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Reading spacetime diagrams

  • Thread starter w3390
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  • #1
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Homework Statement



Two rockets are sent off at t=0, one from x=0 and the other at x=4. The rocket leaving from x=0 is moving at .8c and the rocket leaving x=4 is moving at .2c. When the paths of the two rockets meet, they send a light signal to x=0. Read off the coordinates in the S frame and in the S' frame and check to see that the space and time differences between events 3 and 4 satisfy the invariant rule. Event 3 is the light signal being sent out and event 4 is the light signal arriving at x=0.

The S' frame is moving at .6c.

Homework Equations



x' = [tex]\gamma[/tex](x-vt)

t' = [tex]\gamma[/tex](t - vx/c^2)

invariant rule: (t4 - t3)^2 - (x4 - x3)^2 = (t'4 - t'3)^2 - (x'4 - x'3)^2

The Attempt at a Solution



So after drawing all world lines, I came up with the coordinates (3.5, 2.75) for event 3 and (0, 6.3) for event 4 in the S frame by looking at the graph. I am confident in these coordinates.

In the S' frame, I came up with (2.8, 1.1) for event 3 and (-6, 9.8) for event 4 in the S' frame. This is where I think there may be a mistake. These are just based off reading the graph, so they are approximate.

Now when I check to see if it satisfies the invariant rule,

(6.3-2.75)^2 - (0-3.5)^2 = (9.8-1.1)^2 - (-6 - 2.8)^2

.3525 = -1.75

Clearly this is not correct. I understand there will be some error since I am just eyeballing the coordinates from the graph, but this seems way off. Does anybody see where I went wrong?
 

Answers and Replies

  • #2
vela
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What units are you using? In what direction do the spaceships move? Are the coordinates you're giving (x,t) or (t,x)?
 
  • #3
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Each of the spaceships are moving towards each other. So the ship that launches from x=0 is moving towards x=4 and vice versa. I am giving the coordinates as (x,t).
 
  • #4
vela
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How can the spaceship travel from x=0 to x=3.5 when t goes from 0 to 2.75? Doesn't that mean the ship is moving faster than the speed of light? (I assume you're using units where c=1.)
 
  • #5
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Yes, you're right. I had drawn my line incorrectly. Instead of drawing it as .8c I drew it as 5/4 c. I've got it now. Thanks.
 

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