Aethelwulf Posted June 9, 2012 Share Posted June 9, 2012 (edited) It occurred to me a week past how to variate the energy of a Schwarzschild metric. The Energy changing in a Schwartzschild Metric It is not obvious how to integrate an energy in the Schwartzschild metric unless you derive it correctly. The way this following metric will be presented will be: [math] c^2 d\tau^2 = (1 - 2\frac{GM}{\Delta E} \frac{M}{r_s} c^2 dt^{2}) - \frac{dt}{(1-2\frac{GM}{\Delta E} \frac{M}{r_s})} - r^2 d \phi dt[/math] This will be interpeted as [math] c^2 d\tau^2 = (1 - 2\frac{GM}{E - E'} \frac{M}{r_s} c^2 dt^{2}) - \frac{dt}{(1-2\frac{GM}{E - E'} \frac{M}{r_s})} - r^2 d \phi dt[/math] And this metric is dimensionally-consistent to calculate the energy changes within a metric. Usually, in the spacetime metric, we treat it as a energy efficient fabric. This can be a way to treat a metric with a type of energy variation consistent perhaps with a radiating body. Edited June 9, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

elfmotat Posted June 9, 2012 Share Posted June 9, 2012 The Schwarzschild solution is (as a necessary assumption) t-symmetric, so it doesn't make any sense to talk about changes in the central mass. A changing mass distribution will produce an entirely different solution to the EFE's. I'm also not really sure how you arrived at that metric. It looks pretty nonsensical to me. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 9, 2012 Author Share Posted June 9, 2012 (edited) It occurred to me a week past how to variate the energy of a Schwarzschild metric. The Energy changing in a Schwartzschild Metric It is not obvious how to integrate an energy in the Schwartzschild metric unless you derive it correctly. The way this following metric will be presented will be: [math] c^2 d\tau^2 = (1 - 2\frac{GM}{\Delta E} \frac{M}{r_s} c^2 dt^{2}) - \frac{dt}{(1-2\frac{GM}{\Delta E} \frac{M}{r_s})} - r^2 d \phi dt[/math] This will be interpeted as [math] c^2 d\tau^2 = (1 - 2\frac{GM}{E - E'} \frac{M}{r_s} c^2 dt^{2}) - \frac{dt}{(1-2\frac{GM}{E - E'} \frac{M}{r_s})} - r^2 d \phi dt[/math] And this metric is dimensionally-consistent to calculate the energy changes within a metric. Usually, in the spacetime metric, we treat it as a energy efficient fabric. This can be a way to treat a metric with a type of energy variation consistent perhaps with a radiating body. The Schwarzschild solution is (as a necessary assumption) t-symmetric, so it doesn't make any sense to talk about changes in the central mass. A changing mass distribution will produce an entirely different solution to the EFE's. I'm also not really sure how you arrived at that metric. It looks pretty nonsensical to me. It should be entirely consistent to talk about a changing energy in a metric, considering that energy can be stored in a metric, just like spacetime. Spacetime is made of fields which store energy, the two concepts cannot be separated. Knowing that the metric can be written as [math] c^2 d\tau^2 = (1 - 2\frac{G}{c^2} \frac{M}{r_s} c^2 dt^{2}) - \frac{dt}{(1-2\frac{G}{c^2} \frac{M}{r_s})} - r^2 d \phi dt[/math] Then from the gravitational parameter equation [math]\frac{G}{c^2}E = GM[/math] one can get [math]\frac{G}{c^2} = \frac{GM}{E}[/math] from this you simply replace all the two terms in the metric for the representation [math]\frac{GM}{E}[/math] and variate the energy in the metric. It is certainly possible according to this paper to talk about the energy of a schwartzschild metric http://www.dougweller.com/momentum.pdf I had a couple extra dt's in there which shouldn't have been in there, if that added to the confusion. Sorry about that... it is quite a cumbersome equation to write out. Edited June 9, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

