In the previous post, we saw how the probability distribution p(x,N) after N random steps on a unit lattice is given by, p(x, N) = \dfrac{1}{\sqrt{2 \pi N}} \exp\left(-\dfrac{x^2}{2N}\right) If the average step size is b instead of b=1, then we can generalize, and write the formula as:
p(x, N) = \dfrac{1}{\sqrt{2 \pi Nb^2}} \exp\left(-\dfrac{x^2}{2Nb^2}\right) Now consider a Gaussian random walk in 1D. Suppose the stepsize at each step is drawn from a normal distribution \mathcal{N}(0, 1). While it has the same average stepsize as a walk on the lattice, an individual step may be shorter or longer than b=1.
In polymer physics, where a Gaussian coil is often used as a model for polymer conformations, b is called the Kuhn length, and N is proportional to the molecular weight.
Due to the connection between Brownian motion and random walks, the mean squared distance travelled by a particle in 1D with self-diffusivity D is \langle x^2(t) \rangle = 2Dt. Similarly, the mean end-to-end squared distance of a Gaussian random walk is given by, \langle x^2(N) \rangle = \int_{-\infty}^{\infty} x^2 p(x, N) dx = Nb^2 \equiv 2Dt = \langle x^2(t) \rangle. This allows us to re-parameterize the equation for the position of a Brownian diffusors. p(x, t) = \dfrac{1}{\sqrt{4 \pi Dt}} \exp\left(-\dfrac{x^2}{4Dt}\right) Look at the correspondence between t and N, and b and \sqrt{2D}.
p(x, N) = \dfrac{1}{\sqrt{2 \pi Nb^2}} \exp\left(-\dfrac{x^2}{2Nb^2}\right) Now consider a Gaussian random walk in 1D. Suppose the stepsize at each step is drawn from a normal distribution \mathcal{N}(0, 1). While it has the same average stepsize as a walk on the lattice, an individual step may be shorter or longer than b=1.
In polymer physics, where a Gaussian coil is often used as a model for polymer conformations, b is called the Kuhn length, and N is proportional to the molecular weight.
Due to the connection between Brownian motion and random walks, the mean squared distance travelled by a particle in 1D with self-diffusivity D is \langle x^2(t) \rangle = 2Dt. Similarly, the mean end-to-end squared distance of a Gaussian random walk is given by, \langle x^2(N) \rangle = \int_{-\infty}^{\infty} x^2 p(x, N) dx = Nb^2 \equiv 2Dt = \langle x^2(t) \rangle. This allows us to re-parameterize the equation for the position of a Brownian diffusors. p(x, t) = \dfrac{1}{\sqrt{4 \pi Dt}} \exp\left(-\dfrac{x^2}{4Dt}\right) Look at the correspondence between t and N, and b and \sqrt{2D}.
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