special relativity - 4 dimensional interpretation

Has it ever been hypothetized that, in a 4 dimensional space, being time the 4th D, one body could travel through the dimensions at the combined speed of $c$?

If a body is at rest in the classical 3 dimension, it would travel through time at $c$, but if traveling at $c$ in space, it would be resting in the "time" dimension...

Answer

As it happens, you are absolutely correct.

The velocities we encounter in everyday life are 3D velocities that are vectors defined as:

$$\vec{v} = \left(\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}\right)$$

In special relativity we use a 4D velocity called the four-velocity, and this is a four-vector defined as:

$$\vec{v} = \left(c\frac{dt}{d\tau}, \frac{dx}{d\tau}, \frac{dy}{d\tau}, \frac{dz}{d\tau}\right)$$

where the quantity $\tau$ is called the proper time. The proper time is the time shown on a clock carried by the moving object.

But there's something funny about this four-velocity. Suppose we choose coordinates $(t, x, y, z)$ in which I am not moving. Then $dx/d\tau = dy/d\tau = dz/d\tau = 0$. But I am moving in time, at one second per second, so $dt/d\tau = 1$. In that case my four-velocity is:

$$\vec{v} = (c, 0, 0, 0)$$

And the magnitude of my four velocity is $c$. In other words I am moving at the speed of light even when I am stationary.

In fact you can easily prove that the magnitude of the four-velocity is always $c$. I won't do that here because I suspect the maths is a bit more in depth than you want (shout if you do want the proof and I'll edit it in). But basically when you're moving the $dx/d\tau$ etc are not zero but time dilation changes $dt/d\tau$ to compensate, so the magnitude always remains $c$.