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# INTERM MATH PHYSICS PHYS 4112

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This 24 page Class Notes was uploaded by Elva Fahey on Tuesday October 13, 2015. The Class Notes belongs to PHYS 4112 at Louisiana State University taught by Staff in Fall. Since its upload, it has received 6 views. For similar materials see /class/223001/phys-4112-louisiana-state-university in Physics 2 at Louisiana State University.

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Date Created: 10/13/15

Here are some notes on vector and dyadic tensor notation similar to what I will be using in class with just a couple of changes in notation Notes Dowling X British notation For Cross Product vs American Dowling s A A Notation For Vector vs Standard Dowling s E B Notation For Dyadic Tensor vs Standard Dowling s With these notes you should be able to make sense of the expressions in the vector identities pages below that involve the dyadic tensor T Notes courtesy of Peter Littlewood University of Cambridge The vector identity pages are From the NRL Plasma Formulary order your very own online or download the entire thing at http wwwppdnrna minrlformulamz JPD VECTOR IDENTITIES4 Notation f g are scalars A B etc are vectors Tis a tensor is the unit dyad ABgtltCAXBCBCgtltABgtltCACAgtltBCgtltAB AgtltBgtltCCgtltBgtltAACB ABC AgtltBXCBgtltCgtltACgtltAgtltBO 4 AXBCXDACBD ADBC AXBgtltCgtltDAgtltBDC AgtltBCD Vltf9gt W91 ng gi VfAfVAAVf VgtltfAfVgtltAVfgtltA 9 VAXBBVgtltA AVgtltB 10 VgtltAxBAVB BVABVA AVB AgtltVgtltBVBA AVB VABAgtltVXBBgtltVgtltAAVBBVA v2 VVf 11 12 14 VQA 15 13 16 VVgtltAO If e1 e2 e3 are orthonormal unit vectors a second order tensor T can be written in the dyadic form 17 T Zia Tijeiej In cartesian coordinates the divergence of a tensor is a vector With components 18 VJM ijTjiWa This de nition is required for consistency With Eq 29 In general 19 V AB V AB A VB 20 V fT VfTfVT Let r ix jy kz be the radius vector of magnitude 7 from the origin to the point x y z Then If V is a volume enclosed by a surface S and d8 ndS Where n is the unit normal outward from V 27dVVfde V S 28dVVAdSA V S 29dVVTdST V S 30dVVgtltAdsgtltA V S 31dVfV29 9V2fdSng 9Vf V S 32dVAVgtltVgtltB BVgtltVgtltA V dSBgtltVgtltA AgtltVgtltB s If S is an open surface bounded by the contour C of Which the line element is all 33 dsgtltijdlf S C 34dSVgtltAjdlA s o 35dsgtltVgtltAjdlgtltA s o 36dsVfgtltV9jfdg jgdf s o o EXAMPLES CLASS IN MATHEMATICAL PHYSICS V Introduction to Tensors 3299 Commentary This is the rst of two sheets on tensors the second will come in class VIII Tei39isors have many applications in theore 39 cal physi s not only in polarization prob lems like the ones described below but in dynamics quot quot fluid 39 1 quot quot theory etc The mathematical apparatus for dealing with tensor problems may not yet be familiar so this sheet contains quite a high proportion of formal material Sectioi i A introduces the concept of a tensor in this case a linear operator that transforms one vector into anotl ier in the physical context of studying the polarisation of an aspherical molecule Sectioi i B sui m narims various 139natl 1139natical and notational ideas that are useful in physics tl39iese include suf x notation and coi icepts such the outer product of two vectors s and pro etion operators Tl l are matlmmatical obje hich take the projection or compoimnt of a vector along a particular directioi39i Section C returns to a physical problmn also about polarization and leads to a dis cussion of principal axes ig1 1Vctor s and eigemvalues of tensor quai itities Sectioi i D introduces the properties of the antisyrmnetric third rank tensor eijk Wl liCl l can be used to derive many results in vector calculus much more rapidly than by other methods This ground will be covered again in sheet VIII so it doesn t matter if you run out of time A Introduction We often nd vectors that are linearly related to other vectors for instance the polar isation E induced in a medium by applying a eld E E ox 1 Here E and E are parallel Their cartesian coi39npoi39imits are related to each other by the sai ne constant of proportionality 60X for each component Here we have dm39ioted Vectors