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On page 86 of Theoretical Minimum (part 3 on special relativity) by Susskind, he writes:

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

Where $v$ is the velocity of a moving reference frame relative to the restframe (which is the frame of coordinates $t,x$). He derives this from the equation (I don't know where he is getting it from):

$$d\tau = \sqrt{dt^2-dx^2}$$

But the relation for the proper time should be

$$\tau=\sqrt{t^2-x^2}$$ And if I differentiate this w.r.t. $t$ I get:

$$\frac{d\tau}{dt}=\frac{t-xv}{\sqrt{t^2-x^2}}\neq\sqrt{1-v^2}$$

Why the discrepancy? What am I misunderstanding?

Edit: I now see that the inequality I wrote doesn't hold, because $v=x/t$, which is due to the fact that the path of the reference frame is linear.

IThough, I still don't get how to go from $$\tau=\sqrt{t^2-x^2}$$

To

$$d\tau = \sqrt{dt^2-dx^2}$$

Also, surely the derived relation no longer holds when we are describing a particle that takes a nonlinear trajectory?

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  • $\begingroup$ What discrepancy? $\endgroup$ – WillO Mar 19 '18 at 21:36
  • $\begingroup$ @WillO, i've edited the question. $\endgroup$ – user56834 Mar 19 '18 at 21:58
  • $\begingroup$ But even for a nonlinear trajectory a differential (small change) is essentially linear, right? $\endgroup$ – docscience Mar 19 '18 at 22:08
  • $\begingroup$ BTW, your reference, a three volume set. Which volume in the series? $\endgroup$ – docscience Mar 19 '18 at 22:09
  • $\begingroup$ @docscience part 3 $\endgroup$ – user56834 Mar 19 '18 at 22:30
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You don't "go from $\tau = \sqrt{ t^2 - x^2}$ to $d\tau = \sqrt{dt^2 - dx^2}$" because your starting point itself is incorrect.

Think about what $\tau$ is. It is the proper time interval between two events, $E_1$ and $E_2$. If the two events have coordinates $(t_1,x_1)$ and $(t_2,x_2)$, then the proper time interval between these events is defined by $$ \Delta \tau = \sqrt{ (t_1 - t_2)^2 - (x_1 - x_2)^2 } $$ This is the full and correct formula and this should be our starting point.

Now, if one of the events is at the origin, that is $(t_2,x_2) = (0,0)$ and the other has $(t_1,x_1) = (t,x)$, then the proper time interval is $\Delta\tau= \sqrt{ t^2 - x^2}$. For instance, if I am interested in computing the proper time interval between the origin of my frame and the origin of the moving frame, then I would use the formula above.

On the other hand, if one of the events happens infinitesimally after the other, i.e. if $(t_2,x_2) = (t_1 + dt , x_2 + dx)$, then $d \tau = \sqrt{ dt^2 - dx^2}$ (where now I have written $d\tau$ instead of $\Delta \tau$ since it is an infinitesimal).

The second case applies to the problem you are studying. We take event $E_1$ to be the coordinates of the origin of the moving frame at some instant. Say, this event has coordinates $(t,x)$. We take $E_2$ to be the coordinates of the origin of the moving frame at the next instance. It then has coordinates $(t + dt , x + dt)$. The proper time difference between the location of the origin in the two instances is $d\tau = \sqrt{ dt^2 - dx^2}$. The rate of change of proper time is then $$ \frac{d\tau}{dt} = \sqrt{ 1 - \left( \frac{dx}{dt} \right)^2 } . $$ Now, what is $ \frac{dx}{dt}$? It is precisely the rate of change of the spatial coordinate of the origin of the moving frame (remember our definitions of $x$ and $t$). But this is, by definition, the speed (velocity) of the moving frame. Thus, $$ \frac{d\tau}{dt} = \sqrt{ 1 - v^2 } . $$

You may think I have been a bit too scrupulous here, but making sure that these simple examples are clear reduces confusion when more complicated cases are dealt with.

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I'm not sure $\tau = \sqrt{t^2 + x^2}$ is the relation you're looking for, since you would be looking for $\Delta\tau$ rather than $\tau$. In any case, where his relation comes from is the Minkowski metric, which in one space dimension can be expressed $ds^2 = -c^2dt^2 + dx^2$ where $ds$ is the proper time, where $d\tau = \frac{ds}{c}$, where the sign difference in your expression comes from convention.

A metric is just a way to relate coordinate distance to some proper length for a given geometry; i.e. in 2-D Cartesian space the metric would be $ds^2 = dx^2 + dy^2$. If we 1 space 1 time with 'Galilean' reference frames (non-special relativity), we'd have a metric $ds^2 = c^2dt^2 + dx^2$ (which you can recognize as the triangle inequality $a^2 + b^2 = c^2$); the difference with the Minkowski metric is the sign difference on the $dt$ term. The Minkowski metric describes geometry in flat spacetime, which is another way of saying it handles how length and time change in special relativity.

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