Is the 4-velocity vector in spacetime a unit tangent vector? I've found something interesting that no matter how the function looks, the unit tangent vector along a curve $$\hat{||T||}$$ always has a numerical value of (1), and since
I've learned basic General Relativity on ScienceClic English channel, in episode 2, (Spacetime velocity vector) he told that at a given proper time , an object always travels the same distance through spacetime, therefore the length of the vector $$\vec{||v||}$$ always equals to the speed of light c.
are they related?
Do we actually travel through spacetime at the speed of light or have we forced it to be that way?
 A: Yes the 4-velocity is a unit vector tangent to the worldline.
The fact that it is tangent to the worldline says that it tells you something about the direction in spacetime of the worldline at any given event.
The fact that this 4-vector has a unit size (equal to $c$) tells you simply that the amount of proper time between one event and the next along a worldline is equal to the amount of proper time between those events. Yes I wrote that correctly and yes it is a kind of tautology and yes it says nothing at all about any sort of "speed through spacetime" because the notion of "speed through spacetime" is ill-defined.
A: The notation $||v||$ can be misleading as in four-dimensional space time the "length" of the four-vector $v$ is not the square root of $v_0^2+v_1^2+v_2^2+v_3^2$.
As we know, it is rather the square root of $v_0^2-v_1^2-v_2^2-v_3^2$ or of $-v_0^2+v_1^2+v_2^2+v_3^2$ depending which convention you want to follow.
An object that travels through space time along a curve (world line) $t\mapsto w(t)= (ct,x(t),y(t),z(t))$ has

*

*proper time
$$
\tau=\frac{1}{c}\int_0^t\sqrt{c^2-\dot x^2(s)-\dot y^2(s)-\dot z^2(s)}\,ds
$$


*four-velocity
\begin{align}
\frac{dw}{d\tau}&=\frac{dw}{dt}\frac{dt}{d\tau}=\Big(c,\dot x(t),\dot y(t),\dot z(t)\Big)\frac{dt}{d\tau}\\&=\Big(c,\dot x(t),\dot y(t),\dot z(t)\Big)\frac{c}{\sqrt{c^2-\dot x^2(t)-\dot y^2(t)-\dot z^2(t)}}\,.
\end{align}
It is now easy to see that with the metric convention $(+,-,-,-)$ the lenght of this four-velocity is $c\,$.
This does however not mean that we travel at the speed of light. More precisely, through three-space we travel at the speed
$$
\sqrt{\dot x^2+\dot y^2+\dot z^2}
$$
which (assuming our world line is time like) is always less than $c$.
A: The phrase that "everything moves at the speed of light" is well known and often repeated in popular science. But it is problematic to assign a real meaning to this statement. Indeed, the (pseudo)norm of 4-velocity is equal to the speed of light, but this vector does not relate to any motion in spacetime. Instead it is just another way to specify ordinary, 3D velocity. The only difference that 4-velocity provides a way to specify body movement without using a specific reference frame.
The fact is, there is no movement at all in spacetime. All objects simply exist there, and do not move anywhere anyhow. Spacetime is like a filmstrip, on which movement is statically captured, the whole process of movement including past, present, and future. And unlike a film, there is not supposed to be any kind of projector. The projector would need some other, external time, second time, but we have no reason to believe that such a thing exists.
