Linear momentum

Hey there.

As you might have noticed on the previous post, the one about the mass, its conservation laws, and its mathematical formulations. It enough to replace $\phi$ by some other quantity, and perhaps we can tell something interesting about our fluid, perhaps we can withdraw interesting conservation laws...or not.

If we choose poorly our quantity $\phi$, we won't be able to interpret physically its integral over the fluid volume, and therefore the RHS, even though it is correct, might mean nothing to us. So lets check on another quantity $\phi$ such that its integral over the fluid volume has some importance from the physical point of view.

If we go back to our first lessons on Classical Mechanics, we see that Newtons Second law can be mathematically stated as follows:

\[\frac{d\vec{P}}{dt}=\vec{F}\]

Where $\vec{P}$ is the linear moment of our system of interest, and $\vec{F}$ represents sum of all the forces acting upon it.

If we set our system as our fluid volume Vf, we then can rewrite:

\[\frac{d\vec{P}}{dt}=\frac{d}{dt}\int_{V_f(t)}\rho\vec{u}\,d\Omega\]

According to Newton's second law, this equals the sum of all the forces acting on the fluid. But, what kind of forces can act on our fluid? Well, we have all the forces like the Gravitational, the Lorentz force, or the centripetal force that act on every fluid particle that belongs to Vf. However we also have forces like a direct pressure on one side of the fluid, the air, and the atmospheric pressure that act all over the surface of the fluid, BUT NOT INSIDE OF IT. This distinction is very important because it allows us to classify the forces in Volumetric forces, and Surface forces. the first ones, as their name indicates, act over all the volume Vf, and their magnitude is proportional to the volume Vf. Surface forces are a bit different, they act only on surfaces of the fluid and their final magnitude depends on the surface of the fluid.

Keeping the above classification in mind, we can write the volumetric forces as an integral over Vf(t) of $\rho\vec{a}$ where $\vec{a}$ represents the volumetric acceleration of a given force. We write:

\[\vec{F}_{Volumetric}=\int_{V_f(t)}\rho\vec{a}\,d\Omega\]

For the surface forces we can write this:

\[\vec{F}_{Surface}=\int_{\partial V_f(t)}\vec{f}\,d\sigma\]

Where $f$ is the force per unit area, or pressure. We have to realize that this pressure depends on the specific orientation of the surface over which it is acting upon. In the most general case, at a given point in space the force acting over a differential area section $\vec{n_1}$ is, in general, different to the pressure exerted over another surface element $\vec{n_2}$ even if both $\vec{n_1},\vec{n_2}$ are at the same point in space. If we write the final integral equation, (that is, equating both sides of Newtons second law, and using our version of Reynols Transport Theorem for each scalar components of the vectors involved):

\[\int_{V_f(t)}\rho\vec{a}\,d\Omega+\int_{\partial V_f(t)} \vec{f}\,d\sigma=\frac{d}{dt}\int_{V_c(t)}\rho\vec{u}\,d\Omega+\int_{\partial V_c(t)}\rho \vec{u} [(\vec{u}-\vec{u_c})\cdot\vec{n}]\,d\sigma\]

We see must realize that it is not very general. Its not general int he sense that the surface forces depend on the specific orientation of our surface. This integral formulation, at this stage, assumes we have the following info on our pressures: we know at each point on the fluid, what the pressure would be if the surface of our fluid element at that point would be some $\vec{n}$. Its too much to ask. it might look like it is an infinite amount of information (since there are infinitely many different orientations!!) Like I said, this is not a very useful result. Let's see how we can generalize it, and avoid this awkward result.

The above awkward result is equivalent to the result that arises form the motion of a rigid body: What is the moment of inertia of a given body if we spin-it through an axis $\vec{n}$ with angular velocity $\omega$ ? Again, there are infinitely many possible orientations for this axis, and infinitely many different values for $\omega$. So how do we characterize this situation without turning it into a case-specific type of thing? Lets see, i could also ask: for a given $\vec{\omega} (=\omega\vec{n})$ what the moment of inertia is? the answer is: The tensor of Inertia.

The tensor of inertia allows you to compute the moment of inertia of a body that is being rotated with angular velocity $\vec{\omega}$. That is, exists an entity $\vec{\vec{\Pi}}$ such that:

\[\vec{I}=\vec{\vec{\Pi}}\cdot\vec{\omega}\]

Where $\vec{I}$ is the moment of inertia of the body. Since $\vec{I}$ and $\vec{\omega}$ are not necessarily parallel, our entity $\vec{\vec{\Pi}}$ must, at least, be a 3x3 matrix.

So, one question comes to my mind. How can all the infinite information that we were talking about  be contained in these Tensor? The answer is simple. All the infinite information that we were discussing has been isolated into $\vec{\omega}$. The Tensor only contributes with the specific characteristics of the body itself.

Its components can be calculated explicitly depending on the specific shape of the object. I'm not going to present this computation, Ill simply state that it's construction is possible and depends only on the geometry of the object we are considering.

The cool think is that we can use the same trick for our problem with $f$. We would like to separate the dependence with the normal vector $\vec{n}$ from the physical characteristics of our fluid. The tensor involved will be called, the stress tensor, and we will denote it by $\vec{\vec{T}}$.

With new super tool, we can write our surface forces as:

\[\vec{F}_{Surface}=\int_{\partial V_f(t)} \vec{\vec{T}}\cdot\vec{n}\,d\sigma\]

Finally, our improved and generalized form of Newton's second law can be written in it integral formulations as:

\[\int_{V_f(t)}\rho\vec{a}\,d\Omega+\int_{\partial V_f(t)} \vec{\vec{T}}\cdot\vec{n}\,d\sigma=\frac{d}{dt}\int_{V_c(t)}\rho\vec{u}\,d\Omega+\int_{\partial V_c(t)}\rho \vec{u} [(\vec{u}-\vec{u_c})\cdot\vec{n}]\,d\sigma\]

Now, proceeding in a similar way to the last post, let's derive the differential form of it. 

Using Stokes theorem on the terms involving surfaces integrals, and introducing the time derivative inside the integral (by setting, with no loss of generality Vc(t)=Vc) we can write for the i-th component of the velocity:

\[0=\int_{V_c}\left(\rho a_i + (\nabla\cdot\vec{\vec{T}})_i-\frac{\partial(\rho u_i)}{\partial t}-\nabla\cdot(\rho\vec{u}u_i)\right)\,d\Omega\]

Using the same argument as before, namely that this equality holds for any Vc that you can think off, we equal the integrand to zero. next, we use the identity 

\[\nabla\cdot(\psi\vec{A})=\nabla(\psi)\vec{A}+\psi\nabla\cdot\vec{A}\] 

and identifing $\psi\rightarrow u_i$ and $\vec{A}\rightarrow\rho\vec{u}$, we have:

\[\rho\vec{a}+\nabla\cdot\vec{\vec{T}}=\rho\left(\frac{\partial}{\partial t}+(\vec{u}\nabla)\right)\vec{u}+\vec{u}\left(\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\vec{u})\right)\]

The first parenthesis is nothing but the material derivative operator acting on $\vec{u}$. The second parenthesis on the right, thanks to the continuity equation vanishes. The resulting equation is called the differential formulation of the conservation of liner momentum:

\[\rho\frac{D\vec{u}}{Dt}=\nabla\cdot\vec{\vec{T}}+\rho\vec{a}\]

With this, I finish the $\phi=\rho\vec{u}$ case. Again, I hope you enjoyed as much as I did.