q 1 \u03bd 2 \u2113 q 2 \u03bd 1 \u2113 q 1 \u03bd 2 \u2113 q 2 \u03bd 1 \u2113 \u03bd 2 \u2113 q 1 2 m 2 \u2113 q 2 2 m 2 \u2113 2 m 2 4 m

Q 1 ν 2 ℓ q 2 ν 1 ℓ q 1 ν 2 ℓ q 2 ν 1 ℓ

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q 1 ) ν 2 ( + q 2 ) ν 1 ( q 1 ) ν 2 + ( + q 2 ) ν 1 ν 2 [ ( q 1 ) 2 m 2 ][ ( + q 2 ) 2 m 2 ][ 2 m 2 ] = 4 m f e 2 ϵ μ 1 μ 2 ν 1 ν 2 q 1 ν 1 q 2 ν 2 [ ( q 1 ) 2 m 2 ][ ( + q 2 ) 2 m 2 ][ 2 m 2 ] 31
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f ( ie ) 2 Tr ( γ μ 1 S ( q 1 ) γ 5 S ( + q 2 ) γ μ 2 S ( ) ) = i f e 2 Tr ( γ μ 1 ( ̸ − ̸ q 1 + m ) γ 5 ( ̸ + ̸ q 2 + m ) γ μ 2 ( ̸ + m ) ) [ ( q 1 ) 2 m 2 ][ ( + q 2 ) 2 m 2 ][ 2 m 2 ] = i f e 2 Tr ( ( ̸ + ̸ q 2 + m ) γ μ 2 ( ̸ + m ) γ μ 1 ( ̸ − ̸ q 1 + m ) γ 5 ) [ ( q 1 ) 2 m 2 ][ ( + q 2 ) 2 m 2 ][ 2 m 2 ] using Tr ( γ ν 1 γ ν 2 γ ν 3 γ ν 4 γ 5 ) = 4 i ϵ ν 1 ν 2 ν 3 ν 4 = 4 m f e 2 ϵ μ 1 μ 2 ν 1 ν 2 ν 1 ( q 1 ) ν 2 ( + q 2 ) ν 1 ( q 1 ) ν 2 + ( + q 2 ) ν 1 ν 2 [ ( q 1 ) 2 m 2 ][ ( + q 2 ) 2 m 2 ][ 2 m 2 ] = 4 m f e 2 ϵ μ 1 μ 2 ν 1 ν 2 q 1 ν 1 q 2 ν 2 [ ( q 1 ) 2 m 2 ][ ( + q 2 ) 2 m 2 ][ 2 m 2 ] 32
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f ( ie ) 2 Tr ( γ μ 1 S ( q 1 ) γ 5 S ( + q 2 ) γ μ 2 S ( ) ) = i f e 2 Tr ( γ μ 1 ( ̸ − ̸ q 1 + m ) γ 5 ( ̸ + ̸ q 2 + m ) γ μ 2 ( ̸ + m ) ) [ ( q 1 ) 2 m 2 ][ ( + q 2 ) 2 m 2 ][ 2 m 2 ] = i f e 2 Tr ( ( ̸ + ̸ q 2 + m ) γ μ 2 ( ̸ + m ) γ μ 1 ( ̸ − ̸ q 1 + m ) γ 5 ) [ ( q 1 ) 2 m 2 ][ ( + q 2 ) 2 m 2 ][ 2 m 2 ] using Tr ( γ ν 1 γ ν 2 γ ν 3 γ ν 4 γ 5 ) = 4 i ϵ ν 1 ν 2 ν 3 ν 4 = 4 m f e 2 ϵ μ 1 μ 2 ν 1 ν 2 ν 1 ( q 1 ) ν 2 ( + q 2 ) ν 1 ( q 1 ) ν 2 + ( + q 2 ) ν 1 ν 2 [ ( q 1 ) 2 m 2 ][ ( + q 2 ) 2 m 2 ][ 2 m 2 ] = 4 m f e 2 ϵ μ 1 μ 2 ν 1 ν 2 q 1 ν 1 q 2 ν 2 [ ( q 1 ) 2 m 2 ][ ( + q 2 ) 2 m 2 ][ 2 m 2 ] 33
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= 4 m f e 2 ϵ μ 1 μ 2 ν 1 ν 2 q 1 ν 1 q 2 ν 2 [ ( q 1 ) 2 m 2 ][ ( + q 2 ) 2 m 2 ][ 2 m 2 ] this is symmetric under { q 1 , µ 1 } ↔ { q 2 , µ 2 } so the other diagram is the same M μ 1 μ 2 ( q 1 , q 2 ) = 8 m f e 2 ϵ μ 1 μ 2 ν 1 ν 2 q 1 ν 1 q 2 ν 2 [ ( q 1 ) 2 m 2 ][ ( + q 2 ) 2 m 2 ][ 2 m 2 ] d 4 (2 π ) 4 34
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M μ 1 μ 2 ( q 1 , q 2 ) = 8 m f e 2 ϵ μ 1 μ 2 ν 1 ν 2 q 1 ν 1 q 2 ν 2 [ ( q 1 ) 2 m 2 ][ ( + q 2 ) 2 m 2 ][ 2 m 2 ] d 4 (2 π ) 4 = 8 m f e 2 ϵ μ 1 μ 2 ν 1 ν 2 q 1 ν 1 q 2 ν 2 I ( q 1 , q 2 ) I ( q 1 , q 2 ) = 1 [ ( q 1 ) 2 m 2 + ][ ( + q 2 ) 2 m 2 + ][ 2 m 2 + ] d 4 (2 π ) 4 What do we know about I ( q 1 , q 2 ) ? It is Lorentz invariant - depending only on q 2 1 , q 2 2 and ( q 1 q 2 ) = q 1 μ q μ 2 if q 2 1 , q 2 2 and ( q 1 q 2 ) are large, it should scale like 1 /q 2 if q 2 1 , q 2 2 and ( q 1 q 2 ) m 2 it should go to constant over m 2 we can calculate it exactly by using the standard trick to combine denominators 1 ABC = 2 1 0 1 - α 0 1 ( αA + βB + (1 α β ) C ) 3 dβ dα 35
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M μ 1 μ 2 ( q 1 , q 2 ) = 8 m f e 2 ϵ μ 1 μ 2 ν 1 ν 2 q 1 ν 1 q 2 ν 2 [ ( q 1 ) 2 m 2 ][ ( + q 2 ) 2 m 2 ][ 2 m 2 ] d 4 (2 π ) 4 = 8 m f e 2 ϵ μ 1 μ 2 ν 1 ν 2 q 1 ν 1 q 2 ν 2 I ( q 1 , q 2 ) I ( q 1 , q 2 ) = 1 [ ( q 1 ) 2 m 2 + ][ ( + q 2 ) 2 m 2 + ][ 2 m 2 + ] d 4 (2 π ) 4 What do we know about I ( q 1
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