In one dimension all vector and tensor quantities from the balance equations (2.47)-(2.49) degenerate to scalar quantities, eg.

	$v_i,v_j$	$⟶v_1=v$		(E.1)
	${\sigma }_{ij}^{\prime }$	$⟶{\sigma }_{11}^{\prime }={\sigma }^{\prime }$		(E.2)
	$g_i$	$⟶g_1=g$		(E.3)

The balance equations of fluid dynamics in one dimensions are therefore written like:

	${\displaystyle \frac{\partial }{\partial t}}\rho +{\displaystyle \frac{\partial }{\partial r}}(v\rho )$	$=0$		(E.4)
	${\displaystyle \frac{\partial }{\partial t}}(\rho v)+{\displaystyle \frac{\partial }{\partial r}}(v\rho v)$	$=-{\displaystyle \frac{\partial }{\partial r}}p+{\displaystyle \frac{\partial }{\partial r}}{\sigma }^{\prime }-\rho g$		(E.5)
	${\displaystyle \frac{\partial }{\partial t}}(\rho e)+{\displaystyle \frac{\partial }{\partial r}}(v\rho e)$	$=-{\displaystyle \frac{\partial }{\partial r}}(vp)+{\displaystyle \frac{\partial }{\partial r}}(v{\sigma }^{\prime })-v\rho g$		(E.6)

By using the Euler derivate we can write (E.4)-(E.6) like

	${\displaystyle \frac{d}{dt}}\rho +\rho {\displaystyle \frac{\partial }{\partial r}}v$	$=0$		(E.7)
	${\displaystyle \frac{d}{dt}}(\rho v)+\rho v{\displaystyle \frac{\partial }{\partial r}}v$	$=-{\displaystyle \frac{\partial }{\partial r}}p+{\displaystyle \frac{\partial }{\partial r}}{\sigma }^{\prime }-\rho g$		(E.8)
	${\displaystyle \frac{d}{dt}}(\rho e)+\rho e{\displaystyle \frac{\partial }{\partial r}}v$	$=-{\displaystyle \frac{\partial }{\partial r}}(vp)+{\displaystyle \frac{\partial }{\partial r}}(v{\sigma }^{\prime })-v\rho g$		(E.9)

If we introduce the so called mass coordinate defined by

$\partial m=\rho \cdot \partial r\Rightarrow \partial r={\displaystyle \frac{\partial m}{\rho }}$

(E.10)

we can write (E.4) like

		${\displaystyle \frac{d}{dt}}\rho +{\rho }^2{\displaystyle \frac{\partial }{\partial m}}v$	$=0$		(E.11)
	$\iff$	$-{\displaystyle \frac{1}{{\rho }^2}}{\displaystyle \frac{d}{dt}}\rho -{\displaystyle \frac{\partial }{\partial m}}v$	$=0$		(E.12)
	$\iff$	${\displaystyle \frac{d}{dt}}\left({\displaystyle \frac{1}{\rho }}\right)$	$={\displaystyle \frac{\partial }{\partial m}}v$		(E.13)

The momentum equation (E.5) is transformed like

		${\displaystyle \frac{d}{dt}}(\rho v)+{\rho }^{2}v{\displaystyle \frac{\partial }{\partial m}}v$	$=-\rho {\displaystyle \frac{\partial }{\partial m}}p+\rho {\displaystyle \frac{\partial }{\partial m}}{\sigma }^{\prime }-\rho g$		(E.14)
	$\iff$	${\displaystyle \frac{1}{\rho }}\left(\rho {\displaystyle \frac{d}{dt}}v+v{\displaystyle \frac{d}{dt}}\rho \right)+\rho v{\displaystyle \frac{\partial }{\partial m}}v$	$=-{\displaystyle \frac{\partial }{\partial m}}p+{\displaystyle \frac{\partial }{\partial m}}{\sigma }^{\prime }-g$		(E.15)
	$\iff$	${\displaystyle \frac{d}{dt}}v+v\left({\displaystyle \frac{1}{\rho }}{\displaystyle \frac{d}{dt}}\rho +\rho {\displaystyle \frac{\partial }{\partial m}}v\right)$	$=-{\displaystyle \frac{\partial }{\partial m}}p+{\displaystyle \frac{\partial }{\partial m}}{\sigma }^{\prime }-g$		(E.16)
	$\iff$	${\displaystyle \frac{d}{dt}}v-\rho v\underset{0}{\underbrace{\left(-{\displaystyle \frac{1}{{\rho }^2}}{\displaystyle \frac{d}{dt}}\rho -{\displaystyle \frac{\partial }{\partial m}}v\right)}}$	$=-{\displaystyle \frac{\partial }{\partial m}}p+{\displaystyle \frac{\partial }{\partial m}}{\sigma }^{\prime }-g$		(E.17)
	$\iff$	${\displaystyle \frac{d}{dt}}v$	$=-{\displaystyle \frac{\partial }{\partial m}}p+{\displaystyle \frac{\partial }{\partial m}}{\sigma }^{\prime }-g$		(E.18)

With an analogous transformation the energy equation (E.6) can be written like

${\displaystyle \frac{d}{dt}}e$

$=-{\displaystyle \frac{\partial }{\partial m}}(vp)+{\displaystyle \frac{\partial }{\partial m}}(v{\sigma }^{\prime })-vg$

(E.19)

Summing up the three equations an retransforming them in terms of $dr$ we get the balance equations in one-dimensional lagrangian form as they are often used in numerical fluid dynamics:

	${\displaystyle \frac{d}{dt}}\left({\displaystyle \frac{1}{\rho }}\right)$	$={\displaystyle \frac{1}{\rho }}{\displaystyle \frac{\partial }{\partial r}}v$		(E.20)
	${\displaystyle \frac{d}{dt}}v$	$=-{\displaystyle \frac{1}{\rho }}{\displaystyle \frac{\partial }{\partial r}}p+{\displaystyle \frac{1}{\rho }}{\displaystyle \frac{\partial }{\partial r}}{\sigma }^{\prime }-g$		(E.21)
	${\displaystyle \frac{d}{dt}}e$	$=-{\displaystyle \frac{1}{\rho }}{\displaystyle \frac{\partial }{\partial r}}(vp)+{\displaystyle \frac{1}{\rho }}{\displaystyle \frac{\partial }{\partial r}}(v{\sigma }^{\prime })-vg$		(E.22)