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1. Microscopic picture: In kinetic theory the distribution function has the form $$ f(\vec{v})=f_0(\vec{c}^2)+\delta f(\vec{c}) $$ where $f_0(\vec{c}^2)=\exp(\mu/T)\exp(-m\vec{c}^2/(2T))$ is the equilibrium (Boltzmann) distribution, $\vec{c}=\vec{v}-\vec{u}$ is the velocity of the particles relative to the fluid, and $\delta f$ is a non-equilibrium distribution that leads to dissipation, friction, and entropy generation. In our case $\vec{u}$ is sheared flow between two plates, $u_x(y)=u_0y/L$.

The non-equilibrium distribution $\delta f$ can be obtained from solutions of the Boltzmann equation (for example, in relaxation time approximation). The magnitude of $\delta f$ determines the shear viscosity. We find $$ \delta f(c) = -\frac{\eta}{2PT}\, f_0(c) \sigma_{ij}c^ic^j $$ where $$ \sigma_{ij}=\nabla_iu_j+\nabla_ju_i-\frac{2}{3}\delta_{ij}(\nabla\cdot u) $$ In equilibrium the velocity distribution in the rest frame of the fluid is a circle. Taking dissipation into account we find an ellipsoidal deformation. There is an enhancement of particles with positive $c_y$ and negative $c_x$, as well as negative $c_y$ and positive $c_x$.

This is exactly what we expect, because this type of distribution will tend to equalize the flow velocity.

The corresponding momentum flux is $$ \delta T_{ij} = \int d\Gamma\, p_iv_j \delta f(c) $$ which is obviously symmetric. There is both a flux of $x$ momentum in the $y$-direction, and vice-versa. Again, this is clear from the underlying velocity distribution.

2. Macroscopic picture: The macroscopic stress is $$ \Delta T_{ij}=-\eta\sigma_{ij} $$ which has components $\delta T_{xy}=\delta T_{yx} = -\eta u_0/L$. Note that the stress is constant, so it does not lead to acceleration of the fluid. The momentum equations are $$ \partial_0 \pi_x = -\nabla_y \delta T_{xy} \quad \partial_0 \pi_y = -\nabla_x \delta T_{yx} $$ where $\pi_i=\rho u_i$ is the momentum density of the fluid. One way to say this is that if we consider a fluid cell, the forces on the front and rear, as well as bottom and top face, cancel.

The only place where the forces don't cancel is at the edges, which is the top and bottom plate, where you measure a forces. But this kind of flow cannot have a front and rear boundary, so you cannot measure a force in that direction.

3. Final comment: One way to see that both components of $T_{xy}$ are physical is to compute the dissipative energy current $$ \delta j_i^\epsilon = u_j \delta T_{ij}. $$ In the present case the only component it $\delta j^\epsilon_y=u_x \delta T_{yx}$, which flows into the fluid, orthogonal to the direction of flow. This makes sense: We do work on the boundary, and the energy flows into the fluid and is dissipated as heat.

Also: The total energy dissipated in the flow (which you can measure, by observing the rate of heating) is $$ \dot E = \frac{\eta}{2} \int d^3x \, (\delta T_{ij})^2 $$ which you would obviously get wrong by a factor of 2 if only one of the two components $\delta T_{xy}$ and $\delta T_{yx}$ is not zero.

Postscript: I think I finally understand your main issue. You are asking: "Why is it that I get away with poor-man's kinetic theory in estimating the flux of $x$-momentum in the $y$-direction, but the same argument does not immediately give a flux of $y$-momentum in the $x$-direction?

Poor-man's kinetic theory claims that particles have the velocity of the local flow $u_x$, and superimposed is a random drift velocity $v_y$. This gives an imbalance across planes in the $x$-direction, because $u_x$ depends on $y$.

Now consider a face in the $y$-direction. On average there are as many particles with $+v_y$ and $-v_y$, but the particles with $\pm v_y$ originate from regions with different $u_x$, so they have different fluxes, giving an imbalance in the $y$ momentum. This is obviously a more tedious argument, which is why poor man's theory is used for the first problem, and not the second.

