Transcript Supplementary Fields Notes

Physics 121 - Electricity and Magnetism
Lecture 01 Supplement - Fields - Gravitation
Field concepts:
• Scalar and vector fields in math & physics
• How to visualize fields: contours & field lines
• “Action at a distance” fields – gravitation and electro-magnetics.
• Force, acceleration fields, potential energy, gravitational potential
• Flux and Gauss’s Law for gravitational field: a surface integral of
gravitational field
More math:
• Calculating fields using superposition and simple integrals
• Path integral/line integral
• Spherical coordinates – definition
• Example: Finding the Surface Area of a Sphere
• Example: field due to an infinite sheet of mass
1
Copyright R. Janow – Spring 2015
What’s a “Field” - Mathematical View
•
•
•
•
A FIELD assigns a value to every point in space (2D, 3D, 4D,….)
It may have nice mathematical properties, like other functions:
E.g. superposition, continuity, smooth variation, multiplication,..
A scalar field f maps a vector into a scalar:
f: R3->R1, f: R2->R1 …
• A scalar quantity is assigned to every point in 3D space
ISOBARS
• Temperature, barometric pressure, potential energy
EQUIPOTENTIALS
•
A vector field g maps a vector into a vector:
g: R3->R3, g: R2->R2
• A 2D/3D vector is assigned to every point in 2D/3D space
• Wind velocity, water velocity (flow), acceleration
• Taxing to the imagination, involved to calculate
Example: map of the velocity
of westerly winds flowing
past mountains
Pick single
altitudes and make
slices to create
maps
Copyright R. Janow – Spring 2015
…
FIELD LINES
“FIELD LINES” (streamlines) show wind direction
Line spacing shows speed: dense  fast
Set scale by choosing how many lines to draw
Lines begin & end only on sources or sinks
Scalar field examples
•
•
•
A scalar field assigns a simple number as the field
value at every point in “space”.
Temperature map portrays ground-level
temperature as function of x-y position
Maps R2 -> R1
•
Scalar field: altitude at points
on a mountain as function of x-y
position.
Contours follow constant
altitude
•
Contours far
apart
Contours
closely
spaced
•
Contours
Grade (or slope) is related to
the horizontal spacing of
contours (vector field)
flatter
steeper
Copyright R. Janow – Spring 2015
Side View
Vector Fields
•
The value of a vector field at every point in
space is a vector – with magnitude and direction
•
A vector field (e.g.,gravitational force) can be
generated by taking the gradient of a scalar
field (e.g.,potential energy).
•
Gradient field lines are perpendicular to the
contours (e.g., lines of constant potential
energy)
•
The steeper the gradient (e.g., rate of change
of gravitational potential energy) the larger the
field magnitude is.
DIRECTION
•
Gradient vectors point along the direction of steepest
descent, which is also perpendicular to the contours.
•
Imagine rain flowing down a mountain. The vectors are
also “streamlines.” Water running down the mountain
will follow these streamlines.
Copyright R. Janow – Spring 2015
Side View
Slope, Grade, Gradients (another field) and Gravity
Height contours h can also portray potential energy U = mgh.
The height and potential energy do not change along a contour.
Motion perpendicular to a contour at a point is along the gradient.
•
Slope and grade mean the same
thing. A 15% grade is a slope of
15
lim h / x  dh / dx  0.15
x  0
•
100
Gradient is measured along the path.
For the case above it would be:
lim h / l  dh / dl   15 /101.1  0.148
x  0
•
15%
Gravitational force along path l is the
gradient of potential energy
F  dU / dl  d (mgh) / dl  mg dh / dl
l
dh / dl  sin()
F  mg sin()
•
•
h

x
The steepness (or force) are related to the GRADIENTS of height (or gravitational potential
energy) respectively, and are also fields.
Are the GRADIENTS of scalar fields also scalar fields or are they vector fields?
Copyright R. Janow – Spring 2015
Another scalar field – atmospheric pressure
Isobars: lines
of constant
pressure
How do the isobars affect air motion?
What are the black arrows showing?
Copyright R. Janow – Spring 2015
A related vector field: wind velocity
Wind speed and
direction depend on
the pressure gradient
Copyright R. Janow – Spring 2015
Some fields are used to explain “Action at a Distance”
• Place a test mass, test charge, or test current at some test point in a field
• It feels a force due to the presence of remote sources of the field.
• The sources “alter space” at every possible test point.
• The forces (vectors) at a test point due to multiple sources add up
via superposition (the individual field vectors add & form the net field).
Field Type
Source
Acts on
Definition
Strength
(dimensions)
mass
another
mass
Force per
unit mass at
test point
ag = F g / m
electrostatic
charge
another
charge
Force per
unit charge
at test point
E=F/q
magnetic
electric
current
.length
another
current
.length
Force per unit
current.length
B ~ F/qv or
F/iL
gravitational
Copyright R. Janow – Spring 2015
Summary: Visualizing Physical Fields
Could be:
• 2 hills,
• 2 charges
• 2 masses
Scalar field: lines of constant field magnitude
•
•
•
Altitude / topography – contour map
Pressure – isobars, temperature – isotherms
Potential energy (gravity, electric)
Vector field: field lines show a gradient
•
•
•
•
Direction shown by TANGENT to field line
Magnitude proportional to line density inversely to distance between lines
Lines start and end on sources and sinks of field
(highs and lows)
Forces are fields, with direction related to
gravitational, electric, or magnetic field
Mass or
negative
charge
Magnetic field
around a wire
carrying current
Summary: Scalar and vector fields in mechanics and E&M:
TYPE
MECHANICS (GRAVITY)
ELECTROSTATICS (CHARGE)
FORCE
(vector)
Gravitational Force = GMm / r2
Coulomb Force = kqQ / r2
SCALAR
FIELDS
Gravitational Potential Energy
Gravitational Potential
(PE / UNIT MASS)
Electric Potential Energy
Electric Potential (volts)
(PE / UNIT CHARGE)
Magnetic P. E. (due to a current)
E = Force / Unit Charge
= “Electric Field”
B = Force / Unit Current x Length
= “Magnetic Field”
ag = Force / Unit Mass
VECTOR
= “Gravitational Field”
FIELDS
= Acceleration
of Gravity “g”
Copyright R. Janow
– Spring 2015
MAGNETOSTATICS
(CURRENT)
Magnetic Force = q v X B
Idea of a test mass
•
The field everywhere is proportional to mass M at
the origin
•
The amount of force at some point due to M is also
proportional to the mass m at that point
•
m

