B How To Consistently Explain Electromagnetism With Relativity

[Geocentricist]

Yes

Awesome!

To find the strength of each repulsion or attraction will require some math (Coulombs law, Biot Savart law, Lorentz force law)

Could you define the strength of the magnetic forces in terms of the strength of the electrostatic forces? If top moving proton electrostatically repels bottom moving proton with strength 1, it magnetically attracts it with strength x. What is x? Or is this too much to ask?

I've tried to illustrate the opposite currents scenario below. Is this correct?

 

[Dale]

Could you define the strength of the magnetic forces in terms of the strength of the electrostatic forces? If top moving proton electrostatically repels bottom moving proton with strength 1, it magnetically attracts it with strength x. What is x?

There is no single number x in general. In the Lorentz force law $F=q E + q v \times B$ the first term on the right is the electric force and the second term on the right is the magnetic force.

 

[Geocentricist]

There is no single number x in general. In the Lorentz force law $F=q E + q v \times B$ the first term on the right is the electric force and the second term on the right is the magnetic force.

If I set the current flowing at 87% of c and the distance between the wires was 1 centimeter, would that be enough information to define the magnetic force in terms of the electrostatic force?

 

[jartsa]

If I set the current flowing at 87% of c and the distance between the wires was 1 centimeter, would that be enough information to define the magnetic force in terms of the electrostatic force?

Are we still talking about the proton pair in electron's frame? The magnetic force is 50% of the electric force in that case. Distance does not matter.

(I know that because I know that: $netforce'=netforce/2$)

(The 2 comes from the gamma, which is 2 at speed 0.87 c)

 

[Geocentricist]

Are we still talking about the proton pair in electron's frame? The magnetic force is 50% of the electric force in that case. Distance does not matter.

1A, electric attraction. 1B, electric repulsion, magnetic attraction. 1C, electric repulsion, magnetic attraction.
2A, electric attraction. 2B, electric repulsion, magnetic attraction. 2C, electric repulsion, magnetic attraction.
3A, electric repulsion. 3B, electric attraction. 3C, electric attraction.

• 4 electric attraction
• 5 electric repulsion
• 4 magnetic attraction

Magnetic force is 50% as strong as electric force, so it's effectively 2 attraction.

4 electric attraction + 5 electric repulsion = 1 electric repulsion. 1 electric repulsion + 2 attraction = 1 attraction. This agrees with the observed fact that the wires attract.

Is this correct?

 

[jartsa]

View attachment 215549

1A, electric attraction. 1B, electric repulsion, magnetic attraction. 1C, electric repulsion, magnetic attraction.
2A, electric attraction. 2B, electric repulsion, magnetic attraction. 2C, electric repulsion, magnetic attraction.
3A, electric repulsion. 3B, electric attraction. 3C, electric attraction.

• 4 electric attraction
• 5 electric repulsion
• 4 magnetic attraction

Magnetic force is 50% as strong as electric force, so it's effectively 2 attraction.

4 electric attraction + 5 electric repulsion = 1 electric repulsion. 1 electric repulsion + 2 attraction = 1 attraction. This agrees with the observed fact that the wires attract.

Is this correct?

Yes!

(One line should read: "Magnetic force is 50% as strong as electric force, so it's effectively 1/2 attraction." Right?)

 

[Geocentricist]

Yes!

Yay!

(One line should read: "Magnetic force is 50% as strong as electric force, so it's effectively 1/2 attraction." Right?)

I did intend to mean magnetic attraction is effectively half electrostatic attraction. But I was actually referring to the line,

• 4 magnetic attraction

When I said half would be 2.

Now I have enough information to account for all the forces in the identical currents scenario. But what about the opposite currents scenario?

Maybe I can figure this out on my own. I'll start with the lab frame. In the lab frame, we have two pairs of electrons, one moving right and one moving left. All the electrostatic forces cancel, so there must be a net repulsive magnetic force to account for the observed repulsion between the wires, and since magnetic forces only involve moving charges, it must be between these two pairs of electrons.

