# Derivation for kinetic energy flux

I'm working to derive kinetic energy flux for fluids. I could not find a derivation online. I know from literature the correct answer is $\phi_{kin} = (1/2) \rho v^3$.

The specific context is in snapshots of fluids, so its okay to assume constant acceleration.

I being with the definition of kinetic energy

$$E_k = \frac{1}{2} m v^2 \, .$$

We can assume constant acceleration between each snapshot (each time we can view the fluid). We are interested in calculating the kinetic energy flux of a system with one point, molecule or pixel and how it moves between snapshots.

We begin by placing a single particle in a box. It has energy only due to kinetic energy. As it has kinetic energy, from our previous definition of kinetic energy it must be moving.

Consider this particle in a box moving in an arbitrary direction as depicted in the figure below. We shrink the cube so the particle must pass through it over the duration of the snapshot and measure the flux once the particle has moved through a face of the box. The box has edges which are $\epsilon$ wider then the diameter of the particle.

We describe the particle (with a vector field ) by a Dirac delta with an associated $E_k$ scalar. The vector field therefore looks like

$$\underline{F_{k}} = E_K \ \delta^3(\underline{r} - \underline{r'})$$

But - this isn't actually a vector, because the Dirac delta is a scalar. But, if I include a vector, it screws up the result.

The kinetic energy flux is defined as

$$\phi_k = \oint_S \underline{F_{k}} \cdot \underline{\hat{n}} \,dS$$

We assume the point particle does not go through an edge, so we arbitrarily take the x-y face.

$$\phi_{k} = \int_{x-\epsilon}^{x+\epsilon} \int_{y-\epsilon}^{y+\epsilon} E_k \delta^3(\underline{r} - \underline{r'}) \cdot \underline{\hat{n}} dx dy$$

Taking the normal which will get rid of one dimension. Thus our integral becomes

$$\phi_{k} = \int_{x-\epsilon}^{x+\epsilon} \int_{y-\epsilon}^{y+\epsilon} E_k \delta^2(\underline{r} - \underline{r'}) \cdot \underline{\hat{n}} dx dy$$

Again, I think the fault is in the above step--I can't just get rid of one dimension of the Dirac delta by using the dot product with the normal as an excuse. (also, I'm dotting something that isn't a vector with a vector)--but if I don't do that, I'll end up with a 3D Dirac delta in a 2D integral, which seems dodgy

We invoke the property of the Dirac delta

$$\int_{a-\epsilon}^{a+\epsilon} f(x) \delta(x-a) dx = f(a)$$

And we are left with

$$\phi_{k} = E_{k} \int_{x-\epsilon}^{x+\epsilon} \int_{y-\epsilon}^{y+\epsilon} \delta^2(\underline{r} - \underline{r'}) \cdot \underline{\hat{n}} dx dy = E_{k}$$

Our flux therefore is $\phi_{k} = E_{k} = \frac{1}{2} \ m \ (v_{2}^2 - v_{1}^2)$.

We can set the initial velocity to zero and consider the $E_{k}$ in each snapshot at that instant. Thus

$$E_k = \frac{1}{2} \ m v^{2}$$

This is valid for a single particle. Extrapolating to a fluid and feeding in the density rather than mass, we find it:

$$F_{kin} = \frac{1}{2} \rho v^3$$

Which is the correct expression. (but with a flawed derivation).

Reference (as requested), source: Solar Prominences Vial, Engvold et al • Can you point to a reference in the literature to what you call "kinetic energy flux" here? I am not familiar with the expression $\phi_{kin} = \frac{1}{2} \rho\,v^3$, and cannot see how a scalar expression of this kind would make much sense. – Pirx Feb 14 '17 at 15:31
• I'm confused by your first equation for $\underline{F_k}$. Do you mean to write $F_k=E_k\delta^3(\underline{r}-\underline{r}')$? Recall that the Dirac delta "in 3D" is still a scalar; it just takes a vector as its argument and has dimensions of volume. – Endulum Feb 14 '17 at 15:34
• @Pirx, Reference added. – Tomi Feb 14 '17 at 16:42

