# If something is not moving in space, is it moving on the time axis at the speed of light?

Square the proper time $d \tau = dt / \gamma$, multiply by $c^2$, rearrange, and take the square root: $$ \left(\frac{d\tau}{dt}\right)^2 = \gamma^{-2} = 1 - \left(\frac{\bf{v}}{c}\right)^2 \Rightarrow \sqrt{\left(c \frac{d\tau}{dt}\right)^2 + {\bf{v}}^2} = c $$ The proper time $d\tau$ is the time elapsed in the frame moving with respect to the lab frame, in which the elapsed time is $dt$. So, $c \ d\tau / dt$ is the "speed through time," and

- $\left|d{\bf r}/dt\right| = 0 \Leftrightarrow c \ d\tau / dt = c$. This is why people say "if something is not moving in space, then it is moving on the time axis at the speed of light."
- $d\tau / dt = 0 \Leftrightarrow \left|d{\bf r}/dt\right| = c$. This is why people say light doesn't age, because it always travels at speed $c$ through space.

Yes it is - well, sort of.

The coordinate invariant form of velocity is the four velocity, and the magnitude of any four velocity is always $c$ (or $1$). So even if you are stationary in space in your chosen coordinate system the magnitude of your four velocity is still $c$.

Whether *moving at the speed of light on the time axis* is a good way to state this is debatable. What it actually means is that $dt/d\tau = 1$.

I realize that in essence there is no object which can be considered as "not moving in space".

No object at all is moving in space if you are taking the point of view of its reference frame!

The law of conservation of energy is requiring that the energy of its mass ($e = mc^2$) is "transported through time", or in other words, that time is passing for the massive object.

The contrary is happening for massless particles such like photons: they are travelling at speed of light, but from their (hypothetical) point of view they have no proper time, and their energy is not transported through time.