'Theta function' setting conditions similar to delta function?by Mithra Tags: heaviside function, theta function 

#1
Nov1112, 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}{1z}) [/itex] And apparently this ensures that s > [itex]\frac{4m^2}{z}+\frac{m^2}{1z}[/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 (qpp')[/itex] over p and p' (4momenta). Does anyone have any advice about what this is and how to include it in the integral? Thanks! 



#2
Nov1112, 08:43 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

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 minusinfinity 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}{1z}) [/itex] [itex]=\int^1_0 dz\int^{s\frac{4m^2}{z}\frac{m^2}{1z}}_{\infty} dw~\delta (w) [/itex] 



#3
Nov1112, 09:17 AM

P: 16




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  
[SOLVED] Dirac delta function and Heaviside step function  Advanced Physics Homework  2 