Doubt regarding derivation of relativistic Bernoulli function

Question

The relativistic form of the radial momentum equation for spherical fluid flow is $$\gamma_v^2v\frac{dv}{dr}+\frac{1}{\rho}\frac{dp}{dr}+\frac{d\Phi}{dr}=0$$ where $\gamma_v=\dfrac{1}{\sqrt{1-v^2}}$ is the Lorentz factor and $\Phi$ is the gravitational potential.

The specific enthalpy of the flow is expressed as $h=u+\dfrac{P}{\rho}$, where $u$ is the specific energy. Using the first law of thermodynamics, we obtain $dh=Tds+\dfrac{dp}{\rho}$, so that $ds=\dfrac{dp}{\rho}$ for an isentropic flow. Thus, the above equation reduces to $$\boxed{\gamma_v^2v\frac{dv}{dr}+\frac{dh}{dr}+\frac{d\Phi}{dr}=0}$$

CASE 1:

In the non-relativistic limit, $v\ll1$, and this means $\gamma_v=1$, and the radial momentum equation reduces to $$v\frac{dv}{dr}+\frac{dh}{dr}+\frac{d\Phi}{dr}=0$$ Integrating this equation one obtains the Bernoulli function $$\mathscr{B}=\frac{v^2}{2}+h+\Phi\qquad\qquad\qquad\qquad\qquad\quad(1)$$

CASE 2:

To obtain the Bernoulli function for relativistic flow, we need to retain the Lorentz factor, so that the radial momentum equation is $$\frac{v}{1-v^2}\frac{dv}{dr}+\frac{dh}{dr}+\frac{d\Phi}{dr}=0$$ Thus, the Bernoulli function can be obtained as $$\mathscr{B}=\ln\left(\frac{1}{\sqrt{1-v^2}}\right)+h+\Phi\qquad\qquad\qquad(2)$$

Question:

The non-relativistic Bernoulli function (eqn.$1$) is a well-known expression found in most textbooks, but I am not sure whether the relativistic Bernoulli function (eqn.$2$) that I had derived is correct or not. Am I doing the correct calculations? If not, can someone suggest the correct approach?

Answer 1

I thnk you can compare your work with this page. It's chock-full of rendered equations, ending with this conclusion:

where $v$ is the flow velocity and the Lorentz factor $\gamma$ is given by

$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

The time component of eq. $(9.6)$ is implied by the other three. In the case of isentropic flow, that is, when $\sigma/n=const.$, and assuming the flow to be steady, the spatial components of eq. $(9.6)$ give

$$\gamma(v\cdot grad)(\gamma\omega v/n)+c^2grad(\omega/n)=0$$

Scalar multiplication by $v$ leads to

$$(v\cdot grad)(\gamma\omega/n)=0$$

which implies that along any streamline the quantity

$$\gamma\omega/n=const.$$

This is the relativistic version of Bernoulli's equation.

Another, somewhat longer and deeper analysis can be found here.

Answer 2

Your expression for the lorentz factor is wrong: $$\gamma_v=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\neq\frac{1}{\sqrt{1-v^2}}.$$ The non relativistic case is $v\ll c$ and not $v \ll 1$. This makes your second approach equivalent to your first, which leads to the correct answer.