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Relativistic time, distance and velocity

  1. Nov 7, 2007 #1
    A ship leaves Earth bound for a star 4.5 light-years away. According to the ship's clocks, the trip took 4.5 years. What was the velocity of the ship in the reference frame of the Earth?

    I'm probably wrong, but I say v=c.

    In the reference frame of the ship, the star is 4.5 light-years away and it is approaching you at some velocity v. If it takes 4.5 light-years to get to you, then it is approaching you at a velocity of c. The clocks outside the reference frame of the ship are going slower. In fact, they are stopped when v=c.

    Using the Lorentz equations I alway have one more unknown than I have equations.
     
  2. jcsd
  3. Nov 7, 2007 #2
    Is it? Think in terms of the distance between the Earth and the star as measured from the ship's frame.
     
  4. Nov 7, 2007 #3
    If it isn't, then that just introduces the length contraction equation with three more variables, two of which are unknown.

    If you are on the star, then from your reference from the ship is 4.5 light-years away coming towards you at some speed v.

    If you are on the ship, then the star is 4.5 light-years away coming towards you at some speed v.
     
  5. Nov 7, 2007 #4
    Could you be more explicit? I'm not sure which variables you're referring to.

    That's exactly what's NOT happening. As I stated in my earlier post, the distance between the Earth and the star is not 4.5ly in the ship's frame. The star is travelling a shorter distance while moving towards the ship, as per the measurements made in the ship's frame, in a timespan of 4.5 years.
     
    Last edited: Nov 7, 2007
  6. Nov 7, 2007 #5

    Doc Al

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    Hint: From the Earth frame the ship's clock is a moving clock. All you need is the time dilation formula for moving clocks and the definition of velocity.
     
  7. Nov 7, 2007 #6
    Length contraction.

    YOu have x', x and v. You know x (4.5 light years), but not x' or v.
     
  8. Nov 7, 2007 #7
    Which results in a messy equation. The was a question on the GRE physics exam. It's suppose to be somewhat simple.

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

    [tex]v=\frac{dx}{dt} = \frac{x}{t}[/tex]

    Combining them:

    [tex]t'=\frac{t-\frac{x^2}{tc^2}}{\sqrt{1-\frac{x^2}{t^2c^2}}}[/tex]

    Solving that for t would be too time consuming for a GRE question.
     
  9. Nov 7, 2007 #8

    robphy

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    Can you draw a spacetime diagram of the situation from the earth's frame?
    If you think in terms of intervals, you don't explicitly need Lorentz transformations or time-dilation formulas.
    If you can do [hyperbolic] trigonometry, the solution is very simple.
     
  10. Nov 7, 2007 #9

    Doc Al

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    It is simple. You can solve it using time dilation or length contraction.

    Use the formula for time dilation of a moving clock, not the Lorentz Transformation for time.
     
  11. Nov 7, 2007 #10

    Doc Al

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    To use length contraction, express the distance as seen by the ship in terms of the Earth frame distance. That's leaves only one variable. (Hint: Start by writing the simplest expression for the velocity.)
     
  12. Nov 7, 2007 #11
    It still used the Lorentz factor

    [tex]\Delta{t}=\frac{\Delta{t_0}}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    I get a mess that looks like this:

    [tex]v^4-c^2v^2+c^2\sqrt{\frac{\Delta{x}}{\Delta{t}}}[/tex]

    I can solve that quadratically for v.
     
  13. Nov 7, 2007 #12

    robphy

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    Draw a right triangle.... where you are given a hypotenuse and the [opposite] side. Determine the [adjacent] side... using the interval. Compute the ratio of the (opposite side)/(adjacent side).

