Supernovae & Gravity Waves

[GeeKay]

Can a supernova explosion (or implosion?) on its own create gravity waves? Or do you need to have (say) two stars 'merging' into each other for this effect to happen? In other words, is displacement necessary for the generation of gravity waves? Many thanks.

 

 

[swansont]

If the explosion is asymmetrical, a supernova can produce gravity waves. The necessary component is acceleration.

[Enthalpy]

In the case of a symmetric explosion, that transforms in few days some mass into energy:

 

Take the sphere centered on the supernova, with radius to our postiion. Before the supernova's light reaches us, this sphere contains some amount of mass+energy and creates some attraction where we are. After the light flash has passed, the sphere contains less mass+energy, and the attraction is smaller.

 

Would we call this a gravity wave? It has the wrong polarization (axial), it's dipolar instead of quadripolar...

 

And how does its range compare with a gravity wave? This one would mean a step in the acceleration proportional to R^-2.

[MigL]

Much easier to picture in two dimensions using the common elastic sheet with a weight sitting on it.

A supernova involves collapsing outer shell and eventual rebound to form the neutron core, while some of the mass is ejected into space along with copious amounts of radiation.

Acceleration of this mass, is equivalent to a changing gravitational field, and is represented by the mass on the elastic sheet 'wigglingup and down and making ripples.

These ripples are the gravitational waves.

[GeeKay]

Much easier to picture in two dimensions using the common elastic sheet with a weight sitting on it.

A supernova involves collapsing outer shell and eventual rebound to form the neutron core, while some of the mass is ejected into space along with copious amounts of radiation.

Acceleration of this mass, is equivalent to a changing gravitational field, and is represented by the mass on the elastic sheet 'wigglingup and down and making ripples.

These ripples are the gravitational waves.

 

So the take-home message is this: any high-speed acceleration of stellar material, resulting from a supernova explosion (whether outwards or inwards towards the neutron core) is liable to produce gravitational waves - this occurring despite there being no overall change in the star's centre of gravity, for example?

.

[Enthalpy]

No.

 

Gravitational waves are traditionally described as transverse and quadripolar. Simple outwards acceleration should not produce any. Hence my query, if the step reaching us when light does is called a gravitational wave or not.

 

Change of the centre of mass: if considerig the whole mass, including the ejected one, it won't happen - consistently with the quadripolar nature of these waves. Waves are produced when massive objects get closer or farther apart as seen from our position, which must happen if an unsymmetric explosion ejects more matter to one side.

[GeeKay]

Yes, I understand now. Thank you for clearing this up for me

[MigL]

So a mass wiggling back and forth along a line of sight to me would not produce gravitational waves in my direction ?

 

I beg to differ enthalpy !

 

Gravitational waves travel at the speed of light, so the wave coming from the front is not nullified by the wave coming from the rear. Especially at the astronomical distances of a red giant about to go supernova, i.e. there is no symmetry.

[Enthalpy]

So a mass wiggling back and forth along a line of sight to me would not produce gravitational waves in my direction ?[...]

That's my (very limited) understanding. A movement along the object-to-observer line won't make a transverse wave, as these are said to be. Though, I had nearly the same interrogation as you, asking why a pair of stars shouldn't create a dipolar wave resulting from the difference of propagation time.

[...] Gravitational waves travel at the speed of light, so the wave coming from the front is not nullified by the wave coming from the rear. Especially at the astronomical distances of a red giant about to go supernova, i.e. there is no symmetry.

That would make a longitudinal dipolar wave. Gravitational ones are said to be transverse and quadripolar.

[MigL]

I'd be very interested in reading a source for the claim that gravitational waves are transverse and quadripolar Enthalpy.

 

I would assume that a gravity wave is simply the communication of a changing or even translating gravitational field. As any mass moves, the space-time curvature is continuously re-estabilished along its line of travel. This information travels outward in all directions at the speed of light. Is this not the wave ?

Edited March 12, 2014 by MigL

[Enthalpy]

This exceeds by far my knowledge.

 

Being aware that I don't understand Relativity hence have no possibility to grasp these waves, I just assume nothing at all and stick to what I read.