GMm
F   2 r̂
r
r̂
Use m as a test mass (it could be 1 kg for example)
and measure the force on it as it moves around:

F
GM
  2 r̂   g(r )r̂
m
r
•
g(r) is the “gravitational field”, also called
the gravitational acceleration.
•
The direction (only) is given by
•
g(r) is a vector field, like the force. It’s direction
is conveyed by the unit vector
Same idea for test charges & currents
Copyright R. Janow – Spring 2015
M
Superposition of fields (gravitational)
•
•
•
“Action-at-a-distance”: gravitational field permeates all of space with force/unit mass.
“Field lines” show the direction and strength of the field – move a “test mass” around to map it.
Field cannot be seen or touched and only affects the masses other than the one that created it.
•
What if we have several masses? Superposition—just vector sum the
individual fields.
M
•
M
M
M
The NET force vectors show the direction and strength of the NET field.
The same ideas apply to electric & magnetic fields
Copyright R. Janow – Spring 2015
An important idea called Flux (symbol F):
Basically a vector field magnitude x area
- fluid volume or mass flow
- gravitational - electric - magnetic
Definition: differential amount of flux dFg of field ag crossing vector area dA
ag
“unit normal”
n̂
outward and
perpendicular to
surface dA


dF g  flux of a g through dA

 a g  n̂dA (a scalar)
“Phi”
Flux through a closed or open surface S: calculate “surface integral” of field over S
Evaluate integrand at all points on
surface S
FS 
 dF 

a
 g  n̂dA
S
S
EXAMPLE : FLUX THROUGH A CLOSED, EMPTY,
RECTANGULAR BOX IN A UNIFORM g FIELD
• zero mass inside
• F from each side = 0 since a.n = 0, F from ends cancels
• TOTAL F = 0
• Example could also apply to fluid flow
n̂
n̂
n̂
ag
What if a mass (flux source) is in the box?
Can field be uniform? Can net flux be zero.
Copyright R. Janow – Spring 2015
n̂
FLUID FLUX EXAMPLE: WATER FLOWING ALONG A STREAM
Self
Study
Assume:
•
•
•
•
constant mass density r 
incompressible fluid
constant flow velocity parallel to banks
no turbulence (laminar flow)
n̂'
n̂

A 2
Flux measures the flow (current):
• flow means amount/unit time across area
• either rate of volume flow past a point …or…
rate of mass flow past point

A 1
2 related fluid flow fields (currents/unit area):
• velocity v represents volume flow/unit area/unit time
• J = mass flow/unit area/unit time

A  n̂A
n̂ is the outward unit vector
 to vector area A


J  rv
Flux = amount of field crossing an area per unit time (field x area)

 

V  L  A 
L  vt
The chunk of

volume flux 

 v  A
fluid moves L

t

t
A


in time t:



r


r


mass
of
solid
chunk
m
V
L
A



v

 

m
L

and
\ mass flux 
r


r



 A v A J A
t
t
Continuity: Net flux (fluid flow) through a closed surface = 0
………unless a source or drain is inside
Copyright R. Janow – Spring 2015
Gauss’ Law for gravitational field:
The flux through a closed surface S depends only on the enclosed mass (source of
field), not on the details of S or anything else
Example: spherically symmetric mass distribution, radial gravitational field

GM
2
Field: g   2 r̂ (Newtons/k g or m/s )
r 


 
GM
d(flux )  f ield . dA  g . dA   2 r̂.dA
r
Find total flux through closed surface A


 
GM
GM
F A   g.dA    2 r̂.dA   2  r̂.dA
A
A
A
rA
r
inward force
on test mass m
Spherical
surface of
constant
field & PE
gA
Integral for surface area of sphere
 
