# Supernovae & Gravity Waves

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Just when I thought I was getting a grip - thanks that page MiGL; I will have to do lots more reading

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Please forget (and forgive) my post #25, whose logic only suggests that the gravity step arrives at the same time as the photons (and the neutrinos, said to carry most energy in some types of supernovae).

But I'm still interested in the answer: as apparently the energy emission creates a gravity step along the line of sight, is this step called a gravitational wave, despite not being transverse?

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But I'm still interested in the answer: as apparently the energy emission creates a gravity step along the line of sight, is this step called a gravitational wave, despite not being transverse?

I'm wondering how you would detect it? Would there be a gravitational gradient, detectable using two observers along the line of sight?

And, can you extract energy from it, that is lost from the source? A gravitational wave source radiates energy in the form of those waves. Does such an energy transfer occur here?

From what I've read, it certainly sounds like "gravitational waves" are more than just "change in gravitational field strength." I'm in way over my head, but I'm going to guess that an approaching mass like that of an asymmetrical super nova does not constitute a gravitational wave, but instead might involve movement of the entire gravitational field as if it were static. Or equivalently, the observer is moving through the field. Modelling the supernova as two (or more) masses with static fields that move along with the masses, there would be no need to propagate changes through the field??? Not unless there is a change in acceleration.

Anyway, I think that gravitational waves have a specific definition (they're quadrupole radiation and transverse, as you said, and they require not just acceleration but a change in acceleration). I don't think the longitudinal example is gravitational radiation, nor a gravitational wave.

As an analogy, suppose you have an inertial moving static electric charge, and as it approaches the observer, the measured field strength changes. Does that involve radiation? Would it require photons? I think it doesn't, and that it would be silly to say that the change involves "light waves". There is a detectable change there, but it is not an EM wave, which is something that has a specific definition and describes more than just a field strength gradient.

Edited by md65536
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Thanks for sharing your thoughts! Because I have no chance to understand nor decide an interpretation by myself in this field...

How to detect them: an L-shaped interferometer would have a proper arrangement, and show maximum sensitivity at 45° to its both arms, in a dipole radiation pattern. Though, both arms aligned would be more sensitive with the same arm length. That was the idea behind my question about how quickly such a radiation dampens over distance. I vaguely suppose that true (quadripolar transverse) waves have a longer range, that's why experts hunt these.

In a Newtonian view. energy would be extractible from the wave front. Have two masses, one nearer to the supernova and the other farther. Have a link and a generator between them. Within the propagation time between both positions, the nearer mass is less attracted, so one can exploit a movement and a force that pushes the masses together. What Relativity says to that, I have not the slightest idea.

An inertial mass carrying a charge radiates no electromagnetic field to long distances, hence loses no energy. If one wants to describe it as photons, then as virtual ones, that describe a near field, here electrostatic. Energy can be harvested from a moving charge by an observer, using the proper apparatus, and only then would the moving charge lose energy in the reference frame of this observer - through interaction with the apparatus, not from the near field of the inertial charge alone in space.

Though, a difference with the supernova is that it radiates energy (in the form of neutrinos for instance), which reduces the mass+energy there, as opposed to the inertial charge.

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Well I'm in over my head and not reasoning sensibly. I guess it doesn't matter if you can extract energy at the detector, because you can end up converting between potential and kinetic energy. What really matters is that the source loses energy due to radiation of gravitational energy (independently of other forms of radiation).

Any info I've found leads back to swansont's answer in post #2: A supernova will radiate gravitational waves unless it is (spherically?) symmetrical.

The book, Gravitation (Misner, Thorne and Wheeler) seems to answer all the questions in the thread. Sec 35.5 explains the transverse nature of the wave:

Turn to the special case of a plane wave. Suppose the test-particle separation lies

in the direction of propagation of the wave. Then the wave cannot affect the separa-

tion; there is no oscillation:

{math}

Only separations in the transverse direction oscillate; the wave is transverse not only

in its mathematical description {math}, but also in its physical effects (geodesic deviation)!

