An important particular case of central fields is the field for which U is inversely proportional to the distance r. It is the case of Newtonian gravitation and Coulomb electrostatics. Let's consider the attractive case as
U = -α / r,
where α is a positive constant.
The effective potential becomes
Ueff = -α/r + M²/(2mr²).
While U = -α/r looks like
| U | | | | | |_________________________________________ r | _,.,---'''''''''''''''''' | _.-' | ,-' | ,' | ,' | / | / | | | / || ''
Ueff looks like
.. Ueff || || || |\ | | Ueff = -α/r + M²/(2mr²). | | | \ | | | `. | | | \ +-----+-------------------------------------- r | '. ,.---''''''' | `. ,,-' | | ,' | `. _,' | `-._____.-' | | '
As r → 0, Ueff explodes to ∞. For r → ∞ we get Ueff tending to 0. Ueff also exhibits a minimum value that we calculate as
dUeff/dr=0 => α/r² -M²/(mr³)=0 => => αr = M²/m => r = M²/(mα) => => Ueff(min) = -mα²/M² + M²/(2mM⁴)·m²α² => => Ueff(min) = -mα²/(2M²).
If we apply a "ceiling" E, we see that for E>0 we have infinite motion, while for E<0 we have a bounded motion.
In the previous section we have learned that the shape of the path ϕ(r) is given by
ϕ = C + ∫(M/r²)·dr / (2m[E-U] - M²/r²)¹⸍².
We plug U=-α/r here and then
ϕ = C + ∫(M/r²)·dr / (2m[E+α/r] - M²/r²)¹⸍².
W have now the task of solving this integral. Once again, we pursue an argument of the denominator's square root in the form of 1-(something)².
First we change y=1/r => dy=-dr/r². Then,
ϕ = C - ∫M·dy / (2m[E+αy] - M²y²)¹⸍².
We develop the argument of the square root:
2m(E+αy) - M²y² =
= -M²·y² + 2mα·y + 2mE.
-M²·y² + 2mα·y + 2mE = 0 =>
y = -1/(2M²)·(-2mα ± (4m²α² + 8mEM²)¹⸍²) =>
=> y = mα/M² ± (1/M²)·(m²α² + 2mEM²)¹⸍² =
|_____| |_______________________|
p q
= p ± q =>
=> 2m(E+αy) - M²y² = -M²·(y-p-q)·(y-p+q) =
= -M²·([y-p]² - q²) = M²·(q² - [y-p]²).
The integral is now
C - ∫Mdy / M / ( q² - [y-p]² )¹⸍² = = C - ∫dy / ( q² - [y-p]² )¹⸍².
Now we change z=y-p and dz = dy, so
C - ∫dz / ( q² - z² )¹⸍² = C - ∫ d(z/q) / ( 1 - (z/q)² )¹⸍² =
Now we change z/q = cosθ and then d(z/q) = -dθ·sinθ, and we write
C + ∫ sinθ·dθ / sinθ = C + ∫dθ = C + θ.
We have now to undo all the variable changes (and we choose the origin of ϕ in such a way that C = 0),
θ = arccos(z/q) = arccos[(y-p)/q] = = arccos[ (y - mα/M²)·M² / (m²α²+2mEM²)¹⸍² = = arccos[ (M/r - mα/M) / (m²α²/M² + 2mE)¹⸍² .
Now we do the following transformations
p = M²/(mα) p not a momentum! ε = (1 + 2EM²/(mα²))¹⸍².
This allows us to rewrite the solution as
p/r = 1 + ε·cosϕ .
Here, p is not a momentum, but the *semi-latus rectum* of the orbit and ε its eccentricity.
The latus rectum of a conic curve is the *chord* parallel to the *directrix* and passing through a focus. The half of the latus rectum is usually called the semi-latus rectum.
Recall that a chord of a curve is a straight line segment joining two points of the curve.
The directrix is a fixed line L such that we can define the conic curve as the locus of all points P whose distance to a fixed point (the focus or a focus) is ε multiplied by the distance from P to L.
In other words: we take a straight line L and a fixed point F, and we start finding the points for which the distance from F to L, multiplied by a constant ε, is constant.
Recall that the distance from/to a line like L means taking the perpendicular distance.
If ε=0 we have a circle, and then our construction is just a limit case.
For 0 < ε < 1 we have an ellipse, which has two foci.
For ε = 1 we have a parabola. It has one focus.
