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OverUnity Research > Benches > Smudge (Moderator: Smudge) > Hidden momentum
2024-11-26, 16:38:58
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Author Topic: Hidden momentum  (Read 1662 times)
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img  Smudge
Hidden momentum « on: 2023-07-31, 16:24:37 »

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If you Google "hidden momentum" you will find that a charge Q that is within a vector magnetic field A has an electromagnetic momentum Q*A that is not associated with its mass times velocity, hence it is hidden.  Thus an electron of charge -e would have a hidden momentum -eA.  Force is given by a negative rate-of-change of momentum, so if A is changing with time we get a force F=-Q*dA/dt.  If we assume that this force is from an electric field E then we must assume that this electric field is created by the change in A, thus E=-dA/dt.  There is much debate about the change in A coming from movement through a static non-uniform A field hence influencing electon movement along a conductor and creating anomalous voltage induction.  Relativity theory debunks such claims.  However relativity theory doesn't debunk the possibility that electron movement within such a non-uniform A field might create force at right angles to the conductor if the A field is at right angles but varies in amplitude along the conductor. The images below suggests two possible experiments to discover this possibility.  Both use the A field within a magnetized ring core.  The first image uses a pulse of current into a length of wire that could jump off its posts, while the second image suggests a rotor version.

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img  Grumpy
Re: Hidden momentum « Reply #1 on: 2023-08-01, 01:39:04 »

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tExB=qr
Is the rotor version a variation of the Marinov motor?

Marinov used a toroidal magnet and the rotor was outside, and perpendicular.  Battery rather than pulsed.

See figure 1, here: https://www.semanticscholar.org/paper/Observations-of-the-Marinov-Motor-Phipps/88a937981ae9f9663c591f9aca5db108faf122f0
   
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Re: Hidden momentum « Reply #2 on: 2023-08-01, 14:02:36 »

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No, it is not the Marinov motor.  Marinov did not use a toroidal magnet, he used a bar magnet of circular cross section that had been split longitudinally to create two magnets of semicircular section that were then joined together again with one in reversed position.  This then became a rectangular quadrupole.  His rotor was a circular channel of Mercury around that structure.
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img  chief kolbacict
Re: Hidden momentum « Reply #3 on: 2023-08-03, 19:25:15 »

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And here they consider that there is no moment of momentum.
https://susy.page/2015/09/20/hidden_momentum
It can be translate by google-translate.
I really don’t understand anything about Poynting vectors, but please tell me that https://susy.page/2015/09/20/hidden_momentum is mistaking. :(
   
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Re: Hidden momentum « Reply #4 on: 2023-08-04, 11:03:23 »

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Quote from: chief kolbacict on 2023-08-03, 19:25:15
And here they consider that there is no moment of momentum.
https://susy.page/2015/09/20/hidden_momentum
It can be translate by google-translate.
I really don’t understand anything about Poynting vectors, but please tell me that https://susy.page/2015/09/20/hidden_momentum is mistaking. :(

The Poynting vector, the vector multiplication of E and H, represents EM power flow with the dimensions of watts/square meter.  As a power flow there must be a source and that defines something about the E and H, they must both come from the same source.  It is possible to have E and H from different sources, in which case the E X H product is not a real Poynting vector, it is a pseudo vector with no real meaning.  When we do have a real Poynting vector, from relativity theory the equivalence of mass and energy means the energy flow has a (hidden) momentum.  The pseudo vector does not have hidden momentum since it does not represent energy flow.  The article you refer to with radial E field and circular H field has the pseudo vector so it did not need the long winded to approach to reach its (correct) conclusion, it is obvious right from the start.  (There is a real Poynting vector with radial E field and circular H field within a coaxial cable, and that meets the requirement that E and H come from the same source.)

My point about hidden momentum is not the Poynting vector, but the consideration that charge Q in a vector magnetic potential A has a (hidden) momentum Q*A.  Because in any pratical experiment the forces are so small it may need a superconducting conductor to be used to obtain observable results.

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img  Cortazar
Re: Hidden momentum « Reply #5 on: 2023-08-09, 20:34:16 »
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Hello Smudge,
  Your idea is that the current pulse creates an A field that suddenly negates or amplified the A field of the toroidal magnet, and therefore dA/dt is created and a Force is generated. Am I correct ?
Regards
Cortazar
   
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Re: Hidden momentum « Reply #6 on: 2023-08-10, 09:03:06 »

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Quote from: Cortazar on 2023-08-09, 20:34:16
Hello Smudge,
  Your idea is that the current pulse creates an A field that suddenly negates or amplified the A field of the toroidal magnet, and therefore dA/dt is created and a Force is generated. Am I correct ?
Regards
Cortazar
No,no,no, you have the wrong concept there.  Electrons travel along a radial line at a rate v=dr/dt and they endure a static A field at right angles to their motion.  This A field varies with radial distance so the electron sees dA/dr * dr/dt which by the product rule becomes dA/dt.  So the internal momentum e*A changes with time and with distance yielding a sideways force F = e*dA/dt.

Smudge
   
img  Smudge
Re: Hidden momentum « Reply #7 on: 2023-08-10, 15:19:03 »

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Quote from: Smudge on 2023-08-10, 09:03:06
No,no,no, you have the wrong concept there.  Electrons travel along a radial line at a rate v=dr/dt and they endure a static A field at right angles to their motion.  This A field varies with radial distance so the electron sees dA/dr * dr/dt which by the product rule becomes dA/dt.  So the internal momentum e*A changes with time and with distance yielding a sideways force F = e*dA/dt.

Smudge

Further to this it may be noted that the force on a current carrying wire in a magnetic field, quite often quoted in differential form dF =i.dl X B that comes from the motional induction E = v X B, can also be obtained from the non-uniform A field that "creates" the B field via B = curl(A).  When you do the math you have to take account not only how the A field changes magnitude as the electrons move along the wire but also the change in A field direction.  So doing the same thing in the A field from the toroid where there is no B field because curl(A) = 0 should give a non-zero result.  F6 has pointed out that relativity theory denies the possibility that such a procedure can create voltage induction from a longitudinal non-curl A field along the wire but that does not apply to the force from an A field at right angles to the wire.

Smudge
   
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