1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relativity of time

  1. Apr 3, 2005 #1
    ok im a little stuck on this problem, im hoping some one can help me out

    One way to set up the system of synchronized clocks would be for the chief observer to summon all her helpers to the origin O and syncronize their clocks, and have them travel to their assigned positions very slowly. Prove this claim as follows: Suppose a certain observer is assighned to a position P at a distance d from the origin. If he travels at a constant speed V, when he reaches P how much will his clock differ from the chiefs clock at O? Show that this difference approches 0 as V approches 0.

    So i found the observer that is moving with speed V has a time t=d/v at first i thought this was the difference but the limit is not 0. now i am thinking that there is no difference between the observer and chief because he is moving very slowly, but is that is right how do i show that the limit approches 0 ?

    thanks for the help
     
  2. jcsd
  3. Apr 4, 2005 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Use the Lorentz transformations for time intervals:

    [tex]t' = \gamma t_0 = \frac{t_0}{\sqrt{1 - v^2/c^2}}[/tex]

    What is

    [tex]\lim_{v\rightarrow 0} \frac{t_0}{\sqrt{1 - v^2/c^2}}[/tex] ?

    AM
     
  4. Apr 4, 2005 #3

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Consider the two following events:

    1) the observer leaves the origin at speed v.

    2) the observer arrive at point P (and its speed brutally shuts dow to 0)

    In the chief's frame (i.e. according to its clock), the time elapsed between the two events is [itex]\Delta t=d/v[/itex]. But in the observer's frame, there will be relativistic effects observed (the ground under her feet seems to be moving ==> it is subject to a relativistic lenght contraction). Since as far as the oberver is concerned the 2 events occur at the same place (notably, her own feet), the time interval between the two events is proper ==> it is related to the improper interval observed by her chief by the relation

    [tex]\Delta t' = \frac{\Delta t}{\gamma} = \sqrt{1-(v/c)^2}\Delta t[/tex]

    Therefor the difference between what the chief's clock shows and what her clock shows is

    [tex]\Delta t - \Delta t' = \Delta t - \sqrt{1-(v/c)^2}\Delta t = \Delta t(1-\sqrt{1-(v/c)^2}) = \frac{d}{v}(1-\sqrt{1-(v/c)^2})[/tex]

    Here, let us expand [itex]\sqrt{1-(v/c)^2}[/itex] in its Taylor serie. Remember that if |x|<1, then

    [tex](1+x)^n = 1+nx+\frac{n(n-1)}{2}x^2+...[/tex]

    Here we can set x = (-v/c)² because v < c, making x < 1. Therefor

    [tex]\sqrt{1-(v/c)^2} = (1+(-v/c)^2)^{1/2} = 1-\frac{1}{2}(v/c)^2 + ...[/tex]

    where "..." are terms of higher degrees of v. Back in our original equation, we get

    [tex]\Delta t - \Delta t' = \frac{d}{v}(1-[1-\frac{1}{2}(v/c)^2 + ...]) = \frac{d}{v}(1-1+\frac{1}{2}(v/c)^2 - ...) = \frac{d}{v}(\frac{1}{2}(v/c)^2 - ...) = d(\frac{1}{2}(v/c^2) - ...)[/tex]

    And of course,

    [tex]\lim_{v\rightarrow 0}d(\frac{1}{2}(v/c^2) - ...) = d(\frac{1}{2}(0/c^2) - ...) = 0[/tex]

    as expected. ("..." reprensenting terms of higher degree of v, it also vanishes in the limit)

    This may have seemed somwhat complicated to you, and, in a sense, it is. But you'll find that every textbook problems in relativity are alike and you'll quickly get the hang of it.
     
  5. Apr 4, 2005 #4

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Is this right Andrew? We don't have an expression for t_0, but we doubt it is infinity large as v approches 0, making the limit

    [tex]\lim_{v\rightarrow 0} \left[t_0 - \frac{t_0}{\sqrt{1 - v^2/c^2}}\right][/tex]

    an indeterminate form [itex]\infty\cdot 0[/itex], no? :grumpy:
     
    Last edited: Apr 4, 2005
  6. Apr 4, 2005 #5

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    t' and t_0 are time intervals (not the actual times measured on clocks). t_0 is the time interval measured in the frame of reference of the 'moving' observer and t' is that time interval measured in the frame of the 'stationary' observer.

    I don't see that. The limit of [itex]\sqrt{1-v^2/c^2}[/itex] as [itex]v\rightarrow 0[/itex] is 1. So [itex] t' \rightarrow t_0[/itex]

    AM
     
  7. Apr 4, 2005 #6

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    They're the time intervals between the two events...

    1) the observer leaves the origin at speed v.

    2) the observer arrive at point P (and its speed brutally shuts dow to 0)

    ...??

    In this case t' and t_0 are function of v and you can't put them out of the limit.
     
  8. Apr 4, 2005 #7

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    t_0 is not a function of v. It is just an arbitrary time interval as measured by the moving clock. t' is greater than t_0 by the factor gamma. As v approaches arbitrarily close to 0, t' approaches arbitrarily close to t_0. So the limit of t' as v approaches 0 is t_0.