For ε > 1 we have a hyperbola. It has two foci.
This is not he best place to do a full account on conic sections, but it is a good opportunity to emphasise the importance of mastering conic sections in physics.
The equation
p/r = 1 + ε·cosϕ .
is the equation of a conic section with one focus at the origin.
For ϕ=0 we get p/r = 1+ε => r = p/(1+ε) which is the distance to the perihelion, or the point nearest to the origin.
The word perihelion means nearest to the Sun, and aphelion means closest to the Sun. Both names imply that the orbit is around the Sun (Helios).
For a planet orbiting around its primary body, the Sun or any other star, we call these points apsides, either periapsis or apoapsis.
For a body around the Earth (Gea), we say perigee and and apogee.
For a body around Jupiter, perijove and apojove.
For an exoplanet around its star, periastron and apoastron.
The generic terms are periapsis and apoapsis, although you can also hear pericentre and apocentre.
The word "focus" was coined by Kepler. Focus comes from the word fire, i.e. the Sun.
As we know that this is equivalent to two bodies interacting according to the law U=-α/r, we have *each* particle moving along a conic section, with one focus at the centre of mass of the two particles.
Since we had
ε = (1 + 2EM²/(mα²))¹⸍².
we see that for E<0, ε is < 1, and the orbit is an ellipse. The motion is finite.
According to analytical geometry, the major semi-axis a is
a = p / (1 - ε²) = α / (2|E|),
and it depends on |E| but not on M.
The minor semi-axis b is
b = p / (1-ε²)¹⸍² = M / (2m|E|)¹⸍²,
and it depends on both |E| and M.
The linear eccentricity c is
c = ε·a = (α² - 2|E|M²/m)¹⸍² / (2|E|).
We review some ellipse parameters here:
y
________________________________________ |
∧ _,,.,---'''' `'''---...__ |
| _,.-'' ``-.._ |
| ,-' ________|
b| ,-' ε = c/a = (1 - b²/a²)¹⸍² ∧ |-.
| ,' |p | `.
|/ centre | | \
∨.....F1..........................C.....................∨.....O........x
∧\ | | | /|
| \ |<----------c------------->|<------------c------------>| / |
| `._ | ∨_,' |
b| `.. | ,,' |
| `--._ | _.--' |
| `'--...___ | __,,.,--'' O=F2 |
∨_______________________`'''------+------'''' |
| | |
|<·············a·················>|<···············a···············>|
In simple words, 2a is the length of the ellipse, 2b is its width and 2c is the focus to focus distance. Notice that there is a Phytagorean relation here between these three quantities:
a² = b² + c².
The semi-latus rectum is the (perpendicular to the major axis) distance from a focus to the ellipse.
The minimum value for energy is, as we calculated at the beginning of this section,
Ueff(min) = -mα²/(2M²).
This makes ε=0, so the ellipse is a circle.
The minimum distance to the centre of the field (the focus F2) is
rmin = p / (1+ε) = a(1-ε)
and the maximum distance is
rmax = p / (1-ε) = a(a+ε).
So both rmin and rmax depend on |E| but not on M. They could be obtained as well by setting Ueff=E.
What about the period of revolution T in an elliptical orbit? We recall that
M = 2mḟ
where ḟ is the sectorial velocity. We integrate this equation wrt time from t=0 to t=T,
MT = 2mf,
where now f is the area of the orbit. The area of an ellipse is π·a·b, so we can write
T = 2π·a·b·m/M = 2π·α/(2|E|)·M/(2m|E|)¹⸍²·m/M = π·α·( m / (2|E|³) )¹⸍² => => T² = (π²·α²)·m / (2|E|³).
Since a is proportional to 1/|E| we see here how T² ∝ a³, which is Kepler's 3rd law again. It may be noted that the period depends only on the particle's energy.
If E ≥ 0 the motion is not bounded and becomes infinite. If E > 0 we have ε > 1 and the path is a hyperbola with one focus as the origin.
Hyperbolas have two symmetric branches. Here we show some properties in one branch only:
y
| a
-. |<------------------------->·
↗ `-. |<---------> ·
asymptote `-.| c · ·
|·. · ·
| `:-. · ·
| `.`-._· ·
| \ `._ ·
| `. · `.. ·
| \· `-. ·
| '. `-._ ·
| | `._·
---------------O----------+---------------C.
