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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 $\Gamma^{\mu}_{\alpha\beta}U^{\alpha} U^{\beta}$.

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I'm not really sure what you want. What is your setup? Your terminology is non-standard, and implies to me that you have some space embedded in a higher dimensional space. You can talk about covariant derivatives without any embedding though. The form you quote is the most general thing you can have which is consistent with linearity and the product rule. There are a number of ways to choose the $\Gamma^\mu_{\alpha\beta}$. A standard choice is the Christoffel connection, which is related to the metric. Good notes on GR can be found at: xxx.lanl.gov/abs/gr-qc/9712019 –  Michael Brown Jan 1 '13 at 10:34
    
Thanks for clarifying the question. Are you okay with the formula for the covariant derivative of a vector field: $\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^{\nu}_{\mu\rho} V^\rho$ ? If you are comfortable with that it is a few short steps to the expression you want. If you're not comfortable with that then a revision of covariant derivatives is needed. –  Michael Brown Jan 1 '13 at 12:09
    
@Michael Brown i'm not familiar with covariant derivative. In fact, the university did not teach me anything about it –  vachofski Jan 1 '13 at 13:32
    
covariant derivatives In mathematics have a slightly different formulation with the theory of relativity –  vachofski Jan 1 '13 at 13:37
    
Well there's your problem. :) If you have the time try reading the lecture notes I linked before, a relativity text, or searching the stack exchange: physics.stackexchange.com/search?q=covariant+derivative. The difference between physics and mathematics is mainly notational, I think –  Michael Brown Jan 1 '13 at 13:57
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1 Answer

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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.

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