Skip to main content
Physics

Return to Answer

added 32 characters in body
Source Link
Pirx
Pirx
  • 3.8k
  • 14
  • 24

Your intuition is wrong on this. Consider one-dimensional steady flow, say in the $x$-direction, with a velocity gradient in the $y$-direction. Thus the particles at a given level have average velocity $$\bar{\mathbf u}=(u(y), 0, 0)^T,$$ and fluctuating velocities $${\mathbf u}'=(u', v', w')^T.$$

Let's consider particles that at time $t_0$ are located at $(x,y_0)^T$, which have velocities ${\mathbf u}=(u_0+u', v', w')^T.$ These particles will, on average, travel a distance of the mean free path length $l$ at that velocity, before hitting other particles. The particles will thus have migrated to a different $y$ position, where the average particle velocity will be $$\bar{\mathbf u(y)}=(u(y_0)+(y-y_0)\frac{\partial u}{\partial y}, 0, 0)^T.$$ Notice that the average difference in the $x$-component of the velocity of such particles will therefore be proportional to the mean free path $l$ times an integral $I$ over the distribution of $v'$ and $w'$ velocities which does not matter here: We have $y-y_0=I\,l$. The mean velocity difference for such particles is therefore just $\bar{\Delta u}=I\,l\,(\partial u/\partial y)$.

Since the mean velocity is assumed to stay constant, such particles will have their velocity adjusted to the one at their new $y$-position. Viscous forces correspond to the work required to achieve this. These forces must therefore be proportional to the velocity gradient times the mean free path length.

P.S.: Also see the derivation in the Wikipedia article on viscosity.

Your intuition is wrong on this. Consider one-dimensional steady flow, say in the $x$-direction, with a velocity gradient in the $y$-direction. Thus the particles at a given level have average velocity $$\bar{\mathbf u}=(u(y), 0, 0)^T,$$ and fluctuating velocities $${\mathbf u}'=(u', v', w')^T.$$

Let's consider particles that at time $t_0$ are located at $(x,y_0)^T$, which have velocities ${\mathbf u}=(u_0+u', v', w')^T.$ These particles will, on average, travel a distance of the mean free path length $l$ at that velocity, before hitting other particles. The particles will thus have migrated to a different $y$ position, where the average particle velocity will be $$\bar{\mathbf u(y)}=(u(y_0)+(y-y_0)\frac{\partial u}{\partial y}, 0, 0)^T.$$ Notice that the average difference in the $x$-component of the velocity of such particles will therefore be proportional to the mean free path $l$ times an integral $I$ over the distribution of $v'$ and $w'$ velocities which does not matter here. The mean velocity difference for such particles is just $\bar{\Delta u}=I\,l\,(\partial u/\partial y)$.

Since the mean velocity is assumed to stay constant, such particles will have their velocity adjusted to the one at their new $y$-position. Viscous forces correspond to the work required to achieve this. These forces must therefore be proportional to the velocity gradient times the mean free path length.

Your intuition is wrong on this. Consider one-dimensional steady flow, say in the $x$-direction, with a velocity gradient in the $y$-direction. Thus the particles at a given level have average velocity $$\bar{\mathbf u}=(u(y), 0, 0)^T,$$ and fluctuating velocities $${\mathbf u}'=(u', v', w')^T.$$

Let's consider particles that at time $t_0$ are located at $(x,y_0)^T$, which have velocities ${\mathbf u}=(u_0+u', v', w')^T.$ These particles will, on average, travel a distance of the mean free path length $l$ at that velocity, before hitting other particles. The particles will thus have migrated to a different $y$ position, where the average particle velocity will be $$\bar{\mathbf u(y)}=(u(y_0)+(y-y_0)\frac{\partial u}{\partial y}, 0, 0)^T.$$ Notice that the average difference in the $x$-component of the velocity of such particles will therefore be proportional to the mean free path $l$ times an integral $I$ over the distribution of $v'$ and $w'$ velocities which does not matter here: We have $y-y_0=I\,l$. The mean velocity difference for such particles is therefore just $\bar{\Delta u}=I\,l\,(\partial u/\partial y)$.

Since the mean velocity is assumed to stay constant, such particles will have their velocity adjusted to the one at their new $y$-position. Viscous forces correspond to the work required to achieve this. These forces must therefore be proportional to the velocity gradient times the mean free path length.

P.S.: Also see the derivation in the Wikipedia article on viscosity.

Source Link
Pirx
Pirx
  • 3.8k
  • 14
  • 24

Your intuition is wrong on this. Consider one-dimensional steady flow, say in the $x$-direction, with a velocity gradient in the $y$-direction. Thus the particles at a given level have average velocity $$\bar{\mathbf u}=(u(y), 0, 0)^T,$$ and fluctuating velocities $${\mathbf u}'=(u', v', w')^T.$$

Let's consider particles that at time $t_0$ are located at $(x,y_0)^T$, which have velocities ${\mathbf u}=(u_0+u', v', w')^T.$ These particles will, on average, travel a distance of the mean free path length $l$ at that velocity, before hitting other particles. The particles will thus have migrated to a different $y$ position, where the average particle velocity will be $$\bar{\mathbf u(y)}=(u(y_0)+(y-y_0)\frac{\partial u}{\partial y}, 0, 0)^T.$$ Notice that the average difference in the $x$-component of the velocity of such particles will therefore be proportional to the mean free path $l$ times an integral $I$ over the distribution of $v'$ and $w'$ velocities which does not matter here. The mean velocity difference for such particles is just $\bar{\Delta u}=I\,l\,(\partial u/\partial y)$.

Since the mean velocity is assumed to stay constant, such particles will have their velocity adjusted to the one at their new $y$-position. Viscous forces correspond to the work required to achieve this. These forces must therefore be proportional to the velocity gradient times the mean free path length.