by ul iderlinings one would do if writing this out by hand Usually books dei iote Vectors by bold type in the rest of this sheet we treat the two notations coi39npletely interchangeable It is possible to ii39nagii39ie situations wl391ere the medium is anisotropic and the respoi ise factor 60X in 1 is dil arent for coi39npoi39ients in dil arei it directions The two Vectors E and E are no longer imcessarily parallel and what coi nmcts them is called a tensor We will use sii nple model of molecular polarisability to show the ul iderlying physics and introduce the ideas of tei isors A1 An Electrostatics Problem The diacetylmm molecule H CEC CEC H is highly polarisable along its long axis p 3HEH being the induced dipole for a eld EH applied along the i39nolecule The response to perpei39idicular fields is pi 3JEJ where 3i ltlt 3 Let the cylindrical molecule point in the 1 1 0 direction and apply a eld in the 1 directioi i E Eli i Prove p1 fBLEJ72 2 pg 13H must2 3 P2 0 4 Hence one sees that p and E are not parallel and in general we must write 13 E 5 HQ wl39iere g is the polarisability tensor Tei39isors are written with a double underline as i39natricies often are or in books so139n2ti1m2s doubly bold bold sans serif typface charactt In component form 5 is Pa Xxszx 39l GaryE1 39l XxZEz P1 erx MIEu aTZEZ 7 p2 ZJIEJ 39l zzE7 39l a39zzEz ii What are the cof cints explicitly for the aboVe exai nple Write g a matrix of its COIHpOIl 11CS Note that although the tensor g is the same physical entity in all coordinate systems it relates p with E in wl39iatever basis these are expres its coi npoimi its depend on the coordinate system used Remark If many 139nolculs are present with the same orientatioi39i E is pp where there are p 139nolculs per unit volui39ma A l39iypotl39mtical diacetyleim solid might consist of such rods arranged in an array all directed along 1 1 0 Then the macroscopic X to replace X in 1 would 1391glcting internal field corr tions be pg B Miscellaneous Ideas and Notations B1 Suf x notation Coi ni nonly suf ces such i and j are dui ni ny symbols and when repeated in an expression they are to be summed over this is the Einstein Convention For instance 6 to 8 can be succinctly written Pi WjEj which is Eq5 rewritten in suf x notation Note the order of the indi on the right Thus we have written the vector p simply p and it will be clear from context that a vector is intended and not simply one of its components Likewise Q can be denoted j In suf x notation the dot product also called scalar product or inner product of two vectors a b is written Llbi The operation of a tensor on a vector also involves an irn39ier roduct sum over re eated ind s described above The unit tensor 5m b de nition I U transforms any vector into itself SM j vi B2 The outer product Coi isider the object QM de ned QM E d bj 9 where g and L are vectors i Show using suf x notation that if p is a vector Q p is also a vector nd its magnitude and direction Note that E 25ij We e that QM is a linear operator that transforms one vector E into anotl39ier it is therefore a tensor This defines the outer product or dyadic product of two vectors In i39ioi39i suf x notation sometimes called dyadic notation Eq9 is written Q ab 10 or Q QQ where there is no dot between the two vectors In this notation we have a b c ab c prove this using suf x notation if you did not do so already above This notation is used in various areas of i39natl39iematics and physics but in general suf x notation is more versatile and can always be used instead ii We can write the tensor g in Q A1 in the following form g a 05in 0512 22 6 other terms 11 Cl ieck that 11 indeed agrees with 6 to 8 by finding the result of applying a field in the If direction E Exfc Ill HQ p 39 E 393 Ex xxa Gyma 0521sz7 iii Projection operators In dyadic notation the identity tensor iij is often written the unit operator Consider the action on a vector V of the operators 1111 and 1 1111 where 11 is a unit vector Show that these perform the job of resolving the vector V into vectors V and V i respectively parallel and perpei39idicular to 11 Prove that these projection operators are idempotent The de nition of an idei npotent operator G is that 6767 Show that the polarizability