# Time dilation derivation

1. Mar 1, 2008

### ehj

I have read the derivation of time dilation by using a lightclock, but I can't seem to find an argument that shows why the result applies for anything else but light clocks. Is there perhaps another derivation that can show the same result but perhaps show that it applies for everything and not just light clocks? I would prefer if it didn't require complex numbers or any hyperbolic trigonometry.

2. Mar 1, 2008

### pam

What do you mean by a "light clock"?
There is a simple standard derivation of time dilation from the Lorentz transformation equation.

3. Mar 1, 2008

### robphy

Consider a lightclock [that you have just analyzed] together with a typical wristwatch... (keeping in mind the principle of relativity)...

4. Mar 1, 2008

### tiny-tim

pendulum clock

Hi ehj!

Imagine a pendulum clock, facing South, so the pendulum swings East-West, one swing per second.

Imagine two mirrors that the pendulum just touches, with a ray of light bouncing to and fro, East-West, between the mirrors, N times per second.

Now move away from the clock at speed v, due South.

Then you regard the distance between the mirrors as the same (for the same reason that a train moving at high speed along a straight track will still regard the width of the track and the width of the train as equal).

But the mirrors are now moving away from you with speed v, and so you say the light ray has to travel further, along a zigzag track.

Consider one triangle of that zigzag.

You say that the light still goes at speed c, along the diagonal (hypotenuse) of that triangle. And the long side of the triangle must be v/c times the diagonal.

So (from Pythagoras!) the short side is $$\sqrt{1\,-\,v^2/c^2}$$ times the diagonal.

So you say that the light travels (along the zigzag diagonals) $$\frac{N}{\sqrt{1\,-\,v^2/c^2}}$$ times the distance between the mirrors to go to and fro N times.

Which means that you say that the light takes $$\frac{1}{\sqrt{1\,-\,v^2/c^2}}$$ longer to do it.

Which means that 1 second on the pendulum clock is $$\frac{1}{\sqrt{1\,-\,v^2/c^2}}$$ seconds for you!

(And we've used constancy of the speed of light, and nothing else.)