PBC lecture Notes Series: Transformation of Coordinates by Dr. A. Kar Gupta

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Transformation of Coordinates (Combined Lectures, 1st Ed.)

Lecture Notes prepared by-

Dr. Abhijit Kar Gupta Physics Department, Panskura Banamali College

Panskura R.S., East Midnapore, WB, India, Pin-code: 721152 e-mail: kg.abhi@gmail.com, abhijit_kargupta@rediffmail.com

Lecture-1

Let us start from a Rectangular Cartesian coordinate system.

),,( zyxP is a point in this system where we have the position vector of this to be

zkyjxir ˆˆˆ ++= . We can then write the position vector r to be a function of the coordinates ( ),, zyx :

),,( zyxrr = Now suppose we go over to a new coordinate system ( 321 ,, uuu ) where we write

),,( 321 uuurr = Therefore, it can be written as:

),,( 321 uuuxx = ),,( 321 uuuyy = ),,( 321 uuuzz =

The above are the relations between coordinates in two systems. Example: From Cartesian to spherical Polar system

),,( zyx → ),,( φθr ),,( φθrxx = = φθ cossinr ),,( φθryy = = φθ sinsinr ),,( φθrzz = = θcosr

To express the position vector r in the new coordinate system ),,( 321 uuu we have to

know the unit vectors in that system (like kji ˆ,ˆ,ˆ in Cartesian system) along three new axes. We construct the tangent vectors along the positive directions of 1u , 2u and 3u as:

PBC lecture Notes Series: Transformation of Coordinates by Dr. A. Kar Gupta

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2

1ur

∂∂ ,

2ur

∂∂ ,

3ur

∂∂ respectively.

To understand the above physically, let us imagine the tip of the vector r on a surface curved (or a plane surface) in general. So we can draw tangent on the surface at a particular point in the positive directions of the coordinate axes. We have now three unit vectors:

1e = 11 ur

ur

∂∂

∂∂ =

11

1ur

h ∂∂

2e = 22 u

rur

∂∂

∂∂ =

22

1ur

h ∂∂

3e = 33 u

rur

∂∂

∂∂ =

33

1ur

h ∂∂

Where we introduce the symbols 1h , 2h and 3h in place of 1ur

∂∂ etc.

Now we calculate some important vectors and relevant physical quantities in the new coordinate system. The Displacement Vector: Since, ),,( 321 uuurr = ,

rd = 33

22

11

duurdu

urdu

ur

∂∂

+∂∂

+∂∂

= 333222111 ˆˆˆ eduheduheduh ++ . If 1e , 2e and 3e are the basis vectors of a orthogonal coordinate system then

0ˆˆˆˆˆˆ 133221 =⋅=⋅=⋅ eeeeee . Differential arc length:

23

23

22

22

21

21

2 duhduhduhrdrdds ++=⋅=

PBC lecture Notes Series: Transformation of Coordinates by Dr. A. Kar Gupta

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Infinitesimal Volume element:

( ) ( ) ( )333222111 ˆˆˆ duheduheduhedV ×⋅= = ( ) 321321321 ˆˆˆ dududuhhheee ×⋅ = 321321 dududuhhh . Here we have ( ) 1ˆˆˆ 321 =×⋅ eee when the unit vectors form orthogonal set. Note: The length of a vector, or he arc or the volume etc. all physical quantities remain invariant under any kind of coordinate transformation.

From Cartesian to Spherical Polar Coordinate System

),,(),,( φθrzyx →

φθ cossinrx = φθ sinsinry =

θcosrz = Here, ru =1 , θ=2u and φ=3u . Calculations of Unit Vectors:

=re rr

rr

∂∂

∂∂ , θe =

θθ ∂∂

∂∂ rr and φe =

φφ ∂∂

∂∂ rr

Start from the position vector:

zkyjxir ˆˆˆ ++= = θφθφθ cosˆsinsinˆcossinˆ rkrjri ++

θφθφθ cosˆsinsinˆcossinˆ kjirr

++=∂∂ …………………………….(1)

rhrr

≡=∂∂ 1

θφθφθθ

sinˆsincosˆcoscosˆ rkrjrir−+=

∂∂ …………………………(2)

θθhrr

≡=∂∂

PBC lecture Notes Series: Transformation of Coordinates by Dr. A. Kar Gupta

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φθφθφ

cossinˆsinsinˆ rjrir+−=

∂∂ ………………………………..(3)

φθφ

hrr≡=

∂∂ sin

Therefore, Arc length: 2222222 φθ φθ dhdhdrhds r ++= = 222222 sin φθθ drdrdr ++ Volume element: φθθφθφθ ddrdrddrdhhhdV r sin2== . Now we calculate the unit vectors using the relations (1), (2) and (3):

φφ

θφθφθ

θφθφθ

φ

θ

cosˆsinˆˆsinˆsincosˆcoscosˆˆ

cosˆsinsinˆcossinˆˆ

jie

kjie

kjier

+−=

−+=

++=

In the Matrix form:

φ

θ

eeer

ˆˆˆ

=

−−

kji

ˆˆˆ

0cossinsinsincoscoscos

cossinsincossin

φφθφθφθθφθφθ

φ

θ

eeer

ˆˆˆ

=

kji

Mˆˆˆ

.

