coordenadas esferoidales

8/18/2019 coordenadas esferoidales

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http://www.math.odu.edu/~jhh/part4.PDF

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Examples:

Polar coordinates are an example of a 2-D system of coordinates that are not rectangular:

r = r(x,y) = √(x²+y²) → ! = !(x,y) and x = r cos(") → x = x(!,2)

" = "(x,y) = arctan(y#x) → 2 = 2(x,y) and y = r sin(") → y = y(!,2)$

%nother example in&ol&ing time: rotating polar coordinates

r = r(x,y,t) = √(x²+y²) → ! = !(x,y,t) and x = x(r,",t) = r cos("+ωt)

" = "(x,y,t) = arctan(y#x) - ωt → 2 = 2(x,y,t) and y = y(r,",t) = r sin("+ωt)

'note: sign difference due to fact that if e let = "+ωt, then " = -ωt*

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$ e also need generalized velocities for energy considerations (for use in conser&ation of energy) andfor momentum considerations (for use in etons .econd /a: 0F = dp#dt)$

1or one particle in rectangular form: 3 = 4m&² = 4m(x5²+y5²+652) here x5 = dx#dt$

1or more than one particle in rectangular form: 3 = 7 = 0i84mi'xi5²+yi5²+6i5²*9$

.ince x! = x!(!, 2, $$$ , n, t) e ha&e

dx! = (∂x!#∂!)d! + (∂x!#∂2)d2 + $$$ (∂x!#∂n)dn + (∂x!#∂t)dt

so x!5 = dx!#dt = 8!#dt98(∂x!#∂

!)d! + (∂x!#∂

2)d2 + $$$ (∂x!#∂

n)dn + (∂x!#∂t)dt)

= 0i(∂x!#di)i5 + ∂x!#∂t

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Example: ;otating Polar


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x = r cos(ωt+") → x5 = (∂x#∂r)r5 + (∂x#∂")"5 + (∂x#∂t)

= r5cos(ωt+") + "5(-r sin(ωt+")) + ω(-r sin(ωt+"))

x component of r5 and of r"5 and of r ω $

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$ 7he 3 in&ol&es not >ust x5, ?ut x5²$ hat does this loo@ li@eA

x5² = 80@ '(∂x#∂@ )@ 5 + ∂x#∂t*9 80'(∂x#∂)5 + ∂x#∂t*9

= 80@ 0'(∂x#∂@ )(∂x#∂)@ 55*9 + 80@ '(∂x#∂@ )(∂x#∂t)@ 5*9

+ 80'(∂x#∂)(∂x#∂t)

5*9 + (∂x#∂t)² $

.ince the @ and the are hat are called dummy indices (?ecause they are summed o&er completely) anddo not matter hether they are called @ or , the second and third terms are exactly the same and can ?ecom?ined$ 7hus the expression ?ecomes:

x5² = 80@ 0'(∂x#∂@ )(∂x#∂)@ 55*9 + 280@ '(∂x#∂@ )(∂x#∂t)@ 5*9 + (∂x#∂t)²

Bf e ha&e more than one particle, all the Cx sym?ols a?o&e are replaced ?y Cx i$

e can no rite the expression for 3 (sometimes referred to as 7):

T = 0i'4mi(xi5² + yi5² + 6i5²)* = Σk Σl{½Ak

qk q

! " Σk #k qk " To

here

%@ = 0i'mi8(∂xi#∂@ )(∂xi#∂) + (∂yi#∂@ )(∂yi#∂) + (∂6i#∂@ )(∂6i#∂)9*

E@ = 0i'mi8(∂xi#∂@ )(∂xi#∂t) + (∂yi#∂@ )(∂yi#∂t) + (∂6i#∂@ )(∂6i#∂t)9*

7o = 0i'4mi8(∂xi#∂t)² + (∂yi#∂t)² + (∂6i#∂t)²9* $


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F73: %@ = %@ $ $% %@ = G@ (here G@ = ! if @=, and G@ = H otheriseI it is called the Dirac delta

function), t&en the coordinates are said to ?e ort&ogonal$

.uppose e ha&e a non-orthogonal system$ v = u5u + 5w here the

angle ?eteen u and w is J hich is F7 KH°$ 7hen:

&² = v L v = (u5u + 5w) L (u5u + 5w) = u5² + 5² + 2u55 cos(J) $

otice that this term has an %@ that is not 6ero if @M N


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EXAMPLE :

1or the case of one particle, use the a?o&e to sho hat the @inetic energy loo@s li@e in the

rotating polar coordinates$ Bn particular,

a) find %!!, %!2, %2! and %22I sho that this system is orthogonal (%!2 = %2! = H)I

