Undergrad Vectors and isometries on a manifold

[PeterDonis]

It's easiest if you pick coordinates that match up with the rotation symmetry, e.g., polar coordinates on the 2-d Euclidean plane.

Oops! I just realized that I wrote down the matrices in Cartesian coordinates, not polar. Here are the correct matrices in polar coordinates.

The infinitesimal rotation matrix is

$$
\begin{bmatrix} 1 & 0 \\ \frac{d\theta}{r} & 1 \end{bmatrix}
$$

Applying this to the vector $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ still gives $\begin{bmatrix} 1 \\ d\theta \end{bmatrix}$ as before.

The finite rotation matrix is

$$
\begin{bmatrix} 1 & 0 \\ \frac{\theta}{r} & 1 \end{bmatrix}
$$

Applying this to $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ gives $\begin{bmatrix} 1 \\ \theta \end{bmatrix}$, as expected (in polar coordinates). More generally, applying it to $\begin{bmatrix} r \\ \phi \end{bmatrix}$ gives $\begin{bmatrix} r \\ \phi + \theta \end{bmatrix}$, so it does what it's supposed to, it rotates a vector by an angle $\theta$ without changing its length.

 

[PeterDonis]

Can you write down an expression showing the explicitly use of the Killing vector in doing the rotation of our vector (1,0)?

The action of the infinitesimal rotation matrix is to take $V^\mu = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ to $\begin{bmatrix} 1 \\ d\theta \end{bmatrix}$, which we can rewrite as $\begin{bmatrix} 1 \\ 0 \end{bmatrix} + d\theta \begin{bmatrix} 0 \\ 1 \end{bmatrix} = V^\mu + d\theta \xi^\mu$. That is what we mean when we say the Killing vector generates infinitesimal rotations. You can work this out for a general vector $V^\mu$ to see that it works for any point at all. Note that at the origin, $\xi^\mu$ vanishes, so the origin is left invariant by rotations.

 

[davidge]

I just realized that I wrote down the matrices in Cartesian coordinates, not polar. Here are the correct matrices in polar coordinates.

The action of the infinitesimal rotation matrix is to take Vμ=[10]Vμ=[10]V^\mu = \begin{bmatrix} 1 \\ 0 \end{bmatrix} to [1dθ][1dθ]\begin{bmatrix} 1 \\ d\theta \end{bmatrix}, which we can rewrite as [10]+dθ[01]=Vμ+dθξμ[10]+dθ[01]=Vμ+dθξμ\begin{bmatrix} 1 \\ 0 \end{bmatrix} + d\theta \begin{bmatrix} 0 \\ 1 \end{bmatrix} = V^\mu + d\theta \xi^\mu. That is what we mean when we say the Killing vector generates infinitesimal rotations

I see. Is it possible to find a \xi^{\mu}, other than the one you've found, that is zero at the point we are to perform the rotation (even if that point is not the origin) and such that its (non-zero) covariant derivative equals the ordinary derivative at that point?

 

[PeterDonis]

Is it possible to find a $\xi^{\mu}$, other than the one you've found, that is zero at the point we are to perform the rotation (even if that point is not the origin)

Sure, just transform to new polar coordinates centered on some other point, and $\partial / \partial \theta'$ in those new polar coordinates will be a Killing vector generating rotations about the new point.

such that its (non-zero) covariant derivative equals the ordinary derivative at that point?

Why would you want this?

 

[davidge]

Sure, just transform to new polar coordinates centered on some other point

Ah ok

Why would you want this?

Because, returning to our example... If we want to write the component of a rotated vector V at x as V^{\mu}_{(rotated)} (x) = V^{\mu}(x) + \xi^{\mu}_{,\nu} V^{\nu}(x) and if \nabla_{\nu} \xi ^{\mu} = \xi^{\mu}_{,\nu} at x, these \xi that I used can be the Killing Vectors (or can't they?) because they will satisfy the Killing equation. In our example on 2-d polar coordinates:

\nabla _{r} \xi _{\theta} = \partial _{r} \xi _{\theta} - \xi _{r} {\Gamma}^{r}_{\theta r} - \xi_{\theta}{\Gamma}^{\theta}_{\theta r}. We know (from your post #23) the value the \Gamma's takes, and if we want \nabla _{r} \xi _{\theta} = \partial _{r} \xi _{\theta} at x, so \xi _{\theta} must equals zero at x.

