Exploring Vacuum Energy Calculations

In summary, the vacuum energy can be written as:\int\frac{d^3k}{(2\pi)^3}(\frac{\hbar \omega_k}{2}+\log(1+e^{-\beta\omega_k})) or\int\frac{d^4k}{(2\pi)^4}log(k^2+m^2)
  • #1
robousy
334
1
Hey folks,

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

[tex]\int\frac{d^3k}{(2\pi)^3}(k^2+m^2)[/tex]

and sometimes written:

[tex]\int\frac{d^4k}{(2\pi)^4}log(k^2+m^2)[/tex]

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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  • #2
Well, normally the vacuum energy is just a sum of zero point modes, of the form

[tex]\int\!\frac{d^3k}{(2\pi)^3} \frac{\hbar \omega_k}{2} \,[/tex]

I'm not sure what to make of the dimensionality of the expressions that you wrote down.
 
  • #3
natural units used, and square root forgotten in first eqtn.

:)
 
  • #4
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.,
[tex]
\log(Z)=\frac{-1}{2}\sum_n\sum_{\bf k}\log((\omega_n^2+k^2+m^2)^2/T^2)\;.
[/tex]
You can use the trick
[tex]
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)
[/tex]
(beta=1/T)
to show
[tex]
\log(Z)=V\int\frac{d^3k}{(2\pi)^2}\left(\frac{-\beta\omega_k}{2}-\log(1-e^{-\beta\omega_k})\right)
[/tex]
and then take T to zero to see.
[tex]
E_0=V\int\frac{d^3 k}{(2\pi)^3}\omega_k
[/tex]
(cf. Kapusta "Finite Temperature Field Theory" chap. 2).
 
  • #5
Hmmm, that's 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 [itex]-\beta\omega_k[/itex] diverge and the exponential term just go to 1 leeding to another divergence in the log term??
 
  • #6
robousy said:
Hmmm, that's 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 [itex]-\beta\omega_k[/itex] 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.
 
  • #7
Ahh, that makes sense. Thanks so much for your help!

Richard
 
  • #8
Actually, I'm working through this in detail and I'm still wondering where the [itex]d^4k[/itex] comes from. From Olganrappy's result we see

[tex]E_0=V\int\frac{d^3 k}{(2\pi)^3}\omega_k[/tex]

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

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

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:

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

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

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

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:
  • #9
well... for one thing, it looks like they are working in 5 dimensions not 4 as is typical.
 
  • #10
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
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. :)
 

What is vacuum energy?

Vacuum energy, also known as zero-point energy, is the lowest possible energy that a quantum mechanical physical system may have. It refers to the energy of empty space, even in the absence of any matter or radiation.

Why is vacuum energy important?

Vacuum energy plays a crucial role in many areas of physics, including quantum field theory, cosmology, and the study of the fundamental forces of nature. It is also believed to have contributed to the expansion of the universe during the Big Bang.

How is vacuum energy calculated?

Vacuum energy is typically calculated using quantum field theory, which uses mathematical tools to describe the behavior of quantum particles in space. This involves summing up the zero-point energy contributions of all the virtual particles that can exist in a given space.

What is the significance of exploring vacuum energy calculations?

Exploring vacuum energy calculations can help us better understand the fundamental nature of the universe and the behavior of particles at the quantum level. It can also have practical applications in areas such as energy production and technology.

Are there any practical applications of vacuum energy?

While vacuum energy is still a theoretical concept, there have been some efforts to harness it for practical use. Some proposed applications include using it as a source of clean energy and for propulsion in space travel. However, more research and development is needed in this area.

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