In the previous section we have calculated the period of oscillation T as a function of the energy E. The idea now is to use this result to deduce the knowledge of U(x) from T(E). The equation
T(E) = (2m)¹⸍²∫{x;x₁;x₂}dx / (E-U(x))¹⸍²
can be used if we consider T(E) as known and U(x) as unknown.
For this, we need that U(x) as a single minimum in the spatial region we consider. We don't deal with the question of what happens if there is more than one minimum.
We take the origin x=0 at the minimum U(x) and define U(0)=0.
|U
| U(x)
'. | .'
'. | .'
\| | .'
----+------------------------+-------U=E
|\ | ||
| \ | .'|
| `. | | |
| . | / |
| ·x₁(U) | x₂(U)/ |
| '. | / |
| `. | ,' |
| \ | / |
| `._ | ,' |
U=0 .........+..........::::..........+.............x
x₁ x₂
In the integral above we consider x as x(U), so dx = (dx/dU)·dU.
As we see in the figure, the function x(U) is two-valued, with branches x₁(U) and x₂(U). This means that the integral must be divided in two parts. If the region considered is from x₁ to x₂, then the first part is from x₁ to 0 and the second part from 0 to x₂.
If U is now the variable instead of x, me must adapt the limits of integration. In the first part we integrate from U=E to U=0 (branch 1) and in the second part we integrate from U=0 to U=E. See how the region between x₁ and x₂ is defined by a constant upper bound given by U=E.
The integral is now split in two parts,
T(E) = (2m)¹⸍²∫{U;E;0}(dx₁/dU)·dU/ (E-U)¹⸍² +
+ (2m)¹⸍²∫{U;0;E}(dx₂/dU)·dU / (E-U)¹⸍² =
= (2m)¹⸍²∫{U;0;E}·[(dx₂/dU) - (dx₁/dU)]·dU / (E-U)¹⸍².
Before we start performing tricks on this integral, we should review how to deal with limits of integration when we have a double integral.
Consider the integral
∫{y;0;1}∫{x;1;eʸ}f(x,y)dxdy.
There are two integrals wrt two variables x and y. When facing this, it is strongly recommended to draw a picture of the boundaries given by the integration limits.
y
| x=1
| |
| |
| |
| (1,1) y=1
1|-------+----------------.(e,1)
| |XXXXXXXXXXXX--' ·
| |XXXXXXXX,-' ·
| |XXXXX,' ·
| |XXX/ x=eʸ ·
| |X,' ·
| |,' ·
____|_______,________________·_________ x
O
1 e
The integral wrt y runs from y=0 to y=1, so we cut the plane between these two lines. Then, the integral wrt x runs from x=1, for which we can clearly draw a vertical boundary, to x=eʸ. This last boundary is more difficult. It means that we have to continue our integral until x=eʸ, but this is not very clear. It is better to think this as each value of y having a different x boundary. For y=0, the x boundary runs from x=1 to x=e⁰=1. For a bigger y, the boundary still starts at x=1 but ends a bit later, at x=eʸ. This continues up to the maximum value of y allowed, which is e=1, so the top x-domain runs from x=1 until x=e¹=e.
In this approach, it is clear that we first focus on the y-domain, and then we interpret the x-domain accordingly. But there is another equivalent way! We can first focus on the limits of x, which are x=1 and x=e. We draw two vertical lines to establish this boundary. And then we must focus on y, for which the y-domain will depend on the value of x. The upper bound of the y-domain is clearly given by y=1, so this acts as a ceiling. But the floor will depend on the value of x. For x=1, the y-domain (from top to bottom) runs from y=1 down to y=ln(1)=0, since y=ln(x) is the inverse of x=eʸ. For a higher value of x, the y-domain will run from y=1 down to y=ln(x) until we reach x=e, for which the y-domain runs from y=1 down to y=ln(e)=1, so the y-domain has zero length in this case.
Once we understand all this, we can finally do what we wanted: to swap the order of integration. Instead of writing
∫{y;0;1}∫{x;1;eʸ}f(x,y)dxdy.
we can now write
∫{x;1;e}∫{y;ln(x);1}f(x,y)dydx.
The integral inside is the one to be performed first, while the exterior one is second. With the change we have done, we can now integrate wrt y first.
The exterior integral, although is the last to perform, is the first to think in terms of boundaries.
Let's practise the same process, introducing now inequalities for regions, with the integral
∫{x;0;1}∫{y;x;1}exp(y²)dydx,
in which we need to integrate wrt to y first. But we want to change the order of the integrals, as before. The first thing to do is to draw the regions:
y
|
|
1 |______________________(1,1)
|XXXXXXXXXXXXXXXXXXXX,'|
|XXXXXXXXXXXXXXXXXX,' |
|XXXXXXXXXXXXXXXX,' |
|XXXXXXXXXXXXXX,' |
|XXXXXXXXXXXX,' |
|XXXXXXXXXX,' |
|XXXXXXXX,' ↖ |
|XXXXXX,' y=x |
|XXX,-' |
|X,' |
__,·_____________________|______ x
O
1
The integral wrt to x is here the last to be performed, which means that its limits are the first to be thought. Always follow this rule.
