# 'Theta function' setting conditions similar to delta function?

by Mithra
Tags: heaviside function, theta function
 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: $\int^1_0 dz~\theta (s-\frac{4m^2}{z}-\frac{m^2}{1-z})$ And apparently this ensures that s > $\frac{4m^2}{z}+\frac{m^2}{1-z}$ 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 $\delta (q-p-p')$ over p and p' (4-momenta). Does anyone have any advice about what this is and how to include it in the integral? Thanks!
 PF Patron HW Helper Sci Advisor Thanks P: 25,514 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): $\int^{f(x)}_{-\infty} \delta(y) dy$$\int^1_0 dz~\theta (s-\frac{4m^2}{z}-\frac{m^2}{1-z})$ $=\int^1_0 dz\int^{s-\frac{4m^2}{z}-\frac{m^2}{1-z}}_{-\infty} dw~\delta (w)$
P: 16
 Quote by tiny-tim Hi Mithra! this θ is the usual theta function θ(f(x)) (the Heaviside step function) is the area under a delta function (0 before f(x), 1 after f(x)) … so it is the integral of a delta function from minus-infinity to f(x): $\int^{f(x)}_{-\infty} \delta(y) dy$$\int^1_0 dz~\theta (s-\frac{4m^2}{z}-\frac{m^2}{1-z})$ $=\int^1_0 dz\int^{s-\frac{4m^2}{z}-\frac{m^2}{1-z}}_{-\infty} dw~\delta (w)$
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.

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