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A Flat s-t 4d killing vectors via solving killing equation

  1. Feb 12, 2017 #1
    So I know what these are
    4 translation : ##\frac{\partial}{\partial_ x^{u}} = \partial_{x^u}##
    3 boost: ##z\partial_y - y \partial_z## and similar for ##x,z## and ## y,x##
    3 rotation: ##t\partial_x + x\partial_t ## and similar for ##y , z##

    however I want to do it by solving Killing equation:

    ##\nabla_u V^v + \nabla_v V^u =0 ##
    So in flat space, these ##\nabla_u## reduce to partial derivatives ##\partial_u##

    So Killing equation reduces to : ##\partial_u V^v + \partial_v V^u=0##

    Without writing things out explicitly, e.g the time translation ##\partial x^0 = (1,0,0,0) ## I am confused how to work in index notation. To begin, the translations ##\partial_{x^u}## are covector and not vector, the killing equation works in vector, so rather do I need ##\frac{\partial}{\partial x_u}## instead of ##\frac{\partial}{\partial x^u}##, I don't know what this is explictly?

    Further I am confused with the indices in the boosts and the rotations, so the translations are given as covectors, which we can raise an index to get a vector but isn't something like:

    ##z\partial_y - y\partial_z## a covector multiplied by a vector and so not a vector but a scalar, since ##x^u=x,y,z,t## is a vector but ##\partial_x^u ## is a covector.

    Anyway once I've cleared these up the HINT is to differentiate Killing equation and then solve the ODE.
    Should I do ##\partial_u## or should I choose a different index not already in Killing equation. Does it matter? I dont see how we can convert this PDE itno an ODE since it already has ##\partial_u## and ##\partial_v##, if I hit it with ##\partial_u## I get a ##\partial^2_u## but then also the mixed term ##\partial^2_uv##

    Many thanks
     
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  3. Feb 12, 2017 #2

    stevendaryl

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    That's not the Killing equation. The Killing equation is:

    (with lowered indices on the vector field [itex]V[/itex]). The Killing equation is a relationship between the vector field [itex]V^u[/itex] and the metric tensor, [itex]g_{uv}[/itex]. The equation you wrote down doesn't involve the metric tensor at all.

    The equation I wrote down doesn't appear to involve the metric tensor, either, but it does, because:

    [itex]V_v \equiv g_{av} V^a[/itex]

    So the simple-looking equation
    [itex]\nabla_u V_v + \nabla_v V_u =0 [/itex]

    really means something a little more complicated:

    [itex]\nabla_u (g_{av}V^a) + \nabla_v (g_{au} V^a) =0 [/itex]

    [itex]= (\nabla_u g_{av})V^a + g_{av} \nabla_u V^a + (\nabla_v g_{au}) V^a + g_{au} \nabla_v V^a =0 [/itex]
     
  4. Feb 12, 2017 #3
    hmmm okay, ta.

    however, the fundamental theorem of riemann geometry is that you can write the connection, levita cevita tensor in terms of the metric, so therefore via the covariant derivative i do have dependence on the metric. this is consistent with the last equation you've mentioned, meaning something more complicated, because from the fundamental theorem of riemann geometry the fact arises that ##\nabla_u g_av = 0## enabling you to move the metric outside the expression in the bottom line - so you have here chosen to use of the levita cevita tensor as the connection.

    However even with this , dependence on the metric via the fundamental theorem, it is a relationship between a covector and not a vector so that part of your point i still see.
     
    Last edited: Feb 12, 2017
  5. Feb 12, 2017 #4

    Anyway isnt

    ##z\partial_y - y\partial_z## rank ##(1,1)## object whereas we need a ##(0,1)##?
     
  6. Feb 12, 2017 #5

    stevendaryl

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    I'm not sure about what it is that you are doing. Those 10 operations are symmetries of Minkowsky spacetime, but the relevant quantity is a Killing vector field.

