'Theta function' setting conditions similar to delta function?


by Mithra
Tags: heaviside function, theta function
Mithra
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#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!
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tiny-tim
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#2
Nov11-12, 08:43 AM
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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
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#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.


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