tensor in problem A1 can be written g 39l39 1111 where 11 1 1 C Use of Principal Axes C1 A xed object is placed in a uniform electric eld of 1 kV m VVl iei i the eld is in the 1 direction it acquires a dipole moment with components 4 2 and 1 X 10 12 C m in the f 11 Z directions For the same eld strei39igtl39is in the y 2 directions the dipole moments are 2 4 1 X 10 12 and 1 1 X 103912 C m respectively i Write down the coi39npoi39iei39its of the tensor that descrilms the polarisability of the object using the f 11 Z coordinate frai ne ii What is the torque T p E on the object wl39iei39i it is placed in an electric eld of kVm in the 111 direction Ans 1 10 X109 Nm iii Suppose there are eld directions 110 W for which there is no torque on the object what relation do E and p then have Show that 11 1 and W are the ignvectors of g iv Show that if we use a coordii iate system the unit vectors 111 iv tl39iere are no off diagonal l1 n1 1ts of the tensor g when it is written a matrix of coi39npoi39ients The tlnee diagonal l 13911 11CS A1 A2 and A3 are called the principal polarisabilities of the object and are the eigenvalues of g For this problem one ig1 1val11e is 1 2 X 10391391 nd the m w other two What are the vectors 110 0v in teri ns of Q 1 v In this exai39nple g is syi39ni39netric and hei ice has real eigei ivahms and orthogonal eigen vectors An ei iergy argui nei it shows this to be true for the polarizability of any object the 110 0v coordinate system can hei ice always be found and is called the principal f ame often a very cgt1391v11 1i1 1t frai ne to think about the physics in The same applies to any problem involving a syi39ni39netric tei isor vi In a general coordinate system the decoi npositioi39i of g into prii icipal directions 110 v0 can be accomplislmd using projection operators g A1111 A200 Agv vv v 12 This is a relatioi i between tensors so if it holds in one co ordii iate system it also holds in any other Prove it for a cgt1391v11 1i1 1t choice of co ordii iates system and hei 3e geimrally vii Eq12 expres es in a very direct way the operatioi39i of a syi39ni39netric tensor g on a vector Q To find g Q the recipe is 1 resolve Q into coi39npoimi39its along the prii icipal axes 2 i39nultiply each coi npoi iei it by the corresponding eigei ivalue 3 add to form the resultant Confirm froi39n Eq12 that this is prec ly what operatii39ig with g on Q does C2 Force on a Dipole i By using suf x notation and the ordinary chain rule for derivatives prove the vector relation V 1AA 1W A A VIZ Derive a similar result for VA B It s best to leave the answer in suf x notation ii In a unifori n electric field E a dipole p experiei ices a couple p X E but no net force 23 21 net fOX CQ Fi 1ij 13 and explain why 8a is a tensor This is very gei Jral the gradient of a vector is a second 27 rank tensor iii A i39nore gei ieral expre gtn for the force on a permaimnt dipole derived from the ei iergy is F gradpE or Show that this is equivalent to Eq13 only in the absence of a tii ne varyii ig i39nagi39ietic field iv A speck of i39natter has no peri naimi it dipole 139no139nnt but the field E induces a moment p in it Ui39ider what coi39iditioi39is can we write a p OtijEj b p Write down the force in each ca 1 Under what coi39iditioi39is can this be written F or gradE22 D The antisymmetric three tensor eijk A tensor with two indices such itj is called a tensor of second rank and its linear operation on a vector gives another vector A vector is tcl39n391ically a tei isor of rank one a scalar of rank zero Tei39isors of higher rank can also be delii39ied For exai nple a tensor of rank four Qijkl acts on a rank two tei isor to give another rank two tei isor BU QijklAkl with sui ni natioi i over repeated indic For tei isors higher than rank two dyadic notation becomes extrei nely cui nlersoi39ne and suf x notation is ential D1 An importai it and useful tensor of rank three is the fully antisymmetric three tensor denoted 60k This operates on either a second rank tensor or two vectors to give a vector For i 7 j 7 k it is defined 6mg 1 if ijk is a cyclic permutation of it y z eg 1356 and 1 if the order is ai39iticyclic eg cozy If any two of ijk are the same then 6mg 0 eg ex 0 The tensor 6mg is very