Solution to Killing equations

  • Thread starter PhyPsy
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  • #1
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Homework Statement


Find all Killing vector solutions of the metric
[tex]g_x{_x}=x^2, g_x{_y}=g_y{_x}=0, g_y{_y}=x[/tex]
where [itex](x^a)=(x^0, x^1)=(x, y)[/itex]


Homework Equations


Killing equations:
[tex]L_Xg_a{_b} = X^e\partial_eg_a{_b}+g_a{_d}\partial_bX^d+g_b{_d}{\partial}_aX^d = 0[/tex]

The Attempt at a Solution


[tex]L_Xg_x{_x} = X^a+x\partial_xX^a=0[/tex]
[tex]L_Xg_x{_y}= L_Xg_y{_x}=x\partial_yX^a+\partial_xX^a=0[/tex]
[tex]L_Xg_y{_y}=X^a+2x\partial_yX^a=0[/tex]
In the back of the book, it says the solution is [itex]\frac{\partial}{{\partial}y}[/itex]. I don't really know what they mean by that. I've always seen [itex]\frac{\partial}{{\partial}y}[/itex] as an operation that takes the partial derivative of something with respect to y, not a value. I thought perhaps it meant any function that depends only on y and not on x, but if I plug that into the equations above, I get:
[tex]f(y)\neq0[/tex]
[tex]x\partial_yf\neq0[/tex]
[tex]f(y)+2x\partial_yf\neq0[/tex]
I'm pretty sure I've got the equations correct; I just don't know what they mean when they say the solution is [itex]\frac{\partial}{{\partial}y}[/itex].
 

Answers and Replies

  • #2
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The equations in the first post are wrong. The equations should be:[tex]L_Xg_x{_x}=2xX^x+2x^2\partial_xX^x=0[/tex][tex]L_Xg_x{_y}=L_Xg_y{_x}=x^2\partial_yX^x+x\partial_xX^y=0[/tex][tex]L_Xg_y{_y}=X^x+2x\partial_yX^y=0[/tex]
The first equation restricts you to [itex]X^x=\frac{1}{x}f(y)[/itex], and anything I do with the second and third equations, I come up with [itex]-\frac{1}{2x^3}f(y)=\partial_y\partial_yf(y)[/itex]. Obviously, there is no solution to this except [itex]f(y)=0[/itex], so I'm still finding the only possible solution to be [itex]X^x=0, X^y=c[/itex].

I found out that [itex]\frac{\partial}{{\partial}y}[/itex] indeed means any function of y. I'm not sure why they would offer that as a solution when there are 2 vectors I am solving for. Can anyone figure out why [itex]\frac{\partial}{{\partial}y}[/itex] is a solution in the back of the book?
 
Last edited:
  • #3
Dick
Science Advisor
Homework Helper
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d/dx and d/dy are a basis for the space of tangent vectors. So a general vector is written as X^x*d/dx+X^y*d/dy. Saying d/dy is a solution is just saying X^x=0 and X^y=1 is a solution. And it is, isn't it? You've already shown that. If you think of what a Killing vector means as an isometry of the metric, then d/dy corresponds to translation in the y direction. And it's pretty obvious that's an isometry, yes?
 
Last edited:

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