Here M is called the transformation matrix. Check that the determinant of M is unity: .1=M

Note: Property of Transformation Matrix. We can calculate inverse of the matrix 1−M and thus can determine i , j and k in terms of re , θe and φe :

kji

ˆˆˆ

= 1−M

φ

θ

eeer

ˆˆˆ

.

Thus we can determine the position vector completely in terms of re , θe , φe and r , θ , φ that is the unit vectors and coordinates of the new system.

PBC lecture Notes Series: Transformation of Coordinates by Dr. A. Kar Gupta

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Exercise: #1. Find (....)ˆ(.....)ˆ(....)ˆ φθ eeer r ++= #2. Verify that 0ˆˆˆˆˆˆ =⋅=⋅=⋅ rr eeeeee φφθθ (Orthogonality). #3. Given any vector zyx AkAjAiA ˆˆˆ ++= , determine φφθθ AeAeAeA rr ˆˆˆ ++= .

#4. If we know the position vector r as in #1, we can determine velocity dt

rdv = and

acceleration 2

2

dtrda = . Thus find out:

φφθθ vevevev rr ˆˆˆ ++=

φφθθ aeaeaea rr ˆˆˆ ++= . #5. In the transformation ),,(),,( 321 uuuzyx → the Jacobian of transformation is

333

222

111

321 ,,,,

uz

uy

ux

uz

uy

ux

uz

uy

ux

uuuzyxJ

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=

.

Show that for the transformation to Spherical coordinate system ),,(),,( φθrzyx → the Jacobian is θsin2rJ = . [Note: φθdJdrddxdydzdV == ] #6. Do all the exercise above for Cylindrical coordinate system ),,( zφρ . Lecture-2 Transformation of GRADIENT: Let us have 332211 ˆˆˆ efefef ++=∇φ ……………………………….(1) Here 1e , 2e , 3e are the unit vectors we already know. Now we have to determine the coefficients (or functions) 1f , 2f and 3f .

Since ),,( 321 uuurr = , we can write

PBC lecture Notes Series: Transformation of Coordinates by Dr. A. Kar Gupta

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33

22

11

duurdu

urdu

urrd

∂∂

+∂∂

+∂∂

= = 333222111 ˆˆˆ eduheduheduh ++ …………….(2)

Also ),,( 321 uuuφφ = .

∴ φd 33

22

11

duu

duu

duu ∂

∂+

∂∂

+∂∂

=φφφ ……………………………………(3)

We can also write: rdd ⋅∇= φφ = )ˆˆˆ()ˆˆˆ( 333222111332211 eduheduheduhefefef ++⋅++ = 333222111 dufhdufhdufh ++ …………………….(4) Comparing (3) and (4) we have

111 u

fh∂∂

=φ ⇒

111

1uh

f∂∂

=φ

222 u

fh∂∂

=φ ⇒

222

1uh

f∂∂

=φ

3

33 ufh

∂∂

=φ ⇒

333

1uh

f∂∂

=φ

Hence, 33

3

22

2

11

1 ˆˆˆuh

euh

euh

e∂∂

+∂∂

+∂∂

=∇φφφφ .

Therefore, the Gradient Operator is ∇ ≡33

3

22

2

11

1 ˆˆˆuh

euh

euh

e∂∂

+∂∂

+∂∂ …………(5)

Transformation of DIVERGENCE: Suppose we have a vector, 332211 ˆˆˆ eAeAeAA ++=

Then )ˆˆˆ( 332211 eAeAeAA ++⋅∇=⋅∇

= )ˆ()ˆ()ˆ( 332211 eAeAeA ⋅∇+⋅∇+⋅∇ Let us evaluate )ˆ( 11eA⋅∇ separately. To do that we express the unit vector 1e in terms of the variable 1u and so on.

We have ∇ ≡33

3

22

2

11

1 ˆˆˆuh

euh

euh

e∂∂

+∂∂

+∂∂ as derived above.

PBC lecture Notes Series: Transformation of Coordinates by Dr. A. Kar Gupta

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∴ 1

11

ˆhe

u =∇ ⇒ 111 uhe ∇=

Similarly, 222ˆ uhe ∇= and 333ˆ uhe ∇= .

