Q on general relativity

[Enthalpy]

 

[...] Relativistic mass does not affect the gravitational field. An energy field can also produce gravity, but again, not relativistic energy.

 

Could you (or someone else, I like you all!) tell more about this? These notions are unclear to me.

 

Up to now, I consider that light creates a gravitation field, because mass deflects light, which means some momentum change before/after for light, and conservation of momentum demands that the attracting body gets the opposite momentum, which I imagine results from attraction of mass by light, suggesting gravitation. Or doesn't it?

 

Photon energy E=p*c can also be viewed as a kinetic energy. Or is it just a matter or wording?

 

Now with massive particles: energy, for instance an electric field, creates a gravitation field. The electric field can lose energy transferred to a charged particle that gains kinetic energy. This transfer can be snappy and local enough that a remote measure sees the sum of both gravitation fields. In the case where kinetic energy creates no gravitation, would the remote measure see a quick (although faint!) drop in the gravitation field?

[swansont]

GR uses the Stress-Energy tensor to describe the effect of gravity (curvature) http://en.wikipedia.org/wiki/Stress–energy_tensor

 

It includes both energy and momentum

[Enthalpy]

Thanks!

This seems to confirm that mass, and not just rest mass, creates gravity.

Did I get it properly?

[swansont]

I don't study GR, but I get the impression that the momentum terms cancel out any kinetic effect, as they must — a star doesn't turn into a black hole simply because one observer moves very fast relative to it.

[md65536]

I don't study GR, but I get the impression that the momentum terms cancel out any kinetic effect, as they must — a star doesn't turn into a black hole simply because one observer moves very fast relative to it.

But there are absolute and apparent horizons, and the latter can be different depending on how it observed.

 

Is there any possibility of a "coordinate black hole" where one observer sees a black hole and another doesn't? (I've speculated on related junk and you didn't like it, sorry if I'm going back in that direction.) I assume that an absolute black hole would be measurable with proper time etc, so that if a particle determines that it's in a black hole, then there's not really any debate -- it is in a black hole. So if there are relative black holes, I think it would require that one observer could measure that another is in a black hole, while that other observer does not measure being in a black hole. Is there anything like that in theory, ie. "extreme relative curvature"?

[elfmotat]

I don't study GR, but I get the impression that the momentum terms cancel out any kinetic effect, as they must — a star doesn't turn into a black hole simply because one observer moves very fast relative to it.

 

Ya, momentum usually has a canceling affect to energy. I don't know whether or not you'll find it's a convincing argument, but a moving particle and a stationary particle have the same stress-energy trace [math]T[/math] (as they should, considering they only differ by a coordinate transformation). For example, consider a point particle stress-energy [math]T^{\mu \nu}=mu^\mu u^\nu \delta (x-x_0)[/math] where [math]x_0^\mu (s)[/math] is the path of the particle and [math]u^\mu=dx_0^\mu /ds[/math] is its four-velocity.

 

First we can consider a particle at rest (assuming a diagonal metric), with [math]u^\mu = (a,0,0,0)[/math]. The stress-energy trace is then:

 

[math]T=g_{00}T^{00}=g_{00} ma^2 \delta (x-x_0)[/math]

 

We can then consider a moving particle, with [math]u^\mu = (d,b,0,0)[/math]. We know that [math]a^2 g_{00} = d^2 g_{00} + b^2 g_{11}[/math], so we can rewrite the four-velocity as:

 

[math]u^\mu = (\sqrt{a^2 - b^2 \frac{g_{11}}{g_{00}}},b,0,0)[/math]

 

The SET trace is then:

 

[math]T=g_{00}T^{00}+g_{11}T^{11}=\left [g_{00} m \left ( a^2 - b^2 \frac{g_{11}}{g_{00}} \right ) +g_{11}mb^2 \right ] \delta (x-x_0) = g_{00} ma^2 \delta (x-x_0)[/math]

 

The Ricci tensor is then, by the trace-reversed field equations, compensated for in the 11-component by what was added to the 00-component. This should correspond to a metric with identical geodesic equations to the first (though I don't know if an exact solution for these metrics is possible).

