Description
1. For an object of mass M at a radial distance R, the dimensionless quantity
GM
Rc2 (1)
indicates when relativistic gravity becomes important as this quantity takes on values
near unity.
a) Working in SI units, show that the aforementioned relativistic quantity is in fact
dimensionless.
b) Knowing that c = 2.998 ⇥ 108 m/s and G = 6.67 ⇥ 1011 Nm2/kg2, calculate the
value of the above dimensionless relativistic quantity, to three significant figures,
i. on the surface of the Earth.
ii. on the surface of the Sun.
The Schwarzschild radius is the characteristic length scale for curvature in the Schwarzschild
geometry (the geometry exterior to a static, spherically symmetric object of mass M) and
is calculated via the expression
RS = 2GM
c2 . (2)
c) Calculate the Schwarzschild radius (in meters) of
i. the Earth.
ii. the Sun.
d) Calculate the dimensionless ratios RS,/R and RS,/R where RS is the
Schwarzschild radius for each respective body and R, R are the radii of the Sun,
Earth. Is the Schwarzschild radius, for each respective object, smaller or larger than
the radius of each respective object?
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2. In Cartesian coordinates (x, y), the square of the infinitesimal distance between two
neighboring points is of the form
ds2 = dx2 + dy2
. (3)
Similarly, in polar coordinates (r, ) the square of the distance between two points
separated by an infinitesimal amount is of the form
ds2 = dr2 + r2
d2
. (4)
Since the two coordinate systems are simply di↵erent ways of labeling points in a plane,
these two line elements must be equivalent. The coordinate transformation linking these
two coordinate systems are
x = r cos
y = r sin . (5)
a) Using Eq. (5), calculate the di↵erentials dx and dy in terms of r and .
b) Substitute these di↵erentials into Eq. (3) and simplify to arrive at the line element
given in Eq. (4). Notice that since Eq. (4) can be arrived at from Eq. (3) through a
simple coordinate transformation, this implies that the geometries are equivalent.
3. The line element in 3D spherical-polar coordinates (r, ✓, ) is of the form
ds2 = dr2 + r2
(d✓2 + sin2 ✓d2
). (6)
If we restrict this line element to reside on the surface of a two-dimensional sphere of
radius a, where r = a and dr = 0, the aforementioned line element reduces to
ds2 = a2
(d✓2 + sin2 ✓d2
), (7)
which equates to the expression presented in Eq. (2.15). In Cartesian coordinates (x, y, z),
the line element is of the form
ds2 = dx2 + dy2 + dz2
, (8)
2
with the constraint that
x2 + y2 + z2 = a2
, (9)
if the line element resides on the surface of the sphere of radius a.
The coordinate transformation linking these two coordinate systems are
x = a sin ✓ cos
y = a sin ✓ sin
z = a cos ✓. (10)
a) Using Eq. (10), calculate the di↵erentials dx, dy, and dz in terms of ✓ and .
b) Substitute these di↵erentials into Eq. (8) and simplify to arrive at the line element
given in Eq. (7).
4. Consider the axisymmetric line element
ds2 = a2
(d✓2 + f 2
(✓)d2
), (11)
with the surface specified by
f(✓) = sin ✓ cos2 ✓. (12)
a) Explicitly write the line element on this surface for constant ✓.
b) Calculate the circumference C(✓) of a circle of constant ✓ on this surface.
c) For what values of ✓ do we get a minimum or maximum circumference? Calculate
the circumference for these angles.
5. In the “Map Projections” handout, we focused on understanding the Mercator Projection,
which equated to a mapping of the surface of the Earth onto a 2D map where angles are
preserved. In this problem, we’re interested in understanding the Equal-Area Projection,
which equates to a mapping of the surface of the Earth onto a 2D map where areas are
preserved.
Warning: Do not attempt this problem until you have studied the “Map Projections”
handout!
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a) Construct the infinitesimal patch of area dAmap on the 2D map in terms of the
Cartesian coordinates x and y.
b) Construct the infinitesimal patch of area dAearth on the Earth, projected onto a 2D
map, in terms of the Cartesian coordinates x and y. Hint: you can find this quantity
in the handout!
c) By demanding that these infinitesimal patches are equal, construct the di↵erential
equation that yields y().
d) Integrate the aforementioned expression to find y(). Invert to find (y).
e) Lastly, plug these into (**) in the handout and show that the line element takes the
form
ds2 = 1
f 2(y)
dx2 + f 2
(y)dy2
, (13)
where
f(y) ⌘ a p4⇡2a4/L2 y2
. (14)
