# Explaining time dilation

1. Mar 12, 2009

### bigevil

1. The problem statement, all variables and given/known data

Two planets A and B are at rest with respect to each other and are L apart in this frame. They have synchronised clocks. A spaceship flies at speed v with respect to planet A and synchronises its clock with A-B.

We know that when the spaceship reaches B, B's clock reads L/v and the ship's clock reads L/γv. How would someone account for the fact that B's clock reads L/v, which is more than its own L/γv, considering that the spaceship sees B's clock as running slow.

2. Relevant equations

I'm not sure if I got the correct solution, please help me to check if the reasoning is correct.

Let the frame A-B measure (x,y,z,t).
Let the ship measure coordinates (x',y',z',t').

Using the Lorentz transformation for time,

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

When A-B resets its clock at zero (t=0), t' registers $$- vL\gamma / c^2$$. Relative to the time of A-B, the ship's clock runs at vLγ/(c*c) slower.

When the ship passes B, time at AB is t = L/v. Then,

$$\gamma (\frac{L}{v} - \frac{Lv}{c^2}) = \frac{L}{\gamma v}$$.

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My solution:

When I reset my clock at A, this event is not simultaneous with the time reset at the two planets. In fact, the planets have a "head start" of vLγ/c^2. In the planet's time it takes L/v for me to reach there. In my time, to adjust for the head start, the time to reach B is L/γv.

The resolution of the paradox (that my clock is slower on my ship than the planet's, although due to time dilation, I should read the planet's clock as slower) lies in the assumption that the time reset of the A-B system and the reset of the ship are simultaneous, but they are not.

2. Mar 12, 2009

### tiny-tim

Hi bigevil!

this is all very confusing

what's all this about "resetting"?

i can't see whether you've got it or not …

the important question is whether times on A and B which are simultaneous for A (or B) are also simultaneous for the spaceship