Focusing on the x-domain, we see that we always run from x=0 to x=1. We can write this as
0 ≤ x ≤ 1 .
Once we know about the "exterior" region, we think about the interior, the y-domain. The easy part is the top boundary, which is y=1. The bottom boundary is variable, given by y=x. So the y boundary runs from y=1 down to y=x. We can write it as
x ≤ y ≤ 1 .
See how these two inequality expressions are our limits of integration.
What if we want to change the order of integration? Then, the y integral is the exterior, and its limits must be thought first. We can write the inequalities for y as
0 ≤ y ≤ 1.
And then, the inequalities for x are
0 ≤ x ≤ y.
In other words, from
∫{x;0;1}∫{y;x;1}exp(y²)dydx,
we change to
∫{y;0;1}∫{x;0;y}exp(y²)dxdy.
∫{x;π/2;5π/2}∫{y;sinx;1}f(x,y)dydx
The initial regions are, in order from exterior to interior,
π/2 ≤ x ≤ 5π/2 sinx ≤ y ≤ 1 .
The figure is
y
|
| y=sinx
| '. .'
1 '''''''':'''''''''''''''';''''''''''y=1
| · \XXXXXXXXXXXXXX/ ·
| · \XXXXXXXXXXXX.' ·
| · \XXXXXXXXXXXX| ·
___|______·___\XXXXXXXXXX'___·________ x
| · \XXXXXXXX/ ·
| · \XXXXXXX/ ·
| x=π/2 \XXXXX/ x=5π/2
| `.X,'
-1| ··············'······· y=-1
If we want to swap x and y, the y integral will be the exterior, and its limits are needed first. They seem quite clear in this case:
-1 ≤ y ≤ +1.
What about the x limits? In this case we don't have a fixed limit, unlike the other two examples. Depending on the value of y, the x-domain changes its limits both from the left and from the right. For y=+1, the x-domain runs from x=π/2 to x=5π/2. By symmetry, for y=-1, the x-domain is just a point at x=3π/2. But we want this in general, for any value of y.
But inverting the function y=sinx is not trivial. We cannot just write x = arcsiny, the arcsin is multi-valued. Let's analyse the left branch first: from x=π/2 to x=3π/2. We can write a generic C₁ + C₂·arcsiny = x and then apply the two limit conditions. For x=π/2 we have y=1, so
C₁+C₂(π/2)=π/2
and for x=3π/2 we have y=-1, so
C₁+C₂(-π/2)=3π/2,
where we are considering the principal branch of arcsiny as bounded between -π and π.
Solving the system, we get C₁= π and C₂=-1, so we get x=π-arcsiny.
The right branch needs x= C₁ + C₂·arcsiny again, but this time with different conditions. To be precise, there is a common conditions, the one for the central point. It gives C₁+C₂(-π/2)=3π/2, as before. For the other point, we get C₁+C₂(π/2) = 5π/2. We get C₁=2π and C₂=1, so x=2π+arcsiny. We can conclude our region for x as
π - arcsiny ≤ x ≤ 2π + arcsiny.
Not very easy! But this has trained us to face our integral, which we rewrite here
T = (2m)¹⸍²∫{U;0;E}·[(dx₂/dU) - (dx₁/dU)]·dU / (E-U)¹⸍².
The trick that we use here is very subtle: we divide the whole equation by (α-E)¹⸍² and integrate (exterior integration) wrt E from E=0 to E=α.
These operations seem quite arbitrary. What is the reason behind them? Let's first obey these "rules" and after that asking about them, once we see what we achieve by applying them.
We get
∫{E;0;α} T·dE / (α-E)¹⸍² =
= (2m)¹⸍²∫{E;0;α}∫{U;0;E}·[(dx₂/dU) - (dx₁/dU)]·dU / (E-U)¹⸍² · dE / (α-E)¹⸍².
Now we change the order of integration, a step for which we are more than ready! Let's draw the regions:
E
|
|
|
|
α|·················(α,α)
|xxxxxxxxxxxxx ,'·
|xxxxxxxxxxxx,' ·
|xxxxxxxxxx,' ·
|xxxxxxxx,' ·
|xxxxxx,' E=U ·
|xxxx,' ·
|xx-' ·
__|·_______________·____ U
O α
The integral, as given, has the E integral as exterior, so their inequalities are
0 ≤ E ≤ α .
