# Effective mass of gravitational waves

1. Apr 27, 2012

### GeorgeDishman

Hulse and Taylor's observation of PSR B1913+16 provides indirect evidence for the existence of gravitational waves ("GW") in accordance with GR. Simulations show recoil of merging black holes due to the momentum of the GW they emit during the inspiral.

http://www.black-holes.org/explore2.html

GW propagate at c as disturbances on a flat background and as gravitons are massless, the usual rule E=p should apply. Other than being quadrupole radiation and gravitons being spin 2 accordingly, the picture is analogous to EM waves and photons.

Consider two binary star systems, A-B and C-D with an observer O somewhere between:

A-B O C-D

In the special case of four equal mass stars orbiting in a common plane with equal periods, GW of equal amplitude will travel in both directions from the two source systems resulting in a standing wave pattern at the location of O. More generally, if the periods differ, O can select a motion along the line between the systems which equalises the observed periods via the Doppler effect and if the masses differ, he can select a position closer to the lighter pair to equalise the amplitudes.

We can calculate the effective mass of an aggregate formed from a pair of photons of equal frequency moving anti-parallel. Because the energies add but the momenta cancel so the resulting mass found from m2=E2-p2 is non-zero, i.e. while one photon has no mass, a pair does.

It seems to me that the same applies to the GW scenario, i.e. the standing wave component at location O formed from waves of equal amplitude and frequency moving antiparallel has finite energy but no net momentum and thus should have an effective mass.

Is this correct and if so would that mass also be a source of gravitation, or to put it another way, should the energy density of those waves be included in the stress-energy tensor?

2. Apr 27, 2012

### Mentz114

In the GW metrics I've come across the GWs do not appear in the stress-energy tensor. I find this a bit puzzling if they carry energy.

3. Apr 27, 2012

### Whovian

Define energy in GR. :P

4. Apr 27, 2012

### GeorgeDishman

My guess is that it is ignored as a negigible contribution in order to avoid making the calculation recursive.

From : http://en.wikipedia.org/wiki/PSR_B1913+16#Star_system

"The total power of the gravitational radiation (waves) emitted by this system presently, is calculated to be 7.35×1024 watts. For comparison, this is 1.9% of the power radiated in light by our own Sun."​

;-)

5. Apr 27, 2012

### Mentz114

The pp-wave metric
$$ds^2=\left( -\left( {z}^{2}+{y}^{2}\right) \,C+2\,y\,z\,B+\left( {y}^{2}-{z}^{2}\right) \,A\right){du}^{2}\, +{dz}^{2}+{dy}^{2}+2\,du\,dv$$
with C=0 is a vacuum, but the tidal tensor shows the GW signature Tyy=-Tzz = A(u).

My first thought is that if we superpose two tidal tensors with A(u) in a certain phase then they could cancel or re-inforce each other, so
GW standing waves are not ruled out. I could be wrong.

Last edited: Apr 27, 2012
6. Apr 27, 2012

### GeorgeDishman

My grasp of the maths is minimal but thank you for including the metric, it's something more for me to study.

Whether that means a gravitational standing wave translates to the equivalent of a mass density is still unclear to me but if energy is flowing through the region while the momentum flows cancel, it looks plausible.

7. Apr 27, 2012

### Staff: Mentor

Looks OK to me, as long as by "effective mass" you mean "the system as a whole can be assigned an effective energy-momentum 4-vector in the asymptotically flat coordinate system surrounding it, which 4-vector is timelike and therefore has a non-zero effective invariant mass."

The effective invariant mass, as I just defined it above, is not local; it's global. It's a property that's only "visible" when you look at the global "imprint" of the gravitational waves on the spacetime. You can't look at a particular event and point to it and say "there is gravitational wave energy *here*".

The stress-energy tensor, on the other hand, is *local*; it assigns definite stress-energy to a single event--a *point* in spacetime. The "effective mass" as above can't be localized in this way. That's why it doesn't appear in the SET; it is not a "source" of gravity in the sense of appearing on the RHS of the Einstein Field Equation. As far as the EFE is concerned, gravitational waves are "curvature", and appear on the LHS of the EFE.