elfmotat Posted June 9, 2012 Share Posted June 9, 2012 (edited) It should be entirely consistent to talk about a changing energy in a metric, considering that energy can be stored in a metric, just like spacetime. Spacetime is made of fields which store energy, the two concepts cannot be separated. In general, yes, the metric can be time dependent. It doesn't make any sense to talk about changes in the central mass in the Schwarzschild metric because it is derived using the explicit assumption that it does not change over time. If the metric is t-dependent then it is not the Schwarzschild metric. Knowing that the metric can be written as [math] c^2 d\tau^2 dt = (1 - 2\frac{G}{c^2} \frac{M}{r_s} c^2 dt^{2}) - \frac{dt}{(1-2\frac{G}{c^2} \frac{M}{r_s})} - r^2 d \phi dt[/math] That's not the Schwarzschild metric. I've never seen anything that looks like that before. The Schwarzschild metric is: [math]ds^2=\left (1-\frac{r_s}{r} \right )dt^2-\left (1-\frac{r_s}{r} \right )^{-1}dr^2-r^2d\phi ^2-r^2sin^2\phi d\theta ^2[/math] where [math]r_s=\frac{2GM}{c^2}[/math] Then from the gravitational parameter equation [math]\frac{G}{c^2}E = GM[/math] one can get [math]\frac{G}{c^2} = \frac{GM}{E}[/math] from this you simply replace all the two terms in the metric for the representation [math]\frac{GM}{E}[/math] Yes, you can get the SC metric in terms of rest energy if you wanted to. and variate the energy in the metric. No, you're not allowed to do this. If the energy distribution is changing then you need to go back to the Einstein field equations, insert this information into the stress-energy tensor, then re-solve the equations to obtain a completely new metric. What you're doing doesn't make sense. It is certainly possible according to this paper to talk about the energy of a schwartzschild metrichttp://www.dougweller.com/momentum.pdf That pdf never once mentions any change in central mass in the SC metric. It is certainly possible to talk about the energy of particles in SC spacetime (so long as they're not energetic enough to disturb the metric), but this says nothing about what you're proposing. Edited June 9, 2012 by elfmotat Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 9, 2012 Author Share Posted June 9, 2012 (edited) In general, yes, the metric can be time dependent. It doesn't make any sense to talk about changes in the central mass in the Schwarzschild metric because it is derived using the explicit assumption that it does not change over time. If the metric is t-dependent then it is not the Schwarzschild metric. That's not the Schwarzschild metric. I've never seen anything that looks like that before. The Schwarzschild metric is: [math]ds^2=\left (1-\frac{r_s}{r} \right )dt^2-\left (1-\frac{r_s}{r} \right )^{-1}dr^2-r^2d\phi ^2-r^2sin^2\phi d\theta ^2[/math] where [math]r_s=\frac{2GM}{c^2}[/math] Yes, you can get the SC metric in terms of rest energy if you wanted to. No, you're not allowed to do this. If the energy distribution is changing then you need to go back to the Einstein field equations, insert this information into the stress-energy tensor, then re-solve the equations to obtain a completely new metric. What you're doing doesn't make sense. That pdf never once mentions any change in central mass in the SC metric. It is certainly possible to talk about the energy of particles in SC spacetime (so long as they're not energetic enough to disturb the metric), but this says nothing about what you're proposing. No schwartzschild metric that I have seen written in terms of energy like I have written it, and speaking of a rest mass ... I don't see your point. The point is that my metric is written in terms of energy... can you find me a literature that has written it like I have? Secondly, I never said that the paper did speak of varying energy. I am saying speaking of an energy for a metric is not unheard of. As for how the schwarztschild metric is written, I can find where I copied this one metric down if you give me time. I obviously read many literatures, it will take time. You need to keep in mind I wrote the original terms in metric wrong, I had an extra dt in there on the left hand side of the equation which should not have been there. I made this clear. Does this clear up your problem? I changed that before you even posted that last post>? Right, the metric I have is fine. dτ² where c=1 in some cases, mine is not. The thing which troubled you was the extra dt term which shouldn't have been there, but I explained this. The metric is fine otherwise. Edited June 9, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

