# Length Contraction & Time Dilation

messier992

## Homework Statement

A spaceship of proper length L is moving with respect to the ground with speed v. As measured on the ground, how much time does a light signal need to get from the front to the end of the spaceship.

L'=L/γ
t'=tγ

## The Attempt at a Solution

The right answer is L/[γ(c+v)]
I don't understand how to get the answer from either possibilites. Where, conceptually, I'm misunderstanding the equations.

Distance of Spaceship Viewed From the Ground
L'=L/γ = L/γ

t=L'/(velocity)= [L/γ]/(v+c)]

v+c? But isn't light supposed to travel at the same speed in all reference frames?

Additionally, how would I approach the question using the equation t'=tγ?

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## Answers and Replies

Homework Helper
Gold Member
Distance of Spaceship Viewed From the Ground
L'=L/γ = L/γ

t=L'/(velocity)= [L/γ]/(v+c)]

v+c? But isn't light supposed to travel at the same speed in all reference frames?
Yes. Observers on the ground would measure the light signal as traveling with speed c relative to the ground. Observers in the ship would measure the same signal as traveling with speed c relative to the ship.

v+c in this context represents the speed at which the light signal is traveling relative to the spaceship as measured by observers on the ground. For observers on the ground, the light signal is traveling in one direction with speed c while the back of the ship is moving in the opposite direction with speed v. So, observers on the ground would say that the distance between the light signal and the back of the ship is decreasing at a rate of v+c.

Additionally, how would I approach the question using the equation t'=tγ?
You have to be careful with this equation. The unprimed time t in this equation represents the time between two events as measured in a reference frame for which the two events occur at the same place. In this problem, one event is the light signal leaving the front of the ship and the other event is the signal arriving at the back of the ship. These two events do not occur at the same place for either the ground frame or the ship frame.

PeroK
messier992
Many thanks!

Homework Helper
v+c in this context represents the speed at which the light signal is traveling relative to the spaceship as measured by observers on the ground.
I would emphasize this a bit differently. The "v+c" is not a relative velocity of anyone thing in the rest frame of any other thing. Instead it is a closing velocity -- the rate at which the distance between two moving objects is decreasing [as assessed from a single frame of reference where neither object is at rest].

In this case that would be the rate at which the distance between the rearward moving light and the forward-moving space-ship tail is decreasing as assessed from the Earth frame.

A relative velocity is the rate of change of position per unit time.
A closing velocity is the rate of change of distance per unit time. Similar notions, but not always identical.

In an inertial frame, relative velocities (of objects) cap out at c. Closing velocities (for pairs of objects)
cap out at 2c.

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TSny
Apashanka

## Homework Statement

A spaceship of proper length L is moving with respect to the ground with speed v. As measured on the ground, how much time does a light signal need to get from the front to the end of the spaceship.

L'=L/γ
t'=tγ

## The Attempt at a Solution

The right answer is L/[γ(c+v)]
I don't understand how to get the answer from either possibilites. Where, conceptually, I'm misunderstanding the equations.

Distance of Spaceship Viewed From the Ground
L'=L/γ = L/γ

t=L'/(velocity)= [L/γ]/(v+c)]

v+c? But isn't light supposed to travel at the same speed in all reference frames?

Additionally, how would I approach the question using the equation t'=tγ?
The length of the spaceship seen by the ground observer is L/ϒ .
Time taken by light L/ϒc
Or proper time of the light crossing the spaceship is L/c.
Time taken as seen from ground L/γc as dt=γdΓ ,dΓ is the proper time.