It is conventional to reverse the sign of the right hand sign, making it **(*) s2 = (ct)2 - x2 - y2 - z2**. This is also a geometric invariant, and the same form of the magnitude will be applicable to the derivative of the vector.

Now, the definition of ? is (1 - (v/c)2)-1/2 = c(c2 - v2)-1/2, so ?2 = c2(c2 - v2)-1, v being the speed of the object being observed.

So, finally, we can get an expression for the magnitude of the derivative of the spacetime vector, denoted by s:

**(**)**

**The bold part and the labels (*), (**), (***) is of my doing.**Sorry, but that result

**(***)** is wrong.

If

**(*)** s[sup:1bdyvdq8]2[/sup:1bdyvdq8] = (ct)[sup:1bdyvdq8]2[/sup:1bdyvdq8] - x[sup:1bdyvdq8]2[/sup:1bdyvdq8] - y[sup:1bdyvdq8]2[/sup:1bdyvdq8] - z[sup:1bdyvdq8]2[/sup:1bdyvdq8].

Then taking the derivative wrt time on both sides:

Right-hand side = 2sds/dt

Left-hand side = 2ct - 2x(dx/dt) - 2y(dy/dt) - 2z(dz/dt).

Equating these two, and you can eliminate the 2 on both sides, you get,

s(ds/dt) = ct - x(dx/dt) - y(dy/dt) - z(dz/dt), not your equation which I labelled by

**(**)** From this, you don't get

**(***)** s[sup:1bdyvdq8]2[/sup:1bdyvdq8] = c[sup:1bdyvdq8]2[/sup:1bdyvdq8]