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To prove equation 9.91 globally we will first prove it locally, the generalization is then straitforwardstraightforward.

We take a point $p$ of the spacetime-manifold $M$, at which a tangent space $T_p$ is defined. The scalar fields $\phi(x^{\mu})$ are then defined with respect to a coordinate system constructed from the basis vectors of $T_p$.

Now, in general relativity, curved spacetime looks locally like Minkowski space + a gravitational force (Equivalence Principle). So we can construct Riemann Normal Coordinates $x^{\hat{\mu}}(p)$ satisfying:

$$g_{\hat{\mu} \hat{\nu}}(p) = \eta_{\hat{\mu} \hat{\nu}}, \partial_\hat{\sigma} g_{\hat{\mu} \hat{\nu}}(p) = 0.$$

These coordinates are called locally inertial coordinates (see eq 2.47 in CarrollsCarroll's book).

Next relabel $g_{\hat{\mu} \hat{\nu}}$ to $g_{\mu\nu}$ to avoid confusion. Then, we show that equation 9.91 holds in these coordinates:

$$\pi = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{\mu\nu} \nabla_\mu \phi \nabla_\nu \phi - \cdots \})$$ $$ = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{00} \nabla_0 \phi \nabla_0 \phi + g^{0i} \nabla_0 \phi \nabla_i \phi + g^{i0} \nabla_i \phi \nabla_0 \phi + g^{ij} \nabla_i \phi \nabla_j \phi - \cdots \})$$ by only looking at the first term, as the rest does not depend on $\nabla_0 \phi$, we further derive: $$ \frac{\partial}{\partial(\nabla_0 \phi)} (g^{00} \nabla_0 \phi \nabla_0 \phi) = 2 g^{00} \nabla_0 \phi;$$ just use the Leibniz rule to prove this statement. Then: $$\pi = \sqrt{-g} \nabla_0 \phi.$$ Where the metric is put in its canonical form $g_{\mu \nu} = diag(-1, +1, +1, +1)$ as we are using locally inertial coordinates.

Finally, as $\pi = \sqrt{-g} \nabla_0 \phi$ is a tensorial equation (as the covariant derivative of a scalar field is independent of the used coordinate system) so 9.91 is globally true.

It should be noted that $g^{00} = g_{00} = -1$ and by this equation 9.91 is true.

To prove equation 9.91 globally we will first prove it locally, the generalization is then straitforward.

We take a point $p$ of the spacetime-manifold $M$, at which a tangent space $T_p$ is defined. The scalar fields $\phi(x^{\mu})$ are then defined with respect to a coordinate system constructed from the basis vectors of $T_p$.

Now, in general relativity, curved spacetime looks locally like Minkowski space + a gravitational force (Equivalence Principle). So we can construct Riemann Normal Coordinates $x^{\hat{\mu}}(p)$ satisfying:

$$g_{\hat{\mu} \hat{\nu}}(p) = \eta_{\hat{\mu} \hat{\nu}}, \partial_\hat{\sigma} g_{\hat{\mu} \hat{\nu}}(p) = 0.$$

These coordinates are called locally inertial coordinates (see eq 2.47 in Carrolls book).

Next relabel $g_{\hat{\mu} \hat{\nu}}$ to $g_{\mu\nu}$ to avoid confusion. Then, we show that equation 9.91 holds in these coordinates:

$$\pi = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{\mu\nu} \nabla_\mu \phi \nabla_\nu \phi - \cdots \})$$ $$ = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{00} \nabla_0 \phi \nabla_0 \phi + g^{0i} \nabla_0 \phi \nabla_i \phi + g^{i0} \nabla_i \phi \nabla_0 \phi + g^{ij} \nabla_i \phi \nabla_j \phi - \cdots \})$$ by only looking at the first term, as the rest does not depend on $\nabla_0 \phi$, we further derive: $$ \frac{\partial}{\partial(\nabla_0 \phi)} (g^{00} \nabla_0 \phi \nabla_0 \phi) = 2 g^{00} \nabla_0 \phi;$$ just use the Leibniz rule to prove this statement. Then: $$\pi = \sqrt{-g} \nabla_0 \phi.$$ Where the metric is put in its canonical form $g_{\mu \nu} = diag(-1, +1, +1, +1)$ as we are using locally inertial coordinates.

Finally, as $\pi = \sqrt{-g} \nabla_0 \phi$ is a tensorial equation (as the covariant derivative of a scalar field is independent of the used coordinate system) so 9.91 is globally true.

It should be noted that $g^{00} = g_{00} = -1$ and by this equation 9.91 is true.

