michaelm@bcsaic.UUCP (michael b maxwell) (10/08/85)
Since the funny gravity comments have died down, I decided to inject
something more grave...
First, the disclaimer: I'm not a physicist, and my knowledge of physics
is limited to three semesters in college and what I've read in places
like Scientific American. So-
My question concerns the QED explanation of interactions between charged
particles. In the articles I've read that attempt to explain this to
laymen like me, they often include a Feynman diagram of a scattering
interaction between two electrons, like the following:
e- e-
\ /
\ /
\ /
\ /
\ /
|~~~~~~~~~~~~~~~~|
/ \
/ \
/ \
/ \
/ \
/ \
The explanation runs something like the following. The two electrons
are travelling along, exchange a virtual photon, and are scattered. The
photon therefore acts as the carrier of the electromagnetic force
between the two charged particles, and there is no action at a distance.
Voila.
When I first read this, I tried to expand on the explanation in my mind.
My reasoning went somewhat as follows. The electrons are constantly
emitting and re-absorbing virtual photons (as well as other virtual
particles; e- and e+; anything else?). Each virtual photon carries with it a
bit of momentum away from its e- parent; if it is absorbed by the other
e-, it transfers that momentum to that e-. The time it can "remain" (exist?)
apart from its e- source is dependent on the uncertainty principle--the
more energetic the photon, the larger its equivalent momentum, hence the
shorter time it is "allowed" to rob that momentum from the parent e-
before it is absorbed by the other e- or re-absorbed by its parent.
So (I thought), there is an explanation for why the strength of the
interaction decreases with distance: energetic photons can't "last" long
enough to get very far from their parent.
I then considered the direction of the interaction; the two e-s repel
each other. Simple enough, I thought; the photon has momentum in the
direction of its travel, which is away from its parent (to the right,
say, in the above diagram); so by conservation of momentum, the parent e-
moves in the opposite direction (to the left, in this case).
And when the photon is absorbed by the other e-, it gains the momentum
of the photon in the direction of the photon's travel, i.e. away from
the other e- (to the right in this example). Hence the net result is that
the two e-s are repelled, as observed.
Then came the crunch. How do you explain attractive forces? I.e. what
happens when you replace one of the e-s with a positive particle? It
seems to me that there ought to be an attactive force, but I can't figure
how exchanging a photon can result in a force of attraction! Putting it
naively, when a virtual photon arrives at an e-, how does it "know"
whether it came from an e- or an e+?
I've asked this question of several people who know more physics than I
do, and I'm not satisfied with (or maybe I don't understand) their
replies. Some samples:
(1) This is the wrong way to visualize forces between two particles;
rather like trying to explain the results of a slit experiment with a
particle model rather than a wave model. Along these same lines is the
explanation that scattering is always a repulsive phenomenon: shoot an
electron towards positively charged nuclei, and they are always
scattered away from the nucleus. My reaction: then what does
the exchange of photons have to do with em forces between particles?
Furthermore, what does the exchange of mesons have to do with the
attractive forces between nucleons? (In this case, I'm even more
puzzled, since if I understand rightly, mesons *always* induce an
attractive force.)
(2) A variation of (1): When you have two charged particles attracting
each other, a standing wave of virtual (?) photons is set up between the
particles. Interactions by means of standing waves can't be described
by the simple model of exchange of virutal photons.
If (2) is right, then is scattering always a matter of repulsion? If
so, is there some way of visualizing how the standing waves set up a
force of attraction between two differently charged particles in a non-
scattering environment? I.e. what is the difference between the standing
waves set up between differently charged particles, and the waves between
particles with the same charge?
What am I missing? Is there something "funny" about talking about the
momentum of photons? I've assumed that a photon has momentum in the
direction of its travel; is this wrong? Or do virtual photons not have
a direction of travel, in some sense?
--
Mike Maxwell
Boeing Artificial Intelligence Center
..uw-beaver!{uw-june,ssc-vax}!bcsaic!michaelmethan@utastro.UUCP (Ethan Vishniac) (10/10/85)
Virtual particles can have a momentum oriented against their
direction of motion. Is this weird? You bet.
