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Basic special relativity calculation

  1. Oct 29, 2008 #1
    1. The problem statement, all variables and given/known data

    A spaceship sets off from Earth for a distant destination, travelling in a straight line at uniform speed [tex]\frac{3c}{5}[/tex]. Ten years later a second ship sets off in the same direction at [tex]\frac{4c}{5}[/tex].

    i) Consider the distance travelled by each spaceship as a function of time to work out when (in the Earth frame) the second spaceship will have travelled as far as the first.

    ii) Calculate which of the ship's captains has aged the most during their journey, and by how much.

    2. Relevant equations

    The lorentz transformations, and [tex]speed = \frac{distance}{time}[/tex]


    3. The attempt at a solution

    I have a solution, but it's one of the first SR problems I've done so I am unsure whether it's correct.

    Using c = 1 lightyear per year.

    i) In earth frame:

    distance travelled by first rocket = distance travelled by second rocket

    [tex]\frac{3ct}{5} = \frac{4c(t-10)}{5}[/tex]

    [tex]3t = 4t - 40[/tex]

    [tex]t = 40[/tex]

    So the rockets have travelled the same distance after 40 years in the Earth's frame.


    ii) Where x is distance travelled in Earth's frame when rockets meet:

    x = speed of rocket 1 * time
    [tex] = \frac{3}{5} * 40[/tex]
    = 24 light years

    time elapsed in rocket 1's frame = [tex]\frac{1}{\sqrt{1 - \frac{3^2}{5^2}}} * (40 - \frac{3*24}{5})[/tex]
    [tex]=\frac{5}{4} * 25.6[/tex]
    = 32 years

    time elapsed in rocket 2's frame = [tex]10 + \frac{1}{\sqrt{1 - \frac{4^2}{5^2}}} * (30 - \frac{4*24}{5})[/tex]
    [tex]= 10 + \frac{5}{3} * 10.8[/tex]
    = 28 years

    So the captain of rocket 1 has aged by 4 years more.

    Is this the correct way to approach this problem, and is the answer correct?

    Thanks for your help : )
     
    Last edited: Oct 29, 2008
  2. jcsd
  3. Nov 1, 2008 #2

    Doc Al

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    Staff: Mentor

    Looks good to me.
     
  4. Nov 1, 2008 #3
    Thanks.

    Another question: I have seen the lorentz transform for time, which I know as

    [tex]t new = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} * (t - \frac{x*v}{c^2})[/tex]

    written as

    [tex]t new = t\sqrt{1 - \frac{v^2}{c^2}} [/tex]

    I understand how to get to this form by taking a t out of the brackets, but my question is can I apply the new form in all the same places I use the old one? And if so, why does anyone use the first form, as it seems more complicated?
     
  5. Nov 1, 2008 #4

    Doc Al

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    Staff: Mentor

    That second formula is a special case that only applies when the time is that measured by a single moving clock (what you call t_new). It says that when viewed from another frame, a moving clock runs slow--the so-called time dilation formula.

    No, those formulas are not interchangeable.
     
  6. Nov 1, 2008 #5
    Ok, but for the first formula the time coordinate in the new frame depends on the x coordinate of the event in the old frame. So if I had two events that happened in the rest frame at different x coordinates but the same t, y, and z coordinates coordinates, then their new t values in the moving frame would be different. Is this correct, and if so is there an intuitive way of understanding it?

    edit: OK I think this is the thing about events being simultaneous in some frames but not in others, but I'd still like to understand it intuitively.
     
    Last edited: Nov 1, 2008
  7. Nov 1, 2008 #6

    Doc Al

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    Staff: Mentor

    You are exactly right about it having to do with the relativity of simultaneity. You can deduce from the Lorentz transformation:
    [tex]\Delta t' = \gamma(\Delta t - v\Delta x/c^2)[/tex]

    that two events happening at the same time but different x coordinates in the unprimed ("rest") frame will take place at different times in the primed frame:
    [tex]\Delta t' = \gamma(0 - v\Delta x/c^2) = -\gamma(v\Delta x/c^2)[/tex]

    If you imagine two synchronized (unprimed) clocks present at these two events you can deduce that they will be unsynchronized according to the primed frame.

    In my opinion, there are three ways to understand basic special relativity and all three are important:

    (1) Using the Lorentz transformations. For certain problems, this is the best approach, but often it leaves intuition behind.

    (2) Using the relativistic behavior of clocks and measuring rods. This can be extremely helpful in developing intuition. The three relativistic behaviors are: Length contraction, time dilation, and--the one you brought up--clock desynchronization (or the relativity of simultaneity). Some problems can be solved much quicker if you have a solid understanding of these relativistic effects.

    (3) Using space-time diagrams. This is where you begin to develop a real understanding of relativity.

    You might find this summary helpful: Basic Equations of Special Relativity
     
  8. Nov 2, 2008 #7
    That is really useful. I have started reading about spacetime diagrams, I think it will take a while for me to understand them though.

    Thanks for helping me, it's been really helpful : )
     
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