# How to think of the harmonic oscillator equation in terms of “acceleration = gradient”

This is related to another question I just asked where I learned that the equation of motion of a harmonic oscillator is expressed as:

$$\ddot{x}+kx=0$$

What little physics I grasp centers on geodesics as derived from the principle of stationary action and the Euler-Lagrange equations. I have therefore become accustomed to understanding the equation of motion as the geodesic:

$$\ddot{x}^m+{\Gamma^{\:\:m}_{jk} \dot{x}^j \dot{x}^k}=0$$

which can also be thought of as the covariant derivative of the tangent vector of a particle's path. I guess this second eq. is mostly used for analysis of particle motion in GR, but I also understand it is applicable to any other situations with position-dependent coefficients (like motion of light through opaque substances). (We can get rid of all the indices by the way since the harmonic oscillator is one dimensional)

My question: Is it possible to reduce the second equation to the first? The acceleration term is the same, and (I think) Hooke's constant $k$ is basically like the Christoffel symbol in the second eq., but I don't see the similarity between $x$ and $\dot{x}^2$. I sense I am missing something big. Appreciate your help.

EDIT: --I include here a response to JerrySchirmer in comments section below-- In the Newtonian limit (flat and slow) the $00$ component (or $tt$) of the Chistoffel symbol is the only one that doesn't vanish. I wanted to see if this component could some how be expressed as $-kx$. But (insofar as I understand) this one non-vanishing component is usually of first order (a field gradient), not "0 order" like $-kx$. Is there a way to think of $kx$ as a field gradient--like $$kx=\frac{\partial \phi}{\partial x}$$?

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Lagrangian of a harmonic oscillator would be $\frac{1}{2}m\ddot{x} - \frac{1}{2}k x^2$. I don't follow all the math in the rest of the question, though. – Mark Eichenlaub Oct 9 '12 at 1:49
The question formulation(v3) refers twice to the equation of motion as a Lagrangian density, which is not correct use of terminology. – Qmechanic Oct 9 '12 at 11:10
@Qmechanic Thanks. Duly edited. – ben Oct 9 '12 at 13:34

I) One trick is to associate the coordinates $x^{\mu}$ in the geodesic equation with spacetime (rather than just space). Since the space in OP's harmonic oscillator example is 1-dimensional $x^1:=x$, the corresponding spacetime will be 2-dimensional $(x^0,x^1)$, say with $(+,-)$ signature convention.

Now pick $x^0:=t$ to be time. Then $\dot{x}^0=1$, and we have produced a way to get rid of the single-dot derivatives in the geodesic equation.

II) Next pick a metric as

$$g_{00}:= 1+\frac{2\Phi}{c^2}, \qquad g_{11}~:=~ -1,\qquad g_{01}~\equiv~g_{10}~:=~ 0,$$

where $\Phi \propto x^2$ is the potential of the harmonic oscillator,

Finally, form the corresponding Levi-Civita Christoffel symbols $\Gamma^{\lambda}_{\mu\nu}$ from $g_{\mu\nu}$.

In this way it is possible to reproduce the harmonic oscillator equation as a geodesic equation.

III) In fact, more generally, this is how the weak field limit of General Relativity reproduces the Newtonian theory. See also this Phys.SE post.

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Thanks. I would point out that the $g_{00}$ metric component you "pick" is not as arbitrary as it may seem. If I'm not mistaken $$\bigtriangledown^2 g_{00}=8\pi G T_{00}$$ implies that $g_{00}=2\phi+1$ (where $T$ is the energy-momentum tensor). $\phi$ can be the potential of the harmonic oscillator or of anything. – ben Oct 9 '12 at 16:46
@ben: those things don't couple equally to things of all masses, so a mass of $5kg$ would not follow the "geodesic" of a mass of $2kg$ in this potential. – Jerry Schirmer Oct 9 '12 at 17:15

The second equation doesn't reduce to the first. There's no term proportional to $x$ and without derivatives of $x$ anywhere in the second equation. You could couple GR to some sort of potential and get something like the spring force, but that's rarely done. If you did, you'd have some sort of answer like

$$k\,x^{a} = {\ddot x}^{a} + \Gamma_{bc}{}^{a}{\dot x}^{b}{\dot x}^{c}$$

Models like this don't end up being very relevant in the regime where general relativity is important. You could use this to analyze the simple harmonic oscillator in curvilinear coordinates, though.

(As an added aside, since the Newtonian "force" term arises from GR due to the $\Gamma_{tt}{}^{r}$ term in the Schwarzshild solution in standard coordinates, I suppose you could play the same game and have $\Gamma_{tt}{}^{r} = kr$ for some metric, and have your second equation reduce to the first one.

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Shouldn't your equation be $k\,x^{a} = \Gamma_{bc}{}^{a}{\dot x}^{b}{\dot x}^{c}$ ? – user10001 Oct 9 '12 at 7:01
@dushya: No, because then it would be $x^{a}=0$ in Cartesian coordinates – Jerry Schirmer Oct 9 '12 at 13:17
Sorry, now i understand. Your equation comes from inclusion of some force term in usual geodesic equation. But then if you take $k$ in your equation to be negative of the $k$ in OP's question and if $\Gamma$'s vanish then we'll get harmonic oscillator equation! – user10001 Oct 9 '12 at 13:45
@JerrySchirmer Regarding your aside, now that's what I'm talking about. That's what I was trying to get at. In the Newtonian limit (flat and slow) the $00$ component (or $tt$ as you have it) of the Chistoffel symbol is the only one that doesn't vanish. I wanted to see if this component could some how be expressed as $-kx$. But (insofar as I understand) this one non-vanishing component is usually of first order (a field gradient), not "0 order" like $-kx$. Is there a way to think of $kx$ as a field gradient--like $$kx=\frac{\partial \phi}{\partial x}$$? – ben Oct 9 '12 at 14:31
Sure. If you want to not really care about physicality, you can take the metric to be $ds^{2} = -(1+kr^{2})dt^{2} + dx^{2}+dy^{2}+dz^{2}$, and if you make the approximation that $kr^{2} \ll kr$ for whatever reason (I'm not being too careful since we're not really constructing physical models here), you'll find that, if you cut off angular motion, the radial geodesic equation is ${\ddot r} +kr = 0$ – Jerry Schirmer Oct 9 '12 at 15:12