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Consider the class of Lagrangian known as $f(R)$ Lagrangians where the Lagrangian is some function $f(R)$, \begin{equation} S=\int\sqrt{g}d^4x\ f(R) \end{equation} assuming there are no (or ignoring) boundary terms one finds \begin{equation} \delta S=\int\sqrt{g}d^4x\left(-\frac{1}{2}g_{\mu\nu}f(R)+f'(R)R_{\mu\nu}-\left(\nabla_\mu\nabla_\nu-g_{\mu\nu}\Box\right)f'(R)\right)\delta g^{\mu\nu}. \end{equation} Suppose $f(R)=g^\frac{1}{4}R$.

  1. Is there a way to define the covariant derivative of $g^\frac{1}{4}$?
  2. Is $f(R)=g^\frac{1}{4}R$ a valid choice?
Qmechanic
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Joseph Dewdney
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1 Answers1

3

Comments to the question (v2):

  1. Assuming that the connection $\nabla$ is compatible$^1$ with the metric $\nabla_{\lambda}g_{\mu\nu}=0$, then $\nabla_{\lambda}\det g_{\mu\nu}=0$, and therefore e.g. $\nabla_{\lambda}\left|\det g_{\mu\nu}\right|^{\frac{1}{4}}=0$.

  2. Choosing e.g. $f(R)=\left|\det g_{\mu\nu}\right|^{\frac{1}{4}}R$ would not be a geometrically covariant theory because $f(R)$ would no longer transform as a scalar, i.e. the action would depend on the choice of coordinates.

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$^1$ The Levi-Civita connection is compatible with the metric.

Qmechanic
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