Next we do the same procedure for \nabla _{\theta} \xi _{r}. Unfortunately, this time the equation don't give us a condition for the value of \xi_{r}, because the Christoffel symbols involved in this second equation vanish in our example. But I'm assuming here that \xi_{r} = 0 at x (can I assume this?).

Now the Killing condition (following our condition that covariant and ordinary derivatives are equal) is that \partial _{r} \xi _{\theta} = - \partial _{\theta} \xi _{r} (1) at x. If we multiply both sides of (1) by g^{\theta \theta} (I'm not sure it's allowed), we get \partial_{r} \xi^{\theta} = - g^{\theta \theta} \partial_{\theta}<br /> \xi_{r}. In our example V^{\theta}_{(rotated)} is equal to d \theta, and so \partial_{\theta} \xi_{r} = -g_{\theta \theta}d \theta at x. Finally, we se that the transformed (or rotated) component V^{\theta}_{(rotated)} of our non-rotated vector \begin{bmatrix}1\\0\end{bmatrix} is equal to d \theta, the same you obtained from your rotation matrix in previous posts.

 

[PeterDonis]

If we want to write the component of a rotated vector V at x as $V^{\mu}_{(rotated)} (x) = V^{\mu}(x) + \xi^{\mu}_{,\nu} V^{\nu}(x)$

That's not what I wrote, and it's not correct. Where are you getting this from?

 

[PeterDonis]

The action of the infinitesimal rotation matrix is to take $V^\mu$ to $V^\mu + d\theta \xi^\mu$.

Perhaps the presence of $d\theta$ is confusing you. If so, rewrite this as $V^\mu \rightarrow V^\mu + \epsilon \xi^\mu$, where $\epsilon << 1$. There is no connection assumed between $\epsilon$ and the coordinate $\theta$ or its differential; we find out that the action of the rotation is to move the point through an angle $\epsilon = d\theta$, without changing $r$, by applying the action of the rotation. The key thing is that no derivative of $\xi^\mu$ appears in the action of the rotation. I have said this several times now but it doesn't appear to be getting through to you.

 

[davidge]

The key thing is that no derivative of ξμξμ\xi^\mu appears in the action of the rotation. I have said this several times now but it doesn't appear to be getting through to you.

No. I understood. What I'm asking is if we can also use the derivative of (a possible) another Killing Vector to accomplish this task.

Is it possible to find a ξμξμ\xi^{\mu}, other than the one you've found

Where are you getting this from?

Maybe I'm wrong, but I thought when we perform a change in the components of a vector (keeping the vector the same), one can treat this by saying that the vector changed in \mathbb{R} ^2 from a point x to a point y, where y = x + \epsilon \xi (x), |\epsilon| &lt;&lt; 1 in the infinitesimal case. In the case of a rotation, we require that \xi^{\mu} = 0 at x for any \mu, so that the point x is kept invariant.
I don't know if you've already read Nakahara's book or Weinberg's book, but if you do that, you would understand what I'm trying to say.

 

[PeterDonis]

What I'm asking is if we can also use the derivative of (a possible) another Killing Vector to accomplish this task.

Not to my knowledge.

I thought when we perform a change in the components of a vector (keeping the vector the same)

This is treating the rotation as a coordinate transformation, not a mapping of vectors to vectors. Everything I have said so far has been assuming we are treating the rotation as a mapping of vectors to vectors. A coordinate transformation keeping vectors fixed is not the same thing, although there are similarities.

the vector changed in $\mathbb{R} ^2$ from a point $x$ to a point $y$, where $y = x + \epsilon \xi (x)$, $|\epsilon| << 1$

Now it seems like you can't make up your mind whether you want to talk about a coordinate transformation or a mapping of vectors to vectors. I strongly advise you to take a step back and use very precise language to make sure you are saying exactly what you mean.

