Magnetism seems absolute despite being relativistic effect of electrostatics

[Subplotsville]

Clearly not. That is why continuing to assert it is begging the question. You are asking me to accept as a premise the very point under discussion without any justification other than your continued assertion.

You need to do more than just state that time dilation is "virtually flat", you need to show that it does not account for the magnetic force.

I am doing more: Time dilation is virtually flat over changes in speed at low speeds (presumably you agree with this), while magnetic force is not (presumably you agree with this also). Therefore, time dilation (or the square of it, as it effects acceleration, which is comparably flat) is insufficient to account for the decrease in repulsion between the two electrons which the lab attributes to their magnetic interaction. Unless you expect some surprises to pop up when doing the rudimentary math to make the point rigorous, the case is settled on this basis alone.

They are mutually stationary, but that is not relevant in Maxwell's equations nor the Lorentz force law. There is no "mutual velocity" term in either of those equations. All of the velocity terms in Maxwell's equations and the Lorentz force term are measured wrt an inertial reference frame.

So, in the lab frame you have two moving electrons, each generating an electric and a magnetic field. Each electron then experiences an electric and a magnetic force due to their motion through the field from the other electron.

The fact that they are mutually stationary is relevant to the electrons, because this condition causes neither to see the other as a moving charge generating a magnetic field, though the lab sees them both as doing so.

 

[Dale]

The fact that they are mutually stationary is relevant to the electrons, because this condition causes neither to see the other as a moving charge generating a magnetic field, though the lab sees them both as doing so.

Where in Maxwell's equations or the Lorentz force law does the mutual velocity appear?

 

[Dale]

I am doing more: Time dilation is virtually flat over changes in speed at low speeds (presumably you agree with this), while magnetic force is not (presumably you agree with this also).

I agree that time dilation is virtually flat, with "virtually flat" meaning that the series expansion about v=0 has no first order term. However, the magnetic force in this scenario is also virtually flat.

Therefore, time dilation (or the square of it, as it effects acceleration, which is comparably flat) is insufficient to account for the decrease in repulsion between the two electrons which the lab attributes to their magnetic interaction.

This doesn't follow even if the second premise were correct. We have some function γ(v) and some other function f(γ). From those two functions it is possible by composition to construct a function f(v)=f(γ(v)). You are claiming that the mere fact that γ(v) is virtually flat automatically implies that f(v) is also flat, without any knowledge of f(γ). That is incorrect. Regardless of how flat γ(v) is it is always possible to come up with some f(γ) which makes f(v) arbitrarily "un-virtually flat".

FYI, I just did a quick calculation and the time dilation is definitely of the right order, but I am missing a factor of 2 somewhere.

 

[Dale]

Suppose two point charges each of charge q are at rest separated by a distance r in the y direction, then the three-force, f', is given by Coulomb's law:
f'=\left( 0, \frac{q^2}{4\pi\epsilon_0 r^2}, 0 \right)

This corresponds to a four-force, F', of:
F'=\left( 1 \frac{f' \cdot 0}{c}, 1 f' \right)=\left(0, 0, \frac{q^2}{4\pi\epsilon_0 r^2}, 0 \right)

Boosting in the x direction to a frame where the charges are moving with velocity v gives
F=\Lambda F' = \left(0, 0, \frac{q^2}{4\pi\epsilon_0 r^2} , 0\right)

This corresponds to a three-force in the lab frame of
F=\left( \gamma \frac{f \cdot v}{c}, \gamma f \right)
f=\left(0, \frac{q^2}{4\pi\epsilon_0 r^2} \sqrt{1-v^2/c^2} ,0 \right)
which is nothing more than the relativistic boost of the electrostatic force.

Now, in the lab frame the electric and magnetic fields on the charge due to the other are given by:
E=\frac{q}{4 \pi r^2 \epsilon_0 }\frac{1}{\sqrt{1-v^2/c^2}}
B=\frac{v}{c^2}\frac{q}{4 \pi r^2 \epsilon_0 }\frac{1}{\sqrt{1-v^2/c^2}}

The Lorentz force in the lab frame is given by:
f=q(E+v \times B) = \left( 0, q \frac{q}{4 \pi r^2 \epsilon_0 }\frac{1}{\sqrt{1-v^2/c^2}} - q v \frac{v}{c^2}\frac{q}{4 \pi r^2 \epsilon_0 }\frac{1}{\sqrt{1-v^2/c^2}}, 0 \right)
f=\left(0, \frac{q^2}{4\pi\epsilon_0 r^2} \sqrt{1-v^2/c^2} ,0 \right)

Which is identical to relativistic boost of the electrostatic force.

