# Trouble with metric. Holonomic basis and the normalised basis

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• GR191511
GR191511
TL;DR Summary
##df=\frac {\partial f}{\partial r} dr+\frac {\partial f}{\partial \theta}d\theta\quad
\nabla f=\frac{\partial f}{\partial r}\vec{e_r} +\frac{1}{r}\frac{\partial f}{\partial \theta }\vec{e_\theta }##
On the other hand ## g_{rr}=1\:g_{r\theta}=0\:g_{\theta r}=0\;g_{\theta\theta}=r^2\;##So According to##v_2=v^{\alpha} g_{\alpha 2}\;then\; (df)_{\theta}=\frac{\partial f}{\partial r}*0+\frac{1}{r}\frac{\partial f}{\partial \theta }*r^2=r\frac{\partial f}{\partial \theta }\neq\frac{\partial f}{\partial \theta }##Where have I gone wrong?

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Don't get confused between the holonomic basis ##(\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta})## and the normalised basis ##(\frac{\partial}{\partial r}, \frac{1}{r} \frac{\partial}{\partial \theta})##, or similarly the holonomic covector basis ##(dr, d\theta)## and the normalised covector basis ##(dr, r d\theta)##. If you want to use a normalised basis then remember the metric will be ##g_{\hat{r} \hat{r}} = g_{\hat{\theta} \hat{\theta}} = 1##.

GR191511, jedishrfu and berkeman
ergospherical said:
Don't get confused between the holonomic basis ##(\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta})## and the normalised basis ##(\frac{\partial}{\partial r}, \frac{1}{r} \frac{\partial}{\partial \theta})##, or similarly the holonomic covector basis ##(dr, d\theta)## and the normalised covector basis ##(dr, r d\theta)##. If you want to use a normalised basis then remember the metric will be ##g_{\hat{r} \hat{r}} = g_{\hat{\theta} \hat{\theta}} = 1##.
Thank you! I tried and I found ##g_{\hat{\theta}\hat{\theta}}=1## apply to a transition from the normalised basis ##(\frac{\partial}{\partial r}, \frac{1}{r} \frac{\partial}{\partial \theta})## to the holonomic covertor basis ##(dr, d\theta)## ...But when can I use ##g_{\hat{\theta}\hat{\theta}}=r^2## ?

In the holonomic basis ##g_{\theta \theta} = r^2##. (I used the hat on ##\hat{\theta}## to denote components in the normalised basis, as opposed to no hat for components in the holonomic basis).

GR191511
ergospherical said:
In the holonomic basis ##g_{\theta \theta} = r^2##. (I used the hat on ##\hat{\theta}## to denote components in the normalised basis, as opposed to no hat for components in the holonomic basis).
Thanks! Does ##g_{\theta\theta}=r^2## apply to any transition?from the holonomic basis to the normalised covector basis? I have already checked,it doesn't.

##df = \frac{\partial f}{\partial r} dr + \frac{\partial f}{\partial \theta} d\theta \equiv (df)_r dr + (df)_{\theta} d\theta##
Equivalently
##df = \frac{\partial f}{\partial r} dr + \frac{1}{r} \frac{\partial f}{\partial \theta} (r d\theta) \equiv (df)_{\hat{r}} dr + (df)_{\hat{\theta}} (r d\theta)##
i.e. ##(df)_{\hat{\theta}} = \frac{1}{r} \frac{\partial f}{\partial \theta}## in the normalised basis.

Why are the basis vectors normalised by a factor of ##r##? Consider ##g(\frac{1}{r} \frac{\partial}{\partial \theta}, \frac{1}{r} \frac{\partial}{\partial \theta}) = \frac{1}{r^2} g(\frac{\partial}{\partial \theta}, \frac{\partial}{\partial \theta}) = \frac{1}{r^2} g_{\theta \theta} = \frac{1}{r^2} r^2 = 1##, so this is the correct normalisation.
Meanwhile the corresponding basis covector is ##rd\theta## because ##r d\theta (\frac{1}{r} \frac{\partial}{\partial \theta}) = d\theta(\frac{\partial}{\partial \theta}) = 1##.

GR191511
ergospherical said:
##df = \frac{\partial f}{\partial r} dr + \frac{\partial f}{\partial \theta} d\theta \equiv (df)_r dr + (df)_{\theta} d\theta##
Equivalently
##df = \frac{\partial f}{\partial r} dr + \frac{1}{r} \frac{\partial f}{\partial \theta} (r d\theta) \equiv (df)_{\hat{r}} dr + (df)_{\hat{\theta}} (r d\theta)##
i.e. ##(df)_{\hat{\theta}} = \frac{1}{r} \frac{\partial f}{\partial \theta}## in the normalised basis.

Why are the basis vectors normalised by a factor of ##r##? Consider ##g(\frac{1}{r} \frac{\partial}{\partial \theta}, \frac{1}{r} \frac{\partial}{\partial \theta}) = \frac{1}{r^2} g(\frac{\partial}{\partial \theta}, \frac{\partial}{\partial \theta}) = \frac{1}{r^2} g_{\theta \theta} = \frac{1}{r^2} r^2 = 1##, so this is the correct normalisation.
Meanwhile the corresponding basis covector is ##rd\theta## because ##r d\theta (\frac{1}{r} \frac{\partial}{\partial \theta}) = d\theta(\frac{\partial}{\partial \theta}) =
Thank you!
##df=\frac {\partial f} {\partial r}dr+\frac{1}{r}\frac {\partial f} {\partial \theta}rd\theta## and
##(dr,rd\theta ) \begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix} \begin{pmatrix}
dr \\
rd\theta \\
\end{pmatrix}\Rightarrow g_{rr}=g_{\theta\theta}=g^{rr}=g^{\theta\theta}=1\Rightarrow##
##(\nabla f)^\theta=\frac{1}{r}\frac {\partial f} {\partial \theta}*g^{\theta\theta}=\frac{1}{r}\frac {\partial f} {\partial \theta}\;and\;(\nabla f)^r=\frac {\partial f} {\partial r}##
##\Rightarrow \nabla f=\frac {\partial f} {\partial r}\vec e_r+\frac{1}{r}\frac {\partial f} {\partial \theta}(\vec e_\theta\;or\;\frac{\vec e_\theta}{r})?if\;it\;is\;\frac{\vec e_\theta}{r}\;then##
##\nabla f=\frac {\partial f} {\partial r}\vec e_r+\frac{1}{r^2}\frac {\partial f} {\partial \theta}\vec e_\theta## it is wrong.But if it is##\;\vec e_\theta##...the##\;\vec e_\theta## is not a dual basis vector of##\;rd\theta##.

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