?) find E! and E2I

c) find 7oI

d) rite don the final inetic 3nergyI identify each term in the expression$

1or rotating polar coordinates (2-D):

! = r = r(x,y,t) = √(x²+y²) and x = x(r,",t) = r cos("+ωt)

2 = " = "(x,y,t) = arctan(y#x) - ωt and y = y(r,",t) = r sin("+ωt)

inetic 3nergy: 7 = 0i'4mi(x5² + y5² + 65²)* = 0@ 0l'4%@ @ 55* + 0@ E@ @ 5 + 7o

here

%@ = 0i'mi8(∂xi#∂@ )(∂xi#∂) + (∂yi#∂@ )(∂yi#∂) + (∂6i#∂@ )(∂6i#∂)9*

E@ = 0i'mi8(∂xi#∂@ )(∂xi#∂t) + (∂yi#∂@ )(∂yi#∂t) + (∂6i#∂@ )(∂6i#∂t)9*

7o = 0i'4mi8(∂xi#∂t)² + (∂yi#∂t)² + (∂6i#∂t)²9* $

e ha&e only ! particle, so there is no need to ta@e the sum o&er i, and all the mi5s ill >ust ?e m, all of the

xi5s ill ?e x, and all of the yi5s ill ?e y$

e only ha&e 2-D, so there are no 6 terms$

1or rotating polar coordinates,

(∂x#∂!) = (∂x#∂r) = (∂8r cos("+ωt)9#∂r) = cos("+ωt)

(∂x#∂2) = (∂x#∂") = (∂8r cos("+ωt)9#∂") = -r sin("+ωt)

(∂x#∂t) = (∂x#∂t) = (∂8r cos("+ωt)9#∂t) = -rO sin("+ωt)

(∂y#∂!) = (∂y#∂r) = (∂8r sin("+ωt)9#∂r) = sin("+ωt)


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(∂y#∂2) = (∂y#∂") = (∂8r sin("+ωt)9#∂") = r cos("+ωt)

(∂y#∂t) = (∂y#∂t) = (∂8r sin("+ωt)9#∂t) = rO cos("+ωt)

a) 7herefore, e ha&e:

%!! = m8(∂x#∂!)(∂x#∂!) + (∂y#∂!)(∂y#∂!) = m8cos2("+ωt) + sin2("+ωt)9 = m$

%!2 = m8(∂x#∂!)(∂x#∂2) + (∂y#∂!)(∂y#∂2)9 = %2!

= m8'cos("+ωt)*'-r sin("+ωt)* + 'sin("+ωt)*' r cos("+ωt)*9 = H$

%22 = m8(∂x#∂2)(∂x#∂2) + (∂y#∂2)(∂y#∂2) = m8r 2sin2("+ωt) + r 2cos2("+ωt)9 = mr 2$


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?) E! = m8(∂x#∂!)(∂x#∂t) + (∂y#∂!)(∂y#∂t)9

= m8'cos("+ωt)*' -rO sin("+ωt)* + ' sin("+ωt)*' rO cos("+ωt )* 9 = H$

E2 = m8(∂x#∂2)(∂x#∂t) + (∂y#∂2)(∂y#∂t)9

= m8 '-r sin("+ωt)*' -rO sin("+ωt)* + ' r cos("+ωt)*' rO cos("+ωt)*9

= mr 2O $

c) 7o = 0i'4mi8(∂xi#∂t)² + (∂yi#∂t)² + (∂6i#∂t)²9*

= 4m8 '-rO sin("+ωt)*² + ' rO cos("+ωt)*² 9 = 4mr 2O2 $

d) 7 = 0i'4mi(x5² + y5² + 65²)* = 0@ 0l'4%@ @ 55* + 0@ E@ @ 5 + 7o

= 4 8%!!!5!5 + %!2!525 + %2!25!5 + %2225259 + E!!5 + E225 + 7o

= 4 mr52 + H + H + 4 mr 2"52 + H + mr 2O"5 + 4 mr 2O2

= 4 mr52 + 4 mr 2("5+O)2

here r5 is the radial speed, and r("5+O) is the tangential speed$

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WRITTEN HOMEWORK PROBLEMS #5a and #6a:

(ou should ?e a?le to do part a of pro?lem Q and part a of pro?lem R no$)

Problem #5: (Problem 9-1 in !mon"


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polar orm (r%"7 ind the iner0e tran0ormation0: r(uw" and %(uw"7 then indr8(uwu8w8" and %8(uwu8w8"7 then write the 6ineti& ener2! in term0 o uwu8w8.


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Problem #: (Problem 9- in !mon"

1or plane (2-D) para?olic coordinates f and h defined such that: x = f T h and y = 2(fh)!#2

(see the excel spreadsheet CPara?olic

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coordenadas esferoidales