Transcript Motion in a Straight Line

Lecture 18
3D Cartesian Systems
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• Past exam papers
• Complete solution from last lecture
• Have a look at homework 2 (due in on 12/12/08)
Remember Phils Problems and your notes = everything
http://www.hep.shef.ac.uk/Phil/PHY226.htm
Introduction to PDEs
In many physical situations we encounter quantities which depend on two or
more variables, for example the displacement of a string varies with space
and time: y(x, t). Handing such functions mathematically involves partial
differentiation and partial differential equations (PDEs).
2
1

u
2
 u 2 2
c t
Wave equation
Elastic waves, sound
waves, electromagnetic
waves, etc.
2 2
u

 u  Vu  i
2m
t
1 u
 2u  2
h t
Schrödinger’s
equation
Quantum mechanics
Diffusion
equation
Heat flow, chemical
diffusion, etc.
 u0
Laplace’s
equation
Electromagnetism,
gravitation,
hydrodynamics, heat flow.

2
 u
0
Poisson’s
equation
As (4) in regions
containing mass, charge,
sources of heat, etc.
2
3D Coordinate Systems
We can describe all space using coordinates (x, y, z), each one ranging from -∞ to +∞.
1. PDEs in 3D Cartesian Coordinates
Consider the wave equation. In one dimensional space we had
 2  ( x , t ) 1  2  ( x, t )
 2
x 2
c
t 2
 2 ( x, y, z, t )  2 ( x, y, z, t )  2 ( x, y, z, t ) 1  2 ( x, y, z, t )


 2
In 3D equation becomes
x 2
y 2
z 2
c
t 2
1  2  ( x, y, z, t )
which may be written in shorthand as   ( x, y, z, t )  2
c
t 2
2 ,
Let us look for a solution of the form , ( x, y, z, t )  X ( x)Y ( y)Z ( z )T (t )
i.e. we substitute and separate the variables, as done in 1D ….
 2  ( x, y , z , t )
d 2 X ( x)
 Y ( y ) Z ( z )T (t )
2
x
dx 2
 2  ( x, y , z , t )
d 2 Z ( z)
 X ( x)Y ( y )T (t )
2
z
dz 2
 2 ( x, y, z, t )
d 2Y ( y)
 X ( x) Z ( z )T (t )
2
y
dy 2
 2  ( x, y , z , t )
d 2T (t )
 X ( x)Y ( y ) Z ( z )
2
t
dt 2
.
3D Coordinate Systems
Substituting these into PDE then dividing both sides by
gives …
X ( x)Y ( y) Z ( z )T (t )
1 d 2 X ( x)
1 d 2Y ( y)
1 d 2 Z ( z)
1 d 2T (t )


 2
2
2
2
X ( x) dx
Y ( y) dy
Z ( z ) dz
c T (t ) dt 2
(*)
As before for 1D case we need solutions that can be zero more than once to fulfil
boundary conditions, so we choose each term to equal a negative constant to ensure
we get LHO style solutions.
1 d 2 X ( x)
2


k
Let
x
X ( x) dx2
1 d 2Y ( y)
2
 k y
2
Y ( y) dy
1 d 2 Z ( z)
2
 k z
2
Z ( z ) dz
Comparing with (*) the defined constants, , kx, ky, kz are related by
1 d 2T (t )
  2
2
T (t ) dt
2
c
2
 k x2  k y2  k z2
Each of the ODEs above has the normal harmonic solutions, which we can write in
terms of sines and cosines below.
sin k x x 
X(x) ~ 

cos
k
x
x 

sin k y y 
Y(y) ~ 

cos k y y 
sin k z z 

Z(z) ~ 
cos
k
z
z 

sin t 
T(t) ~ 

cost 
3D Coordinate Systems
sin k x x 
X(x) ~ 

cos
k
x
x 

sin k y y 
Y(y) ~ 

cos k y y 
sin k z z 

Z(z) ~ 
cos
k
z
z 

sin t 
T(t) ~ 

cos

t


sin k x x 
sin k y y 
sin k z z sin t 
Giving special solutions of the form  ( x, y, z, t )  A




cos
k
x
cos
k
y
cos
k
z
cos

t




x 
y 
z 

www.falstad.com/mathphysics.html
Or sometimes it is more convenient to use complex exponentials,
X ( x)  e ikx x ,
Y ( y)  e
iky y
,
Z ( z)  e ikz z ,
T (t )  e it
Then depending on the boundary conditions we can get special solutions such as:
( x, y, z, t )  A exp(ikx x  ik y y  ikz z  i t )  A exp(ik.r i t )
where
k  kxi  k y j  kzk
and
r  xi  yj  zk
As we might have expected, these solutions are plane waves with wavevector k
(which is also the direction of travel of the wave).
3D Coordinate Systems
sin k x x 
sin k y y 
sin k z z sin t 
 ( x, y, z , t )  A




cos
k
x
cos
k
y
cos
k
z
cos

t



x 
y 
z 

A general solution can then be written as a sum over all special solutions. By applying
boundary conditions we can then determine which terms contribute and the allowed
values of kx, ky, kz as before in 1D examples.
For example, suppose we have a box with dimensions L1, L2, L3 in the x, y, z directions
respectively and know that  must vanish at the walls and that it is zero at t = 0. Then
the special solutions after these boundary conditions have been applied will be:
( x, y, z, t )  Asin kx x sin k y y sin kz z sin t
where
kx 
n1
L1
ky 
n2
L2
kz 
n3
L3
So each special solution, or ‘mode’ will be characterized by three integers, n1, n2, n3.
And this mode will have angular frequency   c (k x  k z
2
2
2
2
 n12 n2 2 n3 2 
 k z )   c  2  2  2 
L2
L3 
 L1
2
2 2
A common question is to deduce how many different modes (i.e. unique combinations
of integers n1, n2, n3) exist in a given frequency range  to  + d ? e.g. Planck’s
Law for blackbody radiation.
www.falstad.com/mathphysics.html
3D Coordinate Systems
2. Integrals in 3D Cartesian Coordinates
We have dV = dx dy dz, and must perform a triple integral over x, y and z. Normally
we will only work in Cartesians if the region over which we are to integrate is cuboid.
Example 1 : Find the 3D Fourier transform,
F (k ) 
1
(2 ) 3 / 2

f (r )eik.r dV
if
all space
and
1,  a  x  a,  b  y  b,  c  z  c
f ( x, y, z)  
otherwise
0
k  kxi  k y j  kzk
and
r  xi  yj  zk
The integral is just the product of three 1D integrals, and is thus easily evaluated:
Just integrating over x gives
Mistake in notes
a
I  e
a
ik x x
a

 e 
  e ikx a   eikx a
dx  
  
ik x
 ik x  a 
ik x x
   e
 
 
ik x a
 e ikx a 

ik x

ik b
 ik b
a
b
c
1
1  eikx a  e ikx a  e y  e y  eikz c  e ikz c 
ik y y
ikx x
ikz z




F (k x , k y , k z ) 
e dx  e dy  e dz 
32 
32 


(2 ) a
(2 ) 
ik x
ik y
ik z


b
c

This is therefore a product of three sinc functions, i.e.
8 sin(k x a) sin(k y b) sin(k z c)
8abc
F (k x , k y , k z ) 

sinc(k x a) sinc(k y b) sinc(k z c)
(2 ) 3 2
kx
ky
kz
(2 ) 3 2
Corrections to notes- sorry!!
Homework 2
Lecture 17