Need to find the Ricci scalar curvature of this metric

chinared
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Need to find the Ricci scalar curvature of this metric:

ds2 = e2a(z)(dx2 + dy2) + dz2 − e2b(z)dt2I tried to find the solution, but failed to pass the calculation of Riemann curvature tensor:

<The Christoffel connection> Here a'(z) denotes the first derivative of a(z) respect to z.
\Gamma\stackrel{x}{xz}=\Gamma\stackrel{x}{zx}=a'(z)
\Gamma\stackrel{y}{yz}=\Gamma\stackrel{y}{zy}=a'(z)
\Gamma\stackrel{z}{tt}=b'(z)e2b(z)
\Gamma\stackrel{z}{xx}=\Gamma\stackrel{z}{yy}=-a'(z)e2a(z)
\Gamma\stackrel{t}{tz}=\Gamma\stackrel{t}{zt}=b'(z)
\Gamma\stackrel{}{either}=0

<The Riemann curvature tensor>
R\stackrel{x}{zxz}=R\stackrel{y}{zyz}=-a''(z)-[a'(z)]2
R\stackrel{z}{tzt}=b''(z)+[b'(z)]2

I tried to find the Ricci scalar curvature(R) from current result, but it gave a function depend on z. Is there any problem in my calculation?

Thanks for answering this question~!
 
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I didn't check your calculation, but why do you think the Ricci scalar shouldn't depend on z?
 
I got the same for the Christoffel symbols, but I got a lot more non-zero elements for the Riemann curvature tensor.
 
Sorry for that I did not write down the other non-zero terms of Riemann curvature tensor which can be deduced by symmetry and anti-symmetry properties.
However, I still have a contradiction that
Rt _ztz-b''(z)-[b'(z)]2
but
Rz_tzt=[b''(z)+[b'(z)]2]e2b(z)

Did you also get the same result?
 
I get

Rztzt = -[b''(z)+[b'(z)]2]e2b(z)
 
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Thank you, I will check my result again
 
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