There have been extensive discussions of this in other threads, and it seems to me that part of the difficulty is a natural intuitive desire to have a single, unified concept of "energy" that works like the one we're used to from pre-relativity physics. The problem is that there isn't one; what we intuitively think of as "energy" actually conflates several different concepts that, in GR, are distinct and don't always travel together. One concept is energy as "source" or "stuff", something tangible--this is embodied in the SET. Another is energy as "something that can do work"--this is the sense in which gravitational waves carry energy, and doesn't really have a single representation in GR, since "work" itself does not, it depends on the specific scenario you are looking at. A third is energy as a conserved current arising from time translation symmetry--this is embodied in the timelike 4-vector in the asymptotically flat spacetime, of which energy is the 0-component. Once we start drawing these distinctions, it is easy to see that the different concepts of "energy" don't always have to be correlated with each other.

8. Apr 27, 2012

### pervect

Staff Emeritus
Note that only in Newtonian theory is it true that "mass" causes gravity. In General Relativity, it is the stress-energy tensor, and not mass, which causes gravity. Mass (like energy) is rather hard to define in GR

Gravity waves do not appear in the actual stress-energy tensor. But in linearized theory you can assign them an effective stress-energy tensor - i.e. when you linearize the non-linear Einstein field equations, you get a linear correction term when gravity waves are present which is called the "effective stress energy" tensor of the gravity waves.

As I recall MTW has a section on this.

A rough analogy would be a non-linear electric circuit (say one with a square law component, so the output voltage is something like vout = vin + alpha vin^2, where you make the input voltage a sine wave.

Then if vin = cos(wt), then vout = cos(wt) + alpha cos^2(wt), and the non-linear cos^2(wt) term gives you a DC shift in the output voltage.

.

9. Apr 29, 2012

### GeorgeDishman

Yes, I believe that's the way I'm using the term. Based on my knowledge of SR, the invariant magnitude of the 4-vector is what I think of as the definition of "mass".

OK, this is where I am learning. From cosmology, I thought it could also handle continuous sources such as treating a homogenous distribution of galaxy clusters as a "perfect fluid" as well as an isolated "dust".

Sorry if I'm going over old ground, please just point me to previous threads rather than waste your time typing a repeat explanation if you know of any that cover this.

OK, that helps, thanks. Let me see if I can put those together in two comparative scenarios.

First consider a Dyson sphere one light year in radius around a large single star. Fusion in the star means it loses "mass", the Sun is often quoted as reducing in mass by 4 million tonnes per second (that is mass as "stuff"). That is converted to radiant energy which flows as EM waves to the sphere. Assume the inner surface of the sphere is covered with photovoltaic cells absorbing all the radiation and storing it in batteries all over the surface (that is intended to cover your "something that can do work" variant). I also understand the aggregate of all the photons in transit between the star and the sphere have an effective mass as defined at the top of the post (which I think covers energy as a conserved current). Overall, the star, photons and stored battery energy all contribute to the total "mass" of the Dyson sphere system as seen as a point source (everything is spherically symmetric) by a test particle at some distance outside. If the star exploded as a supernova, a large amount of mass would be lost in a short time (e.g. Type Ia would last about a month). A year later that energy would be stored in the batteries and during the intervening time, the photons have an effective mass so that the remote test particle would not see any change in the system mass, it would be unaware of the destruction of the star.

In the second scenario, consider a pair of orbiting black holes at the centre. They are steadily emitting gravitational radiation at some power level1 and we can envisage some sort of pendulum convertor capturing the energy on the sphere's inner surface (as LIGO should capture a tiny fraction of the energy of a GW) and storing it in the battries. The picture is similar to the steadily shining star but with GW instead of EM. Now what happens when the black holes merge? During the last few minutes of the inspiral, there is a burst of GW radiation produce. That would again travel for a year to the sphere where it would be captured. Our remote test particle would see the same total mass for the system before the merger and after the energy was stored in the batteries (assuming an unrealistic 100% efficiency) but during the time the energy was transported as GW, the aggregate mass would be reduced because the GW don't contribute to the stress-energy tensor.

In the latter case, the remote test particle would detect the merger because the total mass would drop for a year then revert to the original value.

Is that correct?

