# Oscillating masses and gravitational waves

1. Feb 14, 2016

### DaTario

Let me ask one simple question. For less than a thousand dollars one can buy an apparatus which can provide experimental evidence for the gravitational attraction between two masses of one kilogram each, placed at distances of the order of $10^{-1} m$. By making one of these masses to oscillate one time with a very small amplitude ($\approx 0.005 m$) and with a period T of, say, 10 seconds, we will see a change in the gravitational force's direction on the other body. Assuming that the change in gravitational force has propagated according to the predictions of GR, this experiment will also correspond to one which exhibits one GW.

Is it correct?

Best regards,

DaTario

2. Feb 14, 2016

Staff Emeritus
No, that's not radiation. That's near-field.

3. Feb 14, 2016

### DaTario

But isn't the difference between these two cathegories just pure formality? Near the source the field also propagates, isn't it?
It seems to me that this names refer to some practical relevance due to the morphology of the waves or the way the propagation occurs, not due to something that has an intrinsically physical distinction. It seems that this difference has more to do with the mathematical multipole expansion of the field in the two regions than with the essencial physical principles involved.

Please, correct me if I am wrong.

4. Feb 14, 2016

Staff Emeritus
No. Near-field goes as 1/r^2. Radiation goes as 1/r.

5. Feb 14, 2016

### DaTario

I am aware of this fact, but this seems to me as just a mathematical decomposition of the signal. Both sum up to being the complete signal. Near the source we find both.

Are you saying that near my oscilatting one kilogram there is no GW ?

6. Feb 14, 2016

### PAllen

There is GW, but it is like 20 orders of magnitude smaller than the near field effect. This is even true for the inspiralling BH (the factor might be different). The GW become significant at great distance because the near field effect decreases so much faster. Further, the affect of GW on matter is completely different than the effect of pushing a source back and forth.

7. Feb 14, 2016

### DaTario

Ok. But when you say "Further, the affect of GW on matter is completely different than the effect of pushing a source back and forth." are you saying that the physics are totally different? Or is it only a difference in the function F(t) which will act on the body. By the way I manage to convert my question in one gracious figure.

8. Feb 14, 2016

Staff Emeritus
Nevertheless, in the interest of communication, doesn't it make sense to use the words "near-field" and "radiation" the way everyone else does? Otherwise we're in Humpty-Dumpty Land, where "[a word] means just what I choose it to mean, neither more nor less!"

There is also the issue of quadrupole radiation. If you move a mass back and forth in only one direction, there is no gravitational radiation. It needs a changing quadrupole moment, which requires motion in two dimensions. In this respect, gravitational radiation is unlike electromagnetic radiation.

9. Feb 14, 2016

### Staff: Mentor

What is the waveform graph in this picture supposed to be describing? Do you know how the waveforms of GWs are described?

10. Feb 14, 2016

### DaTario

Of course. No problem in using this terminology.
Now considering the figure I posted, it seems to be, in principle, possible to produce the same effect of Strain vesus Time, on those interferometers by artificially moving a considerable piece of mass in the vicinity of the apparatus.

My question address particularly the possibility of imitating the radiation field through the near field produced by the use of a well controlled gravitational source.

best wishes,

DaTario

11. Feb 14, 2016

### lpetrich

Yes indeed. Solar-System tests of general relativity are all near-field tests.
That's not correct. It will still have a quadrupole moment relative to some point, as it does a dipole moment. Monopole (M), dipole (D), and quadrupole (Q):
$$M = \sum m$$
$$D_i = \sum m x_i$$
$$Q_{ij} = \sum m (x_i x_j - (1/3) \delta_{ij} x^2)$$
The lowest-order mass-monopole effect is Newtonian gravity, much like the electrostatic force in electromagnetism. It is quasi-static, meaning that the field seems to propagate instantaneously.

The lowest-order source for electromagnetic waves is the electric dipole. Its gravitational counterpart is the offset of the barycenter, and it doesn't radiate.

The lowest-order source for gravitational waves is the mass quadrupole.

I'll now consider the mass quadrupole moment of an object with mass m displaced in only one direction: {x,0,0}. It is m*x2*{{2/3,0,0},{0,-1/3,0},{0,0,-1/3}}. So it's nonzero.

To lowest order, the energy output of gravitational waves is proportional to the square of the third derivative of the mass quadrupole moment. To vanish, that quadrupole moment must be a polynomial in time with a degree of at most 2. This in turn implies that a single source will have a constant velocity. So a single source accelerating will make gravitational waves, though multiple sources can produce waves that cancel each other out, as with electromagnetism.

12. Feb 14, 2016

### DaTario

Sorry for having inverted the function. I only tried to express the fact that a pendulum tends to stop and it seemed to be reasonable to propose that this would also happen to the oscillations.
Trying to answer your question, I suppose these oscillations are produced due to a time dependent interference pattern caused by local metric oscillations which change differently the arms of the interferometer.