useful for writing down a vector product in suf x notation Show that A Bli th Aj Bk 14 The following relation ijk 6klm ail Sjm Sim Sjl 15 is Well worth remembering you should come across this in maths lectures It saves a great deal of time wl39ien manipulating vector idei itities such A A A ijk Aj 6klm Bl Cm Aj Bl Cm ail Sjm Sim sjl Z BiAjCj 44ij Z AQB ABE Wl liCl l is a familiar result normally derived by far more tedious means The result Eq15 is especially l39ielpful wl39ien dealing with vector idei itities involving the V operator Because eijk is a constant tensor it coi ni nutes with V and expressions can be re ordered accordingly For example V Eo E vi ijk Ej Hk 16 Z ijkViEj Hk Z ijklEjVin HkviEjl 17 Ej 39kvin Hk 18 EV VE 19 Wl liCl l is a result used in in cormection with Poynting s vector in electroi nagnetism i Make sure you understand each step of this derivation ii Show using similar metl iods that VAVAVVA V2A V 1A1V A A VIA Tl iese relations are useful in electromagnetism and other topics in IE and in the parts 11 and 111 courses EXAMPLES CLASS IN MATHEMATICAL PHYSICS VIII Further Applications of Tensor and Vector Calculus 24299 Commentary This is the ond sheet on tei isors It follows on from Sl ieet 5 Section D of that sheet on uses of the ai39itisyi39mnetric 3 tei 1sor eijk is rprinted l39iere an appei39idix and you should complete it before tarting the main part of this sheet if you didn t do so last time As on sheet 5 We use underlining and bold notation intercl iai igeably Suf x notation and the sumi natioi i coi39ivei39itioi39i are also used Carets A denote unit vectors In Sections A and B tensor ideas are used to work through some results in rigid body dynamics You should have covered this material in IE dyi39iamics courses and the formal lEVQlOpIHQHC given here is good practice and will help deepen your understanding of moments of inertia etc well of tei isors Section C introduc the coi icepts of stress and strain tensors in a solid The linear relation between tl39iese is governed by a fourth rank tensor wl39iose properties are ii39ivesti gated Section D contains problems on the trai isformation properties of tei isors in i39natl39iemat ical courses you may have seen tei isors de ned in terms of tl39iese trai isformation properties A Rigid Body Rotations A rigid body rotates with angular velocity g about an axis through its cei itre of mass at the origin 0 The vector g points along the axis of rotation and deteri nii ies the velocity of rotation at r accordii ig to V g r Make sure you can derive this if you are not sure Wl iere it coi nes from i Coi isider the body a collection of point masses 5772f with position vectors r relative to the cei itre of mass 0 The tensor I is called the inertia tensor of the body It is given by I Z r2 ff 57m 1 r Sl iow using Eq1 that the angular i39noi39n139it11139n L and the kinetic ei iergy T of the body are L I g 3 TgIg wiIijwj You will need to expai id the vector triple product f g h see Appei idix Thus I specifies the linear dependei ice of angular momentum L on the angular velocity wj via L U wj Eq2 and replaces the single nui39nber the moment of inertia that would be enough for a very syi ni netrical body such a spl39iere ii Find the coi39npoi39ients of I in a Cartesian system that is the usual basis set 2 2 Ai39iswerz I Era2 225mr IW Zrzry5mr rim iii 1 is a symmetric tensor that is IV V1 for any vector V which should be clear from the delii39iitioi39i 1 Deduce it also from the matrix of compoimnts of U found in ii iv Take the coi itii iuui n limit in Eq1 to obtain formally an ii39itegral the ii39iertia tensor of a body of variable dei isity r v Write down the inertia tei isor of a body coi39isistii39ig of eight masses 1 at the cori39iers of a light cubic frai ne of side a Cl ioose the coordii iate systei39n most coi39ivei iiei it to you VVl lat is the i39noi39nei39it of inertia of this cube VVl iy can we talk of the i39noi39nei39it of inertia wl39iei39i we strictly speaking have a tei isor In fact Well for an asyi39nimrtric body one S1 H Ci1 n speaks of the i39noi39nei39it of inertia about soi39ne specific axis 11 This is written 11111 and gives the ratio of the 11 coi39npoimi39it of angular momentum Lil to the 1 