 

[Subplotsville]

Where in Maxwell's equations or the Lorentz force law does the mutual velocity appear?

Nowhere that I can find. Again, if I said something that appeared otherwise, my clarification is that mutual velocity is relevant because the electrons are at rest and thus do not see each others' magnetic fields, not in any way in the lab's frame of reference.

I agree that time dilation is virtually flat, with "virtually flat" meaning that the series expansion about v=0 has no first order term. However, the magnetic force in this scenario is also virtually flat.

What I mean by "virtually flat" is the slope of the left side of the famous Lorentz function graph showing t' in terms of v. This would be squared to visualize the effect on acceleration (and thus measured force), which varies as the inverse square of time. The result still qualifies as "virtually flat" on the left side. Then we look at the equivalent area of the function graph showing the magnetic field in terms of v of a moving charge. This is not fairly characterizable as "virtually flat." Already, on this first approximation basis, we have cause to doubt that time dilation alone could be responsible for the cancellation of electrostatic repulsion which the lab observer attributes to magnetic interaction but the two moving electrons don't due to their being mutually at rest.

PS: I appreciate the trouble you took to post the math. Sadly this browser is unable to render that format as anything but a tangle of gibberish. So what am I doing on a physics forum with such a browser? Well, this is a good question. I thought I'd just come to chat. You are of course under no obligation to put up with my security-paranoid impairments. Meanwhile, I'll have to let whatever point you made go and step out of the discussion. Thanks for playing.

 

[lugita15]

PS: I appreciate the trouble you took to post the math. Sadly this browser is unable to render that format as anything but a tangle of gibberish. So what am I doing on a physics forum with such a browser? Well, this is a good question. I thought I'd just come to chat. You are of course under no obligation to put up with my security-paranoid impairments. Meanwhile, I'll have to let whatever point you made go and step out of the discussion. Thanks for playing.

Don't get discouraged too quickly, Subplotsville! Attached is a screenshot of Dalespam's post. Tell me if the image resolution is too low.

 

[Dale]

Nowhere that I can find. Again, if I said something that appeared otherwise, my clarification is that mutual velocity is relevant because the electrons are at rest and thus do not see each others' magnetic fields, not in any way in the lab's frame of reference.

You seem to misunderstand the principle of relativity. It states that the laws of physics are the same in all reference frames. In this case, the forces between the charges are governed by Maxwell's equations and the Lorentz force equation in all reference frames.

Applying Maxwell's equations in the lab frame there is clearly a magnetic field, and applying the Lorentz force law there is clearly a magnetic force. So, in the lab frame the electrons do see each others' magnetic fields and are affected by them. Otherwise the principle of relativity would be violated.

What I mean by "virtually flat" is the slope of the left side of the famous Lorentz function graph showing t' in terms of v. This would be squared to visualize the effect on acceleration (and thus measured force), which varies as the inverse square of time. The result still qualifies as "virtually flat" on the left side. Then we look at the equivalent area of the function graph showing the magnetic field in terms of v of a moving charge. This is not fairly characterizable as "virtually flat." Already, on this first approximation basis, we have cause to doubt the claim that time dilation alone is responsible for the cancellation of electrostatic repulsion which the lab observer attributes to magnetic interaction but the two moving electrons don't due to their being mutually at rest.

The math simply doesn't support your reasoning, as described and derived above.

PS: I appreciate the trouble you took to post the math. Sadly this browser is unable to render that format as anything but a tangle of gibberish. So what am I doing on a physics forum with such a browser? Well, this is a good question. I thought I'd just come to chat. You are of course under no obligation to put up with my security-paranoid impairments. Meanwhile, I'll have to let whatever point you made go and step out of the discussion. Thanks for playing.

That is fine, but it seems rather inconsistent of you to request (repeatedly) that I work out the math and then not even be willing to read it when I do.