Consider the velocity vector [itex] v^i=\dot{x}^i \hat{e}_i [/itex].(adsbygoogle = window.adsbygoogle || []).push({});

We know that in the plane polar coordinates,it becomes [itex] \vec{v}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta} [/itex]

We also know that it is a contravariant vector so if we transform it covariantly from cartesian coordinates to plane polar coordinates,we should

get the second equation above from the first.

[itex]

v^1_{polar}=\frac{\partial r}{\partial x} \dot{x} + \frac{\partial r}{\partial y} \dot{y} = \frac{x}{\sqrt{x^2+y^2}} \dot{x}+

\frac{y}{\sqrt{x^2+y^2}}\dot{y}=\cos{\theta} ( \dot{r} \cos{\theta}-r \dot{\theta} \sin{\theta}) + \sin{\theta} ( \dot{r} \sin{\theta} +

r \dot{\theta} \cos{\theta}) = \dot{r} \cos^2{\theta} -r \dot{\theta} \sin{\theta} \cos{\theta} + \dot{r} \sin^2{\theta} + r \dot{\theta} \sin{\theta}\cos{\theta}=\dot{r}

[/itex]

Well,this has no problem

[itex]

v^2_{polar}=\frac{\partial \theta}{\partial x}\dot{x}+\frac{\partial \theta}{\partial y} \dot{y}=-\frac{y}{x^2+y^2} \dot{x}+\frac{x}{x^2+y^2} \dot{y}=

-\frac{\sin{\theta}}{r} ( \dot{r} \cos{\theta}-r \dot{\theta} \sin{\theta})+\frac{\cos{\theta}}{r}( \dot{r} \sin{\theta} + r \dot{\theta} \cos{\theta})=

-\frac{\dot{r}}{r} \sin{\theta} \cos{\theta} + \dot{\theta} \sin^2{\theta} + \frac{\dot{r}}{r} \cos{\theta} \sin{\theta} + \dot{\theta} \cos^2{\theta}=\dot{\theta}

[/itex]

You see?!It lacks r !

It seeme that velocity does not transform contravariantly under transformation from cartesian coordinates to plane polar coordinates!

I don't mean it transforms covariantly because that one gives sth completely different.

So looks like velocity transforms not contravariantly NOR covariantly!

Now consider the displacement vector [itex] \vec{r}=x \hat{i} + y \hat{j} [/itex]

We know that its a contravariant vector,and we have seen the proof so I don't bother writing it.

But maybe someone bored,wants to have some fun and tries to transform it covariantly to plane polar coordinates(Well,that was not my reason)

[itex]

r^1_{polar}=\frac{\partial x}{\partial r} x + \frac{\partial y}{\partial r} y = \cos{\theta} x + \sin{\theta} y = r \cos^2{\theta} + r \sin^2{\theta} =r

[/itex]

[itex]

r^2_{polar}=\frac{\partial x}{\partial \theta} x + \frac{\partial y}{\partial \theta} y = -r \sin{\theta} x + r \cos{\theta} y=-r^2 \sin{\theta} \cos{\theta} + r^2 \cos{\theta}\sin{\theta}=0

[/itex]

You see,it gives the same vector as the contravariant transformation does.Looks like displacement vector transforms contravariantly AND covariantly under transformation from

cartesian to plane polar coordinates.

Sorry to make it too long,but also consider[itex] \phi=xy [/itex]

It seems to be a scalar meaning if we rotate the coordinate system,It remains invariant and we get [itex] \phi'=x'y' [/itex] and [itex] \phi'=\phi [/itex]

[itex]

\phi'=x'y'=(x \cos{\psi} - y \sin{\psi})(x \sin{\psi} + y \cos{\psi} )=x^2 \sin{\psi} \cos{\psi} +xy \cos^2{\psi} - xy \sin^2{\psi} -y^2 \sin{\psi} \cos{\psi}=

(x^2-y^2)\sin{\psi}\cos{\psi}+xy(\cos^2{\psi}-\sin^2{\psi})

[/itex]

So it seems [itex] \phi'=\phi [/itex] only if [itex] \psi=k \pi [/itex] which suggests that [itex] \phi [/itex] is not a scalar.

So what is it?

Also I should say that [itex]\vec{\nabla} \phi [/itex] behaves like what I said about velocity vector!

Thanks

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