张量分析答案完整版

黄克智版张量分析课后习题答案完整版第一章1.1求证：()()()××=•−•uvwuwvuvwuvwuwvuvwuvwuwvuvwuvwuwvuvw并问：)(××与××）（是否相等？、、为矢量证明：因为u=(,,)xyzuuu;v=(,,)xyzvvv;w=(,,)xyz左边=()××uvwuvwuvwuvw=(,,)xyzuuu×[(,,)xyzvvv×(,,)xyz]=(,,)xyzuuu×xyzxyzijkvvv⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦=(,,)xyzuuu×[()yzyzvwwv−,()xzxzwvvw−,(xyxyvwwv−)]=[()()yxyxyzxzxzuvwwvuwvvw−−−,()()zyzyzxxyxyuvwwvuuwwv−−−,()()xxzxzyyzyzuwvvwuvwwv−−−]右边=()()•−•uwvuvwuwvuvwuwvuvwuwvuvw=()xxyyzzuwuwuw++vvvv-()xxyyzzuwuwuw++=()xxyyzzuwuwuw++(,,)xyzvvv-()xxyyzzuwuwuw++(,,)xyz=[()()yxyxyzxzxzuvwwvuwvvw−−−,()()zyzyzxxyxyuvwwvuuwwv−−−,()()xxzxzyyzyzuwvvwuvwwv−−−]所以：()()()××=•−•uvwuwvuvwuvwuwvuvwuvwuwvuvwuvwuwvuvw同理可证：()()()××=•−•uvwuwvvwuuvwuwvvwuuvwuwvvwuuvwuwvvwu所以)(××≠××）（1.11根据上题结果验算公式：ijjig=gggggggg由上题结果：：：：2=gggg,,,,11()2=−++gijkgijkgijkgijk,,,,21()2=−+gijkgijkgijkgijk,,,,312=gggg()+−ijkijkijkijk21rsgrs⎧=⎨≠⎩当r=s当1231231112132ggg++=++gggggggggggggggggggggggg211()()()222=++-i+j+ki-j+ki+j-k-i+j+ki-j+ki+j-k-i+j+ki-j+ki+j-k-i+j+ki-j+ki+j-k1==j+kgj+kgj+kgj+kg及：：：：1231111213ggg=++gggggggggggggggg同理;1231232122232ggg++=++gggggggggggggggggggggggg121()()()222=++-i+j+ki-j+ki+j-k-i+j+ki-j+ki+j-k-i+j+ki-j+ki+j-k-i+j+ki-j+ki+j-k2==ggggi+ki+ki+ki+k及：：：：1232212223ggg=++gggggggggggggggg1231233132332ggg++=++gggggggggggggggggggggggg112()()()222=++-i+j+ki-j+ki+j-k-i+j+ki-j+ki+j-k-i+j+ki-j+ki+j-k-i+j+ki-j+ki+j-k3==ggggi+ji+ji+ji+j及：：：：1233313233ggg=++gggggggggggggggg及验证：：：：ijjig=gggggggg正确1.21试证明若一张量的所有分量在某一坐标系中为零，则它们在任何其他坐标系中亦必为零。证明：不妨取三界张量根据P24页所讲的分量表示法和坐标转换关系知识T=Tijkgigjgk=Tijkgigjgk=Tij..kgigjgk=Ti.jkgigjgk=……其分量为：TijkTijkTij..kTi.jk……他们满足坐标转变关系，先将ijk用rst表示，我们可以得到''''''''''''''''''''''''''''''''''......'''............(,,1,2,3)......ijkijkijkijkrstrstrstrstijkijijtrsrstkkiistrrstjkjkrstrsrrsttstijkijkjTTTTTTTTijkTTTTTTTTββββββββββββ=====∴∵都为零等式左边在新坐标系下的张量分量都为零即''......ik∴全为零n阶张量同理可证当一张量在一个坐标系中所有分量都为零时，则他们在任何坐标系中亦必为零1.31已知：kv为一矢量的协变分量。（根据P31页所讲的张量的对称与反对称知识来证明这个题目。重点(.)(.)nmmnTT=−）求证：mnnmxx∂∂−∂∂νν为一反对称二阶张量的协变分量。证明：令mnnmnmxxT∂∂−∂∂=νν).(则由vvvmmmmmxxmm'''∂∂==β可知:vmnmmnnnmmmnmxxxxxxvxxxv''''''∂∂∂+∂∂∂∂∂∂=∂∂同理可得：vnnmnmmmnnnmnxxxxxxvxxxv''''''∂∂∂+∂∂∂∂∂∂=∂∂则vvnnmnmmmnnnmnmmnnnmmmmnnmnmxxxxxxvxxxxxxxxvxxxvxvT'''''''''''''').(∂∂∂−∂∂∂∂∂∂−∂∂∂+∂∂∂∂∂∂=∂∂−∂∂=由于：vvnnmnmnmmxxxxxx''''∂∂∂=∂∂∂所以)('''''''').(mnnmnnmmmnnmnmxvxvxxxxxvxvT∂∂−∂∂∂∂∂∂=∂∂−∂∂=即).().('''''''''')(nmnnmmmnnmnnmmnmTxvxvTββββ=∂∂−∂∂=所以的证：mnnmnmxxT∂∂−∂∂=νν).(为二阶张量的协变分量。