1 Correction to: Continuum Mech. Thermodyn. https://doi.org/10.1007/s00161-021-01061-9

In Eqs. (18), (20), (23), (24), (26) and (27), “\(\mathbf{I }_{4} \)” should be replaced by “\(\mathbf{i }_{4} \)” where \(\mathbf{i }_{4} =\left[ 0 \,\, 0 \,\, 0 \,\, 1 \right] ^{T}\). The correct expressions for Eqs. (18), (20), (23), (24), (26) and (27) are then:

$$\begin{aligned} \mathbf{f }_{1} (z)= & {} -\frac{q}{2\pi \text{ i }}\left\langle {\ln (z_{\alpha } -p_{\alpha } d)} \right\rangle \mathbf{A }_{1}^{T} \mathbf{i }_{4} -\frac{q}{2\pi \text{ i }}\sum \limits _{j=1}^4 {\left\langle {\ln (z_{\alpha } -\bar{{p}}_{j} d)} \right\rangle \mathbf{KI }_{j} {\bar{\mathbf{A }}}_{1}^{T} \mathbf{i }_{4} } , \end{aligned}$$
(18)
$$\begin{aligned} F= & {} qE_{2}^{C} (0,d) \nonumber \\= & {} \frac{q^{2}}{\pi d}\mathbf{i }_{4}^{T}\text { Im}\left\{ {\mathbf{A }_{1} \sum \limits _{j=1}^4 {\left\langle {\frac{p_{\alpha } }{p_{\alpha } -\bar{{p}}_{j} }} \right\rangle \mathbf{KI }_{j} {\bar{\mathbf{A }}}_{1}^{T} } } \right\} \mathbf{i }_{4} \nonumber \\= & {} \frac{q^{2}}{\pi d}\mathbf{i }_{4}^{T}\text { Im}\left\{ {\sum \limits _{j=1}^4 {\sum \limits _{k=1}^4 {\frac{p_{k} }{p_{k} -\bar{{p}}_{j} }\mathbf{A }_{1} } \mathbf{I }_{k} \mathbf{KI }_{j} {\bar{\mathbf{A }}}_{1}^{T} } } \right\} \mathbf{i }_{4} , \end{aligned}$$
(20)
$$\begin{aligned} F= & {} \frac{q^{2}}{2\pi d}\mathbf{i }_{4}^{T}\text { Re}\left\{ {\sum \limits _{j=1}^4 {\sum \limits _{k=1}^4 {\frac{p_{k} }{p_{k} -\bar{{p}}_{j} }(\mathbf{A }_{1} \mathbf{I }_{k} \mathbf{A }_{1}^{-1} )\mathbf{H }_{1} ({\bar{\mathbf{A }}}_{1} \mathbf{I }_{j} {\bar{\mathbf{A }}}_{1}^{-1} )^{T}} } } \right\} \mathbf{i }_{4} \nonumber \\&-\frac{q^{2}}{\pi d}\mathbf{i }_{4}^{T}\text { Re}\left\{ {\sum \limits _{j=1}^4 {\sum \limits _{k=1}^4 {\frac{p_{k} }{p_{k} -\bar{{p}}_{j} }(\mathbf{A }_{1} \mathbf{I }_{k} \mathbf{A }_{1}^{-1} )({\tilde{\mathbf{H }}}+\text{ i }{\breve{\mathbf{W }}})({\bar{\mathbf{A }}}_{1} \mathbf{I }_{j} {\bar{\mathbf{A }}}_{1}^{-1} )^{T}} } } \right\} \mathbf{i }_{4} . \end{aligned}$$
(23)
$$\begin{aligned} F= & {} \frac{q^{2}}{2\pi d}\mathbf{i }_{4}^{T}\text { Re}\left\{ {\sum \limits _{j=1}^4 {\sum \limits _{k=1}^4 {\frac{-\bar{{p}}_{j} }{p_{k} -\bar{{p}}_{j} }(\mathbf{A }_{1} \mathbf{I }_{k} \mathbf{A }_{1}^{-1} )\mathbf{H }_{1} ({\bar{\mathbf{A }}}_{1} \mathbf{I }_{j} {\bar{\mathbf{A }}}_{1}^{-1} )^{T}} } } \right\} \mathbf{i }_{4} \nonumber \\&+\frac{q^{2}}{\pi d}\mathbf{i }_{4}^{T}\text { Re}\left\{ {\sum \limits _{j=1}^4 {\sum \limits _{k=1}^4 {\frac{\bar{{p}}_{j} }{p_{k} -\bar{{p}}_{j} }(\mathbf{A }_{1} \mathbf{I }_{k} \mathbf{A }_{1}^{-1} )({\tilde{\mathbf{H }}}+\text{ i }{\breve{\mathbf{W }}})({\bar{\mathbf{A }}}_{1} \mathbf{I }_{j} {\bar{\mathbf{A }}}_{1}^{-1} )^{T}} } } \right\} \mathbf{i }_{4} . \end{aligned}$$
(24)
$$\begin{aligned} E= & {} \frac{q}{2}\phi ^{C}(0,d)=-\frac{q^{2}}{2\pi }\mathbf{i }_{4}^{T} \text { Im}\left\{ {\mathbf{A }_{1} \sum \limits _{j=1}^4 {\left\langle {\ln \left[ {d(p_{\alpha } -\bar{{p}}_{j} )} \right] } \right\rangle \mathbf{KI }_{j} {\bar{\mathbf{A }}}_{1}^{T} } } \right\} \mathbf{i }_{4} , \end{aligned}$$
(26)
$$\begin{aligned} F= & {} -\frac{\partial E}{\partial d}=\frac{q^{2}}{2\pi d}\mathbf{i }_{4}^{T} \text { Im}\left\{ {\mathbf{A }_{1} \mathbf{K \bar{\mathbf{A }}}_{1}^{T} } \right\} \mathbf{i }_{4} =\frac{q^{2}}{4\pi d}(\mathbf{H }_{1} -2{\tilde{\mathbf{H }}})_{44} , \end{aligned}$$
(27)

The final expression for the Coulomb force is correct and remains unchanged.