Let me guess, the top, right-moving electron pair repulses the bottom, left-moving electron pair with a magnetic repulsion of 2? One magnetic repulsion per electron?

Thank you very much for your continued help.

 

[jartsa]

Yay!Now I have enough information to account for all the forces in the identical currents scenario. But what about the opposite currents scenario?

View attachment 215561

Maybe I can figure this out on my own. I'll start with the lab frame. In the lab frame, we have two pairs of electrons, one moving right and one moving left. All the electrostatic forces cancel, so there must be a net repulsive magnetic force to account for the observed repulsion between the wires, and since magnetic forces only involve moving charges, it must be between these two pairs of electrons.

Good reasoning there IMO.

Let me guess, the top, right-moving electron pair repulses the bottom, left-moving electron pair with a magnetic repulsion of 2? One magnetic repulsion per electron?

I have no idea. We have not calculated the magnetic attraction of parallel currents yet, in the lab frame. We only calculated it in the electron frame. Okay so let's calculate it then: In the lab frame the magnetic force is 50% of the electric repulsion of the electrons, when electrons co-move at speed 0.87c. I mean the magnetic force is half the electric force and points to opposite direction.

So in this opposite currents case the magnetic force is half the electric force and points to the same direction. Because we know the magnitudes of the magnetic forces should be the same in both cases.I used again this formula to calculate the magnetic force:
total force between electrons co-moving at speed 0.87 c = total force between electrons standing still / gamma

Oh that's wrong . Unless we are talking about same electrons seen from different frames or two electrons side by side. Probably it's gamma squared for electrons in wires, those electrons in the wire see other electrons move away as the electrons gain speed, while in those other cases no such thing occurs.

 

[Geocentricist]

We have not calculated the magnetic attraction of parallel currents yet, in the lab frame. We only calculated it in the electron frame.

Oh yes, I forgot!

Okay so let's calculate it then:

Two wires, two protons in length, all four electrons moving right at 87% c. All electric forces cancel. One top electron exerts 1 magnetic attraction on one bottom electron. There are two top electrons and two bottom electrons so 4 magnetic attraction, equal to 2 electric force. So 2 attraction. This agrees with the calculation in the electron frame.

In the lab frame the magnetic force is 50% of the electric repulsion of the electrons, when electrons co-move at speed 0.87c. I mean the magnetic force is half the electric force and points to opposite direction.

I agree.

So in this opposite currents case the magnetic force is half the electric force and points to the same direction. Because we know the magnitudes of the magnetic forces should be the same in both cases.

At first I thought the magnetic force would be double. But after some thinking I realized I was confusing frames, and now I see why you're correct.

Now on to the final frame, the electron-frame of the opposite currents case.

Top-left proton electrically repulses both bottom protons (2 repulsion) and electrically attracts all four bottom electrons (4 attraction).
Same for top-right proton (2 repulsion, 4 attraction).
Top electron electrically attracts both bottom protons (2 attraction) and electrically repulses all four bottom electrons (4 repulsion).

These are all the electric forces and they sum to 8 repulsion and 10 attraction, which equals 2 attraction. Now for the magnetic forces.

Top-left proton magnetically attracts both bottom protons (1 attraction). Same for top-right proton (1 attraction).

Net force is now 4 attraction. I will assume the force between a left-moving proton and an electron moving left at twice the speed is 1.5 magnetic repulsion. I say 1.5 because it's 1 between two comoving charges, and presumably 2 if you double both their speed, but here we have one moving at original speed and one moving at twice that.

So top-left proton magnetically repulses each of the bottom four electrons with 1.5 magnetic repulsion (6 magnetic repulsion). Top-right proton does the same thing (6 magnetic repulsion). This is 12 magnetic repulsion, equal to 6 repulsion.

