PeterDonis said:
Yes, and the formula already derived for EM is easy to look up.
Ok, that helps to clarify where you're coming from.
for second order correction, you do the same for the gravitational field of the first order correction found earlier...
The energy we found earlier, has it self a gravity. We deduce its energy with the simple EM formula. Third order correction...
Of course, the new energy has it self a gravitational field... We add this again...
And again...
If i didn't messed up, the infinite series should converge.
If i didn't messed up, that's the full energy of the gravitomagnetic field, inside the self consistent framework of the EM-like equations.
Maybe I'm off by a scaling factor?
Were I'm wrong?
PeterDonis said:
Do you have a reference for this, or are you just assuming it happens without having actually tried to do the math?
We defined the source, as ALL the energy/momentum.
So the source is all energy/momentum, except the one of the EM-like gravitational field?
Sounds weird... Why include ALL energy, of everything, every field, true or imaginary, but the gravitational field is excluded? Weird...
This is why i say, it will be inconsistent to just omit the contribution of the EM-like field of gravity.
PeterDonis said:
But see, here's the thing: GR, the full version (i.e., *not* the weak field approximation), the Einstein Field Equation, does *not* include any "energy and momentum in the gravitational field" in the source term on its RHS, the stress-energy tensor; yet it *is* self-consistent.
I start to understand your complaint. You just say, that GR doesn't has the energy of the field as source, so why EM-like equations should? And, is this rigorously derived from GR?
The equations are of different nature.
There's that second term with the ricci scalar that isn't sure if it should be on the right or on the left of the equations. Is it really a curvature? Is it really part of the source?
Also, in the continuity equation of the tensor, you use the covariant divergence. Cristofel symbols all over the place...
Implicitly, space time curvature( gravity, in GR) is part of the source. This is why it get nonlinear.
In GR, you just need to take a covariant derivative, to take into account gravitation.
Obviously, you can't do that in flat space. When you pass to normal derivatives, you need to compensate for what you through out.
By analogy, the EM-like equations, should also have the field it self in its source, in some form.
Or they'll remain linear.
The most natural choice, is to use the energy momentum of the field. It sure has the right units, and can be plugged at the same place as the rest of the sources, and the equations don't get distorted, and become non lineair... The EM field gravitates, why the EM-like gravitational field is not aloud to gravitate?
You want to propose something else?
And no. Its not rigorously derived from GR...
As you can imagine, its a bit difficult.
Is it really important that its not derived rigorously? It's an approximation after all.
What you propose is the simplest way, to render the equations non linear?
Here, you just need to tweak the definition of the source a little...
If its complicated, the whole point of an approximation is pointless...
PeterDonis said:
Now suppose we do what I described in a previous post: we make the weak field approxmation, so that the pressure and stress terms in the SET are negligible compared to the energy and
momentum (which does *not*, as I just pointed out, include any "energy and momentum in the gravitational field"), and we contract the standard EFE with a 4-vector field describing the motion of the matter described by the SET (which, as I noted before, amounts to throwing away the information about pressure and stress). That gives us a 4-vector source ##J^a = ( \rho_g, \vec{J}_g )## on the RHS, which does *not* include any "energy and momentum in the gravitational field". Yet the equation is still self-consistent, because I took a self-consistent, covariant equation and applied a self-consistent, covariant operation (contracting with a 4-vector field) to it.
I don't know that this is actually a way to derive the GEM equations given in the Wiki page; I suspect not (because I suspect there are a bunch more approximations that need to be made first). But it certainly does give a set of self-consistent equations with a 4-vector "source" on the RHS that includes no gravitational field energy-momentum. So I'm extremely skeptical that the GEM equations somehow fail to be self-consistent if the source does not include gravitational field energy-momentum.
GR has implicitly as source, the curvature of space time.
When you did the approximation that way, you just cut off this fact.
The most simple and natural way to have the EM-like gravity, be a source of it self, is to plug in its energy. You want to propose something else?
PeterDonis said:
So what? By your logic, we can only observe a change in the system's mass by being close enough to it for a significant amount of its "gravitational field potential energy" to be above us. So we would have to be *within* the binary pulsar system itself to measure what you were talking about.
