Need to find the Ricci scalar curvature of this metric

chinared

Need to find the Ricci scalar curvature of this metric:

ds² = e^2a(z)(dx² + dy²) + dz² − e^2b(z)dt²


I 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.
[itex]\Gamma\stackrel{x}{xz}[/itex]=[itex]\Gamma\stackrel{x}{zx}[/itex]=a'(z)
[itex]\Gamma\stackrel{y}{yz}[/itex]=[itex]\Gamma\stackrel{y}{zy}[/itex]=a'(z)
[itex]\Gamma\stackrel{z}{tt}[/itex]=b'(z)e^2b(z)
[itex]\Gamma\stackrel{z}{xx}[/itex]=[itex]\Gamma\stackrel{z}{yy}[/itex]=-a'(z)e^2a(z)
[itex]\Gamma\stackrel{t}{tz}[/itex]=[itex]\Gamma\stackrel{t}{zt}[/itex]=b'(z)
[itex]\Gamma\stackrel{}{either}[/itex]=0

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

phyzguy

I didn't check your calculation, but why do you think the Ricci scalar shouldn't depend on z?

DaleSpam

I got the same for the Christoffel symbols, but I got a lot more non-zero elements for the Riemann curvature tensor.

chinared

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]e^2b(z)

Did you also get the same result?

Mentz114

I get

R^z_tzt = -[b''(z)+[b'(z)]2]e^2b(z)

chinared

Thank you, I will check my result again