6 repulsion and 4 attraction equals 2 repulsion.

This agrees with the calculation in the lab frame. Is this right? I know it's a lot to read.

 

[Geocentricist]

I used again this formula to calculate the magnetic force:
total force between electrons co-moving at speed 0.87 c = total force between electrons standing still / gamma

Oh that's wrong . Unless we are talking about same electrons seen from different frames or two electrons side by side. Probably it's gamma squared for electrons in wires, those electrons in the wire see other electrons move away as the electrons gain speed, while in those other cases no such thing occurs.

I wasn't sure how to use your formula so I did my analysis without it. I wonder if you can confirm whether the magnetic repulsion between a proton moving left at 87% c and an electron moving left at double that speed is 1.5 times the magnetic attraction between two electrons moving left at 87% c?

 

[A.T.]

But what about the opposite currents scenario?

DrGreg's diagramm could help here:
https://www.physicsforums.com/threa...agnetism-with-relativity.932270/#post-5886984

 

[jartsa]

I wasn't sure how to use your formula so I did my analysis without it. I wonder if you can confirm whether the magnetic repulsion between a proton moving left at 87% c and an electron moving left at double that speed is 1.5 times the magnetic attraction between two electrons moving left at 87% c?

I can only calculate magnetic attractions between charges moving side by side at the same velocity. Because that is very simple.

There's one thing I should mention: All your pictures are kind of unrealistic, as the two electrons are always perfectly lined up. It's not unphysical, it's just not correct for electrons in wires. I mean, according to an electron the other electrons should disappear into the distance as the current increases. That effect seems to be missing in all of the calculations - I just noticed that.There is also an easy way to calculate these things. That involves going into electron's frame, calculating electric forces on the electron there, there are no magnetic forces, which was the point of the frame change. And then that force can be easily transformed to any frame, by using relativity's force transformation formulas.

 

[Geocentricist]

DrGreg's diagramm could help here:
https://www.physicsforums.com/threa...agnetism-with-relativity.932270/#post-5886984

Not sure how that's supposed to help figure out the force between a moving electron and proton.

I can only calculate magnetic attractions between charges moving side by side at the same velocity. Because that is very simple.

Okay, I hope someone else can help me out here then.

There's one thing I should mention: All your pictures are kind of unrealistic, as the two electrons are always perfectly lined up. It's not unphysical, it's just not correct for electrons in wires. I mean, according to an electron the other electrons should disappear into the distance as the current increases. That effect seems to be missing in all of the calculations - I just noticed that.

Why should the separation between electrons increase?

There is also an easy way to calculate these things. That involves going into electron's frame, calculating electric forces on the electron there, there are no magnetic forces, which was the point of the frame change. And then that force can be easily transformed to any frame, by using relativity's force transformation formulas.

I prefer doing it this way, in the way I can understand.

 

[jartsa]

Okay, I hope someone else can help me out here then.

Hey, why don't I use my wonderful easy calculation method myself.

So, an electron moves at speed 0.87 in a wire, there is another wire where electrons move at the opposite direction at speed 0.87c.

Let's say electron sees the electron formation on the other wire contracted to 1/4 of the normal. That is not the correct number.

So four times more charges produces four time more electric force.

Now the transformation to the lab frame. That is division by two in this case.

Total force in lab frame = 4 * electric_force_in_ lab_frame / 2 = 2 * electric_force_in_ lab_frame

Total force doubled, so magnetic force in lab frame must be same as the electric force in the lab frame. (Not the actual correct result because of the wrong length contraction)This is the force transformation formula: F'=F/gamma

 

[Geocentricist]

(Not the actual correct result because of the wrong length contraction)

But why should the electron spacing increase in their own frame?

 

[Ibix]

Why should the separation between electrons increase?

Because they are accelerating, and their notion of distance is changing as this happens.

I prefer doing it this way, in the way I can understand.

