Relativity, difference in x given t

AI Thread Summary
The discussion revolves around solving a relativity problem involving two events occurring at the same point in frame S, with time intervals of 1.8 seconds and 2.35 seconds in frames S and S', respectively. Participants explore the use of the Lorentz transformations to find the position difference between the events, noting that the invariant spacetime interval can be a helpful approach. A key point raised is the confusion regarding the interpretation of time intervals from different frames, emphasizing that these cannot be directly subtracted for meaningful results. Ultimately, the discussion highlights the importance of understanding the relationship between time, position, and relative motion in the context of special relativity. The conversation concludes with a clarification that time measurements are frame-dependent, which resolves some conceptual misunderstandings.
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Homework Statement



two events are observed in frame of reference S to occur at the same point. the second event is 1.8s after the first. in a second frame S' moving relative to S, the second event is 2.35 s after the first. what is the difference between the positions of the two events as measured in S?

Homework Equations



t'=(t-ux/c2) / sqrt(1 - u2/x2)



The Attempt at a Solution



t = 1.8, t' = 2.35
need to solve for x but the variable u is also unknown?
 
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also set it up so S=0=S' and t=0=t' meaning S and S' begin in the same place at the same time, where and when event 1 occurs.
 
You wrote the equation for t'. Write the corresponding one for x'. That will allow you to solve for gamma and u.

Even easier would be to compare the invariant spacetime interval between the events.
 
what do you mean by invariant spacetime interval? is that the difference between 1.8 and 2.35?

and using the x' formula, solving for u then subbing into t' then gives you the unknown x' so how does that work? cause there's no way to find a numerical answer for u so i need to cancel it out or sub right?
 
ProPatto16 said:
what do you mean by invariant spacetime interval? is that the difference between 1.8 and 2.35?
No, the spacetime interval is given by:

(\Delta s)^2 = (\Delta x)^2 - c^2 (\Delta t)^2 = (\Delta x')^2 - c^2 (\Delta t')^2

See: http://en.wikipedia.org/wiki/Spacetime#Spacetime_intervals"

and using the x' formula, solving for u then subbing into t' then gives you the unknown x' so how does that work? cause there's no way to find a numerical answer for u so i need to cancel it out or sub right?
Using the equation you have for t', you can solve for gamma (and the speed if you like). You'll need the corresponding equation for x' in terms of t and x to find x'.
 
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i must still be missing something. using the formulas:


t'=(t-ux/c2) / sqrt(1 - u2/x2)

and

x'=(x-ut) / sqrt(1 - u2/x2)

the only known quantities are t and t'
that leaves the unknowns to be x, x' and u.

conditions are

at t=0 x=x' and u=0 since S=S'

3 unknowns and 2 formulas. can't solve for them.

im so lost.
forget the spacetime interval way, ill do it the formula way cause its what I've been doing, but i just always seem to end up with 2 unkowns left no matter how i reaarange?
 
ProPatto16 said:
the only known quantities are t and t'
that leaves the unknowns to be x, x' and u.
There's what you're missing. You are given x:
ProPatto16 said:
two events are observed in frame of reference S to occur at the same point.

So what's x?
 
zero. so blind.

so 2.35 = 1.8 / sqrt(1-u2/c2)

so u=0.64289c

sub into eqn for x'

x' = 0 - 0.64289c*1.8 / sqrt(1-0.642892c/c)

= -452935555.9

yes? that would be in metres?
 
that can't be right. that's more distance than light travels in 1 second.
 
  • #10
ProPatto16 said:
that can't be right. that's more distance than light travels in 1 second.
What's that got to do with anything?

(Your answer looks fine. Try using the invariant interval method as a check.)
 
  • #11
the difference between 1.8s and 2.35s is less then 1 second? how can reference frame S' move that far in less than a second?
 
  • #12
ProPatto16 said:
the difference between 1.8s and 2.35s is less then 1 second?
:eek: Those times are from different frames. You can't subtract them and get something meaningful.
how can reference frame S' move that far in less than a second?
It doesn't. According to frame S', frame S moves that distance in 2.35s.
 
  • #13
ohh okay. thanks so much :)
conceptual challenges are not my strongest point.
 
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