    A vector field is a vector [itex]V^\mu(\mathcal{P})[/itex] defined at every point in spacetime. If we choose inertial coordinates (so the connection coefficients are all zero), then the Killing equation just becomes:

    [itex]\partial_\mu V_\nu + \partial_\nu V_\mu = 0[/itex]

    It's obvious that any constant vector field will satisfy that equation. But as you suggest, you can come up with other solutions by taking a second derivative:

    [itex]\partial^2_\mu V_\nu + \partial_\mu \partial_\nu V_\mu = 0[/itex]

    We know that [itex]\partial_\mu V_\mu = 0[/itex] (because that's the [itex]\mu - \mu[/itex] Killing equation). So since partials commute, we have:

    [itex]\partial^2_\mu V_\nu = 0[/itex]

    That's a pretty easy differential equation to solve.
     
  7. Feb 12, 2017 #6

    PeterDonis

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    You have these backwards; what you are calling boosts are actually rotations, and what you are calling rotations are actually boosts.
     
  8. Feb 12, 2017 #7

    PeterDonis

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    No, it's a vector field--a (1, 0) object--whose components depend on the coordinates.
     
  9. Feb 12, 2017 #8
    apologies, indeeed i do.
     
  10. Feb 13, 2017 #9
    apologies I was thinking ##\frac{\partial}{\partial x^u}## was a covector and not a vector. So vector multiplied by a vector gives a vector, and if it was vector multiplied by covector it would be rank ##(1,1)## right?
     
  11. Feb 13, 2017 #10
    Okay thank you, I understand the steps taken to reduce the Killing equation to [itex]\partial^2_\mu V_\nu = 0[/itex]
    however I'm unsure how to get started, explicitly, the indices are throwing me, I havent done any integration/ differential equations which involve indices.

    To try and see things explicitly I wanted to verify that a rotation, for example, is a solution.
    So I have

    ##V^{a}=x^{3}\partial x^1 - \partial x^1 x^3 ##, ##x^{3}=z, x^1=x, x^0=t ..## etc
    And Killing equation as
    ##\partial^2_u g_{av} V^a = 0 ##

    I dont know how to even lower the index on ##V^{a}## written as above as it looks to explicit. I have a general form of all rotations that would probably be easier to work with, is this the expected outcome when you solve the differential equation (for rotations)? the general form that covers all rotations , rather than the more explicit form, thanks
     
  12. Feb 13, 2017 #11

    stevendaryl

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    Well, [itex]V_\nu[/itex] is just a set of 4 (or however many dimensions there are to your space) functions. So this equation is just like the 16 differential equations:

    [itex]\frac{\partial^2}{\partial x^2} V_0(x,y,z,t) = 0[/itex]
    [itex]\frac{\partial^2}{\partial y^2} V_0(x,y,z,t) = 0[/itex]
    [itex]\frac{\partial^2}{\partial z^2} V_0(x,y,z,t) = 0[/itex]
    [itex]\frac{\partial^2}{\partial t^2} V_0(x,y,z,t) = 0[/itex]
    ...
    (same for [itex]V_1, V_2, V_3[/itex])

    [itex]V_0[/itex] is just a function, and the above is just 4 differential equations involving [itex]V_0[/itex]. What's the solution for [itex]V_0[/itex]?
     
  13. Feb 13, 2017 #12

    stevendaryl

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    [itex]V_a[/itex] is defined by:

    [itex]V_a \equiv \sum_b g_{ab} V^b[/itex]

    So if your metric is
    • [itex]g_{ab} = +1[/itex] when [itex]a = b = 0[/itex]
    • [itex]g_{ab} = -1[/itex] when [itex]a = b = 1[/itex] or [itex]a = b = 2[/itex] or [itex]a = b = 3[/itex]
    • [itex]g_{ab} = 0[/itex] when [itex]a \neq b[/itex]
    Then [itex]V_0 = V^0[/itex]
    [itex]V_i = -V^i[/itex] (for [itex]i=1,2,3[/itex])
     
  14. Feb 13, 2017 #13

    PeterDonis

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    No. A vector multiplied by a scalar gives a vector. You can't "multiply" a vector by another vector (at least, not without additional structure on your vector space); but you can form a bivector, which is a rank ##(2, 0)## tensor.

