# I Are gravitational waves subject to gravitational pull?

1. Feb 24, 2016

### greswd

According to the sticky bead argument, gravitational waves do carry energy.

As such, are they, like photons (EM waves), subject to the gravitational pull of planetary bodies?

Last edited: Feb 24, 2016
2. Feb 24, 2016

### PAllen

In the main LIGO thread, several papers are linked providing discussion of gravitational lensing of gravitational waves. The upshot of these, is that GW are lensed the same way as light, by an intervening massive body.

3. Feb 26, 2016

### stevendaryl

Staff Emeritus
The topic of to what extent are gravitational waves just like any other wave (such as electromagnetic) seems a little complicated to answer. In Einstein's field equations

$G_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}$

the left side includes geometric stuff--the metric and its derivatives--and the right side includes ordinary non-gravitational matter, energy and momentum. So the electromagnetic field carries energy and momentum, so it makes an appearance on the right side, as a contribution to $T_{\mu \nu}$. A gravitational wave, however, is a fluctuation of the metric tensor, and so it would appear on the left side, as a contribution to $G_{\mu \nu}$. So in this standard form for General Relativity, it's not at all obvious that a gravitational wave should behave like any other wave under the influence of gravity.

But perhaps the fact that gravitational waves are treated in GR the same as other waves is more apparent in the spin-2 field theory approach to gravity. In this approach, you start with linearized gravity, which describes perturbations $h_{\mu \nu}$ to the flat spacetime metric $\eta_{\mu \nu}$ via a wave equation that uses stress-energy, $t_{\mu \nu}$ as a source. To get to something equivalent to GR, the source $t_{\mu \nu}$ has to include not only the energy/momentum due to nongravitational fields, but also the stress-energy due to $h_{\mu \nu}$ itself. So in this way of formulating gravity, it is maybe clearer that propagating perturbations to the metric act just like any other wave when it comes to gravity.

4. Feb 26, 2016

### PAllen

It is complicated, but the work has been done. Search my posts in the main LIGO thread, and you will find the links to the papers, and from their to earlier ones (the ones I link assume things that are derived in earlier referenced papers).

But the upshot is that my initial intuition was validated in full (I had no prior knowledge of these papers). That is, that if GW travel at c, and there is a realm in which geometric optics approximation applies, they must follow null geodesics, thus lensed the same as light. The body of work in the papers validates this.

5. Feb 26, 2016

6. Mar 2, 2016

### greswd

thanks for the info

7. Mar 7, 2016

### MattRob

It seems like there's a much simpler answer to this: because gravity is the result of curved spacetime, in which the "closest approximation to a straight line" is a geodesic, "going in a straight line" is paramount to following a geodesic (ie, falling under the influence of gravity).

In order to not be subject to gravity's influence, it would have to arbitrarily change its own path in a very specific, non-inertial way, which simply wouldn't make any sense.

So, if it's moving inertially, it will be subject to gravity, end of story, no matter what "it" is.

Is this correct?

8. Mar 7, 2016

### PAllen

Pretty much yes. But to talk about a wave (let alone a metric change) having a world line, you do have to ask about whether a geometric optics approximation is valid (that is, can you model a piece of the wave by a world line)? This is an approximation, never exactly true (for light or GW), but an exceedingly good approximation for many circumstances. My first post on this was that it seemed likely there would be a range of circumstances where this was valid, and, if it was, then GW traveling at c would require their world lines follow null geodesics, just like light.

But it is not quite trivial to claim that GW can be treated with geometric optics approximation. Without doing the math or knowing references, it was only a somewhat educated guess. The references show, in hindsight, it was valid. An example of the pitfalls of assuming simple intuitions are valid is that there are exact solutions where two colliding GW produce a singularity. There is no way to model this with geometric optics. You do have to inquire about validity and not assume it without doing (or reading) the math.

9. Mar 8, 2016

### MattRob

Would it be accurate to say something with a finite extent (a wave) is just an integration of an infinite number of points, thus, since any individual element will follow a geodesic, this sum of all the elements should be effected by geodesics in some way (unless the effects cancel somehow)?

Also, I haven't heard that about GW before. That's neat. So they're modeled as non-planar waves, or it's a non-point singularity (has some extent in one or two dimensions), or something else? The thing that has me curious is that often waves are modeled as planar, and in that case the symmetry would argue that there's no particular, single point that would be any different to have a singularity form there instead of anywhere else. The exception to that would be if the waves are non-planar (emitted from two points), and the singularity forms at a point lying along a line in-between the two emission points?

10. Mar 8, 2016

### PAllen

Look, even for light geometric optics completely ignores interference, phase velocity and group velocity, dispersion, etc. For GW, only part of the wave travels at c when analyzed more exactly. Pieces of wave phenomenon traveling on a world line is always just an approximation.