geordief Posted March 8, 2023 Share Posted March 8, 2023 Suppose we have two massive spherical objects and they are moving wrt each other. They pass each other at one point . Do their respective momentums (in addition to their mass) combine to curve spacetime at that point? Again (a different scenario) ,if the same two massive objects follow the same trajectory as each other do the pair also curve spacetime ,but differently from the earlier scenario? Ps Am I barking up the wrong tree to try and separate curvature effects due to mass from curvature effects due to momentum and are they perhaps two sides of the same coin? Link to comment Share on other sites More sharing options...

Genady Posted March 8, 2023 Share Posted March 8, 2023 The answers to these questions should be independent of reference frame. Perhaps it would be easier to consider the scenarios in reference frame of one of the objects, i.e., where one of them is at rest. Link to comment Share on other sites More sharing options...

geordief Posted March 8, 2023 Author Share Posted March 8, 2023 (edited) 10 minutes ago, Genady said: The answers to these questions should be independent of reference frame. Perhaps it would be easier to consider the scenarios in reference frame of one of the objects, i.e., where one of them is at rest. Is spacetime curvature frame dependent? (In my second scenario both objects are at rest wrt each other,I think) Edited March 8, 2023 by geordief Link to comment Share on other sites More sharing options...

Genady Posted March 8, 2023 Share Posted March 8, 2023 Just now, geordief said: Is spacetime curvature frame dependent? Reimann spacetime curvature is tensor. Its individual components are frame dependent, but they transform all together in a consistent way. So, the tensor is frame independent. Link to comment Share on other sites More sharing options...

geordief Posted March 8, 2023 Author Share Posted March 8, 2023 OK ,so from the frame of either of the two objects in scenario A the other approaches and in scenario B the other is at rest and close proximity. So scenario B curves spacetime less than scenario A? Is the momentum of an object one of the elements in the curvature tensor? Link to comment Share on other sites More sharing options...

joigus Posted March 8, 2023 Share Posted March 8, 2023 Very small curvature in both cases, never mind Riemann tensor components or curvature scalar (some kind of average of all Riemann components). It can be estimated by means of Newtonian gravity. Link to comment Share on other sites More sharing options...

Genady Posted March 8, 2023 Share Posted March 8, 2023 In scenario B, the curvature is static. In A, the curvature changes in time. The moving object will affect the curvature in some non-linear ways. Link to comment Share on other sites More sharing options...

geordief Posted March 8, 2023 Author Share Posted March 8, 2023 12 minutes ago, Genady said: In scenario B, the curvature is static. In A, the curvature changes in time. The moving object will affect the curvature in some non-linear ways. At the exact point in time *then that the two (relatively moving or relatively at rest) objects' effect on the spacetime curvature is calculated (so hopefully stripping out all dynamic effects**)- are the two effects identical? * when the two objects are as close as possible **if it can be done,maybe finding the number value of the tensor resulting from all its components. 34 minutes ago, joigus said: Very small curvature in both cases, never mind Riemann tensor components or curvature scalar (some kind of average of all Riemann components). It can be estimated by means of Newtonian gravity. Well the masses could be very big even though I was imagining them as small in the OP. Are two two curvatures identical or is the one involving approach speed greater ? Link to comment Share on other sites More sharing options...

Genady Posted March 8, 2023 Share Posted March 8, 2023 I think they will not be identical. I don't think reducing them to single numbers for comparison would make sense. In any way, the effects are not linear, so I don't think the effect of the mass can be separated from the effect of the motion. Link to comment Share on other sites More sharing options...

geordief Posted March 8, 2023 Author Share Posted March 8, 2023 9 minutes ago, Genady said: , so I don't think the effect of the mass can be separated from the effect of the motion. Is that because mass is a relativistic concept? (I understand that the bonds that hold an object together are calculated as "mass";does relative motion btw objects also contribute to a system's mass? Link to comment Share on other sites More sharing options...

