Classical Field Theory without Force

[Phrak]

Classical Field Theory without Force

Has anyone seen how this has been approached?

 

[atyy]

Isn't this just GR?

Do you mean something like http://arxiv.org/abs/0905.2391 ?

 

[Phrak]

Isn't this just GR?

Do you mean something like http://arxiv.org/abs/0905.2391 ?

No, not really. He only seems to be rehashing the Lorentz force.

mx''=q(E+ vxB) is neither a field equation nor relativistic. Forcing it to be relativistic is problematically meaningful. Further, forcing it to be generally covariant seems more unlikely.

Could there be anyone that has considered equations consisting of energy, momentum, charge density and electromagnetic fields (or the 4-potential)?

 

[Quickless]

How about this - http://www.phys.unsw.edu.au/einsteinlight/jw/module3_Maxwell.htm#light

 

[Q-reeus]

Is it not the case that simply replacing mx'' with p' on the lhs of mx''=q(E+ vxB) makes it relativistic? Anyway maybe these are about what you're after:
http://peeterjoot.wordpress.com/2011/02/08/energy-term-of-the-lorentz-force-equation/
http://peeterjoot.wordpress.com/201...rength-tensor-lorentz-field-invariants-bianc/

 

[fzero]

Is it not the case that simply replacing mx'' with p' on the lhs of mx''=q(E+ vxB) makes it relativistic?

Yes. I would refer OP to the chapter on Dynamics of Relativistic Particles in Jackson (Ch 12 in the 3rd edition). There is a similar discussion in Landau and Lifschitz, Classical Theory of Fields.

 

[Phrak]

Is it not the case that simply replacing mx'' with p' on the lhs of mx''=q(E+ vxB) makes it relativistic? Anyway maybe these are about what you're after:
http://peeterjoot.wordpress.com/2011/02/08/energy-term-of-the-lorentz-force-equation/
http://peeterjoot.wordpress.com/201...rength-tensor-lorentz-field-invariants-bianc/

Yeah, maybe that's the right direction for a start; it's still about fields acting on little charged balls. It reduces to the Lorentz force, as it should in most cases, if we are to believe the author.

As the Hilbert action implies the Einstein field equations, considering Joot's action, rendered generally covariant, might imply the proper field equations.

 

[jfy4]

Classical Field Theory without Force

Has anyone seen how this has been approached?

I actually studied this for my senior project in undergraduate. The answer seemed to be no, since acceleration for charged particles depends on the charge to mass ratio, whereas, in GR it is essentially independent of the composition of the object in question.

EDIT: It seems you asked if anyone has seen how this is approached (sorry). I saw a few articles, many attempting to formulate E&M using curved space-time as the mechanism for acceleration for the E&M field. In a sense, classical unification of E&M and Gravity.

 

[atyy]

No, not really. He only seems to be rehashing the Lorentz force.

mx''=q(E+ vxB) is neither a field equation nor relativistic. Forcing it to be relativistic is problematically meaningful. Further, forcing it to be generally covariant seems more unlikely.

Could there be anyone that has considered equations consisting of energy, momentum, charge density and electromagnetic fields (or the 4-potential)?

I believe Gralla and Wald do not assume the Lorentz force. They derive it as a good approximation under certain circumstances.

 

[Matterwave]

With E&M, there is no "equivalence principle" to work off of. Not all bodies fall the same way, like with gravity, so it would seem quite difficult to construct a field theory that didn't have any forces in it because the "curavture" must be different for each particle...

 

[jfy4]

With E&M, there is no "equivalence principle" to work off of. Not all bodies fall the same way, like with gravity, so it would seem quite difficult to construct a field theory that didn't have any forces in it because the "curavture" must be different for each particle...

Precisely. This is what I have found on the subject, however, I would be excited and extremely interested to hear other work or thoughts.

