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.