gravity, electricity and qed theory [Archive]

alistair

Sep4-04, 03:05 AM

Jerzy Karczmarczuk

Sep6-04, 05:44 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>\nalistair wrote:\n> There are two protons in the gravitational field of the Earth, one\n> directly above the other.They experience the electric force through\n> exchange of virtual photons.A virtual photon travelling vertically\n> upwards from the lower proton towards the higher proton would be\n> redshifted and so exert a smaller force on the higher proton than it\n> would if there was no gravitational field present.\n> A virtual photon travelling vertically downwards towards the lower\n> proton would be blueshifted and exert a bigger force on the lower\n> proton than it would if there was no gravitational field present.\n> Has gravity given the higher proton a bigger effective charge (since\n> the magnitude of a charge can be determined by the force it exerts on\n> other charges)\n> and the lower proton a smaller effective charge?The sum of the charges\n> would still be the same so we could say that charge is conserved\n> still.\n\nWhile I can\'t say whether the gravitation may or may not *ultimately*\ninfluence the EM interaction (veeeery weakly anyway), the model above\nis plainly inadequate. Macroscopic forces due to the exchange of\nvirtual photons "use" the wee\'est photos possible, with infinite wave-\nlength, this is static. You can\'t really red-shift such photons...\n\nIn close, dynamic interaction between two particles, a virtual photon\nresponsible for the momentum transfer between interacting charges is\noff-shell. It has no \'real\' mass, an \'imaginary\' one if you want to\nabuse the language a bit. The energy is not necessarily positive.\nThere isn\'t much sense in "red-shifting" of such photons either.\n\n==\n\nAnyway, I don\'t see much sense in the statement that a red-shifted\nphoton will "exert a bigger/smaller force" on its target. Why so?\n\nJerzy Karczmarczuk\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>alistair wrote:
> There are two protons in the gravitational field of the Earth, one
> directly above the other.They experience the electric force through
> exchange of virtual photons.A virtual photon travelling vertically
> upwards from the lower proton towards the higher proton would be
> redshifted and so exert a smaller force on the higher proton than it
> would if there was no gravitational field present.
> A virtual photon travelling vertically downwards towards the lower
> proton would be blueshifted and exert a bigger force on the lower
> proton than it would if there was no gravitational field present.
> Has gravity given the higher proton a bigger effective charge (since
> the magnitude of a charge can be determined by the force it exerts on
> other charges)
> and the lower proton a smaller effective charge?The sum of the charges
> would still be the same so we could say that charge is conserved
> still.

While I can't say whether the gravitation may or may not *ultimately*
influence the EM interaction (veeeery weakly anyway), the model above
is plainly inadequate. Macroscopic forces due to the exchange of
virtual photons "use" the wee'est photos possible, with infinite wave-
length, this is static. You can't really red-shift such photons...

In close, dynamic interaction between two particles, a virtual photon
responsible for the momentum transfer between interacting charges is
off-shell. It has no 'real' mass, an 'imaginary' one if you want to
abuse the language a bit. The energy is not necessarily positive.
There isn't much sense in "red-shifting" of such photons either.

==

Anyway, I don't see much sense in the statement that a red-shifted
photon will "exert a bigger/smaller force" on its target. Why so?

Jerzy Karczmarczuk