 

[Ben Niehoff]

One can formulate E&M + GR using an equivalence principle by way of the Kaluza-Klein mechanism. Just append a fifth dimension whose shape is a circle, parametrized by the coordinate $\chi$. If one makes the metric ansatz

$$ds^2 = (d\chi + A)^2 + g_{\mu\nu} \; dx^\mu \; dx^\nu$$

one finds that the 5-dimensional Einstein equations reduce to 4-dimensional Einstein equations plus the Maxwell equations, where A is the 4-vector potential. Also, the geodesic equations on the 5-manifold reduce to the Lorentz force law on the 4-manifold. The momentum conjugate to $\chi$ is a conserved quantity, and is equal to the charge-to-mass ratio (in appropriate units). Hence particles of different q/m ratio merely have different canonical momenta around the circular 5th dimension.

 

[Q-reeus]

Instead of trying to 'geometricize' EM, there is an opposite approach that seeks to turn gravity from geometric to field theory - Yuri Baryshev in particular impresses me (though I'm far from proficient to fully judge) with how many outstanding 'difficulties' in GR that may solve, eg.: http://arxiv.org/abs/0809.2323

 

[Phrak]

One can formulate E&M + GR using an equivalence principle by way of the Kaluza-Klein mechanism. Just append a fifth dimension whose shape is a circle, parametrized by the coordinate $\chi$. If one makes the metric ansatz

$$ds^2 = (d\chi + A)^2 + g_{\mu\nu} \; dx^\mu \; dx^\nu$$

one finds that the 5-dimensional Einstein equations reduce to 4-dimensional Einstein equations plus the Maxwell equations, where A is the 4-vector potential. Also, the geodesic equations on the 5-manifold reduce to the Lorentz force law on the 4-manifold. The momentum conjugate to $\chi$ is a conserved quantity, and is equal to the charge-to-mass ratio (in appropriate units). Hence particles of different q/m ratio merely have different canonical momenta around the circular 5th dimension.

As soon as you roll up a dimension with some fixed circumference, a la Klein's contribution, you have quantized the wave number, in the fifth dimension, of harmonics solutions to field equations. For example, there would be preferred frequencies of light. I spent weeks beating myself over the head trying to get rid of the quantization without success. I dunno--maybe it could be interesting with the quantization left in place. Of course, the circumference doesn't necessarily have to be presumed constant over the spacetime manifold.

And, by the way, any wave propagating in the fifth dimension becomes massive--even light. If you choose to pursue this, you just might rediscover the W and Z particles plus requisite supermassive cousins ad infinitum.

(You've got me going on this. I haven't given Kaluza-Klein much thought in a while. Say we fix the circumference to give us the rest masses of the W+ and W- paricles. However, the Z particle will always have infinite mass in this speculation. This is a problem! It cannot move in R3 space in any inertial frame. Now look at the the first harmonic to get the W+ and W- paticles. How did these helical propagations pick up a charge density? Anyway, I won't pursue this further, because this pushs the limits of speculation allowed here, so you'd be on your own.)

 

[Phrak]

With E&M, there is no "equivalence principle" to work off of. Not all bodies fall the same way, like with gravity, so it would seem quite difficult to construct a field theory that didn't have any forces in it because the "curavture" must be different for each particle...

Yes. That's the nagging question. "What replaces force, because it can't be spacetime curvature, can it?" But this question only serves to argue that it cannot be done. The question should be, "How could force be eliminated if it's not replaced with curvature?" So I think focusing on this argument is nonproductive. No offense intended, and none taken, I hope.

 

[jfy4]

Yes. That's the nagging question. "What replaces force, because it can't be spacetime curvature, can it?" But this question only serves to argue that it cannot be done. The question should be, "How could force be eliminated if it's not replaced with curvature?" So I think focusing on this argument is nonproductive. No offense intended, and none taken, I hope.

I would really like to continue this thread, however, if we are going to be discussing alternative approaches to formulating E&M in a geometric fashion, perhaps we should move the thread to a new location. I'm thinking we start a new thread in "Physics Beyond The Standard Model" and keep it going there. But I really want to continue this thread, it's one of my main passionate ideas in physics.

