how to calculate Ricci tensor from a metric tensor?

Hi,

I am new to Axiom, I am studying theoretical physics (4th grade) in
Prague and I want to use the computer algebra system as a physicist
and I understand that mathematicians are looking at the mathematics
from a completely different angle than physicists. But anyway, I want
this:

I have a diagonal matrix:

gdd=Matrix((
    (-exp(nu(r)),0,0,0),
    (0, exp(lam(r)), 0, 0),
    (0, 0, r**2, 0),
    (0, 0, 0, r**2*sin(theta)**2)
    ))

that is a metric tensor on a 4 dimensional manifold with signature
(-,+,+,+), this corresponds to g_\mu_\nu (i.e. the indices are
lowered). No I want to calculate the Christoffel symbols -> riemann
tensor -> ricci tensor.

the x^\mu vector are variables (t, r, theta, phi). The nu(r) and
lam(r) are unknown functions of "r".  I am interested in the
(differential) equations for the unknown functions nu(r) and lam(r)
that I get by setting:

R_\mu_\nu = 0

If you need some more explanation, I'll be glad to explain the details.

In maple I can use the grtensor

http://grtensor.org/

package, but I find the maple not suitable for me, as I want to use
the symbolic manipulation in my programs and I don't want to use the
ugly maple language.

I found all the other symbolic packages unsuitable for me, so I wrote my own:

http://code.google.com/p/sympy/

And in SymPy I can now do it quite easily:

http://sympy.googlecode.com/svn/trunk/examples/relativity.py

SymPy is just a general package and all I am using from it are just
symbolic matrices. (I am lowering and raising indices by myself in the
relativity.py example).

I was curious - how could I do the same in Axiom?

Thanks,
Ondrej

BTW, the resulting equations are:

-1/4*exp(\nu(r))*exp(\lambda(r))**(-1)*\lambda'(r)*\nu'(r)+1/4*exp(\nu(r))*exp(\lambda(r))**(-1)*\nu'(r)**2+1/2*exp(\nu(r))*exp(\lambda(r))**(-1)*(\nu'(r))'+exp(\nu(r))*r**(-1)*exp(\lambda(r))**(-1)*\nu'(r)
= 0
1/4*\nu'(r)*\lambda'(r)-1/2*(\nu'(r))'+r**(-1)*\lambda'(r)-1/4*\nu'(r)**2 = 0
-1/2*r*exp(\lambda(r))**(-1)*\nu'(r)+1/2*r*exp(\lambda(r))**(-1)*\lambda'(r)-sin(\theta)**(-2)*cos(\theta)**2-(-1-sin(\theta)**(-2)*cos(\theta)**2)-exp(\lambda(r))**(-1)
= 0
-sin(\theta)**2*exp(\lambda(r))**(-1)+sin(\theta)**2-1/2*sin(\theta)**2*r*exp(\lambda(r))**(-1)*\nu'(r)+1/2*sin(\theta)**2*r*exp(\lambda(r))**(-1)*\lambda'(r)
= 0