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The book starts with the equation

\begin{equation} I(D) = \lambda^2 \int \frac{d^Dp_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D} \frac{1}{\mathbf{p}_1^2 + m^2} \frac{1}{\mathbf{p}_2^2+m^2} \frac{1}{(\mathbf{p}_1+\mathbf{p}_2 + \mathbf{q})^2+m^2}\vphantom{\dfrac{a}{b}} \tag{8.74}\label{8.74} \end{equation} then introduce the partial p \begin{equation} 1 = \frac{1}{2D}\left( \frac{\partial p_1^\mu}{\partial p_1^\mu} + \frac{\partial p_2^\mu}{\partial p_2^\mu}\right) \tag{8.75}\label{8.75} \end{equation} and get \begin{equation} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!I(D) = - \frac{\lambda^2}{D-3} \int \frac{d^Dp_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D} \frac{3m^2 + \mathbf{q}(\mathbf{p}_1 + \mathbf{p}_2 + \mathbf{q})}{(\mathbf{p}_1^2 + m^2) (\mathbf{p}_2^2+m^2)[(\mathbf{p}_1+\mathbf{p}_2 + \mathbf{q})^2+m^2]^2} \tag{8.76}\label{8.76} \end{equation}

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  • $\begingroup$ What else have you tried? Did you attempt to put that expression together using a common denominator? $\endgroup$
    – Triatticus
    Commented Sep 8, 2022 at 16:56
  • $\begingroup$ So you should show that work of course, the more work you show the better the help can be of course. $\endgroup$
    – Triatticus
    Commented Sep 8, 2022 at 21:26
  • $\begingroup$ Finding a common denominator is straightforward, it would just be down to whether or not the resulting numerator factors into something that simplifies to the final expression. Though it seems frobenius indicates that there is an error before that part. You still need to show much more work at how you arrived at the expression you did. $\endgroup$
    – Triatticus
    Commented Sep 9, 2022 at 19:33

2 Answers 2

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$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\mb}[1]{\mathbf {#1}} \newcommand{\mc}[1]{\mathcal {#1}} \newcommand{\mr}[1]{\mathrm {#1}} \newcommand{\mf}[1]{\mathfrak{#1}} \newcommand{\gr}{\bl>} \newcommand{\les}{\bl<} \newcommand{\greq}{\bl\ge} \newcommand{\leseq}{\bl\le} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\blr}[1]{\left[#1\right]} \newcommand{\vlr}[1]{\left\vert#1\right\vert} \newcommand{\Vlr}[1]{\left\Vert#1\right\Vert} \newcommand{\lara}[1]{\left\langle#1\right\rangle} \newcommand{\lav}[1]{\left\langle#1\right|} \newcommand{\vra}[1]{\left|#1\right\rangle} \newcommand{\lavra}[2]{\left\langle#1|#2\right\rangle} \newcommand{\lavvra}[3]{\left\langle#1\right|#2\left|#3\right\rangle} \newcommand{\vp}{\vphantom{\dfrac{a}{b}}} \newcommand{\Vp}[1]{\vphantom{#1}} \newcommand{\hp}[1]{\hphantom{#1}} \newcommand{\x}{\bl\times} \newcommand{\ox}{\bl\otimes} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\qqlraqq}{\qquad\bl{-\!\!\!-\!\!\!-\!\!\!\longrightarrow}\qquad} \newcommand{\qqLraqq}{\qquad\boldsymbol{\e\!\e\!\e\!\e\!\Longrightarrow}\qquad} \newcommand{\tl}[1]{\tag{#1}\label{#1}} \newcommand{\hebl}{$\bl{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$}$

We want to check the validity of the equality

\begin{equation} \begin{split} \mathrm A &\bl\equiv\hphantom{+}\frac{m^2}{\hp{^2}(\mathbf{p}_1^2 + m^2)^2 (\mathbf{p}_2^2+m^2)\hphantom{^2}[(\mathbf{p}_1+\mathbf{p}_2 + \mathbf{q})^2+m^2]\hp{^2}}\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ &\hphantom{=} +\:\frac{m^2}{\hp{^2}(\mathbf{p}_1^2 + m^2)\hphantom{^2}(\mathbf{p}_2^2+m^2)^2[(\mathbf{p}_1+\mathbf{p}_2 + \mathbf{q})^2+m^2]\hp{^2}}\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ &\hphantom{=} +\:\frac{m^2 + \mathbf{q}(\mathbf{p}_1 + \mathbf{p}_2 + \mathbf{q})}{\hp{^2}(\mathbf{p}_1^2 + m^2)\hphantom{^2} (\mathbf{p}_2^2+m^2)\hphantom{^2}[(\mathbf{p}_1+\mathbf{p}_2 + \mathbf{q})^2+m^2]^2}\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \\ &\stackrel{\texttt{???}}{\boldsymbol{=\!=}}\;\frac{3m^2 + \mathbf{q}(\mathbf{p}_1 + \mathbf{p}_2 + \mathbf{q})}{\hp{^2}(\mathbf{p}_1^2 + m^2)\hphantom{^2} (\mathbf{p}_2^2+m^2)\hphantom{^2}[(\mathbf{p}_1+\mathbf{p}_2 + \mathbf{q})^2+m^2]^2}\bl\equiv \mr B\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \end{split} \tag{Fr-01}\label{Fr-01} \end{equation}

