Advanced Chemistry - Mono Mole

September 15, 2019October 11, 2024

Buffer capacity (solution chemistry)

The buffer capacity β of a solution is a measure of the solution’s resistance to changes in pH. It is defined as the number of moles of a strong acid or a strong base needed to change one unit of pH of 1.00 L of the solution.

Mathematically,

$\beta=\frac{dc_b}{dpH}\; \; \; \; \; \; \; \; (1)$

$\beta=-\frac{dc_a}{dpH}\; \; \; \; \; \; \; \; (2)$

where c_b is the number of moles per litre of a strong base, and c_ais the number of moles per litre of a strong acid.

Since buffer capacity is defined as a positive value, eq2 has a negative sign because the change in pH is negative when a strong acid is added to the buffer solution.

Question

Is a strong base versus strong acid system a buffer?

Answer

According to the above definition, a strong base versus strong acid system can also be considered a buffer, but with negligible buffer capacity. This is because a strong acid is fully dissociated in water, and any addition of a strong base reduces the concentration of H⁺; whereas an aqueous weak acid is partially dissociated, and any H⁺ removed by the base is partially replenished by further dissociation of the weak acid, thereby minimising the increase in the solution’s pH.

When an acid is added to the strong acid solution, any molecular acid formed by the combination of H⁺ from the added acid and the conjugate base of the strong acid immediately dissociates into the component ions. However, when H⁺from an acid is added to a weak acid solution, it combines with the conjugate base of the weak acid, partially shifts the equilibrium of the weak acid’s dissociation to the left, thus minimising the decrease in the solution’s pH.

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September 15, 2019September 26, 2024

Buffer capacity of an aqueous solution

The buffer capacity of an aqueous solution, β_s, consists of the buffer capacities of the components of the solution:

$\beta_s=\beta_w+\beta_a+\beta_b+...\; \; \; \; \; \; \; \; (7)$

where β_w is the buffer capacity of water, β_ais the buffer capacity of a weak acid and β_bis the buffer capacity of a weak base.

To prove that buffer capacities are additive, we substitute eq1 in eq7

$\frac{dc_{b,s}}{dpH}=\frac{dc_{b,w}}{dpH}+\frac{dc_{b,a}}{dpH}+\frac{dc_{b,b}}{dpH}+...$

$\frac{dc_{b,s}}{dpH}=\frac{d\left (c_{b,w}+c_{b,a}+c_{b,b}+... \right )}{dpH}$

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September 15, 2019January 21, 2025

Buffer capacity of water

What is the formula describing the buffer capacity of water?

Consider a solution containing water and a strong base with the following equilibria:

$H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)$

$NaOH(aq)\rightleftharpoons Na^+(aq)+OH^-(aq)$

With reference to the above equilibria, the sum of the number of moles of H⁺ and Na⁺ must equal to that of OH^–(see this article for details). As the volume of the solution is common to all ions,

$[H^+]+[Na^+]=[OH^-]\; \; \; \; \; \; \; \; (3)$

Substituting c_b = [Na⁺] and K_w= [H⁺][OH^–] in eq3 yields

$c_b=\frac{K_w}{[H^+]}-[H^+]$

$\frac{dc_b}{dpH}=\frac{dK_w[H^+]^{-1}}{dpH}-\frac{d[H^+]}{dpH}\; \; \; \; \; \; \; \; (4)$

Substituting eq1 and pH = –log[H⁺] in eq4 gives

$\beta_w=-\frac{dK_w[H^+]^{-1}}{dlog[H^+]}+\frac{d[H^+]}{dlog[H^+]}\; \; \; \; \; \; \; \; (4a)$

$\beta_w=ln10\left (\frac{K_w}{[H^+]}+[H^+] \right )\; \; \; \; \; \; \; \; (5)$

To determine the maximum or minimum buffer capacity of water, we take the derivative of eq5 with respect to [H⁺] to give:

$\frac{d\beta_w}{d[H^+]}=ln10\left (\frac{dK_w[H^+]^{-1}}{d[H^+]}+\frac{d[H^+]}{d[H^+]} \right )$

$\frac{d\beta_w}{d[H^+]}=ln10\left ( -\frac{K_w}{[H^+]^2}+1 \right )\; \; \; \; \; \; \; \; (6)$

Letting $\frac{d\beta_w}{d[H^+]}=0$ yields

$\frac{K_w}{[H^+]^2}=1$

Substituting K_w= [H⁺][OH^–] in the above equation results in

$[OH^-]=[H^+]$

This means that a stationary point for the function in eq5 occurs when [OH^–] = [H⁺] , i.e. at pH = 7. Let’s investigate the nature of this stationary point by differentiating eq6 again.

$\frac{d^{\, 2}\beta_w}{d[H^+]^2}=2ln10\frac{K_w}{[H^+]^3}$

Substituting K_w= [H⁺][OH^–] in the above equation, and noting that [OH^–] = [H⁺], gives

$\frac{d^{\, 2}\beta_w}{d[H^+]^2}=\frac{2ln10}{[H^+]}$

$\frac{d^{\, 2}\beta_w}{d[H^+]^2}> 0$

If we repeat the above steps and consider the case of the addition of a strong acid to water, we end up with the same conclusion. Hence, water has minimum buffer capacity at pH = 7.

The buffer capacity of water over the entire range of pH can be found by substituting [H⁺] = 10^-pHin eq5 to give

$\beta_w=ln10\left ( \frac{K_w}{10^{-pH}}+10^{-pH} \right )\; \; \; \; \; \; \; \; (6a)$

The plot of the above equation with β_was the vertical axis and pH as the horizontal axis is shown in the diagram below.

