Constitutive Settings with regard to
Energy- and Entropy-Balances
in Non-Equilibrium Thermodynamics:
the Thermodynamical Verification
June 04, 2026
Constitutive equations have to be in agreement with the energy- and entropy-balances. For achieving that, the procedure of thermodynamical verification is introduced: Because heat flux and entropy flux as well as the time differentials of internal
energy and entropy are not independent of each other, energy- and entropy-balances are connected with each other by so-called internal settings laying down the theoretical frame of the applied material description which is characterized by additional
constitutive settings.
Hint: This is a short-note: more details are can be found in the reference below3
For elucidating the procedure of thermodynamical verification, three examples are considered which use different settings.
1. A trivial example: Fourier heat conduction4
Starting out with the energy- and entropy-balances [1] \[\label{a1}
\varrho\stackrel{_\bullet}{u}+\nabla\cdot\boldsymbol{q}\;\stackrel{_1}{=}\;r,
\qquad \sigma\;\stackrel{_2}{=}\;\varrho\stackrel{_\bullet}{s}+\nabla\cdot\boldsymbol{J},\tag{1}\] (mass density \(\varrho\), internal energy \(u\), heat flux \(\boldsymbol{q}\), energy supply \(r\), entropy production \(\sigma\), entropy \(s\), entropy flux \(\boldsymbol{J}\), time derivative \(\bullet\)) four different equal signs are introduced to better distinguish the steps of thermodynamical verification: \[\begin{align}
\tag{2}
\stackrel{_1}{=}\;and\;\stackrel{_2}{=} for the balances,\;&\stackrel{_c}{=}&\;
for a constitutive setting,\\ \tag{3}
&\stackrel{_\bullet}{=}& \;for an internal setting.
\end{align}\] The Fourier heat flux satisfies a constitutive equation, the entropy flux an internal setting \[\label{a4}
\boldsymbol{q}\;\stackrel{_c}{=}\;-\kappa\nabla T,\qquad \boldsymbol{J}\;\stackrel{_\bullet}{=}\;\boldsymbol{q}/T\tag{4}\] (heat conduction \(\kappa\), temperature T). Using (1 )\(_2\) and (4 )\(_2\) results in \[\begin{align}
\tag{5}
\sigma-\varrho\stackrel{_\bullet}{s} &\stackrel{_2}{=}&\nabla\cdot(\boldsymbol{q}/T)\;=\;\boldsymbol{q}\cdot\nabla(1/T)
+(1/T)\nabla\cdot\boldsymbol{q}\;\stackrel{_1}{=}\\ \tag{6} &\stackrel{_1}{=}&\;\boldsymbol{q}\cdot\nabla(1/T)+(1/T)(r-\varrho\stackrel{_\bullet}{u}).
\end{align}\] Now the time differential of the entropy can be determined from (6 ) by an internal setting resulting in the entropy
production \[\label{a7}
-\varrho\stackrel{_\bullet}{s}\;\stackrel{_\bullet}{=}\;-\varrho(1/T)\stackrel{_\bullet}{u},\;\longrightarrow\;
\sigma\;= \boldsymbol{q}\cdot\nabla(1/T)+(1/T)r.\tag{7}\] Inserting the constitutive setting (4 )\(_1\) yields \[\label{a8}
\sigma \stackrel{_c}{=} -\kappa\nabla T\cdot\nabla(1/T)+(1/T)r\;=\;(\kappa /T^2)\nabla T\cdot\nabla T
+(1/T)r.\tag{8}\]
Because the entropy balance (1 )\(_2\) is built up of three terms, two terms have to be chosen by internal settings, here [\(\stackrel{_\bullet}{s},\boldsymbol{J}\)]
are chosen in (7 )\(_1\) and (4 )\(_2\). The same result can be obtained by the internal setting [\(\stackrel{_\bullet}{s},\sigma\)] in (7 ) replacing \(\boldsymbol{J}\) by \(\sigma\). There are three possibilities of internal settings
\[\label{a9}
[\stackrel{_\bullet}{s},\boldsymbol{J}],\;[\stackrel{_\bullet}{s},\sigma],\;[\sigma,\boldsymbol{J}].\tag{9}\] Both balances (1 ) and the constitutive equation (4 )\(_1\) have to be taken into account for generating the entropy production (8 ). If \([\sigma,\boldsymbol{J}]\) is used for the internal setting, the resulting \(\stackrel{_\bullet}{s}\) has to be a total time differential. If not, the internal setting was not suitable and should be replaced and repeated.
