Sometimes, conservation of momentum and energy are enough to calculate important processes in mechanics. When this happens, since the particular type of interaction does not appear in the conservation laws, the result of those processes is independent of the interaction law.
We consider here a spontaneous disintegration of a particle. By this we mean a particle that suddenly becomes fragmented into several (two for now) sub-particles without the intervention of any external force.
The most simple frame to choose is one in which we see the particle at rest before disintegrating. Then, since in absence of external forces momentum must be conserved, after the disintegration we still have a total momentum equal to |0>. This implies that the two particles must move with momenta of equal magnitude and opposite directions.
We can find the magnitude of these momenta with the help of conservation of energy. We call Eᵢ to the internal energy of the original particle, and E₁ᵢ,E₂ᵢ to the internal energies of the resulting particles, so
Eᵢ = E₁ᵢ + p₀²/(2m₁) + E₂ᵢ + p₀²/(2m₂),
where p₀ is the magnitude of the resulting momentum.
The disintegration energy ε is ε = Eᵢ - E₁ᵢ - E₂ᵢ > 0. What is the meaning of this energy? See how the initial system is at rest, so everything is internal energy. The two resulting particles are systems on themselves, each with an internal energy. So if there is a difference between Eᵢ and E₁ᵢ+E₂ᵢ is due to these two systems 1 and 2 moving as a whole. It is important to see that there is no internal motion here acquired by the explosion, so that all the difference in internal energies is devoted to the kinetic terms of the resulting particles as a whole. If part of the disintegration energy would be used to increase the internal motion in one particle (its temperature), then the difference between internal energies would not be equal to the disintegration energy. So
ε = Eᵢ - E₁ᵢ - E₂ᵢ
is only true when neither 1 nor 2 have acquired internal kinetic energy of their components as the result of the explosion. The energy ε is completely devoted here to the increase of the kinetic energy of the systems 1 and 2 as whole.
If the primary particle would explode and produce two resulting particles without any net motion for them, but instead increasing the temperature of the two, then, all the initial internal energy is still internal energy at the end, and Eᵢ-E₁ᵢ-E₂ᵢ would be 0.
Then, we can write ε as a function of p₀,
ε = p₀²/(2) · (1/m₁ + 1/m₂) = p₀²/(2m),
where m is the reduced mass. Remember that 1/m = 1/m₁ + 1/m₂. With this we determine p₀ once we know ε, and then, velocities are as easy as v₁₀=p₀/m₁ and v₂₀=p₀/m₂. You may wonder about the reason of the subscript 0 in v₁₀. We just want to characterise this frame with this subscript, in contrast to what we do next.
We now change to a frame in which the primary particle moves with velocity V before the "explosion". We call this frame the *laboratory system* or L. In a lab, we would always see the particle moving before the event, so the name is easy to remember.
Laboratory system or L: the original particle moves with V. We use no subscript for L.
Centre-of-mass system or C: the original momentum is |0>. We use subscript 0 for C.
It is clear that
|v> = |V> + |v₀> |v> - |V> = |v₀>.
We square the latter expression to get
v² + V² - 2<v|V> = v₀² v² + V² - 2vVcosθ = v₀²,
where θ is the mutual angle of |v> relative to |V>, or vice versa. The latter equation gives a relation between the velocity v and its direction in the L system.
First, consider the vector relation |v> = |V> + |v₀> in a graphical sense. We can draw:
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We always draw |V> horizontally (in the lab, this makes sense). See how θ is the angle between |v> and the horizontal direction given by |V>. Also, θ₀ as the angle of |v₀> wrt to the same horizontal.
Next, consider that since (once ε is fixed), we have a fixed v₀, we can consider |v₀> as a vector defining the points in a circle, since no matter its direction, its magnitude will be v₀. We can draw
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This is a circle of radius v₀ in which vector |v₀> has its base at the centre of the circle. Then, at point A we have the bases of |V> and |v>.
In fact, more than a circle, it should be a sphere if we work in 3d, but the circle representation is very useful, since the two outgoing particles always form a plane, and the circle can be in such a plane.