From the graph, the buffering capacity of water increases infinitesimally from a minimum of 0 at pH 7, to 0.023 at pH 2 on the left and pH 12 on the right, and then drastically to 2.303 at pH 0 on the left and pH 14 on the right. This means that water is only relatively effective as a buffer at extreme pH levels.

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September 15, 2019March 27, 2025

Buffer capacity (monoprotic weak acid and its conjugate base)

What is the formula describing the buffer capacity of an aqueous monoprotic weak acid and its conjugate base?

Consider a solution containing water, a strong base and a weak acid, HA, with the following equilibria:

$H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)$

$NaOH(aq)\rightleftharpoons Na^+(aq)+OH^-(aq)$

$HA(aq)\rightleftharpoons H^+(aq)+A^-(aq)$

With reference to the above equilibria, the sum of the number of moles of H⁺ and Na⁺ must equal to that of OH^– and A^–(see this article for details). As the volume is of the solution is common to all ions,

$[H^+]+[Na^+]=[OH^-]+[A^-]$

Substituting c_b = [Na⁺] and K_w= [H⁺][OH^–] in the above equation gives

$c_b=\frac{K_w}{[H^+]}-[H^+]+[A^-]$

Multiplying the term [A^–] on the RHS of the above equation by $\left (\frac{[HA]+[A^-]}{[HA]+[A^-]} \right )\left (\frac{[H^+]/[HA]}{[H^+]/[HA]} \right )$ and noting that $K_a = \frac{[H^+][A^-]}{[HA]}$ , yields

$c_b=\frac{K_w}{[H^+]}-[H^+]+\frac{[HA]_TK_a}{K_a+[H^+]}$

where [HA]_T=[HA] + [A^–].

$\frac{dc_b}{dpH}=\frac{dK_w[H^+]^{-1}}{dpH}-\frac{d[H^+]}{dpH}+\frac{d\frac{[HA]_TK_a}{K_a+[H^+]}}{dpH}$

Substituting eq1 and pH = –log[H⁺] in the above equation results in

$\beta_s=-\frac{dK_w[H^+]^{-1}}{dlog[H^+]}+\frac{d[H^+]}{dlog[H^+]}-\frac{d\frac{[HA]_TK_a}{K_a+[H^+]}}{dlog[H^+]}$

Substituting eq4a in the above equation gives

$\beta_s=\beta_w-\frac{d\frac{[HA]_TK_a}{K_a+[H^+]}}{dlog[H^+]}\; \; \; \; \; \; \; \; (8)$

Comparing eq8 with eq7 with β_s= 0 (since we are considering the case where a strong base is added to a solution containing water and a weak acid)

$\beta_a=-\frac{d\frac{[HA]_TK_a}{K_a+[H^+]}}{dlog[H^+]}\; \; \; \; \; \; \; \; (9)$

Since K_a is a constant

$\beta_a=\frac{[HA]_TK_a[H^+]ln10}{{\left ( K_a+[H^+] \right )^2}}\; \; \; \; \; \; \; \; (10)$

To determine the maximum or minimum buffer capacity of the weak monoprotic acid buffer,

$\frac{d\beta_a}{d[H^+]}=K_aln10\frac{d\frac{[HA]_T[H^+]}{\left ( K_a+[H^+] \right )^2}}{d[H^+]}$

$\frac{d\beta_a}{d[H^+]}=K_aln10\left \{\frac{\left (K_a+[H^+] \right )^2 [HA]_T-2[HA]_T[H^+]\left ( K_a+[H^+] \right )}{\left ( K_a+[H^+] \right )^4} \right \}\; \; \; \; \; \; \; \; (11)$

Let $\frac{d\beta_a}{d[H^+]}=0$

$K_a=[H^+]\; \; \; \; \; \; \; \; (12)$

A stationary point for the function in eq10 occurs when K_a= [H⁺]. Since $K_a = \frac{[H^+][A^-]}{[HA]}$

$[HA]=[A^-]\; \; \; \; \; \; \; \; (13)$

Let’s investigate the nature of this stationary point by differentiating eq11.

$\frac{d^2\beta_a}{d[H^+]^2}=K_aln10\frac{d\frac{(K_a+[H^+])[HA]_T-2[HA]_T[H^+]}{(K_a+[H^+])^3}}{d[H^+]}$

$\frac{d^2\beta_a}{d[H^+]^2}=K_aln10\frac{2[HA]_T(K_a+[H^+])^2([H^+]-2K_a)}{(K_a+[H^+])^6}$

$\frac{d^2\beta_a}{d[H^+]^2}=K_aln10\frac{2[HA]_T(K_a+[H^+])^2[H^+]\frac{[HA]-2[A^-]}{[HA]}}{(K_a+[H^+])^6}$

Substituting eq13 in the above equation yields

$\frac{d^2\beta_a}{d[H^+]^2}=-\frac{2K_a(ln10)[HA]_T(K_a+[H^+])^2[H^+]}{(K_a+[H^+])^6}$

$\frac{d^2\beta_a}{d[H^+]^2}< 0$

If we consider the case of the addition of a strong acid to the aqueous solution, we end up with the same conclusion. Hence, a monoprotic weak acid has maximum buffer capacity when [HA] = [A^–]. The buffer capacity of the monoprotic weak acid over the entire range of pH can be found by substituting [H⁺] = 10^-pHin eq10 to give

$\beta_a=\frac{[HA]_TK_a10^{-pH}ln10}{(K_a+10^{-pH})^2}\; \; \; \; \; \; \; \; (14)$

Plotting the above equation with β_aas the vertical axis and pH as the horizontal axis results in:

The total buffer capacity of an aqueous monoprotic weak acid is given by the combined equations of eq8 and eq9, i.e.

