Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Time Dilation Question

  1. Feb 16, 2004 #1
    I am reading a book called Simply Einstein...it wonderfully helps a novice understand Einstien's theory or relativity. But I am stuck and need some help here!

    He gives an example in which you have a spaceship uniformly moving at 0.8c and as soon as it passes the earth a clock on the ship reads 0 as does a clock on the earth and a star 20 light years away (to which the ship is travelling.)

    Now, in a discussion about simultaneous time...he says to a person on earth and the star, time will read 0 but from the space ship...the supposedly synchronized time on earth-star time will be different!

    Quoting a part of the book the author says:
    Ship is at rest in its respective frame and objects of interest (earth and star)are moving towards it from the right. Events that are simultaneous in one reference frame (like the Earth-star frame) are not simultaneous in another frame (such as on the ship) and, furthermore, the event that is on the right occurs first. That means the event of the star clock reading 0 occurs BEFORE the Earth clock reads 0 as observed from the ships frame of reference. In other words, the star clock reads a later time. Here, as oberved in the ship frame, Earth (heading to the left) passes the ship at the instant the ship and Earth clocks both read 0. But the star clock is ahead, so it reads a later time. I won't go through the math, but the later time is, in fact, 16 years!

    Somebody please explain!!!
  2. jcsd
  3. Feb 16, 2004 #2

    I'm still trying to get this straight myself, but I think it's actually 26.7 years...

    [tex]t' = \gamma \left(t - \frac{vx}{c^2}\right)[/tex]

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

    [tex] t' = \left(\frac{1}{\sqrt{1 - .8^2}}\right)\left(0 - (-.8)(20)\right) [/tex]

    [tex] t' = \frac{16}{.6} = 26.7[/tex]

    I won't attempt to explain the "why" of it -- this concept of simultaneity is still baffling to me too. So I will join you in eagerly awaiting someone else's explanation.

    Hopefully, at least my math is correct ...
  4. Feb 16, 2004 #3
    a shot at the why

    I don't understand the math but here's my try at the why. Matter and energy are interchangable. Whatever the form, both need representation. To make this clearer, suppose you have a grid in front of you. Horizontal movement is through space and vertical movement through time. Now starting in the bottom left corner, we have to plot a way through this grid. Every electro-magnetic oscillation requires you move your marker either up through time, or right through space. If your marker moves right through space it would be as energy, if it move up through time it would be as mass. If your marker moves up, right, up, right, up, and then right in a diagonal fashion it would be moving at 50% the speed of light. As your marker moves from point a to point b it spends half its oscillations moving through space as opposed to time. It would lose time to the clocks at points A and B, both of which would be oscillating only up through time.

    I made this all up, don't know if any of it is true. If not, hopefully someone will correct me.
  5. Feb 17, 2004 #4

    the formula is correct, but not the substitutions from what i can see with my limited knowledge (so please correct if i am in error).

    time runs slower on the ship therefore it will register a lower value. in other words, the earth-star system time is the t' (the inevitably larger value) and and the ship time should be the t.

    therefore, the equation substitutions become

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

    [tex] 0 = \left(\frac{1}{\sqrt{1 - .8^2}}\right)\left(t - (.8)(20)\right) [/tex]

    [tex] 0 = \left(\frac{1}{\sqrt{1 - .8^2}}\right)\left(t - 16\right) [/tex]

    [tex] 0 = \left(t - 16\right) [/tex]

    [tex] t = 16 [/tex]

    i believe this is how the 16 shows up but what does it all mean?

    well let's see what is happening from the perspectives of the different frames.

    in the earth-star frame, in order to be in sync the stellar beings must be a bit calculative. when the flash of light indicating that the earth has started its clock finally reaches them (20 yrs after it left earth), they must set their clock to 20 and not 0 because that is what the clock on earth would read.

    so how come the spaceship people figure that it is 16 and not 20? time is running slower on the ship frame according to the lorentz transformation formula - each minute on the ship is equivalent to more than a minute in the earth-star frame. so though the light took 20 yrs in the earth-star frame, it would be measured by the ship's clock as only 16 yrs.

    another way tn look at this would be to remember that the people on the ship see themselves as being stationary. it is the earth and the star that are moving towards them. therefore, they see a flash of light at time 0 zooming off towards this star that is admittedly 20 lightyears away but is bearing down upon them at an enormous speed of .8c ! so they figure that if the star weren't moving with respect to them the light would take 20 years to get to the star, but since it is moving in the opposite direction to which the light is travelling, the light will get to the star a good deal sooner and the star clock will have to be adjusted for the calculated 16 years (instead of the 20 if all three bodies were mutually stattonary).

    sure gives a different meaning to 'how time flies'
    Last edited: Feb 17, 2004
  6. Feb 17, 2004 #5
    Well, I agree that time should be going slower on the spaceship (as observed by the people on earth), so my result seems very puzzling. But I don't see why you can set t' = 0 and let t be the unknown in that equation. Isn't t the time separating the two events in the "earth" frame (i.e. 0 since we're defining t=0 on earth and on the distant star), and x the distance between the two events as measured in the "earth" frame (20 ly). If so, isn't t' the unknown we are solving for?
  7. Feb 17, 2004 #6
    Funny, I am reading the same book and have the same problem understanding that oddity, maybe one of the big guns in the forums will come and help us out.
  8. Feb 17, 2004 #7

    good point. i set the x to 20 because at t=0 the ship is aligned with the earth so its distance is the same from the star. admittedly, though, i am somewhat guilty of trying to get the answer the book gave

    what i don't understand is how i cleverly managed to eliminate the lorentz factor which i would have thought would play a prominent part here - unless it really doesn't and my 2nd explanation is what makes sense.

    anyway, i'm off to study this and hope to have a better understanding when i post again - hope to see you then :smile:
  9. Feb 17, 2004 #8
    I should have written:

    [tex]t' = \left(\frac{1}{\sqrt{1-.8^2}}\right)\left(0 - .8(20)\right)[/tex]
    [tex]t' = \frac{-16}{.6} = -26.7[/tex]

    To this I could add:

    [tex]x' = \gamma \left(x - vt\right)[/tex]
    [tex]\gamma = \frac{1}{\sqrt{1-.8^2}}[/tex]
    [tex]x' = \left(\frac{1}{.6}\right)(20 - .8 \times 0)[/tex]
    [tex]x' = 33.3 ly[/tex]

    This is not inconsistent with time dilation or length contraction. These transformations just give the coordinates of the distant star in the spaceship's frame of reference. Thus the coordinates (x,y,z,t) in earth's frame of earth are (0,0,0,0) and the coordinates of the star are (20,0,0,0). In the spaceship's frame of reference the coordinates (x',y',z',t') are (0,0,0,0) while its coordinates for the star are (33.3,0,0,-26.7).

    This is giving a set of coordinates that represent a different place AND time. Not a measurement of the distance (which must be done at a SINGLE time) nor a measurement of the time to travel from one place to the other.

    Very confusing. This should give us all something to chew on for a while.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook