Arbitrariness in Special Relativity

1. Apr 8, 2010

planck42

In an online lecture on Special Relativity, the instructor asserts that in the space-time coordinate system, $$(d{\tau})^{2}=(dt)^{2}-(dx)^{2}$$ with $$\tau$$ representing the proper time in a frame moving with velocity v, t representing a period of time measured from an inertial reference frame, and c being clearly treated as 1. It is known from the Lorentz transformations that x and t are related through hyperbolic angles, but surely this cannot be sufficient to assume that the proper time is as it is given. If there is no clear mathematical basis for $$(d{\tau})^{2}=(dt)^{2}-(dx)^{2}$$, then how can it be possible to derive such an equation willy-nilly?

2. Apr 8, 2010

Staff: Mentor

That is the definition of proper time. You don't derive definitions.

3. Apr 8, 2010

JesseM

I think you can in a sense "derive" it from the fact that coordinate time in any frame is defined in terms of proper time on a network of clocks at rest in that frame at every possible position (synchronized according to the Einstein synchronization convention). If you know that the proper time interval between two events on the worldline of an inertial clock must match up with the coordinate time between those events in the frame where the clock is at rest (i.e. the frame where the two events have the same position coordinate), then the definition of proper time between these events in other frames follows from the Lorentz transformation.

4. Apr 8, 2010

Staff: Mentor

Yes, you can see that simply by setting dx=0 in the above.

5. Apr 8, 2010

yuiop

Willy-nilly derivation

The Lorentz transformation of the time coordinate:

$$t = \tau \cosh(\phi) - x \sinh(\phi)$$

(where $\phi = \tanh^{-1}(v)$)

can be recast as:

$$t = \frac{\tau - vx}{\sqrt{1-v^2}}$$

From this relation it is very easy to obtain the time dilation factor:

$$\frac{dt}{d\tau} = \frac{1}{\sqrt{1-v^2}}$$

It follows that:

$$d\tau^2 = dt^2(1-v^2)$$

and since v = dx/dt,

$$d\tau^2 = dt^2 -dx^2$$

Last edited: Apr 8, 2010
6. Apr 8, 2010

planck42

Clearly I didn't try hard enough to understand the concept.

7. Apr 8, 2010

starthaus

$$(d{\tau})^{2}-(d\zeta)^{2}=(dt)^{2}-(dx)^{2}$$
In the proper frame , there is no motion, so $$d \zeta=0$$
$$(d{\tau})^{2}-0=(dt)^{2}-(dx)^{2}$$