## Thread "Waves hitting a beach, photons exchanged?"

<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>On 2005/02/04 Mike Helland wrote:\n&gt; There is a much better way to explain why you do not observe a photon\n&gt; exchange.\n&gt; It seems as if not many people realize the only way to observe a\n&gt; photon is if it hits your eye (and even then you\'ll need a half dozen\n&gt; or so photons in a certain amount of time to pass the noise filters\n&gt; in your optical gear).\n&gt; It seems people assume you can "see" photons whizzing by you. But\n&gt; that could only be true if other photons bring information about the\n&gt; photon to an observer. I don\'t think that\'s possible in QED. Right?\n\nI\'m sorry I didn\'t make my point clear (English is not my native\nlanguage...) I didn\'t mean "observe" as a synonim to "see", neither\nassumed those photons would be in the visible light range of the EM\nspectrum. So, will you please construe "observe" as "detect somehow"? I\nwas talking about the huge amount of photon exchange when a wave hits\na beach. So, using QED to model that would be unpractical (almost\nimpossible, I would say), as Igor Khavkine pointed out.\n\nAnother thing I want to mention:\n\nOn 2005/01/18 Mike Helland wrote:\n&gt; And a more general question, using particles as a description for any\n&gt; system, is kinetic energy always transfered through photons? If not,\n&gt; how does that work at the particle level?\n\nWell, in a Neutron Beta decay surely some kinetic energy is generated,\nand if there is any photon exchange, it should be attractive between the\nresulting proton and electron, isn\'t it so?\nAnd you should always think of gravity, which has nothing to do with\nQED.\nFinally, I want to point out that, if electron-electron repulsion\nphoton exchange was the only possible interaction between atoms, no\nmolecule whatsoever would be possible.\n\nMy best regards,\nMeia-Noite\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">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>On $2005/02/04$ Mike Helland wrote:
> There is a much better way to explain why you do not observe a photon
> exchange.
> It seems as if not many people realize the only way to observe a
> photon is if it hits your eye (and even then you'll need a half dozen
> or so photons in a certain amount of time to pass the noise filters
> It seems people assume you can "see" photons whizzing by you. But
> that could only be true if other photons bring information about the
> photon to an observer. I don't think that's possible in QED. Right?

I'm sorry I didn't make my point clear (English is not my native
language...) I didn't mean "observe" as a synonim to "see", neither
assumed those photons would be in the visible light range of the EM
spectrum. So, will you please construe "observe" as "detect somehow"? I
was talking about the huge amount of photon exchange when a wave hits
a beach. So, using QED to model that would be unpractical (almost
impossible, I would say), as Igor Khavkine pointed out.

Another thing I want to mention:

On $2005/01/18$ Mike Helland wrote:
> And a more general question, using particles as a description for any
> system, is kinetic energy always transfered through photons? If not,
> how does that work at the particle level?

Well, in a Neutron $\Beta$ decay surely some kinetic energy is generated,
and if there is any photon exchange, it should be attractive between the
resulting proton and electron, isn't it so?
And you should always think of gravity, which has nothing to do with
QED.
Finally, I want to point out that, if electron-electron repulsion
photon exchange was the only possible interaction between atoms, no
molecule whatsoever would be possible.

My best regards,
Meia-Noite

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