Hey ho

@BernardMeurer dude

Everyone failed the test

@0celo7 Sup

Dayum

Call?

Eating with Reb in 10 seconds.

@0celo7 Dammit

Well, I'll be waiting to hear more when you've got ze time

@obe Not especially. But I can give it a shot.

@obe this book will make my library almost complete.

after it I need four more books

five

then I've basically got everything for the next 3 years.

ETA when?

does it ship from Canadia?

wow!

that's terrible because I have work to do tomorrow :p

@obe Numerator of the fraction is $\exp( (2ak)^{-2} )$?

@obe Then it's $\exp( (ak/2)^2 )$?

Yes, that's the sort of trick to use.

Change $d^3k$ to $k^2\, dk \, d\phi \, d(\cos\theta)$

The angular integral just gives you $4\pi$, because your integrand only contains $k^2$

Now you have a one-dimensional integral in $k$, over the real line.

$a$ isn't just a constant?

Okay. So your integral becomes $\frac{4\pi}a \int^\infty_{0} dk \frac{e^{a^2k^2/4}}{1 + (\mu/k)^2}$

in the small $a$ case you can Taylor expand.

In the large...shouldn't that integral explode?

Whoops, wrong limits. Editing rather than reposting

Mod abuse

@0celo7 I actually don't know how the chat works for non-mods. I can do differently if that's unfair somehow.

You can edit messages indefinitely, but we have a 2min limit.

Maybe.

Well, I was within two minutes. Anyway: it's an integral over the half-line because it's over the magnitude of $k$.

@obe So you care about the large-$a$ and small-$a$ limits, but not the tricky middle?

Okay then. The cutoff, you can get from dimensional analysis, is whether $a$ is smaller or larger than $\mu$.

Or is it $1/\mu$.

The cutoff, you can get from dimensional analysis, is whether $a$ is smaller or larger than $1/\mu$.

(since $k$ and $\mu$ have the same unit, and $a$ has the inverse unit)

The numerator is a Gaussian in $k$ with width $1/a$ or thereabouts

and the denominator goes to one for large $k \gg \mu$, blows up for small $k \ll \mu$.

The small-$a$ limit is the big, wide Gaussian. Ignore the denominator's strange behavior for small $k$: you've got an integral over a Gaussian from the middle to infinity. You can do that one, or look it up.

The large-$a$ limit is the same as the small-$\mu$ limit: the integral becomes $\int dk \ k^2 e^{(ak/2)^2}$

Because $(1+(\mu/k)^2)^{-1} = k^2 / (k^2 + \mu^2)$

so ... it isn't

As I said: I'm not especially good at these

That logic seems to suggest that you get just the integral over the Gaussian in both cases. Seems suspect, but plausible.

@obe ?

That appears to be our result. I don't trust it yet.

Oh. Okay then.

Well yes: we still have a $1/a$ outside the integral (in both cases)

Lessons learned:

$d^3k = k^2 dk\ d\phi\ d(\cos\theta)$ (a great and valuable trick)

check units

What was the physical motivation that led to this math problem?

@rob Give me life advice

@BernardMeurer What's your local time?

@rob UTC+1 Lisbon

00:29

@BernardMeurer Go to sleep

@rob lol

@DanielSank Tells me that everyday already

@obe Where $\phi(x,0)$ is ... ?

Okay. So I guess $\phi$ is some scalar field, with particle creation operator $a_k^\dagger$?

@obe

Fun problem. G'night, all.

For next semester

More advnced

But I'll read it instead of the class book