13. Feb 14, 2016

### Staff: Mentor

Before tackling this question with regard to gravity and gravitational waves, it's helpful to consider it for the simpler case of electromagnetism and electromagnetic waves. Suppose we have a source charge that we are wiggling back and forth. The wiggling produces changes in the EM field that propagate outward as wave fronts. But if we have a test charge a fairly short distance away, the field seen by that test charge will not be just the radiation field; instead it will be a combination of the radiation field and the "average" Coulomb field of the source charge. ("Average" here means, heuristically, the Coulomb field that would be produced by the source charge if it were stationary at its average position.) The Coulomb field will be much stronger if we are close enough to the source charge; but it falls off as $1/r^2$ while the radiation field falls off as $1/r$; so as we get farther away from the source charge, the radiation field becomes larger relative to the Coulomb field, and once we are very far away, the Coulomb field is negligible and the radiation field is all that we can detect.

The different dependences on distance ($1/r^2$ vs. $1/r$) are why the two effects, "Coulomb" and "radiation", are physically different; one can't "imitate" the other. Similar remarks apply to gravitational waves vs. the Newtonian gravitational "force".

14. Feb 14, 2016

### DaTario

I am not quite sure about these calculations, but the term which is instantaneous may be seen as an artificiallity of the formalism, once what is measured is always the sum of the terms. And by summing up all the terms, that one which is instantaneous cancels out with other terms, in such a way that the gravitational field cannot be understood as being instantaneous, except if you want to add and take out this term.

15. Feb 14, 2016

### DaTario

I understand this point, but note that I have used a malicious argument with the word "artificially" with respect to the movement.

16. Feb 14, 2016

### Staff: Mentor

Basically, yes. But one waveform can't really describe a GW. GWs have two possible polarizations, as shown here:

https://en.wikipedia.org/wiki/Gravitational_wave#Effects_of_passing

So to fully describe a GW, you need two waveforms, one for each of the polarizations; there is no necessary connection between them. (I don't know if LIGO was able to measure both waveforms; they may only have been able to measure one, since one waveform from each detector is all that I've seen in published info.) Also, you have to bear in mind that even if you have both waveforms, that is not a complete description of spacetime curvature, i.e., it's not a complete description of the effects of gravity; see my previous post.

17. Feb 14, 2016

### Staff: Mentor

You're missing the point. The point isn't that one piece of the field is instantaneous and the other is time delayed. Any changes in the field can only propagate at the speed of light. So what looks like an "instantaneous" Coulomb or Newtonian field (depending on whether we're talking about EM or gravity) isn't really; the observed field isn't responding to the source "right now", it's responding to the source on the past light cone of the event at which the field is being measured. But for the portion of the source which can be considered static (for example, the "average" position of the source in my EM example), the source on the past light cone is the same as the source "right now", so the field looks like an instantaneous field because the source didn't change during the light propagation time.

However, you are ignoring the real point (even though you say you understand it), which is that one piece of the field falls off as $1/r^2$ and the other falls off as $1/r$; in other words, we have two different aspects of the field with two different distance dependences. That is why we separate them conceptually. Of course nature doesn't care how we write the equations; but the conceptual separation of different distance dependences is not "artificial". It isn't put in by hand; it pops right out of the equations.

18. Feb 14, 2016

### DaTario

Ok with this part. The complete information about a wave, in general, rarely can be inferred by monitoring the pertubation in one or two locations.

19. Feb 14, 2016

### DaTario

Ok, so these two terms have important physical differences. But both can be understood as propagating terms, which do not respond to the source " right now", as you said.
What is the relevant difference between them? Specially with respect to this matter, GW?

20. Feb 14, 2016

### Staff: Mentor

To elaborate on this: if you wiggle a mass around (with a quadrupole moment with a nonzero third time derivative), you will, in principle, generate GWs, i.e., fluctuations in spacetime curvature. But those fluctuations will be superimposed on the static field of the static part of the source (heuristically, the average position of the mass, such as the pendulum in your example). Only the fluctuations will produce changes in things like the arm lengths of a GW interferometer; the interferometer won't detect the static part of the field. And if the static part of the field is strong enough (i.e., if you're close enough to the source), it might overwhelm the fluctuating part--heuristically, your interferometer might have to be under such a large strain just to sit at rest in the field that the tiny changes in strain produced by the fluctuating GWs become unmeasurable. (Consider trying to measure small fluctuations in the position of a large electric charge using an antenna that is very close to the charge--the large static electric field might make the small fluctuations in position of the charges in the antenna due to the EM wave unmeasurable.)

So when you talk about "imitating" a radiation field with a near field, that doesn't really make sense. Wiggling a mass produces a radiation field, period. The question is whether that radiation field is detectable against the background field of the source. The further you get from the source, the weaker that background field becomes, and the easier it becomes to detect the radiation field.