elfmotat Posted June 9, 2012 Share Posted June 9, 2012 (edited) No schwartzschild metric that I have has seen written in terms of energy like I have written it... ...The point is that my metric is written in terms of energy... can you find me a literature that has written it like I have? Okay... so what? All you had to do was substitute M=E/c^{2 }to get it in terms of rest energy. I don't see how this is useful. and speaking of a rest mass ... I don't see your point. I never mentioned rest mass. My point is that the energy distribution in the SC metric can't change, or else it ceases to be the SC metric. Secondly, I never said that the paper did speak of varying energy. I am saying speaking of an energy for a metric is not unheard of. He describes the energy associated with test particles in SC spacetime, not some sort of "metric self-energy" like you seem to be implying. Of course, there is automatically an energy-momentum distribution associated with any metric given by the field equations. The SC metric is a vacuum solution, so all of the components of the stress-energy tensor (including energy density) are zero. I also fail to see how this relates to your original post, or any of mine. Right, the metric I have is fine. dτ² where c=1 in some cases, mine is not. The thing which troubled you was the extra dt term which shouldn't have been there, but I explained this. The metric is fine otherwise. Written in the form you have it, it should look like this: [math]c^2 d\tau^2 = (1 - 2\frac{G}{c^2} \frac{M}{r} )c^2 dt^{2} - \frac{dr^2}{(1-2\frac{G}{c^2} \frac{M}{r})} - r^2 d \phi^2-r^2sin^2\phi d\theta^2[/math] Edited June 9, 2012 by elfmotat 1 Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 9, 2012 Author Share Posted June 9, 2012 (edited) Okay... so what? All you had to do was substitute M=E/c^{2 }to get it in terms of rest energy. I don't see how this is useful. I never mentioned rest mass. My point is that the energy distribution in the SC metric can't change, or else it ceases to be the SC metric. He describes the energy associated with test particles in SC spacetime, not some sort of "metric self-energy" like you seem to be implying. Of course, there is automatically an energy-momentum distribution associated with any metric given by the field equations. The SC metric is a vacuum solution, so all of the components of the stress-energy tensor (including energy density) are zero. I also fail to see how this relates to your original post, or any of mine. No I did not just need to replace M with E/c^2... that's rubbish. I replaced G/c^2 with GM/E. That is slightly different. ''Yes, you can get the SC metric in terms of rest energy if you wanted to. '' Sorry, you said rest ''energy''... But we are not talking about a massless system, so I feel you are being pedantic. ''He describes the energy associated with test particles in SC spacetime, not some sort of "metric self-energy" like you seem to be implying. Of course, there is automatically an energy-momentum distribution associated with any metric given by the field equations. '' My approach is much different to his... however, you claimed that you cannot vary such an energy. I challenge that claim from my derivation. I am asking you why any metric, not just this, cannot be varied? It only seems right from the principles of quantum mechanics that you can vary any energy in a metric. If this is a claim as you specified from relativity, relativity is demonstratably wrong since metrics like spacetime are composed of fields which must vary in energy, fields are not generally static, only in special cases. Written in the form you have it, it should look like this: [math]c^2 d\tau^2 = (1 - 2\frac{G}{c^2} \frac{M}{r} )c^2 dt^{2} - \frac{dr^2}{(1-2\frac{G}{c^2} \frac{M}{r})} - r^2 d \phi^2-r^2sin^2\phi d\theta^2[/math] Yes, that is right. I changed it way before you posted look back please. And since you have not provided a reference when asked, NO ONE has ever derived a metric like mine. I have certainly never came across one. Edited June 9, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

