# Vacuum modes

## Main Question or Discussion Point

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!!

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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.

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

:)

olgranpappy
Homework Helper
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).

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??

olgranpappy
Homework Helper
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
there's a beta on the LHS as well (log(Z)=-\beta F)
and the exponential term just go to 1 leeding to another divergence in the log term??
the exponential term e^(-\beta \omega) goes to zero since beta is going to infinity (T to zero) and then log(1+0)=0.

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

Richard

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!

:)

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

Haelfix