--
"Superior firepower is an Ethan Vishniac
important asset when {charm,ut-sally,ut-ngp,noao}!utastro!ethan
entering into ethan@astro.UTEXAS.EDU
negotiations" Department of Astronomy
University of Texasjhc@mtung.UUCP (Jonathan Clark) (10/10/85)
<> Surely you are mixing up 'real' and 'virtual' photons. In the case where the two charged particles act upon (repel or attract) each other a 'real' photon is involved. Hence energy and momentum transfer are allowed. 'Virtual' photons are a mathematical tool (so what are 'real' photons, right?) and only 'exist' for very short periods of time (too short for God to notice). It is not meaningful to ascribe any properties to them, as they are not really there (by definition). They are used to describe a mechanism which explains things like pair production and particle decay (sometimes the 'virtual' ones turn 'real'). Apologies for the extremely sloppy use of sundry terms in this posting. I'm not trying to be rigorous, just to enlighten the original poster a bit. Terms which should be read with a pinch of salt are surrounded with apostrophes. -- Jonathan Clark [NAC]!mtung!jhc Everyone was frightened of Doug. Even Dinsdale was frightened of Doug. 'E used.... sarcasm.
dmcanzi@watdcsu.UUCP (David Canzi) (10/11/85)
I've always been annoyed at the fact that they *never* try to show
attraction by particle exchange. I've thought the matter over, and
come up with a possible explanation of how such a thing can happen.
Disclaimer: This is a half-educated guess, subject to correction by
anybody who really understands QM.
The diagram:
e- e+
/ \
/ \
<--* *<--
| |
| |
The explanation:
The electron on the left emits the virtual photon leftward. The virtual
photon's momentum is fairly well-defined, so, by the uncertainty principle,
its position isn't. As a result, it can be absorbed from the right by
the positron on the right.
--
David Canzi
There are too many thick books about thin subjects.friesen@psivax.UUCP (Stanley Friesen) (10/11/85)
In article <326@bcsaic.UUCP> michaelm@bcsaic.UUCP (michael b maxwell) writes: > >Then came the crunch. How do you explain attractive forces? I.e. what >happens when you replace one of the e-s with a positive particle? It >seems to me that there ought to be an attactive force, but I can't figure >how exchanging a photon can result in a force of attraction! Putting it >naively, when a virtual photon arrives at an e-, how does it "know" >whether it came from an e- or an e+? > Hey, yeah, he's not the only one! How *does* this work? Why can't the physics text cover this case as well? I think the attraction of opposite charges is more interesting than scattering anyway! -- Sarima (Stanley Friesen) UUCP: {ttidca|ihnp4|sdcrdcf|quad1|nrcvax|bellcore|logico}!psivax!friesen ARPA: ttidca!psivax!friesen@rand-unix.arpa
rimey@ucbernie.BERKELEY.EDU (Ken &) (10/13/85)
>Surely you are mixing up 'real' and 'virtual' photons. In >the case where the two charged particles act upon (repel or >attract) each other a 'real' photon is involved. ... > >Jonathan Clark >[NAC]!mtung!jhc No, the force between two charged particles does not involve real photons. Hold two identically charged pith balls near each other. You will not detect any light or radio waves between them. You say that virtual particles exist only for short times, and therefore shouldn't be able to travel large distances. Indeed, this is the reason that the strong nuclear force has a finite range. But moving clocks run slow, and clocks on photons don't run at all. This is why the force between charged particles can be felt even at large distances. Ken Rimey rimey@dali.berkeley.edu ucbvax!dali!rimey
michaelm@bcsaic.UUCP (michael b maxwell) (10/16/85)
In article <36@utastro.UUCP> ethan@utastro.UUCP (Ethan Vishniac) writes: >Virtual particles can have a momentum oriented against their >direction of motion. Is this weird? You bet. Are you *serious*??? Assuming you are, what determines the direction of the momentum of the virtual particle? It seems that in order to explain attractive and repulsive forces between all four combinations of electric charges (i.e. {(--), (-+), (+-), (++)}), the direction of the momentum must be determined by the charge of both the particle emitting the virtual photon, and the particle receiving the virtual photon. Is this right? Can you explain? Similar facts would seem to apply to forces mediated by other particles, too--e.g. the strong nuclear force. BTW, is the nuclear force--or rather, the force between two quarks--ever repulsive? e.g. when the two quarks are the same color? -- Mike Maxwell Boeing Artificial Intelligence Center ..uw-beaver!{uw-june,ssc-vax}!bcsaic!michaelm