Considering again v² + V² - 2vVcosθ = v₀², we see that this drawing is for the particular case V < v₀, since the vector V is drawn with smaller magnitude than the radius, so |V> has its base inside the circle. See how, for this case, we can obtain any value of θ. The outgoing particles can be travelling in any direction.
Imagine the original particle moving horizontally to the right (according to |V>) and then exploding in two. The resulting particles will travel in opposed directions with the same momentum. For V < v₀ we can have the resulting pair along any possible direction.
By contrast, if V > v₀, the base of |V> must lie outside the circle:
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In this figure we see how the base of |V> (A), is outside the circle. Then, not all values of θ are possible. If we reduce the θ initially drawn, we see that there is a maximum value for θ for which |v> becomes tangent to the circle. Consequently, for V > v₀, the outgoing particle can only go forward. We see from trigonometry that
sin(θₘₐₓ) = v₀/V.
We can write the relation between θ and θ₀:
tanθ = v₀sinθ₀ / (V + v₀cosθ₀).
We can develop this equation as
tanθ·(V + v₀cosθ₀) = v₀sinθ₀ tan²θ·(V² + v₀²cos²θ₀+2Vv₀cosθ₀) = v₀²(1-cos²θ₀) v₀²cos²θ₀(tan²θ + 1) + tan²θ·(V² + 2Vv₀cosθ₀) = v₀² v₀²cos²θ₀(tan²θ + 1) = v₀² - tan²θ·(V² + 2Vv₀cosθ₀) cos²θ₀(tan²θ + 1) = 1 - tan²θ·(V² + 2Vv₀cosθ₀)/v₀² cos²θ₀/cos²θ = 1 - tan²θ·(V² + 2Vv₀cosθ₀)/v₀² cos²θ₀ = cos²θ - sin²θ·(V² + 2Vv₀cosθ₀)/v₀² cos²θ₀ + cosθ₀(2Vsin²θ/v₀) = cos²θ - sin²θ·V²/v₀² cos²θ₀ + cosθ₀(2Vsin²θ/v₀) - cos²θ + sin²θ·V²/v₀² = 0 cosθ₀ = -(V/v₀)·sin²θ ± (V²sin⁴θ/v₀² + cos²θ - sin²θ·V²/v₀²)¹⸍² cosθ₀ = -(V/v₀)·sin²θ ± (V²sin²θ/v₀²[sin²θ - 1] + cos²θ )¹⸍² cosθ₀ = -(V/v₀)·sin²θ ± cosθ( 1 - (V/v₀)²sin²θ )¹⸍².
From the latter expression we clearly distinguish the two cases, V>v₀ and V<v₀. For V<v₀ the relation between θ and θ₀ is one-to-one, and we need to take the + solution only, so that we get θ₀=0 for θ=0. But for V>v₀, a single value of θ corresponds to two different values of θ₀. These two values correspond to the points marked as B and C in the last figure. The + and - solutions are valid here.
If we have the disintegration of not only one but many similar particles, we will have even more resulting particles, and we want to know their distribution of direction, energy, etc.
We assume that the primary particles have a random orientation in space, so on average we have an initial isotropic distribution.
First we solve this problem in the C system.
The directions of motion of the resulting particles are isotropically distributed, as a consequence of the random orientation of the primary particles.
We need to use the concept of solid angle here. For a circle, an angle of 2π radians covers an entire turn, so that the circle is completely enclosed. But what about a 3d space? How do we enclose a point? By covering it by a closed surface that intercepts all radial directions emanating from our central point. Here, instead of a radian, an angle coming from a curve enclosing a point in 2d, we use *steradians* (sr), which measure the *solid* angle enclosed by a *surface*.
In short: radians (rad) measure angles covered by curves, and steradians (sr) measure solid angles covered by surfaces.
A whole turn for a curve covers 2π rad, but in general, for a given radius, and since the arc dl is r·dθ, we have that dθ = dl/r. You can think this as the perimeter covered by a arc of a circle of radius 1.