1. For example, the power radiated by a pair of solar mass neutron stars 0.63 light seconds apart is 100 times the EM radiation of the Sun according to Wikipedia which would imply a mass reduction rate for the binary system of 400 million tonnes per second:

10. Apr 29, 2012

### Staff: Mentor

"Dust" as the term is used when describing GR solutions is really just a special case of "perfect fluid" where the fluid pressure is zero; "dust" solutions are not derived by treating the individual dust particles as having individual 4-vectors. Solutions for perfect fluids, including "dust", are derived directly from the Einstein Field Equation: "perfect fluid" is simply a name for a particular form of the stress-energy tensor that is used on the RHS.

The other threads weren't directly about the question you were asking, it just came up in the course of discussion of other topics. You actually asked the question in a much clearer way than it arose in those other threads, so it was easier to just answer it directly.

More or less, yes; there are complications due to the fact that the radiation that is being emitted at the surface of the Sun did not arise directly from fusion reactions deep inside the Sun; IIRC it can take more than a million years for energy produced by fusion in the Sun's core to actually reach the surface and be emitted. But at a gross level, yes, the radiation from the Sun corresponds to "stuff" (nonzero SET) that is lost to the Sun itself.

No problem here. The only unrealistic assumption here is that the Dyson sphere emits nothing outward; any real substance at a temperature greater than absolute zero emits radiation in all directions, so any real Dyson sphere would emit radiation outward--though quite possibly much, much less than the star inside it (if the capacity of the batteries was large enough). For purposes of this thought experiment I'm fine with ignoring this complication.

Yes. That "mass" corresponds to the total "conserved current" with respect to the time translation symmetry of the spacetime as a whole.

Yes, exactly.

Again, no problem here from an energy standpoint. However, there is one key change from the EM scenario: a spacetime with GWs in it *cannot* be spherically symmetric! (And of course two orbiting black holes are *not* spherically symmetric.) This means that the total mass is not the only externally measured quantity involved. See below.

No, this is incorrect. The total mass seen by the outside observer, meaning the effective "conserved current" with respect to the asymptotic time translation symmetry, is *not* in general the same as what you would get by just counting up all the pieces of nonzero SET (and applying any extra factors, such as the "gravitational redshift factor" I noted above, when the field is sufficiently strong to make them significant). It just so happens that, in the particular case of spherical symmetry, it *does* work out that way; but it's not true in general. The case you describe, as noted above, *cannot* be spherically symmetric, because there are GWs present (and even after the final black hole has "settled" into its stationary end state, where no GWs are being emitted, that BH will be rotating rapidly, so it still will not be spherically symmetric). So you can't do what you did in the spherically symmetric case and just "add up" all the pieces of nonzero SET and expect the answer to be conserved. You have to allow for energy stored in GWs that does not appear in the SET but that does appear in the total externally measured mass of the system. As your example shows, allowing for this energy is necessary in order to "balance the books", since otherwise it seems like energy just disappears (when emitted as GWs) and then reappears again (when the GWs do work and are absorbed).

The lack of spherical symmetry also complicates what the external test particle sees; see below.

No; see above. However, that does not necessarily mean that what the remote test particle sees stays the same throughout. For a system that is not spherically symmetric, there are other external "imprints" on the field besides the total mass, and those can potentially change. The two of interest here are the angular momentum and the mass quadrupole moment. Whether these three external imprints on the field (mass, angular momentum, quadrupole moment) change or not depends on how the Dyson sphere reacts as it absorbs the GWs.

If we assume what I think you meant to assume, that the Dyson sphere does not compensate at all for what happens when it absorbs the GWs, but absorbs everything that they carry and changes its own state accordingly, then the sphere will acquire whatever angular momentum and quadrupole moment are carried by the GWs from the original pair of orbiting BHs. That means the sphere might start spinning (if the GWs carry any angular momentum--i.e., if the final AM of the final spinning BH is less than the AM of the original orbiting BHs), and the energy absorbed by the batteries will not be spherically symmetric but will acquire a nonzero quadrupole moment. In that case, the external field seen by the remote test particle *will* stay the same (because the internal components are just transferred to the sphere).

However, if we assume instead that the Dyson sphere always has to remain spherically symmetric, then it can't acquire any angular momentum and it can't acquire any nonzero quadrupole moment. That means that the sphere will only be able to absorb a portion of the GWs; the rest will pass through the sphere and continue outward. As those GWs pass the remote test particle, it will see the field change gradually from the initial state (due to the orbiting BHs) to the final state (due to the single spinning BH). The total mass, angular momentum, and quadrupole moment seen by the remote test particle could all change in this scenario.

