by looking at the blueshift of the hydrogen alpha line. She also sees that the target star is initially at a distance of √3/2 light years (not 1 light year). This initial distance is a calculated value, but is easily confirmed by measurement by observing the increase in apparent magnitude as the spaceship approaches the target star.

    These measurements are confirmed when the spaceship flies by the target star √3 years later.
  12. Dec 4, 2011 #11

    Simon Bridge

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    Lets be careful here - to use Special Relativity, the ship has to fly past both stars ... it does not start at one and end at the other (that would require acceleration).

    If we prepare the stars so there are two clocks in the star-frame, one at Sol and the other at The Destination, which are synchronized, then - as Sol passes the Captain, she looks out and notes the time on the Sol clock and on her own. When the Destination passes, she looks out again, noting both times.

    I think what LatentCorpse is wondering is this:

    In the captains POV
    Speed of the stars is 0.5c
    The trip-time on the ship clock was √3yrs
    Distance between the stars was √3/2ly
    The stars are moving - therefore the star-clock trip time is √3/γ = 1.5yrs.

    So following the above,
    The stars should be only 0.75ly apart from the POV of the star-frame.

    See how easy it is to get messed up with this?

    My fav description is this one... it's geometric and I like geometry. Especially the bits with space-time diagrams: probably the clearest way to sort these ideas out.
  13. Dec 4, 2011 #12

    D H

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    There's nothing wrong with acceleration in special relativity, so long as it's not gravitational acceleration. The only thing that special relativity can't handle is gravitation.
  14. Dec 4, 2011 #13
    So the captain genuinely sees root(3) years elapse on her clock. This is because
    (i) Her clock is running fine i.e. she's not moving in her frame so her proper time is equal to coordinate time
    (ii) The star is moving towards her at 0.5c and so the initial distance undergoes length contraction
    (iii) t=s/v and if s is length contracted i.e. less then t is less

    What about from the perspective of someone at the initial star. They see root(3) years elapse on spaceship clock because:
    (i) Both target and initial start remain a fixed distance of 1 light year apart throughout the experiment
    (ii) Spaceship is flying at 0.5c and so they see time dilation on the spaceship clock
    (iii) t=s/v = 1/0.5 =2 gives the coordinate time elapsed for the journey
    (iv) we put this into the lorentz transformation and see that root(3) of a year is actually elapsed

    What about from the perspective of someone at the target star? This will be the exact same argument as above.

    If you can confirm that all of that is fine then my last concern is how this plays out physically. Consider the following incorrect argument

    Taking the captain and an observer at the initial star:
    (i) It takes root(3) years for the captain to reach the target
    (ii) The observer at star see it take 2 years
    Obviously this is due to special relativistic effects but could we relate it to the time the light takes to travel back to the initial start?

    i.e. spaceship arrives after root(3) years
    it takes 2-(root(3)) years for light to travel back to initial star
    therefore observer at initial start sees spaceship arrive after 2 years

    this argument doesn't work seeing as they are using genuinely different clocks. Is it in fact true to say that the observer at the initial star "CALCULATES" the journey of the spaceship will take 2 years but would in fact actually have to wait much, much longer to see the arrival. In fact, they would have to wait another 2 years for the light to travel back to them

    Finally, suppose the spaceship comes to a halt at target star. This necessitates a deceleration - will this deceleration cause time to essentially "catch up" and realign the spaceship clock with the clock of an observer at the initial star? If so, how does this work?

    Thanks a lot for your help.
  15. Dec 4, 2011 #14

    D H

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    Everything you typed before that looks fine.

    Yes, you could. What the different observers see (as opposed to calculate) is my personal favorite explanation of the twin paradox. It appeals to me. Different explanations appeal to others; it doesn't really matter which explanation is used so as they all yield the same answer.

    Suppose the captain sends a time-stamped video signal back to Earth just as she arrives at the target star. Per the ship's clock, the journey to the target stars takes but √3 (~ 1.732) years. The signal sent back to Earth will have a time stamp of journey start time + 1.732 years, with the captain's apparent age being in agreement with this time stamp. After correcting for the time it took for this signal to reach Earth, the Earth observers will see this as a confirmation of relativity. Her 1.732 years multiplied by her gamma (2/√3) is exactly equal to 2/3 * 3 years.

    I'll anticipate the remark "But wait! The captain should see the Earth-bound clocks as dilated!" Suppose the Earth sends a congratulatory message to the captain. They send this message a year (a year as measured on the Earth, that is) after the start of the journey so that reception of the message coincides with the captain's arrival at the target star. The signal will have a time stamp journey start time + 1 year, with the captain's cohorts' apparent ages being in agreement with this time stamp. After correcting for the time it took for this signal to reach the spaceship, the captain will see this as a confirmation of relativity. Their 1 year multiplied by their gamma (2/√3) is exactly equal to 2/3 * √3 years.

