How Would Gravity Behave with a Four-Component Mass Current as Its Source?

[jostpuur]

What would be wrong with a theory of gravity, which would be analogous to the electromagnetism, and would have a four component mass current as the source? The interaction can be made attractive for like charges with a substitution

$$(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu}) \mapsto -(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu})$$

in the Lagrangian.

 

[genneth]

Do you mean that it would have the same form for the Lagrangian $$F \wedge F$$ (with some constants) where $$F = \mathrm{d}A$$? I'd imagine that it doesn't bear experiments out. For one thing, it would be a linear theory, which I'm fairly sure we can distinguish with just solar system experiments. It is possible to cast GR as a gauge theory, but the Lagrangian isn't exactly that. The original line of research behind loop quantum gravity was to find a set of variables in which looked more like gauge theory, with the hope that quantising it could be analogous.

See http://en.wikipedia.org/wiki/Ashtekar_variables

From a purely aesthetic point of view, your proposed theory would still presume a fixed background. Depending on your precise view, that may or may not be a bad thing.

 

[jostpuur]

I was thinking about a field theory in the special relativity. I should have mentioned it.

 

[genneth]

Like I said, it would probably not match with experiments. There would not be much problem with self-consistency, but measurable deviations from GR tends to invalidate a theory, even if you consider GR to only be an approximation -- it's approximately perfectly accurate in experiments that we can currently do.

 

[jostpuur]

The cursed experiments! Always ruining everything.

 

[rbj]

What would be wrong with a theory of gravity, which would be analogous to the electromagnetism, and would have a four component mass current as the source? The interaction can be made attractive for like charges with a substitution

$$(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu}) \mapsto -(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu})$$

in the Lagrangian.

i think that there already is such a theory. it's called Gravitoelectromagnetism (GEM) and you can wikipedia it. i remember someone named Mashoon is one of the primary authors about it in the lit. they basically show that, in the case where spacetime is sufficiently flat, that an analog to Maxwell's equations can be derived from the Einstein field equation. these are called the GEM equations, i believe, and they look just like Maxwells with mass density replacing charge density (current, in both cases, is density times the vector velocity) and $-G \leftarrow 1/(4 \pi \epsilon_0)$ (the minus sign because instead of charges of identical polarity repelling each other, like-signed mass charges attract).

 

[Mentz114]

jostpuur, have you seen Doug Sweetsers theory in the ind. research section ?

https://www.physicsforums.com/showthread.php?t=87097

 

[jostpuur]

jostpuur, have you seen Doug Sweetsers theory in the ind. research section ?

https://www.physicsforums.com/showthread.php?t=87097

I remember taking a glance at it, but it looked too frightful for me. I don't know GR very much.

 

[Mentz114]

Yes, it is a strange mixture of EM and gravity theory.

He does note that

$$\partial_{\mu}A_{\nu}+\partial_{\nu}A_{\mu}$$

is a symmetric tensor, and so could represent the force tensor of an attractive spin-2 field, from which a Lagrangian may be derived, as in EM theory.

[edit]
For a symmetric tensor
$$\epsilon _{mnij}F^{mn}F^{ij} = 0$$

so no Lagrangian.

 

[OOO]

What would be wrong with a theory of gravity, which would be analogous to the electromagnetism, and would have a four component mass current as the source? The interaction can be made attractive for like charges with a substitution

$$(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu}) \mapsto -(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu})$$

in the Lagrangian.

The reason is, that mass density is the 00-component of a tensor field and not the 0-component of a vector field. See Weinberg, S. - Gravitation and cosmology, chapter 7.1.

 

[jostpuur]

The reason is, that mass density is the 00-component of a tensor field and not the 0-component of a vector field. See Weinberg, S. - Gravitation and cosmology, chapter 7.1.

I don't know how to precisely calculate transformations of mass currents, but I guessed it would be a four current by an analogy to dust. Shouldn't the mass be quite same thing as number of particles in a dust?

 

[pervect]

I don't know how to precisely calculate transformations of mass currents, but I guessed it would be a four current by an analogy to dust. Shouldn't the mass be quite same thing as number of particles in a dust?

Dust doesn't have internal stresses.

If you consider a moving stressed bar, it does not have the same momentum as an unstressed bar moving at the same rate. Note that the moving stressed bar is not an isolated system - you can think of it as part of a bigger system for which you want to find the energy and momentum in "that piece" of the bigger system.

IIRC this is discussed in Rindler's book "Intro. to SR". Furthermore, the relationship between momentum and velocity for such a stressed bar is not even isotropic, i.e. the ratio between velocity and momentum in the direction of motion is different than the ratio between velocity and momentum in the transverse direction.

I'm not quite sure how to demonstrate this without using what we are trying to demonstrate, i.e. that the stress-energy tensor is the correct way to describe energy and momentum. (I don't have / haven't seen OOO's reference).

