# Why Must Pure Force Depend on 4-Velocity?

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• Sagittarius A-Star
In summary: F## to always be orthogonal to ##U## no matter what ##U## is (which is what ##U \cdot F = 0## says), ##F## has to depend on ##U##.
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.

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.

Sagittarius A-Star and vanhees71
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

vanhees71
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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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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Sagittarius A-Star
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".

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.

vanhees71 and Sagittarius A-Star
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.

vanhees71 and Dale
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)##

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.

vanhees71 and Dale

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