{"id":91,"date":"2025-03-17T00:07:08","date_gmt":"2025-03-17T00:07:08","guid":{"rendered":"https:\/\/freepath.info\/?p=91"},"modified":"2025-03-17T00:44:04","modified_gmt":"2025-03-17T00:44:04","slug":"mean-free-path-equation-from-markov-chain-in-a-uniform-medium","status":"publish","type":"post","link":"https:\/\/freepath.info\/?p=91","title":{"rendered":"Markov Chain Derivation of the Mean Free Path"},"content":{"rendered":"

<\/h2>\n

1. Memoryless (Markov) Property<\/h3>\n

A fundamental assumption in deriving the mean free path<\/strong> is the memoryless<\/strong> property. Specifically, let \"X\" be the random variable for the distance traveled by a particle before it collides. The Markov (or memoryless) property<\/strong> states:<\/p>\n

 <\/p>\n

P<\/mi>(<\/mo>X<\/mi>><\/mo>x<\/mi>+<\/mo>\u0394<\/mi>x<\/mi>\u2009<\/mtext>\u2223<\/mo>\u2009<\/mtext>X<\/mi>><\/mo>x<\/mi>)<\/mo>\u2005\u200a<\/mtext>=<\/mo>\u2005\u200a<\/mtext>P<\/mi>(<\/mo>X<\/mi>><\/mo>\u0394<\/mi>x<\/mi>)<\/mo>,<\/mo><\/mspace>\u2200<\/mi>\u2009<\/mtext>x<\/mi>,<\/mo>\u2009<\/mtext>\u0394<\/mi>x<\/mi>\u2009<\/mtext>\u2265<\/mo>0.<\/mn><\/mrow>P\\bigl(X > x + \\Delta x \\,\\big\\vert\\, X > x\\bigr) \\;=\\; P\\bigl(X > \\Delta x\\bigr), \\quad \\forall\\, x, \\,\\Delta x \\,\\ge 0.<\/annotation><\/semantics><\/math><\/p>\n

 <\/p>\n

In words, \u201cthe probability of traveling an additional distance \"\Delta without collision, given that the particle has already traveled \"x\" without collision, does not depend on \"x\".\u201d This memoryless trait is the defining feature of the exponential distribution<\/strong>.<\/p>\n

\n

2. Collision Rate and Exponential Distribution<\/h3>\n

Suppose the medium has:<\/p>\n

    \n
  • Number density<\/strong> of scatterers: \"n\" (scatterers per unit volume),<\/li>\n
  • Effective cross-section<\/strong>: \"\sigma\".<\/li>\n<\/ul>\n

    Then, in a small distance \"\mathrm{d}x\", the probability of collision can be approximated as:<\/p>\n

     <\/p>\n

    Prob<\/mtext>(<\/mo>collision\u00a0in\u00a0<\/mtext>d<\/mi>x<\/mi>)<\/mo>\u2005\u200a<\/mtext>=<\/mo>\u2005\u200a<\/mtext>n<\/mi>\u2009<\/mtext>\u03c3<\/mi>\u2009<\/mtext>d<\/mi>x<\/mi>,<\/mo><\/mrow>\\text{Prob}(\\text{collision in } \\mathrm{d}x) \\;=\\; n\\,\\sigma \\,\\mathrm{d}x,<\/annotation><\/semantics><\/math><\/p>\n

     <\/p>\n

    provided \"n,\sigma,\mathrm{d}x. Consequently, the probability of not<\/em> colliding in that interval is \"1.<\/p>\n

    Let \"S(x) be the survival function<\/strong>, i.e., the probability that the particle has traveled farther than \"x\" without colliding. Because collisions are assumed memoryless,<\/p>\n

     <\/p>\n

    S<\/mi>(<\/mo>x<\/mi>+<\/mo>d<\/mi>x<\/mi>)<\/mo>\u2005\u200a<\/mtext>=<\/mo>\u2005\u200a<\/mtext>S<\/mi>(<\/mo>x<\/mi>)<\/mo>\u2009<\/mtext>[<\/mo>1<\/mn>\u2212<\/mo>n<\/mi>\u2009<\/mtext>\u03c3<\/mi>\u2009<\/mtext>d<\/mi>x<\/mi>]<\/mo>.<\/mi><\/mrow>S(x + \\mathrm{d}x) \\;=\\; S(x)\\,\\bigl[1 – n\\,\\sigma \\,\\mathrm{d}x\\bigr].<\/annotation><\/semantics><\/math><\/p>\n

