Physical significance of the zeroth component of 4-velocity and 4-force Is there any physical significance of the zeroth component of the four velocity vector and four force vector? I understand that the space part of u$^\mu$ is related to ordinary velocity and space part of F$^\mu$ is the usual force. But are there any physical quantity related to the zeroth component of u$^\mu$ and F$^\mu$?  
The zeroth component of four momenta, p$^\mu$ is energy. So, similarly are there any physical significance to u$^0$ and F$^0$component?
 A: The zeroth component of the 4-velocity $u^a=(\gamma ,\gamma \vec v/c)c$ is essentially the time-dilation factor $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$ (multiplied by $c$ for dimensional purposes). Using the rapidity $\theta$ (the Minkowski angle between two timelike vectors), that zeroth component is essentially $\cosh\theta$.
In practice, [in geometric units] the 4-velocity is a unit-timelike vector.
If drawn on a spacetime diagram, the tip would represent "one tick" of that object's clock. Thus, the time-component of the unit 4-velocity would be the apparent duration of that object's tick, namely the time-dilation factor multiplied by "one tick".
The zeroth component of the 4-force is the relativistic power.
A: For the sake of simplicity, consider the case of a particle moving with constant velocity and in a straight line, and a clock that is moving right along with the particle. A second as indicated by the clock is a second as perceived by the particle or, as we say, the clock is measuring the proper time $\tau$ of the particle.
The space components of the 4-velocity of the particle are quite easy to sense fisically. Since they are defined as $$\vec{u}=\frac{d\vec{x}}{d\tau},$$ they precisely represent the distance travelled by the particle (in your frame of reference), in units of its proper time. If the particle is moving with a 4-velocity that has $|\vec{u}|$=3 m/s , that means that every time the clock moving with the particle ticks, it has travelled three meters.
Now for the hard part. The zeroth component is defined as
$$
u^0 = \frac{dt}{d\tau}.
$$
What it precisely represents is the speed at which the particle is moving through the time of your frame of reference, that is how much "time" it has travelled, in units of its proper time. If the particle is moving with a 4-velocity that has $u^0$=6 s/s that means that every time the clock moving with the particle ticks, the particle has travelled 6 seconds in your frame of reference. The way you usually think about this is that the particle's time is dilated: one second as felt by the particle is six seconds long in your frame of reference.
For the force, since the zeroth component of the 4-momentum is energy, the zeroth component of the 4-force ($F^\mu=dp^\mu/d\tau$) is the relativistic analogous of power.
A: The zeroth component of a 4-vector is often referred to as its "time-like" component because it is analogous to the time axis in $(t,x,y,z)$ spacetime. So physically saying, components such as $u^0$ or $F^0$ are simply the same as their spatial cousins, with a difference of a factor of $c$ ($m/s$) for dimensional consistency ($x,y,z$ are measured in metres, whereas $t$ is measured in seconds). On a deeper level, temporal and spatial coordinates can be treated similar due to the fact that the speed of light remains unchanged regardless of the frame of reference. For that to be possible, the length of the vector $u^0$ must remain the same under any spacetime transformation, which implies a shift in time, making time variable and loses its absoluteness.
For a more visual explanation, I'd suggest Henry Reich's YouTube playlist on special relativity: https://www.youtube.com/watch?v=ajhFNcUTJI0&list=PL712E709B05086D32
If you wish to learn more about the mathematics of special relativity, the Lorentz transformation would be a good start: 


*

*Lorentz transformation: https://en.wikipedia.org/wiki/Lorentz_transformation

*Derivation: https://en.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations
I hope this was useful!