Now we can write 321 ˆˆˆ eee ×= = 3232 uuhh ∇×∇ ∴ )()ˆ( 3232111 uuhhAeA ∇×∇⋅∇=⋅∇

= )()()( 3232132321 uuhhAuuhhA ∇×∇⋅∇+∇×∇⋅∇

= 0ˆˆ

)(3

3

2

2321 +×⋅∇

he

he

hhA

= 32

1321

ˆ)(

hhe

hhA ⋅∇

= ( )3211321

1 hhAuhhh ∂∂

Thus =⋅∇ A )ˆ()ˆ()ˆ( 332211 eAeAeA ⋅∇+⋅∇+⋅∇

( ) ( ) ( )

∂∂

+∂∂

+∂∂

=⋅∇ 2133

1322

3211321

1 hhAu

hhAu

hhAuhhh

A ………………(6)

If Φ∇=A

Φ∇⋅∇=⋅∇ A = Φ∇2

We put 33

3

22

2

11

1 ˆˆˆuh

euh

euh

eA

∂Φ∂

+∂Φ∂

+∂Φ∂

= in (6) and get

Φ∇2

=

∂Φ∂

∂∂

+

∂Φ∂

∂∂

+

∂Φ∂

∂∂

33

21

322

13

211

32

1321

1uh

hhuuh

hhuuh

hhuhhh

…………….(7)

In Spherical Polar ( φθ ,,r ) System:

ru =1 , θ=2u , φ=3u

ree ˆˆ1 = , θee ˆˆ2 = , φee ˆˆ3 = 11 == rhh rhh == θ2

θφ sin3 rhh == Putting all the above in (7) we get

PBC lecture Notes Series: Transformation of Coordinates by Dr. A. Kar Gupta

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Φ∇2

=

∂Φ∂

⋅∂∂

+

∂Φ∂

⋅∂∂

+

∂Φ∂

∂∂

φθφθθ

θθ

θ sinsinsin

sin1 2

2 rr

rr

rr

rr

=

∂Φ∂

⋅+

∂Φ∂

∂∂

+

∂Φ∂

∂∂

2

22

2 sin1sinsin

sin1

φθθθ

θθ

θ rr

rr.

Transformation of CURL:

( )332211 ˆˆˆ eAeAeAA ++×∇=×∇

= ( )11eA×∇ + ( )22eA×∇ + ( )33eA×∇ ……………………………..(8) To evaluate the first term: We use 111 uhe ∇= as before.

( ) ( )11111 ˆ uhAeA ∇×∇=×∇

= 111111 )( uhAuhA ∇×∇+∇×∇ [we used: AAA ×∇+×∇=×∇ φφφ )( ]

= 0ˆ

)(1

111 +×∇

he

hA

∴ ( ) =×∇ 11eA 1

111

33

311

22

211

11

1 ˆ)(

ˆ)(

ˆ)(

ˆhe

hAuh

ehA

uhe

hAuh

e×

∂∂

+∂∂

+∂∂

= 1

111

11

1 ˆ)(

ˆhe

hAuh

e×

∂∂ +

1

111

22

2 ˆ)(

ˆhe

hAuh

e×

∂∂ +

1

111

33

3 ˆ)(

ˆhe

hAuh

e×

∂∂

= 0 +

∂∂

− )(ˆ

11221

3 hAuhh

e +

∂∂ )(

ˆ11

331

2 hAuhh

e

= )(ˆ

)(ˆ

11221

311

313

2 hAuhh

ehA

uhhe

∂∂

−∂∂

We can evaluate the other terms of equation (8) in a similar fashion. Thus we can write

=×∇ A )(ˆ

)(ˆ

11221

311

313

2 hAuhh

ehA

uhhe

∂∂

−∂∂ + )(

ˆ)(

ˆ22

332

122

121

3 hAuhh

ehA

uhhe

∂∂

−∂∂ +

)(ˆ

)(ˆ

33113

233

232

1 hAuhh

ehA

uhhe

∂∂

−∂∂

=

∂∂

−∂∂ )()(

ˆ22

333

232

1 hAu

hAuhh

e +

∂∂

−∂∂ )()(

ˆ33

111

313

2 hAu

hAuhh

e +

∂∂

−∂∂ )()(

ˆ11

222

121

3 hAu

hAuhh

e ………(9)

PBC lecture Notes Series: Transformation of Coordinates by Dr. A. Kar Gupta

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Expression (9) can be written as

=×∇ A

∂∂

−∂∂

+

∂∂

−∂∂

+

∂∂

−∂∂ )()(ˆ)()(ˆ)()(ˆ1

112

221

33331

113

22223

332

11321

hAu

hAu

ehhAu

hAu

ehhAu

hAu

ehhhh

=

332211

321

332211

321

ˆˆˆ1

hAhAhAuuu

eheheh

hhh ∂∂

∂∂

∂∂ ………………………………………….(10)

In Spherical Polar Coordinate system:

=×∇ A

φθ

φθ

θφθ

θ

θArrAA

r

erere

rr

r

.sin

ˆ.sinˆˆ

sin1

2 ∂∂

∂∂

∂∂ .