$\beta_s=\beta_w+\beta_a\; \; \; \; \; \; \; \; (15)$

The total buffer capacity of an aqueous monoprotic weak acid over the entire range of pH can be found by substituting eq6a and eq14 in eq15 to give

$\beta_s=ln10\left ( \frac{K_w}{10^{-pH}}+10^{-pH} \right )+\frac{[HA]_TK_a10^{-pH}ln10}{(K_a+10^{-pH})^2}\; \; \; \; \; \; \; \; (16)$

Using eq16 to plot a graph of β_sversus pH for 1.0 M aqueous ethanoic acid (K_a = 1.75 x 10^-5) produces:

The graph shows a maximum point at (4.757,0.576) corresponding to the maximum buffer capacity at pH = 4.757, which is attributed to the ethanoic acid/ethanoate buffer pair; while the increased in buffer capacity below pH 2 and above pH 12 is attributed to water. If we scale the graph by multiplying eq16 throughout by a factor of 10 and plotting 10β_sversus pH, we observe another two stationary points (minimum points), each corresponding to the sum of the minimum buffer capacity of the weak acid and the increased buffer capacity of water:

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January 12, 2019December 21, 2024

Constitutive relation: hypothesis 4

The fourth constitutive relation hypothesis states that the properties of a fluid are isotropic, i.e., independent of direction.

For example, an object moving in a fluid encounters the same resistance regardless of the direction of movement. From eq11, the properties of a fluid are described by β_ijkl, which must be isotropic.

As described in the articles on tensors, the general form of a fourth-order isotropic tensor is:

$\beta_{ijkl}=\lambda\delta_{ij}\delta_{kl}+\mu\delta_{ik}\delta_{jl}+\nu\delta_{il}\delta_{jk}\; \; \; \; \; \; \; (12)$

From eq11 of the previous article, τ_ijis symmetric. Therefore, τ_ij=τ_jiand

$\beta_{ijkl}=\beta_{jikl}$

Substitute eq12 in the above equation, we have

$\left ( \mu-\nu \right )\delta_{ik}\delta_{jl}=\left ( \mu-\nu \right )\delta_{il}\delta_{jk}$

$\mu=\nu\; \; \; \; \; \; \; (13)$

Substitute eq13 in eq12,

$\beta_{ijkl}=\lambda\delta_{ij}\delta_{kl}+\mu\left (\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right )\; \; \; \; \; \; \; (14)$

Substitute eq14 in eq11,

$\tau_{ij}=\frac{1}{2}\lambda\delta_{ij}\delta_{kl}\left ( \frac{\partial u_k}{\partial x_l}+\frac{\partial u_l}{\partial x_k}\right ) +\frac{1}{2}\mu\left (\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right )\left ( \frac{\partial u_k}{\partial x_l}+\frac{\partial u_l}{\partial x_k}\right )$

$\tau_{ij}=\frac{1}{2}\lambda\delta_{ij}\left ( \frac{\partial u_k}{\partial x_k}+\frac{\partial u_k}{\partial x_k}\right ) +\frac{1}{2}\mu\left (\delta_{ik}\frac{\partial u_k}{\partial x_j}+\delta_{ik}\frac{\partial u_j}{\partial x_k} +\delta_{il}\frac{\partial u_j}{\partial x_l}+\delta_{il}\frac{\partial u_l}{\partial x_j} \right )$

$\tau_{ij}=\lambda\delta_{ij} \frac{\partial u_k}{\partial x_k}+\mu\left (\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} \right )\; \; \; \; \; \; \; (15)$

Substitute eq15 in eq3,

$\sigma_{ij}=-p\delta_{ij}+\lambda\delta_{ij} \frac{\partial u_k}{\partial x_k}+\mu\left (\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} \right )\; \; \; \; \; \; \; (16)$

The values of μ and λ can only be determined through experiments. μ is known as the shear viscosity of the fluid while λ is the volume viscosity of the fluid, which is zero for an incompressible fluid. Eq16 becomes:

$\sigma_{ij}=-p\delta_{ij}+\mu\left (\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} \right )\; \; \; \; \; \; \; (17)$

Question

Show that μ in eq16 and eq7 are the same.

Answer

Consider the flow of an incompressible fluid where the flow velocity components are u = u(y), v = 0 and w = 0. Using eq16, the stress components are:

$\sigma_{11}=\sigma_{22}=\sigma_{33}=-p$

$\sigma_{12}=\sigma_{21}=\mu\frac{du}{dy}$

with the remaining components equal to zero.

Clearly, the components of stress in this case include the pressure p and the shear stress $\mu\frac{du}{dy}$ , which is the result of the derivation of Newton’s law of viscosity, eq7. Hence, μ in eq16 and eq7 are the same.

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January 12, 2019December 21, 2024

Constitutive relation: hypothesis 2

The second hypothesis, formulating the constitutive relation for an incompressible and viscous Newtonian fluid, states that the shear stress tensor is a linear function of the deformation tensor.

Consider a shear stress τ that is coplanar to a cross section of a Newtonian fluid ABCD as shown in the diagram below.

