Christoffel Symbols of Vectors and One-Forms in say Polar Coordinates

[Skhaaan]

Hello all,

I've been going through Bernard Schutz's A First Course In General Relativity, On Chapter 5 questions atm.

Should the Christoffel Symbols for a coordinate system (say polar) be the same for vectors and one-forms in that coordinate system?

I would have thought yes, but If you calculate the Christoffel Symbols using the Basis vectors of basis one forms then you get different Christoffel symbols?

I am asking because when you do the Covariant Derivative of a vector or a one form are the Christoffel Symbols the same of different?


For Vectors i.e.

$\vec{V} = v^{\alpha}\vec{e}_{\alpha}$

We can calculate the Christoffel symbols using

$\frac{∂ \vec{e}_{\alpha}}{∂x^{\beta}} = \Gamma^{\mu}_{\alpha \beta} \vec{e}_{\mu}$

Where the Basis vectors for polar coordinates are

$\vec{e}_{r} = Cos(\theta)\vec{e}_{x} + Sin(\theta)\vec{e}_{y}$
$\vec{e}_{\theta} = -r Sin(\theta)\vec{e}_{x} + r Cos(\theta)\vec{e}_{y}$

For One-Forms i.e.

$\tilde{P} = p_{\alpha}\tilde{e}^{\alpha}$

We can calculate the Christoffel symbols using

$\frac{∂ \tilde{e}^{\alpha}}{∂x^{\beta}} = \Gamma^{\alpha}_{\beta \mu} \tilde{e}^{\mu}$

Where the Basis vectors for polar coordinates are

$\tilde{e}^{r} = Cos(\theta)\tilde{e}^{x} + Sin(\theta)\tilde{e}^{y}$
$\tilde{e}^{\theta} = - \frac{Sin(\theta)}{r}\tilde{e}^{x} + \frac{Cos(\theta)}{r}\tilde{e}^{y}$

 

[pervect]

Hello all,

I've been going through Bernard Schutz's A First Course In General Relativity, On Chapter 5 questions atm.

Should the Christoffel Symbols for a coordinate system (say polar) be the same for vectors and one-forms in that coordinate system?

I would have thought yes, but If you calculate the Christoffel Symbols using the Basis vectors of basis one forms then you get different Christoffel symbols?

One minor appearance tip. To make the Christoffel symbols vertical alignment proper, the latex looks llike this: \Gamma^{\alpha}{}_{\beta\mu}, i.e. $\Gamma^{\alpha}{}_{\beta\mu}$. The extra empty {} at the end does the trick.

The Christoffel symbols are the same, but the equation for incorporating them into the covariant derivative changes sign depending on whether you are taking the covariant derivative of a vector or a one-form.

I'm having a bit of trouble following your notation, the way I"m used to writing this is:

eq1 $$\nabla_a t^b = \partial_a t^b + \Gamma^{b}{}_{ac} t^c$$
eq2 $$\nabla_a \omega_b = \partial_a \omega_b - \Gamma^{c}{}_{ab}\omega_c$$

where eq1 represents taking the covariant derivative of a vector $t^b$, and eq2 represents takig the covariant derivative of a one form $\omega_b$.

In the above notation $\partial_a$ represents an ordinary derivative, i.e $$\frac{\partial}{\partial a}$$

$\nabla_a$ represents the corresponding covariant derivative. Thus eq1 and eq2 tell you how to construct the covariant derivative out of the ordinary derivative in some coordinate basis and the Christoffel symbols.

 

[Bill_K]

It looks to me like what the OP is calculating are the Ricci rotation coefficients. They are antisymmetric in the two lower indices rather than symmetric.

 

[Skhaaan]

Hey Guys, thanks a lot for replying

Thanks for the tip with arranging the indices using latex.

After some thought I think I have figured out what I was asking

Cheers guys

 

[haushofer]

Shouldn't these equations in the OP read

$$\nabla_{\mu} e_{(\nu)} \equiv \Gamma_{\mu\nu}^{\rho} e_{(\rho)}$$
stating that the covariant derivative on basis vectors e are linear sums of basis vectors? See e.g. Nakahara.