\begin{equation} \begin{split} &\texttt{Define for convenience}\\ &\mr R_1 \e \mb p_1^2 + m^2\,,\qquad \mr R_2 \e \mb p_2^2 + m^2\\ &\mr S \e \plr{\mb p_1+\mb p_2 + \mb q}^2+m^2\,,\qquad \mr T \e \mb q\plr{\mb p_1+\mb p_2 +\mb q}\\ &\texttt{Then}\\ & \mr A \e\dfrac{m^2}{\mr R^2_1\,\mr R_2\,\mr S}\p\dfrac{m^2}{\mr R_1\,\mr R^2_2\,\mr S}\p\dfrac{m^2\p \mr T}{\mr R_1\,\mr R_2\,\mr S^2}\\ &\mr B \e\dfrac{3m^2\p \mr T}{\mr R_1\,\mr R_2\,\mr S^2}\\ \end{split} \tag{Fr-02}\label{Fr-02} \end{equation} Checking \begin{equation} \begin{split} &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathrm A \stackrel{\texttt{???}}{\bl{\e\!\e}} \mr B \quad \bl\implies \quad \dfrac{m^2}{\mr R^2_1\,\mr R_2\,\mr S}\p\dfrac{m^2}{\mr R_1\,\mr R^2_2\,\mr S}\p\dfrac{m^2\p \mr T}{\mr R_1\,\mr R_2\,\mr S^2}\stackrel{\texttt{???}}{\bl{\e\!\e}}\dfrac{3m^2\p \mr T}{\mr R_1\,\mr R_2\,\mr S^2}\quad \bl\implies \quad\\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\dfrac{1}{\mr R_1\,\mr R_2\,\mr S^2}\blr{\dfrac{m^2\,\mr S}{\mr R_1}\p\dfrac{m^2\,\mr S}{\mr R_2}\p \plr{m^2\p \mr T}}\stackrel{\texttt{???}}{\bl{\e\!\e}}\dfrac{3m^2\p \mr T}{\mr R_1\,\mr R_2\,\mr S^2}\quad \bl\implies \quad\\ & \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\dfrac{m^2\,\mr S}{\mr R_1}\p\dfrac{m^2\,\mr S}{\mr R_2}\p \plr{m^2\p \mr T}\stackrel{\texttt{???}}{\bl{\e\!\e}}3m^2\p \mr T\quad \bl\implies \quad \dfrac{1}{\mr R_1}\p\dfrac{1}{\mr R_2}\stackrel{\texttt{???}}{\bl{\e\!\e}}\dfrac{2}{\mr S}\\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\texttt{or}\\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\dfrac{1}{\mb p_1^2 + m^2}\p\dfrac{1}{\mb p_2^2 + m^2}\stackrel{\texttt{???}}{\bl{\e\!\e}}\dfrac{2}{\plr{\mb p_1+\mb p_2 + \mb q}^2+m^2} \quad \textbf{(invalid !!!)} \end{split} \tag{Fr-03}\label{Fr-03} \end{equation} unless $\:\alpha^2_1\,\mr R_1\e \mr S\e \alpha^2_2\,\mr R_2\:$ with $\:\alpha^2_1\p\alpha^2_2\e 2\:$ that is unless \begin{equation} \alpha^2_1\plr{\mb p^2_1\p m^2}\e\plr{\mb p_1+\mb p_2 + \mb q}^2\p m^2\e\alpha^2_2\plr{\mb p^2_2\p m^2}\,,\quad \alpha^2_1\p\alpha^2_2\e 2 \tag{Fr-04}\label{Fr-04} \end{equation}

So you must check the validity of your expression $\,\mr A$.