[swansont]

Ya, momentum usually has a canceling affect to energy. I don't know whether or not you'll find it's a convincing argument, but a moving particle and a stationary particle have the same stress-energy trace [math]T[/math] (as they should, considering they only differ by a coordinate transformation).

 

Which is what I expected, but I didn't want to overstate the claim since I hadn't actually worked it out. Thanks for that.

[Enthalpy]

[...] the momentum terms cancel out any kinetic effect, as they must — a star doesn't turn into a black hole simply because one observer moves very fast relative to it.

If movement increases the gravity pull, then the fast moving observer will feel the star is heavier, but components of the star - which don't have this fast speed relative to an other - will feel the normal star's gravity and continue their normal life.

 

-----

 

Is it feasible to express the result in words graspable by an engineer? You seem all to consider that the gravitation force results from the rest mass only, with no contribution from the kinetic energy, right?

 

And what happens if an electric field loses energy by accelerating a charged particle, like in alpha radioactivity or in fission: do the set of resulting fragments suddenly produce a weaker gravity field?

 

And what about the kinetic energy of confined particles? Take the 1s electrons in a lead atom. Their energy is -91keV in the nucleus' potential, with -182keV electric energy and +91keV kinetic energy. The resulting atomic mass (1ppm effect) is measureable in mass spectrometers.

 

Such quick particles, for instance electrons confined by a nucleus' electric field, are more difficult to accelerate because they're already fast. Gravitation resulting only from their rest mass would let lead atoms fall slower in an external gravity field than atoms whose electrons have more room.

 

You can imagine systems (quarks and gluons!) where the kinetic energy makes much more than 1ppm of the mass. Shall the gravitational mass of the composite differ from its inertial mass, as inertial mass includes the component's kinetic energy?

Edited April 25, 2013 by Enthalpy

[swansont]

If movement increases the gravity pull, then the fast moving observer will feel the star is heavier, but components of the star - which don't have this fast speed relative to an other - will feel the normal star's gravity and continue their normal life.

 

Which is why we know it doesn't work that way. You can't have an event that occurs in one frame but not in another.

 

Is it feasible to express the result in words graspable by an engineer? You seem all to consider that the gravitation force results from the rest mass only, with no contribution from the kinetic energy, right?

 

Not from the KE of the CoM. If you add internal energy, the mass increases. That's how you balance the books. Coordinate transformations do not change anything, as elfmotat has shown.

[Enthalpy]

The example with the star unimpressed by the moving observer (which I do agree) is a bit complicated, because other forces (radiation pressure and so on) keep its size and they change with speed as well, in a difficult way.

 

Could you try with two stars, on circular orbits around their common center of mass, in a plane perpendicular to us moving observer? Their distance to an other keeps the same to us, their mass is bigger, their angular velocity smaller - and then I botched something (as expected) because the centrifugal force would decrease, so I haven't transformed the time derivative properly.

 

----------

 

In a course, when evaluating the gravitation field by a reasonable mass, Pascal Picard explicitly rejects the Ricci tensor and takes the full Einstein tensor instead. You may ask why, if you wish, but I won't answer that...

 

----------

 

I doubt that we should distinguish between individual kinetic energies of two objects and the kinetic energy of their common center of mass. After all, both add up to give the individual kinetic energy versus us, with which we should be able to make computations.

 

----------

 

Does someone want to claim that a spinning object, whose inertial mass increases due to speed, still feels the same gravitation force because gravitation results from rest mass only?

- To my eyes, it would allow to distinguish between gravitation and acceleration.

- The experiment is easy! Take a toroidal gyroscope of graphite fiber running at 600m/s, its inertial mass increases by 2*10^-12. Put it within enough nested boxes that don't rotate quickly, send everything to low-Earth-orbit. After only one month, the gyroscope will float 20mm behind its original position in the boxes. I bet many experiments (like the recent spacecraft that observed frame-dragging) are more sensitive than that.

- Worse, the kinetic energy of electrons around heavy nuclei. This kinetic energy makes about 2ppm of lead's mass, but only 0.1ppm of carbon's mass. Since our accelerometers are 2*10^-9 accurate, lead falling slower than feathers would be known.