Sec. 36.1 explains the quadrupole (or higher) nature of gravitational waves, and why a spherically symmetric source is impossible. This is the only thing that made sense to me: (p 978)

Closely connected

with this theorem is the "topological fixed-point theorem" [e.g., Lifshitz (1949)],

which distinguishes between scalar, vector, and tensor fields. For a scalar disturbance,

such as a pressure wave, there is no difficulty in having a spherically symmetric

source. Thus, over a sphere of a great radius r, there is no difficulty in having a

pressure field that everywhere, at anyone time, takes on the same value p. In contrast,

there is no way to lay down on the surface of a 2-sphere a continuous vector field,

the magnitude of which is non-zero and everywhere the same ("no way to comb

smooth the hair on the surface of a billiard ball"). Likewise, there is no way to lay

down on the surface of a 2-sphere a continuous non-zero transverse-traceless 2 X 2

matrix field that differs from one point to another at most by a rotation. Topology

thus excludes the possibility of any spherically symmetric source of gravitational

I think that one problem with my own reasoning in this thread is that I'm imagining gravitational waves as unrealistically simple. Perturbations on a rubber sheet can be described with a scalar (ie. the height at a point on the sheet). Longitudinal waves can be described with a vector. So reasoning with these analogies will get us stuck???

I don't think a loss of mass eg. by conversion to energy would itself produce gravitational waves, because wouldn't that be spherically symmetrical?

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It's my guess too. The step or gradient in the line-of-sight attraction by the exploding supernova is not a gravitational wave. Quite simply, it's the attraction by the neutrinos that pass by us.

The hairy ball theorem is tricky... "Spherical symmetry" implies that the polarization is uniform in all direction. If this constraint is relaxed, then power can be radiated in all directions without any zero, and even with nearly-uniform distribution. A banal turnstile antenna does it, improvements bring a more unifrm power density. Examples there, just for fun:

http://arxiv.org/pdf/physics/0312023v1.pdf

the turnstile is on page 6.

--------------------

I've tried to put figures on the gravity step produced by the supernova 1987A, a type II in our neighbour galaxy

http://en.wikipedia.org/wiki/1987A

Computing with Newtonian physics, hence not with waves:

A change of 1046J = 1029kg at 168,000 light-years = 1.6*1021m produces at Earth a change in attraction of 2.6*10-24m/s2 and of gravitational energy of 4.7mJ/kg (wow!), which I call "a".

If (please correct if I botched it) the effect on lengths and times is like sqrt[1-0.5*a/c2], this amounts to 5*10-20 variation.

The neutrino burst lasted "less than 13s" [Wiki], but how short was it? I take these 13s=4Gm.

The nasty bit is that wave detectors (provided they're sensitive to such a gradient) would detect variations like 10-21 but they are much shorter than the gradient, like 3km=10µs:

http://en.wikipedia.org/wiki/Gravitational_wave_detector
http://en.wikipedia.org/wiki/Ligo
http://en.wikipedia.org/wiki/GEO_600
http://www.ego-gw.it/virgodescription/pag_1.html
http://www.ego-gw.it/virgodescription/pag_10.html
that would make a variation about 4*10-26 over the detectors' size, undetectable.

Independent clocks at Earth's antipodes would shift by 10-22s, unmeasureable. It needs a single clock, but fibre transmission isn't accurate enough, and vacuum already exists, with only 3km length.

Our Moon's orbit would change by 10-25m and 10-25m/s, let's forget it.

Space probes on the same orbit as Earth but 150Gm=500s apart would span the whole gradient. The phase shift of a 6.8GHz link would be 30nrd, damned little. At 633nm it would be 2mrd, well over the detectors' noise, but how to extract the information from all contributions to the phase variation?

Hey, if someone knows how to exploit a radio link (polarization shift?) from 90AU to Earth vicinity, sending a probe there in 10 years is feasible, with my Solar thermal rocket engine.

Edited by Enthalpy
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I'll have to have a read through that section of Gravitation ( M/T/W ) and see if it clarifies anything for me md65536.

Didn;t realise it had material on gravity waves.

Edited by MigL
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• 6 months later...