I've not seen it before, but a derivation that pops into my head is, given the energy conservation equation (from the Euler equations), $$\frac{\partial E}{\partial t}+\nabla\cdot\left(\left(E+p\right)\mathbf u\right)=0\tag{1}$$ where $$E=E_{kin}+E_{int}=\frac{1}{2}\rho u^2+\rho e.$$ Assuming, for simplicity, one dimension, then we can write (1) as, \begin{align} \frac{\partial E}{\partial t}&=-\frac{\partial}{\partial x}\left(Eu+pu\right)\\ &=-\frac{\partial}{\partial x}\left(\frac{1}{2}\rho u^2\,u+\rho eu+pu\right)\\ &=-\frac{\partial}{\partial x}\left(\phi_{kin}+\phi_{int}\right) \end{align} so you can see that $\frac{1}{2}\rho u^3$ would be the "kinetic energy flux" and, I guess, the remaining terms would be the internal energy flux.

The magnetic energy flux $\phi_{me}$ in $erg cm^{-2} s^{-1}$ released in an event to the corona from the jet events can be expressed as the sum of the fluxes of kinetic energy $\phi_{kin}$, potential energy $\phi_{pot}$, enthalpy $\phi_{ent}$, wave energy $\phi_{wav}$ and radiative energy $\phi_{rad}$. This is expressed as

\begin{equation} \phi_{me}= \phi_{kin} + \phi_{pot} + \phi_{ent} + \phi_{wav} + \phi_{rad} .\end{equation}

This is determined as all the potential routes for the magnetic energy to be used. It can be derived, in part, from general relativity. Energy transfer due to thermal conduction is neglected because the other energy transfer mechanisms are orders of magnitude more dominant (Pucci 2013, Magyar 2015, Paraschiv 2015).

Mass Flux

To begin the mass flux must be introduced. Mass flux is simply the mass passing through an area per second. Consider an area, $A$, with unit vector $\hat{n}$ on the surface of our volume. The fluid particles which are on the surface at time $t$ will move off the surface at time $t+\Delta t$ and will sweep out a volume given by

\begin{equation} V =\dot{r}_{n} A \Delta t \end{equation}

where $\dot{r}_{n} = \dot{\textbf{r}} \cdot \hat{\textbf{n}}$ is the component of velocity vector normal to the area. The mass of fluid in this swept volume which through the area during the $\Delta t$ interval, is

\begin{equation} \Delta m = \rho \Delta V = \rho \dot{r}_{n} A \Delta t .\end{equation}

The rate of this mass passing through the area can be written as

\begin{equation} \dot{m} = \lim_{\Delta t \to 0} \frac{\Delta m}{\Delta t} = \rho \dot{r}_{n} A .\end{equation}

The mass flux, $\phi_{ma}$ is then defined as the rate of mass passing per unit area,

\begin{equation} \label{eq:mass_flux} \phi_{ma} = \frac{\dot{m}}{A} = \rho \dot{r}_{n} A \end{equation}

and this expression matches with values from literature (Aris 1990).

Kinetic Energy Flux

From thermodynamics we know the rate of work $\dot{W}$ will go towards the kinetic energy $T$ and the total internal energy, $\epsilon \ m$, of the fluid in some unknown proportion. This ambiguity is resolved by defining the total specific energy, $\epsilon_0$ which is the sum of both internal and kinetic specific energy per mass:

\begin{equation} \epsilon_0 \equiv \epsilon + \frac{T}{m} .\end{equation}

This specific energy is the total energy per unit mass and can be thought of as an energy density. We, however, would prefer this in terms of velocity rather than kinetic energy, as this more naturally understood and readily accessible expirementally. We consider the change of energy in time, also known as power. Assuming a constant force, change in energy due a force is

\begin{equation} dE = \textbf{F} \cdot \textbf{dr} \end{equation}

and so we can rewrite power as

\begin{equation} P = \frac{dE}{dt} = \frac{\textbf{F}\cdot\textbf{dr}}{dt} = \textbf{F} \cdot \dot{\textbf{r}} .\end{equation}