What is quoted just above describes mapping a vector $x$ to a vector $y$, holding the coordinate chart fixed. If you want to describe a coordinate transformation, you would say something like: the coordinates of vector $V^\mu$ changed from $x^\mu$ to $y^\mu = x^\mu + \epsilon \xi^\mu$. Here $\xi^\mu$ is describing a coordinate transformation (in the particular case we have been discussing, it would be one that rotates the coordinates about the origin, i.e., changes $\theta$ but not $r$, in the opposite sense to the sense that we would view the mapping of vectors to vectors as rotating a vector).

 

[davidge]

This is treating the rotation as a coordinate transformation

This wouldn't for one thing, we are keeping the point fixed, that is to say we are keeping the coordinates the same as before.

What is quoted just above describes mapping a vector xxx to a vector y

I think you are misunderstanding my notation. Here both x and y denote points in \mathbb{R} ^2 and not vectors. Also, I noticed that you are treating V^{\mu} as a vector. Actually it is the component of a vector $V$, i.e $$V = \sum_\mu V^{\mu}(x) \frac{\partial}{\partial x^{\mu}},$$ where $V^{\mu}$ is allowed to be a function of $x$ at a point $x$ in $\mathbb{R} ^2$.

 

[PeterDonis]

we are keeping the point fixed, that is to say we are keeping the coordinates the same as before.

These two are not the same thing. Keeping a particular point fixed (the origin) does not mean keeping the coordinates fixed; you can still rotate the coordinates without changing the origin (which is what the coordinate transformation you are describing does).

Here both $x$ and $y$ denote points in $\mathbb{R} ^2$ and not vectors.

Doesn't matter: there is a one-to-one correspondence between them once you've picked an origin, and any rotation picks an origin (the point that's left invariant).

I noticed that you are treating $V^{\mu}$ as a vector.

Yes, writing that notation when a vector instead of an individual component is meant is common. In many cases (including the one under discussion), both the vector itself and each of its components obey the same equation, so no harm is done by the notation.

None of this changes anything I was saying.

 

[davidge]

Keeping a particular point fixed (the origin) does not mean keeping the coordinates fixed

That is because I was visualizing a point as a set of real numbers, e.g. $x \doteq \begin{Bmatrix}1&0\end{Bmatrix}$. In this case, keeping $x$ fixed means don't changing either $1$ or $0$. Is my interpretation of a point wrong?

 

[PeterDonis]

keeping $x$ fixed means don't changing either $1$ or $0$.

But the rotations we are talking about don't keep the point $(1, 0)$ fixed. They only keep the origin $(0, 0)$ fixed. If you view the rotation as mapping points to other points (or vectors to other vectors), then the point $(1, 0)$ gets mapped to some other point. If you view the rotation as a coordinate transformation, then it changes the coordinates of the point $x$ that originally had coordinates $(1, 0)$ to different coordinates.

 

[davidge]

If you view the rotation as mapping points to other points (or vectors to other vectors), then the point (1,0)(1,0)(1, 0) gets mapped to some other point

If you view the rotation as a coordinate transformation, then it changes the coordinates of the point xxx that originally had coordinates (1,0)(1,0)(1, 0) to different coordinates.

I see. Are you sure about the infinitesimal rotation matrix in post #31? Would not it be (or could it be replaced with) $\begin{bmatrix}1&-r d\theta\\\frac{d\theta}{r}&1\end{bmatrix}$ or $\begin{bmatrix}1&-r d\theta\\rd\theta&1\end{bmatrix}$ instead of $\begin{bmatrix}1&0\\\frac{d\theta}{r}&1\end{bmatrix}$?

 

[PeterDonis]

Are you sure about the infinitesimal rotation matrix in post #31? Would not it be (or could it be replaced with) $\begin{bmatrix}1&-r d\theta\\\frac{d\theta}{r}&1\end{bmatrix}$ or $\begin{bmatrix}1&-r d\theta\\rd\theta&1\end{bmatrix}$ instead of $\begin{bmatrix}1&0\\\frac{d\theta}{r}&1\end{bmatrix}$ ?

