I Killing Vectors in 2D

1. Feb 22, 2017

davidge

Hi everyone. What are the components of the 2 Translational Killing Vectors in 2-dimensions, in Polar Coordinates? I've solved the Killing equation using Maple, and the solution was $\xi_r = 0$, $\xi_{\theta} = r^2$, but I guess that these are the components for the rotation Killing Vector, because these components correspond to those that @PeterDonis wrote down in another thread when I asked him about the rotational Killing Vector. There should be $N(N+1)/2 = 3$ independent Killing Vectors in $N = 2$ dimensions, in this case the other 2 associated with translations. What would be the general components of them?

2. Feb 23, 2017

Staff: Mentor

Write them down in Cartesian coordinates, where they are just the coordinate basis vectors, and then transform them to polar coordinates. What do you get?

3. Feb 23, 2017

davidge

Would it be $$\xi^1\ _{ (1)}= cos(x_2) - x_{1}sin(x_2), \ \xi^2\ _{(1)} = 0 \\ \xi^1\ _{(2)} = 0, \ \xi^2\ _{(2)} = x_{1}cos(x_2) + sin(x_2)$$

(lower index in $\xi$ indicates the Killing vector; upper index indicates the component of the corresponding Killing vector)

4. Feb 23, 2017

Staff: Mentor

This can't be right, because it is saying that one translation Killing vector only has a radial component and the other only has a tangential component. But that's clearly not true of the Cartesian coordinate basis vectors.

Rather than use your intuition, I would strongly suggest doing the calculation by boring brute force. Using the notation where vectors are partial derivatives really helps here, because it makes transforming vectors just a case of applying the chain rule. For a general vector $\xi = \xi^x \partial_{x} + \xi^y \partial_y$ in Cartesian coordinates, you just substitute the expressions for $\partial_x$ and $\partial_y$ in terms of the polar coordinate basis vectors $\partial_r$ and $\partial_\theta$, which are just the chain rule:

$$\partial_x = \frac{\partial r}{\partial x} \partial_r + \frac{\partial \theta}{\partial x} \partial_\theta$$

$$\partial_y = \frac{\partial r}{\partial y} \partial_r + \frac{\partial \theta}{\partial y} \partial_\theta$$

Then you get $\partial r/\partial x$, $\partial \theta / \partial x$, $\partial r / \partial y$, and $\partial \theta / \partial y$ from the coordinate transformation formulas by taking appropriate derivatives.

Once you have substituted, you just rearrange the expression for the vector $\xi$ until you have it in terms of something times $\partial_r$ plus something times $\partial_\theta$; the two somethings are the components of $\xi$ in polar coordinates. The two particular vectors you need to transform are $\xi_{(1)} = \partial_x$ and $\xi_{(2)} = \partial_y$. These have components $\xi^x_{(1)} = 1$, $\xi^y_{(1)} = 0$, and $\xi^x_{(2)} = 0$, $\xi^y_{(2)} = 1$.

5. Feb 23, 2017

davidge

Since $\xi'^{\mu} = \xi^{\nu} \partial y^{\mu} / \partial x^{\nu} \\ \xi'^{1}\ _{(1)} = \xi^{1}\ _{(1)} \partial y^{1} / \partial x^{1} + \xi^{2}\ _{(1)} \partial y^{1} / \partial x^{2} \\ ...$ But $\xi^{2}\ _{(1)} = \xi^{1}\ _{(2)} = 0$ and $\xi^{1}\ _{(1)} = \xi^{2}\ _{(2)} = 1$. So $\xi'^{1}\ _{(1)} = \partial y^{1} / \partial x^{1} \\ \xi'^{1}\ _{(2)} = \partial y^{2} / \partial x^{2}$ (Primes indicate Polar System.) What's wrong here?

In this case will I have to solve the derivatives $$\frac{\partial r}{\partial [rcos(\theta)]}$$ and $$\frac{\partial \theta}{\partial [rsin(\theta)]}?$$

Last edited: Feb 23, 2017
6. Feb 23, 2017

Staff: Mentor

You are making it much harder for yourself by using abstract notation, instead of just using $x$ and $y$ for the Cartesian coordinates and $r$, $\theta$ for the polar coordinates.

