Method of calculating the vacuum energy divergence

  • #1
Haorong Wu
418
90
TL;DR Summary
When calculating the vacuum energy divergence, an author said, "This divergence can be usefully analysed by performing the integral with n continued away from integral values". What does it mean?
In the book, quantum fields in curved space, when calculating the vacuum energy divergence for scalar fields, it reads:
$$\sum_{\mathbf k}\omega/2=(L^2/4\pi)^{(n-1)/2}\frac 1 {\Gamma ((n-1)/2)} \int_0^\infty (k^2+m^2)^{1/2}k^{n-2}dk .$$ This divergence can be usefully analyzed by performing the integral with ##n## continued away from integral values to obtain $$-L^{n-1}2^{-n-1}\pi^{-n/2}m^n \Gamma(-n/2).$$

I could get the answer by letting ##k=m\tan t ## and using the properties of Beta functions and Gamma functions, but I still do not understand what it means by saying "with ##n## continued away from integral values".

Thanks ahead.
 
Physics news on Phys.org
  • #2
Haorong Wu said:
TL;DR Summary: When calculating the vacuum energy divergence, an author said, "This divergence can be usefully analysed by performing the integral with n continued away from integral values". What does it mean?
Just after your quotation from pg. 16 of Birrell & Davies, Quantum field theory in curved space, it states:
"This method of temporarily making divergent quantities finite by continuing the dimension of spacetime away from integer values forms the basis of dimensional regularization (see chapter 6)."
What don't you understand about this explanation? Are you familiar with dimensional regularization?
 
  • #3
Thanks, @renormalize. I mistakenly thought the integral value was somehow related to the integration in the equation.
 
  • Like
Likes renormalize

Similar threads

Replies
1
Views
947
Replies
7
Views
1K
Replies
0
Views
728
Replies
3
Views
956
Replies
16
Views
2K
Back
Top