2017apho质心大礼包3 2017apho理论第一题

1、Q1-1English (Ocial)TheoryVortices in superuidIntroductionSuperuidity is a property of owing without friction. Everyday experience tells us that motion of an ordinary uid (say, water at room temperature) is always accompanied by viscous dissipation of energy, so that the ow gradually becomes slower u

2、nless it is maintained by external forces. In contrast, super- uid exhibits no loss of kinetic energy: once excited the motion of superuid can continue indenitely. Superuidity was originally discovered experimentally in liquid helium.We study properties of superuid helium at zero temperature. It wil

3、l be treated as an incompressible uid with density . Flow continuity (the fact that the mass owing into and the mass owing out of a given innitesimal volume are equal) implies that the ux of helium velocity through a closed surface is always zero. Superuid velocity in this aspect is analogous to the

4、 magnetic eld intensity. By analogy with the magnetic eld lines, “streamlines” are tangential to the uid velocity at each point and their density is proportional to its magnitude.True superow has an important property of being irrotational: circulation of superuid velocity along any closed path with

5、in helium is zero =(1)This statement must be amended if superuidity is absent along a thin “vortex lament”. The thickness of the lament itself is of approximately atomic dimensions , but the vortex induces long range velocity eld in surrounding superuid: velocity circulation around such lament is th

6、e circulation quantum1 = 2,(2)and zero if the path can be contracted to a single point without crossing the vortex, see Fig. 1. This supports the analogy between superow and the magnetic eld created by wires carrying current: su- perposition of two valid velocity distributions is a valid velocity di

7、stribution and the velocity at any point is equal (up to a dimensional factor) to the magnetic eld produced by the unit currents running through a system of wires representing vortex laments.1Circulation quantization is a macroscopic quantum effect and corresponds to the angular momentum quantizatio

8、n in Bohr model. The circulation quantum can be expressed as = /He, where He is the mass of helium atom.Q1-2English (Ocial)TheoryFig. 1: Vortex lament (red) in superuid (light blue).Velocity circulations along paths 1, 2,5, and 6 are all zero, but those for 3 and 4 are equal to 2. Note that circulat

9、ions along3 and 4 have opposite signs.Part A. Steady lament (0.75 points)Consider a cylindrical beaker (radius 0 ) of superuid helium and a straight vertical vortex lament in its center Fig. 2.Fig. 2: Straight vortex along the axis of a beaker.Part B. Vortex motion (1.4 points)Free vortices move abo

10、ut in space with the ow2. In other words each element of the lament moves with the velocity of the uid at the position of that element.2This is a consequence of momentum conservation, see next section.A.2Work out the free surface shape (height as a function of coordinate () ) around0.5pt the vortex.

11、 Free fall acceleration is . Surface tension can be neglected.A.1Plot the streamlines. Find out the velocity at a point .0.25ptQ1-3English (Ocial)TheoryAs an example, consider a pair of counter-rotating straight vortices placed initially at distance 0 from each other, see Fig. 3. Each vortex produce

12、s velocity 0 = /0 at the axis of another. As a result, the vortex pair moves rectilinearly with constant speed 0 = /0 so that the distance between them remains unchanged.Fig. 3: Parallel vortex laments with opposite circulations.Fig. 4: Parallel vortex laments with equal circulations.A beaker of hel

13、ium (see Part A) is lled with triangular lattice ( 0) of identical vertical vortices, see Fig. 5.Fig. 5: Triangular lattice of vortices in a beaker. The view from above.B.1Consider two identical straight vortices initially placed at distance 0 from each0.25pt other as shown in Fig. 4. Find initial v

14、elocities of the vortices and draw their trajectories.Q1-4English (Ocial)TheoryPart C. Momentum and energy (1.75 points)The long range velocity eld is the major contribution to the energy of a system of vortices, it is insensitive toexact structure ofthe lament. Thelament itself can notbeproperly de

15、scribedbythe macroscopic theory and apparent singularities (innities) are insignicant. Real physical quantities, such as the en- ergy,oftheregioninsideathintubeofradius aroundthelamentshouldbeneglected. Outsideofthis tube the density of superow kinetic energy 2/2 (where = const) is analogous to the

16、energy density of the magnetic eld 2/(20) they are both quadratic in respective variables. This analogy together with thecorrespondence between magnetic eldand superuid velocity generated byvortices (currents) facilitates calculationof the owenergy for agiven system. For instance, giventhe inductanc

17、eofa cir- cular wire loop 0 log(/), where is the loop radius and is wireradius, we getthesuperuid vortex loop energy3 222 log(/)(3)Total uid momentum is also determined by the long range velocity distribution. It is obtained by inte- gration of the momentum density . Again, consider a ow generated b

18、y a circular vortex loop placed in plane. It is obvious from the symmetry considerations, that total momentum has only component:(4) = = ( ) (,)The innermost integration is in fact an integration along appropriate paths parallel to -axis, see Fig. 6. From the circulation identity (2) it follows that

19、(, ) = (,)(5)is piecewise constant. Particularly, it is zero for paths passing outside the ring and 2 for paths inside it. Total momentum is therefore = 2 2 = 222(6)3This expression is also valid only if log / 1.B.5Work out the smoothed out (omitting the lattice structure) free helium surface0.25pt

20、shape () .B.4Find the distance AB() between the vortices A and B at time . Treat AB(0) as0.35pt given.B.3Find velocity () of a vortex positioned at . 0.4ptB.2Draw the trajectories of vortices A, B, and C (located in the center).0.15ptQ1-5English (Ocial)TheoryFig. 6: Velocity eld of a circular vortex

21、 loop and integration paths (green) for (, ) calcula- tion.Fig. 7: A nearly rectangular vortex loop, .C.3Suppose we shift a long straight vortex lament by a distance in direction,0.75pt see Fig. 8. How much does the uid momentum change? Indicate themomen-tum change direction. The lament length (cons

22、trained by the vessel walls) is.C.2Calculate its energy .0.7ptC.1Consider a nearly rectangular vortex loop , , Fig. 7. Indicate the0.3pt direction of its momentum . Find out the momentum magnitude.Q1-6English (Ocial)TheoryFig. 8: Momentum changes whenever the vortex shifts with respect to the uid.Part D. Trapped charges (2.85 points)Electrons, if injected in helium, get trapped in the