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I'm just testing to see if I'm properly editing. $E=m{c}^2$

$H=\int \frac{dP}{\rho }$ .

Try another way: Template:Math

Lagrangian Representation

in terms of velocity:

Among the principal governing equations we have included the

Lagrangian Representation
of the Euler Equation,

[BLRY07], p. 13, Eq. (1.55)

in terms of momentum density:

Multiplying this equation through by the mass density User:Tohline/Math/VAR Density01 produces the relation,

$\rho \frac{d\overrightarrow{v}}{dt}=-\mathrm{\nabla }P-\rho \mathrm{\nabla }\mathrm{\Phi }$ ,

which may be rewritten as,

$\frac{d(\rho \overrightarrow{v})}{dt}-\overrightarrow{v}\frac{d\rho }{dt}=-\mathrm{\nabla }P-\rho \mathrm{\nabla }\mathrm{\Phi }$ .

Combining this with the Standard Lagrangian Representation of the Continuity Equation, we derive,

$\frac{d(\rho \overrightarrow{v})}{dt}+(\rho \overrightarrow{v})\mathrm{\nabla }\cdot \overrightarrow{v}=-\mathrm{\nabla }P-\rho \mathrm{\nabla }\mathrm{\Phi }$ .

Eulerian Representation

in terms of velocity:

By replacing the so-called Lagrangian (or "material") time derivative $d\overrightarrow{v}/dt$ in the Lagrangian representation of the Euler equation by its Eulerian counterpart (see, for example, the wikipedia discussion titled, "Material_derivative", to understand how the Lagrangian and Eulerian descriptions of fluid motion differ from one another conceptually as well as how to mathematically transform from one description to the other), we directly obtain the

Eulerian Representation
of the Euler Equation,

in terms of momentum density:

As was done above in the context of the Lagrangian representation of the Euler equation, we can multiply this expression through by User:Tohline/Math/VAR Density01 and combine it with the continuity equation to derive what is commonly referred to as the,

Conservative Form
of the Euler Equation,

[BLRY07], p. 8, Eq. (1.31)

The second term on the left-hand-side of this last expression represents the divergence of the "dyadic product" or "outer product" of the vector momentum density and the velocity vector, and is sometimes written as, $\mathrm{\nabla }\cdot [(\rho \overrightarrow{v})\otimes \overrightarrow{v}]$.

in terms of the vorticity:

Drawing on one of the standard dot product rule vector identities, the nonlinear term on the left-hand-side of the Eulerian representation of the Euler equation can be rewritten as,

$(\overrightarrow{v}\cdot \mathrm{\nabla })\overrightarrow{v}=\frac{1}{2}\mathrm{\nabla }(\overrightarrow{v}\cdot \overrightarrow{v})-\overrightarrow{v}\times (\mathrm{\nabla }\times \overrightarrow{v})=\frac{1}{2}\mathrm{\nabla }(v^2)+\overrightarrow{\zeta }\times \overrightarrow{v},$

where,

$\overrightarrow{\zeta }\equiv \mathrm{\nabla }\times \overrightarrow{v}$

is commonly referred to as the vorticity. Making this substitution leads to an expression for the,

Euler Equation
in terms of the Vorticity,

Double Check Vector Identities

In a subsection of an accompanying chapter titled, Double Check Vector Identities, we explicitly demonstrate for four separate "simple rotation profiles" that these two separate terms involving a nonlinear velocity expression do indeed generate identical mathematical relations, namely.

$(\overrightarrow{v}\cdot \mathrm{\nabla })\overrightarrow{v}$

$=$

$\overrightarrow{\zeta }\times \overrightarrow{v}+\frac{1}{2}\mathrm{\nabla }(v^2);$

and we explicitly demonstrate that they are among the set of velocity profiles that can also be expressed in terms of the gradient of a "centrifugal potential," $\mathrm{\nabla }\mathrm{\Psi }$.

Relationship Between State Variables

If the two normalized state variables, $\chi$ and $z$, are known, then the third normalized state variable, $p_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}$, can be obtained directly from the above key expression for the total pressure, that is,

$p_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}(\chi ,z)=8(C_g\chi {)}^{3}z+F(\chi )+(\frac{8{\pi }^4}{15})z^4,$

where,

$C_g\equiv (\frac{{\mu }_e{m}_p}{\overline{\mu }{m}_u}{)}^{1/3}.$

If it is the two normalized state variables, $\chi$ and $p_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}$, that are known, the third normalized state variable — namely, the normalized temperature, $z$ — also can be obtained analytically. But the governing expression is not as simple because it results from an inversion of the total pressure equation and, hence, the solution of a quartic equation. As is detailed in the accompanying discussion, the desired solution is,

