The smallest particle?

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vandivx

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Odd little things these photons are... <br />Posted by derekmcd</DIV><br /><br />I believe QMechanical particle-wave duality means that any talk about the&nbsp;size of particles lacks sufficent grounds in reality, it is in the same category as the picture of electorns orbiting atomic nucleus ala the fashion of planets orbiting the Sun, nature abhors repeating itself on different scales</p><p>I understand experiments were made to determine the size of electron and that it was found that electron is nebulous thing, they couldn't find anything there right down to the center of electron <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-surprised.gif" border="0" alt="Surprised" title="Surprised" width="18" height="18" />&nbsp;in some scattering experiments</p><p>&nbsp;photons were never examined 'in transit' but are just detected during&nbsp;their emission/absorbption process via the effects&nbsp;they have on matter, in short they are very elusive&nbsp;drivers on the spatial highways&nbsp;that you would have hard time ticketing for speed violations if they should happen to make&nbsp;any never mind sizing them up and so their size is meaningless</p><p>&nbsp;</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Yep.&nbsp; I give much respect to anyone who can impart enough energy in a stream of protons to make it have the same impact as a 400ton train traveling at 150km/h.&nbsp; That is some serious juice. <br /> Posted by Tsurugi</DIV></p><p>No joke there.&nbsp; I've heard they have to shut down the experiments during the winter time so folks in the area can have enough power to heat their homes.&nbsp;</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Yep.&nbsp; I give much respect to anyone who can impart enough energy in a stream of protons to make it have the same impact as a 400ton train traveling at 150km/h.&nbsp; That is some serious juice. <br />Posted by Tsurugi</DIV></p><p>The energy level that I see published for the LHC is 14Tev or 14 x 10^12 eV.&nbsp; Now 1 eV is 1.602 x 10^-19 J so 14 TeV is 2.242 x 10^-6 J.&nbsp; A 400 metric ton train traveling at 150 km/h has a kinetic energy of 3.472x 10^8&nbsp;J which is about 1.548x 10 ^14 times the mass of an individual particle at 14 Tev.&nbsp; So to match the energy of the train with 14 TeV protons you will need about 155 trillion protons.&nbsp; One might consider that more than simply a "stream" of protons.</p><p>That is still a lot of energy in an elementary particle.&nbsp; A one pound baseball traveling at 60 mph has a kinetic energy of about 164 J.&nbsp; So to that match that would only take about 67.8 million protons.&nbsp;A 10 lb infant crawling across the floor at 3 ft/s has a kinetic energy of about 1.9 J, equivalent to only 784,000 14 Tev protons, which is getting closer to a stream.&nbsp; And a 1 gram insect flying at 1 m/s has a kinetic energy of 5 x 10^-4 J and could be matched by 207 protons at 14 TeV each, and that would certainly qualify as a stream in my book.<br /></p> <div class="Discussion_UserSignature"> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>The energy level that I see published for the LHC is 14Tev or 14 x 10^12 eV.&nbsp; Now 1 eV is 1.602 x 10^-19 J so 14 TeV is 2.242 x 10^-6 J.&nbsp; A 400 metric ton train traveling at 150 km/h has a kinetic energy of 3.472x 10^8&nbsp;J which is about 1.548x 10 ^14 times the mass of an individual particle at 14 Tev.&nbsp; So to match the energy of the train with 14 TeV protons you will need about 155 trillion protons.&nbsp; One might consider that more than simply a "stream" of protons.That is still a lot of energy in an elementary particle.&nbsp; A one pound baseball traveling at 60 mph has a kinetic energy of about 164 J.&nbsp; So to that match that would only take about 67.8 million protons.&nbsp;A 10 lb infant crawling across the floor at 3 ft/s has a kinetic energy of about 1.9 J, equivalent to only 784,000 14 Tev protons, which is getting closer to a stream.&nbsp; And a 1 gram insect flying at 1 m/s has a kinetic energy of 5 x 10^-4 J and could be matched by 207 protons at 14 TeV each, and that would certainly qualify as a stream in my book. <br /> Posted by DrRocket</DIV></p><p>I think the train analogy comes from the wiki article:</p><p>http://en.wikipedia.org/wiki/Large_Hadron_Collider</p><p><em>"The size of the LHC constitutes an exceptional engineering challenge with unique safety issues. While running, the total energy stored in the magnets is <span style="white-space:nowrap">10 <span class="mw-redirect">GJ</span></span>, while each of the two beams carries an overall energy of <span style="white-space:nowrap">362 MJ</span>. For comparison, <span style="white-space:nowrap">362 MJ</span> is the kinetic energy of a TGV running at <span style="white-space:nowrap">157 <span class="mw-redirect">km/h</span></span> (<span style="white-space:nowrap">98 mph</span>), while <span style="white-space:nowrap">724 MJ</span>, the total energy of the two beams, is equivalent to the detonation energy of approximately 173&nbsp;kilograms (380&nbsp;lb) of TNT, and <span style="white-space:nowrap">10 GJ</span> is about <span style="white-space:nowrap">2.4 tons of TNT</span>. Loss of only 10<sup>&minus;7</sup> of the beam is sufficient to quench a superconducting magnet, while the beam dump must absorb an energy equivalent to a <span class="mw-redirect">typical air-dropped bomb</span>." </em></p><p>&nbsp;</p><p>Which seems to fits perfectly with your result of 347 MJ at 150km.