Register to reply

'Theta function' setting conditions similar to delta function?

Share this thread:
Mithra
#1
Nov11-12, 07:56 AM
P: 16
Hi, I'm reading through a paper and have come across what my tutor described as a 'theta function', however it seems to bear no resemblance to the actual 'theta function' I can find online. In the paper it reads:

[itex]\int^1_0 dz~\theta (s-\frac{4m^2}{z}-\frac{m^2}{1-z}) [/itex]

And apparently this ensures that s > [itex]\frac{4m^2}{z}+\frac{m^2}{1-z}[/itex] when that expression is included in a longer integration over s and z, however I've never come across something like this before. That expression above is obtained integrating

[itex]\delta (q-p-p')[/itex]

over p and p' (4-momenta). Does anyone have any advice about what this is and how to include it in the integral? Thanks!
Phys.Org News Partner Mathematics news on Phys.org
Heat distributions help researchers to understand curved space
Professor quantifies how 'one thing leads to another'
Team announces construction of a formal computer-verified proof of the Kepler conjecture
tiny-tim
#2
Nov11-12, 08:43 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,148
Hi Mithra!

this θ is the usual theta function

θ(f(x)) (the Heaviside step function) is the area under a delta function (0 before f(x), 1 after f(x))
so it is the integral of a delta function from minus-infinity to f(x): [itex]\int^{f(x)}_{-\infty} \delta(y) dy[/itex]
[itex]\int^1_0 dz~\theta (s-\frac{4m^2}{z}-\frac{m^2}{1-z}) [/itex]

[itex]=\int^1_0 dz\int^{s-\frac{4m^2}{z}-\frac{m^2}{1-z}}_{-\infty} dw~\delta (w) [/itex]
Mithra
#3
Nov11-12, 09:17 AM
P: 16
Quote Quote by tiny-tim View Post
Hi Mithra!

this θ is the usual theta function

θ(f(x)) (the Heaviside step function) is the area under a delta function (0 before f(x), 1 after f(x))
so it is the integral of a delta function from minus-infinity to f(x): [itex]\int^{f(x)}_{-\infty} \delta(y) dy[/itex]
[itex]\int^1_0 dz~\theta (s-\frac{4m^2}{z}-\frac{m^2}{1-z}) [/itex]

[itex]=\int^1_0 dz\int^{s-\frac{4m^2}{z}-\frac{m^2}{1-z}}_{-\infty} dw~\delta (w) [/itex]
Ah brilliant, thanks very much! I thought it must be related to the Heaviside function, but could only seem to find Theta function. A little embarrassing having not heard of these as a fourth year physicist but hey ho :P. Cheers.


Register to reply

Related Discussions
Composition of a theta function with function General Math 0
Prove that derivative of the theta function is the dirac delta function Advanced Physics Homework 3
A seeming contrdiction in deriving wave function for delta function potential Quantum Physics 2
What is a similar function i can use to prove divergence/convergence of this function Calculus & Beyond Homework 0
Dirac delta function and Heaviside step function Advanced Physics Homework 2