Four velocity with the Schwarzchild metric

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SUMMARY

The discussion focuses on calculating the components of the four-velocity \( U^u \) for a massive particle moving radially above Earth, utilizing the Schwarzschild metric. The metric is given by \( ds^2 = -(1 + 2\Phi)dt^2 + (1 + 2\Phi)^{-1}dr^2 + r^2d\Omega^2 \). The user initially calculated \( U^0 \) as \( (1 + 2\Phi)(-1/2) \) using the normalization condition \( g_{uv} U^u U^v = -1 \). The confusion arose when attempting to isolate \( U^1 \), which was clarified by noting that the equation involves summation over both indices, thus requiring consideration of both \( U^0 \) and \( U^1 \) simultaneously.

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Pogags
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I am trying to solve the following problem but have gotten stuck.
Consider a massive particle moving in the radial direction above the Earth, not necessarily on a geodesic, with instantaneous velocity
v = dr/dt
Both θ and φ can be taken as constant. Calculate the components of the four-velocity Uu in terms of v using the normalization condition for Uu. Do not make any approximations yet.

We are using the metric:

ds2=-(1+2Φ)dt2+(1+2Φ)-1dr2+r22

I have determined that U0 is going to be (1+2Φ)(-1/2) by using the equation

guvUuUv=-1

However when I try to solve for U1 I get

g11*(U1)2=-1

Which is clearly not correct as this would yield

U1=(-(1+2Φ))(-1/2).

What am I doing wrong?

Apologies for formatting.
 
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Pogags said:
I have determined that U0 is going to be (1+2Φ)(-1/2) by using the equation

guvUuUv=-1

Do you realize that this equation contains both ##U^0## and ##U^1##? You can't get an equation for just ##U^0## from it. The notation ##g_{uv} U^u U^v## means you have to sum over all possible values for the indexes ##u## and ##v##.
 
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Likes   Reactions: vanhees71 and Pogags
This clears up the problem thank you. That slipped my mind
 

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