&nbsp; Considering the two beams are 7 TeV each, the stream of photons would be 77.5 trillion photons each which would be a stream, if laid side by side, of only about 5 cm.&nbsp; Seems like a reasonable 'beam' to me.&nbsp; The individual proton might have the energy of a small bug (impressive by itself), but the full impact of the beam is like a train.</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I think the train analogy comes from the wiki article:http://en.wikipedia.org/wiki/Large_Hadron_Collider&quot;The size of the LHC constitutes an exceptional engineering challenge with unique safety issues. While running, the total energy stored in the magnets is 10 GJ, while each of the two beams carries an overall energy of 362 MJ. For comparison, 362 MJ is the kinetic energy of a TGV running at 157 km/h (98 mph), while 724 MJ, the total energy of the two beams, is equivalent to the detonation energy of approximately 173&nbsp;kilograms (380&nbsp;lb) of TNT, and 10 GJ is about 2.4 tons of TNT. Loss of only 10&minus;7 of the beam is sufficient to quench a superconducting magnet, while the beam dump must absorb an energy equivalent to a typical air-dropped bomb." &nbsp;Which seems to fits perfectly with your result of 347 MJ at 150km.&nbsp; Considering the two beams are 7 TeV each, the stream of photons would be 77.5 trillion photons each which would be a stream, if laid side by side, of only about 5 cm.&nbsp; Seems like a reasonable 'beam' to me.&nbsp; The individual proton might have the energy of a small bug (impressive by itself), but the full impact of the beam is like a train. <br />Posted by derekmcd</DIV></p><p>I'm not sure what you are using for the size of the proton, but there is an expression for the radius of a nucleus with an atomic mass A.&nbsp; That expression is</p><p>R = R0 * A^(1/3)</p><p>where R0 is 1.3 * 10^-15 m</p><p>So if you take A = 1, you might call R0 the radius of a proton.&nbsp; If you somehow managed to lay them side by side the length of the string would be about 20 cm, but the string would be kinda thin, and the beads would be repelling one another like crazy.</p><p>Now, if you take all 77.5 * 10^12 protons and put them in a ball (the nucleus of an element of atomic number 77.5 trillion) you get a nucleus of radius 5.54 * 10^-11 m.&nbsp; I'm pretty sure that it is unstable.&nbsp;<img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-wink.gif" border="0" alt="Wink" title="Wink" /> If you concentrated the charge in two points separated by the "nuclear diameter" the repulsive force between the two charges would be 3.13*10^9 N or 7.04*10^8 lb,&nbsp; That is why it takes the&nbsp;strong force to keep a nucleus together.&nbsp; But in this case it woud be kinda hopeless, since we are hypothesizing a really big nucleus.&nbsp; <br /></p> <div class="Discussion_UserSignature"> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I'm not sure what you are using for the size of the proton, but there is an expression for the radius of a nucleus with an atomic mass A.&nbsp; That expression isR = R0 * A^(1/3)where R0 is 1.3 * 10^-15 mSo if you take A = 1, you might call R0 the radius of a proton.&nbsp; If you somehow managed to lay them side by side the length of the string would be about 20 cm, but the string would be kinda thin, and the beads would be repelling one another like crazy.Now, if you take all 77.5 * 10^12 protons and put them in a ball (the nucleus of an element of atomic number 77.5 trillion) you get a nucleus of radius 5.54 * 10^-11 m.&nbsp; I'm pretty sure that it is unstable.&nbsp; If you concentrated the charge in two points separated by the "nuclear diameter" the repulsive force between the two charges would be 3.13*10^9 N or 7.04*10^8 lb,&nbsp; That is why it takes the&nbsp;strong force to keep a nucleus together.&nbsp; But in this case it woud be kinda hopeless, since we are hypothesizing a really big nucleus.&nbsp; <br /> Posted by DrRocket</DIV></p><p> LOL... atomic number of 77.5 trillion.&nbsp; I got a good chuckle out of that. </p><p>Avoiding Wiki to get a good measure, I come up with:</p><p>http://scienceworld.wolfram.com/physics/Proton.html</p><p>http://arxiv.org/abs/hep-ph/9712347 (table on first page)</p><p>Both agree with a diamter of ~1.6 * 10^-15.