Here is the expression for $\xi$ that I wrote down:

$$\xi = \xi^x \partial_x + \xi^y \partial y$$

Translating this into your notation, it becomes

$$\xi = \xi^1 \frac{\partial}{\partial x^1} + \xi^2 \frac{\partial}{\partial x^2}$$

Now let's write the same vector $\xi$ in polar coordinates:

$$\xi = \xi^r \partial_r + \xi^\theta \partial_\theta$$

Translating this into your notation, and using primes, it becomes

$$\xi = \xi'^1 \frac{\partial}{\partial x'^1} + \xi'^2 \frac{\partial}{\partial x'^2}$$

We now have two expressions for the same vector $\xi$, so we can equate them:

$$\xi^1 \frac{\partial}{\partial x^1} + \xi^2 \frac{\partial}{\partial x^2} = \xi'^1 \frac{\partial}{\partial x'^1} + \xi'^2 \frac{\partial}{\partial x'^2}$$

Now we can specialize this general expression to the two particular cases, $\xi_{(1)}$ and $\xi_{(2)}$. Since we know $\xi^2_{(1)} = \xi^1_{(2)} = 0$, we obtain the expressions

$$\frac{\partial}{\partial x^1} = \xi'^1_{(1)} \frac{\partial}{\partial x'^1} + \xi'^2_{(1)} \frac{\partial}{\partial x'^2}$$
$$\frac{\partial}{\partial x^2} = \xi'^1_{(2)} \frac{\partial}{\partial x'^1} + \xi'^2_{(2)} \frac{\partial}{\partial x'^2}$$

In my notation, this would be

$$\frac{\partial}{\partial x} = \xi'^r_{(1)} \frac{\partial}{\partial r} + \xi'^\theta_{(1)} \frac{\partial}{\partial \theta}$$
$$\frac{\partial}{\partial y} = \xi'^r_{(2)} \frac{\partial}{\partial r} + \xi'^\theta_{(2)} \frac{\partial}{\partial \theta}$$

Comparing this to the equations I wrote down earlier for transforming the basis vectors, we can see that

$$\xi'^r_{(1)} = \frac{\partial r}{\partial x}$$
$$\xi'^\theta_{(1)} = \frac{\partial \theta}{\partial x}$$
$$\xi'^r_{(2)} = \frac{\partial r}{\partial y}$$
$$\xi'^\theta_{(2)}= \frac{\partial \theta}{\partial y}$$

Translating this back into your notation, we obtain

$$\xi'^1_{(1)} = \frac{\partial x'^1}{\partial x^1}$$
$$\xi'^2_{(1)} = \frac{\partial x'^2}{\partial x^1}$$
$$\xi'^1_{(2)} = \frac{\partial x'^1}{\partial x^2}$$
$$\xi'^2_{(2)}= \frac{\partial x'^2}{\partial x^2}$$

Note the pattern, which is not the same (at least it does not appear to be) as the one you were assuming.

No. To evaluate $\partial r / \partial x$, you need a formula for $r$ in terms of $x$ and $y$. What is that formula? Once you have it, you just take its derivative with respect to $x$ to get $\partial r / \partial x$. Similarly for the other partial derivatives.

Last edited: Feb 23, 2017
7. Feb 23, 2017

davidge

Ohh I see now. thank you Peter.

So the components would be
$$\xi'^{r}_{(1)} = \frac{\partial}{\partial x} \frac{x}{cos\theta} = \frac {1}{cos\theta}$$
$$\xi'^{\theta}_{(1)} = \ ... \ = - \frac{1}{rsin\theta}$$
and so on?

8. Feb 23, 2017

Staff: Mentor

In other words, you need two formulas: a formula for $r$ in terms of $x$ and $y$, and a formula for $\theta$ in terms of $x$ and $y$. Those are just the coordinate transformation formulas. But that is not what you wrote down.

Last edited: Feb 23, 2017
9. Feb 25, 2017

davidge

Wouldn't these formulas be (since $x = rcos\theta$, $y = rsin\theta$) $r = x/cos\theta$ and $\theta = arccos(x/r)$ and similarly for $y$?

10. Feb 25, 2017

DrGreg

No, $r = x/\cos \theta$ is not in terms of $x$ and $y$, it's in terms of $x$ and $\theta$. You need an expression for $r$ that is independent of $\theta$, and an expression for $\theta$ that is independent of $r$.

The point is that the symbol $\partial/\partial x$ doesn't just mean "differentiate with respect to $x$", it means "differentiate with respect to $x$ while holding $y$ constant". If you do that, neither $r$ nor $\theta$ will remain constant.

11. Feb 25, 2017

Staff: Mentor

As DrGreg pointed out, you need a formula for $r$ in terms of $x$ and $y$ (no $\theta$) and a formula for $\theta$ in terms of $x$ and $y$ (no $r$)--as I have already said repeatedly. Have you looked up the coordinate transformation equations between Cartesian and polar coordinates on a plane, as I have already suggested? This information is easily found online. As I've said before, I think you need to stop trying to guess answers and do things by boring brute force. That means looking up things so you are sure you have the right equations at each step.

12. Feb 25, 2017

davidge

Thanks @DrGreg and @PeterDonis. This is too easy, I do not know how I did not think of it before... $r$ is just the radial distance, so it's equal to $\sqrt{x^2+y^2}$ and $\theta$ is just the angle from the $x$-axis, so it's equal to $arctan(y/x)$.

$$\xi'^{1}_{(1)} = \frac{\partial r}{\partial x} = cos\theta \\ \xi'^{2}_{(1)} = \frac{\partial \theta}{\partial x} = -\frac{sin\theta}{r} \\ ...$$

13. Feb 25, 2017

Staff: Mentor

Yes.