- In a proton or neutron, how much could kinetic energy weigh? Half of the total mass? That would make a difference to other particles!

 

So I continue to believe that any energy, including kinetic energy, is also inertial mass, and that inertial mass is passive gravitation mass, which is active gravitation mass as well. To evaluate GMm/R², where I'm bound to Earth's M hence won't scale G nor R, I take for m the mass, not just the rest mass.

[elfmotat]

The example with the star unimpressed by the moving observer (which I do agree) is a bit complicated, because other forces (radiation pressure and so on) keep its size and they change with speed as well, in a difficult way...

 

My post above assumed a diagonal metric, which in general isn't true of rotating coordinate systems. The metric of a rotating frame will generally have terms like [math]dt d\theta[/math]. If you set a sphere into rotation then you will get a Kerr spacetime, which of course is not equivalent to Schwarzschild spacetime. You get new effects like frame-dragging and such.

 

The example you asked me to do is far too complicated. The Einstein Field Equations can only be solved analytically in the most simple cases. Advanced numerical techniques are required for more complicated situations, like the one you describe.

[Markus Hanke]

In a course, when evaluating the gravitation field by a reasonable mass, Pascal Picard explicitly rejects the Ricci tensor and takes the full Einstein tensor instead. You may ask why, if you wish, but I won't answer that...

 

In four dimensions the Ricci tensor does not fully determine Riemann curvature, so

 

[math]\displaystyle{R_{\mu \nu }=0}[/math]

 

does not necessarily mean that we are deal with a globally flat space-time. All it means that we are considering a vaccum, and that space-time can locally be considered to be Minkowskian. To determine the global geometry one would need additional constraints on the system, as for example the Schwarzschild metric does by demanding that it vanishes at infinity, and that it reduces to Newtonian gravity for weak fields.

[elfmotat]

In four dimensions the Ricci tensor does not fully determine Riemann curvature, so

 

[math]\displaystyle{R_{\mu \nu }=0}[/math]

 

does not necessarily mean that we are deal with a globally flat space-time. All it means that we are considering a vaccum, and that space-time can locally be considered to be Minkowskian. To determine the global geometry one would need additional constraints on the system, as for example the Schwarzschild metric does by demanding that it vanishes at infinity, and that it reduces to Newtonian gravity for weak fields.

 

Assuming there are no singularities, spacetime can always be considered locally Minkowskian, not just in a vacuum.

[Markus Hanke]

Assuming there are no singularities, spacetime can always be considered locally Minkowskian, not just in a vacuum.

 

That's true, but if you are not in a vacuum then [math]\displaystyle{T_{\mu \nu }\neq 0}[/math], and the Ricci tensor would not vanish.

All I was trying to say is that a vanishing Ricci tensor does not automatically imply a flat space-time; it is a necessary, but not a sufficient condition since, as mentioned earlier, the Riemann curvature tensor is not completely determined by its contractions in 4 dimensions.

[Enthalpy]

That kinetic energy creates gravity is observed routinely, not needing to experiment purposely - if I interpret it properly despite understanding so little of it.

 

The kinetic energy of electrons makes around 0.28ppm of aluminium's mass, but only 0.12ppm of nitrogen's one (carbon and oxygen about the same) and about 0.015ppm for hydrogen. The kinetic energy of nucleons must make a bigger difference, and of quarks in protons versus neutrons even more, but let's take just the one from electrons.

 

Satellites are widely made of aluminium, while their liquid propellants comprise generally the other mentioned atoms. Let's imagine briefly that kinetic energy increases the inertial mass hence the centrifugal force, but not the gravitation force: in low-Earth-orbit at 6670km from Earth's center, the orbits with the same angular speed would differ by 0.4m altitude (0.05ppm) for the propellants and their tanks - but for decades we know propellants float freely within the tanks.

 

The kinetic energy fraction would be bigger with iron, copper, lead... than aluminium. Leave cautiously a paperclip floating in the space station, observe 3/4h later (half an orbit) that it's still at the same place, and you've made an experiment of general relativity.

Edited May 4, 2013 by Enthalpy