&nbsp; Round up 155 trillion protons to 1.6 * 10^14 protrons.&nbsp; 1/10th of a meter or 10 cm.&nbsp; Two beams at 5 cm.&nbsp; </p><p>I'm not quite sure how they set up their 'beams', but I'm sure they aren't lined up like I suggested. In the wiki article, they mention they are set up in 2,808 bunches... no clue what that means, but they aren't really beams or anything resembling a nucleus.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'> LOL... atomic number of 77.5 trillion.&nbsp; I got a good chuckle out of that. Avoiding Wiki to get a good measure, I come up with:http://scienceworld.wolfram.com/physics/Proton.htmlhttp://arxiv.org/abs/hep-ph/9712347 (table on first page)Both agree with a diamter of ~1.6 * 10^-15.&nbsp; Round up 155 trillion protons to 1.6 * 10^14 protrons.&nbsp; 1/10th of a meter or 10 cm.&nbsp; Two beams at 5 cm.&nbsp; I'm not quite sure how they set up their 'beams', but I'm sure they aren't lined up like I suggested. In the wiki article, they mention they are set up in 2,808 bunches... no clue what that means, but they aren't really beams or anything resembling a nucleus.&nbsp; <br />Posted by derekmcd</DIV></p><p>Is that 1.6E-15 a diameter or radius?&nbsp; My book shows a radius.&nbsp; That is how I got to 20 cm.</p><p>I know they aren't beams or a nucleus, but those geometries are things that are easy to calculate with.&nbsp; And we are, after all, being a bit silly here, deliberately.</p><p>Wolfram is probably an OK reference, but Wiki usually is too.&nbsp; I don't totally trust Wolfram either, particularly Stephen Wolfram himself.&nbsp; He has some strange ideas about science, at least from my perspective.&nbsp; His book on science was reviewed a while back in the AMS transactions, and the review was pretty scathing.&nbsp; I don't recall who the reviewer was, but I do recall that he has earned some respect.&nbsp; It was certainly enough to prevent me from buying the book (it is fairly expensive).</p><p>Edit:&nbsp; Just checked the books sellers and the book is no longer all that expensive.&nbsp; There are plenty of new and used copies for sale.&nbsp; Apparently it did not sell that well when originally released.</p> <div class="Discussion_UserSignature"> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Is that 1.6E-15 a diameter or radius?<br /> Posted by DrRocket</DIV></p><p>Everything I could find is always at or a bit above a diameter of 1.6 fm.&nbsp; I can't seem to find a clearly defined figure, but I wonder if your source is quoting the 'charge' radius which appears to be more variable and less accurate.</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Everything I could find is always at or a bit above a diameter of 1.6 fm.&nbsp; I can't seem to find a clearly defined figure, but I wonder if your source is quoting the 'charge' radius which appears to be more variable and less accurate. <br />Posted by derekmcd</DIV></p><p>What I find in an expression for the radius of the nucleus based on a determination of binding energy resulting from coulomb energy of the protons in the nucleus assuming they are uniformly distributed over a sphere of radius R with the equation being</p><p>R = R0 * A^(1/3) where R0 =&nbsp;1.2&nbsp;* 10 ^ -15&nbsp; to 1.5 * 10 ^ -15 and A is number of nucleons.</p><p>I find this equation in any of the following books (with the factor varying as noted).&nbsp; <em>Principles of Modern Physics </em>by Roberth Leighton or <em>Concepts of Modern Physics </em>by Arthur Beuser,&nbsp;or <em>Nuclear Physics </em>by Enrico Fermi.</p> <div class="Discussion_UserSignature"> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>What I find in an expression for the radius of the nucleus based on a determination of binding energy resulting from coulomb energy of the protons in the nucleus assuming they are uniformly distributed over a sphere of radius R with the equation beingR = R0 * A^(1/3) where R0 =&nbsp;1.2&nbsp;* 10 ^ -15&nbsp; to 1.5 * 10 ^ -15 and A is number of nucleons.I find this equation in any of books (with the factor varying as noted).&nbsp; Principles of Modern Physics by Roberth Leighton or Concepts of Modern Physics by Arthur Beuser,&nbsp;or Nuclear Physics by Enrico Fermi. <br /> Posted by DrRocket</DIV></p><p>There must be a difference between a free proton and a one that is bound in a nucleus.&nbsp; Maybe the gluons change the shape in the presence of the strong force between two fermions?&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>There must be a difference between a free proton and a one that is bound in a nucleus.&nbsp; Maybe the gluons change the shape in the presence of the strong force between two fermions?&nbsp; <br />Posted by derekmcd</DIV></p><p>Ought not be a big issue wiith hydrogen.<br /></p> <div class="Discussion_UserSignature"> </div>
 