Found a Wiki article (about Liénard-Wiechert potential) that states that:

"static fields (the first term) point exactly at the true instantaneous (non-retarded) position of the charged object if its velocity has not changed over the retarded time delay. This is true over any distance separating objects."

http://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential

in the section "universal speed limit", written in normal language in the second to fourth paragraphs.

An other article that states that, for a source of constant velocity, both the electric field and the gravitation field point to the present position:
http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html

Edited by Enthalpy
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So a gravity wave is not simply a propagating step potential ?

Your second link does, however, state that measurement of the speed of gravity may be possible by the detection of gravity waves from a super nova ( last paragraph ).

I thought we were agreed that the implosion/explosion of a super nova produces a symmetric acceleration of mass resulting in a propagating step potential ?

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My limited understanding (hope more knowledgeable members will jump in):

1. A sudden symmetric transformation of local mass into emitted light or neutrinos creates no change at all in the gravity we observe early, as all mass and energy create the same attraction, and in addition, gravitational waves are not axial.
2. As a supernova's neutrinos and light pass by us, their mass and energy don't attract us any more towards the supernova, so maybe we could observe something (message #31). The speed is of neutrinos and light, not of gravitational waves, and this step is not called a gravitational wave.
3. Only asymmetric events, like a pulsar pair, create gravitational waves, because these are transverse (and quadripolar).
4. These waves are not observed, but the power they would carry away is consistent with the aging of the orbital period that has been observed on a handful of pulsar pairs.
5. This is the only observation of gravitational waves up to now. Compatible with a propagation at light speed. Not a very direct nor accurate observation, but good looking.
6. Direct observation has not succeeded up to now. This is said to be still normal and should improve.
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A super nova is not just a 'symmetric transformation of local mass into emitted light and neutrinos'.

The radiation pressure of the emitted light and neutrinos ( not so much ), results in a sizeable portion of the star's outer mass being blown away. This acceleration of mass should result in a changing gravity field, and since the change in gravity field propagates at c, the change in potential from the near side will arrive before the change in potential from the far side of the star, i.e. not symmetric.

Mow Misner, Thorne and Wheeler say this propagating potential step is NOT a gravitational wave. While the three authors of the article you linked say it IS ( again last paragraph ).

Personally I'd be inclined to trust Misner, Thorne and Wheeler, but I'm bringing this up for the sake of discussion

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A super nova is not just a 'symmetric transformation of local mass into emitted light and neutrinos'. [...] the star's outer mass being blown away. This acceleration of mass should result in a changing gravity field, and since the change in gravity field propagates at c, the change in potential from the near side will arrive before the change in potential from the far side of the star, i.e. not symmetric.

Why do you imagine a change in the gravity field when the outer mass accelerates outwards? The mass+energy is conserved, and if the objetc stays symmetric, the gravity we experience from it repends only on its total mass+energy (which gives the flux) and its distance.

Which is the reason why standard theory precludes axial gravitation waves.

The authors did NOT write a "symmetric" supernova explosion, that's why they write "gravitational wave".

Edited by Enthalpy
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You are looking at the super nova globally, i.e. a distant frame, so it seems fairly symmetric.

But here's another viewpoint...

Consider the edge of the giant star, say about 10x the size of the Sun, about to go super nova, facing us.

Now put the observer in a local frame and let the show begin.

A large amount of mass will fall away ( almost at free fall ) and then it will rebound outwards with great acceleration.

This has the effect of deforming the local gravitational field ( gravitational potential will reduce and then increase, I believe ), and this deformation of the extant gravitational field will propagate outwards at the speed of light.

Now consider the far side of the star. While the 'estabilished' gravitational field is perfectly symmetric as you have noted, the deformation that the field experiences has to travel outwards at the speed limit c, and by necessity, will reach the near side position of the ( original ) star almost a minute later. This near one minute difference in phase will still be there when this deformation reaches us so, many light years away.

GR invalidates the concept of simultaneity, and any symmetries based on events, since an event that changes information has to propagate at a finite spe

I could be wrong in all this, but if not, then the question remains...

