How to find the intrinsic covariant derivative component?
In general relativity the elements of the acceleration four-vector are related to the elements of the four-velocity through a covariant derivative with respect to proper time.
where the covariant derivative is broken into two parts, the extrinsic normal component and the intrinsic covariant derivative component.
$\frac {DU^{\mu}}{d\tau}=\frac {dU^{\mu}}{d\tau}+\delta A^{\mu}$.
infact:
$A^{\mu}_{GR}=A^{\mu}_{SR}+\delta A^{\mu}$.
(GR represent General Relativity and SR represent Special relativity)
I don't know how the $\delta A^{\mu}$ becomes $\Gamma^{\mu}_{\alpha\beta}U^{\alpha} U^{\beta}$.
Best Answer
Very briefly. The line of reasoning is the following: the acceleration $A^{\mu}$ is GR is rather formal construction, it is the covariant derivative of the speed $U^{\mu}$ with respect to some natural parameter $\lambda$ which parameterize a trajectory. For massive particles you can choose this parameter to be proper time $d\tau$ thus $A^{\mu}=DU^{\mu}/d\tau$, although it is not possible for massless particles, for which $d\tau=0$.
Therefore your question is related to the following one: what is $DU^{\mu}$? It turns out that the simplest (and only) way to construct the covariant differential of a vector field is to compare two infinitesimally separated vectors in the same point (it is essential), e.g., the vector $U^{\mu}\left( x^{\alpha}+dx^{\alpha}\right) $ and the vector $U^{\mu}\left( x^{\alpha }\right) $ which should be subject to a parallel translation to the point $x^{\alpha}+dx^{\alpha}$. After the parallel translation we obtain a new infinitisimally close vector $U^{\mu\prime}=U^{\mu}\left( x^{\alpha}\right) +\delta U^{\mu}$, thus $$ DU^{\mu}=U^{\mu}\left( x^{\alpha}+dx^{\alpha}\right) -U^{\mu\prime}=dU^{\mu }-\delta U^{\mu}, $$ where $dU^{\mu}=U^{\mu}\left( x^{\alpha}+dx^{\alpha}\right) -U^{\mu}\left( x^{\alpha}\right) $ is the ordinary differential. Therefore, the small addition $\delta U^{\mu}$ is the result of parallel translation. There are two obvious properties of $\delta U^{\mu}$: it should be linear in $U^{\mu}$ and should vanish with $dx^{\mu}\rightarrow0$. Therefore one can represent $\delta U^{\mu}$ as follows: $$ \delta U^{\mu}=-\Gamma_{\alpha\beta}^{\mu}U^{\alpha}dx^{\beta}, $$ where $\Gamma_{\alpha\beta}^{\mu}$ is the set of some matrices usually referred as “connection coefficients” or “Christoffel symbols”.
Using $\Gamma$, one can generalize the covariant differential $D$ to any tensor quantities. Although, there are no additional mathematical requirements on $\Gamma$, there are physical ones in GR — the equivalence principle requires that $\Gamma$ should be symmetric $\Gamma_{\alpha\beta}^{\mu}=\Gamma_{\beta\alpha}^{\mu}$ and $Dg_{\alpha\beta}=0$. The last condition results in $$ \partial_{\mu}g_{\alpha\beta}=-\left( \Gamma_{\mu,\beta\alpha}+\Gamma _{\beta,\mu\alpha}\right) , $$ where $\Gamma_{\mu,\beta\alpha}=g_{\mu\rho}\Gamma_{\beta\alpha}^{\rho}$. Using the condition that $\Gamma$ is symmetric one can find: $$ \Gamma_{\beta\alpha}^{\rho}=\frac{1}{2}g^{\rho\sigma}\left( \partial_{\beta }g_{\sigma\alpha}+\partial_{\alpha}g_{\sigma\beta}-\partial_{\sigma} g_{\alpha\beta}\right). $$
Let's now consider a parameterized trajectory $x^{\mu}\left( \lambda\right)$, the contravariant vector called speed is $U^{\mu}=dx^{\mu}\left(\lambda\right)/d\lambda$, therefore the contravariant acceleration takes the form: $$ A^{\mu}=\frac{DU^{\mu}}{d\lambda}=\frac{dU^{\mu}}{d\lambda}-\frac{\delta U^{\mu}}{d\lambda}=\frac{dU^{\mu}}{d\lambda}+\Gamma_{\alpha\beta}^{\mu }U^{\alpha}U^{\beta}. $$ If we choose $d\lambda=d\tau$ then in the locally-inertal frame ($\Gamma=0$) for the trajectory $x^{\mu}\left( \lambda\right) $ the acceleration $A^{\mu }$ coincides with the ordinary one $\left( 0,\mathbf{a}\right) $.
And vice versa, $DU^{\mu}=0$ means that $U^{\mu}$ is constant in the locally-internal frame although it does imply that it is constant in any other frame, in fact $dU^{\mu}=\delta U^{\mu}$ implies that a free-fall trajectory is actually a parallel translation in GR, which (for an external observer) looks like the action of gravitational forces.