Philosophical point: if you can't do the maths, do you really understand it? You will always need someone to tell you what the maths says.

Working in the lab frame, the magnetic field from an infinitely long current carrying wire is $B=\mu_0 I/2\pi r$, always in the tangential direction. Defining $\theta$ as the angle in the y-z plane anticlockwise from the y axis, $B_x=0$, $B_y=-B\sin\theta$, $B_z=B\cos\theta$. The electric field is everywhere zero.

Then you can write down the electromagnetic field strength tensor, F$$F=\left (\begin {array}{cccc}
0&-E_x/c&-E_y/c&-E_z/c\\
E_x/c&0&-B_z&B_y\\
E_y/c&B_z&0&-B_x\\
E_z/c&-B_y&B_x&0
\end {array}\right)$$Then you carry out a Lorentz transform to the electron rest frame. This is to a velocity v in the +x direction (v will turn out to be negative when we calculate it later, due to conventional current having its sign defined in an unfortunate way). That's way too much work to typeset for the general case. The result is that in the electron frame there is an electric field whose components are $E'_x=0$, $E'_y=-\gamma v B_z$, $E'_z=\gamma v B_y$ (Edit: removed erroneous factor of c) and a magnetic field whose components are $B'_x=0$, $B'_y=\gamma B_y$, $B'_z=\gamma B_z$, where $\gamma=1/\sqrt {1-v^2/c^2}$.

Now you need to determine the electron drift velocity, v. Wikipedia has the maths (although it uses u instead of v) and even works out a number for you using 1A through a copper wire. Note that they use a slightly inconsistent sign convention - the answer needs to be multiplied by -1 to get the negative velocity I mentioned above. https://en.m.wikipedia.org/wiki/Drift_velocity
Dale already gave you the formula to calculate the force on a charge from the field components.

 

[A.T.]

But why should the electron spacing increase in their own frame?

https://www.physicsforums.com/threa...sm-with-relativity.932270/page-3#post-5888171

 

[Geocentricist]

https://www.physicsforums.com/threa...sm-with-relativity.932270/page-3#post-5888171

The Wikipedia article says spaceships A and B disagree they both accelerated at the same time. But since in S they did accelerate at the same time, and did so with the same acceleration, it seems to me A and B share a frame and definition of simultaneity throughout their acceleration. In which case both spaceships agree they accelerated at the same time.

 

[Dale]

The Wikipedia article says spaceships A and B disagree they both accelerated at the same time.

Actually it doesn’t say that. What it says is subtly different.

In the article neither S nor S’ represent A or B’s perspective. S and S’ are inertial frames and the article is talking about their perspective, not the perspective of A or B which would be non inertial. A and B are only momentarily at rest in S or S’.

 

[Geocentricist]

Actually it doesn’t say that. What it says is subtly different.

In the article neither S nor S’ represent A or B’s perspective. S and S’ are inertial frames and the article is talking about their perspective, not the perspective of A or B which would be non inertial. A and B are only momentarily at rest in S or S’.

Okay, but I still don't see why A and B should disagree they accelerated at the same time, since they always share the same frame and therefore, time.

 

[PAllen]

Okay, but I still don't see why A and B should disagree they accelerated at the same time, since they always share the same frame and therefore, time.

Think about this. In S you have two separated simultaneous events: A started accelerating, B started accelerating. S’ is moving relative to S. It is then impossible for these acceleration events to be simulataneous in S’. Two relatively moving frames can never agree on simultaneity of events separated in the direction of relative motion.

 

[Dale]

Okay, but I still don't see why A and B should disagree they accelerated at the same time, since they always share the same frame and therefore, time.

It isn’t a disagreement between A and B. It is a disagreement between S and S’

 

[PAllen]

Consider a related question. In some S’, there is an event when A is stationary and an event when B is stationary. These necessarily correspond to events described in S as A has speed v and B has speed v. These events are simultaneous by construction in S. Thus the corresponding events in S’ are not simultaneous. Thus, once acceleration has begun, there are no inertial frames where A and B are simultaneously at rest.

