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Mathematics for Quantum Physics
lectures
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6559471ace06dabf0867dc84f9b326e251b52321...d0a2c689c16e68ff15e37f2dedcd94290f1f37c7
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Commits (3)
further rewriting
· 5adebc78
Michael Wimmer
authored
Sep 04, 2019
5adebc78
remove path length example by Jos
· 03993f50
Michael Wimmer
authored
Sep 04, 2019
03993f50
remove path length example by Jos
· d0a2c689
Michael Wimmer
authored
Sep 04, 2019
d0a2c689
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d0a2c689
...
...
@@ 76,16 +76,21 @@ The inverse relation is given as
$$r=
\s
qrt{x^2 + y^2};$$
$$
\v
arphi=
\b
egin{cases}
\a
rctan(y/x) &
\t
ext{$x>0$,}
\\
\p
i +
\a
rctan(y/x) &
\t
ext{$x
<0
$
and
$
y
>
0$,}
\\

\p
i +
\a
rctan(y/x) &
\t
ext{$x<0$ and $y<0$.}
\e
nd{cases}$$
\a
rctan(y/x) &
\t
ext{$x>0$,}
\\
\p
i +
\a
rctan(y/x) &
\t
ext{$x
<0
$
and
$
y
>
0$,}
\\

\p
i +
\a
rctan(y/x) &
\t
ext{$x<0$ and $y<0$.}
\e
nd{cases}$$
The last formula for $
\v
arphi$ warrants a closer explanation: It is easy
to see that $
\t
an(
\v
arphi)=y/x$  but this is not a unique relation, due to
the fact that the $
\t
an$ has different branches. Convince yourself that
the expression above is correct for all the four sectors!
## Distances and areas
Now suppose we want to calculate the distance between two points, one
with polar coordinates $(r_1,
\v
arphi_1)$, and the other with
$(r_2,
\v
arphi_2)$. This looks like a difficult exercise.
A convenient
$(r_2,
\v
arphi_2)$. This looks like a difficult exercise.
One possible
way to perform this is by translating the polar coordinates into
Cartesian coordinates and using the expression given above for this
distance: $$
\D
elta s^2 = (x_1  x_2)^2 + (y_1  y_2)^2,$$ so
...
...
@@ 103,32 +108,34 @@ arc of radius $r_1$ and then we move a small distance radially outward
from $r_1$ to $r_2$. Provided the difference between the angles
$
\v
arphi_1$ and $
\v
arphi_2$ is (very) small, these paths are
approximately perpendicular and we can use Pythagoras’ theorem to find
the distance $
\D
elta s$. Note that the arc is approximately straight –
it has a length $r_1
\D
elta
\v
arphi$, where
$
\D
elta
\v
arphi =
\v
arphi_2
\v
arphi_1$. So we have:
$$
\D
elta s^2 = (
\D
elta r)^2 + (arc~length)^2 = (
\D
elta r)^2 + r_1^2 (
\D
elta
\v
arphi)^2 .$$
For $r
\e
quiv r_1
\a
pprox r_2$, replacing the $
\D
elta$’s by $d$’s to
emphasise the infinitesimal character of the differences, we finally
have: $$ds^2 = dr^2+ r^2 d
\v
arphi^2 .$$ You should recall this formula!
If we traverse a curved path in the plane, we must sum up all the $ds$,
which actually boils down to calculating an integral:
$$Path~length =
\i
nt_{start}^{end} ds =
\i
nt_{start}^{end}
\s
qrt{dr^2 + r^2 d
\v
arphi^2}.$$
Here, the boundaries $start$ and $end$ stand for the starting and end
point of the path.
Note that the path can often be written in the form $r(
\v
arphi)$,
i.e. we write $r$ as a function of $
\v
arphi$. We can then write the
integral as:
$$Path~length =
\i
nt_{start}^{end}
\s
qrt{dr^2 + r^2 d
\v
arphi^2} =
\i
nt_{
\v
arphi_{start}}^{
\v
arphi_{end}}
\s
qrt{
\l
eft(
\f
rac{dr}{d
\v
arphi}
\r
ight)^2 + r^2}
\;
d
\v
arphi.$$
(Note: this parametrisation is strictly speaking only possible when $r$
is a univalued function of $
\v
arphi$; a meandering path generally does
not have this property.)
**Exercise:**
write the same expression as an integral over $dr$ instead
of $d
\v
arphi$. Please sketch two paths, one which is more suitable for
the expression above (the integral over $
\v
arphi$) and the other more
convenient for the expression you obtained in this problem.
the distance $d s$. Note that the arc is approximately straight –
it has a length $r_1 d
\v
arphi$, where
$d
\v
arphi =
\v
arphi_2
\v
arphi_1$. So we have:
$$d s^2 = (d r)^2 + (arc~length)^2 = (d r)^2 + r_1^2 (d
\v
arphi)^2 .$$
We can use the same arguments also for the area: since the different
segments are approximately perpendicular, we find the area by simply
multiplying them:
$$dA = r dr d
\v
arphi.$$
This is an important formula to remember for integrating in polar
coordinates! The extra $r$ that appears here can be intuitively
understood: the area swept by an angle difference $d
\v
arphi$
*increases*
as we move further away from the origin.
!!! check "Example: Integrating over a circular area"
To check the area element we just derived, let us compute a simple
integral. We compute the integral over a circle with radius $r_0$
with a very simple function that equals to $1$. In this case,
we expect to get as a result the are of the region we integrate over.
