Axiomatic Newtonian Mechanics

February 9, 2017 Cornell University, Department of Physics

PHYS 4444, Particle physics, HW # 1, due: 2/2/2017, 11:40 AM

Question 1: Axiomatic Newtonian mechanics

In this question you are asked to develop Newtonian mechanics from simple axioms. We first define the system, and assume that we have point like particles that “live” in 3d space and 1d time. The very first axiom is the principle of minimal action that stated that the system follow the trajectory that minimize S and that Z S = L(x, x,˙ t)dt (1)

Note that we assume here that L is only a function of x and v and not of higher derivative. This assumption is not justified from first principle, but at this stage we just assume it. We now like to find L. We like L to be the most general one that is invariant under some symmetries. These symmetries would be the axioms. For Newtonian mechanics the axiom is that the that system is invariant under: a) Time translation. b) Space translation. c) Rotation. d) Galilean transformations (I.e. shifting your reference frame by a constant velocity in non-relativistic mechanics.).

Note that when we say “the system is invariant” we mean that the EOMs are unchanged under the transformation. This is the case if the new Lagrangian, L0 is the same as the old one, L up to a function that is a total time derivative, L0 = L + df(x, t)/dt.

1. Formally deﬁne these four transformations. I will start you oﬀ: time translation is: the equations of motion are invariant under t → t + C for any C, which is equivalent to the statement L(x, v, t + C) = L(x, v, t) + df/dt for some f. Answer:

b) Space translation: The EOM are invariant under x → x + C for any C, i.e.

df L(x, v, t) = L(x + C, v, t) + . (2) dt

1 c) Rotation: The EOM are invariant under φ → φ + C for any C (here φ is any angle), i.e. df L(x, v, t) = L(R(C)x,R(C)v, t) + . (3) dt d) Galilean transformation: The EOM are invariant under x → x + V t, where V is the relative velocity between any two frames and it does not change with time. I.e. df L(x, v, t) = L(x + Vt, v + V, t) + (4) dt

2. We ﬁrst consider a system with one particle. Let us make the stronger assumption that the Lagrangian itself is invariant under the symmetry transformations (a), (b), (c) 2 2 i (i.e. f = 0 for these transformations). Show that L(x, v, t) = L(v ) where v ≡ viv . Answer: Due to a) we know that L cannot be a function of t. Based on b) we know it is not a function of x, and based on c) we know it can only depend on the magnitude of the velocity, that is on v2.

3. Next we would like to show that L(v2) = C × v2. For this we ﬁrst consider an inﬁnitesimal Galilean transformation, v0 = v + ε, such that

L0 ≡ L[v02] = L[(v + ε)2]. (5)

Expand L0 around v2 to ﬁrst order in ε. Answer:

2 2 2 2 ∂L L[(v + ε) ] = L[(v + 2v · ε + ε )] = L(v ) + 2 2v · + ... (6) ∂v v2

4. Explain why the requirement of invariance under Galilean transformations implies that the ﬁrst order term in ε should be a total time derivative. Answer: Because we want to get the same EOMs in any reference frame. This is ensured if the Lagrangian changes only by the addition of a total time derivative under the symmetry transformation.

5. Show that the ﬁrst order term in ε is a total time derivative only when L(v2) is linear in v2, that is, that L(v2) = Cv2. Answer: We will require that:

∂L df 2 2v · ε = (7) ∂v v2 dt

2 First, recall that the functional form of f is f(x, t). It cannot depend on higher derivatives of x due to our assumption that the Lagrangian depends only on x, v, t, df since dt will introduce derivatives of v if f depends on v. (An alternative argument is to note that the action changes by f(xf , tf ) − f(xi, ti). Since we usually fix x at the start and end times, this change is the same for all paths, so it doesn’t affect the minimisation. On the other hand, v at the start and end points are not fixed, so this change is not the same for all paths, hence affecting the minimisation and the equation of motion!) Therefore: ∂L df(x, t) ∂f 2 2v · ε = = ∇f · v + (8) ∂v v2 dt ∂t Note that since the LHS is only a function of v, the RHS cannot be a function of x ∂f and t, so we demand that ∂t and ∇f be constants. Therefore, we know that the RHS ∂L can at most be a linear term in v plus a constant. This implies that ∂v2 can at most be a constant plus something proportional to 1/(ε·v). But the second term is also not allowed, because why should the partial derivative of L(v2) have anything to do with the arbitrary direction of a boost ? (You can also argue this point using rotational ∂L 2 invariance.) Therefore, we conclude that ∂v2 can only be a constant, so L(v ) is linear in v2.

6. We now move to a two body problem. We further assume that the kinetic term (that is, the term that depend on the velocity) is given by the free particle expression, and we then consider only the potential that depend only on positions. Based on this extra assumption, and on the four axioms, show that:

m v2 m v2 L = 1 1 + 2 2 − V (|x − x |). (9) 2 2 1 2

Answer: The kinetic term is obvious by this extra assumption.

Let’s start with a potential that only depends on position x1 and x2. We can always m1x1+m2x2 deﬁne two new variables X = , r = x1 − x2, and rewrite the potential as a m1+m2 function V (X, r). Translation invariance requires that V (X + C, r) = V (X, r), so V can only be a function of r. Rotation invariance requires that V (R(C)r) = V (r). This can only be true if V is a function of r · r, since this combination is invariant under rotation. Therefore, we

conclude that the only allowed potential is of the form V (|x1 − x2|).

3 7. Usually in mechanics we pose questions that start like “consider a particle in the following potential, V (x).” What is the approximation that is done by moving from the two body problem discussed above to a set-up of a particle in a potential? Answer: There are several cases in which we use it. In some two body problems we use generalized coordinates where one coordinate describes the free-body motion of the center of mass while the other is of a particle with a “reduced mass”. In other cases we assume that we do not care about some of the DOFs, e.g. in a case of a ball that hits the Earth, we assume that the Earth is massive enough that is not aﬀected by the ball.

Question 2: Brief review of relativity

In most of the course we will work in the “mostly negative” metric, and we will use c = 1 (we will also useh ¯ = 1, but it is not relevant for this question).

1. Consider two vectors aµ and bµ. Write down their dot product that is deﬁned as µ µν aµb ≡ g aµbν. Answer:

µ aµb = a0b0 − a1b1 − a2b2 − a3b3 (10)

2. The E-L equation of classical ﬁeld theory is given by ∂L ∂L ∂µ = . (11) ∂(∂µϕ) ∂ϕ Write it explicitly in 1 + 1 dimension. Answer:

∂L ∂L ∂L ∂0 + ∂1 = . (12) ∂(∂0ϕ) ∂(∂1ϕ) ∂ϕ

3. Consider the following physical quantities Aµν, bµ and cµ. Write down all Lorentz scalars you can make out of them that involve at most three quantities. You can assume that A is symmetric. Answer: Here I assume that A is symmetric. If A is not, there will be more possibilities involving swapping the indices of A.

4 µ • Using one quantity: Aµ .

µ µ µ µν µ 2 • Using two quantities: b bµ, c cµ, b cµ, AµνA ,(Aµ) . µ ν µ ν µ ν µρ ν • Using three quantities: Aµνb b , Aµνc c , Aµνb c , AµνA A ρ, any product of a scalar using one quantity with another scalar using two quantities.