Vacuum modes

1. Apr 19, 2008

robousy

Hey folks,

Sometimes I see the calculation of the vacuum energy written as:

$$\int\frac{d^3k}{(2\pi)^3}(k^2+m^2)$$

and sometimes written:

$$\int\frac{d^4k}{(2\pi)^4}log(k^2+m^2)$$

See for example http://arxiv.org/abs/hep-ph/0105021 equation9.

Does anyone know why you can increase the k integral power by one and why this introduces a log???

Thanks!!

Last edited: Apr 19, 2008
2. Apr 19, 2008

lbrits

Well, normally the vacuum energy is just a sum of zero point modes, of the form

$$\int\!\frac{d^3k}{(2\pi)^3} \frac{\hbar \omega_k}{2} \,$$

I'm not sure what to make of the dimensionality of the expressions that you wrote down.

3. Apr 19, 2008

robousy

natural units used, and square root forgotten in first eqtn.

:)

4. Apr 19, 2008

olgranpappy

you are also missing the sum the have in their equation.

It looks like they are calculating the log of the partition function (which gives the free energy divided by the temperature). I.e.,
$$\log(Z)=\frac{-1}{2}\sum_n\sum_{\bf k}\log((\omega_n^2+k^2+m^2)^2/T^2)\;.$$
You can use the trick
$$log((2\pi n)^2+\beta^2\omega_k^2)=\left(\int_1^{\beta^2\omega_k^2}\frac{dx}{x+(2\pi n)^2}\right)+\log(1+(2\pi n)^2)$$
(beta=1/T)
to show
$$\log(Z)=V\int\frac{d^3k}{(2\pi)^2}\left(\frac{-\beta\omega_k}{2}-\log(1-e^{-\beta\omega_k})\right)$$
and then take T to zero to see.
$$E_0=V\int\frac{d^3 k}{(2\pi)^3}\omega_k$$
(cf. Kapusta "Finite Temperature Field Theory" chap. 2).

5. Apr 21, 2008

robousy

Hmmm, thats pretty darn helpful. I hadn't thought of things in terms of statistical mechanics olgranpappy so thanks a lot of the insight.

I'm just working through your mathematics. Sorry if this is a retarded question, but if you take T to zero then won't the term $-\beta\omega_k$ diverge and the exponential term just go to 1 leeding to another divergence in the log term??

6. Apr 21, 2008

olgranpappy

there's a beta on the LHS as well (log(Z)=-\beta F)
the exponential term e^(-\beta \omega) goes to zero since beta is going to infinity (T to zero) and then log(1+0)=0.

7. Apr 21, 2008

robousy

Ahh, that makes sense. Thanks so much for your help!

Richard

8. Apr 22, 2008

robousy

Actually, I'm working through this in detail and I'm still wondering where the $d^4k$ comes from. From Olganrappy's result we see

$$E_0=V\int\frac{d^3 k}{(2\pi)^3}\omega_k$$

which is different to the result from the paper http://arxiv.org/abs/hep-ph/0105021 equation9

$$V=\frac{1}{2}\sum_{n=-\infty}^\infty \int\frac{d^4k}{(2\pi)^4}log(k^2+\frac{\pi^2n^2)}{L^2}$$.

I understand that the 'n' term is a consequence of Kaluza Klein excitations around the extra dimension and appreciate the, its just the 4th power of k and the log that is giving me trouble. I know there is an extra dimension, but why the log?

Weinberg for example (http://ccdb4fs.kek.jp/cgi-bin/img/allpdf?198809190 [Broken]) calculates the vacuum energy (eq 3.5) as:

$$<\rho>=\frac{4\pi}{2(2\pi)^3}\int k^2 dk \sqrt{k^2+m^2}$$

Alternatively Gupta (http://arxiv.org/abs/hep-th/0210069) in eqn. 3 gives the vacuum energy as:

$$<E_{vac}>=\frac{1}{2}\int\frac{d^3k}{(2\pi)^3}\sqrt{k^2+m^2}$$

As you can see, three authors give three different expressions for the vacuum energy, each integrated over different powers of 'k'. As you can imagine I find this quite confusing and am trying to find some principle or reasoning to try and understand why the integrals all a little different. Is it math, or is it in the physics?? I think its in the dimensionality of the spacetime model, but just wanted to check.

Thanks!

:)

Last edited by a moderator: May 3, 2017
9. Apr 22, 2008

olgranpappy

well... for one thing, it looks like they are working in 5 dimensions not 4 as is typical.

10. Apr 22, 2008

Haelfix

Just inspecting the units, implies the equations presented are not the same thing.

1) There is a cutoff dependance, so choice of regularization is important. The exact value will differ between regularizations by one loop, depending on methods.
2) Some of those are energy densities, others are not!

Weinberg's and Gupta's expression differ by a choice of normalization.

11. Apr 23, 2008

robousy

Hmmm, ok. I've found a paper today on finite temperature quantum field theory which explains why its d^4k. Seems like there are a number of regularization schemes. This is making my head spin. :)