Special Relativity, simultaneity and spatial separation

[zhillyz]

Homework Statement

To an observer in frame S, two events are simultaneous and occur 600km apart. What is the time measured by an observer in frame S' who measures their spatial separation to be 1200km? Explain the sign of your answer.

Homework Equations

Lorentz Transformation

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

or

x' = \gamma(x - vt)

The Attempt at a Solution

The question doesn't give a value for v, and as such is making it a little difficult to wrap my head round. My attempt was to insert in the first equation t=0, x=600 and rearrange a little which gave me;

t' = \dfrac{-600v^2}{c^2}

Not sure if this is what the question wants me to stop at or not.

 

[mfb]

The given spatial separation in the second frame allows to calculate v.
The units of your final formula look wrong.

 

[Andrew Mason]

Homework Statement

To an observer in frame S, two events are simultaneous and occur 600km apart. What is the time measured by an observer in frame S' who measures their spatial separation to be 1200km? Explain the sign of your answer.

Homework Equations

Lorentz Transformation

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

or

x' = \gamma(x - vt)

The Attempt at a Solution

The question doesn't give a value for v, and as such is making it a little difficult to wrap my head round. My attempt was to insert in the first equation t=0, x=600 and rearrange a little which gave me;

t' = \dfrac{-600v^2}{c^2}

Not sure if this is what the question wants me to stop at or not.

If you just apply the Lorentz transformation for length: x'1-x'2 you would get:

L' = x'_1 - x'_2 = \gamma(x_1-vt_1) - \gamma(x_2-vt_2) = \gamma(x_1-x_2 + v(t_2-t_1))

Knowing the relationship between t1 and t2 (ie. the times of the two events in the S frame) this simplifies nicely. You know x1-x2 and x'1-x'2 so you can solve for γ and from γ you can find v. Then you can then apply the Lorentz transformation for time to find t1' - t2'.

AM

 

[Chestermiller]

Have you learned yet that if you square the equation for x' and subtract the square of the equation for ct', you get:
(x')^2-(ct')^2=(x)^2-(ct)^2
This equation is independent of the relative velocity v, and is a fundamental characteristic of the geometry of flat space-time. This should help you solve your problem.

 

[zhillyz]

Thank you everyone for your responses. They helped greatly.
So, with the separation in frame S'(x'₂-x'₁) equalling 1200km and applying the Lorentz transformation to each event's position in that frame one would get;

x'_2 - x'_1 = \gamma (x_2 - x_1 - v(t_2 - t_1))

1200 = \gamma (600 - v(0))

\gamma = 200

Now solve for v. Afterwards

\Delta t' = \gamma (\Delta t + \dfrac{v \Delta x'}{c^2})
\Delta t' = \gamma \dfrac{v \Delta x'}{c^2}

Now solve for Δt' using calculated v?

 

[Andrew Mason]

Thank you everyone for your responses. They helped greatly.
So, with the separation in frame S'(x'₂-x'₁) equalling 1200km and applying the Lorentz transformation to each event's position in that frame one would get;

x'_2 - x'_1 = \gamma (x_2 - x_1 - v(t_2 - t_1))

1200 = \gamma (600 - v(0))

\gamma = 200

Now solve for v. Afterwards

\Delta t' = \gamma (\Delta t + \dfrac{v \Delta x'}{c^2})
\Delta t' = \gamma \dfrac{v \Delta x'}{c^2}

Now solve for Δt' using calculated v?

γ=200?? Check that again! The rest is fine except you are using x' in the Lorentz transformation for t'! It should be x.

AM

 

[zhillyz]

2 sorry haha. Thank you.

 

[Chestermiller]

(x')^2-(ct')^2=(x)^2-(ct)^2
1200^2-(ct')^2=600^2
(ct')^2=1200^2-600^2
ct'=600\sqrt{3}

 

[Andrew Mason]

(x')^2-(ct')^2=(x)^2-(ct)^2
1200^2-(ct')^2=600^2
(ct')^2=1200^2-600^2
ct'=600\sqrt{3}

Ok. But it is not so easy to determine v from that. Using the Lorentz transformations you can determine both v and t'.

AM

 

[Chestermiller]

Ok. But it is not so easy to determine v from that. Using the Lorentz transformations you can determine both v and t'.

AM

The original question didn't ask for v.