**Next message:**dinesh: "matrix elements"**Previous message:**Luka Popov: "Symmetric PaVe functions don't cancel out as they should"**In reply to:**Luka Popov: "Symmetric PaVe functions don't cancel out as they should"**Next in thread:**Luka Popov: "Re: Symmetric PaVe functions don't cancel out as they should"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Mail actions:**[ respond to this message ] [ mail a new topic ]

Dear Luka,

so far I don't see a problem with your result. For

res = (PaVe[1, 2, {p10, p12, p20}, {m02, m12, m22}] -

PaVe[1, 2, {p20, p12, p10}, {m02, m22, m12}]) //

PaVeReduce // Simplify

I obtain

((m02 - m12) B0[0, m02, m12] + (-m02 + m22) B0[0, m02,

m22] + (m12 - m22) B0[0, m12, m22])/(p10^2 + (p12 - p20)^2 -

2 p10 (p12 + p20))

Inserting explicit results for B0's

sols = {B0[0, m02, m12] ->

1/(16 Epsilon \[Pi]^4) + (-m02 Log[m02/

m12] + (m02 - m12) (1 + Log[ScaleMu^2/m12]))/(

16 (m02 - m12) \[Pi]^4),

B0[0, m02, m22] ->

1/(16 Epsilon \[Pi]^4) + (-m02 Log[m02/

m22] + (m02 - m22) (1 + Log[ScaleMu^2/m22]))/(

16 (m02 - m22) \[Pi]^4),

B0[0, m12, m22] ->

1/(16 Epsilon \[Pi]^4) + (-m12 Log[m12/

m22] + (m12 - m22) (1 + Log[ScaleMu^2/m22]))/(

16 (m12 - m22) \[Pi]^4)};

via (res/.sols)//PowerExpand//Simplify

I indeed get 0, as it should be. Please let me know, if you get

something else.

Cheers,

Vladyslav

Am 11.06.2015 um 08:37 schrieb Luka Popov:

*> I am a little bit confused with the results I get using PaVeReduce. Naimely, one should expect the Passarino-Veltman function C_{12} to be symmetric with respect to the replacement p1 <-> p2 and m1 <-> m2, according to its definition.
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*>
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*> However, I don't get this result when evaluating it with PaVeReduce:
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*>
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*> PaVe[1, 2, {p10, p12, p20}, {m02, m12, m22}] -
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*> PaVe[1, 2, {p20, p12, p10}, {m02, m22, m12}] // PaVeReduce
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*>
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*> The result of the above line is given by B0 functions and it does not equals zero, even when B0 is exactly calculated and inserted in the result.
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*>
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*> Can you please tell me if I am doing something wrong? Thank you.
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*>
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*> With regards,
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*> Luka Popov
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*>
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