What is the Inner Product in the Schwarzschild Metric?

[parsifal]

I need to write Schwarzschild Metric, that is in spherical coordinates, into the form that has the metric tensor.

Now, if the first the term of the metric is:
$$\Large (ds)^2=f(r)c^2dt^2-...$$ and x⁰=ct,
then the first component g_ij of the metric tensor g is supposed to be:
$$\Large <\frac{\partial}{\partial x^i} \ | \ \frac{\partial}{\partial x^j}> \ ,i=j=0 \Rightarrow (\frac{d}{dx^0}f(r)c^2dt^2 | \frac{d}{dx^0}f(r)c^2dt^2)$$

But I do not actually understand that last statement. I guess dx⁰=cdt, but I do not know how to proceed from that.

So I know this: the component g₀₀ of g is supposed to be f(r), and I know that f(r) should come from the inner product, but I do not understand how. Basically, what does $$\Large \frac{d}{dx^0}f(r)c^2dt^2$$ mean?

I apologize if this should have been in the introductory section, or in the calculus section.

 

[cristo]

I don't see why you need to use the inner products. If you've already written down the metric in the form $$\Large (ds)^2=g_{ij}dx^idx^j$$ then you can simply read off the components of the metric tensor.
i.e. $$(ds)^2=f(r)(dx^0)^2 \Rightarrow g_{00}=f(r)$$

 

[parsifal]

I guess I was trying to do it the hard way, for some unclear reason. I didn't understand that the solution you suggested would do.

Thanks for the answer!

 

[robousy]

I think that f(r) in the schwarzschild metric is:

$$f(r)=1-\frac{2m}{r}$$

You don't need to do any calculus.

 

[cristo]

I think that f(r) in the schwarzschild metric is:

$$f(r)=1-\frac{2m}{r}$$

You don't need to do any calculus.

I'm not sure that the actual form of the function was required in the question.