当nm=时恒有0).(=nmT又有(.)(.)mnnmmnnmTTxxνν∂∂=−+=−∂∂综上可知：mnnmxx∂∂−∂∂νν为一反对称二阶张量的协变分量1.41质量为m、绕定点O以角速度ωωωω转动的质点（见图），其动量矩矢量的定义为m=×LrvLrvLrvLrv，其中，rrrr为定点O至质点的矢径，vvvv为质点的线速度。求证：=•LILILILIωωωω，式中IIII为惯性矩张量，[()]m=•−IrrGrrIrrGrrIrrGrrIrrGrr证明：()mm=×=××LrvrrLrvrrLrvrrLrvrrωωωω=[()()]m•−•rrrrrrrrrrrrrrrrωωωωωωωω此题为书上P34页（1.8）例题[]iimikmkLmrrrrωω=−ωωωωvvvvrrrr[]imikkmkmrrrrδω=−ikkIω=i所以[()]m=•−•=•LrrGrrILrrGrrILrrGrrILrrGrrIωωωωωωωω1.51已知向量1ωωωω与二阶反对称张量1ΩΩΩΩ，矢量2ωωωω与二阶反对称张量2ΩΩΩΩ分别互为反偶。反偶？求证：12121:2•=ωωΩΩωωΩΩωωΩΩωωΩΩ证明：由已知得)(41)(4141)()(41):():(41):21():21(2121212121212121kjjkjkjkstjksktjtksjstjkrstijkrstrstijkijkyxxytsrrstmllmkjiijkggggggggggggΩΩ−ΩΩ=ΩΩ−=ΩΩ∈∈=Ω∈•Ω∈=Ω∈•Ω∈=Ω∈−•Ω∈−=•δδδδωω������������������已知2Ω�为反对称张量，故jkjkkjjk2121ΩΩ−=ΩΩ所以jkjk212121ΩΩ=•ωω��而21212121212::ωωδδ��������•=ΩΩ=ΩΩ=ΩΩ=ΩΩjkjkmkljlmjkmllmkjjkgggg得证第二章2.2已知：二阶张量T与TT互为转置（TijijTT=）求证：T与S具有相同的主不变量。证明：对于T：123()TTaTpaiimmamampJTJtrTTTTGTTJTTTGTTT•••••••••••••••••======对于S：123()TTTTTTaTTTTpajjpmmamamJTJtrTTTTGTTJTTTGTTT•••••••••••••••••======得证。2.3已知：任意二阶张量A,B,且TAB,SBA==ii求证：T与S具有相同的主不变量。....()()()()ijmnabmajinmabammnijabnanmjiabantrTtrABABGTggTgggggTTtrStrBABAGTggTgggggTTTS∗∗==•=•:=•:===•=•:=•:=∴证明：与具有相同的主不变量。T1S1ƒƒ2.4求证：（1）[][][][]φ=⋅+⋅+⋅（2）[][][][]ccccbbbbaaaaccccTTTTbbbbaaaaTTTTccccTTTTbbbbTTTTaaaaccccbbbbTTTTaaaaTTTTT2φ=⋅⋅+⋅⋅+⋅⋅证明：（1）式左边=[]bbbbaaaaiiiiggggggggggggbaijwvjuT⋅+[]ddddjjjjjjjjccccggggggggvvvvggggdijwT⋅cu+[]iiiijjjjffffeeeegggg⋅eu=iabbajijwvuTε⋅+ciddjcijwvuTε⋅+cidjfeijwvuTε⋅=61jabjabiabbajijwvuTεεε⋅+61cjdcjdciddjcijwvuTεεε⋅+61efjefjefijfeijwvuTεεε⋅=[][][]()⋅=[]δ⋅=[]⋅=[]φ，命题得证。（2）式左边=[]cjTTggggggggggggaaaaiiiicbabijcba⋅⋅+[]ababijijdcTbggggggggggggaaaa⋅⋅Td+[]ababeejijcTbggggggggggggaaaa⋅⋅iT=iaccbjcbaTεabijT⋅⋅+diabjdcbaTεabijT⋅⋅+ieabejcbaTεabijT⋅⋅=()jbbjbbieabejdjbdjbdiabjdjbcjbcieacbjcbacbacbaTεεεεεεεεε++⋅⋅abijT61=()[]()[]()[]{}ccccbbbbaaaaccccbbbbaaaaccccbbbbaaaabijabaijbijabaijbijabaijTδδδδδδδδδδδδ−+−+−⋅⋅abijT61=()[]ccccbbbbaaaabijabajiTTδδδδabijabijTT21⋅⋅⋅⋅−=()[]ccccbbbbaaaaaiiaaaiiTT21⋅⋅⋅⋅−TT=[]ccccbbbbaaaaT2φ命题得证。2.5111Naaλ⋅=⋅222Naaλ⋅=⋅21211aNaaaλ⋅⋅=⋅⋅12112aNaaaλ⋅⋅=⋅⋅上式左端相等，1221aNaaNa⋅⋅=⋅⋅故其右端也相等，即()12120aaλλ−⋅=注意到121200aaλλ−≠=同理可得所以123,,aaa互相正交且唯一()()()()()()112212211331'''1232.622(2)2(2)12222020012,,()122122020012222021420iiiijjjjiiiijjjjNeeeeeeeeeeeeNeeeeeeeeeeeeNeeeeeeeeeeeeNeeeeeeeeeeeeNNNNeeeeeeeeeeeeNNNNλλλλλλλ=+−+−+−−⎡⎤⎢⎥⎡⎤=−⎣⎦⎢⎥⎢⎥−⎣⎦−−⎡⎤⎢⎥⎡⎤=−⎣⎦⎢⎥⎢⎥−⎣⎦−−−⎡⎤⎢⎥∆=−−=−+−−⎢⎥⎢⎥−−⎣⎦∴iii