Hot answers tagged

22

None of the interesting equations in physics can be derived from simpler principles, because if they could they wouldn't give any new information. That is, those simpler principles would already fully describe the system. Any new equation, whether it's the Navier-Stokes equations, Einstein's equations, the Schrodinger equation, or whatever, must be ...


19

They are derivable from classical mechanics using either the continuum or molecular points of view. Starting with a continuum view, one applies conservation of mass, momentum, and energy to a control volume and the result is the Navier Stokes equations. The Navier Stokes equations, in the usual form, apply to Newtonian fluids, that is fluids whose stress ...


17

Consider a standard volume of $1\textrm{ m}^3$ of air. This contains on the order of $10^{25}$ molecules of O2 and N2. If you needed to simulate or explain the physics occurring in that volume of air, would you want to model $10^{25}$ molecules and all the interactions between them or, say, 100x100x100 cells based on the Navier-Stokes equations? ...


14

Frank White's Viscous Fluid Flow book contains a good list of these "exact" solutions. I am not sure if it is complete though. I've provided links to a few of the solutions. Steady flow between a fixed and moving plate Axially moving concentric cylinders Flow between rotating concentric cylinders Hagan-Poiseuille flow Combined Couette-Poiseuille flow ...


14

Turbulence is indeed an unsolved problem both in physics and mathematics. Whether it is the "greatest" might be argued but for lack of good metrics probably for a long time. Why it is an unsolved problem from a mathematical point of view read Terry Tao (Fields medal) here : ...


12

I once asked Putterman after a similar colloquium what he meant by this statement, and his answer was "long time tails". Long time tails are fractional powers that appear in the long time behavior of correlation functions, see, for example, here and here. These fractional powers are seen in molecular dynamics (they are more difficult to see experimentally), ...


11

Dear Ondřeji, a good question but a part of the answer is that your equation for the fluid is underdetermined. It treats $p,\rho$ as independent variables. But the physical system only knows how to behave if you also substitute some equation of state, i.e. a function $p=p(\rho)$ or $p=p(\rho,\vec v)$. Note that your Ansatz for the stress-energy tensor ...


10

Actually, there are two different viscosity coefficients. You can see this from the stress tensor $$ \sigma_{ij} = -p_0 \delta_{ij} + \eta \left( \frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} - \frac{2}{3} \delta_{ij} \frac{\partial v_k}{\partial x_k} \right) + \zeta \delta_{ij} \frac{\partial v_k}{\partial x_k} $$ which has the two ...


7

The Reynolds number, with $\rho$ the density, $u$ the velocity magnitude, $\mu$ the viscosity and $L$ some characteristic length scale (e.g. channel height or pipe diameter) is given by $$\text{Re}=\frac{\rho~u~L}{\mu}.$$ This is a dimensionless relation of the ratio of inertial forces ($\rho u u$) to viscous forces ($\mu\frac{u}{L}$). It therefore signifies ...


6

I don't think that such a computation of a theoretical limit of accuracy is possible. There are several sources of uncertainty in weather models: initial and boundary data, parameterizations, numerical instability, rounding and approximation errors of the numerical scheme employed to solve the Navier-Stokes equations for the atmosphere. The term ...


6

I am not sure how useful this "back of the envelope" calculation of reliability of Numerical Weather Prediction is going to be. Several of the assumptions in the question are not correct, and there are other factors to consider. Here are some correcting points: The Weather is 3 dimensional and resides on the surface of the planet up to a height of at ...


6

Turbulence is not one of the great unsolved problems in physics. Physics tells us exactly how turbulence emerges as a direct consequence of local mass and momentum conservation. We can create multiparticle computer models such as lattice gas automata that generate turbulence at large length and time scales. We can write down the equations that govern ...


5

I belive you have it pretty much settled already. If I was to change anything, I would shrink instead of adding more items: Identify the relevant quantities of your system: Energy, Momentum, entropy, electric charge, mass ... Which may or may not be conserved. If you have boundary conditions, most probably you don't have energy and/or momentum ...


5

I don't know a good answer to your first question (I'd be interested in a good text for that myself), but I can answer the second. It's easier to explain if we temporarily imagine $\phi$ represents the concentration of some dye made up of little particles suspended in the fluid. The convective term (aka advective term) is transport of $\phi$ due to the ...