1. Microscopic picture: In kinetic theory the distribution function has the form $$ f(\vec{v})=f_0(\vec{c}^2)+\delta f(\vec{c}) $$ where $f_0(\vec{c}^2)=\exp(\mu/T)\exp(-m\vec{c}^2/(2T))$ is the equilibrium (Boltzmann) distribution, $\vec{c}=\vec{v}-\vec{u}$ is the velocity of the particles relative to the fluid, and $\delta f$ is a non-equilibrium distribution that leads to dissipation, friction, and entropy generation. In our case $\vec{u}$ is sheared flow between two plates, $u_x(y)=u_0y/L$.

The non-equilibrium distribution $\delta f$ can be obtained from solutions of the Boltzmann equation (for example, in relaxation time approximation). The magnitude of $\delta f$ determines the shear viscosity. We find $$ \delta f(c) = -\frac{\eta}{2PT}\, f_0(c) \sigma_{ij}c^ic^j $$ where $$ \sigma_{ij}=\nabla_iu_j+\nabla_ju_i-\frac{2}{3}\delta_{ij}(\nabla\cdot u) $$ In equilibrium the velocity distribution in the rest frame of the fluid is a circle. Taking dissipation into account we find an ellipsoidal deformation. There is an enhancement of particles with positive $c_y$ and negative $c_x$, as well as negative $c_y$ and positive $c_x$.

This is exactly what we expect, because this type of distribution will tend to equalize the flow velocity.

The corresponding momentum flux is $$ \delta T_{ij} = \int d\Gamma\, p_iv_j \delta f(c) $$ which is obviously symmetric. There is both a flux of $x$ momentum in the $y$-direction, and vice-versa. Again, this is clear from the underlying velocity distribution.

2. Macroscopic picture: The macroscopic stress is $$ \Delta T_{ij}=-\eta\sigma_{ij} $$ which has components $\delta T_{xy}=\delta T_{yx} = -\eta u_0/L$. Note that the stress is constant, so it does not lead to acceleration of the fluid. The momentum equations are $$ \partial_0 \pi_x = -\nabla_y \delta T_{xy} \quad \partial_0 \pi_y = -\nabla_x \delta T_{yx} $$ where $\pi_i=\rho u_i$ is the momentum density of the fluid. One way to say this is that if we consider a fluid cell, the forces on the front and rear, as well as bottom and top face, cancel.

The only place where the forces don't cancel is at the edges, which is the top and bottom plate, where you measure a forces. But this kind of flow cannot have a front and rear boundary, so you cannot measure a force in that direction.

3. Final comment: One way to see that both components of $T_{xy}$ are physical is to compute the dissipative energy current $$ \delta j_i^\epsilon = u_j \delta T_{ij}. $$ In the present case the only component it $\delta j^\epsilon_y=u_x \delta T_{yx}$, which flows into the fluid, orthogonal to the direction of flow. This makes sense: We do work on the boundary, and the energy flows into the fluid and is dissipated as heat.

Also: The total energy dissipated in the flow (which you can measure, by observing the rate of heating) is $$ \dot E = \frac{\eta}{2} \int d^3x \, (\delta T_{ij})^2 $$ which you would obviously get wrong by a factor of 2 if only one of the two components $\delta T_{xy}$ and $\delta T_{yx}$ is not zero.

Postscript: I think I finally understand your main issue. You are asking: "Why is it that I get away with poor-man's kinetic theory in estimating the flux of $x$-momentum in the $y$-direction, but the same argument does not immediately give a flux of $y$-momentum in the $x$-direction?

Poor-man's kinetic theory claims that particles have the velocity of the local flow $u_x$, and superimposed is a random drift velocity $v_y$. This gives an imbalance across planes in the $x$-direction, because $u_x$ depends on $y$.

Now consider a face in the $y$-direction. On average there are as many particles with $+v_y$ and $-v_y$, but the particles with $\pm v_y$ originate from regions with different $u_x$, so they have different fluxes, giving an imbalance in the $y$ momentum. This is slightly harder to visualize, which is why you might as well switch to real kinetic theory, but a necessary consequence of the intuitive argument for the flux in $x$ momentum.

1. Microscopic picture: In kinetic theory the distribution function has the form $$ f(\vec{v})=f_0(\vec{c}^2)+\delta f(\vec{c}) $$ where $f_0(\vec{c}^2)=\exp(\mu/T)\exp(-m\vec{c}^2/(2T))$ is the equilibrium (Boltzmann) distribution, $\vec{c}=\vec{v}-\vec{u}$ is the velocity of the particles relative to the fluid, and $\delta f$ is a non-equilibrium distribution that leads to dissipation, friction, and entropy generation. In our case $\vec{u}$ is sheared flow between two plates, $u_x(y)=u_0y/L$.

