6 ms·

Didn't realize that energy could be converted into gravitational waves. It does make sense I suppose, but I always thought gravity and gravitational waves were

• SomeStupidPoint 9y ago
  Can't you radiate potential gravitational energy as gravitational waves without any sort of particle changes, because potential gravitational energy is stored in the configuration of the system? If I drop my bowling ball, it has lower potential energy, but can't some of that be compensated for by the Earth-ball system radiating a gravitational wave (rather than all being converted into kinetic energy)? Similarly, don't orbiting bodies emit gravitational waves? I forget the exact dynamics, but the way that they drag just a little bit as they orbit creates waves that radiate power out of the system -- I think the Sun and Jupiter radiate something like 55 watts of gravitational waves. If you take two things that are each tens of times larger than the Sun and set them spinning an appreciable amount of the speed of light very nearby -- to the point they eventually collide -- it's not surprising they emit quite a bit more than our solar system. Think of it this way: Two large objects start orbiting slowly far apart. But because they're large and the system is big, they have a lot of orbital momentum. They pull each other close, but as they get closer, that orbital momentum has nowhere to dissipate, so they orbit faster and faster!... ...Except that because they're so massive, they slightly tug space around with them as they orbit. And because they're orbiting so fast, space doesn't have a good way to smooth itself out. So some of the momentum from the heavy things spinning far apart gets stored in wrinkles in space as they come together. Which then radiates away from there to us, ...and slightly stretches one direction of reality as it passes by. Which we detect by measuring how long two perpendicular lines are very accurately a few different places on a nearly spherical object. ...So you can model it as just the way that mass interacts with spacetime if you remember to include things like the distribution of your mass as a source of potential energy that can itself be used to generate gravitational waves. However, this does not make the whole situation any less weird to explain. Reports just tend to convert the amount of energy released through mechanisms like that in terms of mass, because it's the only thing vaguely comprehensible. (eg, 3 suns)
  • teraflop 9y ago
    > If I drop my bowling ball, it has lower potential energy, but can't some of that be compensated for by the Earth-ball system radiating a gravitational wave (rather than all being converted into kinetic energy)? Nothing wrong with your comment, but I just want to emphasize that for objects smaller than stars, the amount of energy carried by gravitational waves ranges from tiny to unimaginably tiny. The Earth-Moon system emits about 7 microwatts of gravitational waves. A bowling ball in low earth orbit would radiate about 5*10^-41 W, which is roughly equivalent to one photon of visible light every quadrillion years.
    • deleted 9y ago
      [deleted]
• Florin_Andrei 9y ago
  Well, if it's a wave, it carries energy. That energy has to come from somewhere. And relativity teaches us that energy and mass are the two sides of the same coin.
• WhoBeI 9y ago
  According to Einstein there is no gravity that pulls us to the center of Earth. Instead mass curves spacetime in such a way to make you experience traveling along the curvature as gravity. If the mass is rotating the curvature will also drag on the surrounding spacetime. Imagine something like a whirlpool and then consider what happens if two of those merge. While merging some of their energy will cause waves in the surrounding water that eventually fades away with range. Those waves would be the gravitational waves - ripples in spacetime.
  • lomnakkus 9y ago
    Incidentally, I also find the "geometrical" approach a great way to look at the whole "light vs. black holes" thing: It's not that black holes directly "trap" light or "prevent light from escaping" per se, it's more that spacetime gets curved so much that regardless of which direction a ray of light is traveling it'll always find itself back inside the event horizon. (Not sure where I first heard this explanation, but I can definitely say that I didn't originate it!)
    • nerfhammer 9y ago
      Man, that made me realize that black holes really are literally "holes".
      • lomnakkus 9y ago
        Yup. A "hole" in space in some sense, though I must admit I haven't thought of it such extremely concrete terms[1]. I imagine that realization went into the naming :). [1] I mean, we usually imagine we can retrieve things when they drop into a hole, but that's not really a thing in this case. (Excepting Hawking radiation and the absurd amount of computation/time you would need to reconstruct the original state information from that. If you thought the time-of-evaporation for a Black Hole was long, you've got another thing coming when it comes to reconstructing the state from that evaporated radiation! Not that this has has any practical bearing, but it's fun to think about, right?)
  • raattgift 9y ago
• InclinedPlane 9y ago
  In a system of binary black holes where one is 31 solar masses and the other is 25 they have event horizon radii of 93 and 75 km, respectively. At the point their event horizons start to touch they have a relative distance of 168 km. Each of them have a gravitational potential energy in the other's gravitational field of G m1 m2 / r or 1.22e48 Joules. Both of them together have a total gravitational potential energy of 2.44e48 Joules or 1.36e31 kg (6.8 solar masses). (In reality the mass contribution due to the gravitational potential energy is included in the "mass", this is just illustrative.)