Genady Posted March 8, 2023 Share Posted March 8, 2023 3 minutes ago, geordief said: Is that because mass is a relativistic concept? (I understand that the bonds that hold an object together are calculated as "mass";does relative motion btw objects also contribute to a system's mass? No. In fact, the mass isn't a component in the Einstein field equation. There are densities of energy and momentum, and their various derivatives. Link to comment Share on other sites More sharing options...

geordief Posted March 8, 2023 Author Share Posted March 8, 2023 5 minutes ago, Genady said: No. In fact, the mass isn't a component in the Einstein field equation. There are densities of energy and momentum, and their various derivatives. Thanks.I think I am probably done with the thread now.Quite a bit of info for my level. Link to comment Share on other sites More sharing options...

Markus Hanke Posted March 9, 2023 Share Posted March 9, 2023 17 hours ago, geordief said: Do their respective momentums (in addition to their mass) combine to curve spacetime at that point? All you can really meaningfully say here is that the geometry of these two spacetimes (two stationary masses vs two masses in relative motion) will be different - in particular, in the latter case of relative motion, some or all of the components of the metric will be explicitly time-dependent; it’s essentially a GR 2-body problem (which, btw, can only be solved numerically unless one of the masses is very much smaller than the other). But I think what you are getting at is ultimately whether relative motion in vacuum is in itself a source of gravity, and the answer to that is no, it isn’t. Its presence does, however, have an impact on spacetime geometry, in the sense that it will make the situation less symmetric and thus more complicated. But since you can’t in general meaningfully compare tensors (“tensor X is greater than tensor Y…”), all you can really say is that the spacetimes are different. This “difference” is in itself a non-trivial concept, because you can’t as a rule of thumb tell if spacetimes are different just by looking at the metric - for example, the Schwarzschild metric and the Aichlburg-Sexl metric look very different, but they do in fact describe the same physical spacetime. So there are a lot of subtle issues here. When you are not in a vacuum, ie in the interior of some mass-energy distribution, the situation becomes more complicated, because now the energy-momentum tensor explicitly contains terms that can be interpreted as momentum density. So for example, the motion of plasma currents in the interior of a star will have a gravitational effect within that interior region, as compared to some otherwise identical star without moving currents. But since the field equations are non-linear, it is not possible to neatly separate out these effects and isolate them from the other source terms, such as pressure, strain etc. They all interact and interplay. 16 hours ago, Genady said: No. In fact, the mass isn't a component in the Einstein field equation. There are densities of energy and momentum, and their various derivatives. It should be noted though that in the exterior of the bodies the energy-momentum tensor vanishes, so the masses of the bodies and their relative motion only enters the field equations in the form of boundary conditions. Link to comment Share on other sites More sharing options...

md65536 Posted March 13, 2023 Share Posted March 13, 2023 (edited) On 3/8/2023 at 11:24 PM, Markus Hanke said: But I think what you are getting at is ultimately whether relative motion in vacuum is in itself a source of gravity, and the answer to that is no, it isn’t. I figure it is in this case. (There was another topic about this...) If you have 2 masses moving relative to each other, then there should be a center-of-momentum frame for them. In that frame, the kinetic energy of the masses contributes to the total energy of the system, and its rest mass. It should contribute to gravity just like any mass would? On 3/8/2023 at 5:06 AM, geordief said: Again (a different scenario) ,if the same two massive objects follow the same trajectory as each other do the pair also curve spacetime ,but differently from the earlier scenario? In the case where the two masses are at relative rest with each other, they're at rest in the center-of-momentum frame, so the system has no kinetic energy to contribute to its rest mass. Edited March 13, 2023 by md65536 Link to comment Share on other sites More sharing options...