I think I have an old post that illustrates that if you ignore the container walls and consider only a box with particles bouncing back and forth inside the box, that the mass of the particles inside the box is not covariant, i.e. I compute E^2 - p^2 of the particles (ignoring the walls) in various frames of reference and show that they differ.

But this is something I've worked out, I haven't seen this done in a textbook. (You can find in several sources though that the stress-energy tensor is the correct way to describe energy and momentum, but I don't think anyone else has motivated it by considering the swarm of particles in a box scenario).

The link in question where I worked out the "swarm of particles in a box" is https://www.physicsforums.com/showthread.php?t=117773, I seem to recall there was at least one typo I found when I revisited the thread, (nothing major, I think I transopsed a pair of formula). I'd correct it now that I can edit such old posts, if I could find it again.

 

[jostpuur]

The reason is, that mass density is the 00-component of a tensor field and not the 0-component of a vector field. See Weinberg, S. - Gravitation and cosmology, chapter 7.1.

In fact this a very strange claim. How could energy density and mass density transform similarly, although they are so different? OOO, are you sure you didn't confuse energy with mass. I am following the popular convention, that the mass is the rest mass here.

 

[pervect]

In fact this a very strange claim. How could energy density and mass density transform similarly, although they are so different? OOO, are you sure you didn't confuse energy with mass. I am following the popular convention, that the mass is the rest mass here.

You can really only think of T^00 (which I assume is what OOO had in mind) as a mass density in the center of momentum frame. But in that frame, E=mc^2, or in geometric units, E=m, so energy and mass are more or less interchangable.

 

[jostpuur]

I know that 2nd rank field can be coupled to the energy-momentum tensor, and this is how the usual linear approximation of the gravity works. I then asked, that could we alternatively take 4-component field and couple it to the 4-component mass current.

The answer "no, because you must couple it to the 16-component energy-momentum tensor", is not really an answer.

The mass density does not transform as T^00, so there does not yet seem to be any obvious contradiction in coupling 4-component field to a 4-component mass current.

 

[pervect]

If you don't mean T^00 by "mass density", what do you mean? And how do you think it transforms?

Let me be a bit more specific, so I can be sure we're on the same track. Suppose you have a small piece of a star, a star that's made from an ideal fluid to keep things simple.

The fluid has a density rho, and a pressure P in the rest frame of the fluid. The piece of the star is in the center, so it's not isolated. What do you think the "mass current" associated with this small piece of star is in the rest frame of the fluid?

 

[jostpuur]

If you don't mean T^00 by "mass density", what do you mean? And how do you think it transforms?

Fix $x\in\mathbb{R}^3$, and r>0, and the mass inside a ball B(x,r) is

$$M_r = \sqrt{E^2-|p|^2} = \Big(\big(\underset{B(x,r)}{\int} d^3y\; T^{i0}(y)\big)\big(\underset{B(x,r)}{\int} d^3y\; T_{i0}(y)\big)\Big)^{1/2}$$

The mass density at this point is

$$\rho(x) = \lim_{r\to 0} \frac{3 M_r}{4\pi r^3}$$

I don't know for sure how this transforms, but I guessed it would be a first component of a four vector by analogy to dust, and how mass is a same kind of quantity as the number of particles in a dust is.

 

[timur]

Isn't it a vector theory you are talking about? I remember reading in Straumann that any vector theory gives repulsive force for similar charges.

 

[OOO]

I am following the popular convention, that the mass is the rest mass here.

Using rest mass is not a popular convention in relativity, at least when it comes to conservation of energy-momentum. Total mass is defined as

$$Mc^2=\int T^{00}dV$$

the reason of which being that this quantity is conserved (and I guess you want energy to be conserved).

Anyway we talked about mass density and not mass itself. But then there's at least one additional distinction: you'd have to tell us whether you mean "rest mass density" or "rest mass rest density". Unless you tell us, we can't answer.

As you can read in the textbooks the energy-momentum tensor of non interacting matter (often called dust) is

$$T^{\mu\nu} = \rho v^\mu v^\nu$$

One can also rephrase this in terms of the "rest mass rest density" which I shall write as $$\rho_0$$ in the form

$$T^{\mu\nu} = \rho_0 U^\mu U^\nu$$

But the former is more useful for our consideration because we see from it that $$T^{00}=\rho c^2$$ and consequently

$$M=\int \rho dV$$

as we might have expected. So T00 is indeed the mass density in common parlance. The "rest mass density" or either the "rest mass rest density" is of no immediate physical relevance here because it is not a conserved quantity. Even M is only conserved if there is no interaction.

However it is perfectly legal to try to design one's own theory by considering one of the rest densities (which one you haven't yet explained). But as I assume from your posting you change sign in the "Maxwell tensor" so there's not a chance for the corresponding current to be conserved (remember: current conservation comes from antisymmetry of the field tensor). This is of course good because it shouldn't be conserved, as explained above.

Again, don't complain about us confusing things when you don't precisely tell us what you mean. And remember that there is not enough space here to explain GR to you.