And then, the region for the interior integral (U) is given by
0 ≤ U ≤ E.
In order to change the order of integration, we need the U integral as the external, so we think its region first, which is
0 ≤ U ≤ α.
Once we know this, we can think the region for the interior integral (E), as obtain
U ≤ E ≤ α.
In other words, we change from ∫{E;0;α}∫{U;0:E}·f·dU·dE to ∫{U;0;α}∫{E;U;α}·f·dE·dU.
Now it is time to think about the functions that are inside, here summarised as "f". The first integral *must* be the interior one, in this case the E integral. The function to be integrated must contain all the factors containing E, but we can prefactor those that do not contain E. For example, the factor [(dx₂/dU)-(dx₁/dU)] does not contain E, so it does not need to be in the interior integral. The other two factors, both at the denominator, contain E, so our interior integral is
∫{E;U;α}dE / [(α-E)·(E-U)]¹⸍².
According to the authors, this integral is "elementary". Maybe it is not the most difficult integral to face, but it requires some steps. Let's go one by one. We seem to have joined the club of the integrals with denominators containing square roots! These are important integrals in physics and we have to master them.
Once again, our target is to convert the argument of the square root into 1 - (something)², so that we can make this something a sine of some angle.
First, let's multiply the two binomials
(α-E)·(E-U) = -E² + (U+α)E - αU.
We set this polynomial in E equal to 0 and solve for E, obtaining
E = (U+α)/2 ± (U-α)/2 ≡ p ± q.
Once we have E as a function of p and q, notice that argument of the square root is
-(E-p + q)·(E-p - q) = q² - (E-p)².
The function to integrate becomes
1 / [q² - (E-p)²]¹⸍².
We are almost there. We need to divide up and down by q and obtain
(1/q) / [1 - (E-p)²/q²]¹⸍².
So our "something" is ε ≡ (E-p)/q. This needs a total derivative to deal with dE and dε,
dε = dE/q ,
so the integral becomes
∫{E;U;α}(dE/q) / [1 - (E-p)²/q²]¹⸍² =
= ∫{ε;(U-p)/q;(α-p)/q}dε/[1-ε²]¹⸍².
Our next step is clearly
ε ≡ sinϕ => dε = dϕ·cosϕ.
The integral is now,not considering the integration limits yet,
∫dϕ·cosϕ/cosϕ = ∫dϕ = ϕ.
The initial value is ε = (U-p)/q = (U - [U+α]/2) / (U-α)/2 = 1.
The final value is ε = (α-p)/q = (α - [U+α]/2) / (U-α)/2 = -1.
Then, the initial ϕ is arcsin(1) = π/2 and the final value is arcsin(-1)=-π/2.
So, final ϕ minus initial ϕ gives -π, which is the value of our integral.
Since we are calculating a period, we can disregard the sign, and take it as π. This π goes out as a prefactor for now.
Now we need to perform the external integration, which is "trivial"
∫{U;0;α}dU·[(dx₂/dU)-(dx₁/dU)] = [x₂-x₁]|{U;0;α} = x₂(α)-x₁(α),
where we have used that x₂(0)=x₁(0)=0.
Finally, our result is
∫{E;0;α}T(E)·dE / (α-E)¹⸍² = π·(2m)¹⸍²·[x₂(α)-x₁(α)].
Now we proceed with another step in the trick, which is to substitute α by U. This step sheds light on the reason for these tricks. We can write
x₂(U)-x₁(U) = [1/(2mπ²)¹⸍²·∫{E;0;U}T(E)·dE/(U-E)¹⸍².
Before discussing this result, let's review the tricks we have used. We had T(E) as a U-integral. Then we divided all by (α-E)¹⸍² with the double goal of obtaining an integrable integrand and of changing the order of integration that would leave α as a parameter that we could later fix as we wanted (α=U).
Conclusion: if we know T(E) we can use this to know x₂(U)-x₁(U). But this does not fully determine U(x), since there are infinite U(x) which will satisfy the given difference x₂-x₁. Imagine a curve y(x) and we invert it as x(y), which is multi-valued. So for a given y we get two values x₁(y) and x₂(y). Imagine we are told the value of x₂(y)-x₁(y), but what we want is to know y(x).
We give increasing values of y and we annotate x₂-x₁ for each y. We clearly lack information. If we knew where x₁ was, then we could locate x₂ as well, but it is not the case. We can, however, suppose that our function is symmetrical about the y axis, and then x₂(y)=-x₁(y).
With U(x) we do the same: we assume that U(x) is symmetrical about the U axis, and then x₂(U)-x₁(U) = 2x₂(U) ≡ 2x(U). We can then complete our result as
x(U) = [1/(8mπ²)¹⸍²·∫{E;0;U}T(E)·dE/(U-E)¹⸍².