# Need some help to find the 4-acceleration

• etotheipi

#### 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!

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...)

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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

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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!