Does this 'step' deformation constitute a gravity wave or not ?

Or maybe we're both missing something, and someone else can jump in and point it out ???

Edited by MigL
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Yes, this reasoning puzzles me too, since the beginning of the thread... But

• I have more confidence in the total flux of gravitational attraction through the sphere depending only on the mass+energy contained in the sphere, and this tells "no effect".
• The specialists of waves tell that they're transverse, not longitudinal like the implosion would have produced. I can't do better than believe them.
• The same scenario with electric charges clearly tells "no effect", despite the charges nearer to us apparently should affect us earlier than the ones farther.
• I've become very cautious with retarded potentials. These are not fundamental physics laws. In electromagnetism, they result sometimes (=for a plane wave, at distance) from Maxwell's laws (which are fundamental) and sometimes not (for instance, the electric field made by a particle with constant speed points to the present position of the particle, not to its position one propagation time ago). The attraction field of a massive particle with constant speed also points to its present position, which can't (and isn't) inferred from retarded potentials. So I vaguely suppose that retarded potentials are what misleads us when considering the star implosion.
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And while I would have no problem whatsoever agreeing with you for a 'constant speed' situation, that is not equivalent to an accelerated situation, so I still have my doubts.

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And so do I have my doubts, very much so!

At least, radiation from a spherical source is impossible as soon as we assume the waves are transverse - be they dipolar or quadripolar. The mere symmetry of the sphere lets locations at right compensate the ones at left. Or for a quadripolar wave, the locations up and down compensate the left and right ones. All at the same distance and propagation time, so that's the easy part. Radiation from a spherical source would be longitudinal.

As a sidenote, a hypothetical longitudinal wave would be hard to distinguish from the gravitation resulting from the neutrinos that pass by us. In strong supernovae, neutrinos carry most energy from the explosion, they're nearly as quick as photons and gravitational waves... How to make a difference? It would need an event where much of the released energy stays near the explosion or propagates slowly.

I've so big doubts that I wonder (somewhere on ScienceForums) why a binary star shouldn't emit a dipolar gravitational wave. Both stars have a speed and acceleration that vary over time, and if we see their orbit flat and when they're at 45° and 135° from our direction, their opposed acceleration should have produced waves that we receive with different delays hence shouldn't compensate an other. Very similar to the spherical explosion, isn't it?

I vaguely suppose that the retarded potentials don't apply here neither. It's a clear case where the source is much smaller than the wavelength, and then retarded potentials are often misleading. The step "we observe earlier the influence of nearer parts" would be where gravitation behaves unexpectedly.

For instance in acoustics, near a spherical source or the apex of a cone, the pressure and the speed are out of phase (pressure rises as air accumulates through speed), while they're in phase in a plane wave far from the source, so retarded potentials fail, here in the first geometric half-wave. Similar things happen in electromagnetism, just more complicated.

There must be very few dozen people on Earth who have a rational and critical understanding of gravitational waves. I definitely don't belong to them, so I can only stick to what I read - hoping that these indirect sources make sense and that I read the material properly and that I apply it as it should - which states "transverse quadripolar".

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I guess we're at an impasse.

Like you, I tend to believe the authorities on the subject ( Misner, Thorne, Wheeler ), since it seems neither of us is one.

I still have this little nagging feeling in the back of my head though...

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Let us consider million rocks arranged in a sphere formation. Every rock is simultaneuosly and abruptly pulled outwards by some force. We know that extra energy is used in this pulling procees, because a rock thinks the distance to the other rocks is shorter than the real distance.

But we also know that no gravity waves are generated in this kind of situation, because experts have said so. So we conclude that the extra energy will absorbed by the rocks at some later time.

Now let us remove one rock from the formation. Quite obviously the hole will emit some gravity waves, spherical gravity waves.

If we remove more rocks every hole will be a source of spherical gravity waves.

If we remove all rocks except two at opposite sides of the sphere, there will be 999998 holes that emit spherical gravity waves that interfere with each other. Distant gravity wave detectors will detect a maximum of gravity waves at some place and minimum of gavity waves at some other place.

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