∧ |focus | _,-'centre
| | / _,-'
| | / ,,'
|p | ,' _,-'
| | ,' _.-'
| | ,-',-'
__∨____|;-'
,,,;-'
_.=;-' |
-' |
|
The distance from to focus to the periapsis is
rmin = p/(ε+1) = a(ε-1),
where a=p/(ε²-1) = α/(2E).
If E=0, then ε=1 and we have a parabola, with periapsis rmin = p/2.
If a particle starts from rest at infinity, it will follow a parabola.
In general, if we want the coordinates of the particle as functions of time, we need to integrate
t = ∫dr / ( 2[E+α/r]/m - M²/(m²r²))¹⸍².
For elliptic orbits, we use E=-|E| and then
t = ∫(m)¹⸍²·r·dr / ( 2[-|E|r²+α·r] - M²/m)¹⸍² = = ∫(m/2|E|)¹⸍²·r·dr / ( α·r/|E| - r² - M²/(2m|E|) )¹⸍² = *
Now we use
1/(2|E|) = a / α M²/m = α²(1-ε²) / (2|E|) = aα(1-ε²).
and then we get
* = (m·a/α)¹⸍²·∫r·dr / ( -r² + 2ar - a²(1-ε²) )¹⸍². = (m·a/α)¹⸍²·∫r·dr / ( a²ε² - (r-a)² )¹⸍².
Since we want the argument of the square root at the denominator to be of the form 1 - (something)², we do the change
cosξ = -(r-a)/(a·ε) => r = a(1 - ε·cosξ)
which has a total derivative
dξ·sinξ = dr / (aε) => => dr = aε·dξ·sinξ.
The integral is now
(m·a/α)¹⸍²·∫(1-εcosξ)·a·dξ = = (m·a³/α)¹⸍² · (ξ-ε·sinξ) + C
We choose time so that C=0, and then we obtain t(ξ) and r(ξ),
t = (m·a³/α)¹⸍² · (ξ-ε·sinξ) r = a(1 - ε·cosξ),
where for t=0 the particle is at the periapsis.
If we prefer to use Cartesian coordinates, since x=rcosϕ and y=rsinϕ, we use
p/r = 1 + εcosϕ => p = r + ε·x => => εx = p - r = a(1-ε²) - a(1-εcosξ) = = aε(cosξ - ε).
This implies that (as y² = r² - x²),
x = a(cosξ - ε) y = a·(1-ε²)¹⸍²·sinξ .
A complete revolution around the ellipse corresponds to Δξ = 2π. We know that ξ is the eccentric anomaly, which we have already discussed.
For hyperbolic orbits the process is entirely similar, beginning with
t = ∫dr / ( 2[E+α/r]/m - M²/(m²r²))¹⸍² => => t = ∫(m)¹⸍²·r·dr / ( 2[Er²+α·r] - M²/m)¹⸍² = = ∫(m/2E)¹⸍²·r·dr / ( α·r/E + r² - M²/(2mE) )¹⸍² = *
Now we use
1/(2E) = a / α M²/m = α²(-1+ε²) / (2E) = aα(-1+ε²).
and then we get
* = (m·a/α)¹⸍²·∫r·dr / ( r² + 2ar - a²(-1+ε²) )¹⸍². = (m·a/α)¹⸍²·∫r·dr / ( -a²ε² + (r+a)² )¹⸍².
Since we want the argument of the square root at the denominator to be of the form 1 - (something)², we do the change
coshξ = (r+a)/(a·ε) => r = a(-1 + ε·coshξ)
which has a total derivative
dξ·sinhξ = dr / (aε) => => dr = aε·dξ·sinhξ.
The integral is now
= (m·a/α)¹⸍²·∫a(-1+εcoshξ)·dξ·sinhξ / ( -1 + cosh²ξ )¹⸍² = = (m·a/α)¹⸍²·∫(-1+εcoshξ)·a·dξ = = (m·a³/α)¹⸍² · (-ξ+ε·sinhξ) + C
We choose time so that C=0, and then we obtain t(ξ) and r(ξ),
t = (m·a³/α)¹⸍² · (-ξ+ε·sinhξ) r = a(-1 + ε·coshξ).
If we prefer to use Cartesian coordinates, since x=rcoshϕ and y=rsinhϕ, with r² = x² - y²,
p/r = 1 + εcosϕ => => p = r + ε·x => => εx = p - r = a(-1+ε²) - a(-1+εcoshξ) = = aε(-coshξ + ε).
