# Elementary argument for conservation laws from symmetries *without* using the Lagrangian formalism

It is well known from Noether's Theorem how from continuous symmetries in the Lagrangian one gets a conserved charge which corresponds to linear momentum, angular momentum for translational and rotational symmetries and others.

Is there any elementary argument for why linear or angular momentum specifically (and not other conserved quantities) are conserved which does not require knowledge of Lagrangians? By elementary I mean, "if this is not so, then this unreasonable thing occurs".

Of course, we can say "if we want our laws to be the same at a different point in space then linear conservation must be conserved", but can we derive mathematically the expression for the conserved quantity without using the Lagrangian?

I want to explain to a friend why they are conserved but he doesn't have the background to understand the Lagrangian formalism.

• Have you tried using the symmetry of the metric? Commented Nov 6, 2018 at 10:15
• @safesphere No, I was not aware of that argument. Do you have any resources you can point me to to investigate on my own? Commented Dec 9, 2018 at 14:33

The answer is yes, the essence of Noether's theorem for linear and angular momentum can be understood without using the Lagrangian (or Hamiltonian) formulation, at least if we're willing to focus on models in which the equations of motion have the form $$m_n\mathbf{\ddot x}_n = \mathbf{F}_n(\mathbf{x}_1,\mathbf{x}_2,...) \tag{1}$$ where $$m_n$$ and $$\mathbf{x}_n$$ are the mass and location of the $$n$$-th object, overhead dots denote time-derivatives, and $$\mathbf{F}_n$$ is the force on the $$n$$-th object, which depends on the locations of all of the objects.

(This answer still uses math, but it doesn't use Lagrangians or Hamiltonians. An answer that doesn't use math is also possible, but it would be wordier and less convincing.)

The inputs to Noether's theorem are the action principle together with a (continuous) symmetry. For a system like (1), the action principle can be expressed like this: $$\mathbf{F}_n(\mathbf{x}_1,\mathbf{x}_2,...) = -\nabla_n V(\mathbf{x}_1,\mathbf{x}_2,...). \tag{2}$$ The key point of this equation is that the forces are all derived from the same function $$V$$. Loosely translated, this says that if the force on object $$A$$ depends on the location of object $$B$$, then the force on object $$B$$ must also depend (in a special way) on the location of object $$A$$.

First consider linear momentum. Suppose that the model is invariant under translations in space. In the context of Noether's theorem, this is a statement about the function $$V$$. This is important! If we merely assume that the system of equations (1) is invariant under translations in space, then conservation of momentum would not be implied. (To see this, consider a system with only one object subject to a location-independent force.) What we need to do is assume that $$V$$ is invariant under translations in space. This means $$V(\mathbf{x}_1+\mathbf{c},\mathbf{x}_2+\mathbf{c},...) = V(\mathbf{x}_1,\mathbf{x}_2,...) \tag{3}$$ for any $$\mathbf{c}$$. The same condition may also be expressed like this: $$\frac{\partial}{\partial\mathbf{c}}V(\mathbf{x}_1+\mathbf{c},\mathbf{x}_2+\mathbf{c},...) = 0, \tag{4}$$ where $$\partial/\partial\mathbf{c}$$ denotes the gradient with respect to $$\mathbf{c}$$. Equation (4), in turn, may also be written like this: $$\sum_n\nabla_n V(\mathbf{x}_1\,\mathbf{x}_2,\,...) = 0. \tag{5}$$ Combine equations (1), (2), and (5) to get $$\sum_n m_n\mathbf{\ddot x}_n = 0, \tag{6}$$ which can also be written $$\frac{d}{dt}\sum_n m_n\mathbf{\dot x}_n = 0.$$ This is conservation of (total) linear momentum.

