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Thread: Metric transformation

  1. #1
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    Metric transformation

    I'm trying to understand the paper by Hitchin called ''Polygons and gravitons", Polygons and gravitons - INSPIRE-HEP. I'm stuck at page 471. At this point, he does some computations and obtains a metric:
    $$\gamma dz d\bar{z}+\gamma^{-1}\left(\dfrac{2dy}{y}+\bar{\delta}dz \right)\left(\dfrac{2d\bar{y}}{\bar{y}}+\delta d\bar{z} \right)$$
    where
    \gamma=\sum \dfrac{1}{\Delta_i}=\sum \dfrac{1}{\sqrt{(b-b_i)^2+|\bar{z}+a_i|^2}}
    \delta=\sum \dfrac{(b-b_i)-\Delta_i}{\Delta_i(\bar{z}+a_i)}
    and bis defined implicitly by
    $$\prod ((b-b_i)+\Delta_i)=y\bar{y}$$
    We have that $z,y$are complex coordinates. He proceeds by saying:
    If we return to the form of the metric (4.4), and the description of the space of real quadratic polynomials as Euclidean 3-space with the metric given by the discriminant, the we obtain the metric which describes the gravitationals multi-instantons of Gibbons and Hawking:
    $$\gamma d\vec{x} d\vec{x}+\gamma^{-1}(d\tau+ \vec{\omega}d\vec{x})$$
    where \gamma=\sum \frac{1}{|x-x_i|} and curl $\omega=$grad $\gamma$.
    The form of the metric (4.4) mentioned is
    $$\gamma^2(b'^2+a'\bar{a}')+\left(Im\left(\frac{2A  '}{A}-\delta a' \right) \right)^2$$
    The problem I have is that I don't know how to go from the first metric to the metric of Gibbons and Hawking. Is a change of variables? Which one?
    PS: Maybe this helps to understand (4.4),
    $$A\bar{A}=\prod((b-b_i)+\Delta_i)$$
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  2. #2
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    Re: Metric transformation

    Hey spink.

    If there is indeed a change of co-ordinate system (basically a substitution that retains the information involved) then they usually give these things.

    I guess if you give this information and try and do the change of basis/space we can comment on that.
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  3. #3
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    Re: Metric transformation

    Hi chiro,


    I did tried to do the change, but I'm stuck.


    I think that the first logic step is this:
    x_1=-Re(\bar{z}) \hspace{1cm} x_2=-Im(\bar{z}) \hspace{1cm} x_3= b
    x_{1i}=Re(a_i) \hspace{1cm} x_{2i}=Im(a_i) \hspace{1cm} x_{3i}= b_i
    Doing this, we have that then that
    \gamma=\dfrac{1}{\sqrt{(x_1-x_{1i})^2+(x_2-x_{2i})^2+(x_3-x_{3i})}}
    Like we have in Hawking ansatz. We can obtain also this relation dx_3=db=\gamma^{-1}Re\left(\frac{2d\bar{y}}{\bar{y}}+\delta d\bar{z}\right).

    It will remain to know what is \tau in terms of y,z.

    We could equalize

    d\tau +w_1dx_1+w_2dx_2+w_3dx_3=Im\left( (\frac{2d\bar{y}}{\bar{y}}+\delta d\bar{z})\right)

    But I don't know how to continue. I don't know why we obtain the condition that curl \omega=grad \gamma...
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  4. #4
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    Re: Metric transformation

    Unfortunately I haven't had enough training in vector calculus (presently) to answer the question of your last statement.

    Hopefully others can come along and provide more help than myself.
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  5. #5
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    Re: Metric transformation

    Thank you for your interest anyway chiro

    I hope some will help me...
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