This implies that (as y² = -r² + x²),
x = a(-coshξ + ε) y = a·(-1+ε²)¹⸍²·sinhξ .
The parameter ξ runs from -∞ to +∞.
For a repulsive field, with α > 0, U = α/r.
The effective potential is
Ueff = α/r + M²/(2mr²).
This function decreases monotonically from ∞ to 0 as r goes from 0 to ∞.
E is always >0 and motion is always unbounded.
If we integrate ϕ(r) we obtain
p/r = -1 + ε·cosϕ
which is a hyperbola.
The results are similar:
rmin = p/(ε-1) = a(ε+1) r = a(εcoshξ + 1) t = (ma³/α)¹⸍²·(εsinhξ + ξ) x = a(coshξ + ε) y = a(ε²-1)·sinhξ.
But now the origin of Cartesian coordinates is not at the focus, but at the centre of the hyperbola!
There is an integral of motion which exists *only* in fields U=α/r (with α positive or negative).
The quantity is the following vector
|v> ⨯ |M> + α|r>/r.
To give a proof, we perform its total time derivative. Since
|M> = m|r>⨯|v> => |Ṁ> = m|ṙ>⨯|v> + m|r>⨯|v̇> => => m|v̇>⨯(|r>⨯|v>) + m|v>⨯(|ṙ>⨯|v>) + m|v>⨯(|r>⨯|v̇>) + α|v>/r - αṙ|r>/r².
We see that
ṙ = (d/dt)(<r|r>)¹⸍² = <v|r>/r
and
m|v>⨯(|ṙ>⨯|v>) = |0>.
We then have
m|v̇>⨯(|r>⨯|v>) + m|v>⨯(|r>⨯|v̇>) + α|v>/r - α|r>(<v|r>)/r³ = *
We have to apply now
|A> ⨯ (|B> ⨯ |C>) = -<B|A>|C> + <C|A>|B>.
This is like an associative property in which the vectors between parentheses jump out, the first with negative sign and the second with positive sign. The jumped vector multiplies with · with the vector outside, while the other vector inside remains there. First, |B> jumps outside as -<B| so we get -<B|A>|C>. Second, |C> jumps outside as <C| so we get <C|A>|B>.
Knowing this, we do
|v̇>⨯(|r>⨯|v>) + |v>⨯(|r>⨯|v̇>) = = -<r|v̇>|v> + <v|v̇>|r> - <r|v>|v̇> + <v̇|v>|r> = = 2<v|v̇>|r> - <r|v̇>|v> - <r|v>|v̇>.
We write our partial result as
2<v|v̇>|r> - <r|v̇>|v> - <r|v>|v̇> + α|v>/r - α|r>(<v|r>)/r³.
The key is now to see that m|v̇> = α|r>/r³, and then the whole expression cancels.
The direction of this special vector is along the major axis of the hyperbola, from the focus to the periapsis. Its magnitude is αε.
Find the time dependence of the coordinates x,y of a particle with E=0 moving in a parabola in a field U=-α/r.
Time is found as
t = ∫dr / (2[E-U]/m - M²/(m²r²))¹⸍² + C = = ∫dr / (2[α/r]/m - M²/(m²r²))¹⸍² + C = = ∫r·dr / (2αr/m - M²/m²)¹⸍² + C .
We need to solve this integral now. The trick here is to find a change such that the argument of the square root in the denominator is (something)². So we do
(something)² ≡ s² = 2αr/m - M²/m² => => 2αr/m = M²/m² + s² => => r = m/(2α)·(M²/m² + s²) => => dr = (m/α)·s·ds.
Our integral becomes
t = ∫m²/(2α²)·(M²/m² + s²)·s·ds / s + C = = m²/(2α²)·∫(M²/m² + s²)·ds + C = = M²/(2α²)s + m²/(2α²)·s³/3 + C.
Since p=M²/(mα), we can rewrite
r = m/(2α)·(M²/m² + s²) = = p/2·(1 + s²m/(2α)).
For simplicity, we define η² ≡ s²m/(2α), and then, setting C=0,
r = p/2·(1 + η²)).
Then,
t = (mp³/α)¹⸍²·(η/2)·(1+η²/3).
Since εx = x = p - r, we get
x =(p/2)·(1-η²)
and since x²+y²=r², we have
y = (r²-x²)¹⸍² = ... = pη.
The parameter η goes from -∞ to +∞.
Integrate the equations of motion for a particle in a central field U=-α/r² with α>0.
