Derivative of contravariant metric tensor with respect to covariant metric tensor

[PhyPsy]

Homework Statement

Show that $$\frac{{\partial}g^c{^d}}{{\partial}g_a{_b}}=-\frac{1}{2}(g^a{^c}g^b{^d}+g^b{^c}g^a{^d})$$

Homework Equations

The Attempt at a Solution

It seems like it should be simple, but I just do not see how to come up with the above solution. This is what I am coming up with: $$\frac{\partial[\delta^c{_a}\delta^d{_b}(g_a{_b})^-{^1}]}{{\partial}g_a{_b}}$$$$-\delta^c{_a}\delta^d{_b}(g_a{_b})^-{^2}=-(g^c{^d})^2$$
Maybe I am not applying the Kronecker deltas correctly.

 

[mighty2000]

Can someone please help?

Thank you for your question. I will do my best to explain the steps to arrive at the solution you are looking for.

Firstly, it is important to note that the Kronecker delta is defined as follows: δ^a_b = 1 if a = b, and 0 if a ≠ b. With this in mind, we can rewrite your attempt at a solution as follows:

∂[δ^c_a δ^d_b (g_a_b)^-1]/∂g_a_b - δ^c_a δ^d_b (g_a_b)^-2 = - (g^c_d)^2

Next, we can expand the first term using the product rule for differentiation:

∂[δ^c_a δ^d_b (g_a_b)^-1]/∂g_a_b = δ^c_a δ^d_b (-1) (g_a_b)^-2 + (-1) (g_a_b)^-1 ∂[δ^c_a δ^d_b]/∂g_a_b

Since the Kronecker delta is a constant with respect to g_a_b, the second term becomes 0. Therefore, we are left with:

∂[δ^c_a δ^d_b (g_a_b)^-1]/∂g_a_b = -δ^c_a δ^d_b (g_a_b)^-2

Substituting this back into our original equation, we get:

-δ^c_a δ^d_b (g_a_b)^-2 - δ^c_a δ^d_b (g_a_b)^-2 = - (g^c_d)^2

Simplifying, we get:

-2δ^c_a δ^d_b (g_a_b)^-2 = - (g^c_d)^2

Dividing both sides by -2 and taking the square root, we get:

δ^c_a δ^d_b (g_a_b)^-1 = (g^c_d)/√2

Finally, we can rewrite this in terms of g^a_c and g^b_d as follows:

δ^c_a δ^d_b (g_a_b)^-1 = (g^a_c)^-1 (g^b_d)^-1 = (g^a_c g^b_d)^-1

Therefore, we