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I'm trying to reproduce a calculation done in Birrell & Davies' book Quantum Fields in Curved Space (page 191). Given Klein-Gordon's equation $$(\Box + m^2 + R\xi)\phi = 0$$ and $$T_{\mu \nu} = (1 -2\xi)\phi_{;\mu} \phi_{;\nu} + (2\xi - \frac{1}{2})g_{\mu \nu} g^{\rho \sigma}\phi_{;\rho} \phi_{;\sigma} - 2\xi\phi_{;\mu \nu} \phi + \frac{2}{n}\xi g_{\mu \nu} \phi \Box \phi - \xi\left(R_{\mu \nu} -\frac{1}{2}Rg_{\mu \nu} +\frac{2(n-1)}{n}\xi Rg_{\mu \nu} \right)\phi^2 + 2\left(\frac{1}{4}-(1-\frac{1}{n})\xi \right)m^2g_{\mu \nu} \phi^2,$$ I want to show that $$T^{\mu}_{\mu} = m^2 \phi^2 + (n-1)(\xi - \xi(n))\Box \phi^2,\tag{6.176}$$ where $\xi(n) = \frac{n-2}{4(n - 1)}$. I've done the following

$$ T^{\mu}_{\mu} = g^{\mu \nu} T\_{\mu \nu}= (1 -2\xi )\phi^{;\mu} \phi\_{;\mu} + (2\xi - \frac{1}{2})n\phi^{;\mu} \phi\_{;\mu} -2 \xi \phi \Box \phi +2 \xi \phi \Box \phi - \xi\left(R -\frac{1}{2}Rn +2(n-1)\xi R \right)\phi^2 + 2\left(\frac{1}{4}-(1-\frac{1}{n})\xi \right)m^2n \phi^2,$$ $$ =\left(2\xi (n-1)+ 1 - \frac{n}{2}\right)\phi^{;\mu} \phi\_{;\mu} - \xi R (1 - \frac{n}{2})\phi^2 - 2 \xi (n-1)\xi R \phi^2 + 2 \left(\frac{1}{4}-(1-\frac{1}{n})\xi \right)m^2n \phi^2,$$ using Klein-Gordon's equation, $$T^{\mu}\_{\mu} = \left(2\xi (n-1)+ 1 - \frac{n}{2}\right)\phi^{;\mu} \phi\_{;\mu} + (\Box + m^2)(1 - \frac{n}{2})\phi^2 + \xi (\Box + m^2) 2(n-1) \phi^2 + 2 \left(\frac{1}{4}-(1-\frac{1}{n})\xi \right)m^2n \phi^2,$$ $$ = \left(2\xi (n-1)+ 1 - \frac{n}{2}\right)\phi^{;\mu} \phi\_{;\mu} + m^2 \phi^2 + \left(1 -\frac{n}{2} +2\xi(n-1) \right)\Box \phi^2, $$

using $\Box \phi^2 = 2\phi^{;\mu} \phi_{;\mu} + 2 \phi \Box \phi $,

$$T^{\mu}\_{\mu} = \left(2\xi (n-1)+ 1 - \frac{n}{2}\right)( \frac{1}{2}\Box \phi^2 - \phi \Box \phi) + m^2 \phi^2 + \left(1 -\frac{n}{2} +2\xi(n-1) \right)\Box \phi^2,$$

$$ = m^2 \phi^2 - \left(2\xi (n-1)+ 1 - \frac{n}{2}\right)\phi \Box \phi + \left( \frac{3}{2} - \frac{3n}{4} + 3\xi(n-1) \right)\Box \phi^2 $$

$$= m^2 \phi^2 -2(n-1)(\xi - \xi(n))\phi \Box \phi + 3(n-1)(\xi - \xi(n)) \Box \phi^2.$$ I'm stuck here and I don't know what to do to the $\phi \Box \phi$ factor in order to get a $\Box \phi^2$. Any suggestions?

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  • $\begingroup$ What is $\phi^{;\mu}$? $\endgroup$
    – Mass
    Nov 3, 2023 at 5:15
  • $\begingroup$ Integrate by parts $\endgroup$
    – Eletie
    Nov 3, 2023 at 8:34
  • $\begingroup$ @Mass its the covariant derivative i.e. $\phi^{;\mu} \equiv \nabla^{\mu} \phi$. $\endgroup$ Nov 3, 2023 at 17:46
  • $\begingroup$ @Eletie But why would I want to integrate by parts? I'm looking for $T^{\mu}_{\mu}$, not $\int T^{\mu}_{\mu} dx$. I don't understand. $\endgroup$ Nov 3, 2023 at 17:48
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    $\begingroup$ @SamuelJaramillobo partial derivatives of a scalar is the same as the covariant derivative $\endgroup$
    – Eletie
    Nov 4, 2023 at 13:09

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