Try it and see. A rotation by $d \theta$ should take the vector $\begin{bmatrix} 1 \\ \theta \end{bmatrix}$ to $\begin{bmatrix} 1 \\ \theta + d \theta \end{bmatrix}$. Do either of the alternatives you suggested do that? (Notice that this is a more general requirement than the one I illustrated explicitly in post #31.)

 

[davidge]

Do either of the alternatives you suggested do that?

Yes. Both of them do that at $r = 1$, including the one you wrote down.

 

[PeterDonis]

Both of them do that at $r = 1$, including the one you wrote down.

Do the more general thing that I edited my last post to say?

 

[davidge]

Do the more general thing that I edited my last post to say?

Indeed, they do not satisfy your condition. It's always worth trying something more general to check things out...

 

[davidge]

One more question: does the transformation $\begin{bmatrix}1\\\theta\end{bmatrix} \rightarrow \begin{bmatrix}1\\\theta +d\theta\end{bmatrix}$ takes the same form in all points or only in $r = 1$, i.e. only on $(1 , \theta)$?

 

[PeterDonis]

does the transformation $\begin{bmatrix}1\\\theta\end{bmatrix} \rightarrow \begin{bmatrix}1\\\theta +d\theta\end{bmatrix}$ takes the same form in all points or only in $r = 1$,

What do you think the general form for arbitrary $r$ should be? (Hint: rotations don't change the length of vectors, just the direction in which they point.)

 

[davidge]

What do you think the general form for arbitrary rrr should be?

Since it has an r in one of its entries, I would say its form depends on the point in question.

 

[PeterDonis]

Since it has an r in one of its entries, I would say its form depends on the point in question.

The value of the components does, but the form does not. The general form is that the rotation maps $\begin{bmatrix} r \\ \theta \end{bmatrix}$ to $\begin{bmatrix} r \\ \theta + d \theta \end{bmatrix}$. If you work through the math, you will see that this is why the lower left component of the matrix I gave is $d\theta / r$.

 

[davidge]

The value of the components does, but the form does not

The general form is that the rotation maps [rθ][rθ]\begin{bmatrix} r \\ \theta \end{bmatrix} to [rθ+dθ]

Ok. But will this condition hold for $r = 0$? I've tried here and I found that at $r = 0$ we can't transform the second component from $\theta$ to $\theta + d\theta$. Also, as $r$ tends to zero the derivatives of the Killing Vector tends to infinity.

 

[PeterDonis]

will this condition hold for $r = 0$?

No, and that's ok, because the rotation leaves the origin $r = 0$ fixed, and the coordinate $\theta$ is singular there anyway, so "rotating" from $\theta$ to $\theta + d\theta$ is meaningless.

as rrr tends to zero the derivatives of the Killing Vector tends to infinity

Yes, that's true. So what? As I've said a number of times now, the derivatives of the KVF have nothing to do with generating rotations.

 

[PeterDonis]

Btw, if the rotation matrix being undefined at $r = 0$ bothers you, you can redo the analysis in Cartesian coordinates, which are nonsingular at the origin. In those coordinates you can show explicitly that the rotation matrix does nothing to the origin $x = 0$, $y = 0$.

 

[davidge]

So what? As I've said a number of times now, the derivatives of the KVF have nothing to do with generating rotations

Because I found that at least on $r = 0$ we cand find vectors $\xi^{\rho}$ that leave the point invariant and that obey Killing's equation:
$\xi^{r} = \xi_{r} = \xi^{\theta} = \xi_{\theta} = 0$, $\partial_{\theta} \xi^{r} = 0$, $\partial_{r} \xi^{\theta} = \frac{1}{r}$, and so

$V'^{\theta} = V^{\theta} + d{\theta} (\partial_{r} \xi^{\theta})V^{r} = \theta + d{\theta}$,
$V'^{r} = V^{r} + d{\theta} (\partial_{\theta} \xi^{r})V^{\theta} = V^{r} = r$.