coi39npoi39iei39it of angular velocity 11 Since in gei ieral L has coi39npoimi39its which are not parallel to 11 this does not offer a coi39nplete description vi Angular velocity A body rotates about an axis through its centre of mass with an angular velocity g The velocity of a point in the body is V g r Show that one can equally well expre it V g r wl39iere g is an ai39itisyi39m39netric tensor with coi39npoi ieu its 0 w2 my 12 0 szr wy wx 0 vii Show that any vector W constant with respect to axes fixed in the body cl39iai39iges with tii39ne in a space fixed frai ne by an ai39noui39it dW in tii ne it dW grit W and l l 11 that if W is not body xed but has a rate of cl39iai39ige dWdtgdy the vector W in the space xed frame obeys dWdt5pm dWdthmdy g W 5 B Principal Axes and Precession We now use the above results to solve simply a dif cult problem the rotational motion of an asyi ni netric body i VVl iat is the inertia tensor of four point mas s 1 fixed at the corners of a mas square frai39ne of side a What are the principal axes and prii39icipal moments of inertia Why are two of the axes indeteri39nii39iate This is an example of a nontrivial body unlike the spl39iere or cube wl39iere the prii39icipal moments of inertia are not all the same Now L need not be parallel to g for a body whose angular momentum is C1 l X V d the result is free precession of the angular velocity see iii below ii It is clear from 1 that in general I for a body is constant in body fixed axes and so for an arbitrary motion the inertia tensor 1 in space fixed axes is not constant ul39iless trivially it is proportioi ial to iij Show that in general the motion of a nontrivial body whose angular momentum Llt5m is conserved in a space fixed frame involves an angular velocity g which changes in time Hint consider wj iii The problei39n of lindii ig w for a body of fixed angular momentum not parallel to a principal axis is best solved by adopting a body frai39ne where the ii39iertia tei39isor is constant Show from 5 that in the body frame L obeys dLdt g L 6 iv Deduce that In 033 5ijkwj1klwl and show that by taking the body fixed frame the principal frai39ne of I this can be si139nplilil to 11031 w w313 12 12 032 W3W111 I3 13 W1wz12 11 wl39iere I 1 12 I 3 are the prii39icipal moments of inertia v For the inertia tei isor of the square frai39ne found in questioi39i i 11 12 such a body is known a symmetric top Prove for this case that g preces es in the body frame according to U3 const 11 A cosQt d2 A sinQt wl39iere Q 1 U3 and so coi39iliri n that g is not a constant vector unless it happens to point along one of the principal axes C Stress and strain tensors C 1 Coi isider a small area l 13911 11C dS ei nbelled at a point r in an isotropic liquid If the i39naterial is subject to pressure P tl39iere is a normal force on eacl39i side of the surface l 13911 11C PdS Tl iese point in opposite directions so the net force on the surface l 13911 11C is zero which is it must be sii ice the l 13911 11C has no mass and would otherwise be subject to inlii iite acceleration In the case of a solid there can also be a shear force exerted in equal and opposite directions on the two sides of the l1 n1 1t such a force lies in the plane of the l 13911 11C ratl ier than normal to it In the gei ieral case we can write dF g dS 7 wl39iere is the stress tensor at the point r For exai39nple the elei nei it T of deteri nii is the ctr force per unit area on an elemei39it of surface with its i39iori39nal poii39itii ig in the z direction if szdsz For a surface elei nei it of gei ieral oriei itatioi39i de T de Tagd5 szdsz i Show that a small box of sides dar dy dz experiences a couple 721 Tudzr dy dz about the y axis The correspoi39idii39ig i39noi39nei39it of ii39iertia is of order d32dz2dy Explain why this i39neai39is that 1 must be symi netric Show also that if 5 is a fui39ictioi i of position then the net force on a finite piece of i39naterial is given by F f J W s l drj where in a conventional notation I E VJ 7 ii Strain Suppose a solid material is subjected to a small deformation r gt r ur So long the dfor 1 natio1 1 depends continuously on position we may write dw drx 9 drj J 871 Let us now decompose the tensor UM a into its symimatric and anti