A Variation of the kinetic term in scalar field theory

Click For Summary
Varying the kinetic term in scalar field theory with respect to the metric tensor using two methods yields different results, specifically a discrepancy of a minus sign. Method 1 applies the variation directly to the metric, while Method 2 incorporates the relationship between covariant and contravariant metrics. The issue arises in Method 1, where the variations of the metric within the definitions of the derivatives are not accounted for. This oversight leads to an incomplete variation in the first method. Correctly including these variations resolves the discrepancy between the two methods.
Baela
Messages
17
Reaction score
2
Varying ##\partial_\lambda\phi\,\partial^\lambda\phi## wrt the metric tensor ##g_{\mu\nu}## in two different ways gives me different results. Obviously I'm doing something wrong. Where am I going wrong?

Method 1: \begin{equation}
(\delta g_{\mu\nu})\,\partial^\mu\phi\,\partial^\nu\phi
\end{equation}

Method 2: \begin{align}&\quad\,\, (\delta g^{\mu\nu})\,\partial_\mu\phi\,\partial_\nu\phi \nonumber \\
&=(-g^{\mu\rho}g^{\nu\sigma}\delta g_{\rho\sigma})\,\partial_\mu\phi\,\partial_\nu\phi \quad (\because \delta g^{\mu\nu}=-g^{\mu\rho}g^{\nu\sigma}\delta g_{\rho\sigma} \,\,\text{as can be checked by varying the identity}\,\, g^{\mu\lambda}g_{\lambda\nu}=\delta^\mu_\nu) \nonumber\\
&=-(\delta g_{\rho\sigma})\,\partial^\rho\phi\,\partial^\sigma\phi
\end{align}
The second result differs from the first one by a minus sign. What's going wrong?
 
Last edited:
Physics news on Phys.org
In Method 1 you are missing the variations of the metric inside the definitions ##\partial^\mu \phi = g^{\mu\nu}\partial_\nu \phi##.
 
The Poynting vector is a definition, that is supposed to represent the energy flow at each point. Unfortunately, the only observable effect caused by the Poynting vector is through the energy variation in a volume subject to an energy flux through its surface, that is, the Poynting theorem. As a curl could be added to the Poynting vector without changing the Poynting theorem, it can not be decided by EM only that this should be the actual flow of energy at each point. Feynman, commenting...