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  • $\begingroup$ Not exactly : in that link $$\rm A'=\dfrac{3}{\rm R_1\,\rm R_2\,\rm S}-\dfrac{m^2}{\rm R^2_1\,\rm R_2\,\rm S}-\dfrac{m^2}{\rm R_1\,\rm R^2_2\,\rm S}-\dfrac{m^2+\rm T}{\rm R_1\,\rm R_2\,\rm S^2}=\dfrac{3}{\rm R_1\,\rm R_2\,\rm S}- \rm A$$ $\endgroup$
    – Frobenius
    Commented Sep 12, 2022 at 12:09
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Starting with the integral, \begin{equation} I(d) = \int dq_1dq_2 \frac{1}{(q_1^2+m^2)(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]}, \tag{1}\label{1} \end{equation} and inserting the partial q in (1), we have \begin{equation} I(d) = \frac{1}{2d}\int dq_1dq_2 \left(\frac{\partial q_1^\mu}{\partial q_1^\mu} + \frac{\partial q_2^\mu}{\partial q_2^\mu}\right)\frac{1}{(q_1^2+m^2)(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]}. \tag{2}\label{2} \end{equation} The application of the partial derivative is of the form, $$\frac{\partial x}{\partial x}=1$$ $$\int 1\cdot f(x)dx = \int \frac{\partial x}{\partial x}f(x)dx$$ $$\int \left\{ \frac{\partial}{\partial x}\left[ xf(x)\right] - x\frac{\partial f(x)}{\partial x}\right\}dx$$ the surface term is equal to zero because the contour is infinite! So we have for our case, \begin{equation} \int f(x)dx = -\int x \frac{\partial f(x)}{\partial x}. \tag{3}\label{3} \end{equation} Using (3) equation (2) becomes \begin{equation} I(d) = -\frac{1}{2d}\int dq_1dq_2 \left(q_1^\mu\frac{\partial}{\partial q_1^\mu} + q_2^\mu\frac{\partial}{\partial q_2^\mu}\right)\frac{1}{(q_1^2+m^2)(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]}. \tag{4}\label{4} \end{equation} Differentiating, we get \begin{eqnarray} I(d) &=& \frac{1}{d}\int dq_1dq_2 \Bigg\{\frac{q_1^2}{(q_1^2+m^2)^2(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]}\\ &+& \frac{q_2^2}{(q_1^2+m^2)(q_2^2+m^2)^2[(q_1+q_2+P)^2+m^2]}\\ &+& \frac{(q_1+q_2)(q_1+q_2+P)}{(q_1^2+m^2)(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]^2}\Bigg\}. \tag{5}\label{5} \end{eqnarray} Inserting $m^2 - m^2$ in (5) and $P² - P²$ in the last integral of it \begin{eqnarray} I(d) &=& \frac{1}{d}\int dq_1dq_2 \Bigg\{\frac{q_1^2+m^2-m^2}{(q_1^2+m^2)^2(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]}\\ &+& \frac{q_2^2 + m^2-m^2}{(q_1^2+m^2)(q_2^2+m^2)^2[(q_1+q_2+P)^2+m^2]}\\ &+& \frac{m^2-m^2 + (q_1+q_2+P^2-P^2)(q_1+q_2+P)}{(q_1^2+m^2)(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]^2}\Bigg\}. \tag{6}\label{6} \end{eqnarray} Using only algebra and rewriting in terms of I(d), \begin{eqnarray} I(d) &=& -\frac{1}{d-3}\int dq_1dq_2 \Bigg\{\frac{m^2}{(q_1^2+m^2)^2(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]}\\ &+& \frac{m^2}{(q_1^2+m^2)(q_2^2+m^2)^2[(q_1+q_2+P)^2+m^2]}\\ &+& \frac{m^2+ P(q_1+q_2+P)}{(q_1^2+m^2)(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]^2}\Bigg\}. \tag{7}\label{7} \end{eqnarray} Using the following substitution, $q_1 = -q_2 - P - q$ on the first integral on the right hand side and $q_2 = -q_1 - P - q$ on the second integral on the right hand side of (7), it is easy to see that the three integrals proportions to $m²$ have the same solution. Therefore, we can write (7) as \begin{eqnarray} \boxed{ I(d) = -\frac{1}{d-3}\int dq_1dq_2 \frac{3m^2+ P(q_1+q_2+P)}{(q_1^2+m^2)(q_2^2+m^2)[(q_1+q_2+P)^2+m^2]^2}}. \tag{8}\label{8} \end{eqnarray}

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