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I am getting into general relativity, which of course means getting to grips with curvilinear coordinate systems. Repeatedly, the textbook and lectures have emphasized the relationships $$e_i=g_{ij}e^j\quad \text{and}\quad g_{ij}^{-1}=g^{ij}$$

In a problem set, I have been told that, given $$s(r, \theta, \phi) = r\begin{pmatrix} \sin \theta \cos \phi\\ \sin \theta \sin \phi\\ \cos \theta \end{pmatrix}$$ to find $e_r,e_\theta,e_\phi$ and then their dual, $e^r,e^\theta,e^\phi$.

I have correctly found $e_r,e_\theta,e_\phi$ using the definition in the text, $e_j=\frac{\partial s}{\partial q_j}$, and I have found the correct $g_{ij}=e_i\cdot e_j$, but when I use the above relation, I don't get the correct $e^r,e^\theta,e^\phi$, based on the orthonormality (I get neither orthogonal nor normality) requirement.

Why? As far as I can tell, I should be able to construct the dual basis from the original in this way, but somewhere in the lectures and text I have missed something.

EDIT: to be more explicit on what I have done.

$$e_r=\frac{\partial s}{\partial r}=\begin{pmatrix} \sin \theta \cos \phi\\ \sin \theta \sin \phi\\ \cos \theta \end{pmatrix}$$ $$e_\theta=\frac{\partial s}{\partial \theta}=r \begin{pmatrix} \cos\theta\cos\phi\\ \cos\theta\sin\phi\\ -\sin\theta \end{pmatrix}$$ $$e_\phi=\frac{\partial s}{\partial\phi}=r \begin{pmatrix} -\sin\theta\sin\phi\\ \sin \theta \cos \phi\\ 0 \end{pmatrix}$$ We note that these vectors are all orthogonal, and so the only nonzero components of the metric are the diagonal ones. $e_r\cdot e_r=1$, $e_\theta\cdot e_\theta=r^2$, and $e_\phi\cdot e_\phi=r^2\sin(\theta)$. This gives $$g_{ij}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2\sin^2\theta \\ \end{pmatrix}$$ and we also easily have $$g^{ij}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1/r^2 & 0 \\ 0 & 0 & 1/(r^2\sin^2\theta) \\ \end{pmatrix}$$

Finally, based on the first relation, we have that $$e^r=g^{ij}e_r=\begin{pmatrix} \sin \theta \cos \phi\\ \sin \theta \sin \phi/r^2\\ \cos \theta/r^2\sin^2\theta \end{pmatrix}$$ which already shows the problem. This is not orthonormal with the any of the original basis vectors. I have that the actual answer (constructed another way) is $e^r=e_r$, $e^\theta=e_\theta/r^2$, and $e^\phi=e_\phi/r^2\sin^2\theta$.

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  • $\begingroup$ Are you getting the wrong $e^i$, or do you think they're wrong because they're not orthonormal? The basis vectors will in general not be orthonormal. $\endgroup$
    – Javier
    Commented Nov 2, 2020 at 18:09
  • $\begingroup$ I do have the answer sheet which tells me this is wrong, but I also am basing it being wrong on the lack of orthonormality, as my lectures say explicitly that a basis and it's dual should be orthonormal: $e_i\cdot e^j=\delta_i^j$. $\endgroup$ Commented Nov 2, 2020 at 18:13
  • $\begingroup$ I see, you mean orthonormality between the two dual bases. Well, in that case we can't really help you unless you show your work, but I think your question might be closed anyway for being homework-like. $\endgroup$
    – Javier
    Commented Nov 2, 2020 at 19:56
  • $\begingroup$ It is an ungraded thing, and I feel like writing down a bunch of matrices and vectors isn't particularly useful. I am also using Mathematica to do my arithmetic, so it isn't a wrong calculation, I am just missing something fundamental on why you can't use the metric of a basis to construct the dual, based on that relation that appears in everything I am looking at. $\endgroup$ Commented Nov 2, 2020 at 20:12
  • $\begingroup$ In that case show what you've done, because everything you wrote in your post is correct. $\endgroup$
    – Javier
    Commented Nov 2, 2020 at 20:22

2 Answers 2

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Your problem is simply that you have to use the metric to take inner products; you can't just take the old dot product. The inner product of two vectors is given in components by

$$\mathbf{a} \cdot \mathbf{b} = a^i g_{ij} b^j,$$

which in our case is

$$\begin{aligned} e^r \cdot e_r &= (e^r)^i g_{ij} (e_r)^j \\ &= \left(\sin\theta \cos\phi, \frac{1}{r^2} \sin\theta \sin\phi, \frac{1}{r^2} \frac{\cos\theta}{\sin^2\theta}\right) \begin{pmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2\sin^2\theta \\ \end{pmatrix} \begin{pmatrix} \sin\theta \cos\phi \\ \sin\theta \sin\phi \\ \cos\theta \end{pmatrix} \\ &= 1. \end{aligned}$$

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The definition of the dual basis is to be orthogonal to all the original basis vectors, except one of them, where the dot product is one. $\mathbf e^i.\mathbf e_j = \delta_{ij}$

A vector expressed by contravariant components is: $\mathbf V = V^1\mathbf e_1 + V^2\mathbf e_2 + V^3\mathbf e_3 $ $V^i = \mathbf V.\mathbf e^i$

The same vector expressed by covariant components: $\mathbf V = V_1\mathbf e^1 + V_2\mathbf e^2 + V_3\mathbf e^3 $
$V_i = \mathbf V.\mathbf e_i$

It is true that: $V^i = g^{ij}V_j$

$\mathbf V.\mathbf e^i = g^{ij}(\mathbf V.\mathbf e_j)$

If $\mathbf V$ is one of the dual basis vectors, we get only the definition of the metric component: $\mathbf e^k.\mathbf e^i = g^{ij}(\mathbf e^k.\mathbf e_j) = g^{ik}(\mathbf e^k.\mathbf e_k) = g^{ik}$

The expression: $\mathbf e^r=g^{ij}\mathbf e_r$ is relating 2 vectors instead of components of vectors.

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