Gravitomagnetism: Why isn't mass-current covariant?

[atikof]

"Gravitomagnetism is a widely used term referring specifically to the kinetic effects of gravity, in analogy to the magnetic effects of moving electric charge".

http://en.wikipedia.org/wiki/Gravitoelectromagnetism

Essentially the idea is to take Maxwell's equations and make a substitution of charge with mass, and of current with moving mass. The equations read then:

$\vec{\nabla} \cdot \vec{E_g} = - 4 \pi G \rho_g$
$\vec{\nabla} \cdot \vec{B_g} = 0$
$\vec{\nabla} \times \vec{E_g} = -\partial_t B_g$
$\vec{\nabla} \times \vec{B_g} = -\frac{4\pi G}{c^2} \vec{j_g} - \frac{1}{c^2} \partial_t \vec{E_g}$

(You could also think of a scalar potential $\phi_g$ and a vector potential $\vec{A_g}$ to describe the dynamics, just like in electromagnetism)

However, Wikipedia says that you can't have a full analogy, because the sources $\rho_g$ and $\vec{j_g}$ don't form a 4-vector, but "instead they are merely a part of the stress–energy tensor".

Why? The only difference I can think of between charges and masses is that mass is always positive (and gravity is always attractive because of the minus signs in the equations). Is that reason enough to make the combination $(\rho_g,\vec{j_g})$ not transform as a 4-vector and have to invoke the stress-energy tensor?

For instance, let's have a massive charged particle (its rest mass will be $m_0$ and its charge $q$). If its 4-velocity is $u^\mu$, then the 4-current (to which the electromagnetic field couples) should be $j^\mu=q u^\mu$. And its 4-momentum (to which gravity couples) $p^\mu = m_0 u^\mu$. But those are both 4-vectors! Where's the problem then? What's the precise relationship between the 4-vector $p^\mu=(E,\vec{p})$ and the (not a 4-vector) $j^\mu_g = (\rho_g, \vec{j_g})$?

 

[PeterDonis]

However, Wikipedia says that you can't have a full analogy, because the sources $\rho_g$ and $\vec{j_g}$ don't form a 4-vector, but "instead they are merely a part of the stress–energy tensor".

Why?

A 4-vector has only 4 independent components. A stress-energy tensor has 10 independent components. There's no way to reduce 10 independent numbers to 4.

For instance, let's have a massive charged particle (its rest mass will be m0 and its charge q). If its 4-velocity is uμ, then the 4-current (to which the electromagnetic field couples) should be jμ=quμ. And its 4-momentum (to which gravity couples) pμ=m0uμ.

But this is not a complete model of matter. In this model, matter has only energy and momentum; but a complete model of matter has to include internal stresses as well: pressure, shear stress, etc. That gives an additional six independent numbers (in the general case) beyond the 4 you are capturing here, and all of those additional components can affect the "gravity" that is observed.

 

[bcrowell]

There are probably a lot of different ways of approaching this question.

One way that the analogy between gravity and E&M fails is that the equivalence principle holds for gravity. This leads to a geometrical picture in which gravitational fields are described in terms of curvature. A tensor measuring curvature always has to have an even number of indices, so there is no way that a vector, which only has one index, can act as a source of curvature.

 

[pervect]

Why? The only difference I can think of between charges and masses is that mass is always positive (and gravity is always attractive because of the minus signs in the equations). Is that reason enough to make the combination $(\rho_g,\vec{j_g})$ not transform as a 4-vector and have to invoke the stress-energy tensor?

The total energy momentum 4-vector of a system of particles is a 4-vector with an invariant mass m = E^2-p^2 only if the system is closed.

If the system is open (for instance, if you have a box of light, and you ignore the walls), the total energy-momentum isn't a 4-vector.

In such cases, you don't have a well defined "mass" for the system. But the stress-energy tensor is well defined.

One paper that mentions the fact is http://arxiv.org/abs/physics/0505004

It is known that the volume of an object viewed from distinct inertial frames
are different physical entities that are not connected each other by a Lorentz
transformation. Consequently total energy-momentum of an object in one
frame is not connected to that in another frame, i.e., energy-momentum of
an object with a finite volume is not a covariant entity.
Gamba [6] wrote a paper about the confusion caused by this fact; he states
“... physicists have made the same mistake. The examples are so numerous
that to review them all one should have to write a book, not an article.”

Gamba (A. Gamba, Amer. J. Phys., 35 (1967), 83.) http://adsabs.harvard.edu/abs/1967AmJPh..35...83G might be a better reference if you aren't already familiar with this fact - unfortunately, it's not (as far as I know) publically available.

 

[Bill_K]

atikof, You can't just write down Maxwell's Equations, put a subscript "g" on everything, and expect that to represent gravity! If anything you have to go the other way -- start with the equations of General Relativity, make some approximations (slow motion, weak field, etc) and derive something that looks vaguely like Maxwell.

