Differentiation of vectors In a Cartesian system, i , j , and k - - PowerPoint PPT Presentation

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Differentiation of vectors In a Cartesian system, i , j , and k - - PowerPoint PPT Presentation

Differentiation of vectors In a Cartesian system, i , j , and k are fixed unit vectors If we have a vector A = A x i + A y j + A z k , where the components are functions of t , then we can take a derivative d i + dA


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SLIDE 1

Differentiation of vectors

  • In a Cartesian system, ˆ

i, ˆ j, and ˆ k are fixed unit vectors

  • If we have a vector

A = Axˆ i + Ayˆ j + Azˆ k, where the components are functions of t, then we can take a derivative d A dt = dAx dt ˆ i + dAy dt ˆ j + dAz dt ˆ k

  • For example,

A could be the position of a particle, and then the time derivative is the velocity. If A is the velocity, then the time derivative is the acceleration.

  • What do we do if the vector is described in another coordinate

systems that does not have fixed unit vectors?

Patrick K. Schelling Introduction to Theoretical Methods

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SLIDE 2

Differentiation of vectors in a polar system

  • We can express

A in the xy-plane using unit vectors in a polar system

  • A = Axˆ

i + Ayˆ j = Arˆ er + Aθˆ eθ

  • The ˆ

i and ˆ j unit vectors have a fixed direction, but ˆ er and ˆ eθ do not

Patrick K. Schelling Introduction to Theoretical Methods

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SLIDE 3

Unit vectors in polar coordinate system expressed in terms

  • f Cartesian system
  • We can easily see that,

ˆ er = cos θˆ i + sin θˆ j ˆ eθ = − sin θˆ i + cos θˆ j

  • Then computing the time derivatives is straightforward and easy,

dˆ er dt = − sin θdθ dt ˆ i + cos θdθ dt ˆ j = dθ dt ˆ eθ dˆ eθ dt = − cos θdθ dt ˆ i − sin θdθ dt ˆ j = −dθ dt ˆ er

Patrick K. Schelling Introduction to Theoretical Methods

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SLIDE 4

Differentiation of vectors in polar coordinates continued

  • Now that we have the differentiation of the unit vectors, we can

differentiate A = Arˆ er + Aθˆ eθ d A dt = dAr dt ˆ er + dAθ dt ˆ eθ − Aθ dθ dt ˆ er + Ar dθ dt ˆ eθ

Patrick K. Schelling Introduction to Theoretical Methods

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SLIDE 5

Example: Section 4, problem 8

  • Example: In a polar coordinate system, the position vector of a

particle is r = rˆ

  • er. Find expressions for the velocity and

acceleration.

  • For the velocity, we see that

v = d

r dt , and then Ar = r and

Aθ = 0, hence

  • v = d

r dt = dr dt ˆ er + r dθ dt ˆ eθ

  • For the acceleration, we see

a = d2

r dt2 , and now Ar = dr dt and

Aθ = r dθ

dt , hence

  • a = d2

r dt2 =

  • d2r

dt2 − r dθ dt 2 ˆ er +

  • 2

dr dt dθ dt

  • + r d2θ

dt2

  • ˆ

Patrick K. Schelling Introduction to Theoretical Methods

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SLIDE 6

Fields

  • Any physical quantity that varies in space can be taken as a field
  • Example, the electric field

E(x, y, z) is defined in terms of the electrostatic force on a charge q, E =

  • F

q

  • In the case of electric field or magnetic field, there is a direction

and so we call these vector fields

  • In the case of electrostatic potential, or gravitational potential,
  • r even temperature or density, we have a scalar field since there is

no direction just a magnitude

  • To describe how these fields behave and the observed

phenomena, we need different kinds of differentials of them, which can be called operators

  • Examples of operators include gradient, divergence, curl, and

Laplacian, but there are others

Patrick K. Schelling Introduction to Theoretical Methods

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SLIDE 7

Directional derivative and the gradient

  • Begin by considering a scalar field, for example temperature

T(x, y, z), or electrostatic potential φ(x, y, z) at each point in space

  • We can pick a direction (not necessarily along the Cartesian

coordinate axes!) and determine how φ(x, y, z) changes with a small displacement along that direction

  • Consider a vector

r = xˆ i + yˆ j + zˆ k

  • Get the direction from a unit vector

u = uxˆ i + uyˆ j + uzˆ

  • k. (Since

it is a unit vector. u2

x + u2 y + u2 z = 1)

Patrick K. Schelling Introduction to Theoretical Methods

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SLIDE 8

Total differential of a scalar field

  • Make a displacement of length ds along the

u direction, then

  • ds = ds

u = uxdsˆ i + uydsˆ j + uzdsˆ k = dxˆ i + dyˆ j + dzˆ k

  • We then see the components of the displacement are dx = uxds,

dy = uyds, and dz = uzds

  • Then we can find dφ

dφ = ∂φ ∂x dx + ∂φ ∂y dy + ∂φ ∂z dz = ∂φ ∂x uxds + ∂φ ∂y uyds + ∂φ ∂z uzds

  • The partial derivatives are evaluated at the components of the

vector r = xˆ i + yˆ j + zˆ k

Patrick K. Schelling Introduction to Theoretical Methods

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SLIDE 9

Gradient operator

  • In our calculation of dφ along the vector

ds, we see that it can be described as the scalar product dφ = ∂φ ∂x ˆ i + ∂φ ∂y ˆ j + ∂φ ∂z ˆ k

  • ·
  • uxdsˆ

i + uydsˆ j + uzdsˆ k

  • We take dφ = ∇φ ·

ds = ds∇φ · u and hence define the gradient

  • perator (in a Cartesian system)
  • ∇φ = gradφ = ∂φ

∂x ˆ i + ∂φ ∂y ˆ j + ∂φ ∂z ˆ k

  • We then define the directional derivative as

dφ ds = ∇φ · u

Patrick K. Schelling Introduction to Theoretical Methods