Legendre polynomials, or more accurately, Legendre functions, are just solutions to Legendre’s differential equation – much as we (many earlier blogs ago) saw that Bessel’s functions were solutions to Bessel’s differential equation.
In this case, we are looking at:
d/dx [(1 – x^2) d/dx P_n(x)] + n(n + 1)P_n(x)
Solutions proceed on for n = 0, 1,2 3……..i.e. by substituting the appropriate value for n into the form:
P_n(x) = 1/ 2^n (n!) d^n/dx^n [(x^2 – 1)^n]
Which is the Rodrigues’ formula. In this way, the first two Legendre polynomials are found to be:
P_o(x) = 1 and P_1(x) = x
Let’s check each:
For P_o(x): we have
1/ (2^0)0! [x^2 – 1)^0] = 1/1 x 1 = 1
For P_1(x): we have:
1/ 2^1 (1!) d/dx [(x^2 – 1)] = ½ {2x} = x
Thus, once one follows the operations, obtaining each Legendre function is straightforward- however this doesn’t mean that it can’t get messy!
Legendre functions have multiple uses throughout mathematical physics, but I will only show one – as applied to the issue of perturbation in celestial mechanics.
Celestial mechanics is undoubtedly the prime precision science because it makes use of highly specialized functions and high speed computers to crank out results – say for locating the future position of a planet – that might have taken many hours back in the early 70s using the computers of the era.
Perturbations are important because they show how the positions of some given planet, are influenced by the gravitational attraction of another one. Unless one then reckons the differential location into computations, he will find a spacecraft destined for some destination planet will likely miss it entirely.
As an example, we take the illustration shown which is to quantitatively estimate the affect of Jupiter on the orbital position of Earth. The respective masses are: m1(Sun), m2(Earth) and m3 (Jupiter) and we assign relative radius vectors that approximate to the actual distance in AU or astronomical units where 1 AU = 149 million kilometers. Thus, r = 1.0 and r3 (for Jupiter) = 5. The Greek symbol DELTA defines the separation between Earth and Jupiter and is the key parameter for estimating the magnitude of the perturbation.
The angle S separating the r and r3 vectors can be anything but for working purposes maybe use S = 120 degrees, which yields a value: cos(S) = cos (120) = - ½. This will be useful for computing the first three Legendre polynomials – which factor into the kind of generic perturbations I am looking at.
The first three Legendre polynomials for this context are:
P_o = 1
In this case, we are looking at:
d/dx [(1 – x^2) d/dx P_n(x)] + n(n + 1)P_n(x)
Solutions proceed on for n = 0, 1,2 3……..i.e. by substituting the appropriate value for n into the form:
P_n(x) = 1/ 2^n (n!) d^n/dx^n [(x^2 – 1)^n]
Which is the Rodrigues’ formula. In this way, the first two Legendre polynomials are found to be:
P_o(x) = 1 and P_1(x) = x
Let’s check each:
For P_o(x): we have
1/ (2^0)0! [x^2 – 1)^0] = 1/1 x 1 = 1
For P_1(x): we have:
1/ 2^1 (1!) d/dx [(x^2 – 1)] = ½ {2x} = x
Thus, once one follows the operations, obtaining each Legendre function is straightforward- however this doesn’t mean that it can’t get messy!
Legendre functions have multiple uses throughout mathematical physics, but I will only show one – as applied to the issue of perturbation in celestial mechanics.
Celestial mechanics is undoubtedly the prime precision science because it makes use of highly specialized functions and high speed computers to crank out results – say for locating the future position of a planet – that might have taken many hours back in the early 70s using the computers of the era.
Perturbations are important because they show how the positions of some given planet, are influenced by the gravitational attraction of another one. Unless one then reckons the differential location into computations, he will find a spacecraft destined for some destination planet will likely miss it entirely.
As an example, we take the illustration shown which is to quantitatively estimate the affect of Jupiter on the orbital position of Earth. The respective masses are: m1(Sun), m2(Earth) and m3 (Jupiter) and we assign relative radius vectors that approximate to the actual distance in AU or astronomical units where 1 AU = 149 million kilometers. Thus, r = 1.0 and r3 (for Jupiter) = 5. The Greek symbol DELTA defines the separation between Earth and Jupiter and is the key parameter for estimating the magnitude of the perturbation.
The angle S separating the r and r3 vectors can be anything but for working purposes maybe use S = 120 degrees, which yields a value: cos(S) = cos (120) = - ½. This will be useful for computing the first three Legendre polynomials – which factor into the kind of generic perturbations I am looking at.
The first three Legendre polynomials for this context are:
P_o = 1
P_1 = cos (S)
P_2 = ½ (3 cos^2(S) – 1)
Note that by comparing P_1 above to P_1(x) earlier we can easily discern:
x <-> cos (S)
In other words, x in the Rodrigues’ formula is now replaced by cos (S) for the purposes of this perturbation estimate.
As we are perturbing an inner planet by an outer one, we are expanding (1/ Delta) using the Legendre polynomials. Computing the Legendre polynomials for an Angle S = 120 deg, one finds:P_o = 1 P_1 = - ½ and P_2 = -1/8
The perturbing function is then easily computed, viz.:
1/Delta = 1/ { 1[1 + (1/5)^2 – 2(1/5) (-½)}^1/2
And 1/Delta = 0.18
For next time: Assume r3 = 40 AU for Pluto and r = 0.4 AU for Mercury. Estimate the generic term perturbation of Mercury by Pluto, and also obtain 1/DELTA. Use S = 60 degrees.
Note that by comparing P_1 above to P_1(x) earlier we can easily discern:
x <-> cos (S)
In other words, x in the Rodrigues’ formula is now replaced by cos (S) for the purposes of this perturbation estimate.
As we are perturbing an inner planet by an outer one, we are expanding (1/ Delta) using the Legendre polynomials. Computing the Legendre polynomials for an Angle S = 120 deg, one finds:P_o = 1 P_1 = - ½ and P_2 = -1/8
The perturbing function is then easily computed, viz.:
1/Delta = 1/ { 1[1 + (1/5)^2 – 2(1/5) (-½)}^1/2
And 1/Delta = 0.18
For next time: Assume r3 = 40 AU for Pluto and r = 0.4 AU for Mercury. Estimate the generic term perturbation of Mercury by Pluto, and also obtain 1/DELTA. Use S = 60 degrees.
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