11. May 4, 2012

### GeorgeDishman

Sorry, I wrote a much longer reply yesterday and IE crashed just before I posted, this is severely cut back through limited time.

Oops, thanks for correcting me on that.

OK, that aspect was what I was trying to explore.

Yes, I understand what you say. I was thinking of a test particle on the axis of the binary system so there is rotational symmetry when averaged over a cycle (necessary for handling the GW) but clearly there are going to be effects like Lense-Thirring due to the overall rotation whether in the binary or the Dyson sphere. Also, while there is symmetry for reflection in the binary system's orbital plane, that won't fully cancel at the test particle. There are probably other effects I've missed too, I should have made clear that I was only looking at a restricted aspect.

That's the key point I was querying. I'm aware that energy isn't always conserved in GR so if you said the apparent mass dropped then recovered, I would have considered that this was such a case, though I would be surprised. If the total is constant, life is much simpler :-)

Excellent. That seems to accord with Pervect's point:

What I am gathering is that the effects of GW in the linearised approach can be described by an "effective stress energy tensor" which represents the energy-momentum they transfer, and that this goes back into the RHS to complete the picture. My guess is that we could have a more complex non-linear theory which did it all in one but that taking a linear approximation and then putting this in as a correction is much simpler and an adequate approximation to the non-linear approach.

It also suggests, going back to the original question, that if we calculate the total effective stress energy tensor for the two contributions creating the standing wave and integrate over the wave period in any small volume dV, it should be equivalent to an energy dE in some units) hence give an equivalent mass density c-2dE/dV. Is that correct?

The 3D view of that would be an interesting interference pattern (drifting with time if the binary pairs have different periods) combined with the polar diagram for the GW radiation from each of the binary systems.

12. May 4, 2012

### Staff: Mentor

This is more or less correct; but it's important to understand that the "RHS" here is now *not* the RHS of the standard EFE. What we have essentially done is to separate out a piece of the LHS of the standard EFE, call it the "effective SET due to GWs", and transpose it (or more accurately its time average) over to the RHS, so the new "RHS" is now a combination of the original SET from the standard EFE and the "effective" SET of the GWs--and, correspondingly, the LHS is now no longer "curvature" but rather something like "background curvature", the part of the curvature that is static or very slowly varying, as compared to the rapidly varying GW piece.

The more complex non-linear theory is just standard GR itself.

For many cases, yes; we work with linearized GR because full nonlinear GR is too complicated to solve. However, this is becoming less true as computer power advances; I am not up to speed on the latest work in this area, but my general sense is that it is much more possible now to run computer models using full *nonlinear* GR (or at least going to much higher than linear order in a power series expansion). Even if this is true, though, I still think linearized GR is useful conceptually, as a link back to our Newtonian intuitions as applied to cases where Newtonian theory is a reasonably good approximation, and where the next set of "post-Newtonian" terms can be reasonably conceptualized as small corrections to the Newtonian equations.

13. May 4, 2012

### GeorgeDishman

Yes, I get that. The way I'm looking at it is that the linearised version is an approximation and adding the effective SET of the GW is a first order correction which reduces the discrepancy with the full version.

It's very tempting to see the energy that moves from the stars to the sphere in the example as something being "transported" by the GW but I suspect that owes more to the Newtonian tradition.

You didn't comment on the equivalent density question, can you say if the EFE for a standing GW is equivalent to a such a distribution?

14. May 4, 2012

### Staff: Mentor

Conceptually I think this is OK. I'm not sure if it corresponds mathematically with how the linearized and "post-Newtonian" terms are actually derived from a power series expansion; I'd have to go back and look at the derivations.

Not sure, I'd have to work through it. Conceptually, again, the analogy you're making with photons seems OK. But I haven't worked through it mathematically to see how precisely the analogy holds.

15. May 5, 2012

### GeorgeDishman

Thanks Peter, I wouldn't expect you to do the maths, I should learn enough to do it myself. It's just the general approach I was asking about. If you don't think I'm missing something that means the answer would inevitably be zero, then I can think I have a reasonable qualitative feel for the situation.

Thanks very much for the help to both you and Pervect.

George