  • raattgift 9y ago
    Sure, but the crucial thing is that offsetting the gravitational potential energy is angular momentum; we can treat the former as "negative energy" with the idea that it takes positive energy to pull apart two bodies attracting each other gravitationally. [1] Since the negative energy ultimately wins, there must be less and less positive energy holding the bodies apart. The positive energy has to go somewhere, and that somewhere is gravitational radiation. One could equivalently be very Machian and rotate with the system and consider that the gravitational attraction of "the whirling distant stars" is working to keep the black holes from colliding. The distant matter loses because the binary black holes move positive energy (holding the BHs apart) out towards the distant stars (which also lets the BHs move a bit further away from the distant stars). s ---+ B1 --++ B2 ---+ s -> s ---++ B1 -- B2 ++---s Even if we consider inspiralling stars (no horizons to be found), we're lifting relatively little mass-energy out of them, rather than dealing with the energy that keeps them from colliding. The mass-energy of the system (which you measure in Joules or solar masses) when considered in the cosmological frame is what has the excess ~ 12% that gets radiated away. Finally, in Special Relativity the angular momentum and energy-momentum of localized systems is frame-dependent; the addition of gravitational fields in a linear approximation of GR does not change that, and gravitational waves are a result in linearized gravity. It certainly is useful to discuss dimensionful quantities like the ones you listed, but it's important to remember that different observers are free to disagree violently about the numbers. Thinking about how and why they might do so is often enlightening. - -- [1] Although the convention is mostly there because in Newtonian gravity, one takes T + U = const. for a body felling gravity, and since by virtue of having been described earlier T (kinetic energy) is positive so U (gravitational potential) must be negative, with the result that F = - \nabla U. We break the conservation of T + U in General Relativity quite happily, but retain the notion of the gravitational interaction as negative in many useful frames, so in the footnoted paragraph we have a generalized T and U available by a suitable choice of frame of reference in the weak limit.
• raattgift 9y ago
  > Didn't realize that energy could be converted into gravitational waves. I'll try an explanation, hopefully it's useful to you or another reader. General Relativity gives us $\nabla_\mu T^{\mu\nu} = 0$, which reads that the covariant derivative of the contravariant stress-energy tensor (the matter tensor) generally vanishes. When remembering the heart of General Relativity: G_{\mu\nu} = T_{\mu nu}, that is, the gravitational interaction arises from moving matter, we observer that changes in the gravitational field match changes in the arrangements and/or motions of matter and vice-versa. This is a fancy way of saying that there is a global conservation of stress-energy-momentum, even when you work in coordinates or gauges in which there is no global conservation of energy. For simplicity, here and below I am setting some constants to 1, only considering classical behaviour, completely ignoring strain-squash modes, and otherwise being only very mildly post-Newtonian. In a system like a rotating barbell with a very thin, almost massless stick, like in a Cavendish apparatus ( http://en.wikipedia.org/wiki/File:Cavendish_Torsion_Balance_Diagram.svg http://en.wikipedia.org/wiki/File:Cavendish_Torsion_Balance_... ) the masses on the end of each bar will influence the motions of masses near them, and vice-versa, because their masses attract one another. We are free to choose whatever coordinates we want to describe this sort of setup, including one in which the two grey M balls move in response to the proximity of the two red m balls, with the point where the bar is suspended taken as an origin of some useful coordinates (e.g. spherical polar ones). In particular, as the ms pass close to the Ms, the rotation of the bar is dragged on, and one is free to model that as angular momentum leaving the rotating barbell under a gravitational interaction. We can treat two massive objects (stars, for example) in a binary orbit as a barbell-like system; the bar itself simply has no mass and is in free-fall rather than suspended on a wire. However, just like in the cavendish experiment, concentrations in the distant mass on either end of the bar will cause the bar to stretch and the system's rotation to slow, while sparsenesses in the distant mass on either end of the bar will cause the bar to contract (the "bar" itself is just the gravitational interaction between the binary masses). With a clever change of point of view, we may ignore the influence of nonuniformities in the distant masses and just talk about the momentum leaving the rotating system in response to defined (or discovered) changes in the length of the "bar". That momentum is gravitational radiation, and its amplitude and frequency depend on the masses and orbital periods of the ends of the shrinking "invisible bar" of their mutual gravitational interaction. Near the very end of the inspiral, the bar shrinks rapidly, and the amount of angular momentum that is shed is consequently enormous. Spacetime enters into the picture here via a linerization of the General Relativity picture of inspiralling massive objects: we start by picking out a flat spacetime (\eta [1]) and a field of perturbations on that spacetime (h) and then get the real metric tensor g via g = \eta + h (+ h^2 + h^3 ...). We drop the higher order perturbations on the spacetime (h^2 + h^3 + ...) and have a linear theory that is approximately General Relativity and is very very very close practically everywhere outside of event horizons and the extremely early universe. Gravitational waves are simply excitations in h that obey the massless wave equation. Spacetime is pretty much meaningless without a measure of lengths and durations between bits of matter, or equivalently, spacetime is the measure (i.e., geometry) of distances in space and time between bits of matter. Since gravitational waves are waves in an addend of the metric of spacetime, they are equally waves of spacetime. We can consider a region of spacetime in which we find the "barbell" of merging compact massive objects (black holes, neutron stars) and discuss how, since the momentum of what's inside it decreases as the "bar" shortens, that momentum must flow outside the region to elsewhere. Matter elsewhere in turn picks up some of that momentum, just as the grey M balls pick up some momentum inwards as the red m balls pass near them. Gravitational waves are seen when one makes some deliberate choices of coordinates and a linearizable approximation of the "real" metric. On that basis it is perfectly reasonable to treat them as physical, even though you could use a very different description the gravitational interactions between the inspiralling bodies and all the other masses in the universe, in a Machian sort of way. The descriptions are to all practical purposes equivalent, and this can be demonstrated mathematically. - -- [1] g_\mu\nu, \eta_mu\nu, and h_\mu\nu all have covariant indices, but I'll just stop writing them now because they are distracting. g is a component of G in the G = T mentioned near the top of this comment, where G is the gravitational field and T is all the stress-energy (the non-gravitational fields, or just "matter").