# I Gravitational wave like a harmonic oscillator?

1. Aug 9, 2016

### exponent137

Electromagnetic wave behaves like a harmonic oscillator. Similarly a photon behaves like a quantum harmonic oscillator.
http://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf
$dA/dt$ and $A$ behaves like $dx/dt$ and $x$ at a harmonic oscillator.

I suppose that gravitational wave also behaves like a harmonic oscillator. (Let us assume that gravitational fields are very weak.) What are analogies here for the harmonic oscillator, Maybe $g$ and $dg/dt$?

2. Aug 9, 2016

### Jonathan Scott

If you are just looking at analogies, the choice is not unique, as for a continuous harmonic oscillator each further derivative with respect to time relates to the previous one in the same way as $dx/dt$ relates to $x$.

3. Aug 9, 2016

### Staff: Mentor

If by "g" you mean the metric, then yes, more or less; what oscillates in a gravitational wave is the spacetime geometry, which is described by the metric. If by "g" you mean the "acceleration due to gravity", then no.

4. Aug 9, 2016

### Jonathan Scott

Are you sure? That's effectively the gradient of the metric in a chosen coordinate system and I'd expect it to fluctuate too at the same frequency (about its average value, not about zero). I'd also expect its time derivatives to fluctuate at the same frequency. Of course, the amplitude for any of those derivatives would be even more negligible than the basic fluctuation in the geometry.

5. Aug 9, 2016

### Staff: Mentor

No, it's the gradient of "gravitational potential". But that concept is only well-defined in a particular set of spacetimes. A spacetime with GWs present is not one of them.

6. Aug 9, 2016

### exponent137

Let us look equation
$(\partial^2/\partial t^2-\nabla^2)h^{\mu\nu}=0$
$h$ is a little deviation from $g$.
How we can see equation for harmonic oscillator from this equation?
Analogous equation for electromagnetic wave gives equation for the harmonic oscillator, as I wrote above.

7. Aug 9, 2016

### Jonathan Scott

In both cases, this is simply a wave equation which says that something propagates at a fixed speed. The oscillation is driven by the source of the wave.

8. Aug 9, 2016

### exponent137

Is this the correct equation of the source:
$(\partial^2/\partial t^2-\nabla^2)h^{\mu\nu}=16 \pi G^{\mu\nu}$
I think about equation in page 5 in http://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf :
$H=\int d^3 k (\dot{|A|}^2/c^2+k^2 |A|^2)$
In the further derivation it is evident that this is an analogy for a harmonic oscillator. I ask, how to find parallels with gravitational field instead of electromagnetic field?

9. Aug 9, 2016

### Jonathan Scott

I don't recall off hand the details of the equation for a gravitational wave source, so I can't comment on that.

One can consider a general electromagnetic wave to be composed of oscillations at particular frequencies. However, the pattern and frequencies of oscillations is still due to whatever is emitting (and perhaps receiving) the waves. For electromagnetic waves, such waves are often created and absorbed in quantized form as photons, for example when energy levels change in atoms. For gravitational waves, the source is typically bodies orbiting one another. In the case of the recent LIGO observations, the bodies were spiralling in to a collision so the frequency was increasing.

10. Aug 9, 2016

### pervect

Staff Emeritus
The equation you wrote, $(\partial^2/\partial t^2-\nabla^2)h^{\mu\nu}=0$, is the wave equation. I don't quite see the relationship here to harmonic oscillators, which (looking it up to be sure) has the form $d^2 x / dt^2 + k x = 0$. They both have a second derivative with respect to time, but there isn't anything like the term $\nabla^2$ in the harmonic oscillator equation that I see. Perhaps I'm missing something - but I do believe you'd do better to research the wave equation than the harmonic oscillator equatiion.

The solution for the wave equation in one dimension, by the way , is just f(t,x) = f'(t-x). i.e. any function of (t-x) will satisfy the wave equation. You can check this with the chain rule, the starting point for working this out is noting that $(\partial / \partial t) f(u) = (df / du) \partial u / \partial t$ with u = t-x.

If you want to get into some of the finer details, it's actuall $\bar{h}_{\mu\nu}$ that satisfies the wave equation (with the right gauge choices), but once you have $\bar{h}$ it's easy to find h - and vice versa. $h_{\mu\nu} = \bar{h}_{\mu\nu} - \bar{h}^a{}_a \eta_{\mu\nu}$ where $\eta_{\mu\nu}$ is just the diagonal metric of flat space-time. Note that $h^a{}_a$ often written just as h, it is found by taking the trace of $h_{\mu\nu}$, for the Minkowskii metric of flat space-time it's $h = h^a{}_a = h_{00} - h_{11} - h_{22} - h_{33}$ with your implied choice of a (+---) signature.

In electromagnetism, there is a potential function $\phi$ and a magnetic vector potential A, which are often combined into a single 4-potential (also usually called A). The analogy is that the E&M potential A satisfies the wave equation in the Lorentz gauge for E&M, and in linearized gravity, $\bar{h}$ also satisfies the wave equation (in the appropriate gauge). The difference is that A only has 4 components, while h has (in general) 4 x 4 = 16 components.

