Rubber Physics (Part 7)

Classical Mechanics Level 3

Now consider the case where a one-dimensional linear chain of N N chain links of length l l is connected to a heat bath at room temperature T = 293 K T = 293 \,\text{K} , so that heat exchange with the environment takes place. In addition, a external force f ⃗ = f e ⃗ x \vec f = f \vec e_x acts on the chain. As already shown in part 4 , the alignments l ⃗ i = ± l e ⃗ x \vec l_i = \pm l \vec e_x of the chain links are binomially distributed. However, the probabilities p + p_+ and p − p_- for the alignment in positive and negative x-direction, respectively, are now different and obey a Boltzmann distribution: p + = A exp ⁡ ( + l f k B T ) p − = A exp ⁡ ( − l f k B T ) A = 1 exp ⁡ ( + l f k B T ) + exp ⁡ ( − l f k B T ) = 1 2 cosh ⁡ l f k B T \begin{aligned} p_+ &= A \exp \left(+ \frac{l f}{k_B T} \right)\\ p_- &= A \exp \left(- \frac{l f}{k_B T} \right)\\ A &= \frac{1}{\exp \left(+ \frac{l f}{k_B T} \right) + \exp \left(- \frac{l f}{k_B T} \right)} = \frac{1}{2 \cosh \frac{lf}{k_B T}} \end{aligned} with the Boltzmann constant k B ≈ 1.38 ⋅ 1 0 − 23 J / K k_B \approx 1.38 \cdot 10^{-23}\,\text{J}/\text{K} . The pre-factor A A is chosen so that the normalization condition p + + p − = 1 p_+ + p_- = 1 is met. Calculate the mean value ⟨ x ⟩ \langle x \rangle of the displacement for the vector r ⃗ = x e ⃗ x \vec r = x \vec e_x that connects the beginning and the end of the chain. Derive an equation for the linear force constant k k of the chain with k = ( ∂ f ∂ ⟨ x ⟩ ) ⟨ x ⟩ = 0 = ( ∂ ⟨ x ⟩ ∂ f ) f = 0 − 1 k = \left( \frac{\partial f}{\partial \langle x \rangle} \right)_{\langle x \rangle = 0} = \left( \frac{\partial \langle x \rangle}{\partial f} \right)_{f = 0}^{-1} Give as result the numerical value for the force constant k k with two significant digits in units of newtons per meter. Assume a unit length l = 0.35 nm l = 0.35 \,\text{nm} and a number N = 1 0 4 N = 10^{4} of chain links.


The answer is 0.0000033.

Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

Markus Michelmann
Jan 21, 2018

We want to find out the following mean value ⟨ x ⟩ = ⟨ r ⃗ ⟩ ⋅ e ⃗ x = ⟨ n ⟩ l = ( 2 ⟨ N + ⟩ − N ) l \langle x \rangle = \langle \vec r \rangle \cdot \vec e_x = \langle n \rangle l = (2 \langle N_+ \rangle - N) l with the number N + = 0 , 1 , … , N N_+ = 0,1,\dots, N chain links aligned in positive x-directions, that is bionomially distributed. We have already shown in part 4, that this mean value results ⟨ N + ⟩ = N p + \langle N_+ \rangle = N p_+ Therefore, we get a displacement ⟨ x ⟩ = ( 2 N p + − N ) l = N l ( 2 e β l f e β l f + e − β l f − 1 ) = N l 2 e β l f − ( e β l f + e − β l f ) e β l f + e − β l f = N l e β l f − e − β l f e β l f + e − β l f = N l sinh ⁡ ( β l f ) cosh ⁡ ( β l f ) = N l tanh ⁡ ( β l f ) ⇒ ∂ ⟨ x ⟩ ∂ f = β N l 2 sinh ⁡ ′ ( β l f ) cosh ⁡ ( β l f ) − sinh ⁡ ( β l f ) cosh ⁡ ′ ( β l f ) cosh ⁡ 2 ( β l f ) = β N l 2 cosh ⁡ 2 ( β l f ) − sinh ⁡ 2 ( β l f ) cosh ⁡ 2 ( β l f ) = β N l 2 1 cosh ⁡ 2 ( β l f ) ⇒ k = ( ∂ ⟨ x ⟩ ∂ f ) f = 0 − 1 = 1 β N l 2 = k B T N l 2 \begin{aligned} & & \langle x \rangle &= (2 N p_+ - N) l = N l \left(\frac{2 e^{\beta l f}}{e^{\beta l f} + e^{-\beta l f} } - 1\right) \\ & & &= N l \frac{2 e^{\beta l f} - (e^{\beta l f} + e^{-\beta l f})}{e^{\beta l f} + e^{-\beta l f} } \\ & & &= N l \frac{e^{\beta l f} - e^{-\beta l f}}{e^{\beta l f} + e^{-\beta l f} } \\ & & &= N l \frac{\sinh(\beta l f)}{\cosh(\beta l f)} = N l \tanh(\beta l f) \\ \Rightarrow & & \frac{\partial \langle x \rangle}{\partial f} &= \beta N l^2 \frac{\sinh'(\beta l f) \cosh(\beta l f) - \sinh(\beta lf )\cosh'(\beta l f) }{\cosh^2(\beta l f)}\\ & & &= \beta N l^2 \frac{\cosh^2(\beta l f) - \sinh^2(\beta l f) }{\cosh^2(\beta l f)} = \beta N l^2 \frac{1}{\cosh^2(\beta lf)} \\ \Rightarrow & & k &= \left(\frac{\partial \langle x \rangle}{\partial f}\right)_{f=0}^{-1} = \frac{1}{\beta N l^2} = \boxed{\dfrac{k_B T}{N l^2}} \end{aligned} with the reciprocal value β = 1 k B T \beta = \frac{1}{k_B T} of the thermal energy. Numerical evaluation yields k = 1.38 ⋅ 1 0 − 23 J / K ⋅ 293 K 1 0 4 ( 0.35 ⋅ 1 0 − 9 m ) 2 ≈ 3.3 ⋅ 1 0 − 6 N m k = \frac{1.38 \cdot 10^{-23} \,\text{J}/\text{K} \cdot 293 \,\text{K}}{10^4 (0.35 \cdot 10^{-9} \,\text{m})^2} \approx 3.3 \cdot 10^{-6} \,\frac{\text{N}}{\text{m}}

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