To prove equation 9.91 globally we will first prove it locally, the generalization is then straightforward.

We take a point $p$ of the spacetime-manifold $M$, at which a tangent space $T_p$ is defined. The scalar fields $\phi(x^{\mu})$ are then defined with respect to a coordinate system constructed from the basis vectors of $T_p$.

Now, in general relativity, curved spacetime looks locally like Minkowski space + a gravitational force (Equivalence Principle). So we can construct Riemann Normal Coordinates $x^{\hat{\mu}}(p)$ satisfying:

$$g_{\hat{\mu} \hat{\nu}}(p) = \eta_{\hat{\mu} \hat{\nu}}, \partial_\hat{\sigma} g_{\hat{\mu} \hat{\nu}}(p) = 0.$$

These coordinates are called locally inertial coordinates (see eq 2.47 in Carroll's book).

Next relabel $g_{\hat{\mu} \hat{\nu}}$ to $g_{\mu\nu}$ to avoid confusion. Then, we show that equation 9.91 holds in these coordinates:

$$\pi = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{\mu\nu} \nabla_\mu \phi \nabla_\nu \phi - \cdots \})$$ $$ = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{00} \nabla_0 \phi \nabla_0 \phi + g^{0i} \nabla_0 \phi \nabla_i \phi + g^{i0} \nabla_i \phi \nabla_0 \phi + g^{ij} \nabla_i \phi \nabla_j \phi - \cdots \})$$ by only looking at the first term, as the rest does not depend on $\nabla_0 \phi$, we further derive: $$ \frac{\partial}{\partial(\nabla_0 \phi)} (g^{00} \nabla_0 \phi \nabla_0 \phi) = 2 g^{00} \nabla_0 \phi;$$ just use the Leibniz rule to prove this statement. Then: $$\pi = \sqrt{-g} \nabla_0 \phi.$$ Where the metric is put in its canonical form $g_{\mu \nu} = diag(-1, +1, +1, +1)$ as we are using locally inertial coordinates.

Finally, as $\pi = \sqrt{-g} \nabla_0 \phi$ is a tensorial equation (as the covariant derivative of a scalar field is independent of the used coordinate system) so 9.91 is globally true.

It should be noted that $g^{00} = g_{00} = -1$ and by this equation 9.91 is true.

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Rewriting the argument, major clarifications and many corrections.
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Silverwhale
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Looking at eq 2.46 in Carrolls book; The metric is Lorentzian in General Relativity so that $g^{\mu \nu} = diag(-1,+1, ...,+1)$. This is the canonical form of the metric, where the signatures are, as written, -1 for the time coordinate and +1, ..To prove equation 9.91 globally we will first prove it locally,+1 for the spatial coordinates and where the metricgeneralization is diagonalthen straitforward.

Next, we say that at anyWe take a point p element$p$ of the manifold M there exists a coordinate system where the metric is canonical and further the following equation holdsspacetime-manifold (eq 2.47 in Carrolls book): $$g_{\hat{\mu} \hat{\nu}}(p) = \eta_{\hat{\mu} \hat{\nu}}.$$ We say the coordinate system$M$, at p iswhich a tangent space locally intertial. This$T_p$ is very important because we are assuming that locally curved space looks like flat space. An assumption which, in General Relativity, must be fulfilled.

Following equation 3defined.66 in Carrolls book, we have: $g_{\hat{\mu} \hat{\nu}} = g(\hat{e}_{(\hat{\mu})},\hat{e}_{(\hat{\nu})})$, and by using formulae 3.64, 3.65 and 3.66 we have an actual algorithm on how to rewrite the metric into oneThe scalar fields $\phi(x^{\mu})$ are then defined onwith respect to a coordinate system that is locally inertial. It should be noted thatconstructed from the $\hat{e}_{(\hat{\mu})}$ are basis vectors of the tangent space $T_p$.

Now, by taking equationin general relativity, curved spacetime looks locally like Minkowski space + a gravitational force (Equivalence Principle). So we can construct Riemann Normal Coordinates $x^{\hat{\mu}}(p)$ satisfying:

$$g_{\hat{\mu} \hat{\nu}}(p) = \eta_{\hat{\mu} \hat{\nu}}, \partial_\hat{\sigma} g_{\hat{\mu} \hat{\nu}}(p) = 0.$$

These coordinates are called locally inertial coordinates (see eq 2.47, it is easy in Carrolls book).