    That won't happen. It is velocity and velocity only (i.e., not acceleration or any higher order derivative) that causes time dilation. This is the clock hypothesis, and it has been tested time and time again.

    Suppose that the spaceship starts on a return journey to Earth after making a brief stop at the target star. Ignoring the time spent at the target star, the captain will have aged a total of 3.464 years, but her Earth-bound will have aged 4 years. This is the twin paradox. There are lots and lots of web pages that describe this phenomenon.
  16. Dec 4, 2011 #15

    Simon Bridge

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    However - if the captain started from Sol and stopped at the Destination star, that would change the math we've been doing.
  17. Dec 5, 2011 #16
    I don't get what you are doing here. Looking at your first paragraph. You have the captain sending the message after ~1.7 years.
    The light takes 1 year to reach Earth
    Surely it is the 1 year that we have to Lorentz transform (giving 1* 2/root(3) = 2/root(3))
    And so the video message would be received on Earth ~1.7+2/root(3) years later, no?

    Where am I going wrong in my understanding?

    So the captain is genuinely younger? Or is this just from the perspective of the Earth observer?

    Surely then, seeing as we said that from the captain's frame the clocks on Earth run slow, the captain will see the Earth observer as being younger than her when she returns?
  18. Dec 5, 2011 #17

    D H

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    What you are doing wrong is that you are mixing and matching frames without realizing it.

    Let's start with Newtonian universe. The spaceship travels the 1 light year distance at 0.5c. From the perspective of an Earth-bound observer, the outward journey takes two years, and the signal announcing the arrival at the target takes another year to return to Earth. So, three years total pass from launch to reception of the arrival signal. In this Newtonian universe, the time stamp on that signal is 2 years mission elapsed time (MET).

    Now let's go to our relativistic universe. From the perspective of an Earth-bound observer, the outward journey still takes two years, and the signal announcing the arrival at the target still takes another year to return to Earth. There is no difference between our universe and the Newtonian universe in this regard. Our universe and the Newtonian one depart when it comes to the time stamp on that announcement message. It's 2 years MET in the Newtonian universe, but only 1.732 years MET in our relativistic universe.

    The captain is definitely younger on return than her Earth-bound cohorts. I suggest you google "twin paradox". You'll find plenty of explanations on the 'net, including several discussions right here at this site.
  19. Dec 5, 2011 #18
    Got it I think.
    And the light takes the same time in both universes because it travels on null geodesics and these are immune to the effects of relativity i.e. c is the universal speed limit and can't be changed regardless of frame of reference?

    Captain is definitely younger. From perspective of Earthling, this is because she is moving at a velocity close to speed of light and therefore spaceship clock will run slow i.e. less time will elapse and so it takes her 1.7 years to get there and 1.7 years back i.e. she is 3.4 years older whilst Earthling has aged 4 years.

    What about from captain's perspective? Well her clock is running fine (from her perspective at least). However, she is travelling at 0.5c and so there will be some length contraction at play. In fact she will have a shorter journey at the same speed and so less time will elapse. In fact, only 1.7 years will elapse in each direction. Again, she returns to Earth having aged only 3.4 years whilst the Earthling has aged 4(*)

    Is this all correct? I think it is!

    The only bit I don't get is the (*) bit at the end of the last paragraph I wrote. From the captain's frame of reference it is the Earth that is receeding at 0.5c and so surely she should see Earth clocks undergo time dilation. From this perspective, less time should elapse on Earth clocks and in fact the Earthling should age 3.4 also instead of 4 years.

    To sum up, I think I've got it all except the Earth time from captain's perspective!
  20. Dec 5, 2011 #19

    D H

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    The key to understanding the twin paradox is to realize that the spaceship turns around. If the spaceship didn't turn around, you would instead have the seeming paradox that the captain and her Earth-bound cohorts each see the other's clock as running slow.

    Before explaining the twin paradox yet again, try reading this post that discusses a spaceship traveling at 0.8c to a star 16 light years away: [post]2696480[/post].
  21. Dec 5, 2011 #20
    Hi again.

    That's a good explanation but....

    (i) Are you assuming the transmission of signals between A and B is instantaneous? Presumably so otherwise a signal sent upon birth of A's child 4 years after take off wouldn't arrive at target start until 18 years have elapsed by A's calculation (4+9.6=13.6 years by B's calculation).

    (ii) How can there ever be any speeding up of clocks?
    The Lorentz transformation is [itex] \Delta t = \sqrt{1-\frac{v^2}{c^2}} t[/itex] and since [itex]v \leq c[/itex], we have that [itex]\Delta t \geq t[/itex] always. So it appears to me that we are only able to generate time dilations not speeding up of clocks?
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