Using Newton's Second Law, $\textbf{F} = m\ddot{\textbf{r}}$ we can rewrite the above Equation as

\begin{equation} P = \dot{\textbf{r}} \cdot \textbf{F} = \dot{\textbf{r}} \cdot m \ddot{\textbf{r}} = \frac{m}{2} \frac{d(\dot{\textbf{r}} \cdot \dot{\textbf{r}})}{dt} = \frac{d}{dt} \Big(\frac{1}{2} m \dot{\textbf{r}}^{2} \Big) \end{equation}

We define the kinetic energy of motion as the quantity which corresponds to this power as

\begin{equation} T = \frac{1}{2}m \dot{\textbf{r}}^{2} .\end{equation}

Returning to above listed equation and substituting the kinetic energy per unit mass, we are left with an expression for the total specific energy,

\begin{equation} \epsilon_0 \equiv \epsilon + \frac{1}{2}\dot{\textbf{r}}^{2} .\end{equation}

So, to summarise, the $\epsilon$ relates to the molecular internal energy and the $\frac{\dot{\textbf{r}}^{2}}{2}$ relates to the bulk motion of the molecules. From the definition of flux given, the kinetic energy flux is defined as the mass flux times the kinetic energy per unit mass divided by the area:

\begin{equation} \phi_{ke} =\frac{ \dot{m} \times \frac{T}{m}}{A} .\end{equation}

We can now see kinetic energy divided by mass is $\frac{T}{m} = \dot{\textbf{r}}^2/2$, and the mass flux is given by the mass flux and thus the kinetic energy flux is given by

\begin{equation} \phi_{ke} = \frac{\dot{m} \frac{1}{2} \dot{\textbf{r}}^2}{A} = \frac{ \rho (\dot{\textbf{r}}\cdot \hat{\textbf{n}}) A \frac{1}{2} \dot{\textbf{r}}^2}{A} .\end{equation}

The kinetic flux can then be further simplified as we can approximate $\dot{\textbf{r}} \cdot \hat{\textbf{n}} = \dot{r}$ because most of the particles in solar eruptions are all moving in a uniform direction and we are interested in the flux perpendicular to the area bisecting the motion of the eruption. Using this simplification we are then left with the expression for kinetic energy flux:

\begin{equation} \phi_{ke} = \rho \frac{1}{2} \dot{r}^3 .\end{equation}

This is the expression which is used when discussing coronal jets in literature (Johnston 2017, Pucci 2013, Paraschiv2015).

Your question is a little confusing. When we calculate flux of something, we mean flux across a plane area element. We take a cube if we are going to do a control volume analysis, and then differences in fluxes across parallel faces of the cube comes into play. If $\phi$ is any extensive scalar property carried by the fluid/particle then its flux (per unit area) is simply $\phi \mathbf{v}$, where $\mathbf{v}$ is velocity vector of fluid/particle under consideration. If you are considering kinetic energy of a particle, then $\phi=\frac{1}{2}m\mathbf{v}\cdot\mathbf{v}=\frac{1}{2}mv^2$, and the corresponding flux is $\frac{1}{2}mv^2\mathbf{v}$. If you now take an area element $\delta A~\mathbf{n}$, then the flux across it is $(\frac{1}{2}mv^2\mathbf{v})\cdot(\delta A~\mathbf{n})$. In the case of fluid $m$ is replaced by $\rho$. If the area normal is oriented parallel to velocity vector, i.e. $\mathbf{n}=\frac{\mathbf{v}}{|\mathbf{v}|}$, only then do you have a flux (per unit area) equal to $\frac{1}{2}\rho v^3$.

• Yes - this confusion was central to my confusion. You are pretty much correct - I worked out a workable general derivation - will post it. I was confusing flux density with flux (the surface integral). Looking between mathematical formulations and physics, one must remember the definition changes. – Tomi Feb 15 '17 at 17:39