Why Must Pure Force Depend on 4-Velocity?

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SUMMARY

The discussion centers on the necessity for a pure force to depend on the four-velocity of a test particle, as articulated in Wolfgang Rindler's "Introduction to Special Relativity," 2nd edition. A pure force, defined as one that does not alter the rest mass of the particle, must satisfy the condition that the four-force is orthogonal to the four-velocity. The standard electromagnetic force, represented by the equation ##f_\mu = q F_{\mu \nu} u^\nu##, is deemed pure due to the antisymmetry of the electromagnetic field tensor ##F_{\mu \nu}##, which ensures that ##f_\mu u^\mu = 0##. However, this formulation does not encompass all electromagnetic interactions, particularly those involving absorption or emission of electromagnetic waves that affect mass.

PREREQUISITES
  • Understanding of four-vectors and four-velocity in special relativity
  • Familiarity with the electromagnetic field tensor ##F_{\mu \nu}##
  • Knowledge of the Lorentz force equation ##f_\mu = q F_{\mu \nu} u^\nu##
  • Concept of pure forces and their implications in relativistic physics
NEXT STEPS
  • Study the derivation and implications of the Lorentz force in classical electromagnetism
  • Explore the role of the electromagnetic field tensor in quantum electrodynamics (QED)
  • Investigate the limitations of classical electromagnetism in modeling mass-changing interactions
  • Examine the relationship between charge-current density and electromagnetic forces in different frames
USEFUL FOR

This discussion is beneficial for physicists, particularly those specializing in relativity and electromagnetism, as well as students seeking to deepen their understanding of the interplay between force, mass, and electromagnetic interactions.

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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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