The fluid is continuously deformed by the force with different layers of the fluid having different velocities, for example, u(y) at point D and u(y + Δy) at point A. At time t, the body of fluid moves from ABCD to A’B’CD. Let’s define the shear strain in the fluid as $\gamma=\frac{\Delta x}{\Delta y}$ where

$\frac{\Delta x}{\Delta y}=\frac{\left [ u\left ( y+\Delta y \right )-u\left ( y \right ) \right ]\Delta t}{\Delta y}$

If Δy → 0,

$\gamma=\frac{du}{dy}\Delta t$

If t is small,

$\frac{d\gamma}{dt}=\frac{du}{dy}\; \; \; \; \; \; \; (5)$

We call $\frac{d\gamma}{dt}$ the shear strain rate and $\frac{du}{dy}$ the velocity gradient of the fluid. Clearly, the greater the shear stress applied, the higher the value of the shear strain rate, i.e.:

$\tau \propto \frac{d\gamma}{dt} = \mu\frac{d\gamma}{dt}\; \; \; \; \; \; \; (6)$

where μ is the constant of proportionality, which is attributed to the resistance of the fluid to the shear stress.

We call this resistance, the viscosity of the fluid. Substituting eq5 in eq6, we have the Newton’s law of viscosity,

$\tau=\mu\frac{du}{dy}\; \; \; \; \; \; \; (7)$

When extended to three dimensions (see above diagram), eq7 becomes:

$\tau_{yx}=\mu\frac{du_x}{dy}\; \; \; \; \; \; \; (7a)$

From eq4 of the previous article, shear stress is a second-order tensor with nine components τ_ij. Hence, the general form of eq7 is:

$\tau_{ij}=\mu\frac{\partial u_j}{\partial x_i}$

where for a Cartesian system, i = 1, 2, 3, j = 1, 2, 3 and x₁ = x, x₂ = y, x₃ = z.

$\frac{\partial u_j}{\partial x_i}$ can therefore be expressed as a second-order tensor (deformation tensor) of nine components. Since each of the nine components of τ_ijis linearly proportional to each of the nine components of $\frac{\partial u_j}{\partial x_i}$ , there are a total of 9² = 81 proportionality constants. In other words,

$\tau_{ij}=\alpha_{ijkl}\frac{\partial u_k}{\partial x_l}\; \; \; \; \; \; \; (8)$

where $\alpha_{ijkl}$ is a fourth-order tensor of eighty one components, which can be representation by a 3x3x3x3 matrix for a Cartesian system. This satisfies hypothesis 2.

The fourth-order tensor, which is composed of the proportionality constants, characterises the properties of the fluid.

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January 12, 2019December 21, 2024

Isotropic tensors

An isotropic Cartesian tensor is one where its components are identical in any orthogonal Cartesian system.

An isotropic property is one that is independent of direction, e.g. thermal expansion of a solid, and therefore has a quantity that is independent of the reference frame. An isotropic property can therefore be described by a Cartesian tensor like $T'_{j_1j_2...j_n}$ in eq15.

To evaluate the properties of an isotropic tensor, we begin by considering a generic fourth-order tensor $T_{jlnp}$ , whose transformation is expressed by eq14:

$T'_{ikmo}=a_{ij}a_{kl}a_{mn}a_{op}T_{jlnp}\; \; \; \; \; \; \; (16)$

Since j, l, n and p, each of which ranges from 1 to 3, the 81 components of $T'_{ikmo}$ are classified in four groups as follows:

1. Components with four equal subscripts, i.e. $T'_{1111}$ , $T'_{2222}$ and $T'_{3333}$ .
2. Components with three equal subscripts, e.g. $T'_{1211}$ , $T'_{2223}$ , etc.
3. Components with two different pairs of equal subscripts, e.g. $T'_{1212}$ , $T'_{3322}$ , $T'_{1331}$ , etc.
4. Components with one pair of equal subscripts, e.g. $T'_{1213}$ , $T'_{2213}$ , etc.

Next, consider the transformation of $T_{jlnp}$ to $T'_{ikmo}$ as a rotation about the x₃-axis of $\theta=180^o$ where x₃coincides with x’₃(x’₃is equal to x₃and is perpendicular to the plane of the page). From eq12, the transformation matrix is:

$a_{ij}\: or\: a_{kl}\: or\: a_{mn}\: or\: a_{op}=\begin{pmatrix} cos\theta &sin\theta &0 \\ -sin\theta &cos\theta &0 \\ 0 &0 &1 \end{pmatrix}=\begin{pmatrix} -1 &0 &0 \\ 0 &-1 &0 \\ 0 &0 &1 \end{pmatrix}$

Therefore, $a_{11}=a_{22}=-1$ and $a_{33}=1$ , with the rest of the components of the transformation matrix equal to zero. From eq16, a component of group 2 of the above classification is:

$T'_{1113}=a_{1j}a_{1l}a_{1n}a_{3p}T_{jlnp}=a_{11}a_{11}a_{11}a_{31}T_{1111}+a_{11}a_{11}a_{11}a_{32}T_{1112}$

$+a_{11}a_{11}a_{11}a_{33}T_{1113}+...\; \; \; \; \; \; \; (17)$

Substituting $a_{11}=a_{22}=-1$ , $a_{33}=1$ and $a_{i\neq j}=0$ in eq17, we have:

$T'_{1113}=-T_{1113}\; \; \; \; \; \; \; (18)$

If $T_{jlnp}$ is an isotropic tensor, $T'_{1113}=T_{1113}$ (since an isotropic Cartesian tensor is one where its components are identical in any orthogonal Cartesian system). Hence, the only way to satisfy eq18 is for $T'_{1113}=T_{1113}=0$ . Otherwise, the value of $T'_{1113}$ varies with the value of $T_{1113}$ , e.g. when $T_{1113}=-7$ , $T'_{1113}=7$ .