The non-equilibrium distribution $\delta f$ can be obtained from solutions of the Boltzmann equation (for example, in relaxation time approximation). The magnitude of $\delta f$ determines the shear viscosity. We find $$ \delta f(c) = -\frac{\eta}{2PT}\, f_0(c) \sigma_{ij}c^ic^j $$ where $$ \sigma_{ij}=\nabla_iu_j+\nabla_ju_i-\frac{2}{3}\delta_{ij}(\nabla\cdot u) $$ In equilibrium the velocity distribution in the rest frame of the fluid is a circle. Taking dissipation into account we find an ellipsoidal deformation. There is an enhancement of particles with positive $c_y$ and negative $c_x$, as well as negative $c_y$ and positive $c_x$.

This is exactly what we expect, because this type of distribution will tend to equalize the flow velocity.

The corresponding momentum flux is $$ \delta T_{ij} = \int d\Gamma\, p_iv_j \delta f(c) $$ which is obviously symmetric. There is both a flux of $x$ momentum in the $y$-direction, and vice-versa. Again, this is clear from the underlying velocity distribution.

2. Macroscopic picture: The macroscopic stress is $$ \Delta T_{ij}=-\eta\sigma_{ij} $$ which has components $\delta T_{xy}=\delta T_{yx} = -\eta u_0/L$. Note that the stress is constant, so it does not lead to acceleration of the fluid. The momentum equations are $$ \partial_0 \pi_x = -\nabla_y \delta T_{xy} \quad \partial_0 \pi_y = -\nabla_x \delta T_{yx} $$ where $\pi_i=\rho u_i$ is the momentum density of the fluid. One way to say this is that if we consider a fluid cell, the forces on the front and rear, as well as bottom and top face, cancel.

The only place where the forces don't cancel is at the edges, which is the top and bottom plate, where you measure a forces. But this kind of flow cannot have a front and rear boundary, so you cannot measure a force in that direction.

3. Final comment: One way to see that both components of $T_{xy}$ are physical is to compute the dissipative energy current $$ \delta j_i^\epsilon = u_j \delta T_{ij}. $$ In the present case the only component it $\delta j^\epsilon_y=u_x \delta T_{yx}$, which flows into the fluid, orthogonal to the direction of flow. This makes sense: We do work on the boundary, and the energy flows into the fluid and is dissipated as heat.

Also: The total energy dissipated in the flow (which you can measure, by observing the rate of heating) is $$ \dot E = \frac{\eta}{2} \int d^3x \, (\delta T_{ij})^2 $$ which you would obviously get wrong by a factor of 2 if only one of the two components $\delta T_{xy}$ and $\delta T_{yx}$ is not zero.

1. Microscopic picture: In kinetic theory the distribution function has the form $$ f(\vec{v})=f_0(\vec{c}^2)+\delta f(\vec{c}) $$ where $f_0(\vec{c}^2)=\exp(\mu/T)\exp(-m\vec{c}^2/(2T))$ is the equilibrium (Boltzmann) distribution, $\vec{c}=\vec{v}-\vec{u}$ is the velocity of the particles relative to the fluid, and $\delta f$ is a non-equilibrium distribution that leads to dissipation, friction, and entropy generation. In our case $\vec{u}$ is sheared flow between two plates, $u_x(y)=u_0y/L$.

The non-equilibrium distribution $\delta f$ can be obtained from solutions of the Boltzmann equation (for example, in relaxation time approximation). The magnitude of $\delta f$ determines the shear viscosity. We find $$ \delta f(c) = -\frac{\eta}{2PT}\, f_0(c) \sigma_{ij}c^ic^j $$ where $$ \sigma_{ij}=\nabla_iu_j+\nabla_ju_i-\frac{2}{3}\delta_{ij}(\nabla\cdot u) $$ In equilibrium the velocity distribution in the rest frame of the fluid is a circle. Taking dissipation into account we find an ellipsoidal deformation. There is an enhancement of particles with positive $c_y$ and negative $c_x$, as well as negative $c_y$ and positive $c_x$.

This is exactly what we expect, because this type of distribution will tend to equalize the flow velocity.