    As above, this is not really "multiplication", but the analogue of a bivector with a vector and a covector gives a rank ##(1, 1)## tensor, yes.
     
  15. Feb 13, 2017 #14
    So then aren't the rotations in the form ##z\partial x - x \partial z## a vector multiplied by a vector a ##(2,0)## not a ##(1,0)##?

    [Moderator's note: edited to correct formatting of quote.]
     
  16. Feb 13, 2017 #15

    PeterDonis

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    No. The Killing vector ##z \partial_x - x \partial_z## is a vector whose components depend on the coordinates, as I said before. The ##z## and ##x## coefficients are functions of the coordinates, not components of a vector.
     
  17. Feb 13, 2017 #16
    Okay thanks, 16 differential equations allows me to see things clearer.
    Solving the equations for ##V_0##..

    the first one for example, on the first integration I get ##c(y,z,t) ##, a constant that wrt ##x## depending on ##y,z,t##, and then on the second integration I get ##x c(y,z,t)## , similarly for the other three equations.

    So I have from each equation:
    ##V_0 = x c(y,z,t)##
    ##V_0 = y c(x,z,t)##
    ##V_0 = z c(y,x,t)##
    ##V_0 = t c(y,z,x)##

    well the constant ##c## shouldn't be the same, but anyway, I now need to solve these simulatenously? or have I done something wrong? no idea how to solve simultaneously? thanks
     
  18. Feb 13, 2017 #17
    ah yes apologies thank you, individual components of a vector are scalar
     
  19. Feb 13, 2017 #18

    stevendaryl

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    Well, you haven't chosen the most general solution. If

    [itex](\partial_x)^2 V_0 = 0[/itex], then [itex]V_0 = a(y,z,t) + c(y, z, t) x[/itex]

    Now, plug that answer into the next differential equation:
    [itex](\partial_y)^2 V_0 = 0 \Rightarrow (\partial_y)^2 (a(y,z,t) + c(y, z, t) x) = 0[/itex]

    So solve that for [itex]a(y,z,t)[/itex] and [itex]c(y,z,t)[/itex]. Etc.

    We can skip to the answer, unless you want to work it out yourself. Do you?
     
  20. Feb 13, 2017 #19
    Okay so doing that I get ##a''=\frac{-c''}{x}## where ##c' = \partial c / \partial y ##, if I now integrate this to get ##c## in terms of ##a## I end up introducing to more constants that now depend on ##x,z,t## instead, so this cant be the right move?
     
  21. Feb 13, 2017 #20

    stevendaryl

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    I don't understand what you're doing, but let's work it out in 2-D space, and maybe you can see how it would generalize:

    If [itex]x, y[/itex] are Cartesian coordinates, then the Killing equation [itex]\nabla_\mu V_\nu + \nabla_\nu V_\mu = 0[/itex]
    becomes 3 equations:
    [itex]\partial_x V_x = 0[/itex]
    [itex]\partial_y V_y = 0[/itex]
    [itex]\partial_x V_y + \partial_y V_x = 0[/itex]

    We take an additional derivative to get an equation for just [itex]V_x[/itex]:

    [itex]\partial^2_x V_x = 0 \Rightarrow V_0 = a(y) + c(y) x[/itex]
    [itex]\partial^2_y V_x = 0 \Rightarrow \partial^2_y [a(y) + c(y) x] = 0 \Rightarrow a(y) = b + dy, c(y) = e + fy[/itex]

    So [itex]V_x = b + dy + e x + f x y[/itex] where [itex]b, d, e, f[/itex] are constants.

    Similarly, [itex]V_y = g + hy + i x + j x y[/itex] where [itex]g, h, i, j[/itex] are constants.

    Now, plug these back into the original 3 Killing equations, to get:
    [itex]\partial_x V_x = 0 \Rightarrow e = f = 0[/itex]
    [itex]\partial_y V_y = 0 \Rightarrow h = j = 0[/itex]

    [itex]\partial_x V_y + \partial_y V_x = 0 \Rightarrow i = -d[/itex]

    So the solution in 2D is: [itex]V_x = b + d y[/itex], [itex]V_y = g - d x[/itex]
     
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