 

Beware about moving objects and retarded potential, though. At least for charged particles and electric fieds, the observed electric field results from the particle's position at the time of observation, not one propagation time prior to it.

 

I've also read that this holds for gravitational waves, too. This being necessary to keep stable planet orbits.

[imatfaal]

This exceeds by far my knowledge.

 

Being aware that I don't understand Relativity hence have no possibility to grasp these waves, I just assume nothing at all and stick to what I read.

 

Beware about moving objects and retarded potential, though. At least for charged particles and electric fieds, the observed electric field results from the particle's position at the time of observation, not one propagation time prior to it.

 

I've also read that this holds for gravitational waves, too. This being necessary to keep stable planet orbits.

 

I cannot quite understand what you are getting at - are you saying that gravitational and electromagnetic fields act such that a test object reacts to the charged particle/massive object simultaneously? ie that these is no lag for any change to propagate?

 

Surely changes in em fields are by definition transmitted to surroundings at light speed and no reaction by a test particle can be seen till that propagation has taken place. We might still use Newtons/Keplars laws to understand why the earth orbits the sun - and they imply instantaneous action at a distance , but this is a convenient fiction; we or any test particle react to the field that they are in and that field is not that which is simultaneously being generated, it is the field generated in the past that is propagating outwards, ie any change in the sun's em or gravitational field would not be noticed here for 8 minutes.

[swansont]

 

I cannot quite understand what you are getting at - are you saying that gravitational and electromagnetic fields act such that a test object reacts to the charged particle/massive object simultaneously? ie that these is no lag for any change to propagate?

 

Surely changes in em fields are by definition transmitted to surroundings at light speed and no reaction by a test particle can be seen till that propagation has taken place. We might still use Newtons/Keplars laws to understand why the earth orbits the sun - and they imply instantaneous action at a distance , but this is a convenient fiction; we or any test particle react to the field that they are in and that field is not that which is simultaneously being generated, it is the field generated in the past that is propagating outwards, ie any change in the sun's em or gravitational field would not be noticed here for 8 minutes.

 

That's why there is a retarded potential for E&M, and why the light we see from the sun comes from where the sun was 8 minutes ago. But AFAIK gravity points to where the sun is now, because the sun has already had a chance to warp the space we are in. (some people take this to mean it travels instantly but that's a misinterpretation because they treat gravity and E&M as being identical interactions) So, as you say, changes in gravity would take 8 minutes to propagate.

[MigL]

Had an interesting discussion concerning this effect with someone on another forum.

 

Consider two gravitationally interacting masses moving parallel to each other at, say, 50% light-speed. If they interact instantaneously, there is only the usual force between them orthogonal to their direction of motion, i.e. they will approach each other.

If however, we consider the propagation delay ( finite light-speed ), the force is no longer orthogonal to their direction of motion but has a rear-ward component, i.e. they don't only come together, they also slow each other down.

 

This is where my thought experiment ends, as it contradicts what I expect to be true.

Perhaps a better mind than mine can shed some insight.

[DimaMazin]

 

That's why there is a retarded potential for E&M, and why the light we see from the sun comes from where the sun was 8 minutes ago. But AFAIK gravity points to where the sun is now, because the sun has already had a chance to warp the space we are in. (some people take this to mean it travels instantly but that's a misinterpretation because they treat gravity and E&M as being identical interactions) So, as you say, changes in gravity would take 8 minutes to propagate.

Then gravity has communication with own source.The communication doesn't change speed of propagation, but changes direction of propagation.

[imatfaal]

Had an interesting discussion concerning this effect with someone on another forum.

 

Consider two gravitationally interacting masses moving parallel to each other at, say, 50% light-speed. If they interact instantaneously, there is only the usual force between them orthogonal to their direction of motion, i.e. they will approach each other.

If however, we consider the propagation delay ( finite light-speed ), the force is no longer orthogonal to their direction of motion but has a rear-ward component, i.e. they don't only come together, they also slow each other down.

 

This is where my thought experiment ends, as it contradicts what I expect to be true.

Perhaps a better mind than mine can shed some insight.

