Why Must Pure Force Depend on 4-Velocity?

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Discussion Overview

The discussion revolves around the nature of pure forces in the context of relativistic physics, particularly focusing on why a pure force must depend on the four-velocity of a test particle. Participants explore concepts from Wolfgang Rindler's work on special relativity, including the definitions of pure forces and their implications for electromagnetic forces.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants note that a pure force is defined as one that does not change the rest mass of a particle, leading to the condition that the four-velocity and the force must be orthogonal.
  • One participant argues that if the force does not depend on the four-velocity, it cannot remain orthogonal to all possible four-velocities unless the force itself is zero.
  • Another participant challenges the assumption that the electromagnetic force is always pure, citing examples where electromagnetic interactions can change the mass of an atom or material.
  • Some participants discuss the standard form of the electromagnetic force and its antisymmetry, which they argue ensures that the force is pure by definition.
  • There is a suggestion that the assumption of a pure force may limit the types of interactions that can be modeled, particularly in cases of absorption or emission of electromagnetic radiation.
  • Participants express skepticism about deriving conclusions regarding electromagnetism solely based on the assumption of a pure force.

Areas of Agreement / Disagreement

Participants express differing views on the nature of pure forces and their applicability to electromagnetic interactions. There is no consensus on whether the assumption of a pure force is valid in all contexts, particularly regarding electromagnetic forces and their effects on mass.

Contextual Notes

Participants highlight limitations in modeling certain physical processes, such as absorption or emission of electromagnetic waves, using the standard electromagnetic force. The discussion reflects ongoing debates about the implications of definitions and assumptions in relativistic physics.

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  • #33
Dale said:
I am not sure why you think I am missing that. His equation 72 is the equation I wrote in post 8.
I wanted to make you aware, that he covers what you wrote. Is something else missing in his book to come to the conclusion "So such a theory describes nothing"?
 
  • #34
Sagittarius A-Star said:
Rindler meant specifically the Lorentz force on a charged particle ##F_\mu = \frac {q}{c} E_{\mu \nu} U^\nu##. An this is pure.
Ok. Now write down an equation describing a resistor using the Lorentz force.
 
  • #37
PeterDonis said:
As @Dale already pointed out way back in post #8, the force in that equation is not a pure force. So it can't be the Lorentz force, which is pure.
Yes, but you asked only for "using the Lorentz force". I did this. This "effective" force is a sum of (pure) Lorentz forces, but is not pure itself. Then I can't guess, what you mean else.
 
  • #38
Sagittarius A-Star said:
Is something else missing in his book to come to the conclusion "So such a theory describes nothing"?
No, it is not something missing in his book. It is something missing in nature. There are in nature no classical point particles. So the sum in the middle of equation 72 is a sum over nothing. This is why I disagree with this approach.
 
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  • #39
  • #40
Sagittarius A-Star said:
sum over Lorentz forces on unit proper (co-moving) volumes of fluid, and thus on charges ρ0, not on particles
Then it is not a pure force, violating one of the foundational assumptions of his approach
 
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  • #41
Dale said:
Then it is not a pure force, violating one of the foundational assumptions of his approach
The sum is not a pure force, but the partial forces that are summed-up are pure, because they each fulfill ##\mathbf F \cdot \mathbf U = 0##.

A (partial) pure Lorentz-force on a unit proper (co-moving) volume of fluid with charge ρ0 preserves mass of the moving fluid (the temporal component is zero in it's rest-frame), but not system mass in the rest-frame of the wire. See the following equation (73).

http://www.scholarpedia.org/article...romagnetism#The_electromagnetic_energy_tensor
 
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  • #42
Sagittarius A-Star said:
the partial forces that are summed-up are pure, because they each fulfill ##\mathbf F \cdot \mathbf U = 0##.
And now we are back to post 28 and post 38. There is no such object in nature. There is nothing that is large enough to be classical and yet has only pure-force interactions with EM
 
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  • #43
Dale said:
And now we are back to post 28 and post 38. There is no such object in nature. There is nothing that is large enough to be classical and yet has only pure-force interactions with EM
The sum result on the right side of equation (72) is an impure-force "interaction with EM".
 
  • #44
Sagittarius A-Star said:
The sum result on the right side of equation (72) is an impure-force "interaction with EM".
And the things that are summed over don’t exist. As I already said.

Look, this is getting repetitive. If you like this approach, you are free to take it. I don’t like it at all for the reasons that I have stated (repeatedly) already. So I will use other approaches that I think are based on good foundations. But you do you, just don’t look to me for approval on this.
 
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  • #45
Sagittarius A-Star said:
you asked only for "using the Lorentz force". I did this.
You're quibbling. The point is that the force in this case is not pure.

Sagittarius A-Star said:
This "effective" force is a sum of (pure) Lorentz forces
@Dale has already responded to this. I share his skepticism regarding this approach.
 
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  • #46
PeterDonis said:
Ok. Now write down an equation describing a resistor using the Lorentz force.
PeterDonis said:
I share his skepticism regarding this approach.

This "effective" force is a sum of (pure) Lorentz forces. I split in the following the example of @Dale into two (pure) Lorentz-forces, separately for the 4-currrents of positive and negative charges. Adding both forces gives his result.

Dale said:
$$F^{\mu \nu }
=\left(
\begin{array}{cccc}
0 & -{E_x} & {0} & {0} \\
{E_x} & 0 & {0} & {0} \\
{0} & {0} & 0 & {0} \\
{0} & {0} & {0} & 0 \\
\end{array}
\right)$$
And if that field is applied to a resistive material then we have $$J^{\mu }=\left(\rho,{j_x},{j_y},{j_z}\right) =\left(0,{\sigma E_x},{0},{0}\right) $$ Then $$f_\mu = F_{\mu\nu}J^\nu =\left(\sigma {E_x}^2 ,0 ,0,0\right)$$
##f_{\mu+} = \rho_+F_{\mu\nu}U_+^\nu = F_{\mu\nu}J_+^\nu= F_{\mu\nu} \left(\rho,{0},{0},{0}\right)=\left(0,-\rho E_x,0,0\right)##
##f_{\mu-} = \rho_-F_{\mu\nu}U_-^\nu = F_{\mu\nu}J_-^\nu= F_{\mu\nu} \left(-\rho,{\sigma E_x},{0},{0}\right)=\left(\sigma {E_x}^2 ,+\rho E_x ,0,0\right)##
 
  • #47
Sagittarius A-Star said:
This "effective" force is a sum of (pure) Lorentz forces.
We're going around in circles. @Dale has already responded to this. You're not addressing his concerns at all. You're just repeating the same assertions.
 
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