Lorentz Transformation

  • #1

Homework Statement



Two events occur at the same place in an inertial reference fram S, but are separated in time by 3 seconds. In a different frame S', they are separated in time by 4 seconds.

(a) What is the distance between the two events as measured in S'?
(b) What is the speed of S relative to S'?

Homework Equations



I'm presuming:

t' = gamma*(t-ux/c^2)

The Attempt at a Solution



I have the answer, and a hint saying to use the interval S^2, but I have no idea what that means, and where I start. When I look and the relevant Lorentz equations, they involve velocity, and I do not have a velocity here.

Could you please point me in the right direction?

Thank you in advance.
 

Answers and Replies

  • #2
G01
Homework Helper
Gold Member
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The hint is implying that you use the invariant space-time interval to solve this problem. The following quantity is called the space time interval:

[tex](\Delta s)^2= (\Delta x)^2 + (\Delta y)^2 + (\Delta z^2) - (c\Delta t)^2[/tex]

This quantity is a Lorentz scalar and is thus invariant over Lorentz transformations (it is the same in all inertial frames). So, in one dimension this means:

[tex](\Delta x)^2- (c\Delta t)^2=(\Delta x')^2- (c\Delta t')^2[/tex]

Can you use this the solve the problem?
 
  • #3
The hint is implying that you use the invariant space-time interval to solve this problem. The following quantity is called the space time interval:

[tex](\Delta s)^2= (\Delta x)^2 + (\Delta y)^2 + (\Delta z^2) - (c\Delta t)^2[/tex]

This quantity is a Lorentz scalar and is thus invariant over Lorentz transformations (it is the same in all inertial frames). So, in one dimension this means:

[tex](\Delta x)^2- (c\Delta t)^2=(\Delta x')^2- (c\Delta t')^2[/tex]

Can you use this the solve the problem?

Do I find [tex](\Delta t)[/tex] by doing SQRT[(4^2)-(3^2)] = ROOT 7

Then at they are both at the same coordinates in the inertial reference frame, we can ignore x , y and z. Therefor S equals the root of (c^2)*(ROOT 7) = 7.9x10^8m

Is this correct?
 
  • #4
G01
Homework Helper
Gold Member
2,682
16
Do I find [tex](\Delta t)[/tex] by doing SQRT[(4^2)-(3^2)] = ROOT 7

Then at they are both at the same coordinates in the inertial reference frame, we can ignore x , y and z. Therefor S equals the root of (c^2)*(ROOT 7) = 7.9x10^8m

Is this correct?

You should only have one factor of c in your final line, since you take the square root of c^2 when solving for the answer.
 

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