Why is normal differentiation correct here when the function has 2 variables? Background:
I was studying about Four velocity (Special Relativity) and was trying to understand the derivation of relation between 3-dimensional velocity and Four velocity from Proff. Leonard Susskind's 3rd lecture. While doing this I came at 26:40 and immediately got confused due to my reasoning clashing with what was occuring.
Question:
The variable $\tau$ depends on 2 variables $t$ and $x$. With the relation $\tau^2 = t^2 - |\vec{x}|^2$, where $|\vec{x}|^2 = x^2+y^2+z^2$. We want to find the term $\frac{d\tau}{dt}$.
Now my confusion starts with me seeing $\tau^2=t^2-|\vec{x}|^2$ as multivariable function and so $\frac{d\tau}{dt}$ doesn't make a lot of sense to me. Then secondly when I ignored the $d$ and do my (in my mind) more logical $\delta$. I get the wrong result...
Calulation:
$$\tau = \sqrt{t^2 - |\vec{x}|^2}$$
$$\frac{\delta \tau}{\delta t} = \frac{\frac{1}{2} \times 2t}{\sqrt{t^2 - |\vec{x}|^2}}$$
$$\frac{\delta \tau}{\delta t} = \frac{1}{\sqrt{1 - \frac{|\vec{x}|^2}{t^2}}}$$
$$\frac{\delta \tau}{\delta t} = \frac{1}{\sqrt{1 - |\vec{v}|^2}}$$
So this $\frac{\delta \tau}{\delta t} = \frac{1}{\sqrt{1 - |\vec{v}|^2}}$ is what I have got but in the video lectures proffesor got the inverse result through these calculations:
$$\tau = \sqrt{t^2 - |\vec{x}|^2}$$
$$d\tau = \sqrt{{dt}^2 - |\vec{dx}|^2}$$
$$d\tau = dt\sqrt{1 - \frac{|\vec{dx}|^2}{{dt}^2} }$$
$$ \frac{d\tau}{dt} = \sqrt{1 - |\vec{v}|^2 }$$
Now I kinda understand that the first $|\vec{v}|$ is fixed and represents average velocity but does $|\vec{v}|$ in second one represent instantaneous? I only have some vague idea that in second method we say that we are comparing very small changes in all three variables.
I summary why is my method wrong, and what are we exactly doing in the second method?
 A: The primary issue is that $\tau$ (proper time) is actually a scalar representation of the minkowski 4-vector. In minkowski space we are interested in a vector $u = <t, x, y, z>$ under the "size-measure" $\eta(u,u) = t^2 - (x^2 + y^2 + z^2)$, i.e. we construct some vector space that represents the spatial and temporal coordinates of a particle $u$ and use $\eta$ to compute distance (or size of the vector in this case) which is analogous to distance in euclidean space $d(x,x) = x^2 + y^2 + z^2$.
So if we take the differential of our 4-vector we see $du = <dt, dx, dy, dz>$ and using our distance function we see the distance $\eta(du,du)  = d\tau^2 =  dt^2 - (dx^2 + dy^2 + dz^2)$. Thus, $d\tau = \sqrt{dt^2 - (d|\vec{x}|)^2} \implies \frac{d\tau}{dt} = \sqrt{1 - \frac{d|\vec{x}|^2}{dt^2}}$.
Essentially, the real issue is that we aren't differentiating $\tau$ since $\tau$ is just the size of the 4-vector we're interested in. Instead we are differentiating the components of the vector itself then plugging them into our size function / metric tensor.
Unfortunately, this isn't very clear in linked lecture.