$z(\chi ,p_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}})={\theta }_{\chi }{\varphi }^{-1/3}[(\varphi -1{)}^{1/2}-1],$

where,

${\theta }_{\chi }$	$\equiv$	$(\frac{3\cdot 5}{2^2{\pi }^4}{)}^{1/3}{C}_g\chi ,$
$\varphi$	$\equiv$	$2^{3/2}[1+(1+{\lambda }^3{)}^{1/2}{]}^{1/2}\left\{\right[1+(1+{\lambda }^3{)}^{1/2}{]}^{2/3}-\lambda {\}}^{-3/2},$
$\lambda$	$\equiv$	$\left(\frac{{\pi }^4}{2\cdot 3^4\cdot 5}{)}^{1/3}\right[\frac{p_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}-F(\chi )}{(C_g\chi {)}^4}].$

It also would be desirable to have an analytic expression for the function, $\chi (z,p_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}})$, in order to be able to immediately determine the normalized density from any specified values of the normalized temperature and normalized pressure. However, it does not appear that the above key expression for the total pressure can be inverted to provide such a closed-form expression.

STEP #2

As viewed from the tipped coordinated frame, the curve that is identified by this intersection should be an

Off-Center Ellipse
$1$	$=$	$[\frac{x^{\prime }}{x_{\mathrm{m}\mathrm{a}\mathrm{x}}}{]}^2+[\frac{y^{\prime }-y_c}{y_{\mathrm{m}\mathrm{a}\mathrm{x}}}{]}^2$
	$=$	$[\frac{x^{\prime }}{x_{\mathrm{m}\mathrm{a}\mathrm{x}}}{]}^2+[\frac{(y^{\prime }{)}^2-2y^{\prime }{y}_c+y_c^2}{y_{\mathrm{m}\mathrm{a}\mathrm{x}}^2}],$

that lies in the x'-y' plane — that is, $z^{\prime }=0$. Let's see if the intersection expression can be molded into this form.

$1-\frac{z_0^2}{c^2}-\frac{(x^{\prime }{)}^2}{a^2}$	$=$	$(y^{\prime }{)}^2[\frac{c^2+b^2{\tan }^2\theta }{b^2{c}^2}]{\cos }^2\theta +2y^{\prime }[\frac{z_0\sin \theta }{c^2}]$
	$=$	$\left[\frac{c^2+b^2{\tan }^2\theta }{b^2{c}^2}\right]{\cos }^2\theta \{(y^{\prime }{)}^2-2y^{\prime }\left[\frac{-z_0\sin \theta }{c^2{\cos }^2\theta }\right]\left[\frac{b^2{c}^2}{c^2+b^2{\tan }^2\theta }\right]\}$
	$=$	${\kappa }^2[(y^{\prime }{)}^2-2y^{\prime }\underset{y_c}{\underbrace{\left(\frac{-z_0\sin \theta }{c^2{\kappa }^2}\right)}}],$

RESULT 3 (same as Result 1, but different from Result 2, below)

$\frac{y_c}{z_0}$ $=$ $-\frac{\sin \theta }{c^2{\kappa }^2}$

where,

${\kappa }^2$

$\equiv$

$\frac{c^2{\cos }^2\theta +b^2{\sin }^2\theta }{b^2{c}^2}.$

Dividing through by ${\kappa }^2$, then adding $y_c^2$ to both sides gives,

$(y^{\prime }{)}^2-2y^{\prime }{y}_c+y_c^2$

$=$

$\underset{y_{\mathrm{m}\mathrm{a}\mathrm{x}}^2}{\underbrace{[\frac{1}{{\kappa }^2}-\frac{z_0^2}{c^2{\kappa }^2}+y_c^2]}}-\frac{(x^{\prime }{)}^2}{a^2{\kappa }^2}.$

Finally, we have,

$\frac{1}{y_{\mathrm{m}\mathrm{a}\mathrm{x}}^2}[(y^{\prime }{)}^2-2y^{\prime }{y}_c+y_c^2]$

$=$

$1-(x^{\prime }{)}^2\underset{1/x_{\mathrm{m}\mathrm{a}\mathrm{x}}^2}{\underbrace{\left[\frac{1}{a^2{\kappa }^2{y}_{\mathrm{m}\mathrm{a}\mathrm{x}}^2}\right]}}.$

So … the intersection expression can be molded into the form of an off-center ellipse if we make the following associations:

$\frac{y_c}{z_0}$ $=$ $-\frac{\sin \theta }{c^2{\kappa }^2},$ $y_{\mathrm{m}\mathrm{a}\mathrm{x}}^2$ $=$ $\frac{1}{{\kappa }^2}[1-\frac{z_0^2}{c^2}-\frac{z_0\sin \theta }{c^2}],$ $x_{\mathrm{m}\mathrm{a}\mathrm{x}}^2$ $=$ $a^2[1-\frac{z_0^2}{c^2}-\frac{z_0\sin \theta }{c^2}].$ Note as well that, $(\frac{x_{\mathrm{m}\mathrm{a}\mathrm{x}}}{y_{\mathrm{m}\mathrm{a}\mathrm{x}}}{)}^2$ $=$ $a^2{\kappa }^2=\frac{a^2}{b^2{c}^2}[c^2{\cos }^2\theta +b^2{\sin }^2\theta ].$