Magnetism seems absolute despite being relativistic effect of electrostatics

[Subplotsville]

No, at low speeds the gamma factor is virtually flat, that does not imply that the less acceleration effect is virtually flat.

Acceleration (the second derivative of position versus time) varies inversely with the square of time. The slowed clock effect is so flat at low speeds (see Lorentz equation) that the corresponding change in measurements of acceleration is also virtually flat. The equations are what imply the virtual flatness. But this is beside the point anyway, since it turns out that a slower clock measures greater acceleration, while the lab observer is trying to account for (the electrons' observation of) the decrease in acceleration he ascribes to their magnetic fields.

 

[Dale]

The slowed clock effect is so flat at low speeds (see Lorentz equation) that the corresponding change in measurements of acceleration is also virtually flat.

Begging the question again.

 

[Subplotsville]

Begging the question again.

See my previous post. See also the relevant Lorenz equation, as mentioned. It is not begging the question. All the information you need is there. I don't know how else to explain it. Do you not understand how changing measurement of time effects measurement of acceleration, or how squaring next-to-nothing yields next-to-nothing in the present case, or how this is an irrelevant digression anyway because time dilation increases measured acceleration while the lab observer is looking for a decrease?

 

[Dale]

But this is beside the point anyway, since it turns out that a slower clock measures greater acceleration, while the lab observer is trying to account for (the electrons' observation of) the decrease in acceleration he ascribes to their magnetic fields.

This is incorrect. The time dilation is in the correct "direction". Consider that the charges are initially at rest wrt each other and some apparatus with distances marked along the path they will travel. Suppose that in the apparatus' frame the electrons leave the initial mark at t=0 and reach the first mark at t=1. In the moving frame (moving perpendicular to the apparatus) they were released at t'=0 and reached the first mark at t'=γ>1. So since it took more time to get to the same location that corresponds to a reduced acceleration in the frame where the charges are moving. This is the correct "direction" for the magnetic force.

 

[Dale]

It is not begging the question. All the information you need is there.

Maybe, but you haven't proved it.

Similarly, I can simply point you to Maxwell's equations and the Lorentz transform and the writings of Einstein and Purcell and others. All the information you need is there.

 

[Dale]

Oh, one other thing. The magnetic force does not increase linearly with speed at low speeds. For a constant external magnetic field the magnetic force on a charge is proportional to the speed of the charge. However, in this scenario the external magnetic field is not constant but depends on the speed of the other charge.

 

[Subplotsville]

This is incorrect. The time dilation is in the correct "direction". Consider that the charges are initially at rest wrt each other and some apparatus with distances marked along the path they will travel. Suppose that in the apparatus' frame the electrons leave the initial mark at t=0 and reach the first mark at t=1. In the moving frame they were released at t'=0 and reached the first mark at t'=γ>1. So since it took more time to get to the same location that corresponds to a reduced acceleration in the frame where the charges are moving. This is the correct "direction" for the magnetic force.

This does not fit the present case. The electrons have the slower clock and therefore measure less time for whatever increase in distance due to electrostatic repulsion takes place between them. (Since this distance is perpendicular to their velocity according to the lab observer, it is unaffected by their motion.) Time dilation is doing the opposite of what the lab observer seeks to explain how the two electrons account for the decreased repulsive force between them.

Oh, one other thing. The magnetic force does not increase linearly with speed at low speeds. For a constant external magnetic field the magnetic force on a charge is proportional to the speed of the charge. However, in this scenario the external magnetic field is not constant but depends on the speed of the other charge.

What? The electrons are at rest wrt each other. The only speed we're looking at is the speed of the two electrons wrt the lab observer. That is the cause of their magnetic fields.

 

[Dale]

This does not fit the present case. The electrons have the slower clock and therefore measure less time for whatever increase in distance due to electrostatic repulsion takes place between them. (Since this distance is perpendicular to their velocity according to the lab observer, it is unaffected by their motion.) Time dilation is doing the opposite of what the lab observer seeks to explain how the two electrons account for the decreased repulsive force between them.

Yes, that is what I said above. The electrons measure less time (greater acceleration, electrostatic force only). The lab measures more time (less acceleration, electrostatic force minus magnetic force). That is correct, not opposite.

What? The electrons are at rest wrt each other. The only speed we're looking at is the speed of the two electrons wrt the lab observer. That is the cause of their magnetic fields.

Understood, and so the magnetic force in the lab is not proportional to the velocity in the lab because the magnetic force depends both on the velocity of the charge in the lab and the strength of the external field, which in turn depends on the velocity of the other charge in the lab.

 

[Subplotsville]

Yes, that is what I said above. The electrons measure less time (greater acceleration, electrostatic force only). The lab measures more time (less acceleration, electrostatic force minus magnetic force). That is correct, not opposite.

This is correct. The previous poster who brought this up originally was right to do so. Nevertheless, the fact that this effect is virtually flat at low speeds per the Lorentz equation, compared to how the magnetic field varies directly with speed, remains relevant. Do you not accept this discrepancy? I'm still not sure exactly what you disagree with. If you look at the Lorentz equation (which this browser has trouble with, so I won't try to post it), you will see how time dilation varies with speed. Though acceleration does vary with the square of time, you're still nowhere near how the magnetic field varies with the speed of the moving charge generating it. If you could clarify precisely where you differ with this reasoning rather that just saying "begging the question," that would be great.

Understood, and so the magnetic force in the lab is not proportional to the velocity in the lab because the magnetic force depends both on the velocity of the charge in the lab and the strength of the external field, which in turn depends on the velocity of the other charge in the lab.

This is not about an electric charge moving through a magnetic field. The lab sees the two electrons as 1) mutually stationary like charges causing a repulsion and 2) mutually stationary magnets with fields aligned to cause an attraction. On the other hand, the electrons only see each other as like charges.

 

[Dale]

Nevertheless, the fact that this effect is virtually flat at low speeds per the Lorentz equation, compared to how the magnetic field varies directly with speed, remains relevant. Do you not accept this discrepancy?

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.

Though acceleration does vary with the square of time, you're still nowhere near how the magnetic field varies with the speed of the moving charge generating it.

This is the part that you haven't shown. You are simply asserting that we are "nowhere near" the required force without showing it.

This is not about an electric charge moving through a magnetic field. The lab sees the two electrons as 1) mutually stationary like charges causing a repulsion and 2) mutually stationary magnets with fields aligned to cause an attraction.

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.

 

[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.