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I have read on web that:

Proper time $\tau$ is a time interval measured by a clock which at rest relative to the observer.

But lets consider this problem:

The plane is flying with a speed $u=300m/s$. How much time $\Delta t'$ would have to pass for a passanger on that plane for time difference $1s$ to occur between the time $\Delta t'$ on the plane and the time $\Delta t$ on the Earth. We postulate that speed of light equals to the speed of sound $c=343m/s$.

Ok so lets say i set myself in coordinate system $x'y'$ and check the whole situation from a perspective of a passenger on the plane.

I first calculate the Lorenz factor:

$$\gamma=\frac{1}{\sqrt{1-\frac{300^2}{343^2}}} = 2.06$$

If everything is checked from the perspective of the observer on the plane, he is not moving according to his clock so his time is the proper time and i set $\boxed{\Delta t'\equiv\tau}$. Because the proper time is the shortest it follows that:

\begin{align} \Delta t = \gamma \Delta t' = \Delta t' + 1s \xrightarrow{\text{i use the RHS}}\gamma\Delta t' &= \Delta t' + 1s\\ \Delta t' \left( \gamma - 1 \right) &=1s\\ \Delta t' = \frac{1s}{\gamma - 1}\\ \Delta t' = 0.94s \end{align}

So now i reverse the situation and set myself in the coordinate system $xy$ on Earth.

Here the Lorentz factor is the same as before even if i had to use negative speed $u=-300m/s$ (squaring does its trick hehe). So it again holds $\gamma=2,06$.

If now everything is checked from the perspective of an observer on Earth, he is not moving accordint to his clock which means now his time is the proper time and i can set $\boxed{\Delta t=\tau}$. Because the proper time is the shortest it follows that:

\begin{align} \Delta t' = \gamma \Delta t = \Delta t + 1s \xrightarrow{\text{i use the RHS}}\gamma\Delta t &= \Delta t + 1s\\ \Delta t \left( \gamma - 1 \right) &=1s\\ \Delta t = \frac{1s}{\gamma - 1}\\ \Delta t = 0.94s \end{align}

Well this means that the proper time definition needs an add which is:

"In the situation where we have more clocks the proper time is measured by the clock which is situated in the coordinate system from which we observe."

and

"we are free to choose to observe from whichever coordinate system we want."

Qmechanic
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1 Answers1

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The proper time is a fundamental invarient. By this I mean that all observers, regardless of their co-ordinate system, will measure the same value for the proper time. This applies to General Relativity as well as Special Relativity.

In SR the proper time is defined by:

$$ c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 $$

This immediately explains why the proper time is the time measured in the rest frame, because in the rest frame $dx$, $dy$ and $dz$ are all zero so the equation reduces to:

$$ c^2d\tau^2 = c^2dt^2 $$

or:

$$ d\tau = dt $$

The point is that I wouldn't start from the definition that the proper time is the time measured in the rest frame because this is a consequence of a much deeper principle. The equation I gave for $d\tau^2$ may seem a bit arbitrary, but it's a special case of the much deeper statement that:

$$ d\tau^2 = \eta_{\alpha\beta} dx^\alpha dx^\beta $$

where the object $\eta$ is the metric, which describes the structure of space time (strictly speaking we use the above equation to calculate the line element not the proper time, but this is a complication we can gloss over for now).

In the example you give, you can calculate the elapsed time on the plane without using the Lorentz factor. Indeed, you can derive the Lorentz transformations from the expression for the proper time.

John Rennie
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