Markus Hanke Posted March 13, 2023 Share Posted March 13, 2023 1 hour ago, md65536 said: If you have 2 masses moving relative to each other, then there should be a center-of-momentum frame for them. In that frame, the kinetic energy of the masses contributes to the total energy of the system, and its rest mass. It should contribute to gravity just like any mass would? Yes, the motion does of course “contribute” to the geometry of this spacetime, as I said in my post - but not really in an intuitive Newtonian way of making it “stronger”, which is an ill-defined concept in GR. This is the classic “does a body become a black hole if it moves fast enough” question, which I don’t think we need to go over again. The rest (!) mass of the orbiting bodies doesn’t change, irrespective of how they move, and in any case we are looking for a vacuum solution here, so \(T_{\mu \nu}=0\) everywhere outside these bodies. Thus the equation you are solving is simply \(R_{\mu \nu}=0\), without any source terms at all. The angular motion makes an appearance only as part of the initial/boundary conditions once you solve that system of DGLs. So in practical terms, what would be the difference between the spacetime surrounding two bodies momentarily at relative rest at some distance wrt to each other, and the same system with the two bodies orbiting around a common center of gravity? In the former case you have a stationary spacetime that looks more and more like Schwarzschild the further away from the two bodies you get, and eventually becomes Newtonian; in the latter case you get a gravitational radiation field, where frequency and amplitude depend in some way on the rest masses of the binary system, the distance between the orbiting bodies, and the angular frequency. In the full (non-linearised) description these dependencies are pretty non-trivial, especially in the region very close to and in between the orbiting bodies. When you are located further away from the binary system, the amplitude will fall off with distance, again in a specific way. So the crucial difference is that the former case is stationary (i.e. the components of the metric do not depend on time, thus there is a time-like Killing field on this spacetime), whereas the latter is not. It’s a difference in symmetries, more so than degrees. What is true though is that the amplitude of the wave field at a given distance from the binary system will depend on the angular frequency of the bodies - the faster the bodies orbit each other, the higher the amplitude will be at any given distance. So if you wanted to, I guess you could - in this very particular sense - say that the gravitation of the binary system gets “stronger” if they orbit faster, since you’ll have more powerful tidal forces at a given distance. Personally though I would say that such a concept isn’t helpful, because it very easily lends itself to misinterpretation and various misconceptions. In general there just isn’t any way to meaningfully say that one metric is somehow “stronger” than another metric, because you are comparing tensors, not numbers. They are just different. But I guess you can always pick out specific local measurements and compare those. What you can absolutely not do though is take a known metric in a specific coordinate system - such as Schwarzschild -, and simply replace rest mass by relativistic mass to reflect relative motion, while leaving everything else the same, and expect the result to be a valid metric again. That doesn’t work, because such an operation is not in general a valid diffeomorphism. Link to comment Share on other sites More sharing options...

md65536 Posted March 13, 2023 Share Posted March 13, 2023 13 hours ago, Markus Hanke said: but not really in an intuitive Newtonian way of making it “stronger”, which is an ill-defined concept in GR. This is the classic “does a body become a black hole if it moves fast enough” question, which I don’t think we need to go over again. The rest (!) mass of the orbiting bodies doesn’t change, irrespective of how they move, Yes, the rest masses don't change, but together the rest mass of the system includes the kinetic energy of the bodies in the system's COM frame. For example, the rest mass of an atom includes the (small) contribution of the kinetic energy of its electrons. This additional mass affects both the curvature around the atom, and the atom's Newtonian gravitational strength. 13 hours ago, Markus Hanke said: What you can absolutely not do though is take a known metric in a specific coordinate system - such as Schwarzschild -, and simply replace rest mass by relativistic mass to reflect relative motion, while leaving everything else the same, and expect the result to be a valid metric again. That doesn’t work, because such an operation is not in general a valid diffeomorphism. No one's talking about doing that. I think a more appropriate analogy is comparing the Schwarzschild metric and the Kerr. If a non-rotating spherically symmetric mass (not necessarily a BH) is rotated, what changes? Its shape, and kinetic energy? But spacetime curvature is different, and so is its mass. Can you still say that the motion of the mass's components doesn't contribute to its mass or gravitational influence? It's similar for non-symmetric systems, such as 2 masses orbiting each other. Its mass and gravitational influence can include more than just the rest mass of its components. It's true for 2 orbiting bodies or for an atom or anything else that has a COM frame. Link to comment Share on other sites More sharing options...