Next relabel $g_{\hat{\mu} \hat{\nu}}$ to derive eq$g_{\mu\nu}$ to avoid confusion. Then, we show that equation 9.91 holds in Carrolls bookthese coordinates:

$$\pi = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{\mu\nu} \nabla_\mu \phi \nabla_\nu \phi - \cdots \})$$ $$ = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{00} \nabla_0 \phi \nabla_0 \phi + g^{0i} \nabla_0 \phi \nabla_i \phi + g^{i0} \nabla_i \phi \nabla_0 \phi + g^{ij} \nabla_i \phi \nabla_j \phi - \cdots \})$$$$ = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{00} \nabla_0 \phi \nabla_0 \phi + g^{0i} \nabla_0 \phi \nabla_i \phi + g^{i0} \nabla_i \phi \nabla_0 \phi + g^{ij} \nabla_i \phi \nabla_j \phi - \cdots \})$$ by only looking at the first term, as the rest does not depend on $\nabla_0 \phi$, we further derive: $$ \frac{\partial}{\partial(\nabla_0 \phi)} (g^{00} \nabla_0 \phi \nabla_0 \phi) = 2 g^{00} \nabla_0 \phi.$$$$ \frac{\partial}{\partial(\nabla_0 \phi)} (g^{00} \nabla_0 \phi \nabla_0 \phi) = 2 g^{00} \nabla_0 \phi;$$ Justjust use the Leibniz rule to prove this statement. Then put this into the starting equation to get: $$\pi = \sqrt{-g} \nabla_0 \phi.$$

We are actually done. But by assuming that Where the metric is Minkowskian locally we can calculate easilyput in its determinant, which should be -1. Such that thecanonical form $\sqrt{-g}$ term is equal to 1. So finally$g_{\mu \nu} = diag(-1, +1, +1, +1)$ as we get: $$\pi = \nabla_0 \phi.$$are using locally inertial coordinates.

Please note that $g^{\mu\nu}$ is defined here to beFinally, as $g^{\hat{\mu} \hat{\nu}}$ which$\pi = \sqrt{-g} \nabla_0 \phi$ is just a shorthand fortensorial equation $g^{\hat{\mu} \hat{\nu}}(p)$. So to make(as the computation clearcovariant derivative of a scalar field is independent of the used coordinate system) so 9.91 is globally true.

As a notice, it is not clear why the author wrote the termIt should be noted that $\sqrt{-g}$ explicitly in$g^{00} = g_{00} = -1$ and by this equation 9.91 .is true.

Looking at eq 2.46 in Carrolls book; The metric is Lorentzian in General Relativity so that $g^{\mu \nu} = diag(-1,+1, ...,+1)$. This is the canonical form of the metric, where the signatures are, as written, -1 for the time coordinate and +1, ...,+1 for the spatial coordinates and where the metric is diagonal.

Next, we say that at any point p element of the manifold M there exists a coordinate system where the metric is canonical and further the following equation holds (eq 2.47 in Carrolls book): $$g_{\hat{\mu} \hat{\nu}}(p) = \eta_{\hat{\mu} \hat{\nu}}.$$ We say the coordinate system at p is locally intertial. This is very important because we are assuming that locally curved space looks like flat space. An assumption which, in General Relativity, must be fulfilled.

Following equation 3.66 in Carrolls book, we have: $g_{\hat{\mu} \hat{\nu}} = g(\hat{e}_{(\hat{\mu})},\hat{e}_{(\hat{\nu})})$, and by using formulae 3.64, 3.65 and 3.66 we have an actual algorithm on how to rewrite the metric into one defined on a coordinate system that is locally inertial. It should be noted that the $\hat{e}_{(\hat{\mu})}$ are basis vectors of the tangent space $T_p$.

Now, by taking equation 2.47, it is easy to derive eq 9.91 in Carrolls book:

$$\pi = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{\mu\nu} \nabla_\mu \phi \nabla_\nu \phi - \cdots \})$$ $$ = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{00} \nabla_0 \phi \nabla_0 \phi + g^{0i} \nabla_0 \phi \nabla_i \phi + g^{i0} \nabla_i \phi \nabla_0 \phi + g^{ij} \nabla_i \phi \nabla_j \phi - \cdots \})$$ by only looking at the first term, as the rest does not depend on $\nabla_0 \phi$, we further derive: $$ \frac{\partial}{\partial(\nabla_0 \phi)} (g^{00} \nabla_0 \phi \nabla_0 \phi) = 2 g^{00} \nabla_0 \phi.$$ Just use the Leibniz rule. Then put this into the starting equation to get: $$\pi = \sqrt{-g} \nabla_0 \phi.$$

We are actually done. But by assuming that the metric is Minkowskian locally we can calculate easily its determinant, which should be -1. Such that the $\sqrt{-g}$ term is equal to 1. So finally we get: $$\pi = \nabla_0 \phi.$$

Please note that $g^{\mu\nu}$ is defined here to be $g^{\hat{\mu} \hat{\nu}}$ which is just a shorthand for $g^{\hat{\mu} \hat{\nu}}(p)$. So to make the computation clear.