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derekmcd

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<p>The value you provided is the Compton wavelength and the value I gave is the charge radius.&nbsp; The link I provide give values for a multitude of different constants in which you can find both values for the proton.&nbsp; I think NIST is a pretty reliable source.&nbsp;</p><p>Link to atomic and nuclear constants<br /></p><p>I think the difference between the two is dependent on what the context of the discussion is.&nbsp; The Compton wavelength is probably a better value when describing the behavior of particles and their interactions at a quantum level.&nbsp; The charge radius is probably better when talking about physical description and/or dimensions (whether they exist or not) of isolated particles.</p><p>My interpretation above could be completely wrong and which of the two values to use in context of this discussion (or any other discussion) is beyond my scope.&nbsp; I would <strong><em>guess</em></strong> my value is likely better for the thought experiment of lining them up side by side, but yours is better when talking about the structure of a nucleus.&nbsp; Though, I'm still not quite clear why there would actually be a difference.</p><p>Even if I'm wrong, I hope I'm at least on the right track.&nbsp; I think I learned something, but not quite sure what it is I learned. <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-laughing.gif" border="0" alt="Laughing" title="Laughing" /> </p><p>&nbsp;</p><p>[edit:&nbsp; fixed link]&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>The value you provided is the Compton wavelength and the value I gave is the charge radius.&nbsp; The link I provide give values for a multitude of different constants in which you can find both values for the proton.&nbsp; I think NIST is a pretty reliable source.&nbsp;http://physics.nist.gov/cgi-bin/cuu/Value?rp%7Csearch_for=all!I think the difference between the two is dependent on what the context of the discussion is.&nbsp; The Compton wavelength is probably a better value when describing the behavior of particles and their interactions at a quantum level.&nbsp; The charge radius is probably better when talking about physical description and/or dimensions (whether they exist or not) of isolated particles.My interpretation above could be completely wrong and which of the two values to use in context of this discussion (or any other discussion) is beyond my scope.&nbsp; I would guess my value is likely better for the thought experiment of lining them up side by side, but yours is better when talking about the structure of a nucleus.&nbsp; Though, I'm still not quite clear why there would actually be a difference.Even if I'm wrong, I hope I'm at least on the right track.&nbsp; I think I learned something, but not quite sure what it is I learned. &nbsp; <br />Posted by derekmcd</DIV></p><p>The link you provided did not work for me.&nbsp; But the values that I was using are determined from scattering experiments (referred to as Rutherford scattering experiments), I think with neutrons and electrons and are used in&nbsp;nuclear physics to describe both interactions with such particles and for calculation of things like density of&nbsp;the nucleus.&nbsp; I think it is reasonable to use them to describe the size of a nucleus in classic terms.&nbsp; Compton scattering, in my experience, is a phenomena in which a photon and a massive particle collide.&nbsp; The scattered photon has a different wavelength than the incident photon and the difference in frequencies can be determined by the mass of the scattering particle and the angle of scattering.&nbsp; The relationship is</p><p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l1 - l2 = h/(m0*c)[ 1 - cos phi )</p><p>Where l1 is the wavelength of the scattered photon, l2 is the wavelength of the incident photon, phi is the angle of scattering of the scattered photon, h is Planck's constant, m0 is the mass of the scattering particle and c is the speed of light.&nbsp; The quantity h/m0 *c is the Compton wavelength.&nbsp;</p><p>For a proton with mass 1.672 *10^-27 kg I compute the Compton wavelength as 1.322 * 10^-15 m.&nbsp; I don't know how that converts to a radius, a diameter, or even how to think about&nbsp;it in terms of size.&nbsp;But it is intriguing that&nbsp;the Compton wavelength&nbsp;is the same as the number R0 that comes from Rutherford scattering experiments, which involve scattering of charged particles under forces that behave according to Coulomb's Law.&nbsp; I can see why that value might be called a charge radius.</p><p>I'm not sure that the distinction, for our purposes, between the radius and diameter is all that important.&nbsp; I think the order of magnitude is far more important than the factor of 2, particularly when talking about something in the quantum realm for which the notion of size itself is a bit cloudy.&nbsp; For purposes of calculating the energies within nuclei it would be important.<br /></p> <div class="Discussion_UserSignature"> </div>
 
