Angular momentum and magnetic monopoles [Archive]

Gerard Westendorp

Sep29-04, 02:25 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nI was thinking about magnetic monopoles, and I thought there\nmight be some trouble with angular momentum.\n\nIf you calculate the angular momentum density rXEXB for\na point charge (e) at (0,0,d/2) and a magnetic monopole\n(m) at (0,0,-d/2), you get for the z-component something like:\n\n\nLz = e*m*d*R^2 /(r_e^3 r_m^3)\n\nwhere R = (x^2+y^2)^(1/2)\nand r_e is the distance from the e_pole, and r_m is the distance from\nthe m_pole.\n\nIf the 2 poles were fired at each other along the z-axis, suppose\nthey would miss each other and continue along the z-axis, then\nthe total field angular momentum would undergo changes as this\nhappens.\n\nA striking feature would be a sign flip as (d) goes from positive\nto negative.\n\nIs this a problem?\n\nThe poles e and m could be spinless particles, so where does\nthe angular momentum go/ come from?\n\nGerard\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>I was thinking about magnetic monopoles, and I thought there
might be some trouble with angular momentum.

If you calculate the angular momentum density rXEXB for
a point charge (e) at (0,0,d/2) and a magnetic monopole
(m) at (0,0,-d/2), you get for the z-component something like:

Lz = e*m*d*R^2 /(r_e^3 r_m^3)

where R = (x^2+y^2)^(1/2)
and r_e is the distance from the e_{pole}, and r_m is the distance from
the m_{pole}.

If the 2 poles were fired at each other along the z-axis, suppose
they would miss each other and continue along the z-axis, then
the total field angular momentum would undergo changes as this
happens.

A striking feature would be a sign flip as (d) goes from positive
to negative.

Is this a problem?

The poles e and m could be spinless particles, so where does
the angular momentum go/ come from?

Gerard

Douglas Natelson

Sep29-04, 12:18 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nHello -\n\nGerard Westendorp wrote:\n> I was thinking about magnetic monopoles, and I thought there\n> might be some trouble with angular momentum.\n[snip]\n> If the 2 poles were fired at each other along the z-axis, suppose\n> they would miss each other and continue along the z-axis, then\n> the total field angular momentum would undergo changes as this\n> happens.\n>\n> A striking feature would be a sign flip as (d) goes from positive\n> to negative.\n>\n> Is this a problem?\n>\n> The poles e and m could be spinless particles, so where does\n> the angular momentum go/ come from?\n\nCaveat: I haven\'t really thought this through. However,\nis it reasonable kinematically to think that the charge and\nmagnetic monopoles will fly straight past each other along\nthe z-axis? i.e. In a frame co-moving with the charged\nparticle, the magnetic field from the monopole will be\ntransformed into some electric field that will push on\nthe charge, and vice versa. Presumably, when the final\npost-pass "orbital" angular momentum around the z-axis is\ncalculated, the "orbital" plus "field" total will be\nconserved....\n\nToo lazy to do the real calculation,\nDoug\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hello -

Gerard Westendorp wrote:
> I was thinking about magnetic monopoles, and I thought there
> might be some trouble with angular momentum.
[snip]
> If the 2 poles were fired at each other along the z-axis, suppose
> they would miss each other and continue along the z-axis, then
> the total field angular momentum would undergo changes as this
> happens.
>
> A striking feature would be a sign flip as (d) goes from positive
> to negative.
>
> Is this a problem?
>
> The poles e and m could be spinless particles, so where does
> the angular momentum go/ come from?

Caveat: I haven't really thought this through. However,
is it reasonable kinematically to think that the charge and
magnetic monopoles will fly straight past each other along
the z-axis? i.e. In a frame co-moving with the charged
particle, the magnetic field from the monopole will be
transformed into some electric field that will push on
the charge, and vice versa. Presumably, when the final
post-pass "orbital" angular momentum around the z-axis is
calculated, the "orbital" plus "field" total will be
conserved....

Too lazy to do the real calculation,
Doug

Gerard Westendorp

Oct26-04, 12:55 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Douglas Natelson wrote:\n\n[..]\n\n\n> Caveat: I haven\'t really thought this through. However,\n> is it reasonable kinematically to think that the charge and\n> magnetic monopoles will fly straight past each other along\n> the z-axis? i.e. In a frame co-moving with the charged\n> particle, the magnetic field from the monopole will be\n> transformed into some electric field that will push on\n> the charge, and vice versa. Presumably, when the final\n> post-pass "orbital" angular momentum around the z-axis is\n> calculated, the "orbital" plus "field" total will be\n> conserved....\n\n\nHmm, maybe they could orbit each other... interesting.\n\nBut then, why should there be a force between the 2 poles?\n\nA moving electric charge feels a force from the magnetic\nfield proportional to its speed. But the speed (v) could be\nslow, and because the forces are then small, remain slow.\n\nhmm, OK, slow speed means a long time for the force to act,\nand\nForce*time = change in momentum\n~ e v B s/v ~ e B s ~ independent of speed (v)\n\nOK, yeh.\n\nThinking about it, the 2 poles will not orbit each\nother. This is clear if you think about the fields,\nthe velocities and the forces using the "right hand\nrule" ( or was it the left hand?). What will happen:\nSuppose the particles move toward each other along\nthe z-axis, but very slightly offset along the y-axis\nso that they narrowly miss each other. The the\nLorentz force on the e-pole, and its magnetic analog\non the m-pole will be in the direction. This will\nproduce a momentum change in the z-direction.\n\n\nI agree that to do the actual calculation would be\na bit too laborious. But I am starting to believe\nthat the solution is that the 2 poles will do something like:\n\n^ ^\n/ |\nee X\nee\n-> eeeeeeeee mmmmmm <- Z ->\nmm\n\nmm\n\n/\nV\n\nThey need a very slight offset to initiate this, but the\nsmaller the offset, the bigger the force, because of the 1/r^2\nCoulomb law.\n\nGerard\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Douglas Natelson wrote:

[..]

> Caveat: I haven't really thought this through. However,
> is it reasonable kinematically to think that the charge and
> magnetic monopoles will fly straight past each other along
> the z-axis? i.e. In a frame co-moving with the charged
> particle, the magnetic field from the monopole will be
> transformed into some electric field that will push on
> the charge, and vice versa. Presumably, when the final
> post-pass "orbital" angular momentum around the z-axis is
> calculated, the "orbital" plus "field" total will be
> conserved....

Hmm, maybe they could orbit each other... interesting.

But then, why should there be a force between the 2 poles?

A moving electric charge feels a force from the magnetic
field proportional to its speed. But the speed (v) could be
slow, and because the forces are then small, remain slow.

hmm, OK, slow speed means a long time for the force to act,
and
Force*time = change in momentum
~ e v B s/v ~ e B s ~ independent of speed (v)

OK, yeh.

Thinking about it, the 2 poles will not orbit each
other. This is clear if you think about the fields,
the velocities and the forces using the "right hand
rule" ( or was it the left hand?). What will happen:
Suppose the particles move toward each other along
the z-axis, but very slightly offset along the y-axis
so that they narrowly miss each other. The the
Lorentz force on the e-pole, and its magnetic analog
on the m-pole will be in the direction. This will
produce a momentum change in the z-direction.

I agree that to do the actual calculation would be
a bit too laborious. But I am starting to believe
that the solution is that the 2 poles will do something like:

^ ^/ |
ee X
ee
-> eeeeeeeee mmmmmm <- Z ->
mm

mm

/
V

They need a very slight offset to initiate this, but the
smaller the offset, the bigger the force, because of the 1/r^2
Coulomb law.

Gerard