Markus Hanke Posted March 14, 2023 Share Posted March 14, 2023 6 hours ago, md65536 said: Yes, the rest masses don't change, but together the rest mass of the system includes the kinetic energy of the bodies in the system's COM frame. Where and how does the kinetic energy in the system’s COM appear in the field equations, exactly? 6 hours ago, md65536 said: Can you still say that the motion of the mass's components doesn't contribute to its mass or gravitational influence? The dynamics of a system of moving masses, as in this example, most definitely has a gravitational influence - evidently a stationary system and a non-stationary one have very different spacetime geometries. I’ve tried to explain this in some detail in my previous post. My point though was specifically that you won’t find any explicit “mass” or “kinetic energy” terms in the vacuum field equations that somehow increase when there’s more motion, and that act as gravitational sources. That’s how it works in Newtonian gravity, but in GR it’s much more subtle. In fact, if you want to work out the geometry of the vacuum region outside the orbiting bodies, you’ll find that there are in fact no source terms at all. All you have is a rather complicated system of coupled differential equations for the relevant components of the metric, but nowhere does this system contain any terms for either masses or momenta (or energies of any kind). This being a set of differential equations, you then have to supply boundary and initial conditions to obtain a specific solution - and it’s only here that a “description” of the 2-body system enters the picture. The form these boundary conditions will take will be the position of the bodies in question at a specific time (in your chosen coordinate system), two free parameters, and the velocity 4-vectors of each body at that time, plus perhaps some global constraint or another. Starting with this initial “freeze frame”, the solution to the system of DGLs then describes the future evolution of this system - this will be a 2-parameter family of metrics, and these parameters are just the rest masses of the two bodies. So yes, motion (in vacuum) most certainly has a gravitational influence - but it enters as a boundary/initial condition, and not as a source term containing energies of any kind. The difference is subtle but very important. Also, because the system of equations is coupled, you cannot mathematically isolate the effects of motion from the effects of the other boundary conditions; so you can’t write the overall metric as the sum of a purely static metric, plus a metric purely for motion (which is, I believe, what geordief’s question was originally about). The effects of mass and motion are both intertwined in the overall metric, you can’t neatly separate them. Link to comment Share on other sites More sharing options...

md65536 Posted March 14, 2023 Share Posted March 14, 2023 (edited) 1 hour ago, Markus Hanke said: Where and how does the kinetic energy in the system’s COM appear in the field equations, exactly? I don't see how that is relevant so I wonder if we're even talking about the same thing? Where is my mistake in the following? All else being equal, a rotating object will have a higher invariant mass than the same object when not rotating. Curvature is related to invariant mass, in particular a higher invariant mass corresponds with a stronger Newtonian gravitational field. An object made of two massive bodies moving around each other, will have different spacetime curvature and "stronger" Newtonian gravity compared to the same bodies not moving (all else being equal). This answer's OP's main question with "yes." This also seems to contradict your statements. As far as I can tell, you're saying that because you can't plug the relevant values directly into the field equations, the result will be more complicated? But that won't change the statements above. 1 hour ago, Markus Hanke said: you cannot mathematically isolate the effects of motion from the effects of the other boundary conditions; so you can’t write the overall metric as the sum of a purely static metric, plus a metric purely for motion (which is, I believe, what geordief’s question was originally about). The effects of mass and motion are both intertwined in the overall metric, you can’t neatly separate them. That answer's OP's last question with "yes". I don't see how that matters. If you change the conditions that give you a different mass or different motion, you get different curvature, a different metric. Why would that imply anything about combining separate metrics? Edited March 14, 2023 by md65536 Link to comment Share on other sites More sharing options...

