Quarta Avaliação de Métodos de Física Teórica 2

UFS Universidade Federal de Sergipe Centro de Ciências Exatas e Tecnologia Departamento de Física Curso de Licenciatura em Física Modalidade EAD A4 20242 FISI0173 1 Mostre que 𝑒𝑧𝐶𝑜𝑠𝜃 𝐼𝑚𝑧 𝑚 𝑒𝑖𝑚𝜃 𝐼𝑚 é a função de Bessel modificada 2 No cálculo do deslocamento quadrático médio do oscilador harmônico temos a integral 𝑥2 𝑒𝑥2 𝐻𝑛2𝑥𝑑𝑥 Obtenha o valor da integral em termos de 𝑛 3 Mostre que para o átomo de H o valor médio da variável 𝑟1 𝑎01 𝑛2 sendo 𝑎0 o raio de Bohr 4 Usando a transformação de Laplace resolva a equação diferencial 𝑡2𝑥 4𝑡𝑥 𝑡2 2𝑥 𝑆𝑒𝑛𝑡 ignore condições iniciais 5 Mostre com o formalismo tensorial que 𝐴 𝐴 𝐴 Dúvidas ejrplazaacademicoufsbr Prof Edison Plaza 15032025 1 Mostre que ezCos heta summinftyinfty Imzeim heta Im é a função de Bessel modificada ① Função gamma Jpn1 pn ezt2 summ0infty fraczm tn2m m 1 ez2t sumn0infty fraczn2n tn n 2 Multiplicando 1 por 2 ez2t frac1t summ0infty fraczm tm2m m cdot sumn0infty fraczn2n tn n m o pn tp sumn0infty fracz2p2n frac1pn n Ipz n o pm tp summ0infty fracz2p 2m frac1pmm Ipz ez2t frac1t sumpinftyinfty tp Jpz Fazendo t ei heta t1 frac1t ei heta zfrac12 t frac1t z cos heta Assim ez2t frac1t ez cos heta summinftyinfty Imzeim heta 2 No cálculo do deslocamento quadrático médio do oscilador harmônico temos a integral intinftyinfty x2 ex2 Hn2 x dx Obtenha o valor da integral em termos de n ② Vamos usar a relação de recorrência Hn1x 2 x Hn x 2n Hn1 x 2 x Hn x Hn1 x 2n Hn1 x x Hn x frac12 Hn1 x 2n Hn1 x x Hn x2 frac12 Hn1 x 2n Hn1 x2 x2 Hn x Hn x frac14 Hn1 x Hn1 x 4n2 Hn1 x Hn1 x 4n Hn1 x Hn1 x x2 Hn x Hn x frac14 Hn1 x Hn1 x n2 Hn1 x Hn1 x n Hn1 x Hn1 x intinftyinfty x2 ex2 Hn2 x dx intinftyinfty ex2 x2 Hn x Hn x dx intinftyinfty ex2 frac14 Hn1 x Hn1 x n2 Hn1 x Hn1 x n Hn1 x Hn1 x dx ex2 x2 Hn2 x dx 14 ex2 Hn1 x Hn1 x dx m2 ex2 Hn1 x Hn1 x dx m ex2 Hn1 x Hn1 x dx Vamos usar a relação ex2 Ha x Hb x dx π 2b b δab Delta de Kronecker a b n 1 14 ex2 Hn1 x Hn1 x dx 14π 2n1 n1 δn1n1 1 a b n 1 m2 ex2 Hn1 x Hn1 x dx m2 π 2n1 n1 δn1n1 1 a b δab 0 m ex2 Hn1 x Hn1 x dx 0 x2 ex2 Hn2 x dx π4 2n1 n1 m2 π 2n1 n1 π 2n12 n1 m n1 m2 π 2n1 n1 x2 ex2 Hn2 x dx π 2n1 n1 n n1n π 2n1 n n n1n x2 ex2 Hn2 x dx 2n π n n12 n2 n 12 x2 ex2 Hn2 x dx 2n π n n 12 3 Mostre que para o átomo de H o valor médio da variável r1 a0n2 sendo a0 o raio de Bohr r1 1r ψ 1r ψ dσ ψ enra0 π a03 ψ dσ 4π r2 dr r1 4π 1π a032 0 e2nra0 r2 r dr r1 4 a03 0 r e2nra0 dr 4 a0 0 r e2nra0 dr Integração por partes u r du dr dv e2nra0 v dv a0 e2nra0 2n u dv uv v du 4 a03 a0 r e2nra0 2n 0 a0 2n 0 e2nra0 dr 4 a03 0 0 a0 2n a0 2n e2nra0 0 4 a03 a02 4 n2 0 