As a notice, it is not clear why the author wrote the term $\sqrt{-g}$ explicitly in equation 9.91 ..

To prove equation 9.91 globally we will first prove it locally, the generalization is then straitforward.

We take a point $p$ of the spacetime-manifold $M$, at which a tangent space $T_p$ is defined. The scalar fields $\phi(x^{\mu})$ are then defined with respect to a coordinate system constructed from the basis vectors of $T_p$.

Now, in general relativity, curved spacetime looks locally like Minkowski space + a gravitational force (Equivalence Principle). So we can construct Riemann Normal Coordinates $x^{\hat{\mu}}(p)$ satisfying:

$$g_{\hat{\mu} \hat{\nu}}(p) = \eta_{\hat{\mu} \hat{\nu}}, \partial_\hat{\sigma} g_{\hat{\mu} \hat{\nu}}(p) = 0.$$

These coordinates are called locally inertial coordinates (see eq 2.47 in Carrolls book).

Next relabel $g_{\hat{\mu} \hat{\nu}}$ to $g_{\mu\nu}$ to avoid confusion. Then, we show that equation 9.91 holds in these coordinates:

$$\pi = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{\mu\nu} \nabla_\mu \phi \nabla_\nu \phi - \cdots \})$$ $$ = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{00} \nabla_0 \phi \nabla_0 \phi + g^{0i} \nabla_0 \phi \nabla_i \phi + g^{i0} \nabla_i \phi \nabla_0 \phi + g^{ij} \nabla_i \phi \nabla_j \phi - \cdots \})$$ by only looking at the first term, as the rest does not depend on $\nabla_0 \phi$, we further derive: $$ \frac{\partial}{\partial(\nabla_0 \phi)} (g^{00} \nabla_0 \phi \nabla_0 \phi) = 2 g^{00} \nabla_0 \phi;$$ just use the Leibniz rule to prove this statement. Then: $$\pi = \sqrt{-g} \nabla_0 \phi.$$ Where the metric is put in its canonical form $g_{\mu \nu} = diag(-1, +1, +1, +1)$ as we are using locally inertial coordinates.

Finally, as $\pi = \sqrt{-g} \nabla_0 \phi$ is a tensorial equation (as the covariant derivative of a scalar field is independent of the used coordinate system) so 9.91 is globally true.

It should be noted that $g^{00} = g_{00} = -1$ and by this equation 9.91 is true.

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Silverwhale
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I was going through the same computation right now. Looking at equation (2eq 2.46), in Carrolls book; The metric as being defined to beis Lorentzian in General Relativity so that $g^{\mu \nu} = diag(-1,+1, ...,+1)$. This is the canonical form of the metric, where the signatures are, as written, -1 for the time coordinate and +1, ...,+1 for the spatial coordinates and where the metric is diagonal.

Next, we say that at any point p element of the manifold M there exists a coordinate system defined to be inertial locally wewhere the metric is canonical and further the following equation holds (eq 2.47 in Carrolls book): $$g_{\hat{\mu} \hat{\nu}}(p) = \eta_{\hat{\mu} \hat{\nu}}.$$ We say the coordinate system at p is locally intertial. This is very important because we are assuming that $g^{00} = diag(-1,+1,+1,+1)$locally curved space looks like flat space. Which shouldAn assumption which, in General Relativity, must be rightfulfilled. BUT how can it be that

Following equation 3.66 in Carrolls book, the determinant is not explicitly computed if we make such an assumptionhave: $g_{\hat{\mu} \hat{\nu}} = g(\hat{e}_{(\hat{\mu})},\hat{e}_{(\hat{\nu})})$, and by using formulae 3. I am confused64, 3. IF65 and 3.66 we have an actual algorithm on how to rewrite the metric into one defined on a coordinate system that is locally inertial. It should be noted that the $\hat{e}_{(\hat{\mu})}$ are basis vectors of the tangent space $T_p$.