Delete gravitationally interacting replace with oppositely charged particles. Does this change the outcome? Does it change your expectations? I am at a loss - it is a good gedanken

[md65536]

Delete gravitationally interacting replace with oppositely charged particles. Does this change the outcome? Does it change your expectations? I am at a loss - it is a good gedanken

Also what happens when you consider aberration of light?

 

Suppose the two observers are traveling the length of an infinity long football field, crossing some number of transverse lines, and they send out such a particle at every line. The field lines will appear warped toward the forward direction.

 

Since the two objects initially share an inertial frame, they should appear to not be moving, so no relativistic effects. They should appear directly to each other's side. The pull of the particle should be directly to the side, so not slowing nor speeding up the mass in the forward direction, relative to their rest frame. But this is a backward pull in the football field's frame.

 

 

 

But what does it mean that gravitational effects appear in the direction that the mass is "now"? If the objects are massive, and one mass crosses a given line, it sees the line aberrated and the other mass not yet appearing to have reached that line, due to delay of light. But "now" the two synchronized masses are both crossing the line at the same time. Does that mean that the pull of gravity is along the aberrated line? This would make sense, because in the football field's frame, the pull of either mass is directly perpendicular, meaning that they would neither speed up nor slow down???

 

Definitely puzzling. It seems that gravitational effects shouldn't change the speed of the masses, yet charged particles should slow them down relative to the football field.

 

 

Since the masses are not accelerating with respect to their direction of travel, there is no need for a graviton to be transmitted; there's no need for a delay of gravitational effect? Another puzzling aspect is that if one of the masses were to suddenly stop well before the line, such that the other would cross the line before seeing the other stop, then the other would "feel" the gravitational pull in the direction of the line as it was crossing. It would not be pulled toward where the other mass "is now" until it is able to know where the mass actually should be now.

 

 

 

 

Edit: On second thought... why would one mass feel a slightly forward pull from the other mass, if the other is essentially at rest directly off to the side? Why would the existence of another massless moving frame of reference make any difference (it mustn't). When we say "where something is now" we must be referring to the observer's rest frame? I think so far I've only made myself more confused.

 

My current feeble understanding is that the masses would slow relative to the football field, whether due to gravitational attraction or reception of opposite charged particles, but that this is consistent with the difference between photons and gravitons. I think you would need relative motion of the gravitational mass before detecting any difference in the direction of light and gravity, so I don't think this thought experiment is explanatory.

Consider two gravitationally interacting masses moving parallel to each other at, say, 50% light-speed. If they interact instantaneously, there is only the usual force between them orthogonal to their direction of motion, i.e. they will approach each other.

If however, we consider the propagation delay ( finite light-speed ), the force is no longer orthogonal to their direction of motion but has a rear-ward component, i.e. they don't only come together, they also slow each other down.

Okay, but they're moving relative to what?

 

In that other frame, yes the particles seem to come from behind, but the two masses seem twisted due to relativistic effects, so the particles don't seem to hit the backs of the masses. The particles hit the sides.

 

Consistent with that, in the frames of the masses, the other mass doesn't have relative motion. The direction of the particles, and of gravitational attraction, is the same. The particles hit the sides.

 

These two frames must be consistent. From the "other frame's" perspective, the particles come from behind and hit the twisted object on its side. Meanwhile if gravitational attraction is directly to the side then it affects the object as if it isn't twisted at all.

 

 

So, just like the direction of light and gravity are different, the relativistic effects on the shape of the mass are different (it would be twisted by measure of light, but not twisted by measure of gravitational attraction).

 

 

[This still doesn't make sense because it seems to imply that the particles would slow the masses in the "other" frame, while gravitation wouldn't, even though the acceleration is in the same direction in the masses' frame, so I must be missing something important. :S Perhaps "frame dragging" provides a solution.]

Edited March 14, 2014 by md65536

[Enthalpy]

Delete gravitationally interacting replace with oppositely charged particles. Does this change the outcome? Does it change your expectations?

I've read as well that, for charged particles with constant velocity, the electric field points to the direction where they're now, and not where they were one propagation time earlier.

 

Some math is, I believe, on en.wiki.

 

Though, we compute antennas with good results by using delayed (not "retarded" as I wrote) fields, which apparently contradicts the previous statement.

