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Finding a function whose double covariant derivative is delta function

+ 2 like - 0 dislike
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I have a function $f$ defined as $\frac{\overline{z}-\overline{w}}{(z-w)(1 + z\overline{z})(1+w\overline{w})}$.

I want to prove that $$\partial_{\overline{z}}^2 f + \frac{2z}{1+z\overline{z}}\partial_{\overline{z}}f = \pi \frac{2}{(1+ z\overline{z})^2}\delta(z-w).$$

This is basically a double covariant derivative wrt to $\overline{z}$ taking the metric in unit sphere in stereographic coordinates.

I found that $\partial_{\overline{z}}f = \delta(z-w)\frac{(1+z\overline{z})(\overline{z}-\overline{w})}{1+w\overline{w}} + \frac{1}{z-w}\frac{(1+2z\overline{z}-z\overline{w})}{1+w\overline{w}}$.

If I differentiate again I am encountering a derievative of a delta function which I do not know how to do.

But since I know the result I calculated back what the derivative of delta function should be a found out to be $\frac{2\delta(z-w)}{\overline{z}-\overline{w}}\frac{w\overline{w}-z\overline{z}}{1+z\overline{z}}$.

Basically during the calculation I used the product rule and took $\frac{1}{z-w}$ as one function and rest as another and did the diffrentiation.I also used the fact that $\partial_{\overline{z}}\frac{1}{z-w}=\delta(z-w)$.

I would like to know whether the procedure I have done is correct.Is the derivative of the delta function I have mentioned is correct or not.Pls help me.

This problem is basically involved during computing the BMS charge

This post imported from StackExchange Mathematics at 2017-05-05 20:53 (UTC), posted by SE-user Anupam Ah
asked Apr 30 in Mathematics by Anupam Ah (10 points) [ no revision ]
Notice that the $\delta$-function is being multiplied by ${\bar z} - {\bar w}$ so it is zero.

This post imported from StackExchange Mathematics at 2017-05-05 20:53 (UTC), posted by SE-user Prahar

1 Answer

+ 1 like - 0 dislike

Note

$$ D_{\bar z}^2f = \gamma_{z\bar z} \partial_{\bar z}( \gamma^{z\bar z} \partial_{\bar z} f ) $$ We are solving the equation \begin{align} D_{\bar z}^2f &= \pi \gamma_{z\bar z} \delta^2(z-w)\\ \gamma_{z\bar z} \partial_{\bar z}( \gamma^{z\bar z} \partial_{\bar z} f ) &= \pi \gamma_{z\bar z} \delta^2(z-w) \\ \partial_{\bar z}( \gamma^{z\bar z} \partial_{\bar z} f ) &= \frac{1}{2} \partial_{\bar z} \frac{1}{z-w} \\ \gamma^{z\bar z} \partial_{\bar z} f &= \frac{1}{2} \frac{1}{z-w} + g ( z,w,\bar w)\\ \partial_{\bar z} f &= \frac{1}{2} \frac{\gamma_{z\bar z} }{z-w} + \gamma_{z\bar z} g( z,w,\bar w) \end{align}

Note that we have

$$ f = \frac{ \bar z - \bar w}{ (z-w)(1+z\bar z)(1+w\bar w)} $$

so that

$$ \partial_{\bar z} f = \frac{ 2\pi \delta^2 ( z - w )( \bar z - \bar w )}{ (1+z\bar z)(1+w\bar w)} + \frac{1}{2} \frac{\gamma_{z\bar z} }{z-w} + \frac{1}{2} \gamma_{z\bar z} \frac{ \bar w }{ ( 1 + w \bar w ) } $$ The first term is zero. The rest of this is precisely the of the required form.

This post imported from StackExchange Mathematics at 2017-05-05 20:53 (UTC), posted by SE-user Prahar
answered Apr 30 by prahar21 (535 points) [ no revision ]
thank you prahar.This was exactly what I was looking for.i was confused when the dereivative of the $\delta$ came.But I didnt notice the fact that it is with factor of $\overline{z-w}$ so it will be zero.I guess the reason why this should be zero is when you are acting that on a test function you you get the integrand as zero right ?

This post imported from StackExchange Mathematics at 2017-05-05 20:53 (UTC), posted by SE-user Anupam Ah
@AnupamAh yep, that's right.

This post imported from StackExchange Mathematics at 2017-05-05 20:53 (UTC), posted by SE-user Prahar
I have another doubt .How do you get the last piece from $f$.How does the $\frac{\overline{w}}{(1+w\overline{w})^2}$

This post imported from StackExchange Mathematics at 2017-05-05 20:53 (UTC), posted by SE-user Anupam Ah
@AnupamAh - There was a typo. I fixed it.

This post imported from StackExchange Mathematics at 2017-05-05 20:53 (UTC), posted by SE-user Prahar
Thanks,I got the exact form.

This post imported from StackExchange Mathematics at 2017-05-05 20:53 (UTC), posted by SE-user Anupam Ah

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