 

[Mentz114]

An interesting twist in GR is that an electric field can produce curvature and so affect uncharged matter. But the effect on charge is entirely through the electric field, not the curvature.

Given that one can't replace the Lorentz force with curvature - what remains ? Something that causes proper acceleration is a force by any name. In QFT the force is quantised and we get the virtual particles carrying energy and momentum but the end effect is the same.

 

[Ben Niehoff]

As soon as you roll up a dimension with some fixed circumference, a la Klein's contribution, you have quantized the wave number, in the fifth dimension, of harmonics solutions to field equations. For example, there would be preferred frequencies of light. I spent weeks beating myself over the head trying to get rid of the quantization without success. I dunno--maybe it could be interesting with the quantization left in place. Of course, the circumference doesn't necessarily have to be presumed constant over the spacetime manifold.

Yes, there is an infinite tower of massive modes around the circle. However, we may always get around that by simply assuming that all these possible particles don't actually exist (an inelegant assumption, I know, but consistent).

Remember that the momentum canonically conjugate to $\chi$ is a conserved quantity. Therefore if it starts out equal to zero, it will stay that way! So the photon causes no problems: it's $\chi$-momentum is always zero, making it both uncharged and massless.

While Kaluza-Klein may have other problems, it is certainly able to reproduce all known (classical!) E&M + GR physics. Its only problem is that it can produce extra things that we don't observe. But there is no issue with light propagation...light simply does not propagate around the circle!

 

[jfy4]

An interesting twist in GR is that an electric field can produce curvature and so affect uncharged matter. But the effect on charge is entirely through the electric field, not the curvature.

Surly an electron will follow space-time curvature. I think the case is: an electric field can cause curvature and hence accelerate uncharged matter, and space-time curvature can accelerate charged matter also. There is acceleration is both cases, for both parties, but what the mechanism is in each case is certainly different, as opposed to what establishes the mechanism.

EDIT: I suppose in the wide view of unification, it's all the same, so the scope here is a little narrow.

 

[Phrak]

Given that one can't replace the Lorentz force with curvature - what remains ? Something that causes proper acceleration is a force by any name. In QFT the force is quantised and we get the virtual particles carrying energy and momentum but the end effect is the same.

I've been reviewing the Dirac wave function of the electron, where the vector potential effects frequency and wave number--which we might rename eneregy and momentum. The action of the vector potential on charge replaces the Lorentz force, "steering" the electron field. This hints at generally covariant fields, all in terms of the metric, the 4-vector potenial and Planck's constant. Apparently this involves skew symmetric tensors in energy-momentum density

 

[Mentz114]

Surely an electron will follow space-time curvature. I think the case is: an electric field can cause curvature and hence accelerate uncharged matter, and space-time curvature can accelerate charged matter also. There is acceleration is both cases, for both parties, but what the mechanism is in each case is certainly different, as opposed to what establishes the mechanism.

EDIT: I suppose in the wide view of unification, it's all the same, so the scope here is a little narrow.

Actually I said the effect on charge is purely through the electric field, but obviously charge is always attached to some mass so an electron will of course 'feel' the curvature as well.

I've been reviewing the Dirac wave function of the electron, where the vector potential effects frequency and wave number--which we might rename energy and momentum. The action of the vector potential on charge replaces the Lorentz force, "steering" the electron field. This hints at generally covariant fields, all in terms of the metric, the 4-vector potenial and Planck's constant. Apparently this involves skew symmetric tensors in energy-momentum density

Do you mean the QED Lagrangian, where the interaction energy of the electron/positrons with the field is

$$H_Q=\int A^\mu J_\mu\ d^3x$$

I always understood this is the energy of the virtual photons. Dirac showed that Maxwell's equation follow from this interaction.

I'm not familiar with the Dirac field as a classical field, but presumably the same Lagrangian can also define a classical field.

 

[Phrak]

I'm not really sure, Mentz. But what do you think of this?

http://en.wikipedia.org/wiki/Lagrangian#Electromagnetism_in_general_relativity" and also the the following section.

