Cross Product Force: Exploring Magnetism's Role

In summary, magnetism can be formulated in such a way that it is only a fictitious force in some rotating frame of reference. This is because magnetism is an inseparable part of the electromagnetic force (E,B) and can be "framed" away in certain conditions. However, in special relativity, any ordinary force can develop a cross-product component in a different frame, as seen in the example of a bowling ball connected to a rocket by a string.
  • #1
snoopies622
846
28
To me the idea of a force that pushes an object only in a "sideways" direction - that is, always at a right angle to the direction of motion - has always seemed strange. I can see how it comes about in the case of the Coriolis force because there the frame of reference is rotating and the force is "fictitious". But what about magnetism? Can magnetism be formulated in such a way that it is only a ficititous force in some rotating frame of reference?
 
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  • #2
snoopies622 said:
Can magnetism be formulated in such a way that it is only a ficititous force in some rotating frame of reference?

Hi snoopies622! :smile:

Magnetism is an inseparable part of the electromagnetic force (E,B)

If you change frames, E.B and E2 - B2 remain constant.

You can "frame" magnetism away (at a particular point) only if E.B = 0 and E2 - B2 is negative.
To me the idea of a force that pushes an object only in a "sideways" direction - that is, always at a right angle to the direction of motion - has always seemed strange.

It's a variety of Murphy's law …

anything that can mathematically happen will physically happen …

in this case, cross product forces are mathematically possible, so they must be physically possible! :biggrin:
 
  • #3
Thanks, tiny-tim.

Three questions:

1. By E.B do you mean [tex]

\vec E \cdot \vec B

[/tex] or [tex]

| \vec E | | \vec B | ?

[/tex] (I'm guessing the former but I just want to be sure.)

2. Are there any other forces that take the form

[tex]

\vec F = a ( \vec v \times \vec x )

[/tex]

where [tex] a [/tex] is a scalar, [tex] \vec v [/tex] is velocity and [tex] \vec x [/tex] is some other vector?

3. A long time ago a grad student showed me how the magnetic force between two adjacent charged particles moving on parallel paths could be accounted for by combining the Coulomb force with special relativity. Is this what you mean by "framing" it away? He then told me that this combination can be used to explain/replace all magnetic forces. I now take it that this is not correct?
 
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  • #6
Hi snoopies622! :smile:

1. Yes, the former.

(btw, on this forum, bold is preferable to arrows for vectors, partly because it's easier to type, and partly because LaTeX uses a lot of server power :wink:)

2. In special relativity, any ordinary (push-or-pull) force develops a cross-product part in a different frame …

for example (if you ignore general relativity), the force of gravity develops a "gravitomagnetic" component, with a v x, in a frame in which the gravitational source is moving.

3. Yes, that's exactly what I mean by "framing".

So far as we know, there are no magnetic monopoles (the equivalent of a point charge),

and so the only way nature can create a magnetic field is from a moving electric charge.

So basically he's right.

The Earth's magnetic field, for example, is created from moving electrons, and the tiny effect of one electron for one tiny instant can be explained without electric force by "framing" (though of course there's no frame that will explain away the electric force of all the electrons, or all the time).
 
  • #7
I am puzzled by the orginal post. F= ma so the acceleration is in the direction of the force but there is no requirement that it be in the direction of motion. If an object is swung in a circle, say a rock moves at constant speed at the end of a string, the force is the tension in the rope, toward the center of the circle, and perpendicular to the direction of motion.
 
  • #8
atyy: thanks for the physics.weber.edu reference. That does address the question.

tiny-tim said:
In special relativity, any ordinary (push-or-pull) force develops a cross-product part in a different frame …

I am not quite following this. Suppose I am somewhere in outer space, a bowling ball is ten meters away from me and a ten-meter string connects the ball to my hand. If I pull the string in my direction, in my reference frame a = F /m on the ball. Are you saying that in a different reference frame the acceleration of the ball will have a component that is at a right angle to the direction of force?


HallsofIvy: I was considering only forces of nature.
 