These are not the Killing Vectors you wrote down in a previous post, because those were obtained from metric relations.

you can redo the analysis in Cartesian coordinates

I will

 

[PeterDonis]

I found that at least on $r = 0$ we cand find vectors $\xi^{\rho}$ that leave the point invariant and that obey Killing's equation

This doesn't make sense; $r = 0$ is just one point, and the vector $\xi$ vanishes at that point; but if you are only taking into account one point, derivatives are meaningless (you need at least a neighborhood). And at the point $r = 0$ it is also meaningless to talk about "rotating" $\theta$ to $\theta + d\theta$, since the coordinate $\theta$ is singular there.

Why are you persisting with trying to use derivatives of Killing vectors instead of Killing vectors themselves? That is not done anywhere in the literature that I'm aware of. If it's just your personal thing, then (a) it's not going to work, and (b) PF has rules about personal theories, which you should review.

 

[PeterDonis]

These are not the Killing Vectors you wrote down

Not quite, as you wrote it, because you used partial derivatives instead of covariant derivatives. But if you alter your definition to use covariant derivatives, you will see that, if we consider a neighborhood of $r = 0$ instead of just that point (so that derivatives are meaningful), and if you the components of $\xi$ at $r = 0$ plus the derivatives you gave there define a vector field $\xi$ on the neighborhood that is identical to the one I wrote down earlier.

(Also, your definition using partial derivatives is not well-defined at $r = 0$, since $1 / r$ is undefined there; so it doesn't work anyway. Whereas the definition in terms of covariant derivatives works fine at $r = 0$.)

 

[davidge]

Why are you persisting with trying to use derivatives of Killing vectors instead of Killing vectors themselves? That is not done anywhere in the literature

As I said before, I'm following what I learned from Weinberg's book and some two or three notes I found on web. (Also from a lecture I attended last month at the university.) Maybe I misunderstood what I've read?

at the point r=0r=0r = 0 it is also meaningless to talk about "rotating" θθ\theta to θ+dθθ+dθ\theta + d\theta, since the coordinate θθ\theta is singular there

I don't see why it is singular there.

you used partial derivatives instead of covariant derivatives

I did not show my complete derivation in post #56, but as I mentioned, I imposed the condition that partial derivatives of them are equal to the covariant ones.

Whereas the definition in terms of covariant derivatives works fine at r=0r=0r = 0.

I don't think so. Consult your post on the derivation of the covariant derivatives to see that at $r = 0$, $\nabla_{r}\xi^{\theta}$ is undefined.

if you are only taking into account one point, derivatives are meaningless (you need at least a neighborhood)

I agree.

 

[PeterDonis]

Maybe I misunderstood what I've read?

I think you must be. Unfortunately I don't have Weinberg's book to check the relevant sections for myself.

I don't see why it is singular there.

The metric is $ds^2 = dr^2 + r^2 d\theta^2$. This metric has no inverse at $r = 0$ (its determinant is zero), because $g_{\theta \theta} = 0$ there. Saying that $\theta$ is singular at $r = 0$ is a (somewhat sloppy) shorthand for that.

I imposed the condition that partial derivatives of them are equal to the covariant ones.

You can't "impose" this condition; it is either satisfied by the coordinates you've chosen or it isn't. For polar coordinates, it isn't.

Consult your post on the derivation of the covariant derivatives to see that at $r = 0$, $\nabla_{r}\xi^{\theta}$ is undefined.

Yes, but that doesn't stop $\xi^\theta$ itself from being well-defined (and nonzero) at $r = 0$ by my definition; it's just $\xi^\theta = 1$, like it is everywhere else (and $\partial r \xi^\theta = 0$ everywhere, including $r = 0$). What vanishes according to my definition at $r = 0$ is the norm of $\xi$, i.e., $\sqrt{g_{\mu \nu} \xi^\mu \xi^\nu}$. From the metric you will see that this norm is just $r$. Whereas by your definition, you tried to make $\xi^\theta$ itself vanish at $r = 0$, which doesn't work.