elfmotat Posted June 9, 2012 Share Posted June 9, 2012 No I did not just need to replace M with E/c^2... that's rubbish. I replaced G/c^2 with GM/E. That is slightly different. They are equivalent and equivalently useless. ''Yes, you can get the SC metric in terms of rest energy if you wanted to. '' Sorry, you said rest ''energy''... But we are not talking about a massless system, so I feel you are being pedantic. You're the one who decided to write the SC metric in terms of energy, not me. As I've said before, I don't see the point. ''He describes the energy associated with test particles in SC spacetime, not some sort of "metric self-energy" like you seem to be implying. Of course, there is automatically an energy-momentum distribution associated with any metric given by the field equations. '' My approach is much different to his... however, you claimed that you cannot vary such an energy. I challenge that claim from my derivation. I am asking you why any metric, not just this, cannot be varied? I've already tried to explain this to you: the metric is related to the energy-momentum distribution in spacetime by the Einstein field equations. The SC metric is solved from the EFE's using the assumption that it does not change over time. If the central mass is changing, then the metric produced by that mass is not the SC metric. This doesn't mean that the central mass isn't allowed to change, it just means that the metric it produces will be different from the SC metric. You can't just say "so now we vary the energy..." It doesn't work that way. It only seems right from the principles of quantum mechanics that you can vary any energy in a metric. If this is a claim as you specified from relativity, relativity is demonstratably wrong since metric are composed of fields which must vary in energy, fields are not generally static, only in special cases. Agreed. Yes, that is right. I changed it way before you posted look back please. They still look significantly different to me. Mine: Yours: See if you can spot the differences (there are quite a few). Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 9, 2012 Author Share Posted June 9, 2012 (edited) They are equivalent and equivalently useless. You're the one who decided to write the SC metric in terms of energy, not me. As I've said before, I don't see the point. [/size] I've already tried to explain this to you: the metric is related to the energy-momentum distribution in spacetime by the Einstein field equations. The SC metric is solved from the EFE's using the assumption that it does not change over time. If the central mass is changing, then the metric produced by that mass is not the SC metric. This doesn't mean that the central mass isn't allowed to change, it just means that the metric it produces will be different from the SC metric. You can't just say "so now we vary the energy..." It doesn't work that way. [/size] Agreed. They still look significantly different to me. Mine: Yours: See if you can spot the differences (there are quite a few). It's not useless. You said it yourself, it is a self-energy. Are you going to try and wheel the idea that any metric does not have an energy? If that is your position, I can tell you right now that physics rightly disagree's in that predication. As for the metric, I see the last terms seem different, this is because the phi --- to make this clearer because you are stuck on a notational problem, my phi can be replaced with [math]r^2 d\Omega[/math] It means the same thing. My term is absorbing the rest of your terms up. Ok? Now, for the rest... where [math]d\Omega[/math] is basically [math]d\phi^2 + (sin\phi)^2d\theta[/math] That is hands up, my fault for making that confusing but I am well aware of your presentation. Now, the metric you say is ''I've already tried to explain this to you: the metric is related to the energy-momentum distribution in spacetime by the Einstein field equations. The SC metric is solved from the EFE's using the assumption that it does not change over time. If the central mass is changing, then the metric produced by that mass is not the SC metric. '' I feel something is wrong with this statement. Any metric, including the SC will change over time. Even especially in the case of an integral over the time, even a SC metric should not remain constant. That goes against the statement you said yourself you agreed with. I have not ever heard of relativity deriving the metric from non-changing values of energy for the metric, but I can say right now that I believe this to be a faulty premise. I think that any metric will change in time, whether it is your normal spacetime metric to even a Schwarztschild metric, to even this latter metric solved for a local flat neighbourhood which... by the way... is basically mathematically the same thing as the kind of flat spacetime you deal with when looking in any direction of this metric. Edited June 9, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

juanrga Posted June 9, 2012 Share Posted June 9, 2012 (edited) The Schwarzschild metric is: [math]ds^2=\left (1-\frac{r_s}{r} \right )dt^2-\left (1-\frac{r_s}{r} \right )^{-1}dr^2-r^2d\phi ^2-r^2sin^2\phi d\theta ^2[/math] where [math]r_s=\frac{2GM}{c^2}[/math] Either you lack a [math]c^2[/math] factor in [math]dt^2[/math] or you are using a [math]c=1[/math] system of units and then [math]r_s=2GM[/math], but both expressions cannot be true. When corrected, the above is the Schwarzschild metric only in Schwarzschild coordinates. Edited June 9, 2012 by juanrga 1 Link to comment Share on other sites More sharing options...