Another component of group 2 of the above classification is $T'_{1211}$ where after expanding and substituting with $a_{11}=a_{22}=-1$ , $a_{33}=1$ and $a_{i\neq j}=0$ ,

$T'_{1211}=T_{1211}\; \; \; \; \; \; \; (20)$

Eq20 shows that $T'_{1211}=T_{1211}$ but does not indicate the value of either component. If the rotation is instead made about the x₂-axis at $\theta=180^o$ , the transformation matrix is:

$\begin{pmatrix} cos\theta & 0 &sin\theta \\ 0 & 1 &0 \\ -sin\theta &0 &cos\theta \end{pmatrix}=\begin{pmatrix} -1 &0 &0 \\ 0 &1 &0 \\ 0 &0 &-1 \end{pmatrix}$

We have $a_{11}=a_{33}=-1,a_{22}=1$ , and $a_{i\neq j}=0$ , giving $T'_{1211}=-T_{1211}$ . If $T_{jlnp}$ is an isotropic tensor,

$T'_{1211}=T_{1211}=0\; \; \; \; \; \; \; (22)$

Repeating the above steps and assuming that $T_{jlnp}$ is an isotropic tensor, we find that all the components of $T'_{ikmo}$ in group 2 and group 4 are zero, which means that all the components of $T_{jlnp}$ in group 2 and group 4 are zero.

We shall now analyse the components of group 1 by a rotation about the x₃-axis at $\theta=90^o$ with the transformation matrix:

$\begin{pmatrix} cos\theta &sin\theta &0 \\ -sin\theta &cos\theta &0 \\ 0 &0 &1 \end{pmatrix}=\begin{pmatrix} 0 &1 &0 \\ -1 &0 &0 \\ 0 &0 &1 \end{pmatrix}$

We have $T'_{1111}=T_{2222},T'_{2222}=T_{1111},T'_{3333}=T_{3333}$ .

A rotation about the x₂-axis at $\theta=90^o$ involves the transformation matrix:

$\begin{pmatrix} cos\theta & 0&sin\theta \\ 0 &1 &0 \\ -sin\theta &0 & cos\theta\end{pmatrix}=\begin{pmatrix} 0 &0 &1 \\ 0 &1 &0 \\ -1 &0 &0 \end{pmatrix}$

and gives $T'_{1111}=T_{3333},T'_{2222}=T_{2222},T'_{3333}=T_{1111}$ .

Comparing the six equations $T'_{1111}=T_{2222},T'_{2222}=T_{1111},T'_{3333}=T_{3333}$ and $T'_{1111}=T_{3333},T'_{2222}=T_{2222},T'_{3333}=T_{1111}$ , all the components of $T'_{ikmo}$ in group 1 are equal to one another, which means that all the components of $T_{jlnp}$ in group 1 are equal to one another.

Finally, let’s look at the components in group 3. Under a rotation about the x₃-axis at $\theta=90^o$ , we have $T'_{1122}=T_{2211},T'_{2233}=T_{1133},T'_{3311}=T_{3322}$ .

A rotation about the x₂-axis at $\theta=90^o$ gives $T'_{1122}=T_{3322},T'_{2233}=T_{2211},T'_{3311}=T_{1133}$ .

Comparing the six equations $T'_{1122}=T_{2211},T'_{2233}=T_{1133},T'_{3311}=T_{3322}$ and $T'_{1122}=T_{3322},T'_{2233}=T_{2211},T'_{3311}=T_{1133}$ , components in group 3 of $T'_{ikmo}$ (which is equal to the components in group 3 of $T_{jlnp}$ ) with i = k and m = o are equal to one another. Repeating the above steps, we can show that components in group 3 with i = m and k = o are equal to one another and those with i = o and k = m are equal to one another.

Therefore, if $T_{jlnp}$ is an isotropic tensor:

i) components of $T'_{ikmo}$ with i = k = m = o are equal and therefore $T_{jlnp}$ with j = l = n = p are equal

ii) components of $T'_{ikmo}$ with i = k ≠ m = o are equal and therefore $T_{jlnp}$ with j = l ≠ n = p are equal

iii) components of $T'_{ikmo}$ with i = m ≠ k = o are equal and therefore $T_{jlnp}$ with j = n ≠ l = p are equal

iv) components of $T'_{ikmo}$ with i = o ≠ k = m are equal and therefore $T_{jlnp}$ with j = p ≠ l = n are equal

v) all other components not mentioned are equal to zero

Note that components of parts ii, iii and iv belong to group3 of the isotropic tensor. Now, consider a case of $T_{jlnp}$ where components of $T_{jlnp}$ in iii and iv are set to zero ( $T_{jlnp}$ is still an isotropic tensor but with more components equal to zero). A rotation about the x₃-axis at $\theta=45^o$ gives the transformation matrix:

$\begin{pmatrix} cos\theta &sin\theta &0 \\ -sin\theta &cos\theta &0 \\ 0 & 0 &1 \end{pmatrix}=\begin{pmatrix} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0\\ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} &0 \\ 0 &0 &1 \end{pmatrix}$

Since components in part iii and part iv are set to zero and the conditions of part v applies, we have $T'_{1111}=\frac{1}{4}T_{1111}+\frac{1}{4}T_{1122}+\frac{1}{4}T_{2211}+\frac{1}{4}T_{2222}$ . Furthermore, part i states that $T_{1111}=T_{2222}$ and part ii states that $T_{1122}=T_{2211}$ . So,

$T'_{1111}=\frac{1}{2}\left ( T_{1111}+T_{1122}\right )\; \; \; \; \; \; \; (23)$

If $T_{jlnp}$ is an isotropic tensor, $T'_{1111}=T_{1111}$ and therefore, $T_{1111}=T_{1122}$ . This means that all non-zero components of this particular $T_{jlnp}$ have the same value, say, λ, where:

$T_{jlnp}=\lambda\delta_{jl}\delta_{np}\; \; \; \; \; \; \; (24)$

where $\delta_{jl}$ and $\delta_{np}$ are Kronecker deltas.