A whole sphere for a surface covers 4π sr. You can think this as the area covered by a spherical surface of radius 1. Since in 3d, with spherical coordinates, we use two angles ϕ and θ, we can write
dΩ = dϕ·sinθ·dθ,
where dΩ is the increment of solid angle Ω, and where ϕ runs from 0 to 2π and θ runs from 0 to π. If we consider that there is azimuthal symmetry, then we can write dΩ = 2π·sinθ·dθ.
Now, in our C system, in which we will write quantities with a 0 subscript, we can count the number of particles entering a solid angle element dΩ₀. We need to imagine a fictitious spherical surface and think how particles come inside this sphere or go out from it. Then, we imagine how particles enter and scape the sphere but only through a given patch of the spherical surface. A very small patch with solid angle dΩ₀ = 2π·sinθ₀·dθ₀. If the patch were the entire spherical surface, then ΔΩ₀ would be 4π. This means that the *relative* fraction of particles entering a solid angle dΩ₀ wrt the total solid angle is dΩ₀/(4π), so the fraction is
dΩ₀/ΔΩ₀ᵀ = dΩ₀/(4π) = (1/2)·sinθ₀·dθ₀.
This fraction is also the fraction of all particles given by dΩ₀, and in this case, the fraction of particles being with a given θ₀. So we have the distribution of particles as a function of the latitude angle θ₀.
The corresponding distribution for the lab system is obtained by using this result and perform a transformation of frame. Again we have |v> = |v₀> + |V>, and we square it to obtain v² = v₀²+V²+2v₀Vcosθ₀. Since we consider the "equator" defined by the direction of |V>, the angle of |v₀> wrt |V> is given by θ₀. (See our two big diagrams above.)
This squared expression can be rewritten as
( v²-v₀²-V² ) / (2v₀V) = cosθ₀ d(cosθ₀) = d(v²-v₀²-V²) / (2v₀V) => d(cosθ₀) = d(v²) / (2v₀V) = sinθ₀·dθ₀ = 2·dΩ₀/(4π),
since v₀² and V² are killed by the d operator. V is clearly killed by d since it is constant, but what about v₀? Why is it constant? Since we have isotropy in the C system, the value v₀ cannot depend on direction. And on the distance?
We use T = (m/2)·v², where m is either m₁ or m₂ (one of the two resulting particles). Then,
(1/m)dT / (2v₀V) = dΩ₀/(4π).
In other words, the distribution of particles is
dT / (2mv₀V).
While kinetic energy was the same for every direction in the case of the C system, in our L system we see that T depends on the angle θ. Specifically, Tmin = (m/2)·(v₀-V)² for θ=π and Tmax = (m/2)·(v₀+V)² for θ₀=0. But, since the prefactor in the distribution does not depend on the angle, the particles are uniformly distributed over this range.
If in the C system we see all particles with the same properties along every direction, now in the L system we break such a symmetry, because we are having a vector |V> that offers a privileged direction. It is natural, then, that if we look in the direction θ₀=0 (moving like |V>) we see particles with less kinetic energy. And if we move against |V>, with θ₀=π, then we see particles with greater kinetic energy. However, it is not trivial that the ratio of particles we see in a solid angle wrt to the total solid angle is independent of the direction.
What if a primary particle disintegrates into more than two parts? Then, although the conservation of energy and momentum still apply, the equations for those conservations cannot provide a big constraint on the numerous degrees of freedom that we get with so many resulting particles.
The freedom is so great now that even in the C system, the energies of the resulting particles do not have a fixed value. If you have a string pinned at x=0 and you need to form a round-trip path with it, if the string has length L, your only choice is to go to x=L/2 and come back, so you end up with two equal legs. However, if you have the freedom of performing three legs with the string, as describing a triangle, then you can form many triangles all having the same total perimeter L.
We can calculate an upper bound for the T of a chosen particle like m₁. Consider the system formed by all the resulting particles except the chosen one. The internal energy of such system (in the C system) is written as Eᵢ'. Then, the kinetic energy of 1 is T₁₀ = p₀²/(2m₁).
By the conservation of energy, Eᵢ = E₁ᵢ + Eᵢ' + p₀²/(2m). This means that the initial energy with the centre of mass at rest, is just the total internal energy. And this must be equal to the internal energies of the resulting particles plus the kinetic energy of each resulting particle, here condensed in a single term for the reduced mass.