elfmotat Posted June 9, 2012 Share Posted June 9, 2012 It's not useless. You said it yourself, it is a self-energy. Are you going to try and wheel the idea that any metric does not have an energy? If that is your position, I can tell you right now that physics rightly disagree's in that predication. I still don't see your point. As for the metric, I see the last terms seem different, this is because the phi --- to make this clearer because you are stuck on a notational problem, my phi can be replaced with [math]r^2 d\Omega[/math] It means the same thing. My term is absorbing the rest of your terms up. Ok? Now, for the rest... where [math]d\Omega[/math] is basically [math]d\phi^2 + (sin\phi)^2d\theta[/math] That is hands up, my fault for making that confusing but I am well aware of your presentation. There are still quite a few differences. Now, the metric you say is ''I've already tried to explain this to you: the metric is related to the energy-momentum distribution in spacetime by the Einstein field equations. The SC metric is solved from the EFE's using the assumption that it does not change over time. If the central mass is changing, then the metric produced by that mass is not the SC metric. '' I feel something is wrong with this statement. Any metric, including the SC will change over time. Even especially in the case of an integral over the time, even a SC metric should not remain constant. That goes against the statement you said yourself you agreed with. Too bad, math doesn't take into account your feelings. Take the time derivative of the SC metric components and you'll see that ∂_{t}g_{μv}=0. I have not ever heard of relativity deriving the metric from non-changing values of energy for the metric, but I can say right now that I believe this to be a faulty premise. I think that any metric will change in time, whether it is your normal spacetime metric to even a Schwarztschild metric, to even this latter metric solved for a local flat neighbourhood which... by the way... is basically mathematically the same thing as the kind of flat spacetime you deal with when looking in any direction of this metric. Whether or not it is reflective of reality is irrelevant. The point is that the SC metric doesn't change over time. This is a fact. Either you lack a [math]c^2[/math] factor in [math]dt^2[/math] or you are using a [math]c=1[/math] system of units and then [math]r_s=2GM[/math], but both expressions cannot be true. Yes, you're right. I'm not used to working in conventional units, and I forgot the extra c^{2} factor on the tt component. When corrected, the above is the Schwarzschild metric only in Schwarzschild coordinates. Yep. 1 Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 9, 2012 Author Share Posted June 9, 2012 Well, I am going to be frank, either way in writing the metric is fine. That's is really not the point however. The whole of this discussion is whether you can vary the metric energy. I have shown you can. You say what is the point? I say it is a prediction of quantum mechanics. That's the point. Show me some precise calculations of your approach and prove to me you cannot vary the energy. Then I will decide how valid a statement it is. Link to comment Share on other sites More sharing options...

elfmotat Posted June 9, 2012 Share Posted June 9, 2012 This is becoming increasingly frustrating; it's starting to feel like I'm talking to a wall. I've already tried to explain to you (multiple times) why what you suggest is invalid. You need to specify how the mass is changing FIRST, and then you can determine the metric it produces. Talking about a changing SC metric is an oxymoron. I don't know how else I can explain it to you. Also, GR is a classical theory, and I don't see at all how this ties in with QM. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 10, 2012 Author Share Posted June 10, 2012 This is becoming increasingly frustrating; it's starting to feel like I'm talking to a wall. I've already tried to explain to you (multiple times) why what you suggest is invalid. You need to specify how the mass is changing FIRST, and then you can determine the metric it produces. Talking about a changing SC metric is an oxymoron. I don't know how else I can explain it to you. Also, GR is a classical theory, and I don't see at all how this ties in with QM. You said something about the time derivatives being zero, I'd like to see some math so I can work with it. As for quantum mechanics, I believe it would say that a metric like a Schwartzschild metric would vary in time. Just like a spacetime metric, energies in metrics generally change. Link to comment Share on other sites More sharing options...

elfmotat Posted June 11, 2012 Share Posted June 11, 2012 (edited) You said something about the time derivatives being zero, I'd like to see some math so I can work with it. Yes, ∂_{t}g_{μv}=0, i.e. it is constant in time. As for quantum mechanics, I believe it would say that a metric like a Schwartzschild metric would vary in time. Just like a spacetime metric, energies in metrics generally change. This is the last time I'm going to explain it because I'm sick of repeating myself: First of all, GR is a classical theory which doesn't take into account QM effects. There is currently no good theory of quantum gravity, so talking about how quantum effects will change the metric is impossible at this point. The best you can currently do is specify a background metric and describe quantum effects within this background. If these effects are large enough to affect the background, then we have no idea how to deal with them. The Schwarzschild metric is the metric which describes a non-rotating, uncharged, spherically symmetric mass that doesn't change in time. That is all it can and does describe. If, for example, you want to describe a spherically symmetric body with angular momentum then you need an entirely different metric: the Kerr metric. If you want to describe a cloud of dust, you need something like the Lemaître–Tolman metric. If you want to describe really specific situations, you might not be able to solve the EFE's perfectly and will have to resort to numerical methods. The SC metric is not necessarily an accurate representation of reality if you're taking into account small details like radiation. It is useful, for example, to model the Sun with the SC metric. It's obviously not a 100% accurate model since the Sun is not perfectly symmetrical, it's giving off radiation, it's rotating a bit, etc. Nonetheless, the SC metric is a good approximation. The SC metric itself cannot and does not change over time (like I said above: take the time derivatives of its components if you don't believe me). It is derived by specifying the fact that it is t-symmetric. If you want to see a specific derivation, I like Carroll's: http://ned.ipac.calt...3/Carroll7.html. He talks about how the metric is static after equation (7.20). Edited June 11, 2012 by elfmotat 2 Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 11, 2012 Author Share Posted June 11, 2012 All I asked for was a more in-depth explanation and you have now provided it. I will certainly follow the link... I could be very wrong in my approach which is why I questioned it. You do mentioned the radiation a few times: ''The SC metric is not necessarily an accurate representation of reality if you're taking into account small details like radiation. It is useful, for example, to model the Sun with the SC metric. It's obviously not a 100% accurate model since the Sun is not perfectly symmetrical, it's giving off radiation, it's rotating a bit, etc. Nonetheless, the SC metric is a good approximation.'' This was the kind of idea I had in mind... except for a non-rotating body. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 13, 2012 Author Share Posted June 13, 2012 Right, so I have had time to read it. Generally, it is saying the metric possesses a timelike killing vector, which is actually makes the metric static - or t-symmetric in your case example. I find it hard to believe still, a metric as such does not gives up energy... as you yourself established, the SC metric is not an accurate representation of reality. My case in the OP would highlight that. So it seems, there is definitely a problem with its accuracy - perhaps the way it is treated mathematically is the problem. Link to comment Share on other sites More sharing options...

