Homework 6 -...

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Unformatted text preview: Physics
601
Homework
6­­­Due
Friday
October
15
 
 
 1. In
class
starting
with
the
action
 S = ∫ dτ (−m − S ) we
used
covariance
to
show
that
 d (( m + S ) u ) = ∂ S 
where

 for
a
particle
moving
in
a
Lorentz
scalar
field
 µ µ dτ ∂ µ ≡ g µν ∂ .
 ∂x ν € d( u ) € = ∂ µ S − (∂ α S ) uα u µ 
 a. Show
that
this
can
be
rewritten
in
the
form
 ( m + S ) dτ b. Show
that
this
equation
of
motion
automatically
satisfies
the
condition
 d ( uµ u µ ) = 0 .

This
indicates
that
imposition
of
covariance
yielded
a
self‐ dτ € consistent
result
that
respects
the
condition
 uµ u µ = 1.
 c. Show
that
in
the
non‐relativistic
limit
where
all
of
the
velocities
are
much
 less
than
the
speed
of
light
and

 S << m 
the
Lagrangian
for
the
system
 € ˙ reduces
to
 L = 1 mx 2 − S 
plus
an
irrelevant
constant
and
the
equation
of
 2 € x motion
reduces
to
 m˙˙ = −∇S .

 
 € 2. Start
from
the
action
 S = ∫ dτ (−m + V µ uµ ) where
 Aµ 
is
a
four
vector
field
that
 € depends
on
space‐time.

Show
that
the
equation
of
motion
is
 € d ( muµ ) ∂Vµ ∂Vν ν = ν − µ u .
 dτ x € ∂€ ∂x µ € 
 € 3. In
electro‐magnetism,
one
can
write
the
scalar
and
vector
potentials
in
a
form
that
 Φ Ax µ looks
like
a
4‐vector:
 A = .

Because
one
can
make
arbitrary
gauge
 Ay Az µ transformations
 A 
need
not
transform
as
a
4‐vector.


 
 € a. Show
that
a
gauge
transformation
can
be
written
in
the
form
of
the
form
 € €µ → A' µ = Aµ + ∂G 
where
G
is
an
arbitrary
function
of
space‐time
which
 A ∂x µ need
not
transform
as
a
4‐scalar
under
Lorentz
transformations
.
 b. A
sufficient
condition
to
show
that
 A' µ 
transforms
as
a
4‐vector
is
to
show
 ∂A'µ µ that

 µ = 0 
with
 A → 0 

as
 x → ∞ 

Explain
briefly
why.
 ∂x 
 c. Show
that
it
is
always
possible
to
make
a
gauge
transformation
(i.e.
to
 € choose
G)
to
ensure
that
 A' µ 
does
transform
as
a
4‐vector
by
picking
 € € € € choosing
Λ
to
satisfy
the
condition
 ∂µ A µ = −∂µ∂ µ G .

This
is
called
the
Lorentz
 gauge.
 d. Show
that
the
field‐strength
tensor
 Fµν ≡ ∂µ Av − ∂ v A µ is
gauge
invariant.

 That
is
show
that
 F 'µν ≡ ∂ µ A' v −∂v A' µ = Fµν for
any
transformation
of
the
form
 € given
in
part
a.
 e. Show
that
 Fµν 
transforms
under
Lorentz
transformations
as
a
4‐tensor:
 € α € β Fµν → Λ€ Λ v Fαβ 
where
Λ
is
the
matrix
which
specifies
the
Lorentz
 µ transformation.
 f. Using
the
result
problem
1,

plus
the
preceding
parts
of
this
problem
show
 € that
the
equation
of
motion

for
a
particle
of
mass
m
and
charge
q
moving
in
 € d ( mu µ ) = qF µν uν .
 an
electro
magnetic
field
is
given
by
 dτ 
 d ( mu µ ) = qF µν uν .

Staring
with
this
 4. In
the
preceding
section
you
showed
that
 dτ € d ( uµ u µ ) = 0 .