  • #9
snoopies622 said:
I am not quite following this. Suppose I am somewhere in outer space, a bowling ball is ten meters away from me and a ten-meter string connects the ball to my hand. If I pull the string in my direction, in my reference frame a = F /m on the ball.
Are you saying that in a different reference frame the acceleration of the ball will have a component that is at a right angle to the direction of force?

ah … good point …

suppose you are generating your force in the x-direction, causing both you and the bowling ball to accelerate in the x-direction, by rockets expelling mass in the minus-x direction (and somehow missing the bowling ball, of course! :rolleyes:) …

in a frame moving in the y-direction, the mass will be moving at an angle (which will be less the faster your speed increases), but the string will still (initially) be in the x-direction …

in that frame, there will appear to be a force perpendicular to the string, changing the angle of the string. :smile:
 
  • #10
Hi again, tiny-tim. Let me see if I understand this.

Using a Cartesain coordinate system, let's say there is a bowling ball at (0,0), a rocket at (10, 0), and a string connecting them (0<x<10,y=0). The rocket begins to expel mass in the -x direction, and the rocket, string and ball accelerate uniformly in the +x direction.

From the perspective of an observer moving at a constant rate in the +y direction:

Using classical mechanics only, the rocket and bowling ball move on parabolic paths, identical except that they are separated from each other by 10 units in the x direction. The string remains horizontal, and the acceleration of (and force on) the ball is still completely in the +x direction.

But applying special relativity, the string does not remain horizontal because there is an additional force acting on the ball which is at a right angle to the direction in which the string is pointing. This force is a cross product of the ball's velocity vector and another vector.

Is this correct? If so, what is the other vector?
 
  • #11
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snoopies622 said:
… But applying special relativity, the string does not remain horizontal because there is an additional force acting on the ball which is at a right angle to the direction in which the string is pointing. This force is a cross product of the ball's velocity vector and another vector.

Is this correct? If so, what is the other vector?

Hi snoopies622! :smile:

I've changed my mind! :redface:

I don't think my example works at all … the expelled mass still goes parallel to the string.

I think the correct analysis is that if it's a conservative force arising from a vector potential, (of the form (A,φ)), then it has to have these v x components …

but if it isn't, it doesn't!​
 
  • #12
tiny-tim said:
I think the correct analysis is that if it's a conservative force arising from a vector potential, (of the form (A,φ)), then it has to have these v x components …

A magnetic field is conservative? I thought a conservative field had to have a curl of zero.
 
  • #13
snoopies622 said:
A magnetic field is conservative? I thought a conservative field had to have a curl of zero.

mmm … why did I write that? :redface:

ignore the word "conservative"!​
 
  • #14
So I guess the only remaining question for me is - are there any forces of nature besides magnetism that take the form F = a (v x b) where a is a scalar, v is the velocity of the particle subject to the force, and b is another vector?

Does anyone out there know of any?
 

Related to Cross Product Force: Exploring Magnetism's Role

1. What is a cross product force?

A cross product force is a type of force that occurs when two magnetic fields interact with each other at a perpendicular angle. This results in a force being exerted on a charged particle, causing it to move in a circular path.

2. How does magnetism play a role in the cross product force?

Magnetism plays a crucial role in the cross product force because it is the magnetic fields of two objects that interact and create the force. The strength and orientation of the magnetic fields determine the magnitude and direction of the force.

3. What are some real-world applications of the cross product force?

The cross product force has many practical applications, such as in electric motors and generators, particle accelerators, and magnetic levitation trains. It is also essential in understanding the behavior of charged particles in space and the Earth's magnetic field.

4. How is the cross product force calculated?

The cross product force can be calculated using the formula F = qvBsinθ, where q is the charge of the particle, v is its velocity, B is the magnetic field strength, and θ is the angle between the velocity and magnetic field vectors. This formula is known as the Lorentz force equation.

5. What happens if the angle between the velocity and magnetic field vectors is 0 degrees?

If the angle between the velocity and magnetic field vectors is 0 degrees, the cross product force will be 0, and the particle will not experience any force. This is because the sine of 0 degrees is 0, and any number multiplied by 0 is 0. Therefore, the particle will continue to move in a straight line without being affected by the magnetic fields.

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