Then, if mᵣ is the mass of the rest of resulting particles and M the total mass,
T₁₀ = p₀² / (2m) · (m/m₁) m/m₁ = 1 / (1 + m₁/mᵣ) = mᵣ/(mᵣ+m₁) = mᵣ/M T₁₀ = (Eᵢ - E₁ᵢ - Eᵢ') ·(mᵣ/M).
Then, T₁₀ is greatest when Eᵢ' is smallest. When is the smallest Eᵢ' achieved? When all resulting particles except the chosen are moving with the same velocity (the same vector velocity). It is as if the rest of the particles are behaving as a single resulting particle. Then, the difference between Eᵢ and E₁ᵢ+Eᵢ' is fully devoted to produce net motion of 1 and the system of the rest as whole bodies. Only then we can write ε=Eᵢ-E₁ᵢ-Eᵢ'. If the system of the other particles would have motion around their centre of mass, then, part of the difference of internal energies would be devoted to such net motion of particles around their centre of mass, and the last equality would not be valid.
Conclusion: T₁₀(max) = (mᵣ/M)·ε.
A particle disintegrates into two. Find the relation between θ₁ and θ₂ (angles of each resulting particle in the L system).
In the C system, their relation is so easy, since θ₁₀ + θ₂₀ = π.
Now recall
tanθ = v₀sinθ₀ / (V + v₀cosθ₀).
We can apply this to both angles in C. But first, we define θ₁₀ as just θ₀ and then θ₂₀ is just π - θ₀. Then, as
sin(π-θ₀) = -sin(θ₀-π) = sinθ₀ cos(π-θ₀) = cos(θ₀-π) = -cosθ₀,
we can write
tanθ₁ = v₁₀sinθ₀ / (V + v₁₀cosθ₀) tanθ₂ = v₂₀sinθ₀ / (V - v₂₀cosθ₀).
Here we just need to eliminate θ₀ and relate θ₁ with θ₂. We can divide both equations to get
tanθ₁/tanθ₂ = (v₁₀/v₂₀) ·(V - v₂₀cosθ₀)/ (V + v₁₀cosθ₀).
Recall that v₁₀/v₂₀ = m₂/m₁, and we define tanθ₁/tanθ₂ ≡ t₁/t₂, so
t₁/t₂ = (m₂/m₁) ·(V - v₂₀cosθ₀)/ (V + v₁₀cosθ₀) m₁t₁/(m₂t₂) = (V - v₂₀cosθ₀)/ (V + v₁₀cosθ₀) (m₁t₁)/(m₂t₂)·(V + v₁₀cosθ₀) = (V - v₂₀cosθ₀) [(m₁t₁)/(m₂t₂)·v₁₀ + v₂₀]·cosθ₀ = V( 1 - (m₁t₁)/(m₂t₂)) cosθ₀ = V( 1 - (m₁t₁)/(m₂t₂))/[(m₁t₁)/(m₂t₂)·v₁₀ + v₂₀].
Now we rewrite our two expressions to be able to cancel the cosines instead of the sines,
tanθ₁·(V + v₁₀cosθ₀) = v₁₀sinθ₀ tanθ₂·(V - v₂₀cosθ₀) = v₂₀sinθ₀ tanθ₁·v₁₀cosθ₀ = v₁₀sinθ₀ - V·tanθ₁ -tanθ₂·v₂₀cosθ₀ = v₂₀sinθ₀ - V·tanθ₂ ,
so we divide them,
-(t₁/t₂)·(m₂/m₁) = (v₁₀sinθ₀ - V·tanθ₁) / (v₂₀sinθ₀ - V·tanθ₂) -(t₁/t₂)·(m₂/m₁)·(v₂₀sinθ₀ - V·tanθ₂) = (v₁₀sinθ₀ - V·tanθ₁) sinθ₀·[-(t₁/t₂)·(m₂/m₁)·v₂₀ - v₁₀] = (-V·t₁ - V·t₂·(t₁/t₂)·(m₂/m₁)) sinθ₀ = (-V·t₁ - V·t₂·(t₁/t₂)·(m₂/m₁))/[(t₁/t₂)·(m₂/m₁)·v₂₀ - v₁₀]
In summary, we have for now
cosθ₀ = V( 1 - (m₁t₁)/(m₂t₂))/[(m₁t₁)/(m₂t₂)·v₁₀ + v₂₀]. sinθ₀ = (-V·t₁ - V·t₂·(t₁/t₂)·(m₂/m₁))/[(t₁/t₂)·(m₂/m₁)·v₂₀ - v₁₀]
Now we add the squares of these expressions to get 1,
1/V² = (m₂t₂ - m₁t₁)² / (m₁t₁v₁₀ + m₂t₂v₂₀)² + t₁²/m₁²·(m₁+m₂)²/(t₁m₂v₂₀ - t₂m₁v₁₀)².