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derekmcd

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<p>I fixed the link above.</p><p>Same link here</p><p>&nbsp;</p><p>I started on thread to maybe get some clarification here:</p><p>http://www.physicsforums.com/showthread.php?t=241465</p><p>&nbsp;</p><p>I'm not questioning your values.&nbsp; I've recently read a few arxiv paper that are in complete agreement with what you are saying.&nbsp; I'm just trying to figure out the significance of each.</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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derekmcd

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<p>Ok.&nbsp; I think I finally have it figured out.&nbsp; You're right in that the compton wavelength has nothing to do with the physical dimension.&nbsp; If someone were to ask, "what is the radius of a proton", I would agree with you and cite 1.2 fm.&nbsp; </p><p>I'm still trying to figure out what exactly the significance of the rms charge radius is, but I think doubling that figure to get the diameter of the proton makes no sense.</p><p>The Wiki article has the charge radius correct in the diagram, but in the first paragraph, they have the diameter at 1.65 fm.&nbsp; In the discussion section, they simply doubled the rms charge radius to come up with the diameter.&nbsp; This is what let to my confustion thinking the charge radius what a physical dimension of the proton... it isn't.&nbsp; Obviously Wikipedia isn't bullet proof yet.&nbsp; I'm gonna do some more digging just to make sure I'm making sense and then will set up a Wiki account and go get that fixed.</p><p>Thanks for making me question myself <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-laughing.gif" border="0" alt="Laughing" title="Laughing" />.</p><p>&nbsp;</p><p>[edited out bad info concerning exchange forces]&nbsp;</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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derekmcd

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I'm still trying to work this out.&nbsp; It's obviously a bit more complicated than I had anticipated.&nbsp; Lack a formal education in 'nuculer' physics doesn't help...<br /> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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Tissa_Perera

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<p>I would say, what is the size of a naked electron. If the electric field is stripped off</p><p>what is left is I believe is a one dimensional point particle in our 3D space. Having said that,</p><p>what if there is a 4th dimension and a bulk dimension. Then what we percieve as a point particle</p><p>of zero dimensions is one end of a one dimensional string attached to our 3D world. So the answer</p><p>is, the smallest size is the very small length of a string. Of course I have worked out the formula</p><p>of the string vibrational origins of fundamental baryon mass. See cosmicdarkmatter.com&nbsp; &nbsp; &nbsp;</p>
 
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DrRocket

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Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I would say, what is the size of a naked electron. If the electric field is stripped offwhat is left is I believe is a one dimensional point particle in our 3D space. Having said that,what if there is a 4th dimension and a bulk dimension. Then what we percieve as a point particleof zero dimensions is one end of a one dimensional string attached to our 3D world. So the answeris, the smallest size is the very small length of a string. Of course I have worked out the formulaof the string vibrational origins of fundamental baryon mass. See cosmicdarkmatter.com&nbsp; &nbsp; &nbsp; <br />Posted by Tissa_Perera</DIV><br /><br />cosmiddarkmatter.com apparently does not exist.&nbsp; The rest of your post appears to be gibberish.&nbsp; Do you have a point&nbsp; ? <div class="Discussion_UserSignature"> </div>
 
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