11. Aug 9, 2016

### Staff: Mentor

The operator on the LHS is applied to $\bar{h}^{\mu \nu}$, as pervect said.

The tensor on the RHS should be $\tau^{\mu \nu}$, not $G^{\mu \nu}$. Basically, this equation is what you get when you start with the Einstein Field Equation, make an appropriate choice of gauge (transverse-traceless linearized), and rearrange some terms. The LHS is what's left of the LHS of the EFE (the Einstein tensor $G^{\mu \nu}$) after doing all this. The RHS $\tau^{\mu \nu}$ is the stress-energy tensor plus some terms from the Einstein tensor that do not involve derivatives (these terms are sometimes interpreted as "energy stored in the gravitational field", though that interpretation has significant limitations).

12. Aug 9, 2016

### exponent137

I heard that, mathematically, graviton exist for small gravitational fields. So I suppose that gravitational wave can be quantized (for small gravitational fields) similarly, as electromagnetic field can be quantized, i.e. with the analogy with the harmonic oscillator.

But I do not see how to find this analogy.
Do you maybe claim, that this analogy for gravitational waves does not exist?

13. Aug 9, 2016

### Staff: Mentor

I'm not sure where you are "hearing" this, but it's not the way I would put it. There is a hypothesis that gravity should be quantized like everything else is, and the graviton would be the quantized particle associated with gravity if this hypothesis is true. But we don't have any evidence for it.

Why not? You just described it, and you gave the wave equation for it earlier in this thread. It's just another case of quantizing a harmonic oscillator.

The mathematics works just fine. Whether it's actually true is something we will hopefully find out eventually by experiment.

14. Aug 9, 2016

### George Jones

Staff Emeritus
Taking the Fourier transform (with respect to space) of the wave equation results in equations that have the same form as the harmonic oscillator equation.

15. Aug 9, 2016

### pervect

Staff Emeritus
In n spatial dimensions, or just in one? The OP's question was about n dimensions, my post really only addressed the 1d aspect, which is I suppose where I'm somewhat concerned I could be missing something.

The harmonic oscillator has a characterstic frequency $\omega$ built write into the solution of the equation, the natural period of oscillation of the system modelled by the equation. The wave equation doesn't really have any natural period, or frequency, rather it has a natural speed of propagation, the speed of the wave, equal to c for light. One can decompose a plane-wave solution of the wave equation into a fourier series (if it's a solution periodic in time) or a fourier transform (if it's not), because any function $f(\psi)$ can be written as a fourier series. When each of the plane waves travels at c, the sum of the plane waves also satisfies the wave equation. While one can impose a period on the solution artificially, the solution to the wave equation doesn't have a natural period like the harmonic oscillator. The Ligo chirp is an example, the Ligo chirp isn't a sine wave, though it is convenient to analyze it's spectrum via foruier transform methods.

My thinking on the topic is oriented towards plane waves, which have a single direction of propagation. It might be helpful to consider a more general non-plane wave solution of the wave equation, but at the moment I don't have one to write down. Recently, I've been thinking that I should try and write one down for the E&M case, to better understand (by analogy) non-plane GW's.

I used to think that any non-plane wave solution could be decomposed into the sum of an infinite number of plane wave solutions, but attempting to carry out exactly such a decomposition makes me wonder if this general idea was misguided.

16. Aug 18, 2016

### exponent137

1. Does this means that the first eq. in page 5 in http://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf is completely equivalent to the wave equation?

2. So, $h$ in my above eq. for gravitational wave equation is equivalent to $A$ in EM wave equation.

3. Does exist something equivalent to magnetic field in gravitational field?

Last edited: Aug 18, 2016
17. Aug 18, 2016

### Jonathan Scott

Yes, in the form of a rotation field, which relates to acceleration in a very similar way to that in which a magnetic field relates to an electric field. A rapidly rotating gravitational source can induce rotational "frame dragging" where an object can be rotating as seen from a distance even though locally it does not experience any of the effects associated with rotation. This effect is very weak and difficult to observe, and it was hoped that the Gravity Probe B experiments would confirm it, but although the results were generally consistent with predictions, experimental problems meant that it was not possible to obtain a very accurate result.

There is also an additional frame-dragging effect related to the normal linear acceleration, which is that a gravitational source undergoing acceleration causes a small additional induced gravitational field. This is analogous to the $\partial A/\partial t$ induction contribution to the electric field. The effect is so tiny compared with other gravitational effects that it does not seem likely that any experiment can be done in the foreseeable future to test it directly, but there is an interesting link with Mach's Principle in that it appears that if the whole universe were conceptually accelerated in this way relative to a test particle, then the total effect would be that the test particle would require a force to keep it from being accelerated along with it, proportional to the mass of the test particle, which from the point of view of the universe is simply normal inertia. This "sum for inertia" of the gravitational induction effects of all masses seems to be at least around the right order of magnitude, and it seems that this could provide an explanation of inertia, but so far there is no complete consistent theory as to how this could work, and so far it doesn't seem fully compatible with General Relativity.