# Is this calculus on the Lorentz factor right?

Can the following process be justified?

The Lorentz factor is given by $$\gamma (V^i)=\left(1-\frac{V^2}{c^2}\right)^{-1/2}$$ where $$V^2=\sum_{i=1,2,3}{V^iV^i}$$

So, the partial derivative of the Lorentz factor with respect to a four velocity $$V^\alpha$$ will be written as follows: $$\frac{\partial\gamma(V^i)}{\partial V^\alpha}=\frac{\gamma^3}{c^2}V^i,$$ where $$\alpha = 0,1,2,3$$.

Q1: The above equation states that the result should be written in the three vector form. Is this right?

Q2: Due to the above result, I think that the following four vector quantity is reduced to the three-vector quantity. Is this right? (Please note $$\alpha \rightarrow i$$.)

$$\sum_{\alpha=0,1,2,3}{\frac{\partial\gamma(V^i)}{\partial V^\alpha}\frac{dV^\alpha}{d\tau}}=\frac{\gamma^3}{c^2}\sum_{i=1,2,3}V^i\frac{dV^i}{d\tau},$$

Thank you.

• Q2: Which vector quantity? The expression below is scalar (which is built from indexed quantities that are probably not tensorial). – Cryo Feb 27 at 15:05
• Could you state the purpose of your question? What are you trying to proove? – Cryo Feb 27 at 15:06
• Why do you want to consider $\gamma$ as a function of four velocity? It's a function of the magnitude of the three-velocity, it doesn't see the time component at all. – jacob1729 Feb 27 at 16:16