We need to consider the two integrals:
t = ∫dr / ( 2[E-U(r)]/m - M²/(m²r²) )¹⸍² + constant ϕ = ∫ (M/r²) · dr / ( 2m[E-U(r)] - M²/r² )¹⸍² + constant.
For our potential, we get
t = ∫dr / ( 2[E+α/r²]/m - M²/(m²r²) )¹⸍² + constant ϕ = ∫ (M/r²) · dr / ( 2m[E+α/r²] - M²/r² )¹⸍² + constant.
We can rewrite them as
t = ∫rdr ·(m/2)¹⸍² / ( Er² + α - M²/(2m) )¹⸍² + constant ϕ = ∫ (M/r) · dr·(m/2)¹⸍² / ( Er² + α - M²/(2m) )¹⸍² + constant.
We need to be concerned with E>0 and E<0. But we need to consider the sign of α - M²/(2m) as well. What cannot happen is that both E<0 and α - M²/(2m) < 0, since we get a negative argument of the square root. In order to be positive we can have three cases:
This means we need to integrate for the three cases. But in fact, cases 1 and 3 can be combined in one integration if we recall that cosh(iz)=cos(z) and cos(iz)=cosh(z).
The integrals are very similar to others we have already solved.
When a small correction δU(r) is added to the potential energy U=-α/r, the paths of finite motion are no longer closed, and at each revolution the perihelion is displaced through a small angle δϕ. Find δϕ for a) δU = β/r³ , b) δU = γ/r³.
Befire General Relativity, the planet Mercury was known to show a precession of its perihelion that could not be explained by Newtonian gravitation. Einstein's theory introduces (for Schwarzschild's metric), a correction factor of δU = -G(m₁+m₂)M²/(mc²)·(1/r³), so it is the case of b).
We know that the angle covered between rmin and rmax and back is
Δϕ = 2∫{r;rmin;rmax} (M/r²) · dr / ( 2m[E-U(r)] - M²/r² )¹⸍² .
We can perform a trick here, seeing that the factor (M/r²) can be obtained by differentiating the 1/()¹⸍² wrt M:
Δϕ = -2(∂/∂M)∫{r;rmin;rmax} · dr · ( 2m[E-U(r)] - M²/r² )¹⸍² .
Thanks to this trick, we avoid "spurious divergences".
Now we place U = -α/r + δU there:
Δϕ = -2(∂/∂M)∫{r;rmin;rmax} · dr · ( 2m[E + α/r + δU)] - M²/r² )¹⸍² .
Now, since δU is supposed to be a correction, we expand the expression as powers of δU, only for 1st order. The 1st order is easy, since for a revolution we get 2π. Then, we only write the correction δϕ
δϕ = (∂/∂M)∫{r;rmin;rmax}·dr·(2mδU) / (2m[E + α/r)] - M²/r² )¹⸍² = *
The integration is now along the unperturbed motion. We change the integration over r to integration over ϕ between ϕ=0 and ϕ=π. The trick is to see that the integral contains now r²dϕ, so we can write
* = (∂/∂M)[ (2m/M) · ∫{ϕ;0;π}r²·δU·dϕ ].
The case a) gives
δϕ = (∂/∂M)[ (2m/M) · β ·∫{ϕ;0;π}·dϕ ] =
= 2πmβ·(∂/∂M)(1/M) = -2πmβ / M² = -2πβ / (αp),
where in the last step we have used the semi-latus rectum p = M²/(mα).
The case b) gives
δϕ = (∂/∂M)[ (2m/M) · γ ·∫{ϕ;0;π}·(1/r)·dϕ ].
We clearly need r(ϕ), which is given by
p/r = 1 + ε·cosϕ,
which is the equation of a conic section with one focus at the origin. Then.
δϕ = (∂/∂M)[ (2m/M) · γ ·∫{ϕ;0;π}·(1/p + ε·cosϕ/p)·dϕ ] =
= 2mγ·(∂/∂M)[ (1/M)·(ϕ/p + ε·sinϕ/p)|{ϕ;0;π} ] =
= 2mγ·(∂/∂M)[ (1/M)·(π/p)|{ϕ;0;π} ] .
Now be careful not to take p out of the integral, since p = M²/(mα), and then
δϕ = 2πm²αγ·(∂/∂M)·(1/M³) = 2παγm²·(-3·M⁻⁴) = = -6παγm²/M⁴ = -6παγm² / (p²·m²α²) = = -6πγ / (αp²).