     <\/p>\n

    Taking the limit \"\mathrm{d}x, we get the differential equation:<\/p>\n

     <\/p>\n

    d<\/mi>S<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac>\u2005\u200a<\/mtext>=<\/mo>\u2005\u200a<\/mtext>\u2212<\/mo>\u2009<\/mtext>n<\/mi>\u2009<\/mtext>\u03c3<\/mi>\u2009<\/mtext>S<\/mi>.<\/mi><\/mrow>\\frac{\\mathrm{d}S}{\\mathrm{d}x} \\;=\\; -\\,n\\,\\sigma \\,S.<\/annotation><\/semantics><\/math><\/p>\n

     <\/p>\n

    Its solution is the familiar exponential<\/strong> function:<\/p>\n

     <\/p>\n

    S<\/mi>(<\/mo>x<\/mi>)<\/mo>\u2005\u200a<\/mtext>=<\/mo>\u2005\u200a<\/mtext>exp<\/mi>\u2061<\/mo>\u2009\u2063<\/mtext>(<\/mo>\u2212<\/mo>\u2009<\/mtext>n<\/mi>\u2009<\/mtext>\u03c3<\/mi>\u2009<\/mtext>x<\/mi>)<\/mo>.<\/mi><\/mrow>S(x) \\;=\\; \\exp\\!\\bigl(-\\,n\\,\\sigma\\,x\\bigr).<\/annotation><\/semantics><\/math><\/p>\n

     <\/p>\n

    Thus, the probability density function<\/strong> (PDF) for \"X\" is:<\/p>\n

     <\/p>\n

    f<\/mi>X<\/mi><\/msub>(<\/mo>x<\/mi>)<\/mo>\u2005\u200a<\/mtext>=<\/mo>\u2005\u200a<\/mtext>d<\/mi>d<\/mi>x<\/mi><\/mrow><\/mfrac>[<\/mo>1<\/mn>\u2212<\/mo>S<\/mi>(<\/mo>x<\/mi>)<\/mo>]<\/mo>\u2005\u200a<\/mtext>=<\/mo>\u2005\u200a<\/mtext>n<\/mi>\u2009<\/mtext>\u03c3<\/mi>\u2009<\/mtext>exp<\/mi>\u2061<\/mo>\u2009\u2063<\/mtext>(<\/mo>\u2212<\/mo>\u2009<\/mtext>n<\/mi>\u2009<\/mtext>\u03c3<\/mi>\u2009<\/mtext>x<\/mi>)<\/mo>,<\/mo><\/mspace>x<\/mi>\u2265<\/mo>0.<\/mn><\/mrow>f_X(x) \\;=\\; \\frac{\\mathrm{d}}{\\mathrm{d}x}\\bigl[1 – S(x)\\bigr] \\;=\\; n\\,\\sigma \\,\\exp\\!\\bigl(-\\,n\\,\\sigma\\,x\\bigr), \\quad x \\ge 0.<\/annotation><\/semantics><\/math><\/p>\n

     <\/p>\n

    \n

    3. Mean Free Path<\/h3>\n

    From the properties of the exponential distribution, the expected value of \"X\" is the reciprocal of the rate:<\/p>\n

     <\/p>\n

    \u27e8<\/mo>X<\/mi>\u27e9<\/mo>\u2005\u200a<\/mtext>=<\/mo>\u2005\u200a<\/mtext>\u222b<\/mo>0<\/mn>\u221e<\/mi><\/msubsup>x<\/mi>\u2009<\/mtext>(<\/mo>n<\/mi>\u2009<\/mtext>\u03c3<\/mi>\u2009<\/mtext>e<\/mi>\u2212<\/mo>\u2009<\/mtext>n<\/mi>\u2009<\/mtext>\u03c3<\/mi>\u2009<\/mtext>x<\/mi><\/mrow><\/msup>)<\/mo>\u2009<\/mtext>d<\/mi>x<\/mi>\u2005\u200a<\/mtext>=<\/mo>\u2005\u200a<\/mtext>1<\/mn>n<\/mi>\u2009<\/mtext>\u03c3<\/mi><\/mrow><\/mfrac>.<\/mi><\/mrow>\\langle X \\rangle \\;=\\; \\int_0^\\infty x \\, \\bigl(n\\,\\sigma \\,e^{-\\,n\\,\\sigma\\,x}\\bigr)\\, \\mathrm{d}x \\;=\\; \\frac{1}{n\\,\\sigma}.<\/annotation><\/semantics><\/math><\/p>\n

