Derivation of the Laplace-Operator: Derivation of Coordinates by Partial Derivative [email protected] Ahmed.S.Arife Tel: +20102958263 Abstract : The Laplace operator is a second order differential operator often used in theoretical physics applications. We wish to find a method to derive coordinates by partial derivative using the Laplace operator. I will discuss curvelinear coordination in the following chapters : 1- Cartesian Coordinates ( x , y , z) 2- Polarity Coordinates ( r, θ ) 3- Cylindrical Coordinates ( ρ ,φ , z) 4- Spherical Coordinates ( r , θ , φ ) 5- Parabolic Coordinates ( u, v , θ ) 6- Parabolic Cylindrical Coordinates (u , v , z) 7- Curvilinear Coordinates, this general coordination And we can use this coordination to derive more Laplace operators in any coordinates. The question is, why is the Laplace operator used in more application in physics electricity, and in wave functions 1 , for example, A wave function can be written, ψ= ψ(x ++ ct )( ψ x − ct ) Now we can derive the wave function, ∂ψ ∂ ψ =ψψ(x ++ ct )( x −→= ct ) c (( ψψ x +− ct )( x − ct )) ∂x ∂ t ∂2ψ ∂ 2 ψ =++−→ψψ(x ct )( x ct ) = c2 (( ψψ x ++− ct )( x ct )) ∂x2 ∂ t 2 The last equation can give, ∂2ψ1 ∂ 2 ψ = ∂x2 c 2 ∂ t 2 In x,y,z coordinates, ∂∂∂222ψψψ1 ∂ 2 ψ ++− = 0 ∂∂∂x2 y 2 yct 222 ∂ where the Laplace operator gives in 3D x,y,z : ∂2 ∂ 2 ∂ 2 ∇=2 + + ∂y2 ∂ y 2 ∂ y 2 In the quantum mechanical and Schrödinger equation, ∂ψ h2 ih =−[ ∇+2 V ( x )] ψ ∂t2 m And we can write d'Alembertian in spacetime : 1 ∂ 2 ≡ ∇2 − c2∂ t 2 some propertities of crul: 1−∇⊗ ()(kA = k ∇⊗ A ) 2−∇⊗ ()(fAf = ∇⊗ A ) +∇ f ⊗ A Af(∇⊗ A ) −∇ fA ⊗ 3−∇⊗ ( ) = f f 2 And the properties of the divergence: 1−∇ ()kA = k ( ∇ A ) 2−∇ ()fA = f ( ∇ A ) +∇ fA A f(∇ A ) − ∇ fA 3− ∇ ( ) = f f 2 4(−∇ABB ⊗ )( = ∇⊗ AA )( − ∇⊗ B ) What the Laplace operator in Cartestian coordination The ∇ is given by ∂ ∂ ∂ ∇=i + j + k ∂x ∂ y ∂ z And the ∇2ψ =∇ ∇ ψ ∂∂∂∂∂ψ ψ ∂ ψ ∂∂∂2 ψψψ 2 2 ∇∇=ψ (i ++ jk )( i + j + k ) =++ ∂∂∂xyz ∂∂ xy ∂ z ∂∂∂ xyz2 2 2 Laplace operator derivative by partial derivation: The Laplace operator is given by ∇2V where V is the function in x,y I will assume the function V in x,y, derivative of Laplace in polar coordinates ( r, θ ) VVxy=(,) →= xr cosθ →= yr sin θ By partial derivation, ∂V ∂∂ Vx ∂∂ Vy = + ∂r ∂∂ xr ∂∂ yr ∂V ∂ V ∂ V =cosθ + sin θ ∂r ∂ x ∂ y And the same in θ , ∂V ∂∂ Vx ∂∂ Vy = + ∂θ ∂∂x θ ∂∂ y θ ∂V ∂ V ∂∂ Vy = −rsinθ + r cos θ ∂θ ∂x ∂∂ y r We can solve the 2 equations by matrix operator or any methods given that, ∂V ∂ Vsinθ ∂∂ V ∂ sin θ ∂ =cosθ − ⇒ = cos θ − ∂x ∂ rr ∂∂θ x ∂ rr ∂ θ and in they dimension ∂V ∂ Vcosθ ∂∂ V ∂ cos θ ∂ =sinθ + ⇒ = sin θ + ∂y ∂ rr ∂∂θ x ∂ rr ∂ θ ∂2V ∂∂∂sinθ V sin θ ∂ V =(cosθ − ) (cos θ − ) ∂x2 ∂∂ rrθ ∂ rr ∂ θ ∂2V ∂∂ VVsinθ ∂ sin θ ∂∂ VV sin θ ∂ =(cosθ (cos θ −− ) (cos θ − )) ∂x2 ∂∂ rrr ∂θ r ∂∂ θ rr ∂ θ ∂2V ∂∂ V ∂∂sinθθ V sin ∂∂ V sin θθ ∂∂ sin V =(cosθθθ cos − cos − cos θ + ∂x2 ∂∂ rr ∂∂ rrθθ r ∂∂ rrr ∂∂ θθ 2 ∂∂2VV 2 cosθθ sin ∂ V sin θθ cos ∂∂∂ VVV sin2 θ sin 2 θ =+cos2 θ 2 − 2 ++ ∂x2 ∂ rr 2 2 ∂θ rrr ∂∂∂ θ2 θ 2 rr ∂ And in the y dimension ∂2V ∂cosθ ∂∂ V cos θ ∂ V =(sinθ + )(sin θ − ) ∂y2 ∂∂ rrθ ∂ rr ∂ θ The same algbera derivative gives, 2 ∂∂2VV 2 cosθθ sin ∂ V sin θθ cos ∂∂∂ VVV cos2 θ cos 2 θ =−sin2 θ 2 + 2 ++ ∂y2 ∂ rr 2 2 ∂θ rrr ∂∂∂ θ2 θ 2 rr ∂ By addition, gives Laplace in x,y ∂∂∂22VVV 21 ∂ 2 V 1 ∂ V + =+ + ∂∂∂x2 y 2 rr 222 ∂θ rr ∂ Laplace in r, θ 1∂ ∂V 1 ∂ 2 V ∇2V =( r ) + rr∂ ∂ r r ∂ θ 2 Cylindrical Coordinates ( ρ ,φ , z) The same method can we derive the Laplace operator in cylindrical coordinates VVxyzx=(,,) →=ρφ cos →= y ρφ sin →= zz By partial derivative ∂V ∂∂ Vx ∂∂ Vy ∂∂ Vz = + + ∂ρ ∂∂x ρ ∂∂ y ρ ∂∂ z ρ ∂V ∂ V ∂∂ Vy =cosφ + sin θ ∂r ∂ρ ∂∂ y ρ And the same in φ : ∂V ∂∂ Vx ∂∂ Vy ∂∂ Vz = + + ∂φ ∂∂x φ ∂∂ y φ ∂∂ z φ ∂V ∂ V ∂ V = −ρφsin + ρφ cos ∂φ ∂x ∂ y And the same in z ∂V ∂∂ Vx ∂∂ Vy ∂∂ Vz = + + ∂z ∂∂ xz ∂∂ yz ∂∂ zz ∂V ∂ V = ∂z ∂ z By the same method in r, θ can derivative in ρ, φ , z in the last give : 1∂∂V 1 ∂2 V ∂ 2 V ∇=2V (ρ ) + + ρρ∂ ∂ ρ ρφ ∂2 ∂ z 2 Spherical Coordinates Derivative by Walton 1967, Arfken 1985, it is called the Spherical Coordinates ψψ=(xyzxr , , ) →= cos θφ sin →= yr sin θφ sin →= zr cos φ By partial derivative ∂ψψ ∂∂x ∂∂ ψ y ∂∂ ψ y = + + ∂r ∂∂ xr ∂∂ yr ∂∂ yr ∂ψ ∂ ψ ∂∂ ψψ =cosθφ sin + sin θφ sin + cos φ ∂r ∂ x ∂∂ yy ∂ψψ ∂∂x ∂∂ ψ y ∂∂ ψ y = + + ∂θ ∂∂x θ ∂∂ y θ ∂∂ y θ ∂ψ ∂ ψ ∂ ψ = −rsinθφ sin + r sin θφ sin ∂θ ∂x ∂ y ∂ψψ ∂∂x ∂∂ ψ y ∂∂ ψ y = + + ∂φ ∂∂x φ ∂∂ y φ ∂∂ y φ ∂ψ ∂ ψ ∂∂ ψψ =rcosθφ cos + r sin θφ cos − r sin φ ∂r ∂ x ∂∂ yy We can solve the system liner equation 1,2,3 by matrix inverse methods which give ∂ψ ∂ ψ ∂r cosθφ sin sin θφ sin cos φ ∂ x 1 ∂ψ ∂ ψ = − sinθφ sin cos θφ sin 0 r ∂θ ∂ y cosθφ cos sin θφ sin− sin φ 1 ∂ψ ∂ψ r ∂φ ∂z By matrix properties, gives the inverse, 1 A−1 = A % t DetA cosθφ sin sin θφ sin cos φ A = − sinθφ sin cos θφ sin 0 cosθφ cos sin θφ sin− sin φ By algebra solving the detriment ADetA = − sin φ give −cosθφ sin2 sin θ − cos θφφ sin cos % t 2 A =− sinθφ sin − cos θ − sin θφφ sin cos 2 −sincosφ φ 0 sin φ The inverse matrix gives, sin θ cosθφ sin− cos θφ cos sin φ −1 cos θ A = sinθφ sin sin θφ cos sin φ cosφ 0 sin φ Now we can find the equation solution by matrix ∂ψ sin θ ∂ ψ cosθφ sin− cos θφ cos sin ∂x φ ∂ r ∂ψ cos θ 1 ∂ ψ = sinθφ sin sin θφ cos y r ∂ sin φ ∂ θ ∂ψ cosφ 0 sin φ 1 ∂ψ ∂z r ∂φ ∂ψ ∂∂ ψsin θψ cos θφψ cos ∂ =cosθ sin φ − + ∂x ∂∂ rrsin φ θ r ∂ φ ∂ψ ∂∂ ψcos θψ sin θφψ cos ∂ =sinθ sin φ + + ∂y ∂∂ rrsin φ θ r ∂ φ ∂ψ ∂ ψsin φψ ∂ =cos φ − ∂x ∂ r r ∂ φ Gasiorowicz 1974, pp. 167-168; Arfken 1985, p. 108.