GM
\ F A   g.dA   2 x 4 rA2   4GM
A
rA
rA
M
Flux depends only on the enclosed mass
(same flux for any closed surface enclosing M)
FLUX measures the strength of a field source that
is inside a closed surface - “GAUSS’ LAW”
Copyright R. Janow – Spring 2015
gA
Shell Theorem follows from Gauss’s Law
Self Study
1. The force (field) on a test particle OUTSIDE a uniform SPHERICAL shell
of mass is the same as that due to a point mass concentrated at the
shell’s mass center (use Gauss’ Law & symmetry or see section 13.6)
m
r
x
r
m
x
Same for a solid sphere (e.g., Earth, Sun) via nested shells
m
r
r
x
x
+
r
x
+
2. For a test mass INSIDE a uniform SPHERICAL shell of mass m,
the shell’s gravitational force (field) is zero
• Obvious by symmetry for center point
• Elsewhere, integrate over sphere (painful) or
apply Gauss’ Law & Symmetry
m
x x
3. Inside a solid sphere combine the above. The force on a test mass
INSIDE depends only on mass closer to the CM than the test mass.
x
• Example: On surface, measure acceleration
distance
r
from center
• Halfway to center,
Copyright R. Janow – Spring 2015
ag = g/2
g
a
4
3
Vsphere  r 3
Superposition Example: Calculate the field (gravitational) at a
special point due to two point masses
Find the field at point P on x-axis due to
two identical mass chunks m at +/- y0
• Superposition says add fields created at P
by each mass chunk (as vectors!!)
• Same distances r0 to P for both masses
y
m
r0
+y0
r02  x 02  y02

+x0
• Same angles with x-axis
cos()  x 0 / r0
•
Gm
x 02  y02

P
ag
-y0
r0
• Same magnitude ag for each field vector
ag 
ag
(from Newtons law of gravitatio n)
m
y components of fields at P cancel, x-components reinforce each other
a tot  a x  
2Gm cos()
r02
 
2Gm x 0
r03
where
• Result simplified because problem had a lot of symmetry
Copyright R. Janow – Spring 2015
r03  [ x 02  y02 ]3 / 2
Direction: negative x
x
Example: Calculate gravitational field due to an infinitely long line of
uniformly distributed mass. Find the field at point P on x-axis
to y  
• Integrate over the source of field holding P fixed
•
Add differential amounts of field created at P by
differential point mass chunks at y (vectors!!)
dm = ldy
•
Include mass from y = – infinity to y =+ infinity
• For symmetrically located point mass chunks dm:
r
y
•
y-components of fields cancel,
•
x-components of fields reinforce
• Mass per unit length l is uniform, find dm in terms of :
-y

a x  da x where
Gdm
da

da
cos(

)


cos()
-
x
g
r2

dm  ldy  lx[1  tan2 ()] d
\ da x  
Glx[1  tan2 ()]
cos() d  
x 2 [1  tan2 ()]
• Integrate over  from –/2 to +/2
\ ax  
Gl
cos() d
x
Gl   / 2
Gl
cos(

)
d



2
x   / 2
x
Field of an infinite line falls off as 1/x not 1/x2
Copyright R. Janow – Spring 2015
x
dag


P
x
l = mass/unit length
to y
 
y  x tan()
dy
dtan()
x
 x [1  tan2 ()]
d
d
r 2  x 2  y2  x 2 [1  tan2 ()]
 /2
 cos()d  2
- /2
Example of a “line integral” (path integral)- Work done on
a mass traversing a gravitational field
Self Study
How much work is done on a test mass
 as it traverses a particular path


dW  dU  F  ds  mag  ds
through a field?
B
B

\ U    F  ds
evaluate along path
test mass
A
Gravitational field is conservative, meaning U is
independent of path chosen
A
 

F

d
s

F

d
s


B
A
for any path between A & B
B
 
U   F  ds  0 for any closed path that is chosen
S
circulation,or path integral
EXAMPLE
uniform field
U= - mgh
Copyright R. Janow – Spring 2015
U= + mgh
Surface Integral Example: Show that the surface area of
a sphere A= 4R2 by integrating over the sphere’s surface Self
Study
Find dA – an area segment on the surface of the sphere,
then integrate on angles f (azimuth) and  (co-latitude).
z
dA  dl  dh

dA  r̂ dA
Where:
• dl is a curved length segment of the circle around
the z-axis (along a constant latitude line)
• dh is a segment along the  direction (along a
constant longitude line)
dl  r sin() df

r
f
y
dh  r d
x
  [0,  ]
Angle range for a full sphere:

A   r̂.dA 
surface
f  [0, 2 ]
Factors into 2 simple 1 dimensional integrations
2

0
0
 
2

0
0
r sin() ddf  r {  df} {  sin() d }
2
2


 2r 2  sin() d  2r 2 [  cos()] 0  2r 2 ()[ 1  1]
0
\ A  4 r 2
Copyright R. Janow – Spring 2015
Gravitational field due to an infinite sheet of mass
Simple 2 dimensional
Copyright R. Janow – Spring 2015
integration
Self Study
Constant - does not depend
on distance from plane!