<html><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space; "><font class="Apple-style-span" size="5"><span class="Apple-style-span" style="font-size: 18px;">Dear Kirk, </span></font><div><font class="Apple-style-span" size="5"><span class="Apple-style-span" style="font-size: 18px;"><br></span></font></div><div><font class="Apple-style-span" size="5"><span class="Apple-style-span" style="font-size: 18px;"> If you employ (1.5.29) and (1.5.30) in my book, you will find that</span></font></div><div><font class="Apple-style-span" size="5"><span class="Apple-style-span" style="font-size: 18px;"><br></span></font></div><div><font class="Apple-style-span" size="5"><span class="Apple-style-span" style="font-size: 18px;">H=\gamma mc^2+q\psi.</span></font></div><div><font class="Apple-style-span" size="5"><span class="Apple-style-span" style="font-size: 18px;"><br></span></font></div><div><font class="Apple-style-span" size="5"><span class="Apple-style-span" style="font-size: 18px;">Then. from (1.6.5), it follows that</span></font></div><div><font class="Apple-style-span" size="5"><span class="Apple-style-span" style="font-size: 18px;"><br></span></font></div><div><font class="Apple-style-span" size="5"><span class="Apple-style-span" style="font-size: 18px;">p_t=-H=-\gamma mc^2-q\psi,</span></font></div><div><font class="Apple-style-span" size="5"><span class="Apple-style-span" style="font-size: 18px;"><br></span></font></div><div><font class="Apple-style-span" size="5"><span class="Apple-style-span" style="font-size: 18px;">in agreement with Rob's slide 16.</span></font></div><div><font class="Apple-style-span" size="5"><span class="Apple-style-span" style="font-size: 18px;"><br></span></font></div><div><font class="Apple-style-span" size="5"><span class="Apple-style-span" style="font-size: 18px;">Best,</span></font></div><div><font class="Apple-style-span" size="5"><span class="Apple-style-span" style="font-size: 18px;"><br></span></font></div><div><font class="Apple-style-span" size="5"><span class="Apple-style-span" style="font-size: 18px;">Alex<br></span></font><div> </div><div>If you </div><div><br><div><div>On Mar 11, 2011, at 8:10 PM, Kirk T McDonald wrote:</div><br class="Apple-interchange-newline"><blockquote type="cite"> <div style="PADDING-LEFT: 10px; PADDING-RIGHT: 10px; WORD-WRAP: break-word; PADDING-TOP: 15px; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space" id="MailContainerBody" leftmargin="0" topmargin="0" bgcolor="#ffffff" canvastabstop="true" name="Compose message area"> <div><font size="2" face="Arial">Rob,</font></div> <div><font size="2" face="Arial"></font> </div> <div><font size="2" face="Arial">Thanks for this comment.</font></div> <div><font size="2" face="Arial"></font> </div> <div><font size="2" face="Arial">Your slide 16 seems to imply that the canonical momentum associated with coordinate t, when using (x,y,t) as coordinates, is</font></div> <div><font size="2" face="Arial">p_t = - (E_mech + q V).</font></div> <div><font size="2" face="Arial"></font> </div> <div><font size="2" face="Arial">This does not quite match what I infer from Alex Dragt that</font></div> <div><font size="2" face="Arial"></font> </div> <div><font size="2" face="Arial">p_t = - H</font></div> <div><font size="2" face="Arial">= - { sqrt[ m^2 c^4 + (p_mech - q A / c)^2 ] + q V }</font></div> <div><font size="2" face="Arial"></font> </div> <div><font size="2" face="Arial">How did you arrive at your simplification?</font></div> <div><font size="2" face="Arial"></font> </div> <div><font size="2" face="Arial">Your result matches Dragt's if the vector potential is zero.....</font></div> <div><font size="2" face="Arial"></font> </div> <div><font size="2" face="Arial">Are you saying that we can ignore the vector potential, but not the scalar potential?</font></div> <div><font size="2" face="Arial"></font> </div> <div><font size="2" face="Arial">--Kirk</font></div> <div style="FONT: 10pt Tahoma"> <div><br></div> <div style="BACKGROUND: #f5f5f5"> <div style="font-color: black"><b>From:</b> <a title="mailto:rdryne@lbl.gov
CTRL + Click to follow link" href="mailto:rdryne@lbl.gov">Robert D Ryne</a> </div> <div><b>Sent:</b> Saturday, March 12, 2011 12:07 AM</div> <div><b>To:</b> <a title="mailto:kirkmcd@Princeton.EDU
CTRL + Click to follow link" href="mailto:kirkmcd@Princeton.EDU">Kirk T McDonald</a> </div> <div><b>Cc:</b> <a title="mailto:dragtnb@comcast.net
CTRL + Click to follow link" href="mailto:dragtnb@comcast.net">alex dragt</a> ; <a title="mailto:map-l@lists.bnl.gov
CTRL + Click to follow link" href="mailto:map-l@lists.bnl.gov">MAP List</a> </div> <div><b>Subject:</b> Re: [MAP] Use of x,y,t as Hamiltonian coordinates.</div></div></div> <div><br></div><span style="FONT-SIZE: 14px" class="Apple-style-span">Kirk,</span> <div style="FONT-SIZE: 14px"><br></div> <div style="FONT-SIZE: 14px">The 6-vector of canonical variables is shown on slide 16 of my presentation at the MAP meeting. These are the variables that should be used for eigen-emittance calculations. Of course this happens "for free" in a code that uses canonical variables. When I calculate eigen-emittances from a non-canonical code, the diagnostic subroutine does the conversion to canonical variables.</div> <div style="FONT-SIZE: 14px"><br></div> <div style="FONT-SIZE: 14px">Rob</div> <div style="FONT-SIZE: 16px"><span style="FONT-SIZE: medium" class="Apple-style-span"><font class="Apple-style-span" size="4"><span style="FONT-SIZE: 16px" class="Apple-style-span"><br></span></font></span></div> <div><br></div></div></blockquote></div><br></div></div></body></html>