 

One possibility: whether the particle accelerates or not changes everything.

Consider two gravitationally interacting masses moving parallel to each other at, say, 50% light-speed.[...]

If however, we consider the propagation delay ( finite light-speed ), the force is no longer orthogonal to their direction of motion but has a rear-ward component, i.e. they don't only come together, they also slow each other down. [... contradicts]

 

In addition to thought experiments, we can check what would happen to our Moon if Earth's gravity there pointed to the direction where Earth was one propagation time earlier.

 

http://en.wikipedia.org/wiki/Moon

 

The Moon is 0.0123 times as heavy as Earth and 384Mm away, so Earth wobbles by R=4.7Mm due to the Moon's attraction; light takes 16ms to cover this distance, and the orbit takes 27.32days=2.36Ms.

 

Pointing to an older position would mean an angle of 42nrd.

 

Earth's gravity lets the Moon orbit at 1022m/s, which reveals an acceleration of 2.72mm/s². The hypothetical angle would impart an azimutal acceleration of 114pm/s², which would perturbate the orbit by 1000m/s within 278,000 years, obviously not the case.

 

Besides numbers, we could have a thought experiment here as well, since the Moon and Earth would get energy from nothing.

[imatfaal]

A citation to the wiki article would be useful. I am not sure about your calculations - the earth wobbles by 4.7e6 metres??? And your thought experiment begs the question.

[MigL]

Well since we're already off topic ...

If we consider just the Earth-Moon system I would think the gravitational well of the Earth could be considered static, i.e. already existing and not propagating at c, as it is much larger than the moon's.

What would be the result of your analisys Enthalpy, if we consider the Earth-Moon system as they orbit about the Sun, such that the Earth's field is constantly moving around the sun ?

 

Maybe I should have been clearer in my thought experiment md65536. I should have stated the two masses have the same velocity, i.e. same speed, same direction, so that they are in effect int he same frame. The 'retarded' potential creates a rearward vector as well as the expected perpendicular. The masses 'drag' each other back !

That is the part I have trouble swallowing.

[Enthalpy]

Earth can't be static while the Moon orbits it. Both have to orbit around they common center of mass. Our Moon is interesting because it has an important fraction of its planet's mass: 1.23%.

 

I've taken the ean distance to the Moon, 384Mm, and multiplied by 0.0123, to get 4.7Mm as the radius of Earth's wobble. Anything wrong in that?

[imatfaal]

Earth can't be static while the Moon orbits it. Both have to orbit around they common center of mass. Our Moon is interesting because it has an important fraction of its planet's mass: 1.23%.

 

I've taken the ean distance to the Moon, 384Mm, and multiplied by 0.0123, to get 4.7Mm as the radius of Earth's wobble. Anything wrong in that?

 

Yes - it is not a wobble. A wobble implies a regular change - the moon and earth(any bodies in fact) orbit around the centre of mass. It has been there since the event that created the moon. I presume - but do not know - that the microscopic effects that you would get if you did these calculations properly as a two body problem would be exactly those anomalies that newtonian celestial mechanics had; the precession of mercury and other finer problems that were sorted with GR. If you are actually working on estimating the effect of the moon on the earth within its orbit around the sun - this is damn close to a three body problem and not guessable even with newtonian physics (its solvable but not easy nor always intuitive)

 

Had an interesting discussion concerning this effect with someone on another forum.

 

Consider two gravitationally interacting masses moving parallel to each other at, say, 50% light-speed. If they interact instantaneously, there is only the usual force between them orthogonal to their direction of motion, i.e. they will approach each other.

If however, we consider the propagation delay ( finite light-speed ), the force is no longer orthogonal to their direction of motion but has a rear-ward component, i.e. they don't only come together, they also slow each other down.

 

This is where my thought experiment ends, as it contradicts what I expect to be true.

Perhaps a better mind than mine can shed some insight.

 

I have requoted the original. The only way I can begin understand it is that the ease of use of a frame of reference within which both are travelling together on parallel tracks - together with the old newtonian force model - makes us think that this frame is special and requires no extra thought. But put yourself on the left meteor - where do you look for the righthand meteor? To the right and backward. From the right hand meteor you look left and backward. Whatever answers you get from other frames of reference must tally with this. The force is directed straight at where the meteor-based observer is seeing the other meteor.

