How to handle squares of delta distributions

[Bartolius]

Homework Statement

I have to write equations of motion for a field, namely $A$.
The full action has actually three terms, but my problem is with a part of the action reading:

$$ S =\int d^{10}x \sqrt{-g} [ f(x_1, ... , x_{10}) + \delta (y) A ]^2 $$

In the 10 x's there is of course the coordinate called y. Also I omitted various irrelevant numerical factors.

Homework Equations

My problem is that the product of $\delta$'s , in particular $[\delta (y)]^2$ is not defined for they are distributions and not simple functions, so I don't actually know how to write down my equations.

The Attempt at a Solution

The only idea I have right now is to replace the deltas with something approaching the delta in a certain limit and then take the limit at the end of the calculations, but having done some research on the internet it seems to turn out that the result would be dependent on my choice of the function approaching delta. I can recall having found a similar problem in one of my first courses on QFT, but can't recall how it was solved. However it seems a rather common problem, so I hope that someone here already has a solution and can lead me to find it.

 

[Geofleur]

If you go ahead and vary the functional anyway, does $\delta(x)^2$ show up inside the resulting integral?

 

[Bartolius]

Yes because I have to differentiate on $A$, so there is actually no reason for the $\delta (y)$ to disappear.
Also, I would like to use the $\delta$ whenever I have them to integrate away the y coordinate, for the other terms of the action are 9 dimensional (the field $A$ lives on a 9-dimensional space embedded in a 10 dimensional one)

 

[JorisL]

You will look at the variations $\delta S$

Doesn't this mean you can look at

$\delta S = \int d^{10} x \delta \sqrt{-g}\left[ f(x_1, \ldots, x_{10}) +\delta(y)A\right]^2 +2 \int d^{10} x \sqrt{-g} \delta\left[ f(x_1, \ldots, x_{10}) +\delta(y)A\right]$

Now for the Equations of Motion for the field A you find them by using the functional derivative $\frac{\delta S}{\delta A} = 0$
To me it seems that in that case only a factor $\delta(y)$ remains because of the remark above.

Of course you'll need to flesh out the details.

Another thing I've read a little bit about is that sometimes delta-distributions get replaced by smeared out functions.

 

[Bartolius]

ok, maybe I was doing the calculations too fast, I'll check later, thank you! as for the smeared out function: yes it was the starting point of the calculation, the delta distribution was smeared out on a circle (the directions y and z combine to give a subspace with a circle as a boundary) but most of the papers I am using just use the localized version, so it turns out easier to carry the confrontation between the results I obtain and the ones already found by others.

 

[Bartolius]

You will look at the variations $\delta S$

Doesn't this mean you can look at

$\delta S = \int d^{10} x \delta \sqrt{-g}\left[ f(x_1, \ldots, x_{10}) +\delta(y)A\right]^2 +2 \int d^{10} x \sqrt{-g} \delta\left[ f(x_1, \ldots, x_{10}) +\delta(y)A\right]$

Now for the Equations of Motion for the field A you find them by using the functional derivative $\frac{\delta S}{\delta A} = 0$
To me it seems that in that case only a factor $\delta(y)$ remains because of the remark above.

Didn't you leave a factor in the second part of the variation? shouldn't it be
$$ \delta S = \int d^{10} x \delta \sqrt{-g}\left[ f(x_1, \ldots, x_{10}) +\delta(y)A\right]^2 +2 \int d^{10} x \sqrt{-g} \left[ f(x_1, \ldots, x_{10}) +\delta(y)A\right]\delta\left[ f(x_1, \ldots, x_{10}) +\delta(y)A\right]$$

since it should be $$ \delta(F^2) = 2F\delta F $$

 

[JorisL]

Indeed, sorry for the confusion. I knew I should've grabbed a piece of paper.

Looking at it mathematically, $\delta^2(y)$ seems to entail all kinds of trouble.
I found this older topic https://www.physicsforums.com/threads/dirac-delta-function.207661/
Where the final post might be worth a shot.