 

[jostpuur]

Using rest mass is not a popular convention in relativity, at least when it comes to conservation of energy-momentum. Total mass is defined as

$$Mc^2=\int T^{00}dV$$

the reason of which being that this quantity is conserved (and I guess you want energy to be conserved).

I'm surprised. This sounds like you are saying that the relativistic mass being the mass would be the popular convention. I thought this camp is the minority.

However it is perfectly legal to try to design one's own theory by considering one of the rest densities (which one you haven't yet explained).

I haven't put much effort into developing such theory, although that was the idea behind all this. I'm merely asking if there are already known issues related to such attempt.

In fact some days earlier I was asking if EM-field could be 2nd rank tensor field, in the QM forum.

Again, don't complain about us confusing things when you don't precisely tell us what you mean. And remember that there is not enough space here to explain GR to you.

No need to get on the war path. I explained what I meant with the mass density once it became clear that I hadn't made myself clear.

 

[jostpuur]

Isn't it a vector theory you are talking about? I remember reading in Straumann that any vector theory gives repulsive force for similar charges.

I remember hearing this as a rumour, but wasn't sure. I don't understand what's the argument behind this claim. Suppose we start with a Lagrange's function

$$L = n \int d^3x' (\partial_{\mu} A_{\nu}(x'))(\partial^{\mu} A^{\nu}(x')) - \sum_{k=1}^N \Big(q_k A^0 (x_k) - q_k \boldsymbol{A}(x_k)\cdot\boldsymbol{v}_k + m\sqrt{1-|v_k|^2}\Big)$$

where n=1 or n=-1. So the interaction term in the Lagrange's density has been $-A_{\mu} j^{\mu}$, and the point charges

$$j^{\mu}(x') = \sum_{k=1}^N q_k \delta^3(x'-x_k)(1,\boldsymbol{v}_k)^{\mu}$$

have been substituted. If I solve the equations of motion for this system, starting with the Hamilton's principle, they turn out to be

$$\partial_{\mu}\partial^{\mu} A^{\nu}(x') = -n\sum_{k=1}^N q_k \delta^3(x'-x_k)(1,\boldsymbol{v}_k)^{\nu}$$

$$\boldsymbol{F}_k = -q_k (\nabla A^0(x_k) + \partial_0\boldsymbol{A}(x_k)) + q_k\boldsymbol{v}\times(\nabla\times\boldsymbol{A}(x_k))$$

From here we can solve that with choice n=1 we get a theory where like charges attract each others, and with n=-1 they repel each others. If you put charges in scalar field, or 2nd-rank field, it still goes the same way. The signs in the Lagrange's function determine whether the like charges will attract or repel.

 

[OOO]

I'm surprised. This sounds like you are saying that the relativistic mass being the mass would be the popular convention. I thought this camp is the minority.

As long as you talk about discrete particles the rest mass is of course the preferred quantity. But if you consider continuous quantities then you won't get around conservation laws and this is why mass density as the T00-component is more important in this context.

I haven't put much effort into developing such theory, although that was the idea behind all this. I'm merely asking if there are already known issues related to such attempt.

As I have indicated I don't want to discourage you about developing your own theories. But it is definitely easier to sort out alternatives when you master the established theories first.

No need to get on the war path. I explained what I meant with the mass density once it became clear that I hadn't made myself clear.

Peace .

 

[rbj]

What would be wrong with a theory of gravity, which would be analogous to the electromagnetism, and would have a four component mass current as the source? The interaction can be made attractive for like charges with a substitution

$$(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu}) \mapsto -(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu})$$

in the Lagrangian.

i think that there already is such a theory. it's called Gravitoelectromagnetism (GEM) and you can wikipedia it. i remember someone named Mashoon is one of the primary authors about it in the lit. they basically show that, in the case where spacetime is sufficiently flat, that an analog to Maxwell's equations can be derived from the Einstein field equation. these are called the GEM equations, i believe, and they look just like Maxwells with mass density replacing charge density (current, in both cases, is density times the vector velocity) and $-G \leftarrow 1/(4 \pi \epsilon_0)$ (the minus sign because instead of charges of identical polarity repelling each other, like-signed mass charges attract).

a regular PF poster here (that i have a lot of respect for) has asked me to correct my "mistaken claim". i don't know what is mistaken (i know about the factor of 2 in the Lorentz force equation, but i wasn't referring specifically to it) but i acknowledge that this answer may not have been applicable to what the OP is bringing up. as far as i know, i didn't misrepresent what Mashoon et.al. did and the result of it.

so, i'll just say "i dunno" and leave it at that.

 

[pervect]

I don't know for sure how this transforms, but I guessed it would be a first component of a four vector by analogy to dust, and how mass is a same kind of quantity as the number of particles in a dust is.

Well, I think your first step should be to demonstrate that it is a 4-vector. I'm very skeptical that it is, for several reasons, which I've already mentioned. I'll add one more to the list: see http://arxiv.org/abs/physics/0505004