# Proper time

1. Jun 2, 2009

### thoughtgaze

is proper time invariant? proof? thanks...

2. Jun 2, 2009

### neopolitan

Proper time is invariant by definition. It's the time read on a clock, for that clock.

It's similar in a way to the probability of throwing a six on a die. Before you start, the chances of getting a six are 1/6. Once you have thrown the die and got a six, the chances of having a six are 1/1.

Similarly, you can manipulate your scenario to get whatever proper time you want, and to that extent it is variable. But once you have your proper time (which has been read from a clock, for that clock), it's invariant.

cheers,

neopolitan

3. Jun 2, 2009

### sylas

Yes, it is invariant; by almost by definition. Proper time is the time measured by a clock in a given world line. What you can prove are that the co-ordinate transforms conserve proper time.

Here is a proof for special relativity, that proper time is invariant under the Lorentz transformation.

The increment in proper time du for a clock moving a small distance dx in a small time dt is, by definition
$$du^2 = dt^2 - (dx/c)^2$$​

That is for any inertial co-ordinate system.

Suppose we transform to a new co-ordinate system. The Lorenz transformations are
\begin{align*} t' & = \gamma ( t - vx/c^2 ) \\ x' & = \gamma ( x - vt ) \\ \intertext{Where} \gamma & = \frac{1}{\sqrt{1-(v/c)^2)}} \\ \intertext{Hence} dt' & = \gamma ( dt - (v/c^2).dx ) \\ dx' & = \gamma ( dx - v.dt ) \\ \intertext{Hence} dt'^2 - (dx'/c)^2 & = \gamma^2 ( ( dt - (v/c^2).dx )^2 - ( dx/c - (v/c).dt )^2 ) \\ & = \gamma^2 ( dt^2 - 2(v/c)(dx/c).dt + (v/c)^2.(dx/c)^2 - (dx/c)^2 + 2(v/c).(dx/c).dt - (v/c)^2.dt^2) \\ & = \gamma^2 ( 1 - (v/c)^2 ) ( dt^2 - (dx/c)^2 ) \\ & = dt^2 - (dx/c)^2 \end{align}​

4. Jun 2, 2009

### thoughtgaze

bravo, this is all i needed to see thank you very much :)