The corresponding momentum flux is $$ \delta T_{ij} = \int d\Gamma\, p_iv_j \delta f(c) $$ which is obviously symmetric. There is both a flux of $x$ momentum in the $y$-direction, and vice-versa. Again, this is clear from the underlying velocity distribution.

2. Macroscopic picture: The macroscopic stress is $$ \Delta T_{ij}=-\eta\sigma_{ij} $$ which has components $\delta T_{xy}=\delta T_{yx} = -\eta u_0/L$. Note that the stress is constant, so it does not lead to acceleration of the fluid. The momentum equations are $$ \partial_0 \pi_x = -\nabla_y \delta T_{xy} \quad \partial_0 \pi_y = -\nabla_x \delta T_{yx} $$ where $\pi_i=\rho u_i$ is the momentum density of the fluid. One way to say this is that if we consider a fluid cell, the forces on the front and rear, as well as bottom and top face, cancel.

The only place where the forces don't cancel is at the edges, which is the top and bottom plate, where you measure a forces. But this kind of flow cannot have a front and rear boundary, so you cannot measure a force in that direction.

3. Final comment: One way to see that both components of $T_{xy}$ are physical is to compute the dissipative energy current $$ \delta j_i^\epsilon = u_j \delta T_{ij}. $$ In the present case the only component it $\delta j^\epsilon_y=u_x \delta T_{yx}$, which flows into the fluid, orthogonal to the direction of flow. This makes sense: We do work on the boundary, and the energy flows into the fluid and is dissipated as heat.

Also: The total energy dissipated in the flow (which you can measure, by observing the rate of heating) is $$ \dot E = \frac{\eta}{2} \int d^3x \, (\delta T_{ij})^2 $$ which you would obviously get wrong by a factor of 2 if only one of the two components $\delta T_{xy}$ and $\delta T_{yx}$ is not zero.

Postscript: I think I finally understand your main issue. You are asking: "Why is it that I get away with poor-man's kinetic theory in estimating the flux of $x$-momentum in the $y$-direction, but the same argument does not immediately give a flux of $y$-momentum in the $x$-direction?

Poor-man's kinetic theory claims that particles have the velocity of the local flow $u_x$, and superimposed is a random drift velocity $v_y$. This gives an imbalance across planes in the $x$-direction, because $u_x$ depends on $y$.

Now consider a face in the $y$-direction. On average there are as many particles with $+v_y$ and $-v_y$, but the particles with $\pm v_y$ originate from regions with different $u_x$, so they have different fluxes, giving an imbalance in the $y$ momentum. This is obviously a more tedious argument, which is why poor man's theory is used for the first problem, and not the second.

1. Microscopic picture: In kinetic theory the distribution function has the form $$ f(\vec{v})=f_0(\vec{c}^2)+\delta f(\vec{c}) $$ where $f_0(\vec{c}^2)=\exp(\mu/T)\exp(-m\vec{c}^2/(2T))$ is the equilibrium (Boltzmann) distribution, $\vec{c}=\vec{v}-\vec{u}$ is the velocity of the particles relative to the fluid, and $\delta f$ is a non-equilibrium distribution that leads to dissipation, friction, and entropy generation. In our case $\vec{u}$ is sheared flow between two plates, $u_x(y)=u_0y/L$.

The non-equilibrium distribution $\delta f$ can be obtained from solutions of the Boltzmann equation (for example, in relaxation time approximation). The magnitude of $\delta f$ determines the shear viscosity. We find $$ \delta f(c) = -\frac{\eta}{2PT}\, f_0(c) \sigma_{ij}c^ic^j $$ where $$ \sigma_{ij}=\nabla_iu_j+\nabla_ju_i-\frac{2}{3}\delta_{ij}(\nabla\cdot u) $$ In equilibrium the velocity distribution in the rest frame of the fluid is a circle. Taking dissipation into account we find an ellipsoidal deformation. There is an enhancement of particles with positive $c_y$ and negative $c_x$, as well as negative $c_y$ and positive $c_x$.

This is exactly what we expect, because this type of distribution will tend to equalize the flow velocity.

The corresponding momentum flux is $$ \delta T_{ij} = \int d\Gamma\, p_iv_j \delta f(c) $$ is obviously symmetric. There is both a flux of $x$ momentum in the $y$-direction, and vice-versa. Again, this is clear from the underlying velocity distribution.