Markus Hanke Posted March 14, 2023 Share Posted March 14, 2023 1 hour ago, md65536 said: I don't see how that is relevant so I wonder if we're even talking about the same thing? Well, what I want to know is specifically what the spacetime geometry around a binary system looks like, and on what physical parameters it depends. The metric that describes that geometry must necessarily be a valid solution to the Einstein equations, so it can depend only on quantities that appear in these equations, either as a source term, a boundary condition, or as an integration constant when solving them. So I think it is very relevant for what I wish to do here. Newtonian concepts do not come into this for me at all, because they do not correctly describe the external radiation field and thus the dynamics of these bodies, nor even the geometry close to the binary system. I’m thinking purely in GR terms, whereas you appear to be mixing GR and Newtonian terms, so it seems we are unfortunately not talking about the same thing. As a word of warning - it’s rarely a good idea to mix Newton with GR, because there are subtle but extremely important differences in basic concepts - most notably mass, since in GR this will need to incorporate non-linear contributions from gravitational self-energy, and is usually a global property of the entire spacetime, rather than something that an isolated objected “possesses”. See link further below for a general overview of this. My understanding of geordief’s original question was that it was about GR, so I don’t use Newtonian physics here. You can of course try to analyse this using Newtonian gravity, but that’s not my goal; you will also find that for this kind of scenario the predictions from the two models will differ substantially. 1 hour ago, md65536 said: All else being equal, a rotating object will have a higher invariant mass than the same object when not rotating. That’s Newtonian physics (and correct in that context only), not GR. Invariant mass never appears anywhere in this GR calculation. It can’t, because it is a Newtonian concept that does not straightforwardly generalise to GR. 1 hour ago, md65536 said: Curvature is related to invariant mass The curvature in the exterior vacuum does not arise from any source terms. Counterintuitively, the masses (however defined) of the two bodies do not even enter as boundary conditions. All that happens is that, while solving the equations, you are left with two integration constants that can’t be eliminated, and these are - through some subtle argumentation - identified as the “masses” of these bodies, but not in the same sense as would be done in Newtonian physics. These are parameters in a 2-parameter family of metrics, and thus global properties of the entire spacetime. Unlike for the case of - say - Schwarzschild, this spacetime is not asymptotically flat, so even far from the binary system you can’t straightforward identify these parameters as mass in the Newtonian sense, as you would do for the Schwarzschild case. Also, the above statement I quoted you on, when taken as a general statement, is highly problematic - Newtonian gravity has no concept of curvature, and GR has no concept of invariant mass, so saying that the two are related strictly speaking doesn’t have any meaning. In exterior vacuum, the Einstein equations have no source term; and for interior spacetimes, the source is the energy-momentum tensor, which also does not contain invariant mass in the Newtonian sense. So either way, invariant mass never comes into this. 1 hour ago, md65536 said: and "stronger" Newtonian gravity Again, I am treating this as a pure GR problem. 1 hour ago, md65536 said: As far as I can tell, you're saying that because you can't plug the relevant values directly into the field equations, the result will be more complicated? What I am essentially saying is that this is a scenario where you cannot mix GR and the Newtonian concept of invariant mass. You’ll end up in a mess. And why would you? You can either analyse this purely in GR terms (that’s what I am trying to do), or purely in Newtonian terms (if you are prepared to ignore the radiation field, and the evolution of the system). But don’t mix them. 2 hours ago, md65536 said: This answer's OP's main question with "yes." In Newtonian gravity you can ascribe a gravitational force to each point of the surrounding space (iff you can define a potential field, which, btw, you can’t do here), which you can then compare - so if you analyse this in Newtonian terms, then the answer is probably yes. But in GR the question itself is essentially meaningless - all you have to compare are two metric tensors, and there is no meaningful way to say that one is greater/equal/less than the other. They are just different. What I suggest you could do though is look at a small region somewhere outside the binary system, take two test particles which are initially at relative rest, and measure by how much their relative separation changes as each wave front passes. A GW detector, essentially. You’ll find that the faster the bodies orbit, the larger the local wave amplitudes - so in that very particular and local sense, you could say the gravitation gets “stronger” with increasing angular momentum, in that region where you perform the measurement. Link to comment Share on other sites More sharing options...

Markus Hanke Posted March 14, 2023 Share Posted March 14, 2023 P.S. I’ve forgotten to mention that not only mass, but also angular momentum in binary systems such as this one is a concept that does not easily generalise from Newtonian to GR physics, because angular momentum in GR radiation fields is subject to a mathematical issue called “supertranslation ambiguity”. This has in fact long been a very difficult problem, that has only recently been resolved. So not only is “mass” a problematic concept, but “angular momentum” is too - that’s why neither of these are used when solving the Einstein equations for such systems. You just work with initial and boundary conditions, and let the non-linearity of the equations themselves take care of the rest. Link to comment Share on other sites More sharing options...