We find:
$$ \int_0^{2\pi} d\varphi \int_0^r_0 r dr = \\
2\pi \int_0^r_0 r dr = 2 \pi \frac{1}{2} r_0^2 = \pi r_0^2$$
which is indeed the area of a circle with radius 0.
## Cylindrical coordinates
...
...
@@ 142,10 +149,10 @@ choose the coordinate system ourselves  this is not imposed by the
problem.
Cylindrical coordinates are defined straightforwardly: we use polar
coordinates $
\r
ho
$ and $
\v
arphi$ in the $xy$ plane, and the distance $z$
coordinates $
r
$ and $
\v
arphi$ in the $xy$ plane, and the distance $z$
along the symmetryaxis as the third coordinate. If the axis system is
chosen in physical space, we have two coordinates which have the
dimension of a distance: $
\r
ho
$ and $z$. The other coordinate,
dimension of a distance: $
r
$ and $z$. The other coordinate,
$
\v
arphi$, is of course dimensionless.
What is the distance travelled along a path when we express this in
...
...
@@ 153,15 +160,15 @@ cylindrical coordinates? Let’s consider an example (Figure).
![
image
](
figures/Coordinates_13_0.svg
)
We want to find the length of the (small) red segment $
\D
elta
s$. By
We want to find the length of the (small) red segment $
d
s$. By
inspecting the figure, we see that the horizontal (i.e. parallel to the
$xy$plane) segment $
\D
elta
l$ is perpendicular to the vertical segment
$dz$. Using for $
\D
elta
l$ the length we obtained before for a line
$xy$plane) segment $
d
l$ is perpendicular to the vertical segment
$dz$. Using for $
d
l$ the length we obtained before for a line
segment in the $xy$ plane, expressed in polar coordinates, we
immediately find:
$$
\D
elta s^2 =
\D
elta l^2 +
\D
elta z^2 =
\D
elta
\r
ho^2 +
\r
ho^2
\D
elta
\v
arphi^2 +
\D
elta
z^2.$$
Using $d$ instead of $
\D
elta$ for infinitesimal changes, this reads:
$$d
s^2 = dl^2 + dz^2 = d
\r
ho^2 +
\r
ho^2 d
\v
arphi^2 + dz^2
.$$
$$
d s^2 = d l^2 + d z^2 = d r^2 + r^2 d
\v
arphi^2 + d
z^2.$$
The volume element is consequently given as
$$d
V = r dr d
\v
arphi dz
.$$
## Spherical coordinates
...
...
@@ 178,12 +185,11 @@ The position of a point on the sphere is specified using the two angles
$
\t
heta$ and $
\p
hi$ indicated in the figure.
!!! warning
Note that in mathematics,
usually
the angles are labelled the other way
Note that in mathematics,
often
the angles are labelled the other way
round: there, $
\p
hi$ is used for the angle between a line running from
the origin o the point of interest and the $z$axis, and $
\t
heta$ for
the angle of the projection of that line with the $x$axis. The
convention used here is custom in physics. IT IS IMPORTANT TO BE AWARE
OF THIS!!
convention used here is custom in physics.
The relation between Cartesian and coordinates is defined by
$$x = r
\c
os
\v
arphi
\s
in
\v
artheta$$
...
...
@@ 215,6 +221,10 @@ displacement.
![
image
](
figures/Coordinates_19_0.svg
)
From these arguments we can again also find the volume element, it is
here given as
$$dV = r^2
\s
in
\t
heta dr d
\t
heta d
\v
arphi.$$
### summary
We have discussed four different coordinate systems:
...
...
@@ 232,12 +242,14 @@ We have discussed four different coordinate systems:
given in terms of these coordinates.
Infinitesimal distance: $$ds^2 = dr^2 + r^2 d\phi^2.$$
Infinitesimal area: $$dA = r dr d\varphi.$$
3.
*Cylindrical coordinates*
: $${
\b
f r} = (
\r
ho
,
\p
hi, z).$$ Can be
3.
*Cylindrical coordinates*
: $${
\b
f r} = (
r
,
\p
hi, z).$$ Can be
used in three dimensions. Suitable for systems with axial symmetry
or functions given in terms of these coordinates.
Infinitesimal distance: $$ds^2 = d\rho^2 + \rho^2 d\phi^2 + dz^2.$$
Infinitesimal distance: $$ds^2 = dr^2 + r^2 d\phi^2 + dz^2.$$
Infinitesimal volume:: $$dV = r dr d\varphi dz.$$
4.
*Spherical coordinates*
: $${
\b
f r} = (r,
\t
heta,
\p
hi).$$ Can be
used in three dimensions. Suitable for systems with spherical
...
...
@@ 245,6 +257,8 @@ We have discussed four different coordinate systems:
Infinitesimal distance:
$$ds^2 =r^2 (\sin^2 \theta d\phi^2 + d\theta^2) + dr^2 .$$
Infinitesimal volume:
$$dV = r^2 \sin(\theta) dr d\theta d\varphi.$$
Problems
========
...
...
@@ 279,11 +293,3 @@ Problems
\frac{\partial}{\partial \vartheta}\left( \sin\vartheta \frac{\partial\psi(r,\vartheta, \varphi)}{\partial \vartheta}\right).$$
This is however even more tedious (you do not have to show this).
4.
[:smirk:] A particle moves over a cone which is characterized
by the condition $rho = z$. Calculate the kinetic energy expressed
in terms of the cylindrical coordinates $z$ and $
\v
arphi$ and their
first order time derivatives.
5.
[:smirk:] The same question for a particle moving on a
spherical shell of radius $R$. (Use coordinates $
\v
arphi$ and
$
\v
artheta$.)