In the previous section we obtained the conservation of energy from the homogeneity of time. It is one of the most fascinating aspects of physics that for each symmetry there is an associated conservation law. So, if we know consider the homogeneity of space, which is the associated conservation law?
An homogeneous space means that if we globally translate all positions by a translation vector, the Lagrangian cannot be affected. For now, let's consider a small translation vector |ε>, for which positions |r> are transformed into |r>+|ε>.
Recall that a variation δ, called a virtual displacement, consisted in translate positions by a constant and small amount, while keeping time constant. Then, δL develops only with respect to changes in position. We consider δ|r>=|ε>, so
δL = ∑ (∂L/∂|rₐ>)·δ|rₐ> = |ε>∑∂L/∂|rₐ>.
Homogeneity of space implies that δL=0. But the vector |ε> is arbitrary, so this 0 cannot be due to |ε> being |0> or being perpendicular to the other vector. What this implies is that
∑∂L/∂|rₐ> = |0>.
This expression which is equal to |0> appears in E-L equations, so we rewrite them now as
∑(d/dt)(∂L/∂|vₐ>) = (d/dt)∑(∂L/∂|vₐ>) = |0>.
This clearly means that
∑∂L/∂|vₐ> = |constant> ≡ |p> = |momentum>.
This is how we define momentum and how we see that it is constant during motion.
If we use L = ∑(mₐ/2)vₐ² - U, we see that |p> = ∑mₐ|vₐ>.
The additivity of momentum is clear from the linearity of the operator on L that produces it.
In the case of momentum, the total momentum is equal to the sum of the momentum of each constituent particle. So that |p> = ∑|pₐ>.
This last property doest not apply to energy! Recall that the energy of the whole system is only the sum of the energy of each particle if the interaction between them can be neglected. For momentum, this is always valid, with interaction or not.
Notice something very important: we are converting concepts like energy and momentum into operators that operate on L.
Energy can be viewed as a quantity E and as an operator (∑̇qᵢ(∂/∂q̇ᵢ) - 1) that acts upon L.
Momentum can be as a vector quantity |p> and as an operator ∑∂/∂|vₐ> acting upon L. For the momentum, we also have the operator for a single particle (just drop the ∑).
All three components of |p> are conserved, but unlike the energy, momentum is not conserved with a uniform field. We need to be in a closed system, with no fields.
However, since momentum consists in three separate quantities, there are circumstances in which we have a field and one or two components of momentum are conserved while the others are not.
For example, consider a potential energy that depends on x and y but not on z. This means that the properties of the system cannot change upon a translation along z. This implies that the z component of momentum will be conserved. If a field depends on x only, then the y and z components of the momentum will be conserved.
Recall Newton's law
∂L/∂|rₐ> = -∂U/∂|rₐ> = |Fₐ>,
where |Fₐ> is the force suffered by particle a.
If momentum is conserved, ∑∂L/∂|rₐ> = |0>, which means that ∑|Fₐ> = |0>. Conservation of momentum is equivalent to the sum of the forces acting on a system being equal to |0>.
For a system of two particles, |F₁>+|F₂>=|0>, so |F₁> = -|F₂>, which means that the force suffered by 1 is equal in magnitude but opposed in direction to the force suffered by 2. This is Newton's 3rd law, of action and reaction. If 1 performs a force (the action) that 2 suffers, 2 will perform a reaction suffered by 1 in such a way that both action and reaction cancel out.
If we work with generalised coordinates qᵢ, then we will obtain *generalised momenta* pᵢ, with
pᵢ = ∂L/∂q̇ᵢ.
And, if we apply the operator ∂/∂qᵢ to L we obtain a generalised force Fᵢ= ∂L/∂qᵢ. We can relate both concepts as
ṗᵢ = Fᵢ.
In summary, ∂/∂qᵢ is the force operator and ∂/∂q̇ᵢ is the momentum operator (both need to be acting upon L).
While for Cartesian coordinates we obtain our familiar |p> = m|v>, for generalised coordinates this is not true in general.
A particle of mass m and moving with |v₁> moves in a very special space: half of it has a constant potential energy U₁ and the other half has U₂, also constant but of different value. Our particle moves in the first half and then enters the second half. Determine the change in direction due to the sudden change of U.
This problem is reminiscent of that problem in which a ray of light changes between two media with different refraction indices, so that we get Snell's law in which sinθ₁/sinθ₂ = n₂/n₁, where n's are the indices of refraction and θ's are the angles wrt to the surface normals.
In our case, the potential energy only depends on x if the surface between the two regions is parallel to the yz plane. This means that the momentum y and z components must be conserved in the process. Only the x component, px, will change.
We call the normal directions (incident and outgoing) as θ₁ and θ₂, respectively. The velocities are |v₁> and |v₂>.
region 1, U₁ yz (surface) region 2, U₂
| .'
| .'
| .'
| .'|θ₂
_____________________|.'__|__________________ x (normal)
| /|
θ₁| / |
|/ |
/ |
/ |
/ |
/ |
Since p1_yz = p2_yz, and since mass does not change in the process, v1_yz=v2_yz, so v1·sinθ₁=v2·sinθ₂. This is already Snell's law, if you consider that v2/v1 = n1/n2.
But in this case we can go beyond and see that energy is conserved, so
mv₁²/2 + U₁ = mv₂²/2 + U₂.
This establishes a relation between v1 and v2:
v₂ = ( v₁² + 2(U₁-U₂)/m )¹⸍².
Then,
sinθ₁/sinθ₂ = ( 1 + 2(U₁-U₂)/(mv₁²) )¹⸍².
Could energy conservation be applied to the Snell's law case, in order to get a relation between n₁ and n₂?