omidj said:
So, if we don't make all four dimensions same, then it shouldn't have any effect on invariancy of products, but plz try it, calculate the inner product of 4-velocity to itself (concluding C^2 as the result is impossible).
The vacuum-speed-of-light ##c## is a scalar invariant under Lorentz transformations, while the 4-velocity ##v^\mu## is a 4-vector and is therefore not invariant (although the scalar product ##v_\mu v^\mu## of two 4-vectors is invariant).
If you define 4-velocity as a derivative ##v^\mu = \frac{\mathrm{d} x^\mu}{\mathrm{d} s}## of the position vector ##x^\mu = (x^0, \mathbf{x})## with respect to the interval ##s## (where ##\mathrm{d}s = c\sqrt{1-\frac{v^2}{c^2}}##), then you obtain the 4-velocity of the form
$$
v^\mu = \left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}, \frac{\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}}\right) \rm{,}
$$
which gives you the product ##v_\mu v^\mu = 1## (which is constant, but not equal ##c^2##).
However, if you decided to define the 4-velocity as a derivative ##u^\mu = \frac{\mathrm{d} x^\mu}{\mathrm{d} \tau}## of ##x^\mu## with respect to the
proper time ##\tau## (in order to make the expression more similar to the 3-dimensional case ##\mathbf{v} = \frac{\mathrm{d} \mathbf{x}}{\mathrm{d} t}##) and use the fact that ##\mathrm{d} \tau = \mathrm{d} s / c##, then you will obtain that ##u^\mu = c v^\mu## where ##v^\mu## was defined above. Then since ##v_\mu v^\mu = 1## now you get that ##u_\mu u^\mu = c^2## like you wanted.
The difference is that you define the 4-velocity either by taking the derivative with respect to the space-time interval or with respect to the proper time.
