- #1

TomServo

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## Homework Statement

The problem is from D'Inverno's book on GR, problem 5.6. We're using the Jacobian/transformation matrix to convert the tangent to a circle centered at the origin of radius A from Cartesian to polar coordinates. I can do the problem and get the book answer, that's okay. What confuses me is when I try to do the conversion without using the matrix and just do it the old-fashioned calculus class way I get an extra factor of A.

## Homework Equations

Polar coordinate system is primed, Cartesian is unprimed.

[tex]x=a cos(\phi)\\

y=a sin(\phi)\\

\vec{x}=x\hat{x}+y\hat{y}\\

\vec{x'}=a\hat{r}\\

\vec{T}=\dot{x}\hat{x}+\dot{y}\hat{y}=-y\dot{\phi}\hat{x}+x\dot{\phi}\hat{y}[/tex]

[tex]

\vec{T'}=\frac{d}{dt}a\hat{r}=a\dot{\phi}\hat{\phi}\\

\hat{r}=cos(\phi)\hat{x}+sin(\phi)\hat{y}\\[/tex]

[tex]\hat{\phi}=-sin(\phi)\hat{x}+cos(\phi)\hat{y}[/tex]

$$\dot{\hat{r}}=\dot{\phi}\hat{\phi}$$

I don't know why this last one won't render, I've tried everything I could think of.

**Mod note**: It's fixed now. Instead of \dot\hat{r}, use \dot{\hat{r}}

## The Attempt at a Solution

When I try to convert [tex]

\vec{T}[/tex] to polar coordinates by substituting x and y in terms of [tex]r, \phi[/tex] I get:[tex]

\vec{T}=-(a sin(\phi))\dot{\phi}\hat{x}+(a cos(\phi))\dot{\phi}\hat{y}=a\dot{\phi}\hat{\phi}=\vec{T'}\\[/tex]

The problem is that using the transformation matrix I get [tex]\vec{T'}=\dot{\phi}\hat{\phi}[/tex], no factor of a, and this is what the book has as an answer too. Where am I going wrong? Thanks.

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