Using the same logic by considering the cases of $T_{jlnp}$ where components of $T_{jlnp}$ in part ii and part iv are set to zero and where components of $T_{jlnp}$ in part ii and part iii are set to zero, we have:

$T_{jlnp}=\mu\delta_{jn}\delta_{lp}\; \; \; \; \; \; \; (25)$

and

$T_{jlnp}=\nu\delta_{jp}\delta_{ln}\; \; \; \; \; \; \; (26)$

respectively.

Therefore, we can write the general form of the fourth-order isotropic tensor as:

$T_{jlnp}=\lambda\delta_{jl}\delta_{np}+\mu\delta_{jn}\delta_{lp}+\nu\delta_{jp}\delta_{ln}\; \; \; \; \; \; \; (27)$

Question

Show that $T_{jlnp}$ in eq27 continues to satisfy parts i to v, all of which are needed to define a fourth-order isotropic tensor.

Answer

In eq27,

i) the components of $T_{jlnp}$ with j = l = n = p have values of $\lambda+\mu+\nu$ and are therefore still equal to one another;

ii) the components of $T_{jlnp}$ with j = l ≠ n = p have values of $\lambda$ and are therefore still equal to one another;

iii) the components of $T_{jlnp}$ with j = n ≠ l = p have values of $\mu$ and are therefore still equal to one another;

iv) the components of $T_{jlnp}$ with j = p ≠ l = n have values of $\nu$ and are therefore still equal to one another;

v) all other components not mentioned are equal to zero.

Therefore, the general form of the fourth-order isotropic tensor is valid.

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Tensor and tensor transformation

A tensor is an array of numbers or functions that can be used to describe properties of a body, such as the scalar $x$ for temperature, the vector $\begin{pmatrix} a_1\\a_2 \\a_3 \end{pmatrix}$ for velocity, and the matrix $\begin{pmatrix} \sigma_{11} &\sigma_{12}&\sigma_{13}\\ \sigma_{21} &\sigma_{22}&\sigma_{23}\\ \sigma_{31} &\sigma_{32}& \sigma_{33}\end{pmatrix}$ for stress.

Tensors are categorised by their ranks (also known as order), where

1. A zeroth-order tensor is an array representing a quantity that only has magnitude and no direction, i.e. a scalar.
2. A first-order tensor is an array representing a quantity that has magnitude and a direction, i.e. a vector.
3. A second-order tensor is an array representing a quantity that has magnitude and two directions.

The above implies that the number of elements of a tensor is 3ⁿ, where n is the order of the tensor.

Next, we shall investigate how the elements of a tensor, particularly a 2^nd-order tensor, transform between two sets of coordinates. From an earlier article, u_j, which is a set of three quantities, is known as a first order tensor (i.e. a vector). Consider the product of two first order tensors u_jand v_l, whose elements with respect to another set of axes are given by:

$u'_iv'_k=\left ( a_{ij}u_j \right )\left ( a_{kl}v_l \right )\; \; \; \; \; \; \; \; 12a$

Since each term on the RHS of eq12a is a scalar, we can rewrite eq12a as

$u'_iv'_k=a_{ij}a_{kl}u_jv_l=\sum_{j}\sum_{i}a_{ij}a_{kl}u_jv_l\; \; \; \; \; \; \; \; 12b$

There are 9 possible products of $u'_iv'_k$ because i = 1,2,3 and k = 1,2,3. In other words, $u'_iv'_k$ is a set of 9 elements, each being a sum of nine quantities , e.g.

$u'_1v'_1=a_{11}a_{11}u_1v_1+a_{11}a_{12}u_1v_2+a_{11}a_{13}u_1v_3$

$+a_{12}a_{11}u_2v_1+a_{12}a_{12}u_2v_2+a_{12}a_{13}u_2v_3$

$+a_{13}a_{11}u_3v_1+a_{13}a_{12}u_3v_2+a_{13}a_{13}u_3v_3$

We can therefore represent the elements of $u'_iv'_k$ by a 3×3 matrix $\begin{pmatrix} u'_1v'_1 &u'_1v'_2&u'_1v'_3\\ u'_2v'_1&u'_2v'_2&u'_2v'_3\\ u'_3v'_1&u'_3v'_2&u'_3v'_3 \end{pmatrix}$ . Similarly, $u_jv_l$ in eq12b is a multiplication of two first order tensors and can be represented by a 3×3 matrix $\begin{pmatrix} u_1v_1 &u_1v_2&u_1v_3\\ u_2v_1&u_2v_2&u_2v_3\\ u_3v_1&u_3v_2&u_3v_3 \end{pmatrix}$ . This results in:

$T'_{ik}=a_{ij}a_{kl}T_{jl}\; \; \; \; \; \; \; (13)$

Eq13 indicates that each element of $u'_1v'_1$ in one reference frame is a result of the transformation of $u_jv_l$ in a different reference frame by $a_{ij}a_{kl}$ . In general, a set of nine quantities $T_{jl}$ (as is the case of a second-order tensor) with reference to an orthogonal set of axes is transformed to another set of nine quantities $T'_{ik}$ with reference to another orthogonal set of axes by the transformation matrix $a_{ij}a_{kl}$ .