    AM
     
  9. Apr 4, 2005 #8

    learningphysics

    User Avatar
    Homework Helper

    quasar is right. Andrew, using your variables:

    t'=d/v (chief observer's frame) and

    t_0=(d/v)(sqrt(1-v^2/c^2)) (moving frame)

    Both are functions of v.
     
  10. Apr 4, 2005 #9

    learningphysics

    User Avatar
    Homework Helper

    I'd use L'Hopital's rule for the limit above as v->0.
     
  11. Apr 4, 2005 #10

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I see what you mean. But in this particular situation, it seems that t_0 is not just an arbitrary time interval. It is the time it took observer to get from O to P at the speed v.

    The way I see it is that we have [itex]t' = \gamma t_0[/itex] and [itex]t' = d/v \Rightarrow t_0 = d/ \gamma v[/itex]. So [itex]t' - t_0 = d/v - d/ \gamma v = d/v(1- 1/\gamma) = d/v(1- \sqrt{1-(v/c)^2})\rightarrow \infty\cdot 0[/itex]
     
  12. Apr 4, 2005 #11

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yeah, hadn't tought of that :blushing: .
     
  13. Apr 4, 2005 #12
    Is this a little simpler?

    Let distances be in light seconds => v = beta
    Let the distance to P be 1 light second
    Let Y = gamma (because I'm no good at Latex!)
    The chief is in S and the helper is in S'

    The event "helper arrives at P" has a time coordinate of t = d/v => t = 1/beta

    => t' = t/Y

    => t' - t = t(1/Y - 1)

    => t' - t = (1/Y - 1)/beta

    1/Y is a function of beta squared => the Taylor expansion will only have even powers of beta

    Y(beta = 0) = 1 => the constant term in the expansion will be 1

    => t' - t = (homogeneous polynomial in beta squared)/beta

    QED

    ***************
    I just thought of this while I was jogging; it's even simpler!

    ct' =Y(ct - d*beta)

    QED
     
    Last edited: Apr 4, 2005
  14. Apr 4, 2005 #13

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    In that case,you would also have to transform the distance measurement of the moving observer.

    (1) [itex]t_0 = x'/\gamma v[/itex]

    So:

    [tex]t' = \gamma t_0 = \gamma x'/\gamma v = x'/v[/tex]

    [tex]t' -t_0 = \gamma t_0 - t_0 = x'/v - x'/\gamma v[/tex]

    [tex]t' -t_0 = \frac{x'}{v}(1 - 1/\gamma)[/tex]

    (2) [tex]t' -t_0 = \frac{x'}{v}(1 - \sqrt{1-v^2/c^2})[/tex]

    Using the binomial expansion of [itex]\sqrt{1-v^2/c^2}[/itex]:

    [tex]\sqrt{1-v^2/c^2} \rightarrow 1 - \frac{v^2}{2c^2}[/tex] for small v^2/c^2

    so (2) reduces to:

    [tex]t' -t_0 = \frac{x'}{v}(\frac{v^2}{2c^2})[/tex]

    [tex]t' -t_0 = \frac{x'v}{2c^2}[/tex]

    So as [tex]v \rightarrow 0, t' - t_0 \rightarrow 0[/tex]

    AM
     
  15. Apr 4, 2005 #14

    learningphysics

    User Avatar
    Homework Helper

    I don't see how you get the above. How is x' related to d? Is x'=d?
     
  16. Apr 4, 2005 #15

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Right. x' = d.

    AM
     
  17. Apr 4, 2005 #16

    learningphysics

    User Avatar
    Homework Helper

    Cool. Sorry about that. :blushing:
     
  18. Apr 5, 2005 #17

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Isn't this exactly the same procedure I meant through in my first post?!

    (with the exception that I kept d instead of x')
     
  19. Apr 5, 2005 #18

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    I don't think so. I was a little confused by your reasoning. You have

    [tex]\Delta t' = \frac{\Delta t}{\gamma}[/tex] and then you assume that

    [tex]\Delta t = x/v[/tex]

    Neither of these is correct (the correct relations are: [itex]\Delta t' = \gamma \Delta t[/itex] and [itex]\Delta t = x/v\gamma[/itex])
    ----------
    Edit:Quasar, I owe you an apology. I misunderstood your reference frames. Mine are the reverse of yours. Your [itex]\Delta t[/itex] is my t'.
    Sorry for doubting your excellent work..! :frown:
    ----------
    If you apply the Lorentz transformation to relate the proper clock times (T) and proper distances (X):

    [tex]T' = \gamma (T_0 - vX_0/c^2)[/tex]

    [tex]X_0 = \gamma(X' + vT')[/tex]

    Now X_0 is 0 in the rest frame of the moving observer. So:

    [tex]X' = -(-v)T' = v\gamma (T_0 - vX_0/c^2) = v\gamma T_0[/tex]

    So: [tex]T_0 = X'/v\gamma[/tex]

    Since: [tex]T' = X'/v[/tex],

    [tex]T'-T_0 = \Delta t = \frac{X'}{v}(1-\frac{1}{\gamma})[/tex]

    AM
     
    Last edited: Apr 6, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Relativity of time
  1. Relativity of time (Replies: 5)

Loading...