In this lecture we will finally arrive at the full Navier Stokes Equation (or at least all the major players) by introducing viscosity.
To understand what viscosity is it is effectively momentum diffusion. It takes time for the interaction with the surface to transfer upwards through the fluid. This takes time and at different length scales. Consider the fluid moving with velocity \(v_y\) in the \(y\) direction. This is parallel to the surface. The height is in the \(x\) direction. The slab can be defined between \(x, x+\Delta x\) thickness with a \(yz\) planar area of \(A\). We define the mean free path as \(l\) which is the average distance a particle travels before hitting another particle.
With these variables in place we can define the momentum transfer \[p_y = m (v_y(x+l/2) - v_y(x-l/2))\] This is the momentum coming from above and from below. Here \(m\) is the mass of each particle \[p_y = m (v_y(x+l/2) - v_y(x-l/2))\] \[p_y = m \frac{\partial v_y}{\partial x}l\] Now we need to consider the speed at which the momentum is being transfered. We know that each particle has an average \(v_{therm}\) thermal velocity which is approximately \(v = \sqrt{k_BT/m}\) and since the number of particles crossing the slab is \(nAv_{therm}\) we can define the \[\dot p_y(x) = m \frac{\partial v_y}{\partial x}l (nAv_{therm})\] We can define the net force then as \[F_y = \dot p_y(x+\Delta x) - \dot p_y(x) \approx \frac{\partial \dot p_y}{\partial x} \Delta x\] We compute this derivative \[F_y \approx \frac{\partial }{\partial x} (m \frac{\partial v_y}{\partial x}l (nAv_{therm}))\Delta x\] \[F_y \approx A\Delta x\frac{\partial }{\partial x}(m \frac{\partial v_y}{\partial x}l (nv_{therm}))\] We can define a few new terms \[\nu = l v_{therm} \tag{kinematic visc}\] \[\eta = \rho \nu \tag{dynamic visc}\] \[\sigma_{yx} = \eta \frac{\partial v_y}{\partial x}\tag{visc. shear}\] This gives us the force per unit volume as \[\frac{F_y}{A \Delta x} = \frac{\partial \sigma_{yx}}{\partial x}\] This is specific to our problem setup so its time to make this more generic
Let us assume that the viscuous tensor is spanned by a linear combination of various velocity gradients. Which are the individual cross gradients (the first two) and the individual gradient. We want to find what these linear combinations are: \[\sigma_{ij} = \eta(\partial_j v_i + \alpha \partial_i v_j + \beta \delta_{ij}\partial_k v_k)\] Note we used the Einstein summation notation for the last index. \[\] To find what these are we need to look at limiting cases. \[\]
We have thus arrived at the viscous tensor \[\boxed{\sigma_{ij} = \eta(\partial_j v_i + \partial_i v_j - \frac{2}{3} \delta_{ij}\partial_k v_k)}\]
Let us arrive at the full Navier Stokes equation in all of its glory! We can now define the stress tensor as the following, recall from before \[T_{ij} = (\rho v_iv_j + \delta_{ij}P - \sigma_{ij})\] The first componet is the ram pressure and the second component is the thermal pressure the last component is the viscous stress. To rewrite this in a more familiar notation we can say: \[\sigma_{ij} = \eta (\frac{\partial v_i}{\partial v_j} +\frac{\partial v_j}{\partial v_i} - \frac{2}{3}\delta_{ij}\nabla v )\] Since if we wish to add this into the momentum equation we need to take the gradient of this \[\partial_j \sigma_{ij} = \eta ( \nabla\frac{\partial v_i}{\partial v_j} +\frac{\partial v_j}{\partial v_i} - \frac{2}{3}\delta_{ij}\nabla v )\] \[\partial_j \sigma_{ij} = \nu(\nabla^2 v+ \frac{1}{3}\nabla (\nabla \cdot v))\] We now add this to the momentum equation \[\boxed{\frac{\partial v}{\partial t} + v \nabla \cdot v = \frac{-1}{\rho}\nabla P - \nabla \Phi + \nu(\nabla^2 v+ \frac{1}{3}\nabla (\nabla \cdot v))}\] Physics
Reynolds Number: is SUPER important in understanding when the viscous term is important. At low Reynolds number \(\mathcal{R} < 1\) viscous term dominates over the advection term and likewise when its greater the advection term dominates \[\mathcal{R} = \frac{vl}{\nu} \approx \frac{v \nabla\cdot v}{\nu \nabla^2 v}\]
When Reynolds number is higher > 3000 then we have turbulence.
