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I've been trying to follow the article Kaluza-Klein for Kids where the author derives the lagrangian density in the Kaluza-Klein theory. He takes scalar function $\Phi =1$, then he uses the "ansatz" to obtain bunch of relations on Christoffel symbols. I used the relations to find the expression of the Ricci tensor $\widetilde{R}_{55}$. However, later on I got stuck in the calculation of: $$\widetilde{R}_{\mu5}= -\frac{1}{2}F^{\nu}_{\hphantom{\nu}\mu,\nu} -\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}A_{\mu} $$ That's what I got so far: $$\begin{align} \widetilde{R}_{\mu5}&= \widetilde{\Gamma}^{\nu}_{\mu\nu,5}-\widetilde{\Gamma}^{\nu}_{\mu5,\nu}+ \widetilde{\Gamma}^{\nu}_{\mu\rho} \widetilde{\Gamma}^{\rho}_{5\nu}- \widetilde{\Gamma}^{\nu}_{\nu\rho} \widetilde{\Gamma}^{\rho}_{\mu5}\\ &= -\frac{1}{2}F^{\nu}_{\hphantom{\nu}\mu,\nu} \\ &+\frac{1}{4}F^{\nu}_{\hphantom{\nu}\mu}\Gamma^{\nu}_{\mu\rho} +\frac{1}{4}F^{\nu}_{\hphantom{\nu}\mu}F^{\nu}_{\hphantom{\nu}\rho}A_{\mu} +\frac{1}{4}F^{\nu}_{\hphantom{\nu}\mu}F^{\nu}_{\hphantom{\nu}\mu}A_{\mu}\\ &-\frac{1}{2}F^{\rho}_{\hphantom{\rho}\mu}\Gamma^{\nu}_{\nu\rho} -\frac{1}{4}F_{\nu\rho}F^{\rho}_{\hphantom{\rho}\mu}A^{\nu} \end{align}$$ I cant figure out how do Christoffel symbols in 4D, in the above expression, cancel each other out, more generally I'm not sure if I multiply Christoffel symbols in 5D correctly.

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  • $\begingroup$ Index picture of your last equation looks totally wrong. $\endgroup$ – lalala Feb 15 '18 at 20:56
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To answer your first question, the 4D Christoffel symbols aren't supposed to cancel out. The result that you're aiming for is actually supposed to be

$$\widetilde{R}_{\mu5}= -\frac{1}{2}F^{\nu}_{\hphantom{\nu}\mu;\nu} -\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}A_{\mu} $$

(notice the semicolon instead of a comma). In other words, the equation is supposed to be taking the covariant derivative of $F^{\nu}_{\hphantom{\nu}\mu}$ instead of the ordinary derivative. The 4D Christoffel symbols are needed to form the covariant derivative, i.e.,

$$-\frac{1}{2}F^{\nu}_{\hphantom{\nu}\mu;\nu}= -\frac{1}{2}F^{\nu}_{\hphantom{\nu}\mu,\nu} +\frac{1}{2}\Gamma^{\rho}_{\nu\mu}F^{\nu}_{\hphantom{\nu}\rho} -\frac{1}{2}\Gamma^{\nu}_{\nu\rho}F^{\rho}_{\hphantom{\rho}\mu}\ \ .$$

To answer your second question, no, you're not multiplying the 5D Christoffel symbols correctly, in a couple different ways.

First, when doing a summation over components of the 5D Christoffel symbols, you need to make sure that each summation index takes on all five dimensions in the sum. Some of the resulting terms may turn out to be zero due to one identity or another, but you're missing some terms that are nonzero. In particular, in the Ricci tensor

$$\widetilde{R}_{AB} = \widetilde{\Gamma}^{C}_{AC,B} - \widetilde{\Gamma}^{C}_{AB,C} + \widetilde{\Gamma}^{C}_{AD}\widetilde{\Gamma}^{D}_{BC} - \widetilde{\Gamma}^{C}_{CD}\widetilde{\Gamma}^{D}_{AB}\ \ ,$$

the nonzero terms for the $A=\mu$, $B=5$ component are

$$\widetilde{R}_{\mu5} = - \widetilde{\Gamma}^{\nu}_{\mu5,\nu} + \left(\widetilde{\Gamma}^{\nu}_{\mu\rho}\widetilde{\Gamma}^{\rho}_{5\nu} + \widetilde{\Gamma}^{\nu}_{\mu5}\widetilde{\Gamma}^{5}_{5\nu}\right) - \left(\widetilde{\Gamma}^{\nu}_{\nu\rho}\widetilde{\Gamma}^{\rho}_{\mu5} + \widetilde{\Gamma}^{5}_{5\rho}\widetilde{\Gamma}^{\rho}_{\mu5}\right)\ \ .$$

Second, when multiplying an $F$ with another $F$ or a $\Gamma$, you aren't handling the indices properly. Generally, when multiplying, all contravariant indices remain contravariant, and all covariant indices remain covariant. Summation indices appear as contravariant in exactly one place in a term, and as covariant in exactly one place in the term. It's OK to raise or lower summation indices on tensors, but only if you both raise the covariant index and lower the corresponding contravariant index. And you can of course rename a summation index, as long as you rename it everywhere within a term and don't cause a naming conflict. For example,

$$\begin{align} \widetilde{\Gamma}^{\nu}_{\mu\rho}\widetilde{\Gamma}^{\rho}_{5\nu} &= \left[\Gamma^{\nu}_{\mu\rho} + \frac{1}{2}\left(A_{\mu} F^{\nu}_{\hphantom{\nu}\rho} + A_{\rho} F^{\nu}_{\hphantom{\nu}\mu} \right)\right] \frac{1}{2}F^{\rho}_{\hphantom{\rho}\nu} \\ &=\frac{1}{2}\Gamma^{\nu}_{\mu\rho}F^{\rho}_{\hphantom{\rho}\nu} + \frac{1}{4}F^{\nu}_{\hphantom{\nu}\rho}F^{\rho}_{\hphantom{\rho}\nu}A_{\mu} + \frac{1}{4}F^{\nu}_{\hphantom{\nu}\mu}F^{\rho}_{\hphantom{\rho}\nu}A_{\rho}\\ &=\frac{1}{2}\Gamma^{\rho}_{\mu\nu}F^{\nu}_{\hphantom{\nu}\rho} + \frac{1}{4}F^{\alpha}_{\hphantom{\alpha}\beta}F^{\beta}_{\hphantom{\beta}\alpha}A_{\mu} + \frac{1}{4}F^{\nu}_{\hphantom{\nu}\mu}F^{\rho}_{\hphantom{\rho}\nu}A_{\rho}\\ &=\frac{1}{2}\Gamma^{\rho}_{\mu\nu}F^{\nu}_{\hphantom{\nu}\rho} + \frac{1}{4}F_{\alpha\beta}F^{\beta\alpha}A_{\mu} + \frac{1}{4}F^{\nu}_{\hphantom{\nu}\mu}F^{\rho}_{\hphantom{\rho}\nu}A_{\rho}\\ &=\frac{1}{2}\Gamma^{\rho}_{\nu\mu}F^{\nu}_{\hphantom{\nu}\rho} - \frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}A_{\mu} + \frac{1}{4}F^{\nu}_{\hphantom{\nu}\mu}F^{\rho}_{\hphantom{\rho}\nu}A_{\rho} \end{align}$$

On the last line above, I'm using the symmetry of $\Gamma$, and the antisymmetry of $F$. The third term will cancel with another term elsewhere.

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