elfmotat Posted June 13, 2012 Share Posted June 13, 2012 (edited) Right, so I have had time to read it. Generally, it is saying the metric possesses a timelike killing vector, which is actually makes the metric static - or t-symmetric in your case example. I find it hard to believe still, a metric as such does not gives up energy... as you yourself established, the SC metric is not an accurate representation of reality. My case in the OP would highlight that. So it seems, there is definitely a problem with its accuracy - perhaps the way it is treated mathematically is the problem. It's not a problem so much as a simplifying approximation. Physics itself really only deals with the study of simple models. Taking the radiation of a star into account when making gravitational calculations is like taking into account dust particles when measuring the diameter of a basketball (actually, the dust would almost definitely be more significant in relation to the basketball than the radiation from the Sun would be gravitationally). Edited June 13, 2012 by elfmotat Link to comment Share on other sites More sharing options...

juanrga Posted June 13, 2012 Share Posted June 13, 2012 Right, so I have had time to read it. Generally, it is saying the metric possesses a timelike killing vector, which is actually makes the metric static - or t-symmetric in your case example. I find it hard to believe still, a metric as such does not gives up energy... as you yourself established, the SC metric is not an accurate representation of reality. My case in the OP would highlight that. So it seems, there is definitely a problem with its accuracy - perhaps the way it is treated mathematically is the problem. The SC metric is enough to study light-bending, mercury perihelion anomaly, and other so-named classic tests of GR up to PN-order in complete agreement with observations. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 14, 2012 Author Share Posted June 14, 2012 (edited) It's not a problem so much as a simplifying approximation. Physics itself really only deals with the study of simple models. Taking the radiation of a star into account when making gravitational calculations is like taking into account dust particles when measuring the diameter of a basketball (actually, the dust would almost definitely be more significant in relation to the basketball than the radiation from the Sun would be gravitationally). Here I will assume by simple, we really really mean a first approximation in energy - and higher perturbations could not be calculated because the metric is static - though the metric surely would change over time, so it seems the SC is an oversimplification of the real dynamics going therefore the problem lies within the consistency over long periods of time. How accurate would it be for a system which was radiating fast, perhaps in some kind of plasma-like bursts? Given enough time, I'd say it was pretty inaccurate; which leads me to my own question, is there any non-rotating metrics out there which can describe systems more accurately than a SC metric? I don't believe there is. The SC metric is enough to study light-bending, mercury perihelion anomaly, and other so-named classic tests of GR up to PN-order in complete agreement with observations. Sure, but these things aren't being question. All I am questioning is how a metric like this can remain effectively at the same energy level over short or long periods of times. Even after long periods of time, the SC metric becomes inaccurate in explaining your system because no doubt in many cases, the system will have given up energy in the form of radiation. My studies right now is trying to find a way round that problem - I guess it doesn't help that the Schwartzschild metric is derives from principles which require a timelike killing vector making it essentially static and t-symmetric. Edited June 14, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

elfmotat Posted June 14, 2012 Share Posted June 14, 2012 which leads me to my own question, is there any non-rotating metrics out there which can describe systems more accurately than a SC metric? I don't believe there is. You mean like this?: http://en.wikipedia.org/wiki/Vaidya_metric Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 14, 2012 Author Share Posted June 14, 2012 Thank you, never heard of it... a new subject for me to learn the next few days. One difference though is that it is not static. Link to comment Share on other sites More sharing options...