Using the same logic, a third order tensor transforms as follows:

$u'_iv'_kw'_m=\left ( a_{ij}u_j \right )\left ( a_{kl}u_l \right )\left ( a_{mn}u_n \right )=a_{ij}a_{kl}a_{mn}u_jv_lw_n$

$T'_{ikm}=a_{ij}a_{kl}a_{mn}T_{jln}$

and a fourth order tensor:

$u'_iv'_kw'_mx'_o=\left ( a_{ij}u_j \right )\left ( a_{kl}u_l \right )\left ( a_{mn}u_n \right )\left ( a_{op}x_p \right )=a_{ij}a_{kl}a_{mn}a_{op}u_jv_lw_nx_p$

$T'_{ikmo}=a_{ij}a_{kl}a_{mn}a_{op}T_{jlnp}\; \; \; \; \; \; \; (14)$

In general, a tensor of rank n is a quantity that transforms from one set of orthogonal axes to another set of orthogonal axes according to:

$T'_{j_1,j_2...j_n}=a_{j_1k_1}a_{j_2k_2}...a_{j_nk_n}T_{k_1k_2...k_n}\; \; \; \; \; \; \; (15)$

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Change of basis

A change of basis is a method to convert vector coordinates with respect to one basis to coordinates with respect to another basis.

Consider a vector u with reference to orthogonal unit basis vectors x₁, x₂and x₃.

$\textbf{\textit{u}}=px_1+qx_2+rx_3\; \; \; \; \; \; \; (1)$

where p, q and r are the scalar components of the unit basis vectors x₁, x₂and x₃ respectively.

Clearly,

$p=\textbf{\textit{u}}\cdot x_1\; \; \; \; q=\textbf{\textit{u}}\cdot x_2\; \; \; \; r=\textbf{\textit{u}}\cdot x_3\; \; \; \; (2)$

If we define u with respect to another orthogonal reference frame with unit basis vectors x’₁, x’₂and x’₃ (see above diagram), we have:

$\textbf{\textit{u}}=p' x'_1+q'x'_2+r'x'_3\; \; \; \; (3)$

and

$p'=\textbf{\textit{u}}\cdot x'_1\; \; \; \; q'=\textbf{\textit{u}}\cdot x'_2\; \; \; \; r'=\textbf{\textit{u}}\cdot x'_3\; \; \; \; (4)$

To find the relationship between the two reference frames, we substitute eq1 in eq4 to give

$p'=\left ( px_1+qx_2+rx_3 \right )\cdot x'_1=px_1\cdot x'_1+qx_2\cdot x'_1+rx_3\cdot x'_1\; \; \; \; \; \; \; (5)$

$q'=\left ( px_1+qx_2+rx_3 \right )\cdot x'_2=px_1\cdot x'_2+qx_2\cdot x'_2+rx_3\cdot x'_2\; \; \; \; \; \; \; (6)$

$r'=\left ( px_1+qx_2+rx_3 \right )\cdot x'_3=px_1\cdot x'_3+qx_2\cdot x'_3+rx_3\cdot x'_3\; \; \; \; \; \; \; (7)$

Since the dot product of two vectors gives a scalar, we can write eq5, eq6 and eq7 as

$p'=a_{11}p+a_{12}q+a_{13}r$

$q'=a_{21}p+a_{22}q+a_{23}r$

$r'=a_{31}p+a_{32}q+a_{33}r$

which can be expressed in the following matrix equation:

$\begin{pmatrix} p'\\ q'\\ r' \end{pmatrix}=\begin{pmatrix} a_{11} &a_{12} &a_{13} \\ a_{21} & a_{22} &a_{23} \\ a_{31} &a_{32} & a_{33} \end{pmatrix}\begin{pmatrix} p\\ q\\ r \end{pmatrix}\; \; \; \; \; \; \; (8)$

where $a_{ij}=x'_i\cdot x_j=\left | x'_i \right |\left | x_j \right |cos\theta_{x'_i,x_j}=cos\theta _{x'_i,x_j}$ .

The matrix containing a_ij is known as the change of basis matrix . We can rewrite eq8 as:

$u'_i=\sum_{j=1,2,3}a_{ij}u_j\; \; \; \; \; \; \; (9)$

where $u'_1=p',u'_2=q',u'_3=r'$ and $u_1=p,u_2=q,u_3=r$ .

Eq9 is often written as a short form by omitting the summation symbol:

$u'_i=a_{ij}u_j\; \; \; \; \; \; \; (10)$

For example, the change of basis from the orthogonal basis x_j to the orthogonal basis x’_ias a rotation about the x₃-axis, where x₃coincides with x’₃ (x’₃= x₃ and is perpendicular to the plane of the page), is depicted as:

To determine the change of basis matrix, we have

$x'_1\cdot x_3=x'_3\cdot x_1=x'_2\cdot x_3=x'_3\cdot x_2=cos90^o=0$

$x'_3\cdot x_3=cos0^o=1$

$x'_1\cdot x_1=x'_2\cdot x_2=cos\theta$

$x'_1\cdot x_2=cos\left ( 90^o-\theta \right )=sin\theta$

$x'_2\cdot x_1=cos\left ( 90^o+\theta \right )=-sin\theta$

Therefore,

$a_{ij}=\begin{pmatrix} a_{11} & a_{12}&a_{13}\\ a_{21}& a_{22}&a_{23}\\ a_{31}& a_{32}& a_{33}\end{pmatrix}=\begin{pmatrix} x'_1\cdot x_1 & x'_1\cdot x_2&x'_1\cdot x_3\\ x'_2\cdot x_1&x'_2\cdot x_2 &x'_2\cdot x_3\\ x'_3\cdot x_1& x'_3\cdot x_2& x'_3\cdot x_3\end{pmatrix}=\begin{pmatrix} cos\theta &sin\theta &0 \\ -sin\theta & cos\theta &0 \\ 0 &0 &1 \end{pmatrix}\; \; \; \; \; \; \; (12)$

Consider again the vector u with reference to a coordinate system that is defined by the orthogonal unit basis vectors x₁, x₂and x₃ (see diagram below).