 

The force is directed at the point at which the gravitationally interacting object appears - both the interaction and the EMR have taken the same amount of time to get there (nasty phrasing but...) Talking about where the object "really is" makes trouble with the relativity of simultaneity - there is no god-frame in which positions and timings are absolute. Because of the clever nature of the problem we are falling into the trap of believing that there is.

 

In the frame of the LH meteor the attraction is towards where the RH meteor is observed and vice versa. In the frame of the moon the attraction is towards where the earth is observed.

[md65536]

No worries MigL, your example was clear enough but I completely misread it and got lost in my own reasoning. I'll stick to citable references here...

 

The force is directed at the point at which the gravitationally interacting object appears - both the interaction and the EMR have taken the same amount of time to get there (nasty phrasing but...) Talking about where the object "really is" makes trouble with the relativity of simultaneity - there is no god-frame in which positions and timings are absolute.

This isn't true, neither for gravity nor for EM.

 

http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html :

For weak fields, though, one can describe the theory in a sort of newtonian language. In that case, one finds that the "force" in GR is not quite centralit does not point directly towards the source of the gravitational fieldand that it depends on velocity as well as position. The net result is that the effect of propagation delay is almost exactly cancelled, and general relativity very nearly reproduces the newtonian [instantaneous gravity propagation] result.

 

This cancellation may seem less strange if one notes that a similar effect occurs in electromagnetism.

... the gory mathematical details of which can be found in the reference:

Aberration and the Speed of Gravity -- http://arxiv.org/abs/gr-qc/9909087

EM case:

Note that every term in eqns. (1.7)(1.8) is retarded, and that nothing depends on the

instantaneous position or direction of the source. [...] The second term in this expression is essentially a linear extrapolation from the retarded

direction n^i toward the instantaneous direction. In particular, for a charge in uniform

motion it is easy to check that n^i − v_R points toward the instantaneous position, so the

effects of aberration are exactly canceled.

GR case, and explanation of how the lack of a "god-frame" doesn't matter:

In other words, the gravitational acceleration is directed toward

the retarded position of the source quadratically extrapolated toward its instantaneous

position, up to small nonlinear terms and corrections of higher order in velocities.

 

Does eqn. (2.4) imply that gravity propagates instantaneously? As in the case of elec-

tromagnetism, it clearly does not. Every term in the connection Γ^ρ_μν depends only on the

retarded position, velocity, and acceleration of the source; despite Van Flanderns claim to

the contrary [15], there is no dependence, implicit or explicit, on the instantaneous di-

rection to the source. Indeed, the vector (2.5) does not point toward the instantaneous

position of the source, but only toward its position extrapolated from this retarded data.

In particular, as in Maxwells theory, if a source abruptly stops moving at a point z(s0 ), a

test particle at position x will continue to accelerate toward the extrapolated position of the

source until the time it takes for a signal to propagate from z(s0 ) to x at light speed.

[arc]

Much easier to picture in two dimensions using the common elastic sheet with a weight sitting on it.

A supernova involves collapsing outer shell and eventual rebound to form the neutron core, while some of the mass is ejected into space along with copious amounts of radiation.

Acceleration of this mass, is equivalent to a changing gravitational field, and is represented by the mass on the elastic sheet 'wigglingup and down and making ripples.

These ripples are the gravitational waves.

 

MigL, thank you for the simple explanation, it is greatly appreciated.

[Enthalpy]

Gravitational waves travel at the speed of light, so the wave coming from the front is not nullified by the wave coming from the rear. Especially at the astronomical distances of a red giant about to go supernova, i.e. there is no symmetry.

 

This seems logical, but if we consider the sphere centered on the exploding star, with radius to us, I see this argument against:

- Until light from the explosion reaches us, the sphere contains a constant amount of mass+energy

- This amount makes a constant flux of gravitational attraction through the sphere's surface

- And if the explosion is symmetric, the distribution of the flux must be symmetric too

- Which results in a gravitational field constant everywhere, until light from the explosion reaches us.

 

Or is there something wrong in this attempt?