{"id":118,"date":"2006-08-26T00:13:00","date_gmt":"2006-08-26T00:13:00","guid":{"rendered":"https:\/\/scottaaronson.blog\/?p=118"},"modified":"2006-08-26T00:13:00","modified_gmt":"2006-08-26T00:13:00","slug":"why-im-not-a-physicist-reason-4328","status":"publish","type":"post","link":"https:\/\/scottaaronson.blog\/?p=118","title":{"rendered":"Why I&#8217;m not a physicist: reason #4328"},"content":{"rendered":"<p>There&#8217;s a trivial question about particle accelerators that bugged me for a while.   Today I finally figured out the answer, and I&#8217;m so excited by my doofus &#8220;discovery&#8221; that I want to tell the world.<\/p>\n<p>In Ye Olde Times, accelerators used to smash particles against a fixed target.  But today&#8217;s accelerators smash one particle moving at almost the speed of light against another particle moving at almost the speed of light &#8212; that&#8217;s why they&#8217;re called particle colliders (duhhh).  Now, you&#8217;d think this trick would increase the collision energy by a constant factor, but according to the physicists, it does asymptotically better than that: it squares the energy!<\/p>\n<p>My question was, how could that be?  Even if both particles are moving, we can clearly imagine that one of them is stationary, since the particles&#8217; motion with respect to the Earth is irrelevant.  So then what&#8217;s the physical difference between a particle hitting a fixed target and two moving particles hitting each other, that could possibly produce a quadratic improvement in energy?<\/p>\n<p>[Warning: Spoiler Ahead]<\/p>\n<p>The answer pops out if we consider the rule for adding velocities in special relativity.  If in our reference frame, particle 1 is headed left at a v fraction of the speed of light, while particle 2 is headed right at a w fraction of the speed of light, then in particle 1&#8217;s reference frame, particle 2 is headed right at a (v+w)\/(1+vw) fraction of the speed of light.  Here 1+vw is the relativistic correction, &#8220;the thing you put in to keep the fraction less than 1.&#8221;   If v and w are both close to 0, then of course we get v+w, the Newtonian answer.<\/p>\n<p>Now set v=w=1-\u03b5.  Then (v+w)\/(1+vw) = 1 &#8211; \u03b5<sup>2<\/sup>\/(2-2\u03b5+\u03b5<sup>2<\/sup>), which scales like 1-\u03b5<sup>2<\/sup>.  Aha!<\/p>\n<p>To finish the argument, remember that relativistic energy increases with speed like 1\/sqrt(1-v<sup>2<\/sup>).  If we plug in v=1-\u03b5, then we get 1\/sqrt(2\u03b5-\u03b5<sup>2<\/sup>), while if we plug in v=1-\u03b5<sup>2<\/sup>, then we get 1\/sqrt(2\u03b5<sup>2<\/sup>-\u03b5<sup>4<\/sup>).  So in the case of a fixed target the energy scales like 1\/sqrt(\u03b5), while in the case of two colliding particles it scales like 1\/\u03b5.<\/p>\n<p>In summary, nothing&#8217;s going on here except relativistic addition of velocities.  As with Grover&#8217;s algorithm, as with the quantum Zeno effect, it&#8217;s our intuition about linear versus quadratic that once again leads us astray.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>There&#8217;s a trivial question about particle accelerators that bugged me for a while. Today I finally figured out the answer, and I&#8217;m so excited by my doofus &#8220;discovery&#8221; that I want to tell the world. In Ye Olde Times, accelerators used to smash particles against a fixed target. But today&#8217;s accelerators smash one particle moving [&hellip;]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"advanced_seo_description":"","jetpack_seo_html_title":"","jetpack_seo_noindex":false,"jetpack_seo_schema_type":"","_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_hide_featured_image":false,"_jetpack_featured_image_duplicate_dismissed":0,"_jetpack_feature_clip_id":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"{title}\n\n{excerpt}\n\n{url}","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2},"_wpas_customize_per_network":false,"jetpack_post_was_ever_published":false},"categories":[18],"tags":[],"class_list":["post-118","post","type-post","status-publish","format-standard","hentry","category-embarrassing-myself"],"jetpack_publicize_connections":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/scottaaronson.blog\/index.php?rest_route=\/wp\/v2\/posts\/118","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/scottaaronson.blog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/scottaaronson.blog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/scottaaronson.blog\/index.php?rest_route=\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/scottaaronson.blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=118"}],"version-history":[{"count":0,"href":"https:\/\/scottaaronson.blog\/index.php?rest_route=\/wp\/v2\/posts\/118\/revisions"}],"wp:attachment":[{"href":"https:\/\/scottaaronson.blog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=118"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/scottaaronson.blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=118"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/scottaaronson.blog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=118"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}