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The nuclear fluid-dynamical Hamiltonian which takes the distortion of the local Fermi surface into account predicts the Twist mode (T = O, J = 2 ) at a_finite frequency (about 7.5 MeV for 20~pb).The validity of hydrodynamics in its usual Newtonian form for the nuclear fluid is highly questionable because the mean free nucleon path (for energies less than about 50 MeV/particle) is comparable to nuclear dimensions and therefore the concept of local equilibrium seems doubtful. As we have discussed in detail elsewhere /i/, the non-equilibrium effect leads to a distortion of the local Fermi surface throughout the whole nucleus, which, however, can be incorporated into a fluid dynamical formulation, i.e. a formalism involving only local densities and currents. The essential difference to hydrodynamics is a non-local contribution to the local kinetic energy density. In order to calculate this contribution for a twisting (T=O) mode, we define the "Twist" through the operator = e -i~zs = e ~'~, ~ = (yz,-xz,o) , i.e., a rotation around the body-fixed z-axis with rotation angle proportional to z. The operator zl z has spin-parity 2because the scalar ~art of the tensor product ~ ~ ~, namely r-~, vanishes identically. Although for axially symmetric nuclei there evidently is no change in the local density, the Twist still creates a distortion of the local Fermi surface which in lowest order is given by PF(r,p) = PF(r)N(e) (i-~2~/15" (Y(Y21 (~)-Y2-t (P))+ix(Y21 (P) +Y2-I (P)) ) ) with N(~) = 1 -a2(x2+y2)/15. In Thomas-Fermi approximation we then obtain for the kinetic energy P~ where <~2> is the mean square twisting amplitude in the ground state = I%/(2(BC) I/2)Of course, this model is not able to predict how much of the M2-strength is actually concentrated into one single collective 2 state. This question must be

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