How to find 4-acceleration scalar product in terms of $ds$ spacetime interval?

Question

We know 4-velocity $$U^i =dx^i/ds$$ where $$ds=\sqrt{dx^idx_i}$$ so we have 4-acceleration $$A^i=dU^i/ds$$ Then we have $$A^iA_i=\dfrac{dU^i}{ds}\dfrac{dU_i}{ds}$$ How should I proceed to find this scalar product if ds is converted to $$A^iA_i=\dfrac{dU^i}{\sqrt{dx^idx_i}}\dfrac{dU_i}{\sqrt{dx^idx_i}}$$ Defining in terms of proper time gives a nice result but if we also go through the above process will it also give similar result?

Answer 1

Both the 4-velocity and 4-acceleration are defined as derivatives in terms of the proper time. Consider a massive particle moving through a geodesic $\mathbf{x}(\tau)$ in spacetime in its own reference frame. In $c=1$ units and assuming a Minkowski metric $\eta = \mathrm{diag}(-1, 1, 1, 1)$, the Local Inertial Frame (LIF) spacetime interval will be $ds^2 = -d \tau^2 + dx^2 + dy^2 + dz^2$, but $dx^2 + dy^2 + dz^2 = 0$ in its own reference frame, so you will have $ds^2 = -d \tau^2$, hence your formula is correct, up to a sign convention: $$ u^\mu = \frac{dx^\mu}{d \tau} = -\frac{dx^\mu}{d s}. $$ Now, the spacetime interval (and proper time) are frame invariant, so if for some reason you wanted to express the 4-velocity in terms of a set of coordinates which are not the LIF, where $ds^2 = g^{\mu\nu} dx_\mu dx_\nu$, which I guess is your question, apply the chain rule w.r.t the new coordinate time, $$ u^\mu = -\frac{dx^\mu}{d s} = -\frac{dx^\mu}{d t} \frac{dt}{d s} = -\frac{dx^\mu}{d t} \frac{1}{ds/dt} = -\frac{dx^\mu}{d t} \left[g^{\nu \lambda} \frac{dx_\nu}{dt}\frac{dx_\lambda}{dt} \right]^{- \frac{1}{2}}. $$ The obvious upside to this approach is that it enables you to work in arbitrary coordinates, although it's more cumbersome.