This expression can be further simplified, but it is a long algebraic process.
Find the angular distribution of the resulting particles in the L system.
First we use this
cosθ₀ = -(V/v₀)·sin²θ ± cosθ( 1 - (V/v₀)²sin²θ )¹⸍².
To get sinθ₀·dθ₀ we differentiate (taking V and v₀ as constant)
sinθ₀dθ₀ = 2(V/v₀)·sinθcosθdθ ± sinθdθ/(1-(V/v₀)²sin²θ )¹⸍² ± ± cos²θV²/v₀²·sinθdθ/(1-(V/v₀)²sin²θ )¹⸍² = sinθdθ·[2V/v₀·cosθ ± 1/√·(1+V²/v₀²cos(2θ))].
For v₀ > V we take the positive solution, and then the distribution is
(1/2)·sinθdθ·[2V/v₀·cosθ + 1/√·(1+V²/v₀²cos(2θ))]
for 0 ≤ θ ≤ π and with √ = (1-(V/v₀)²sin²θ )¹⸍². See how the signs of dθ and dθ₀ are the same in the corresponding diagram.
For v₀ < V we must consider both the positive and negative solutions. In the corresponding diagram, we see that when dθ increases, the dθ₀ going to C increases, but the dθ₀ going to B decreases. Then, since we have
sinθ₀dθ₀ = sinθdθ·[2V/v₀·cosθ ± 1/√·(1+V²/v₀²cos(2θ))],
we must do the sum of the two (sub-)distributions
distr= sinθ₀dθ₀/2 + sinθ₀'dθ₀'/2
with non-primed corresponding to the + branch and primed to the - branch, and dθ₀'=-dθ₀, so we get
distr = sinθdθ·[1/√·(1+V²/v₀²cos(2θ))],
for 0 ≤ θ ≤ θₘₐₓ.
Determine the range of possible values of θ between the directions of motion of the two resulting particles in the L system.
Consider again diagram 1 but with now explicit θ₁₀=θ₀ and θ₂₀=π-θ₀. See how the two angles are considered positive. This means that the relative angle between v₁₀ and v₂₀ is θ₁₀+θ₂₀. Beware: the vector with angle θ₂₀ is not |v₂₀> but has its same direction. If m₁=m₂ then it is |v₂₀>.
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We know that
tanθ₁·(V + v₁₀cosθ₀) = v₁₀sinθ₀ tanθ₂·(V - v₂₀cosθ₀) = v₂₀sinθ₀
and we define θ ≡ θ₁ - θ₂. We know that
tan(θ₁+θ₂)=tanθ= (tanθ₁+tanθ₂)/(1-tanθ₁tanθ₂) = = [v₁₀sinθ₀/(V+v₁₀cosθ₀) + v₂₀sinθ₀/(V-v₂₀cosθ₀)] / [1 - v₁₀v₂₀sin²θ₀/(V+v₁₀cosθ₀)/(V-v₂₀cosθ₀)] = = [(v₁₀+v₂₀)Vsinθ₀+2v₁₀v₂₀cosθ₀] / [(V+v₁₀cosθ₀)(V+v₂₀cosθ₀) - v₁₀v₂₀sin²θ₀] = = V(v₁₀+v₂₀)sinθ₀ / (V²+V(v₁₀-v₂₀)cosθ₀-v₁₀v₂₀).