The usual approach is to split the Riemann tensor into space and time parts: E_ij ≡ R⁰_ij0, B_ij ≡ R⁰_imnε_jmn. Then you can write the linearized Bianchi Identities to look like Maxwell's vacuum equations except that E, B are rank two symmetric tensors.

Here, it looks like what they are doing instead is to start with the geodesic equations, d²x^μ/dτ² + Γ^μ_νσ dx^ν/dτ dx^σ/dτ = 0. Approximate dx⁰/dτ = 1, dxⁱ/dτ = vⁱ, and you are led to a "Lorentz force" equation in which Γⁱ₀₀ = Eⁱ, Γⁱ_j0ε_ijk = B_k.

To find the field equations obeyed by these definitions of "E" and "B", approximate the Einstein equations. Discard quadratic terms in the Einstein tensor and get (approximately!) Γⁱ_00,i = T₀₀, that is, ∇·E = ρ. And of course we see that ρ = T₀₀ is in fact not part of a 4-vector, it's part of the stress-energy tensor.

 

[bcrowell]

atikof, You can't just write down Maxwell's Equations, put a subscript "g" on everything, and expect that to represent gravity!

But there is nothing a priori obviously wrong with that, if you're starting from scratch. Gravity and electrical forces are both 1/r^2 forces. The OP is asking why this prescription, which on the surface seems as though it ought to work, doesn't work. Saying that one needs GR instead is not an answer. The OP is asking why you need GR in the first place.

 

[Bill_K]

Sorry, there is everything wrong with that! If the assignment is, "Take Newton's theory of gravitation and make it relativistic", then this approach is wrong on the face of it. It neglects what we do know about gravity (principle of equivalence) and adds something that has never been observed (gravitomagnetism.)

And immediately one encounters a fatal problem - even in Newtonian physics both energy and momentum are conserved, whereas in electromagnetism only one quantity, the charge, is conserved. If the source of gravity (like the source of electromagnetism) is a vector density, you can't possibly conserve all four components. For example, dipole radiation would be possible.

 

[atikof]

But there is nothing a priori obviously wrong with that, if you're starting from scratch. Gravity and electrical forces are both 1/r^2 forces. The OP is asking why this prescription, which on the surface seems as though it ought to work, doesn't work. Saying that one needs GR instead is not an answer. The OP is asking why you need GR in the first place.

Thank you, that's exactly my point! I'm not asking if this approach to gravitomagnetism is right for describing gravity (of course GR is better and more beautiful), but wether it's fully consistent (i.e. Lorentz covariant) or not.

The total energy momentum 4-vector of a system of particles is a 4-vector with an invariant mass m = E^2-p^2 only if the system is closed.

If the system is open (for instance, if you have a box of light, and you ignore the walls), the total energy-momentum isn't a 4-vector.

In such cases, you don't have a well defined "mass" for the system. But the stress-energy tensor is well defined.

I didn't know that! Could you please explain it a bit more or give me a link? I've still plenty of questions regarding the differences between charge and mass. For instance:

-How is the 4-current $j^\nu$ built from the charge $Q$? If you have a single particle I understand that it would be $j^\nu=qu^\nu$ (or wouldn't?). But if you have a set of charges inside a volume, would it be simply $J^\nu = \Sigma j^\nu= \Sigma qu^\nu$, or something more complex? (perhaps that sum isn't well-defined?)

-The same for mass. Here I know that the 4-momentum of a particle is $p^\mu=m_0 u^\mu$. Why isn't in general $\Sigma p^\mu$ a 4-vector?

 

[bcrowell]

Sorry, there is everything wrong with that! If the assignment is, "Take Newton's theory of gravitation and make it relativistic", then this approach is wrong on the face of it. It neglects what we do know about gravity (principle of equivalence) and adds something that has never been observed (gravitomagnetism.)

And immediately one encounters a fatal problem - even in Newtonian physics both energy and momentum are conserved, whereas in electromagnetism only one quantity, the charge, is conserved. If the source of gravity (like the source of electromagnetism) is a vector density, you can't possibly conserve all four components. For example, dipole radiation would be possible.

It's not true that gravitomagnetism has never been observed. Gravity Probe B has confirmed effects that are predicted correctly by the gravitomagnetic approximation.

Re "It neglects what we do know about gravity (principle of equivalence)," yes, this is the kind of thing that the OP was asking to have pointed out.

 

[Bill_K]

The Wikipedia article on Gravity Probe B, and the NYT article it refers to, say that the experiment has confirmed the deSitter effect to within one percent or less, but the much smaller frame dragging due to rotation of the Earth only to within about 15 percent. The latter would be a gravitomagnetic effect. I guess this counts as a detection of the effect, but not yet a confirmation of GR.

 

[TrickyDicky]

Bill is right, GPB was not conclusive for frame dragging, it only confirmed the de Sitter effect.

 

[pervect]

Thank you, that's exactly my point! I'm not asking if this approach to gravitomagnetism is right for describing gravity (of course GR is better and more beautiful), but wether it's fully consistent (i.e. Lorentz covariant) or not.