Now, by taking equation 2.47, it is easy to derive eq 9.91 in Carrolls book:

$$\pi = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{\mu\nu} \nabla_\mu \phi \nabla_\nu \phi - \cdots \})$$ $$ = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{00} \nabla_0 \phi \nabla_0 \phi + g^{0i} \nabla_0 \phi \nabla_i \phi + g^{i0} \nabla_i \phi \nabla_0 \phi + g^{ij} \nabla_i \phi \nabla_j \phi - \cdots \})$$ by only looking at the first term, as the OP saidrest does not depend on $\nabla_0 \phi$, generic and we do not make any further assumption then yesderive: $$ \frac{\partial}{\partial(\nabla_0 \phi)} (g^{00} \nabla_0 \phi \nabla_0 \phi) = 2 g^{00} \nabla_0 \phi.$$ Just use the determinant is "generic" tooLeibniz rule. Then put this into the starting equation to get: $$\pi = \sqrt{-g} \nabla_0 \phi.$$

We are actually done. But ifby assuming that the metric is set, then theMinkowskian locally we can calculate easily its determinant, which should be set too-1. Such that the $\sqrt{-g}$ term is equal to 1. So I am really confused by equationfinally we get: $$\pi = \nabla_0 \phi.$$

Please note that (9$g^{\mu\nu}$ is defined here to be $g^{\hat{\mu} \hat{\nu}}$ which is just a shorthand for $g^{\hat{\mu} \hat{\nu}}(p)$.91) So to make the computation clear.

As a notice, it is not clear why the author wrote the term $\sqrt{-g}$ explicitly in equation 9.91 ..

I was going through the same computation right now. Looking at equation (2.46), The metric as being defined to be Lorentzian in General Relativity and the coordinate system defined to be inertial locally we say that $g^{00} = diag(-1,+1,+1,+1)$. Which should be right. BUT how can it be that in Carrolls book, the determinant is not explicitly computed if we make such an assumption. I am confused. IF the metric is, as the OP said, generic and we do not make any further assumption then yes the determinant is "generic" too.. But if the metric is set, then the determinant should be set too.. So I am really confused by equation (9.91)...

Looking at eq 2.46 in Carrolls book; The metric is Lorentzian in General Relativity so that $g^{\mu \nu} = diag(-1,+1, ...,+1)$. This is the canonical form of the metric, where the signatures are, as written, -1 for the time coordinate and +1, ...,+1 for the spatial coordinates and where the metric is diagonal.

Next, we say that at any point p element of the manifold M there exists a coordinate system where the metric is canonical and further the following equation holds (eq 2.47 in Carrolls book): $$g_{\hat{\mu} \hat{\nu}}(p) = \eta_{\hat{\mu} \hat{\nu}}.$$ We say the coordinate system at p is locally intertial. This is very important because we are assuming that locally curved space looks like flat space. An assumption which, in General Relativity, must be fulfilled.

Following equation 3.66 in Carrolls book, we have: $g_{\hat{\mu} \hat{\nu}} = g(\hat{e}_{(\hat{\mu})},\hat{e}_{(\hat{\nu})})$, and by using formulae 3.64, 3.65 and 3.66 we have an actual algorithm on how to rewrite the metric into one defined on a coordinate system that is locally inertial. It should be noted that the $\hat{e}_{(\hat{\mu})}$ are basis vectors of the tangent space $T_p$.

Now, by taking equation 2.47, it is easy to derive eq 9.91 in Carrolls book:

$$\pi = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{\mu\nu} \nabla_\mu \phi \nabla_\nu \phi - \cdots \})$$ $$ = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{00} \nabla_0 \phi \nabla_0 \phi + g^{0i} \nabla_0 \phi \nabla_i \phi + g^{i0} \nabla_i \phi \nabla_0 \phi + g^{ij} \nabla_i \phi \nabla_j \phi - \cdots \})$$ by only looking at the first term, as the rest does not depend on $\nabla_0 \phi$, we further derive: $$ \frac{\partial}{\partial(\nabla_0 \phi)} (g^{00} \nabla_0 \phi \nabla_0 \phi) = 2 g^{00} \nabla_0 \phi.$$ Just use the Leibniz rule. Then put this into the starting equation to get: $$\pi = \sqrt{-g} \nabla_0 \phi.$$

We are actually done. But by assuming that the metric is Minkowskian locally we can calculate easily its determinant, which should be -1. Such that the $\sqrt{-g}$ term is equal to 1. So finally we get: $$\pi = \nabla_0 \phi.$$

Please note that $g^{\mu\nu}$ is defined here to be $g^{\hat{\mu} \hat{\nu}}$ which is just a shorthand for $g^{\hat{\mu} \hat{\nu}}(p)$. So to make the computation clear.

As a notice, it is not clear why the author wrote the term $\sqrt{-g}$ explicitly in equation 9.91 ..

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