 

[Mentz114]

I'm not really sure, Mentz. But what do you think of this?

http://en.wikipedia.org/wiki/Lagrangian#Electromagnetism_in_general_relativity" and also the the following section.

That is recognizably the generalization to curved spacetime of the classical EM Lagrangian except the sign on the AⁿJ_n. From this Maxwell's equation will follow on application of the least action principle.

It's much neater written with forms but doesn't change the physics.

But how is this relevant to the question about field theory without forces ?

 

[Phrak]

That is recognizably the generalization to curved spacetime of the classical EM Lagrangian except the sign on the AⁿJ_n. From this Maxwell's equation will follow on application of the least action principle.

It's much neater written with forms but doesn't change the physics.

But how is this relevant to the question about field theory without forces ?

Maybe nothing. dA^*dA - A^S can be rewritten as dA^G - A^dG, putting both terms on a somewhat more equal basis. This motivates a replacement of the poynting theorem as a means to properly formulate dynamics in terms of energy and momentum rather than force and mass.

 

[Mentz114]

Maybe nothing. dA^*dA - A^S can be rewritten as dA^G - A^dG, putting both terms on a somewhat more equal basis.

OK, I'm learning. Is this, dA^*dA - A^S the integrand in the action ? ( there's a constant factor missing compared with the Wiki page). What is G ? I think it's a 2-form ( assuming a 4D manifold) but that's all I can deduce.

This motivates a replacement of the poynting theorem as a means to properly formulate dynamics in terms of energy and momentum rather than force and mass.

Sounds like a worthy aim. But you'd still get dp/dt as a dynamic variable wouldn't you ?

 

[Ben Niehoff]

S for the current density is non-standard notation. I usually see (S for action here):

$$S = \frac{1}{4\pi} \int_{\mathcal{M}} \Big( -\frac12 F \wedge * F + A \wedge * J \Big)$$

where J is the current density 1-form and F = dA. Note that F is NOT the fundamental field, here. You don't vary the action with respect to F to get the equations of motion; you vary it with respect to A.

Maybe nothing. dA^*dA - A^S can be rewritten as dA^G - A^dG, putting both terms on a somewhat more equal basis. This motivates a replacement of the poynting theorem as a means to properly formulate dynamics in terms of energy and momentum rather than force and mass.

I assume you mean $G = * F$. There is no reason to use a separate letter for it...you will quickly run out of letters that way.

At any rate, you can't just plug the equations of motion back into the action that way. Your new action is different. For example, it doesn't contain the current density J. Therefore by varying your new action, you can never get the (important!) equation of motion

$$d * F = * J$$

 

[Phrak]

OK, I'm learning.

Me too!

Is this, dA^*dA - A^S the integrand in the action ? ( there's a constant factor missing compared with the Wiki page). What is G ? I think it's a 2-form ( assuming a 4D manifold) but that's all I can deduce.

Sorry. Ben Niehoff has explained it; G=*F. I think the 8pi factor is an artifact of forcing some particular units on J--it's an anthropamorphic scale factor.

With the substitution of the results back into the integrand (There's nothing wrong or immoral about this, but as Ben Niehoff has pointed out, you can't do the variation anymore.), I get

-dA^G + A^dG.

dA = ∂^A. The partial derivative operator is a 1-form, and commutes under the wedge like any other 1-form. You can see this by comparing the definition of the exterior derivative with the definition of the wedge product of a 1-form acting on any differential form.

dA^G = (∂^A)^G = -(A^∂)^G = -A^(∂^G)

dA^G = -A^dG.

So -dA^G + A^dG = 2A^dG.

The integral is stationary about 2A^dG.

Now this is really wierd. It says the electromagnetic energy--the sum dA^G + A^dG, rather than the difference, is everywhere zero!??! Did I make a sign error somewhere?

 

[Ben Niehoff]

dA = ∂^A. The partial derivative operator is a 1-form, and commutes under the wedge like any other 1-form. You can see this by comparing the definition of the exterior derivative with the definition of the wedge product of a 1-form acting on any differential form.

dA^G = (∂^A)^G = -(A^∂)^G = -A^(∂^G)

dA^G = -A^dG.