juanrga Posted June 14, 2012 Share Posted June 14, 2012 (edited) Sure, but these things aren't being question. All I am questioning is how a metric like this can remain effectively at the same energy level over short or long periods of times. Even after long periods of time, the SC metric becomes inaccurate in explaining your system because no doubt in many cases, the system will have given up energy in the form of radiation. My studies right now is trying to find a way round that problem - I guess it doesn't help that the Schwartzschild metric is derives from principles which require a timelike killing vector making it essentially static and t-symmetric. Everyone knows that Sun will disappear, finally, and then the SC metric will be not working anymore. Everyone knows that the SC metric is not valid for long distances, because it is ignoring the cosmological constant term. Everyone knows that. If one want to obtain a more general metric, e.g. a time-dependent metric valid at large scales as well, one solves the Hilbert & Einstein equations and obtain g_{ab}. Edited June 14, 2012 by juanrga Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 14, 2012 Author Share Posted June 14, 2012 (edited) Everyone knows that Sun will disappear, finally, and then the SC metric will be not working anymore. Everyone knows that the SC metric is not valid for long distances, because it is ignoring the cosmological constant term. Everyone knows that. If one want to obtain a more general metric, e.g. a time-dependent metric valid at large scales as well, one solves the Hilbert & Einstein equations and obtain g_{ab}. Saying ''just the metric'' [math]g_{\mu \nu}[/math] is a massive oversimplification, but... your approach would be consistent when you proved [math]g_{\mu \nu} \ne 0[/math]. Anyway, I don't really think you are following my problems very well -- I am well aware of the Einstein Equations and the metric. This is not what I am talking about. I was concerned strictly with the SC metric would does have inconsistencies. Elfmotat had the right suggestion for me http://en.wikipedia.org/wiki/Vaidya_metric I'm wondering now if Hawking radiation would account for loss of radiation in a non-rotating black hole, but since the Schwartzschild metric is static and t-symmetric, I am wondering if Hawking's work has been able to solve equations for a static black hole. If one cannot do it in any way, I'd dare say the Schwartzschild metric does not really purport to a real type of object. Edited June 14, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

juanrga Posted June 14, 2012 Share Posted June 14, 2012 Everyone knows that Sun will disappear, finally, and then the SC metric will be not working anymore. Everyone knows that the SC metric is not valid for long distances, because it is ignoring the cosmological constant term. Everyone knows that. If one want to obtain a more general metric, e.g. a time-dependent metric valid at large scales as well, one solves the Hilbert & Einstein equations and obtain g_{ab}. Saying ''just the metric'' [math]g_{\mu \nu}[/math] is a massive oversimplification, but... your approach would be consistent when you proved [math]g_{\mu \nu} \ne 0[/math]. Anyway, I don't really think you are following my problems very well -- I am well aware of the Einstein Equations and the metric. This is not what I am talking about. I was concerned strictly with the SC metric would does have inconsistencies. Elfmotat had the right suggestion for me http://en.wikipedia....i/Vaidya_metric I'm wondering now if Hawking radiation would account for loss of radiation in a non-rotating black hole, but since the Schwartzschild metric is static and t-symmetric, I am wondering if Hawking's work has been able to solve equations for a static black hole. If one cannot do it in any way, I'd dare say the Schwartzschild metric does not really purport to a real type of object. No. The SC metric is not inconsistent. It is a perfectly admisible solution to the Hilbert & Einstein equations under well-defined physical assumptions. And, as already said before, the SC metric has been well tested in half dozen of different tests. In fact, the SC metric is probably the best tested metric of general relativity. Your claim that "the Schwartzschild metric does not really purport to a real type of object" is unfounded. Modifications to the SC metric are known as well. Link to comment Share on other sites More sharing options...

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