1 1 n2 a0 r1 1 n2 a0 a01 n2 4 Usando a transformação de Laplace resolva a equação diferencial t2 x 4t x t2 2x Sent 4 t2 x 4t x t2 2 x Sent Lt2 x 4 Lt x Lt2 x 2 Lx LSent Lf tk ft 1k dk dsk Lft Para Lt2 x 12 d2 ds2 Lx d2 ds2 Lx Lt2 x d2 ds2 s2 Xs s x0 x0 d ds d ds s2 Xs s x0 x0 2s Xs s2 dds xs x0 dds 2s Xs s2 dds xs x0 2 Xs 2 s dds xs 2s dds xs s2 d2ds2 xs s2 d2ds2 xs 4 s dds xs 2 Xs Lt x 11 dds Lx dds s Xs x0 Lt x Xs s dds xs Lt2 x 12 d2 ds2 Lt x d2 ds2 xs LSent 1s2 1 Substituindo na equação Lt2 x 4 Lt x Lt2 x 2 Lx LSent s2 d2ds2 xs 4 s dds xs 2 Xs 4 Xs s dds xs d2ds2 xs 2 Xs 1 s2 1 Para facilitar a notação vamos usar d2 ds2 xs x dds xs x Xs X s2 x 4 s x 2 X 4 X 4 s x x 2 X 1 s2 1 x s2 1 x 4 s 4 s x 2 4 2 1s2 1 0 0 s2 1 x 1 s2 1 x 1 s2 12 x d ds x 1 s2 12 dds x ds ds s2 12 No relevant text to extract from the third image as it cuts off midequation dx x ds s² 1² sec²u du sec²u² Integração por partes u arctgs s tgu ds sec²u du s² 1 sec²u x 1 sec²u du cos²u du 12 du cos2u du x 12 u sen2u2 u2 sen2u4 x arctgs2 s 2s² 1 c₁ dxds X 12 arctgs ds 12 s s² 1 ds c₁ ds Xs 12 dds sarctgs ds c₁ s c₂ sarctgs2 c₁ s c₂ xt 12 𝓛¹ sarctgs c₁ 𝓛¹ s³ c₂ 𝓛¹ 1 12 ddt 𝓛¹ arctgs c₁ ddt 𝓛¹ s³ c₂ 𝓛¹ 1 12 ddt sintt c₁ ddt δt c₂ δt 12 costt 1sint t² c₁ ddt δt c₂ δt 12 sint tcost t² 0 0 t 0 Xt sint tcost 2t² 5 Mostre com o formalismo tensorial que A A A 5 A A A Gradiente A A xi êi Ajk êj êk xi êi Ajk xi êj êk êi Divergente A xi A êi Ajk xi êj êk êi δki Aji xi êj Rotacional A A xi x êi Ajk xi êj êk x êi Ajk xi εkim êj êm Laplaciano divergente do gradiente A A xm Ajk xi êj êk êi êm ² Ajk xi xm êj êk êi êm A ² Ajk xi xm êj êk Rotacional do Rotacional x x A Xₙ AᵢₖXᵢ εₖᵢₘ êⱼ êₘ x εₖₙₘ êₙ êₘ ²AᵢₖXₙ Xᵢ εₖᵢₘ εₖₙₘ êⱼ êₘ x êₙ êₘ ²AᵢₘXₘ Xᵢ εᵢₘₖ êₖ êⱼ εₖₘₙ êⱼ êₖ A gradiente do divergente A Xₘ AᵢⱼXᵢ êⱼ êⱼ êₖ êᵢ ²AᵢⱼXₘ Xᵢ êᵢ êⱼ x x A A A ²AᵢⱼXₘ Xᵢ êᵢ êⱼ ²AⱼₖXᵢ Xₘ êⱼ êₖ ²AᵢₘXₘ Xᵢ εᵢₘₖ êₖ êⱼ εₖₘ ₙ êⱼ êₖ ²AᵢⱼXₘ Xᵢ êᵢ êⱼ ²AⱼₖXᵢ Xₘ êⱼ êₖ 1 Mostre que ezCosθ ₘ Iₘz eimθ Iₘ é a função de Bessel modificada ① Função gamma Γpn1 pn ezt2 ₘ₀ zm tn 2m m 1 ez2t ₙ₀ zn 2n tn n 2 Multiplicando 1 por 2 ez2 t 1t ₘ₀ zm tm 2m m ₙ₀ zn 2n tn n m pn tp ₙ₀ z2p2n 1pn n Iₚz n pm tp ₘ₀ z2p2m 1pm m Iₚz ez2 t 1t ₚ tp Jₚz Fazendo t eiθ t1 1t eiθ z12 t 1t z cosθ Assim ez2 t 1t ezcosθ ₘ Iₘz eimθ 2 No cálculo do deslocamento quadrático médio do oscilador harmônico temos a integral x2 ex2 Hn2x dx Obtenha o valor da integral em termos de n 2 Vamos usar a relação de recorrência Hn1x 2x Hnx 2n Hn1x 2x Hnx Hn1x 2n Hn1x x Hnx 12 Hn1x 2n Hn1x x Hnx2 12 Hn1x 2n Hn1x2 x2 Hnx Hnx 44 Hn1x Hn1x 4n2 Hn1x Hn1x 4n Hn1x Hn1x x2 Hnx Hnx 14 Hn1x Hn1x n2 Hn1x Hn1x n Hn1x