md65536 Posted March 14, 2023 Share Posted March 14, 2023 (edited) 8 hours ago, Markus Hanke said: In Newtonian gravity you can ascribe a gravitational force to each point of the surrounding space (iff you can define a potential field, which, btw, you can’t do here), which you can then compare - so if you analyse this in Newtonian terms, then the answer is probably yes. But in GR the question itself is essentially meaningless - all you have to compare are two metric tensors, and there is no meaningful way to say that one is greater/equal/less than the other. They are just different. Then what is the correct way to describe their difference? Say you have 2 metrics, one for a system with little mass, and another for one with greater mass. The one with greater mass fits all intuitive notions of "stronger gravity", but how would that be expressed correctly? Greater mass corresponds with greater curvature, wouldn't the latter have greater curvature? Edited March 14, 2023 by md65536 Link to comment Share on other sites More sharing options...

md65536 Posted March 15, 2023 Share Posted March 15, 2023 I re-read the thread looking for why this is not at all making sense to me. One thing I see is I was only talking about curvature far away from the masses, while OP mentions only the curvature at the point they meet. Other replies talk about both, and it's only the stuff about curvature far away that's not making sense to me. Another thing is I might not understand what is meant by a "source of gravity". I'll try again to rephrase OP's question to see if I can figure out where I'm going wrong. Say that you have two identical massive objects far apart from each other, and a test particle at rest at the midpoint between them. Then say you add a relatively massive amount of energy to one of the objects by spinning it. A very basic understanding of relativity is that the test particle will fall toward the object with more mass (energy), and that it is due to curvature of spacetime that it falls. Even though that additional energy is equivalent to mass, it is not what you're calling a "source of gravity"? Does this experiment make sense with what's already been said in the thread? The only other difference I can think of is that I'm talking about a huge contribution of kinetic energy and maybe other replies are talking about insignificant amounts? I'm also not sure if we're not in agreement about what would happen in an experiment like this, or if I'm just failing to understand the formal descriptions in GR. Link to comment Share on other sites More sharing options...