symmetric parts 7 U0 RM 1 in g UM Uj i 1 RM g UM Uri Show that Q is syrmmtric and R antisyrmmtric Show also that R corresponds to a pure rotation ie without distortion of the bit of matter near r Hintz compare Eq4 in section A What type of deformation does 6 represent Why may we assume that the stress tensor 1 d2pends on the strain tensor Q but has no dependence on E iii Elasticity For small dfor139nations we expect the stress and strain to be lii iearly 4 h V related but since both are ed by secoi39id rai39ik tensors the relation between them generally involves a tensor of the fourth rank m ijkl 6k 10 which is the gmmralizatioi i of Hooke s law to an arbitrary threc dimensioi39ial i39naterial Since 1 is syi ni netric Eijkl must be syi39nimatric under exchange of i and j various other symmetries can be found from considering the form of the stored elastic energy but evei i when these are taken into accoul lt Eijkl has 21 independent COIHpOIl 11CS iv Isotropic elastic solids are i39nuch sii npler l39iowever and have only two indepei39ident constai its Tlmse are the Young s modulus Y and Poisson s ratio a The definition of tl iese is that in the prii icipal axes of TM the following relation applies Yen 7271 073971 39i 722 11 with similar relatioi39is for Guy 22 all off diagonal elei nei its of Q are zero in this coordinate system By rarranging Eq11 obtain the result YGJIJI 1 39i aszxrzx ITii wl39iere 1 indicies are not sui rn39mad over but T is sumnmd with similar relatioi39is for yy and 22 compoimnts Note that T Trace under the sumi39natioi39i coi39iV1 itio1 i Show that by adding Eq11 to its ai39ialogues W 7 1 20 Coi39nbii39ie the above two results to give expressioi39is for 7 etc in teri ns of the gl 1 11 11CS of Q and hence derive the following result in the prii icipal frame Y 0 TH m Edi3339 Bear in mind that 0 for i 7 j in this frame Hence derive the general frai39ne independent form for the elasticity tensor in an isotropic solid Y 5k 5x1 5l xk Eijkl 7 1 I 1 20 C 2 i Apply EU to suitable dfor1 nations to recover expressioi39is for the bulk modulus and shear modulus of an isotropic solid in terms of Y and 0 ii Show that the displacement g inside a l39ioi39nogeneous isotropic solid subject to external forces only at its surface obeys grad dng 1 20V2g 0 D Transformation Laws D 1 Here We consider trai isformations of vectors and tensors btrtween two se s of base vectors ach of which is ortl39ioi39iormal We call our two sets of base vectors 61 62 63 and 6 1 9 63 so that we can use suf x notation eg the fact that the 9 basis is ortl39ioi39iormal can be writtei39i e ej 57 12 The unit vectors 6 are linear combii39iatioi39is of the 91 I ez Tijej 15 Suppose that x ante scie is any vector and y Prove that i e ek T ii The ii39iverse trai isformation el T 2192 has Tm iii 3 Tm 1V in Tici552 V y TiszkAjkCUE 4 in the rotated coordii iate This means you have found the compoimnts of the tei isor system A TiszkAjk hide this property ui ider orthogonal coordinate tranaformations T is matlmmatically often taken the delii iii ig property of a tensor For tei isors of l39iigl ier rank than 2 one factor of the trai isformation matrix T is needed for each index Trai39isformations with detT 1 are usually called proper rotations and those with detT 1 improper rota tions Show that these are the only possible values of the determii iai it for trai isformations l CW 11 ortl39ioi39iormal bases de ned above Appendix The antisymmetric threetensor eijk Reprise of Sheet V section D A tei isor with two indi such itj is called a tensor of second rank and its linear operatioi39i on a vector gives another vector A vector is tcl39n 1ically a tensor of rank one a scalar of rank zero Tei39isors of higher rank can also be defined For example a tei39isor of rank four Qijkl acts on a rank two tensor to give anotl ier rank two tei39isor BU QijklAkl with sui m natioi i over repeated indi For tensors higher than rank two dyadic notation becomes extrei nely cumbersome and suf x notation is c ential 1 An important and useful tensor of rank three is the fully antisymmetric three tensor denoted ijk This operates on either a second rank tensor or two vectors to give a vector For i 7 j 7 k it is defined 6mg 1 if ijk is a