No, it doesn't work that way at all! You can't just go moving the d around. It's a derivative operator!

However, what you can do is write down the product rule:

$$d(A \wedge G) = dA \wedge G - A \wedge dG$$

And then integrate over all spacetime:

$$\int_\mathcal{M} d(A \wedge G) = \int_\mathcal{M} dA \wedge G - \int_\mathcal{M} A \wedge dG$$

Then if the manifold has no boundary (or if the manifold is noncompact but A and G vanish sufficiently fast at infinity), then by Stoke's theorem the left-hand side is zero, and hence

$$\int_\mathcal{M} dA \wedge G = \int_\mathcal{M} A \wedge dG$$

But this is ONLY true when the forms are integrated over all spacetime! Now since the action IS integrated over all spacetime, your result still holds, but your manipulations are quite misleading.

The integral is stationary about 2A^dG.

Now this is really wierd. It says the electromagnetic energy--the sum dA^G + A^dG, rather than the difference, is everywhere zero!??! Did I make a sign error somewhere?

You are not using the word "stationary" correctly. Remember that the fundamental field is A. And since G = *F, what you really have is (killing the factor of 2, because the original action had a 1/2 in it):

$$S = \int_\mathcal{M} A \wedge d * dA$$

If you vary this action with respect to A (and make careful use of the product rule), then you get the equation of motion

$$d * dA = 0$$

or

$$d * F = 0$$

As you can see, this just gives you Maxwell's equations without sources. This makes sense, because when you concocted this action, you removed the source term J.

Note that the action is NOT the energy, so I'm not sure how you decided to make conclusions about the energy from it. To discuss the energy, you need to look at the stress-energy tensor.

 

[Phrak]

No, it doesn't work that way at all! You can't just go moving the d around. It's a derivative operator!

The action of the exterior derivatrive on a q-form amounts to multiplication. The definition of the wege product of a 1-form acting on a q-form is

$$(V \wedge T)_{\mu\nu_1...\nu_q} \equiv (q+1) V_{[\mu}T_{\mu\nu_1...\nu_q]}$$​

The exterior derivative symbol "d" is the shorthand notation for the action of the partial derivative 1-form, $\partial = \partial_\mu dx^\mu$ on a q-form.

$$(dT)_{\mu\nu_1...\nu_q} \equiv (q+1) \partial_{[\mu}T_{\nu_1...\nu_q]}$$

$$dT_{\nu_1...\nu_q} = \partial\wedge T_{\nu_1...\nu_q}$$

(U^V)^W = (U^(V^W), U^V=-V^U -associative and commutative laws.

(dA)^B = (∂^A)^B = (-A^∂)^B = -A^(∂^B) = -A^(dB)​

A^B^...^Z^∂^ is a linear operator.

A^B^...^Z^∂^W = A^B^...^Z^(dW)​

However, what you can do is write down the product rule:

$$d(A \wedge G) = dA \wedge G - A \wedge dG$$

How is this product rule obtained?

 

[Ben Niehoff]

The exterior derivative symbol "d" is the shorthand notation for the action of the partial derivative 1-form, $\partial = \partial_\mu dx^\mu$ on a q-form.

$$(dT)_{\mu\nu_1...\nu_q} \equiv (q+1) \partial_{[\mu}T_{\nu_1...\nu_q]}$$

$$dT_{\nu_1...\nu_q} = \partial\wedge T_{\nu_1...\nu_q}$$

(U^V)^W = (U^(V^W), U^V=-V^U -associative and commutative laws.

(dA)^B = (∂^A)^B = (-A^∂)^B = -A^(∂^B) = -A^(dB)​

I'm sorry, this line of reasoning is totally wrong, and reaches a completely wrong conclusion. I'm not sure where you got this from, but it is clear you do not understand what 'd' is and how to apply it. Sounds like you should do some more basic reading.​