The rotation of u with respect to this fixed coordinate system is expressed as:

$\begin{pmatrix} p_j\\ q_j\\ r_j \end{pmatrix}=\begin{pmatrix} cos\theta &-sin\theta &0 \\ sin\theta & cos\theta &0 \\ 0 &0 & 1 \end{pmatrix}\begin{pmatrix} p_i\\ q_i\\ r_i \end{pmatrix}\; \; \; or\; \; \; \; \boldsymbol{\mathit{u'}}=R\boldsymbol{\mathit{u}}$

Compared to the change of basis of u that was described earlier, the rotation matrix R is the inverse of the change of basis matrix (similarly, a reflection matrix is the inverse of the change of basis matrix by a reflection). In other words, the rotation of u by θ can be perceived a change of basis, with the coordinate system rotated by –θ. This implies that we can analyse a problem in two ways.

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Constitutive relation: hypothesis 3

The third constitutive relation hypothesis states that the shear stress tensor is zero if the flow involves no shearing of the fluid-body.

A square matrix A is symmetric if a_ij = a_jifor all i and j, i.e. A = A^T, e.g.

$\begin{pmatrix} 1 &5 &-3 \\ 5 &8 &0 \\ -3 &0 &2 \end{pmatrix}$

A square matrix is anti-symmetric (or skew symmetric) if a_ij = –a_jifor all i and j, i.e. –A = A^T, e.g.

$\begin{pmatrix} 0 &-7 &3 \\ 7 &0 &-4 \\ -3 &4 &0 \end{pmatrix}$

If a_ij = –a_ji , the diagonal components of the matrix must all be zero.

In general, we can write a square matrix in the following form:

$A=\frac{1}{2}\left ( A+A^T \right )+\frac{1}{2}\left ( A-A^T \right )=B+C$

where B = (A + A^T)/2, C = (A – A^T)/2 and A^Tis the transpose of A.

Using the transpose identity of (X + Y)^T = X^T+ Y^T (which can be verified by substituting any matrix with any components into the equality), we can show that B is symmetric (B = B^T) and C is anti-symmetric (C = –C^T).

This means that a square matrix can be decomposed into a symmetric and an anti-symmetric component. From the previous section, $\frac{\partial u_k}{\partial x_l}$ is a second-order tensor that can be represented by a 3×3 matrix. Therefore,

$\frac{\partial u_k}{\partial x_l}=\left [ \frac{1}{2}\left ( \frac{\partial u_k}{\partial x_l}+\frac{\partial u_l}{\partial x_k}\right )+\frac{1}{2}\left ( \frac{\partial u_k}{\partial x_l}-\frac{\partial u_l}{\partial x_k}\right )\right ]\; \; \; \; \; \; \; (9)$

where $\frac{1}{2}\left ( \frac{\partial u_k}{\partial x_l}+\frac{\partial u_l}{\partial x_k}\right )$ is symmetric and $\frac{1}{2}\left ( \frac{\partial u_k}{\partial x_l}-\frac{\partial u_l}{\partial x_k}\right )$ is anti-symmetric.

Substitute eq9 in eq8,

$\tau_{ij}=\alpha_{ijkl}\left [ \frac{1}{2}\left ( \frac{\partial u_k}{\partial x_l}+\frac{\partial u_l}{\partial x_k}\right )+\frac{1}{2}\left ( \frac{\partial u_k}{\partial x_l}-\frac{\partial u_l}{\partial x_k}\right )\right ]\; \; \; \; \; \; \; (10)$

Consider a two-dimensional body of fluid ABCD as depicted in the above diagram. The term $\left ( \frac{\partial u_1}{\partial x_2}+\frac{\partial u_2}{\partial x_1}\right )$ tends to shear the fluid. If we consider the rotation of the fluid in the clockwise direction, the term $\left ( \frac{\partial u_1}{\partial x_2}-\frac{\partial u_2}{\partial x_1}\right )$ tends to rotate the fluid. This means that $\frac{\partial u_k}{\partial x_l}$ in eq9 can be non-zero even if the shear component $\left ( \frac{\partial u_1}{\partial x_2}+\frac{\partial u_2}{\partial x_1}\right )$ is zero, as the rotation component $\left ( \frac{\partial u_1}{\partial x_2}-\frac{\partial u_2}{\partial x_1}\right )$ is non-zero. However, hypothesis 3 states that the shear stress tensor τ_ij in eq10 is zero if the flow involves no shearing of the fluid-body. To satisfy hypothesis 3, eq10 becomes:

$\tau_{ij}=\beta_{ijkl}\: \frac{1}{2}\left ( \frac{\partial u_k}{\partial x_l}+\frac{\partial u_l}{\partial x_k}\right )\; \; \; \; \; \; \; (11)$

β_ijkl, which is composed of eighty-one components, characterizes the properties of the fluid. Since $\frac{1}{2}\left ( \frac{\partial u_k}{\partial x_l}+\frac{\partial u_l}{\partial x_k}\right )$ is symmetric, and β_ijklconsists of scalar components, τ_ijis also symmetric.

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