5

Strictly speaking, turbulence doesn't exist in two dimensions. The energy cascade required for turbulence to develop (transfer energy from large scales to small scales) is due to the (incompressible for illustration) vorticity equation: $\frac{D\vec{\omega}}{Dt} = \left(\vec{\omega}\cdot\nabla\right)\vec{v} + \nu\nabla^2\vec{\omega}$ specifically the ...


5

For a single-component fluid, the conservation of mass follows $$ \left(\begin{array}{c}\text{mass of fluid } \\ \text{in volume }\Delta V\end{array}\right)=\left(\begin{array}{c}\text{flux of fluid } \\ \text{in/out of volume }\Delta V\end{array}\right)+\left(\begin{array}{c}\text{sources or} \\ \text{sinks in }\Delta V\end{array}\right) $$ In terms of a ...


5

Gravity acts as a source term in the equations, and it is a source term on the energy and momentum equations. The mass conservation equation is not modified by gravity. So, looking at the momentum equation with gravity, we have: $$ \frac{\partial \rho u_i}{\partial t} + \frac{\partial \rho u_i u_j}{\partial x_j} = -\frac{\partial p}{\partial x_i} + \mu ...


4

As Thomas has commented, the trick is that we only assume first order terms and this convective acceleration would be small of the second order. In fact, that is one of the first assumptions to drop when you consider more general cases. See e.g. Burger's equation for first generalizations and/or Lighthill's equation for source terms arising in the wave ...


4

The problem is that you are not allowed to use Gauss law in this way. The force is given by $$ F_i= \int_{S} n_j\sigma_{ij} dS $$ where $S$ is the surface of the sphere. In order to use Gauss law you have to interpret $S$ as the boundary of some volume. You cannot use the interior of the sphere, because the velocity field is not defined there. But you ...


4

The question you ask is actually the central question of a huge sub-discipline of fluid dynamics. Some have even referred to it as "the last great unsolved problem in classical physics." If you get a complete answer, please let me know! (And don't tell anyone else. Just keep between us, eh?) Generally, there are always small fluctuations in any flow, ...


4

The only thing I can think of is that this is done in cylindrical coordinates: $$\partial_t \rho + \partial_x \rho u_x + \frac{1}{r}\partial_r r \rho u_r=0$$ Using the product rule the last term can be written as: $$\frac{1}{r}\partial_r r \rho u_r=\partial_r \rho u_r + \frac{1}{r}\rho u_r$$ Update: The reference in vol. ii indicates that my feeling about ...


4

It's not that the random motion decreases when the flow rate increases. It is only that the random motion stays the same but the coherent motion dominates. If the diffusion velocity in a gas is 1 and the convective velocity of the flow is 1000 (units don't matter), then the diffusive action can be pretty safely ignored. The important thing to remember is ...


4

First, Navier-Stokes governs the fluid in your setup. So, anything apart from the fluid will be an external force in N-S equation. Body-force means an external force that applies in the bulk of the fluid, like gravity or a magnetic force. Interaction with a "body", as a wing, which is external to the fluid domain, is done through boundary conditions : the ...


4

For Newtonian fluids (such as water and air), the viscous stress tensor, $T_{ij}$, is proportional to the rate of deformation tensor, $D_{ij}$: $$D_{ij} = \frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right)$$ $$T_{ij} = \lambda\Delta\delta_{ij} + 2\mu D_{ij}$$ where $\Delta \equiv D_{11} + D_{22} + D_{33}$. The ...


4

Solutions of the form $$ cos(x_i)e^{-x_j}$$ are common specific solutions of the Navier-Stokes equations in simplified (not simple) problems. These are however problems where inertia is ignored, which you include. (Please note that I am using index notation, with $i,j\in\{1,2,3\}$). $x_j$ is then the wall normal direction. This is actually quite well ...


4

My answer will focus just on the mathematical parts pertaining to partial differential equations. Scale invariance is the fact that some partial differential equations stay the same if you appropriately scale the variables. For example the heat equation (where $\boldsymbol{x}$ is the position vector in 1, 2 or 3D, doesn't matter) $$ \partial_t ...


3

Let me here just derive the equation (6.11) that follows the sentence, you mention. The Navier-Stokes equation (6.6a) reads $$\partial_t v_i + v_j\partial_j v_i = -\partial_i p + f_i + \nu~\partial_j\partial_j v_i.$$ The incompressibility condition (6.6b) reads $$\partial_jv_j=0. $$ Hence we have in the unprimed and the primed point that $$\partial_t ...



Only top voted, non community-wiki answers of a minimum length are eligible