- #1
Kreizhn
- 714
- 1
Hey,
This may seem like a simple question, but hopefully someone can answer it quickly.
Consider the Euclidean 2-metric ds^2 = dx^2 + dy^2. There are three killing fields, two translations
K_1 = \frac{\partial}{\partial x}, \qquad K_2 = \frac{\partial}{\partial y}
and a rotation. Now my issue is this, if K_3 is the rotational Killing field, with coordinate decomposition
K_3 = K_3^x \frac{\partial}{\partial x} + K_3^y \frac{\partial}{\partial y}
We can show that K_3's components must satisfy the differential question
\frac{\partial K^y}{\partial x} + \frac{\partial K^x}{\partial y} = 0
Hence we can have two solutions (K^x,K^y) = (-y,x) and (K^x,K^y) = (y, -x). Clearly these are just a multiple of -1 different. But if one were to plot these, they would give rotations in opposite directions. One would move clockwise, the other counter-clockwise.
Do we care about the orientation of the field? Or just it's general flow? Do these both represent the same field all the same?
This may seem like a simple question, but hopefully someone can answer it quickly.
Consider the Euclidean 2-metric ds^2 = dx^2 + dy^2. There are three killing fields, two translations
K_1 = \frac{\partial}{\partial x}, \qquad K_2 = \frac{\partial}{\partial y}
and a rotation. Now my issue is this, if K_3 is the rotational Killing field, with coordinate decomposition
K_3 = K_3^x \frac{\partial}{\partial x} + K_3^y \frac{\partial}{\partial y}
We can show that K_3's components must satisfy the differential question
\frac{\partial K^y}{\partial x} + \frac{\partial K^x}{\partial y} = 0
Hence we can have two solutions (K^x,K^y) = (-y,x) and (K^x,K^y) = (y, -x). Clearly these are just a multiple of -1 different. But if one were to plot these, they would give rotations in opposite directions. One would move clockwise, the other counter-clockwise.
Do we care about the orientation of the field? Or just it's general flow? Do these both represent the same field all the same?