14. Feb 26, 2017

davidge

Just one more question: Can these components of the Translation Killing Vectors be derived directly from Killing's equation? If so, why did I find only the components of the Rotation Killing Vector by solving the Killing equation (and I'm sure the software I used correctly solved that partial differential equation)?

15. Feb 26, 2017

vanhees71

I think we had this question just a while ago concerning 4D Minkowski space. As is a priori clear to find all Killing vectors for flat spaces it's most simple to work in Cartesian coordinates. For the Euclidean $\mathbb{R}^2$ then $g_{jk}=\delta_{jk}$ and $\nabla_j=\partial_j$. The Killing equation thus simply reads
$$\partial_j \xi_k+\partial_k \xi_j=0.$$
The trick is to take one more derivative,
$$\partial_a \partial_j \xi_k + \partial_a \partial_k \xi_j=0.$$
Then write down the two other equations following from cyclically permuting the indices,
$$\partial_j \partial_k \xi_a + \partial_j \partial_a \xi_k=0,\\ \partial_k \partial_a \xi_j + \partial_k \partial_j \xi_a=0.$$
Now add the first two equations and subtract the third, which leads to
$$\partial_a \partial_k \xi_j=0,$$
i.e., all 2nd derivatives of $\xi_j$ vanish, which means that $\xi_j$ must be a linear function of the coordinates,
$$\xi_j=a_j+ \omega_{jk} \xi_k.$$
Now we have to check the original 1st order equation, since the 2nd order equations as more solutions than the 1st-order one. This leads to
$$\partial_a \xi_j + \partial_j \xi_a=\omega_{ja}+\omega_{aj}=0,$$
which means that $\omega_{jk}=-\omega_{kj}$.

So you have three independent killing-vector fields. The constant ones refer to the translations, and the $\xi_j=\epsilon_{jk} x_k$ (where $\epsilon_{jk}$ is the Levi-Civita tensor with $\epsilon_{12}=-\epsilon_{21}=1, \quad \epsilon_{11}=\epsilon_{22}=0$ refers to the rotations. The Killing equation gives you the full isometry group ISO(2) of the Euclidean plane, as it must be.

16. Feb 26, 2017

davidge

yea @vanhees71 . I saw the thread where you explain it to the opener. I understood the solution you presented for the Minkowski metric. What I would like to know is how to find it when the metric is not Minkowskian, e.g. using polar coordinates. One way is as @PeterDonis have shown, that is by converting the Cartesian components to Polar components. Is there a way to find them out directly from the Killing Equation?

17. Feb 26, 2017

Staff: Mentor

I think you mean "when the coordinates are not Cartesian", not "when the metric is not Minkowskian". First, the metric, meaning the actual geometry of the manifold, is independent of the choice of coordinates. Second, a 2-D plane does not have a Minkowskian metric (geometry), it has a Euclidean metric (geometry).

That said, the general method vanhees71 describes in post #15 (take one more derivative of Killing's equation) works in any coordinates, not just Cartesian. It looks simpler in Cartesian coordinates on the 2-D plane because there are fewer terms to evaluate. But you can do it in polar coordinates on the 2-D plane.

18. Feb 26, 2017

davidge

Ah, ok
In this case the analogue step of post #15 would be "covariant" differentiate once more the Killing equation? Like
$$\nabla_{\sigma}\nabla_{\mu}\xi_{\nu} + \nabla_{\sigma}\nabla_{\nu}\xi_{\mu} = 0$$

19. Feb 26, 2017

Staff: Mentor

Yes.

20. Feb 26, 2017

davidge

I did it.
$$\nabla_{\sigma}\nabla_{\mu}\xi_{\nu} + \nabla_{\sigma}\nabla_{\nu}\xi_{\mu} =$$
$$\xi_{\nu,\mu,\sigma} + \xi_{\mu,\nu,\sigma} - \xi_{\rho,\mu}\Gamma^{\rho}_{\ \sigma \nu} - \xi_{\rho,\nu}\Gamma^{\rho}_{\ \sigma \mu} - 2(\xi_{\rho,\sigma}\Gamma^{\rho}_{\ \mu \nu} + \xi_{\rho}\Gamma^{\rho}_{\ \mu \nu,\sigma}) + \xi_{\kappa}(\Gamma^{\kappa}_{\ \mu \rho}\Gamma^{\rho}_{\ \sigma \nu} + \Gamma^{\kappa}_{\ \nu \rho} \Gamma^{\rho}_{\ \sigma \mu}) = 0$$

Is this result correct?