Accreation disks are everywhereeeeeeeee in Astrophysics. They drive the formation of planets to edges of black holes. These disks form when gas have angular momentum. Consider a disk moving around in \(\Omega\) angular frequency and that the disk is razor thin with density \(\Sigma\). We can write down our fluid equations: \[\] Mass conservation \[\frac{\partial \Sigma}{\partial t} + \nabla \cdot (\Sigma v) = 0\] Recall that \(v = Rv_r\) \[\boxed{\frac{\partial \Sigma}{\partial t} + \nabla \cdot (\Sigma Rv_R) = 0}\] Let us then define accreation with \(\dot M(R) = -2\pi R \Sigma v_R\). We know that \(2\pi R\) is the thin ring and thus the matter along the ring is being sucked inwards at \(v_R\) speed and \(\Sigma\) gives the amount of mass thats sucked in. \[\] Momentum Equation \(\phi\) direction \[\] The angular momentum of ring per unit area is \(\Sigma R^2 \Omega\) because the \(I\) per unit area is \(\Sigma R^2\) for the ring and the \(L = I\Omega\). \[\] We can also define the angular momentum loss from this unit area due to advection with radial flow which is just taking the spherical divergence \( \frac{1}{R}\frac{\partial }{\partial R}(R L) = \frac{1}{R}\frac{\partial }{\partial R}(\Sigma R^3 \Omega)\) \[\] Lastly we define Torque per unit volume which is \( T = \frac{\partial L}{\partial t} = \frac{\partial }{\partial t}\Sigma R^2 \Omega\) Thus viscous torque is \[T = 2\pi \Sigma R^3 \frac{\partial \Omega}{\partial R}\] We can now write the momentum equation \[\boxed{\frac{\partial }{\partial t}\Sigma R^2 + \frac{1}{R}\frac{\partial }{\partial R}(\Sigma R^3 \Omega) = \frac{1}{2\pi R} \frac{\partial T}{\partial R}}\] Combine Momentum Eq. and Mass Eq. \[\] With significant amount of algebraic pain that I do not wish to endure because I want to still see my family during the holidays, shows that \[\frac{\partial \Sigma}{\partial t} = -\frac{1}{2\pi R} \frac{\partial}{\partial R}\left[ \frac{1}{\partial_R (R^2\Omega) \frac{\partial T}{\partial R}}\right]\] Let us assume that we are looking at a Keplerian disk. This means that \(\Omega = \sqrt{\frac{GM}{R^3}} \propto R^{-3/2}\). \[\frac{\partial }{\partial R} R^2\Omega = \frac{1}{2}R \Omega\] \[T = 2\pi \Sigma R^3 \frac{\partial \Omega}{\partial R} = -3\pi \Sigma R \Omega\] \[\frac{1}{\partial_R (R^2\Omega) }\frac{\partial T}{\partial R} = -6\pi R^{1/2}\frac{\partial}{\partial R}(\nu \Sigma R^{1/2}) \] We can obtain the time evolution equation for the density: \[\boxed{\frac{\partial \Sigma}{\partial t} = \frac{3}{R} \frac{\partial}{\partial R}\left[R^{1/2} \frac{\partial}{\partial R}(\nu \Sigma R^{1/2})\right]}\] Note that we define the accreation disk with some distribution INITIALLY \[\Sigma = \frac{\delta }{2\pi R_0}\delta (R-R_0)\] One can then solve the unfriendly PDE with the age old technique called "just knowing it" or my favourite "using wolfram alpha". \[\Sigma = \boxed{\frac{m}{\pi R_0^2}\frac{1}{\tau x^{1/4}} e^{-\frac{1+x^2}{\tau}} \mathbb{I}_{1/4}(2x/\tau)}\] Here the \(\mathbb{I}\) is a modified bessel function. And we define dimensionless length and time \(x = \frac{R}{R_0},\quad \tau = \frac{12 \nu t}{R_0}\). I know what you are thinking, this is incredibly trivial and obvious. \[\] We can also then get the radial velocity: \[v_R = -\frac{3}{R_0} \frac{\partial}{\partial x} \left[ \frac{1}{4} \ln x - \frac{1+x^2}{\tau} + \ln \mathbb{I}_{1/4}(2x/\tau)\right]\] Physics