- #1

Calu

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## Homework Statement

The example below illustrates the relativistic phenomenon that synchronicity of events is not absolute but it depends on the reference frames.

Spaceships A and B, while moving away from each other with a constant speed of v = 0.553c, are watching a competition between spaceships C and D. Spaceship C is heading towards planet C and spaceship D is approaching planet D. The winner is the spaceship that reaches its target planet first.

The astronauts on spaceship B find, to their great surprise, that the spaceships C and D reached their planets at the same time. At that moment, planet C was at

__r__

_{C}= (-250, 130, -130) ls, and planet D was at

__r__

_{D}= (160, -290, -170) ls , where the xyz coordinate system is attached to spaceship B and the first, x, axis is parallel to the velocity vector of spaceship B relative to spaceship A.

According to spaceship A, however, the race has a definite winner. According to spaceship A, how many seconds were between reaching planet C by spaceship C and reaching planet D by spaceship D?

## Homework Equations

Δτ = γ(Δτ' + βΔx')

Where ΔT is the difference in time, given in light seconds, β = v/c. The ' notation represents the reference frame.

γ = 1/(√(1-(v

^{2}/c

^{2}))

## The Attempt at a Solution

Let the view from spaceship A be reference frame s.

Let the view from spacesip B bence frame s'.

x

_{C}' = |

__r__

_{C}| = √((250

^{2}) + (130

^{2}) + (130

^{2}))

x

_{D}' = |

__r__

_{D}| = √((160

^{2}) + (290

^{2}) + (170

^{2}))

Δτ' = 0

Δτ = γ(Δτ' + βΔx')

Δτ = (1/1-(0.533))(βx

_{D}' - βx

_{C}')

Δτ = β(1/1-(0.533))(x

_{D}' - x

_{C}')

Δτ = 76679.6992

Which I'm told is incorrect. Could anyone help me out?