Need some help to find the 4-acceleration

[etotheipi]

Homework Statement

Find the 4-acceleration of a particle at rest on the surface of a non-rotating planet of mass $M$

Relevant Equations

N/A

I'd just like a bit of guidance here, because I'm not sure if what I'm doing is correct. First, the equation of the motion,$$A^{\mu} = \frac{dU^{\mu}}{d\tau} + \Gamma^{\mu}_{\sigma \rho} U^{\sigma}U^{\rho}$$I decided to use the Schwarzschild coordinates $(t,r,\theta, \phi)$, and in these coordinates I can take $U^r = U^{\theta} = U^{\phi} = 0$. Since the coordinate acceleration is zero we get$$A^{\mu} = \Gamma^{\mu}_{tt} U^t U^t$$The relevant Christoffel symbol$$\Gamma^{\mu}_{tt} = \frac{1}{2} g^{\mu m} \left( \frac{\partial g_{mt}}{\partial x^t} + \frac{\partial g_{mt}}{\partial x^t} - \frac{\partial g_{tt}}{\partial x^m} \right) = - \frac{1}{2} g^{\mu r} \frac{\partial g_{tt}}{\partial r}$$where I used that the metric is time-independent and that $g_{tt}$ depends only on $r$, i.e.$$g_{tt} = - \left(1- \frac{2GM}{r} \right) \iff \frac{\partial g_{tt}}{\partial r} = - \frac{2GM}{r^2}$$and so$$A^{\mu} = g^{\mu r} \frac{GM}{r^2}$$So for instance, if $\mu = r$, we get$$g^{rr} = \left(\frac{2GM}{r} - 1 \right)^{-1} = - \left( 1 + \frac{2GM}{r} \right) + \mathcal{O}(r^{-2})$$and then$$A^{\mu} = - \left( 1 + \frac{2GM}{r} \right) \cdot \frac{GM}{r^2} = - \frac{GM}{r^2} + \mathcal{O}(r^{-3})$$I wondered, why did I get a minus sign? I expected the four-acceleration to have a positive radial component. I wonder if I've mixed up the metric signature somewhere, but don't know where. thanks!

 

[Ibix]

Not quite. You seem to have assumed that $U^t=1$, but $g_{\mu\nu}U^\mu U^\nu=-1$, so $U^t=1/\sqrt{|g_{tt}|}$ (edit: and not $U^t=1/\sqrt{g_{tt}}$ as I originally wrote) in this case. (Edit: if I finally kept the sign conventions straight, anyway...)

 

[etotheipi]

Thanks! I also can't seem to get the signs of the metric components right. Starting from where it went wrong,$$A^{\mu} = g^{\mu r} \frac{GM}{r^2} U^t U^t = \frac{g^{\mu r}}{g_{tt}} \frac{GM}{r^2}$$then with$$g^{rr} = (g_{rr})^{-1} = \left(1- \frac{2GM}{r} \right)$$and$$g_{tt} = -\left( 1- \frac{2GM}{r} \right)$$and in that case $$A^{r} = - \frac{GM}{r^2}$$ exactly, but now still with an incorrect sign

 

[Ibix]

I made a mistake in my last post - I said that $U^t=1/\sqrt{g_{tt}}$, but it's $1/\sqrt{|g_{tt}|}$, so $g_{tt}U^tU^t=\mathrm{sgn}(g_{tt})=-1$. When I correct that I get the right answer. My Christoffel symbols agree with yours, so I think you've just made the same mistake.

Sign conventions are an unbelievable pain. I like +--- because four velocities have positive moduli, and that fits in my brain better.

 

[etotheipi]

Ah, okay, gotcha. Yeah, that's annoying. In the future I'll make more of an effort to make an even number of sign errors.

Also, have a nice new year's eve!