# How does the derivative of time with respect to proper time equals gamma?

So, I want to really get a deep understanding of everything that goes beneath special relativity. Since I teach myself with books, I have no teacher to ask to which makes things a bit harder most of the times.

Proper time is great, I finally get how it is indeed a clock moving through a given worldline.

I have two questions (which are related to using the proper time to get the 4-Velocity and further 4-vectors...)

If we use (-,+,+,+) metric, how can we avoid the imaginary numbers when: $$d\tau=\sqrt{-ds^2}=idt; \text{When the particle is at rest}$$ And, how is this result obtained: $$\frac{dt}{d\tau}=\gamma$$ Thanks!

• Do you mead $d\tau = \sqrt{-ds^2}$, where $ds^2$ is the invariant spacetime interval? Commented Feb 9, 2021 at 1:45
• Yeah, my mistake! But not really. I was doing the Momentarily Comoving Reference Frame method in which dx=dy=dz=0 Commented Feb 9, 2021 at 2:58

The two questions are related. First, proper time is defined as the spacetime interval between timelike separated events.

$$d\tau = \sqrt{-ds^2}$$

If the interval is timelike, then $$ds^2<0$$. This means $$-ds^2$$ is positive and the proper time is a real number.

The spacetime interval is $$ds^2 = d\vec{s}\cdot d\vec{s} = -dt^2 + dx^2 + dy^2 +dz^2$$

If we define the spatial separation of the events as the 3-vector $$d\mathbf{r}$$, then we can write the interval as $$ds^2 = -dt^2 + d\mathbf{r}^2.$$

We can do some sloppy, physicist math and combine the interval with the definition of proper time to see $$d\tau = dt \sqrt{1 - \left(\frac{d\mathbf{r}}{dt}\right)^2}.$$ and eventually arrive at your second equation.

• I see! That 3-vector trick does simplify a lot of things... I was trying to do a different method shows by a youtube video (in which as usual many steps regarding calculations are omited). It's interesting how gamma appears there. Commented Feb 9, 2021 at 3:05