This
indicates
that
equation
 equation
of
motion
show
that
of
necessity
 dτ of
motion
self‐consistently
respects
the
condition
 uµ u µ = 1.
 € 
 
 € 5. Suppose
that
one
has
a
charged
particle
with
mass
m
and
charge
q
interacting
with
 € external
electromagnetic
fields
specified
by
the

potentials

 Φ = 0, Ax = − E 0 t, Ay = Az = 0 .

Note
that
potentials
are
independent
of
special
 position.
 a. Verify
that
these
potentials
satisfy
the
Lorentz
gauge
condition
 ∂µ A µ = 0 
.
 b. Construct
the
field
strength
tensor
 Fµν .
 c. Suppose
that
the
particle
starts
from
rest
at
t=0,
find
the
position
of
the
 particle
as
a
function
of
time.

(Hint,
you
may
find
your
solution
of
problem
 € 4.d
of
homework
5
to
be
useful.)

 € 
 6. In
class
we
found
the
Green’s
function
for
the
harmonic
oscillator.

In
this
problem,
 I
want
you
to
find
and
use
the
analogous
one
for
a
damped
oscillator.

The
damped
 2 ˙ driven
oscillator
satisfied
the
equation:
 m˙˙ + 2βmx + mω 0 x = f ( t ) 
where
β
is
a
 x damping
parameter.

The
solution
is
 x ( t ) = ∫ ∞ −∞ dt ' G( t, t ' ) f ( t ' ) where
the

Green’s
 2 function
satisfies
 (∂ t2 + 2β∂ t + ω 0 )G( t, t ' ) = δ ( t − t ' ) .

A
useful
first
step
in
constructing
 € this
is
to
exploit
the
known
solution
for
steady
state
motion
with
a
harmonic
 2 2 driving
force:
 (∂ t + 2β∂ t + ω 0 ) x ( t ) = f 0e iω t 
has
a
solution
of
the
form
of
the
form
 € f 0e iΩ( t − t ' ) e iΩ( t − t ' ) 2 x(t) = € 2 = e iΩ( t − t ' ) .


Let
us
now
 .

Thus
 (∂ t2 + 2β∂ t + ω 0 ) 2 2 2 ω 0 − Ω + 2iβΩ ω 0 − Ω + 2iβΩ integrate
both
sides
with
respect
to
Ω and

divide
by
2π: € € € ∞ ∫−∞ dΩ (∂t2 + 2β∂t + ω 02 ) 2π ∫ ∞ −∞ dω e iω ( t − t ' ) e iω ( t − t ' ) 2 ω 0 − ω 2 + 2iβω = ∫ ∞ −∞ dω e iω ( t − t ' ) 2π .

We
know
that
 = δ ( t − t ' ) .



Moreover
on
the
left
hand
side
we
can
pull
 2π 2 (∂ t2 + 2β∂ t + ω 0 ) 
out
of
the
integral
as
it
does
not
depend
on
ω.

Thus
 ∞ e iω ( t − t ' ) ∫−∞ dω 2 ω 0 − ω 2 + 2iβω 2 2 (∂ t + 2β∂ t + ω 0 ) = δ ( t − t ' ) 
and
the
object
in
the
parenthesis
 2π 
 is
a
Green’s
function.

 a. Evaluate
the
integral
above
using
contour
integration
to
find
an
explicit
 expression
for
 G( t, t ' ) .


Note
that
the
complex
exponential
implies
that
the
 ½
plane
in
which
the
contour
is
to
be
closed
depends
on
the
sign
of
t­t’.

 2 ˙ b. Use
this
Green’s
function
to
find
a
solution
of
 m˙˙ + 2βmx + mω 0 x = f 0 e−Γtθ ( t ) .
 x c. Consider
the
result
in
b.
in
the
regime
where
Γ>>ω>>β.


In
that
regime
the
 € system
should
look
like
an
underdamped
oscillator
getting
a
delta‐function‐ like
impulse
at
t=0.

Does
it?
 € 
 € € € € 
 
 ...
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This note was uploaded on 12/29/2011 for the course PHYSICS 601 taught by Professor Hassam during the Fall '11 term at Maryland.

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