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It is known that the classical equation of motion for a scalar field wave packet on a curved spacetime background gives the geodesic trajectory (the e.o.m. is $(\nabla_\mu \nabla^\mu + m^2) \Phi=0$). However, I couldn't see that.

How can one derived the geodesic equation from the above e.o.m. ?

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  • $\begingroup$ Hint: assume $\phi = \delta(t(\tau))\delta(x(\tau))\delta(y(\tau))\delta(z(\tau))$ $\endgroup$ – Jerry Schirmer Jun 6 '14 at 16:04
  • $\begingroup$ Actually, that's how I first did the problem. No, I don't see how one can get it from there, but there might be away to see that by choosing $\Phi \sim e^{ik^\mu x_\nu}$, then if from something like $k^2 = m^2$ condition one gets $k^\mu \nabla_\mu k^\nu = 0$ (this is the geodesic equation, as $k_\mu \sim u_\mu$, in classical limit), then problem solved. $\endgroup$ – user109798 Jun 6 '14 at 16:11
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    $\begingroup$ in curved spacetime? $\endgroup$ – Valter Moretti Jun 6 '14 at 16:17
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    $\begingroup$ I was referring to that formula for the delta fuction based on Fourier transform. It generally fails since no global coordinates exist. $\endgroup$ – Valter Moretti Jun 6 '14 at 17:59
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    $\begingroup$ @user39158 This has been discussed on usenet back in the day, but you may find similar discussion in, say, Aspects of Quantum Field Theory in Curved Spacetime by Stephen A. Fulling...or any introductory book on QFT in curved spacetime. $\endgroup$ – Alex Nelson Aug 6 '15 at 3:46
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This is covered (in two pieces) in Chapter 4 of Wald's General Relativity for the Maxwell field; I'll adapt his proof here.

We start with the equation $(\nabla^a \nabla_a + m^2) \Phi = 0$. Let us look for a solution to this equation of the form $$ \Phi = \Phi_0 e^{iS}, $$ where $S$ is a real-valued function and $\Phi_0$ is "slowly varying" compared to $S$. This is the standard sort of "geometrical optics" approximation; in practical terms, it means that we can ignore any derivatives of $\Phi_0$. Thus, we have $$ \nabla_a \Phi = \Phi_0 e^{iS} (i \nabla_a S)\quad \Rightarrow \quad \nabla^a \nabla_a \Phi + m^2 \Phi = \Phi_0 e^{iS} \left[ m^2 - (\nabla^a S) (\nabla_a S) + i \nabla^a \nabla_a S \right] = 0. $$ Both the imaginary and real parts of the quantity in square brackets above must vanish: $$ m^2 = (\nabla^a S) (\nabla_a S), \qquad \nabla^a \nabla_a S = 0. $$ If we let $k_a = \nabla_a S$, the first equation implies that $k_a k^a = m^2$, as we expect. Moreover, if we take the derivative of both sides of the first equation, we get \begin{align*} 0 &= \nabla_b \left[ (\nabla^a S) (\nabla_a S) \right] \\ &= 2 \nabla^a S \nabla_b \nabla_a S && \text{(product rule)} \\ &= 2 \nabla^a S \nabla_a \nabla_b S && \text{(derivatives commute on scalars)} \\ &= 2 k^a \nabla_a k_b, \end{align*} and so the vector $k_a$ satisfies the geodesic equation.

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As in your comment, make an Ansatz $$ \phi = \exp(i f)$$ where $f$ is an unknown scalar function such that $(df)_\mu (df)^\mu = m^2$. Then the equation of motion for $\phi$ is equivalent to the geodesic equation for $(df)_\mu$.

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  • $\begingroup$ Sorry, but I don't see why $k^2=m^2$ must imply it's the geodesic equation for $k_\mu$ $(*)$. I think that's where I stuck -- see my later comments above. Would you please show me how $(*)$ can be true? $\endgroup$ – user109798 Jun 6 '14 at 19:30
  • $\begingroup$ You're right. I got ahead of myself and actually what one obtains is that $(df)_\mu$ should have zero divergence. $\endgroup$ – Robin Ekman Jun 6 '14 at 20:44
  • $\begingroup$ Sorry; it might be too trivial for you, but it's not for me. Could you please be more specific about "$(df)_\mu$ has $0$ divergence" and its connection to the geodesic? $\endgroup$ – user109798 Jun 6 '14 at 22:03
  • $\begingroup$ With a $\phi$ of this form the equation of motion is $\exp(if) [ i^2 (df)_\mu (df)^\mu + m^2 + \nabla^\mu (df_\mu) ] = 0$ where $(df)_\mu = \partial_\mu f = \nabla_\mu f$, the latter equality from $f$ being a scalar. When the first two terms cancel, what is left is that $\nabla_\mu \nabla^\mu f = 0$. There is not necessarily a connection to geodesics here. As I said I got ahead of myself and made a mistake. However, many spacetimes admit vector fields that are both divergence-free and geodesic, so in these you can find solutions of the form specified, and the gradient of $f$ will be geodesic. $\endgroup$ – Robin Ekman Jun 6 '14 at 22:09
  • $\begingroup$ Thank you, but it doesn't make sense to me why the first 2 term must cancel. I mean, why don't $f$ has a form so that $\nabla^\mu \nabla_\mu f$ is non-zero and both of them cancel? Moreover, I still don't see why the last equation goes to $0$ indicates geodesic equation yet. $\endgroup$ – user109798 Jun 6 '14 at 22:17

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