We differentiate wrt θ₀ and get the condition
cosθ₀ = V(v₂₀-v₁₀)/(V²-v₁₀v₂₀),
where the sign of the derivative comes from
(d/dθ₀)tanθ = ( cosθ₀(V²-v₁₀v₂₀) + V(v₁₀-v₂₀) ) / ()².
The formula for cosθ₀ can only be valid if the values of V, v₁₀, v₂₀ are such that they don't give a rhs greater than 1 or smaller than -1. If we set the cosine = 1, we get V = -v₁₀ and V = v₂₀, so we choose the latter. For cosine = -1, we get V = -v₂₀ and V = v₁₀ and again we choose the latter. Since for V=0 and for V → ∞ we get a valid cosθ₀ this means that the formula for cosθ₀ is only valid for V ≤ v₁₀ and V ≥ v₂₀.
This means that between v₁₀ and v₂₀ we cannot have an extremum, and then the function tanθ cannot change sign.
So, between v₁₀ and v₂₀ we get cosθ₀ between 1 and -1, which means that 0 ≤ θ₀ ≤ π.
We still need to analyse what happens in the two other cases.
Now we substitute the extremum value into tanθ and, after some steps,
tanθ' = V(v₁₀+v₂₀)sinθ₀ / (V²+V(v₁₀-v₂₀)cosθ₀-v₁₀v₂₀),
where θ' indicates the value for the extremum.
We can calculate sinθ' from here and get
sinθ' = V(v₁₀+v₂₀) / (V²+v₁₀v₂₀) = = (v₁₀+v₂₀) / (V + v₁₀v₂₀/V) = = (v₁₀/V + v₂₀/V) / (1 + v₁₀v₂₀/V²).
Consider the sign of th numerator: always positive. The denominator is always positive as well. So sinθ' is always positive.
Our last analysis is to consider
tanθ = V(v₁₀+v₂₀)sinθ₀ / (V²+V(v₁₀-v₂₀)cosθ₀-v₁₀v₂₀)
and evaluate its signs. The numerator is always positive, which means it nevers goes to 0. Now we focus on the denominator,
V²+V(v₁₀-v₂₀)cosθ₀-v₁₀v₂₀.
V² is always positive. Also, -v₁₀v₂₀ is always negative.
The term V(v₁₀-v₂₀) is always negative if we consider v₁₀ < v₂₀ (as we do). But it is multiplied by cosθ₀, which runs from -1 to 1.
If v₁₀ > V, then the competition V² - v₁₀v₂₀ is always negative. Since the numerator is always positive and it never goes to 0, tanθ will not change sign. Then, as for cosθ₀=0 we can see that the denominator is negative, it will be negative for every value of cosθ₀. Conclusion: tanθ is always negative for these conditions.
If v₂₀ < V, then the competition V² - v₁₀v₂₀ is always positive. Again, as tanθ will not change sign, it will be always positive.
After all these calculations, we can discuss all the ranges, always considering v₁₀ < v₂₀. We can consider three cases here, depending on the value of V wrt to v₁₀ and v₂₀. Also take into account that θ₀ goes from 0 to π:
1) if v₁₀ ≤ V ≤ v₂₀, which is already discussed: 0 ≤ θ₀ ≤ π.
2) if v₂₀ ≤ V, we have seen how tanθ is always positive, and we can see how the limits θ₀ going to 0 and π give tanθ=0. This leads to the conclusion that the extremum is a maximum. Here, as tanθ is positive, we can interpret θ=arctan(tanθ) in an direct way, since a positive tanθ here means that the angle θ is in the first quadrant. It cannot be in the third because we don't allow angles beyond π. Then, the range for the angle θ is 0 ≤ θ ≤ θ'.
3) if V ≤ v₁₀, we see that for θ₀ tending to 0 or π, tanθ always tends to zero, while tanθ being always negative. This means that the extremum is a minimum. Also, be careful when tanθ is negative, since here it means the angle being in the second quadrant. So the angles we are considering, are not ArcTan[tanθ] but θ = π - ArcTan[tanθ], and then the minimum is at π-θ'. Conclusion: the range is π-θ' ≤ θ ≤ π.