 

[Geocentricist]

It isn’t a disagreement between A and B. It is a disagreement between S and S’

Consider a related question. In some S’, there is an event when A is stationary and an event when B is stationary. These necessarily correspond to events described in S as A has speed v and B has speed v. These events are simultaneous by construction in S. Thus the corresponding events in S’ are not simultaneous. Thus, once acceleration has begun, there are no inertial frames where A and B are simultaneously at rest.

Any S' equidistant to A and B will consider the events simultaneous. So I don't see how a possible disagreement between S and some hypothetical S' proves A and B will move relative to one another.

 

[PAllen]

Any S' equidistant to A and B will consider the events simultaneous. So I don't see how a possible disagreement between S and some hypothetical S' proves A and B will move relative to one another.

S’ is a reference frame, not an observer, so equidistant is a nonsequiter. There is no reference frame in which A and B are both at rest at the same time except the original reference frame for the times before acceleration starts. If you can’t see this, then you need to review the basics of relativity of simultaneity. You are making statements comparable to 1 + 1 = 3 and insisting they are right.

 

[Geocentricist]

S’ is a reference frame, not an observer, so equidistant is a nonsequiter. There is no reference frame in which A and B are both at rest at the same time except the original reference frame for the times before acceleration starts. If you can’t see this, then you need to review the basics of relativity of simultaneity. You are making statements comparable to 1 + 1 = 3 and insisting they are right.

I didn't say A and B were both at rest in S' and I'm not aware anyone else did either. I thought everyone said S' is a frame moving inertially relative to S, which would of course mean A and B are moving inertially relative to S' when they are at rest in S.

I still maintain it seems A and B share a frame at all times and this means they also agree on what is simultaneous. What part of this sentence is wrong?

 

[PAllen]

I didn't say A and B were both at rest in S' and I'm not aware anyone else did either. I thought everyone said S' is a frame moving inertially relative to S, which would of course mean A and B are moving inertially relative to S' when they are at rest in S.

I still maintain it seems A and B share a frame at all times and this means they also agree on what is simultaneous. What part of this sentence is wrong?

Do you understand that in any frame S’, moving with respect to S, in which A is momentarily at rest at some time t, then B will not be at rest at that time t; and further, A and B will have started accelerating at different times in this frame?

As to your second question, all of it is wrong. That is what I am trying to get you to see by statements about S’. What do you think sharing a frame means? It isn’t standard usage, but I am guessing you think it means there is a frame in which they are both at rest at some given time. There is no such frame after acceleration begins.

 

[Geocentricist]

Do you understand that in any frame S’, moving with respect to S, in which A is momentarily at rest at some time t, then B will not be at rest at that time t; and further, A and B will have started accelerating at different times in this frame?

No.

As to your second question, all of it is wrong. That is what I am trying to get you to see by statements about S’. What do you think sharing a frame means? It isn’t standard usage, but I am guessing you think it means there is a frame in which they are both at rest at some given time. There is no such frame after acceleration begins.

I know A and B aren't at rest while they're accelerating, but they are still sharing the same accelerating frame.

 

[Ibix]

What's going on is the relativity of simultaneity. In S, the initial rest frame, the ships always have the same velocity at the same time, and that velocity is always changing. But any other frame, S', does not share the same definition of "at the same time". So in S' the ships are always moving at different speeds once at least one of them is accelerating, because what it calls "at the same time" is what S calls "at different times".

 

[pervect]

Ibix is right. To understand how special relativity consistently explains electromagnetism, one needs to understand special relativity. And how special relativity is consistent. This requires undestanding the whole of the theory, which includes understanding the relativity of simultaneity. There are a lot of threads on the topic, either under the name relativity of simlutaneity or "Einsteins train".