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Phrak
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Classical Field Theory without Force
Has anyone seen how this has been approached?
Has anyone seen how this has been approached?
atyy said:
Q-reeus said:Is it not the case that simply replacing mx'' with p' on the lhs of mx''=q(E+ vxB) makes it relativistic?
Q-reeus said: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/
Phrak said:Classical Field Theory without Force
Has anyone seen how this has been approached?
Phrak said: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)?
Matterwave said: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...
Ben Niehoff said: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 [itex]\chi[/itex]. If one makes the metric ansatz
[tex]ds^2 = (d\chi + A)^2 + g_{\mu\nu} \; dx^\mu \; dx^\nu[/tex]
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 [itex]\chi[/itex] 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.
Matterwave said: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...
Phrak said: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.
Phrak said: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.
Mentz114 said: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.
Mentz114 said: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.
jfy4 said: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.
Phrak said: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
Phrak said: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 said:That is recognizably the generalization to curved spacetime of the classical EM Lagrangian except the sign on the AnJn. 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 said:Maybe nothing. dA^*dA - A^S can be rewritten as dA^G - A^dG, putting both terms on a somewhat more equal basis.
Sounds like a worthy aim. But you'd still get dp/dt as a dynamic variable wouldn't you ?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.
Phrak said: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 said: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.
Phrak said:dA = ∂^A. The partial derivitive 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.
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 said: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:
[tex]d(A \wedge G) = dA \wedge G - A \wedge dG[/tex]
Phrak said:The exterior derivative symbol "d" is the shorthand notation for the action of the partial derivative 1-form, [itex]\partial = \partial_\mu dx^\mu[/itex] on a q-form.
[tex](dT)_{\mu\nu_1...\nu_q} \equiv (q+1) \partial_{[\mu}T_{\nu_1...\nu_q]}[/tex]
[tex]dT_{\nu_1...\nu_q} = \partial\wedge T_{\nu_1...\nu_q}[/tex]
(U^V)^W = (U^(V^W), U^V=-V^U -associative and commutative laws.
(dA)^B = (∂^A)^B = (-A^∂)^B = -A^(∂^B) = -A^(dB)
Ben Niehoff said: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.
Phrak said:You keep saying this sort of thing, and I appreciate it greatly. Can you be more specific in what you find erronious?
What do you recommend for more basic reading?
Point 3. Could you please answer me about how you arrived at d(A^G) = dA^G - A^dG?
WannabeNewton said:Hey Phrak if you still need to know the general product rule for differential forms is: given an n - form [tex]\Psi[/tex] and an m - form [tex]\Phi [/tex]
[tex]d(\Psi \wedge \Phi) = d\Psi \wedge \Phi + (-1)^{n}\Psi \wedge d\Phi[/tex]
If you want to prove it then just work with the general equation for the wedge product of two differential forms and the general equation for the exterior derivative of a differential form.
Phrak said:Commuting the wege product in each term and negating gives
[tex]d(\Psi \wedge \Phi) = \Psi \wedge d\Phi + (-1)^{m}d\Psi \wedge \Phi \ .[/tex]
Thank's for the correction.element4 said:You cannot do that!
Classical Field Theory without Force is a branch of physics that studies the behavior of fields without the influence of external forces. It is based on the principle of least action, where the field follows a path that minimizes the action or energy of the system.
Classical Field Theory without Force describes the dynamics of continuous fields, such as electromagnetic fields or fluid flow, while Classical Mechanics deals with the motion of discrete particles. Additionally, Classical Field Theory without Force does not take into account external forces, whereas Classical Mechanics does.
Classical Field Theory without Force has many applications in physics, including describing the behavior of electromagnetic fields, fluid dynamics, and the motion of particles in a magnetic field. It is also used in the study of general relativity and quantum field theory.
The fundamental principles of Classical Field Theory without Force include the principle of least action, which states that the field follows a path that minimizes the action or energy of the system, and the principle of superposition, which states that the total field at a point is the sum of the individual fields at that point.
Current research in Classical Field Theory without Force includes the study of topological defects in field theories, the application of field theory to cosmology and astrophysics, and the development of new mathematical techniques for solving field equations.