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Also, since only the leading term or terms in the small-t expansion have been retained above, the neglected terms must increasingly contribute as t increases and the above explicit results must become increasingly inapplicable. Hence these results hold for t in the < t < tlong . intermediate regime, tshort $ $ The matrix Lð^tÞ is called the transport matrix, and it satisﬁes Lð^tÞ ¼ E Lð^tÞT E ¼ LðÀ^tÞT ð55Þ This follows because, in the grouped representation, Qþ contains nonzero blocks only on the diagonal and is symmetric, and QÀ contains nonzero blocks only off the diagonal and is asymmetric.
The entropy of the latter gives the additional weight to be factored into the weight of the disordered macrostate. the second law of nonequilibrium thermodynamics 41 The static probability places the subsystem in a dynamically disordered state, _ 1 Þ ¼ 0. If the system is G1 so that at t ¼ 0 the ﬂux most likely vanishes, xðG constrained to follow the adiabatic trajectory, then as time increases the ﬂux will become nonzero and approach its optimum or steady-state value, _ xðtÞ ! Lðx1 ; þ1ÞX1 , where x1 xðG1 Þ and X1 Xs ðG1 Þ.
Consider two variables, x ¼ fA; Bg, where A has even parity and B has odd parity. Then using the terminal velocity it follows that Lð^tÞ ¼ ^tkBÀ1 _ þ tÞAðtÞi hAðt 0 _ þ tÞAðtÞi0 hBðt _ þ tÞBðtÞi hAðt 0 _ þ tÞBðtÞi0 hBðt ð58Þ 20 phil attard The matrix is readily shown to be antisymmetric, as it must be. In the intermediate regime, the transport matrix must be independent of jtj, which means that for nonzero t, € q À1 hAðt þ tÞAðtÞi0 0 ¼ Lð^tÞ ¼ ^tkB € þ tÞAðtÞi0 hBðt qt € þ tÞBðtÞi hAðt 0 € þ tÞBðtÞi0 hBðt ð59Þ That the terminal acceleration should most likely vanish is true almost by deﬁnition of the steady state; the system returns to equilibrium with a constant velocity that is proportional to the initial displacement, and hence the acceleration must be zero.