To put it another way: in order to see the mass of a system decrease significantly by your logic, we ourselves would have to be gravitationally bound within the system. But we are certainly not gravitationally bound to any of the binary pulsar systems we observe; the gravity of those systems is negligible, by many orders of magnitude, to us here on Earth. Yet we *do* detect the masses of these systems decreasing (by, as you say, measuring the change in orbital parameters which shows the systems becoming more tightly bound). So whatever the reason for the observed decrease, it can't be what you're suggesting.
What I've just said also makes clear, btw, that it doesn't take binary pulsar observations to show that the mass of a system doesn't decrease as you get closer to it because of more "gravitational field energy" being above you. If that were the case, orbits of satellites and spacecraft near the Earth, and indeed anywhere in the solar system, would be substantially changed from what we observe. All of our observations confirm that, for example, the mass of the Earth is the same whether you measure it from just above its surface or from the distance of the Moon or further; it certainly doesn't get smaller as you get closer to the Earth, which is what your reasoning would imply.
The pulsars are inside each other's gravitational fields. We can observe there orbits decaying. So, i assumed, we could detect how there mutual attractions with each other varies with distance. And from there deduce, if what i explained about was true or not. I didn't mean the gravitational pull with us on earth, i was trying to explain the phenomenon back then.
The effect is very very small. Thats why, we could in theory observe it with a binary system of pulsars orbiting each other, not with our moon...
PeterDonis said:
too many degrees of freedom to be captured in a single 4-vector. You need a tensor, the stress-energy tensor, with 10 independent components.
Pressure and stress are macroscopic constructs.
Not the smartest way to do it. but...
you take the stress energy tensors of the gabilion of particles. They all have 0 in the space space components. You add them. You get the stress energy tensor of the whole system, with 0 in the space space components. Its not the smartest way to built the tensor of the system, but, here it is, nothing in the space space components, and you truly have all the energy of the system.
That proves, you can somehow re-express the space space components into the time components. Its not necessarily trivial in practice ...
PeterDonis said:
No, the corrections needed to get the right precession for Mercury's perihelion are not "SR corrections"; they are "GR corrections". That's why the GEM model, even though it is covariant, doesn't (I believe--see below) get them right.
I said GEM + SR > for calculating the precession
You are saying ... GEM + GR, in the context of what i meant.
The hyper relativistic limit, the way i use the term...
I'm pretty sure that GEM is used just like that. They just put mercury in the equation, and that's it. They don't consider the apparent field mercury sees. Per bellow...
PeterDonis said:
I agree that this ought to be true based on the analogy with EM; but the Wiki article's expression for ##\vec{B}_g## doesn't give any help. That's why I said I wanted to see more explicit expressions for the fields. (I would also like to see a similar computation for the straight EM case of, say, an electron with ##v \rightarrow c## flying by a proton or some other positively charged object and being deflected.)
If i had found that, i wouldn't be asking nonsense questions here...
PeterDonis said:
Not really; the effect on light should be similar to the effect on a particle like an electron with ##v \rightarrow c##. (It certainly is in the full GR calculation.)
We talk about GEM, not GR.
Light going at the speed of light, should be used with SR if considered in the context of GEM .
PeterDonis said:
No, I think they just don't bother pointing out that, just like the EM equations, the GEM equations are covariant. I suspect this is because, if the GEM framework gets any practical use, it is for problems where none of the issues we've been discussing arise.
I don't think so. They usually are very literal. If they didn't, most likely, they didn't do it.
I think they just use GR instaid, not GEM/SR
PeterDonis said:
Remember I said "the Newtonian force", which does not include the magnetic component, will produce zero precession. Before coming to a conclusion about the GEM prediction, I would want, as I said above, to see more explicit expressions for the fields. I would also, for this case, want to see a computation for the EM case of, say, an electron in an elliptical orbit about a proton or other positively charged object, to see if the EM equations predict perihelion precession. If Bill_K is right that a linear theory has to predict zero precession, then the EM equations will predict zero precession since they are linear. But the presence of the magnetic field is a significant difference between the EM case and the straight Newtonian (not GEM) case, so I think it's worth having a separate computation for the EM case.
I would like to see that too.I knocked at the right door i hope...
Bill is wrong. The "magnetic" part of the GEM field will induce rotations.
I just, don't know what it looks like...