2. Macroscopic picture: The macroscopic stress is $$ \Delta T_{ij}=-\eta\sigma_{ij} $$ which has components $\delta T_{xy}=\delta T_{yx} = -\eta u_0/L$. Note that the stress is constant, so it does not lead to acceleration of the fluid. The momentum equations are $$ \partial_0 \pi_x = -\nabla_y \delta T_{xy} \quad \partial_0 \pi_y = -\nabla_x \delta T_{yx} $$ where $\pi_i=\rho u_i$ is the momentum density of the fluid. One way to say this is that if we consider a fluid cell, the forces on the front and rear, as well as bottom and top face, cancel.

The only place where the forces don't cancel is at the edges, which is the top and bottom plate, where you measure a forces. But this kind of flow cannot have a front and rear boundary.

3. Final comment: One way to see that both components of $T_{xy}$ are physical is to compute the dissipative energy current $$ \delta j_i^\epsilon = u_j \delta T_{ij}. $$ In the present case the only component it $\delta j^\epsilon_y=u_x \delta T_{yx}$, which flows into the fluid, orthogonal to the direction of flow. This makes sense: We do work on the boundary, and the energy flows into the fluid and is dissipated as heat.

1. Microscopic picture: In kinetic theory the distribution function has the form $$ f(\vec{v})=f_0(\vec{c}^2)+\delta f(\vec{c}) $$ where $f_0(\vec{c}^2)=\exp(\mu/T)\exp(-m\vec{c}^2/(2T))$ is the equilibrium (Boltzmann) distribution, $\vec{c}=\vec{v}-\vec{u}$ is the velocity of the particles relative to the fluid, and $\delta f$ is a non-equilibrium distribution that leads to dissipation, friction, and entropy generation. In our case $\vec{u}$ is sheared flow between two plates, $u_x(y)=u_0y/L$.

The non-equilibrium distribution $\delta f$ can be obtained from solutions of the Boltzmann equation (for example, in relaxation time approximation). The magnitude of $\delta f$ determines the shear viscosity. We find $$ \delta f(c) = -\frac{\eta}{2PT}\, f_0(c) \sigma_{ij}c^ic^j $$ where $$ \sigma_{ij}=\nabla_iu_j+\nabla_ju_i-\frac{2}{3}\delta_{ij}(\nabla\cdot u) $$ In equilibrium the velocity distribution in the rest frame of the fluid is a circle. Taking dissipation into account we find an ellipsoidal deformation. There is an enhancement of particles with positive $c_y$ and negative $c_x$, as well as negative $c_y$ and positive $c_x$.

This is exactly what we expect, because this type of distribution will tend to equalize the flow velocity.

The corresponding momentum flux is $$ \delta T_{ij} = \int d\Gamma\, p_iv_j \delta f(c) $$ which is obviously symmetric. There is both a flux of $x$ momentum in the $y$-direction, and vice-versa. Again, this is clear from the underlying velocity distribution.

2. Macroscopic picture: The macroscopic stress is $$ \Delta T_{ij}=-\eta\sigma_{ij} $$ which has components $\delta T_{xy}=\delta T_{yx} = -\eta u_0/L$. Note that the stress is constant, so it does not lead to acceleration of the fluid. The momentum equations are $$ \partial_0 \pi_x = -\nabla_y \delta T_{xy} \quad \partial_0 \pi_y = -\nabla_x \delta T_{yx} $$ where $\pi_i=\rho u_i$ is the momentum density of the fluid. One way to say this is that if we consider a fluid cell, the forces on the front and rear, as well as bottom and top face, cancel.

The only place where the forces don't cancel is at the edges, which is the top and bottom plate, where you measure a forces. But this kind of flow cannot have a front and rear boundary, so you cannot measure a force in that direction.

3. Final comment: One way to see that both components of $T_{xy}$ are physical is to compute the dissipative energy current $$ \delta j_i^\epsilon = u_j \delta T_{ij}. $$ In the present case the only component it $\delta j^\epsilon_y=u_x \delta T_{yx}$, which flows into the fluid, orthogonal to the direction of flow. This makes sense: We do work on the boundary, and the energy flows into the fluid and is dissipated as heat.

Also: The total energy dissipated in the flow (which you can measure, by observing the rate of heating) is $$ \dot E = \frac{\eta}{2} \int d^3x \, (\delta T_{ij})^2 $$ which you would obviously get wrong by a factor of 2 if only one of the two components $\delta T_{xy}$ and $\delta T_{yx}$ is not zero.