2. The procedure: Thermodynamical Verification
The formal procedure presented in sect.1 is applicable to arbitrary materials. For short communication this scheme between (1 ) and (8 ) should be called thermodynamical verification of the entropy balance
equation. Now this is inspected in more detail.
As an example we consider pure heat coduction with no power and material fluxes. Consequently, the first internal setting is as (4 )\(_2\) (resulting in (5 )\(_2\)) \[\label{a10}
\boldsymbol{J}\;\stackrel{_\bullet}{=}\;(1/T)\boldsymbol{q}\;\longrightarrow\;\nabla\cdot\boldsymbol{J}\;=\;
\boldsymbol{q}\cdot\nabla(1/T)+(1/T)\nabla\cdot\boldsymbol{q.}\tag{10}\] Taking the balances (1 ) into account (10 )\(_2\) results in (6 )
\[\begin{align}
\tag{11}
\sigma - \varrho\stackrel{_\bullet}{s}\;&\stackrel{_{2,1}}{=}&\;\boldsymbol{q}\cdot\nabla(1/T)
+(1/T)(r-\varrho\stackrel{_\bullet}{u}),\\ \tag{12}\;
\sigma &=&\;\boldsymbol{q}\cdot\nabla(1/T)+(1/T)r
+\varrho\Big(\stackrel{_\bullet}{s}-(1/T)\stackrel{_\bullet}{u}\Big).
\end{align}\] The second internal setting in (12 ) is needed. Instead of (7 )\(_1\), we now assume that \(s\) beyond \(u\) and \({\boldsymbol{q}}\) also depends of additional internal variables \(\boldsymbol{\xi}\), that means \[\label{a15}
\stackrel{_\bullet}{s}\;\stackrel{_\bullet}{=}\;(1/T)\stackrel{_\bullet}{u}+\boldsymbol{\alpha}\cdot\stackrel{_\bullet}{\boldsymbol{q}}
+\boldsymbol{\beta}\cdot\stackrel{_\bullet}{\boldsymbol{\xi}}\tag{13}\] is valid. Thus the entropy production (12 ) becomes instead of (7 )\(_2\)
\[\label{a14}
\sigma\;=\;\boldsymbol{q}\cdot\nabla(1/T)+(1/T)r+\varrho\boldsymbol{\alpha}\cdot\stackrel{_\bullet}{\boldsymbol{q}}
+\varrho\boldsymbol{\beta}\cdot\stackrel{_\bullet}{\boldsymbol{\xi}}.\tag{14}\]
Now we check, if the thermodynamical verification was performed completely: There are two internal settings in (10 )\(_1\) and (13 ) [\(\stackrel{_\bullet}{s},\boldsymbol{J}\)]. The scheme of thermodynamical verification is the same as in sect. 1. and 2., but the result depends on the special internal settings. The balances (1 ) are
introduced in (11 ). A special constitutive equation such as (4 )\(_1\) was up to now not taken into account and will be added later on.
The time differential of the entropy (13 ) is in contrast to (7 )\(_1\) defined on an enlarged set of variables \(\{u,\boldsymbol{q},\boldsymbol{\xi}\}\) which span a state space [1], [2], if the variables are indepedent of each other. This is not evident because \(u\) and \(\boldsymbol{q}\) occur jointly in the energy balance. Clear is
that \(\stackrel{_\bullet}{\boldsymbol{q}}\) and \[\label{a16}
\nabla\cdot\boldsymbol{q}\;\stackrel{_1}{=}\;r-\varrho\stackrel{_\bullet}{u}\;\dashv\;\stackrel{_\bullet}{\boldsymbol{q}}\tag{15}\] are independent \(\dashv\) of each other [1]. The opposite assumption that \(u\) and \(\boldsymbol{q}\) are connected with each other
\[\label{a17}
u\;\dashv\;\boldsymbol{/}{\boldsymbol{q}}\;\longrightarrow\;
u\;=\;f(\boldsymbol{q})\;\longrightarrow\;\stackrel{_\bullet}{u}\;=\;(df/d\boldsymbol{q})\cdot\stackrel{_\bullet}{\boldsymbol{q}}\tag{16}\] results in (16 )\(_3\). Inserting \(\stackrel{_\bullet}{u}\) into (15 ) yields an expression which depends on \(\stackrel{_\bullet}{\boldsymbol{q}}\) \[\label{a17a}
