Undergrad Trouble with metric. Holonomic basis and the normalised basis

[GR191511]

TL;DR

about metric

$df=\frac {\partial f}{\partial r} dr+\frac {\partial f}{\partial \theta}d\theta\quad \nabla f=\frac{\partial f}{\partial r}\vec{e_r} +\frac{1}{r}\frac{\partial f}{\partial \theta }\vec{e_\theta }$
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?

 

[ergospherical]

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]

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 $g_{\hat{\theta}\hat{\theta}}=r^2$ ?

 

[ergospherical]

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]

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.

 

[ergospherical]

$df = \frac{\partial f}{\partial r} dr + \frac{\partial f}{\partial \theta} d\theta \equiv (df)_r dr + (df)_{\theta} d\theta$
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]

$df = \frac{\partial f}{\partial r} dr + \frac{\partial f}{\partial \theta} d\theta \equiv (df)_r dr + (df)_{\theta} d\theta$
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$\;rd\theta$.