# Calculating time dilation for photon traveling towards a moving spaceship

Suppose a spaceship is moving away from the Earth at $0.5c$. When the spaceship is one light-year away from Earth, an observer on Earth sends a photon toward the spaceship. According to the observer on Earth, the photon is traveling at $c$ and the spaceship is traveling in the same direction at $0.5c$, so it takes 2 years for the photon to reach the spaceship. From the frame of reference of someone inside the spaceship, the spaceship is not moving and the photon is traveling towards it at $c$. So it takes 1 year for the photon to reach the spaceship. So why isn't 1 year inside the spaceship equal to 2 years on Earth, i.e. why is this method of deriving the Lorentz factor invalid?

• I&#39;m not an expert, but I believe you also have to account for length contraction in the figures for the ship: in your example, Major Tom on the spaceship not only records the time between leaving Earth and receiving the signal to be less than two years, but also measures his traveled distance as less than 2 ly, which will effect the transformation. – Asher Jan 4 '15 at 18:26

So it takes 1 year for the photon to reach the spaceship

This isn't correct.

Assume, for simplicity, that both the Earth's clock and spacecraft's clock read $t=0=t'$ when the spacecraft passes Earth.

Then, when the photon is sent, Earth's clock reads $t=2\, \mathrm {y}$. Both observers agree on this. Also, both observers agree that the spacecraft's clock reads $t' = 3.464\, \mathrm {y}$ when the photon is received.

However, due to the relativity of simultaneity, the observers don't agree on the reading of the spacecraft's clock when the photon is sent.

• According to the observer on Earth, the spaceship's clock reads $t' = 1.732\, \mathrm {y}$ when the photon is sent.
• But, according to the observer on the spacecraft, the spacecraft's clock reads $t' = 2.309\, \mathrm {y}$ when the photon is sent.

And, the observers don't agree on the reading of Earth's clock when the photon is received.

• According to the observer on Earth, the Earth's clock reads $t = 4\, \mathrm {y}$ when the photon is received.
• But, according to the observer on the spacecraft, the Earth's clock reads $t = 3\, \mathrm {y}$ when the photon is received.

In summary, according to the observer on Earth, 2 Earth years elapses between the emission and reception events while 1.732 spacecraft years elapses.

According to the observer on the spacecraft, 1.155 spacecraft years elapses between the emission and reception events while 1 Earth years elapses.

From either perspective, the moving clock's elapsed time is smaller and by the same proportion:

$$\gamma = \frac{2}{1.732} = \frac{1.155}{1} = \frac{1}{\sqrt{1 - (0.5)^2}}$$

If you will draw the spacetime diagram for this, the above results will be clear.

The spaceship crosses your path at location $0$, at time $0$.

How things appear to you:

At time $2$, when the spaceship has reached location $1$, you release the photon. At time $4$, the photon and the ship both reach location $2$ and collide.

How things appear to the other guy:

At time $2.3$ (when your slow clock was reading $2$), you (who, along with the rest of the earth, had been traveling backward at speed $1/2$) had reached location $-1.15$, where you released your photon. Since he's never moved from location $0$, the photon reaches him $1.15$ minutes later, at time $3.45$ by his ship's clock.

More of how things appear to you:

Huh. When the photon hit the spaceship --- that is, at time 4 --- the ship's clock showed 3.45. That clock must be running slow.