cyclic permutation of guy 2 eg 11251 and 1 if the order is anticyclic eg stay If any two of ijk are the same then eijk 0 eg ex 0 The tensor eijk is very useful for writing down a vector product in suf x notation Show that A 13 Z 6 Aj Bk 14 The following relation ijk 6klm ail Sjm Sim Sjl 15 is tedious to prove using coi npoi iei its you should come across this in maths lectures but is worth rememlmring It saves a great deal of time wl39iei39i manipulating vector identities such A B Q Z 6 Aj 6mm Bl Cm Aj Bl Cm ail Sjm Sim sjl Z BiAj C39j CiAj 3 Z AQB AEW Wl liCl l is a familiar result nori nally derived by more tedious means The result Eq15 is especially helpful wl39iei39i dealing with vector idei itities involving the V operator Because eijk is a constant tensor it coi ni nuts with V and expressions can be re ordered accordingly For example V Eo E vi ijk Ej Hk 16 ijkViEj Hk ijklEjVin HkviEjl 17 Ej 39kvin Hk 18 EV VE 19 Wl liCl l is a result used in in coi39nmction with Poynting s vector in electromagimtisi n i Make sure you ul iderstai id each step of this derivation ii Show using similar i39rmtl39iods that VAVAVVA V2A V 1A1V A A VIZ Tl iese relatioi39is are useful in electromagimtism and other topics in the 1B and parts 11 and 111 courses 10 As in the notesthe scaled time independent Schrodinger equatio h n the particle bouncing off t e floor in a grav1tational field is for DSolvelw z 5 7 mm wlzl 2 226 c1 AiryBiz2 c2 z z aAiryAi Thus by choosing 71 we can rescale the mass so in the scaled units a is the scaled energy and z is the scaled he39ght of the atom off the floor The general solution is then WzaAiz ebBiz e where e is the scaled energy 39s the scaled height of the atom off the floor Ai and Bi are the Airy function solutions to Bessel39s equation and a and b are constants of integration Let39s plot the Airy Functions PlotAiryAix AiryBix x 6 5 x 2 2 4 w v L The Ai in blue converges for large 2 like Exp z but the Bi in red diverges like Hence to get a physical solution we must impose the boundary condition b n and hence WezaAiz e i t demand that the wavefunc ion vani h at the floor where e po we0aAi e0 This quantizes the energies to be the roots of the A1 function Looking at the above blue plot we can g ess the first three roots and use the root finder to get the exact values for the first three roots tential is infinite FindRootAiryAix x 2 x gt 233811 FindRootAiryAix x 4 x gt 408795 FindRootAiryAix x 6 x gt 552056 Using these results we can write down the first three quantized energy eigenvalues 51 2 338107410459767 233811 52 408794944413097 408795 2 12 18 nb 53 5520559828095552 552056 These are all positive since we take Vz to be zero at the floor where z0 Corresponding to these three energies there will be three eigenfunctions 1 2 and 3 with corresponding normalization constants a1 a2 and a To find the normalization constants we integrate lll2 from z 0 to zm and then divide by the square root of the result al l Sqrt IntegrateAiryAiz el quot2 z 0 Infinity 14261 a2 l Sqrt IntegrateAiryAiz 52 quot2 z 0 Infinity 124516 a3 l Sqrt IntegrateAiryAiz e3 quot2 z 0 Infinity 11558 We now plot the three normalized eignefunctions and the potential which in these scaled units has the form Vzz As is usually done we displace the wavefunctions vertically by an amount of 51 for clarity llll z 51 al AiryAiz el n112z 52 a2 AiryAiz 52 113 z 53 a3 AiryAiz e3 P1 tilll21 illzlz w3z z z o 8 AxesLabel gt z 11 III w w w w z 2 4 6 8 The blue red and yellow curves are w1z w2z and W3z T e green curve is Vzz Classically a bouncing ball is confined between the y azis and the green line The green line is the turn around point where the ball reaches its maximum height The quantum particle tunnels into the forbidden non classical regime to the right of the green line where it becomes a decaying exponential Hence if you drop the atom rom 20 microns there is a non zero probability of finding it at 21 microns later on This is impoosible classically Between the y axis and the green potential line the wavefunction oscillates like a standing wave Also classically any ene gy is allowed where quantum mechanically only energies that are roots of Ai are allowed

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