You'll find a lot of textbooks that tell you that the stress-energy tensor is the correct way to model the contributions of matter - and none (that I'm aware of) that talk about a mass current.

As far as references go, I've already given a few. You might want to re-read the original post to make sure you spotted them. I'll give a few more scattered ones that are online, but they won't necessarily address your qeustion head on.

Textbooks will tell you that the stress energy tensor is the correct approach if you pick one more or less at random. This leaves you with two possibilities - 1) that your approach is somehow equivalent to the standard approach or 2) that your approach is not equivalent, in which case it's up to you to understand both approaches well enough to make predictions from them and then go to experiment to decide which one agrees.

As far as I know your approach falls in category 2), i.e. it' makes different predictions, but I suppose I haven't actually worked that out in great detail. Mostly I say this because the stress-energy tensor has more degrees of freedom. Where I'd start looking for specific predictions is that, the standard approach predicts that pressure can influence gravity, and I don't think there's any room for that in your approach.

For a reference for this point (that pressure can cause gravity) that might also help you understand how the GR works, see http://math.ucr.edu/home/baez/einstein/, "The Meaning of Einstein's equation".

There's one more textbook resource that might be helpful - Rindler does a short calculation that shows how pressure (on a non-closed system) affects it mass. I believe this is in "Relativity: Special, General , and Cosmological" https://www.amazon.com/dp/0198567324/?tag=pfamazon01-20. As far as I recall (I glanced at it once I don't own the box) he does this by assuming the stress-energy tensor is correct, however.
If you're not aware that the mass of an open system can be affected by the pressure, it might be worth reading Rindler.

So in short - textbooks currently do a good job of telling you that the stress-energy tensor approach is the right one, I'm not sure if they do as good a job of motivating it as they do telling it. I suspect that it's one of those things you have to accept, and maybe revist later. It'd be interesting to know more about the history, as far as I know Einstein already knew that the Stress energy tensor was the right thing to use, but it's not clear how he knew.

 

[atikof]

You'll find a lot of textbooks that tell you that the stress-energy tensor is the correct way to model the contributions of matter - and none (that I'm aware of) that talk about a mass current.

Okay, I'll try to understand more deeply what the stress-energy tensor means (I only understand it from a variational principle, taking the variation of your lagrangian w.r.t the inverse metric, which doesn't give me intuitive insight)

However, I'd like to insist on my previous questions, which remain unanswered:

-How is the 4-current $j^\nu$ built from the charge $Q$? If you have a single particle I understand that it would be $j^\nu=qu^\nu$ (or wouldn't?). But if you have a set of charges inside a volume, would it be simply $J^\nu = \Sigma j^\nu= \Sigma qu^\nu$, or something more complex? (perhaps that sum isn't well-defined?)

-The same for mass. Here I know that the 4-momentum of a particle is $p^\mu=m_0 u^\mu$. Why isn't in general $\Sigma p^\mu$ a 4-vector?

 

[pervect]

In general, there's a 4-vector associated with a particle density. Try looking up "particle density 4-vector" or perhaps "Number Flux 4-vector".

You should be able to find similar discussions for the charge-current 4-vector, it's simply a matter of replacing "number' with "charge".

I'll try and provide a quick description, then some references.

Quick description:

The density of particles (or charges) depends on the reference frame you use. If you imagine that the particles are all co-moving, you can see that the density is a minimum in the rest frame of the particles. In other non-rest frames the density scales by a factor of gamma = 1/sqrt(1-v^2/c^2) due to Lorentz contraction.

If we consider that the velocity v has components v_x,. v_y, and v_z, such that v = sqrt(v_x^2 + v_y^2 + v^z^2) we can see that

(rho*gamma, v_x * rho*gamma, v_y*rho*gamma, v_z*rho*gamma)

a) represents density and current in that frame
b) transforms as a 4-vector

References:

http://books.google.com/books?id=UMDurS6HSl4C&pg=PA368&lpg=PA368&dq#v=onepage&q&f=false has a good brief discussion (I'm not sure how persistent google books results are).

They point out the interesting fact that the stress-energy tensor can be considered as the tensor product of the number density 4-vector and the energy-momentum 4-vector.

http://web.mit.edu/edbert/GR/gr2b.pdf also looks reasonable, though perhaps not as clear as I'd like.

For textbooks, try MTW's gravitation, pg 138, if you have it.

For the second part of your invention, "mass current 4-vector" isn't well defined (as being able to be looked up in a textbook) as far as I know. I suppose you could try constructing one from the invariant mass and the particle density 4-vector, if you explained what you were doing (using he relativisitic mass won't give you a tensor I think) - but pretty much everyone else uses the tensor product of the number 4-vector and the energy-momentum 4-vector known as the "stress energy tensor".

I'd really advise not getting too tied up in the "mass current" approach - I think it can be made to work in the weak-field limit (though I'm not sure of the details). But for the general strong field limit you probalby need the full stress-energy tensor.