r-\varrho (df/d\boldsymbol{q})\cdot
\stackrel{_\bullet}{\boldsymbol{q}}\;\dashv\;\boldsymbol{/}\stackrel{_\bullet}{\boldsymbol{q}},\tag{17}\] a statement which is in contradiction to (15 ). Consequently, the assumption of contradiction (16 )\(_1\) that \(u\) and \(\boldsymbol{q}\) depend on each other is wrong: they are independet of each other
\[\label{a18}
u\;\dashv\;\boldsymbol{q}.\tag{18}\]
Consequently, \(\{u,\boldsymbol{q},\boldsymbol{\xi}\}\) can be used as a state space, and the time differential of the entropy \(s(u,\boldsymbol{q},\boldsymbol{\xi})\) (13 ) is a total one \[\label{a19} (\partial s/\partial u)\;=\;(1/T),\;\;(\partial s/\partial\boldsymbol{q})\;=\;\boldsymbol{\alpha},\;\; (\partial s/\partial\boldsymbol{\xi})\;=\;\boldsymbol{\beta}.\tag{19}\] Now two additional constitutive settings are introduced \[\label{d20} \boldsymbol{\alpha}\;\stackrel{_c}{=}\;{\boldsymbol{c}}onst,\;\;\boldsymbol{\beta}\;\stackrel{_c}{=}\;{\boldsymbol{k}}onst.\tag{20}\] Taking (19 ) into account, the second mixed differentials result in \[\begin{align} \tag{21} (\partial /\partial\boldsymbol{q})(1/T) &=& (\partial /\partial u)\boldsymbol{\alpha}\;=\;\boldsymbol{0}, \\ \tag{22} (\partial /\partial\boldsymbol{\xi})(1/T) &=& (\partial /\partial u)\boldsymbol{\beta}\;=\;\boldsymbol{0}, \\ \tag{23} (\partial /\partial\boldsymbol{\xi})\boldsymbol{\alpha} &=& (\partial /\partial \boldsymbol{q})\boldsymbol{\beta}\; =\;\boldsymbol{0}. \end{align}\]
The zeros in (21 ) to (23 ) are generated by the constitutive settings (20 ). From (21 )\(_1\) and (22 )\(_1\) follows \[\label{d24}
T\;=\;F(u)\;\longrightarrow\;\stackrel{_\bullet}{T}\;=\;\stackrel{_\bullet}{u}(\partial /\partial u)F\;
\longrightarrow\;\frac{\stackrel{_\bullet}{T}}{T(\partial /\partial u)F}\;=\;\stackrel{_\bullet}{u}/T,\tag{24}\] the possibility to change the state space \[\label{d25}
\{u,\boldsymbol{q},\boldsymbol{\xi}\}\;\longrightarrow\;\{T,\boldsymbol{q},\boldsymbol{\xi}\}.\tag{25}\] The time differential of the entropy (13 ) becomes \[\label{d26}
\stackrel{_\bullet}{s}\;=\;\frac{\stackrel{_\bullet}{T}}{T(\partial /\partial u)F}+\boldsymbol{\alpha}\cdot\stackrel{_\bullet}{\boldsymbol{q}}
+\boldsymbol{\beta}\cdot\stackrel{_\bullet}{\boldsymbol{\xi}}.\tag{26}\] This trivial example shows that state spaces represent an essential tool of material description.
3. A special [\(\stackrel{_\bullet}{s},\boldsymbol{J}\)]-verification using an extra entropy flux
The first internal setting introduces the extra entropy flux \(\boldsymbol{K}\) generalizing (4 )\(_2\) and (10 )\(_1\). Taking (6 ) into consideration, one obtains (29 ) \[\begin{align}
\tag{27}
\boldsymbol{J} &\stackrel{_\bullet}{=}& (1/T)\boldsymbol{q}+\boldsymbol{K},
\\ \tag{28}
\nabla\cdot\boldsymbol{J} &=& \boldsymbol{q}\cdot\nabla(1/T)+(1/T)\nabla\cdot\boldsymbol{q}
+ \nabla\cdot\boldsymbol{K}\;\stackrel{_2}{=}\\ \tag{29}
&\stackrel{_2}{=}& \sigma-\varrho\stackrel{_\bullet}{s}\;\stackrel{_1}{=}\;
\boldsymbol{q}\cdot\nabla(1/T)+(1/T)(r-\varrho\stackrel{_\bullet}{u})
+ \nabla\cdot\boldsymbol{K}.