     <\/p>\n

    We call this value the mean free path<\/strong> \"\lambda\":<\/p>\n

     <\/p>\n

    \u03bb<\/mi>\u2005\u200a<\/mtext>=<\/mo>\u2005\u200a<\/mtext>1<\/mn>n<\/mi>\u2009<\/mtext>\u03c3<\/mi><\/mrow><\/mfrac>.<\/mi><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{\\lambda \\;=\\; \\frac{1}{n\\,\\sigma}.}<\/annotation><\/semantics><\/math><\/p>\n

     <\/p>\n

    \n

    4. Markov Chain Interpretation<\/h3>\n
      \n
    1. In a discrete Markov chain<\/strong>, the system moves from one state to another with a transition probability depending only on the current state.<\/li>\n
    2. Here, in a continuous<\/strong> sense, the position of the particle changes incrementally, but the chance of collision in the next infinitesimal interval is constant\u2014independent of how far it has already traveled.<\/li>\n<\/ol>\n

      Thus, the distance to collision<\/strong> is a waiting-time-like variable with the memoryless property, which is exactly what defines the exponential distribution. Hence, viewing collision events as a Markov process<\/strong> directly yields:<\/p>\n

       <\/p>\n

      \u03bb<\/mi>\u2005\u200a<\/mtext>=<\/mo>\u2005\u200a<\/mtext>1<\/mn>n<\/mi>\u2009<\/mtext>\u03c3<\/mi><\/mrow><\/mfrac>.<\/mi><\/mrow>\\lambda \\;=\\; \\frac{1}{n\\,\\sigma}.<\/annotation><\/semantics><\/math><\/p>\n

       <\/p>\n

      \n

      5. Summary<\/h3>\n
        \n
      • Memoryless assumption<\/strong> \"\rightarrow\" exponential free-path distribution.<\/li>\n
      • Rate parameter<\/strong> \"n,\sigma\" \"\rightarrow\" mean free path \"\lambda.<\/li>\n
      • Markov chain viewpoint<\/strong> \"\rightarrow\" collisions occur in a Poisson-like manner with no dependence on prior \u201ccollision-free\u201d distance.<\/li>\n<\/ul>\n

        This completes the Markov chain derivation<\/strong> of the mean free path<\/strong> in a uniform medium.<\/p>\n","protected":false},"excerpt":{"rendered":"

        1. Memoryless (Markov) Property A fundamental assumption in deriving the mean free path is the memoryless property. Specifically, let be the random variable for the distance traveled by a particle before it collides. The Markov (or memoryless) property states:   P(X>x+\u0394x\u2009\u2223\u2009X>x)\u2005\u200a=\u2005\u200aP(X>\u0394x),\u2200\u2009x,\u2009\u0394x\u2009\u22650.P\\bigl(X > x + \\Delta x \\,\\big\\vert\\, X > x\\bigr) \\;=\\; P\\bigl(X > \\Delta x\\bigr), […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_crdt_document":"","_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[1],"tags":[],"class_list":["post-91","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"brizy_media":[],"_links":{"self":[{"href":"https:\/\/freepath.info\/index.php?rest_route=\/wp\/v2\/posts\/91","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/freepath.info\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/freepath.info\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/freepath.info\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/freepath.info\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=91"}],"version-history":[{"count":5,"href":"https:\/\/freepath.info\/index.php?rest_route=\/wp\/v2\/posts\/91\/revisions"}],"predecessor-version":[{"id":101,"href":"https:\/\/freepath.info\/index.php?rest_route=\/wp\/v2\/posts\/91\/revisions\/101"}],"wp:attachment":[{"href":"https:\/\/freepath.info\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=91"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/freepath.info\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=91"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/freepath.info\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=91"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}