I How to work out the tetrads for given functions?

[etotheipi]

Given functions $f=f(r)$ and $h = h(r)$, author has defined a tetrad$$(e_0)_a = f^{1/2} (dt)_a, \quad (e_1)_a = h^{1/2}(dr)_a, \quad (e_2)_a = r(d\theta)_a, \quad (e_3)_a = r\sin{\theta} (d\phi)_a$$where $(t,r,\theta,\phi)$ are coordinates. It is required to work out $\partial_{[a} (e_\mu)_{b]}$ for each of these, with $\partial_a$ the normal derivative operator associated with these coordinates. So e.g. for the first one I would have thought$$
\begin{align*}
\partial_{[a} (e_0)_{b]} = \partial_{[a} f^{1/2} (dt)_{b]} = \frac{1}{2}f^{-1/2} (\partial_{[a} f) (dt)_{b]} + f^{1/2} \partial_{[a} (dt)_{b]} \ \ \ (1)
\end{align*}
$$I figured that $\partial_{\mu} f = \delta_{\mu r} f'$, i.e. zero for $\mu = t,\theta,\phi$ and $f'$ for $\mu = r$. But solution is$$\partial_{[a} (e_0)_{b]} = \frac{1}{2} f^{-1/2} f' (dr)_{[a} (dt)_{b]} \ \ \ (2)$$How to get from $(1)$ to $(2)$? Thanks

 

[PeterDonis]

author

Are you referring to a particular textbook or paper? If so, please give a reference.

 

[etotheipi]

Are you referring to a particular textbook or paper? If so, please give a reference.

it's chapter 6.1, page 121 of wald

 

[PeterDonis]

it's chapter 6.1, page 121 of wald

Ah, ok, I'll take a look.

 

[PeterDonis]

I figured that $\partial_{\mu} f = \delta_{\mu r} f'$

Remember that Wald's Latin indexes are abstract indexes, i.e., $\partial_a$ is not $\partial_r$ in this case, it's a set of four derivative operators, which in this case are the coordinate derivative operators: $\partial_t$, $\partial_r$, $\partial_\theta$, $\partial_\phi$. We happen to know that, for the function we are interested in, which is a scalar function of $r$ only, three of these four derivative operators give zero; but Wald is deliberately staying with abstract index notation to make clear what is happening, geometrically: we are taking the derivative of a scalar and the result is a vector.

Now, in what direction does that vector point, when we take the derivative of a scalar function of $r$ only? Obviously, it points in the direction of increasing $r$. So, if we were just taking the derivative of $f(r)$ by itself, we would have $\partial_a f(r) = f^\prime(r) \left( dr \right)_a$. The rest is just the chain rule and the extra factor of $\left( dt \right)_b$.

 

[etotheipi]

Ah, I think the $\partial_a f = f' (dr)_a$ part makes sense now. But also, how can we show that the other term $f^{1/2} \partial_{[a} (dt)_{b]}$ vanishes?

 

[PeterDonis]

how can we show that the other term $f^{1/2} \partial_{[a} (dt)_{b]}$ vanishes?

All of the vectors $\left ( dx^\mu \right)_a$ are constants, since they're just the coordinate basis vectors. The variation in the tetrads is contained entirely in the functions that multiply them. So all quantities of the form $\partial_{[a} \left( dx^\mu \right)_{b]}$ vanish.

 

[etotheipi]

All of the vectors $\left ( dx^\mu \right)_a$ are constants, since they're just the coordinate basis vectors. The variation in the tetrads is contained entirely in the functions that multiply them. So all quantities of the form $\partial_{[a} \left( dx^\mu \right)_{b]}$ vanish.

Nice! Thanks for your help ☺

 

[PeterDonis]

Thanks for your help ☺

You're welcome!

 

[PeterDonis]

$\partial_a$ is not $\partial_r$ in this case, it's a set of four derivative operators, which in this case are the coordinate derivative operators: $\partial_t$, $\partial_r$, $\partial_\theta$, $\partial_\phi$.

And to be strictly correct and capture the "which direction does the vector point" issue, we should really write this as:

$$
\partial_a = \left( dt \right)_a \partial_t + \left( dr \right)_a \partial_r + \left( d \theta \right)_a \partial_\theta + \left( d \phi \right)_a \partial_\phi
$$