\end{align}\]
Presupposing that the extra entropy flux has two parts, one which is not parallel to the heat flux and the other which is parallel to the time differential of the heat flux, we obtain the constitutive setting \[\begin{align} \tag{30} \boldsymbol{K} &\stackrel{_c}{=}& \mathbb{Q}\cdot\boldsymbol{q}+a\stackrel{_\bullet}{\boldsymbol{q}}, \\ \tag{31} \nabla\cdot\boldsymbol{K} &=& \nabla\cdot(\mathbb{Q}\cdot\boldsymbol{q})+\nabla\cdot (a\stackrel{_\bullet}{\boldsymbol{q}})\;=\\ \tag{32} &=&\boldsymbol{q}\nabla:\mathbb{Q}^{\top}+\mathbb{Q}:\nabla\boldsymbol{q}\;+ \stackrel{_\bullet}{\boldsymbol{q}}\cdot\nabla a + a\nabla\cdot\stackrel{_\bullet}{\boldsymbol{q}}. \end{align}\] 5 Now the second internal setting follows from (29 ) by inserting (32 ) \[\tag{33} -\varrho\stackrel{_\bullet}{s}\;\stackrel{_\bullet}{=}\;-\varrho(1/T)\stackrel{_\bullet}{u}+ \stackrel{_\bullet}{\boldsymbol{q}}\cdot\nabla\sf{a}\;\longrightarrow\;\tag{34}\; \stackrel{_\bullet}{s}\;\stackrel{_\bullet}{=}(1/T)\stackrel{_\bullet}{u}-(1/\varrho) \stackrel{_\bullet}{\boldsymbol{q}}\cdot\nabla\sf{a},\] establishing the state space \(\{u,\boldsymbol{q}\}\)6: \[\label{d34} a\;=\;a(u,\boldsymbol{q}),\;\;\;\mathbb{Q}\;=\;\mathbb{Q}(u,\boldsymbol{q}).\tag{35}\] Taking (32 ) and (33 )\(_1\) into account, (29 ) results in the entropy production \[\label{a31} \sigma\;=\;\Big(\nabla(1/T)+\nabla\cdot\mathbb{Q}\Big)\cdot\boldsymbol{q} +(1/T)r+\mathbb{Q}:\nabla\boldsymbol{q}\;+ a\nabla\cdot\stackrel{_\bullet}{\boldsymbol{q}}\tag{36}\] which has an interesting shape in comparison to (7 ): the gradient of the reciprocal temperature is supplemented by the divergence of that tensor \(\mathbb{Q}\) which destroys the parallelism of the extra entropy flux with the heat flux.
Additional constitutive settings \[\label{d36}
a\;\stackrel{_c}{=}\;const,\;\;\;\mathbb{Q}\;\stackrel{_c}{=}\;\mathbb{K}onst,\tag{37}\] change the state space, the time differential of the entropy (34 )\(_2\) and the entropy
production (36 ) \[\label{d37}
\stackrel{_\bullet}{s}\;\stackrel{_\bullet}{=}\;(1/T)\stackrel{_\bullet}{u},\;\;\;\sigma\;=\;\nabla(1/T)\cdot\boldsymbol{q}
+(1/T)r+\mathbb{Q}:\nabla\boldsymbol{q}\;+ a\nabla\cdot\stackrel{_\bullet}{\boldsymbol{q}}.\tag{38}\] If aditionally \(a\) and \(\mathbb{Q}\) vanish, Fourier heat conduction of
sect. 1 emerges.
Acknowledgment
My warm thanks to Prof. Dr. Karl Heinz Hoffmann for helpful discussions on the utilization of thermodynamical verifications and for reviewing a former version of this short note.
Corresponding author: muschik@physik.tu-berlin.de↩︎
In memory of Bogdan Maruszewski↩︎
see e.g.: Restuccia, L.; Jou, D.; Pavelka, M. On the identification of two internal tensorial variables and a heat transport equation with internal, thermal viscosity and vorticity terms. J. Non-Equilib. Thermodyn. 2026, 51, 1-18.↩︎
for a warm up↩︎
\(\mathbb{A}:\mathbb{B}\;\rightarrow\;A_{kl}B_{kl}\)↩︎
remember: u and \(\boldsymbol{q}\) are independent of each other↩︎