Markus Hanke Posted March 15, 2023 Share Posted March 15, 2023 12 hours ago, md65536 said: Then what is the correct way to describe their difference? Say you have 2 metrics, one for a system with little mass, and another for one with greater mass. That’s a good and valid and very tricky question, md65536. Unfortunately it’s not possible to do this globally - there simply is no standard by which you can take an entire spacetime geometry and say “this spacetime contains more gravity than the other one”, because you can’t meaningfully compare tensor fields as “more/less” or “smaller/greater”. The only thing you can in fact do is check whether two different metrics might describe the same physical spacetime - there’s a standard procedure for that. What you can do though - and that’s how I would answer your question - is make the issue local. Pick a specific small region within that spacetime, and then evaluate your curvature tensors in that specific region - for example you can look at the scalar invariants of the tensors there, or go for broke and explicitly calculate the tidal forces between test particles within that local region. You can then vary the gravitational sources (the binary system, in this example), and check what physical consequences this has in your test region. So while a direct comparison is pretty much meaningless globally, it can easily be done locally around a specific point in your spacetime. Just remember that if gravity gets stronger in your test region, that doesn’t mean that the same happens everywhere else in your spacetime - perhaps at other points nothing changes, or gravity even gets weaker there. Curvature can “shift around”, or “radiate away”. 12 hours ago, md65536 said: Greater mass corresponds with greater curvature This statement is based on a Newtonian intuition (where it is absolutely valid), but unfortunately it isn’t this straightforward in GR at all, because here the curvature arises from several distinct things: 1. The actual local source term in the field equations, which is the energy-momentum tensor 2. Gravitational self-interaction, which is encoded in the non-linear structure of the equations themselves (no explicit source term) 3. Boundary/initial conditions, which you must manually supply to obtain a specific solution for the field equations (this roughly represents distant sources) 4. Integration constants, which appear in the process of solving the equations, and are given a physical meaning (if applicable) by comparison with other known solutions in a given region. If they appear in the final metric, they are global properties of the entire spacetime. Counterintuitively, the “mass” we are familiar with does not appear as a source at all - it is not part of the energy-momentum tensor (which only contains densities, fluxes, and pressures for the interior of bodies/fields), nor is it part of the boundary conditions you supply. Unlike in Newtonian gravity, in GR the mass simply isn’t in the picture until you get to step (4) - here, integration constants appear, and these can often (but not always!) be interpreted as concepts of mass, charge, angular momentum. One very important difference to Newton is that these constants are properties of the entire spacetime, not just some isolated body within it - so they contain their Newtonian equivalents, but also contributions from initial/boundary conditions (such as motion), as well as gravitational self-interaction. One could nearly say that mass/charge/angular momentum are not inputs into the equation, but rather that they arise from the equation, once it has been properly set up. So this is very different from Newton. And then there’s of course the issue of what “greater curvature” even means, because that’s not a straightforward concept either. For example, at the EH of a BH with the size of the earth, tidal forces are so great that any ordinary material body would immediately get ripped to shreds. On the other hand, at the EH of a supermassive BH, tidal forces are vanishingly small, to the degree that it would be an engineering challenge to detect them at all. So which curvature is “greater”, and how does this relate to “mass”? It’s not a straightforward comparison. As a simply example of where the Newtonian intuition fails completely, consider a particular spacetime called the Bonner beam - it’s essentially two parallel, very long beams of light. One is free to give each of these beams arbitrarily much energy. Because they contain lots of energy, and energy is equivalent to mass, we should be able to ascribe some notion of mass to them - meaning these beam should gravitationally attract one another, because of curvature etc. Right? What you will actually find is that, if you shoot these beams parallel in the same direction, they will not attract at all; but if you shoot them in opposite directions (all other things equal), they will attract, but not according to Newtonian inverse square laws, but something much more complicated. So, mass alone won’t always work well in GR - it sometimes gives the right intuition, but in other circumstances it can fail really badly. 2 hours ago, md65536 said: Other replies talk about both, and it's only the stuff about curvature far away that's not making sense to me. Far away from the binary system you simply have a radiation field with a succession of wave fronts of a particular wavelength and amplitude - so if you were to place a bunch of test particles there, then their separations would oscillate at a certain frequency and with a certain amplitude, transverse to the direction of propagation of the waves. Note that you can’t replicate this finding with Newtonian gravity. This is an example of a spacetime that has three “hairs” - mass, angular momentum, and a quadrupole moment. 2 hours ago, md65536 said: Another thing is I might not understand what is meant by a "source of gravity". It’s “source” in the field-theoretic sense. In GR, sources are the energy-momentum tensor, gravitational self-interaction, and boundary conditions. The first appears explicitly in the equations, whereas the other two are encoded in the structure and nature of the equations themselves. Do also note that tensors are local objects - so the energy-momentum tensor is non-zero only in the interior of objects and fields, and vanishes in vacuum. So for example, if you are looking for the curvature around the binary system, you are in fact solving the vacuum equations \(R_{\mu \nu}=0\), absent of any explicit source term. 2 hours ago, md65536 said: Even though that additional energy is equivalent to mass, it is not what you're calling a "source of gravity"? Yes, the angular momentum you are adding to one of the bodies is a source of gravity, in the sense that it changes the geometry of this spacetime. It definitely has an influence. It’s just that in GR it doesn’t simply appear as a contribution to the mass of the body, in the same way as you might do that in Newtonian theory. Rather, adding angular momentum takes away one of the symmetries of your spacetime - any free fall into such a body will not only have a radial, but also an angular component (frame dragging), so the overall geometry is no longer spherically symmetric. In practice, you would start with a different metric ansatz that reflects these symmetries when you set out to solve the field equations. Fewer symmetries in your spacetime generally leads to more free parameters in your final solution - here, you’d get a spacetime with two global properties instead of one, which can be identified with mass and angular momentum. Again, remember that these are properties of the entire spacetime, not just the isolated body within it. As to where your test particle will fall, I can only make an educated guess - in GR a free fall test particle will trace out that world line which maximises its proper time (principle of extremal ageing). If you add angular momentum to one the planets, the curvature tensors will take on additional terms that reflect this, and the particle will fall not just radially but also sideways. I think this should translate to geodesics with longer geometric length in spacetime, so the particle should fall towards the rotating body, rather than the non-rotating one. It should be clearly noted though that this is an example of where Newtonian gravity gets the actual free-fall trajectory wrong, because it would just predict a more rapid but purely radial in-fall - whereas in GR the in-fall takes on an angular component as well, so you’d get a segment of a spiral. In practice one would have to run the actual numbers to be sure of what happens - which would be very non-trivial, because the influence of the other planet cannot really be neglected here, and curvatures combine in non-linear ways. So I would not be at all surprised if it turned out that my educated guess is in fact wrong. There are just too many subtleties involved to be sure without doing the maths. 3 hours ago, md65536 said: or if I'm just failing to understand the formal descriptions in GR. The fundamental concepts are completely different. With Newton, you start with sources (distribution of mass densities), put them into the field equation, and get potentials/forces as a result. In GR, everything is geometry - you start with a metric ansatz that reflects the global symmetries of the spacetime you are looking for, and then you constrain that ansatz more and more until you obtain a specific metric: you first combine the ansatz with any local sources but putting it into the field equations; you then combine it with distant sources by supplying boundary conditions; and finally you solve the system and find physical interpretations for any remaining integration constants. As a result you get a specific metric, from which you can derive curvatures, geodesics etc etc. The fundamental difference is that in GR the gravitational field also interacts with itself, so even in weak-field scenarios you might get phenomena that run counter to Newtonian intuition. It also means that many of the fundamental concepts like mass, angular momentum etc don’t straightforwardly carry over from Newton, because they can’t account for the self-interaction of gravity, not even in principle. Sometimes such differences can be neglected, and sometimes not - it really depends on the scenario. But I think the most important thing is to not try and mix concepts, because that is almost guaranteed to go wrong in some way or another. Link to comment Share on other sites More sharing options...