Hn1x x2 ex2 Hn2x dx ex2 x2 Hnx Hnx dx ex2 14 Hn1x Hn1x n2 Hn1x Hn1x n Hn1x Hn1x dx ex2 x2 Hn2x dx 14 ex2 Hn1x Hn1x dx n2 ex2 Hn1x Hn1x dx n ex2 Hn1x Hn1x dx Vamos usar a relação ex2 Hax Hbx dx π 2b b δab Delta de Kronecker a b n1 14 ex2 Hn1x Hn1x dx 14 π 2n1 n1 δn1n1 1 a b n1 n2 ex2 Hn1x Hn1x dx n2 π 2n1 n1 δn1n1 1 a b δab 0 n ex2 Hn1x Hn1x dx 0 x2 ex2 Hn2x dx π4 2n1 n1 n2 π 2n1 n1 π 2n12 n1 m n1 n2 π 2n1 n1 x2 ex2 Hn2x dx π 2n1 n2 n n1n π 2n1 n n n1n x2 ex2 Hn2x dx 2n π n n12 n 12 x2 ex2 Hn2x dx 2n π n n 12 3 Mostre que para o átomo de H o valor médio da variável r1 a0n2 sendo a0 o raio de Bohr r1 1r ψ 1r ψ dv ψ enra0 sqrtπ a03 ψ dv 4π r2 dr r1 4π 1 sqrtπ a032 0 e2nra0 r2 r dr r1 4 a03 0 r e2nra0 dr 4 a03 0 r e2nr a0 dr Integração por partes u r du dr dv e2nra0 v du a0 e2nr a0 2n u dv uv v du 4 a03 a0 r e2nr a0 2n 0 a0 2n 0 e2nr a0 dr 4 a03 0 0 a0 2n a0 2n e2nr a0 0 4 a03 a02 4 n2 01 1 n2 a0 r1 1 n2 a0 a01 n2 Blank page 4 Usando a transformação de Laplace resolva a equação diferencial t2 x 4 t x t2 2 x Sent Lt2 x 4 t x t2 2 x LSent Ltk ft 1k dk dsk Lft Para Lt2 x 12 d2 ds2 Lx d2 ds2 Lx Lt2 x d2 ds2 s2 Xs s x0 x0 dds dds s2 Xs s x0 x0 dds 2 s Xs s2 dds Xs x0 2 Xs 2 s dds Xs 2 s dds Xs s2 d2 ds2 Xs s2 d2 ds2 Xs 4 s dds Xs 2 Xs Lt x Xs s dds Xs Lt2 x 12 d2 ds2 Lx d2 ds2 Xs LSent 1 s2 1 Substituindo na equação L2t2x 4L1t x L0t2 x 2L0x L0sent s2 d2ds2 xs 4s dds xs 2 xs 4 xs s dds xs d2ds2 xs 2 xs 1s21 Para facilitar a notação vamos usar d2ds2 xs X dds xs X xs X s2 X 4s X 2 X 4 X 4 s X X 2 X 1s21 X s21 X 4s4s X 2 4 2 1s21 0 0 s21 X 1s21 X 1s212 X dds X 1s212 dds X ds dss212 dX X dss212 sec2u dusec2u2 Integração por partes u arctgs s tgu ds sec2u du s2 1 sec2u X 1sec2u du cos2u du 12 du cos2u du X 12 u sen2u2 u2 sen2u4 X arctgs2 s2s21 C1 dXds X 12 arctgs ds 12 ss2 1 ds C1 ds Xs 12 dds s arctg s ds C1 s C2 s arctgs2 C1 s C2 xt 12 L1 s arctgs C1 L1 s3 C2 L11 12 ddt L1 arctgs C1 ddt L11 C2 L11 12 ddt sintt C1 ddt δt C2 δt 12 cost t 1sintt2 C1 ddt δt C2 δt 12 sint t costt2 0 0 t 0 Xt sint t cost2 t2 5 Mostre com o formalismo tensorial que    Gradiente  AXi êi Ajk êj êkXi êi δAjk êj êk êi Divergente  Xi A êi AjkXi êj êk êi δAjiXi êj Rotacional  AXi x êk AjkXi êj êk x êiεkim êm AjkXi εkim êj êm Laplaciano divergente do gradiente   Xm AjkXi êj êk êi êm ² AjkXi Xm êj êk êi êmδim A ² AjkXi Xm êj êk Rotacional do Rotacional  Xn AjkXi εkim êj êm x εknm ên êm ² AjkXn Xi εkim εknm êj êm x ên êm ² AjmXm Xi εmk êi êj εkmn êj êk  gradiente do divergente  Xm AijXi êj êj êk êi ² AijXm Xi êi êj    ² AijXm Xi êi êj ² AjkXi Xm êj êk ² AimXm Xi εimu êi êj εkm êj êk ² AijXm Xi êi êj ² AjkXi Xm êj êk