- #1

Sandra Conor

- 9

- 0

##\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2}##

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Sandra Conor
- Start date

- #1

Sandra Conor

- 9

- 0

##\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2}##

- #2

Svein

Science Advisor

- 2,274

- 785

I am thinking polar coordinates - but what does the summation sign stand for?

- #3

Sandra Conor

- 9

- 0

Initially, I want to evaluate this integral in spacetime.

$$\int_{\Sigma} \frac{dydz}{[a_{o}(y^{2}+z^{2})+2f_{o}y+2g_{o}z+c_{o}]^{2}}$$ where $$a_{o}c_{o}-f_{o}^{2}-g_{0}^{2}=\frac{1}{4}.$$

My way is to define

$$y':=y+\frac{f_0}{a_0},\,z':=z+\frac{g_0}{a_0}$$

Then the quadratic becomes $$a_0(y'^2+z'^2)+c_0-\frac{f_0^2+g_0^2}{a_0}=a_0\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg).$$

So now, I will need to evaluate$$\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2}\cdot$$ But this seems complicated. So I am thinking of using polar coordinate and if I do so, it becomes:

$$\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2} = \frac{1}{a_0^2}\int_\Sigma\frac{rd\phi dr}{\bigg(r^2+\tfrac{1}{(2a_0)^2}\bigg)^2}$$

Since this is in reference to spacetime, I am stuck as to how I should continue. If it is about a disk with radius R then I know what to do but for a spacelike 2 surface, I am still wandering.

- #4

Svein

Science Advisor

- 2,274

- 785

- I don't know why, but Stokes' theorem keeps swirling around in my head (https://en.wikipedia.org/wiki/Stokes'_theorem)
- I have a tendency to think complex variables - have you considered that
- If I haven't done a blunder, I think [itex]\frac{d}{dr}\frac{1}{r^{2}+C}=\frac{-2r}{(r^{2}+C)^{2}} [/itex]

- #5

mathman

Science Advisor

- 8,100

- 559

- #6

Sandra Conor

- 9

- 0

Yes, mathman. The summation sign represent a spacelike 2 surface in spacetime. Its mentioned in the third post. Sorry about that.

Share:

- Replies
- 3

- Views
- 660

- Last Post

- Replies
- 0

- Views
- 330

- Last Post

- Replies
- 16

- Views
- 500

- Last Post

- Replies
- 1

- Views
- 199

- Replies
- 5

- Views
- 1K

- Replies
- 6

- Views
- 785

- Last Post

- Replies
- 3

- Views
- 94

- Replies
- 3

- Views
- 119

- Last Post

- Replies
- 8

- Views
- 1K

- Last Post

- Replies
- 4

- Views
- 746