geordief Posted March 16, 2023 Author Share Posted March 16, 2023 Thanks so much @md65536 and @Markus Hanke That was quite some detail and probing I did "sign off " half way through this (my) thread but am glad I was able to (come back and (re)read it all the way through. Hopefully some of it will sink in. And ,yes the OP was all about GR,in my mind anyway Link to comment Share on other sites More sharing options...

md65536 Posted March 16, 2023 Share Posted March 16, 2023 On 3/13/2023 at 3:33 AM, Markus Hanke said: Yes, the motion does of course “contribute” to the geometry of this spacetime, as I said in my post - but not really in an intuitive Newtonian way of making it “stronger”, which is an ill-defined concept in GR. I think my main confusion was taking this to mean that adding "mass" energy generally doesn't correspond with any "stronger" Newtonian gravity. I'm also not interested in Newtonian gravity except as an approximation of the predictions of GR. But as you pointed out, in my example the strong frame-dragging effect has no approximation in Newtonian gravity. It should be possible to contrive an example where the Newtonian forces are small, and the purely GR aspects dominate and give results that are the exact opposite of what Newtonian gravity predicts. 20 hours ago, Markus Hanke said: As a simply example of where the Newtonian intuition fails completely, consider a particular spacetime called the Bonner beam - it’s essentially two parallel, very long beams of light. One is free to give each of these beams arbitrarily much energy. Because they contain lots of energy, and energy is equivalent to mass, we should be able to ascribe some notion of mass to them - meaning these beam should gravitationally attract one another, because of curvature etc. Right? What you will actually find is that, if you shoot these beams parallel in the same direction, they will not attract at all; but if you shoot them in opposite directions (all other things equal), they will attract, but not according to Newtonian inverse square laws, but something much more complicated. So, mass alone won’t always work well in GR - it sometimes gives the right intuition, but in other circumstances it can fail really badly. Maybe, but in this example mass alone and the notion of energy having mass in the COM frame seems to work fine and fits with intuition. With 2 beams in the same direction, there's no COM frame. The energy of the beams is frame-dependent, and as total momentum approaches zero, so does the energy. Intuitively there is no energy that could make up "rest mass." With beams in opposite directions there is energy in the COM frame. True, that's not enough to guess how much they'd attract. I wouldn't even be certain the beams would attract (because of the uncommon geometry of the "mass"). As you've said such a notion of mass would be "problematic" in GR. But I think at least it's intuitive that the energy that doesn't vanish with a choice of coordinates, curves spacetime around the beams, similar to how any other equivalent mass would (and I don't know how to say that without using the word 'mass' even if it's just in a vague way). Link to comment Share on other sites More sharing options...

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