    [FYI.. but not needed: the adjacent side is related to the time-dilation formula]
     
  14. Nov 7, 2007 #13
    Now the problem is this:

    [tex]a=1[/tex]
    [tex]b=-c^2=-1[/tex]
    [tex]c=c^2\sqrt{\frac{\Delta{x}}{\Delta{t}}}=\sqrt{\frac{4.5}{4.5}}=1[/tex]

    Now when you put that into a quadratic forumla:

    [tex]\sqrt{b^2-4ac}=\sqrt{-4}[/tex]
     
  15. Nov 7, 2007 #14

    Doc Al

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    That's fine.

    Just write the velocity as distance over time, as measured by the Earth. You'll get an easy equation for v.
     
  16. Nov 7, 2007 #15

    Doc Al

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    Try this:
    [tex]v = \frac{\Delta x}{\Delta t}[/tex]

    [itex]\Delta x[/itex] is given; figure out [itex]\Delta t[/itex] from time dilation.
     
  17. Nov 8, 2007 #16
    I did. Everytime I work it I get the square root of a negative number.
     
  18. Nov 8, 2007 #17

    Doc Al

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    What did you put for [itex]\Delta x[/itex] and [itex]\Delta t[/itex]?
     
  19. Nov 8, 2007 #18
    Perhaps somebody can show me the error

    Distance to star in reference frame of Earth: [tex]\Delta{x_0}=4.5[/tex] light years
    Time to reach star in reference frame of ship: [tex]\Delta{t}=4.5[/tex] years
    Distance to star in reference frame of ship: [tex]\Delta{x}=x[/tex]
    Time to reach star in reference frame of Earth: [tex]\Delta{t_0}=t_0[/tex]
    Velocity of ship in reference frame of Earth: [tex]v=\frac{\Delta{x_0}}{\Delta{t_0}}=\frac{x_0}{t_0}=\frac{4.5}{t_0}[/tex]
    Solving for [tex]t_0[/tex]: [tex]t_0=\frac{4.5}{v}[/tex]

    Ship's clock in reference frame of Earth: [tex]\Delta{t}=\frac{\Delta{t_0}}{\sqrt{1-\frac{v^2}{c^2}}}=t=4.5=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    Combining the previous two equations:

    [tex]4.5=\frac{\frac{4.5}{v}}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    The 4.5s cancel, and rearranging:

    [tex]v\sqrt{1-\frac{v^2}{c^2}}=\sqrt{v^2-\frac{v^4}{c^2}}=1[/tex]

    Sqiaring both sides:

    [tex]v^2-\frac{v^4}{c^2}=1[/tex]

    [tex]c=1[/tex] so [tex]c^2=1[/tex] therefore:

    [tex]v^2-v^4=1[/tex]

    There are no real values for [tex]v[/tex] that can satisfy that equation.

    Rewriting it: [tex]v^4-v^2+1=0[/tex]

    Using quadratic formula:

    [tex]a=1[/tex]
    [tex]b=-1[/tex]
    [tex]c=1[/tex]

    [tex]\sqrt{b^2-4ac}=\sqrt{1-4}=\sqrt{-3}[/tex]

    Any questions?
     
  20. Nov 8, 2007 #19

    robphy

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    This says that [tex] v =\gamma [/tex]... but [tex] v<1[/tex] and [tex] \gamma=\frac{1}{\sqrt{1-v^2/c^2}} \geq 1[/tex]!

    I suspect this line:
    Can you justify it?

    (By the way... this method is unnecessarily complicated... but I'll hold off on my solution (by the method I suggested above) until you get it with your method.)
     
    Last edited: Nov 8, 2007
  21. Nov 8, 2007 #20
    http://en.wikipedia.org/wiki/Time_dilation

    [tex]\Delta{t}[/tex] is the time interval between two colocal events (i.e. happening at the same place) for an observer in some inertial frame (e.g. ticks on his clock),
    [tex]\Delta{t_0}[/tex] is the time interval between those same events, as measured by another observer, inertially moving with velocity v with respect to the former observer,
    [tex]v[/tex] is the relative velocity between the observer and the moving clock,

    I see. I got my [tex]t[/tex] and [tex]t_0[/tex] reversed.
     
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