Now consider angular momentum. For this, we need to assume that $$V$$ is invariant under rotations. To be specific, assume that $$V$$ is invariant under rotations about the origin; this will lead to conservation of angular momentum about the origin. The analogue of equation (5) is $$\sum_n\mathbf{x}_n\wedge \nabla_n V(\mathbf{x}_1\,\mathbf{x}_2,\,...) = 0 \tag{7}$$ where the components of $$\mathbf{x}\wedge\nabla$$ are $$x_j\nabla_k-x_k\nabla_j$$. (For three-dimensional space, this is usually expressed using the "cross product", but I prefer a formulation that works in any number of dimensions so that it can be applied without hesitation to easier cases like two-dimensional space.) Equation (7) expresses the assumption that $$V$$ is invariant under rotations about the origin. As before, combine equations (1), (2), and (7) to get $$\sum_n \mathbf{x}_n\wedge m_n\mathbf{\ddot x}_n = 0, \tag{8}$$ and use the trivial identity $$\mathbf{\dot x}_n\wedge \mathbf{\dot x}_n = 0 \tag{9}$$ (because $$\mathbf{a}\wedge\mathbf{b}$$ has components $$a_jb_k-a_kb_j$$) to see that equation (8) can also be written $$\frac{d}{dt}\sum_n \mathbf{x}_n\wedge m_n\mathbf{\dot x}_n = 0. \tag{10}$$ This is conservation of (total) angular momentum about the origin.

• A fascinating aside: If this doesn't work, then it says something about the universe -- either it is not well described using the equation you started with, or it does not have translational symmetry. I find thinking about this case and what that would imply about how our world would have to work gives me a great deal of appreciation for the fact that that equation and those symmetries sure seem to work! Commented Nov 6, 2018 at 17:58

Try the Hamiltonian formalism: If the symmetry generator $$Q$$ commutes with the Hamiltonian $$[Q,H]=0$$ then $$Q$$ is a conserved quantity.

• Sure, but that's QM. What about classical physics? Commented Nov 6, 2018 at 12:47
• This also works in classical mechanics where $[\cdot,\cdot]$ is the Poisson bracket. Commented Nov 6, 2018 at 12:48
• I guessed so, but since I'm used to the {} notation... haha. Anyways, you should add this to your answer. Commented Nov 6, 2018 at 12:50
• @Qmechanic Is it accurate to, in classical mechanics, interpret the Poisson bracket $\{f,H\}$ of two functions $f$,$H$ as follows? "$\{f,h\}$ measures the derivative of $f$ along the curve lines of $H$, meaning the partial derivative along the vector $X$ which is tangent to the curve lines of $H$ ($X$ being a Hamiltonian vector field)." This is the rough idea I got from reading CM but I don't know if it is rigorous. Since $X$ gives the time flow, essentially $f$ is conserved in time if it's constant along the curve lines of $H$, which are the constant energy trajectories in phase space. Commented Dec 9, 2018 at 19:51
• Yes, $\frac{df}{dt}=\{f,H\} +\frac{\partial f}{\partial t}$. Commented Dec 9, 2018 at 19:57

Is there any difference from saying that those are just the laws and there really isn't an explanation, that that's just how nature works because there aren't experiment where we ever observed non conservation of those quantities? Symmetry under the action of the Poincaré group implies those conservation laws, but I'm not sure you're giving a deeper explanation in this. We impose that physical system have those symmetries because we want something to be conserved, and how do you prove the correctness of the hypothesis of the existence of some symmetries? Performing experiments that show conservation of certain quantities. So we're back at the start, they are conserved because they are conserved, that is, experiments tells us they are conserved.

• The question is "can we derive mathematically the expression for the conserved quantity without using the Lagrangian?" Your answer doesn't address the question at all. It's not about the philosophy of why conservation happens. Commented Nov 6, 2018 at 22:29
• @Beanluc In the original post it ask: "Is there any elementary argument for why linear or angular momentum specifically (and not other conserved quantities) are conserved which does not require knowledge of Lagrangians?" And I gave an argument that does not require the use of Lagrangians, saying that there are conservation laws cause that's how it works. Commented Nov 7, 2018 at 13:46
• I agree that it is circular in the end, so rigorously speaking you cannot completely justifiy the symmetries. But, you can frame the question in different ways and this might give you insights. Lagrangians make it really easy to generalise but to me they are not very enlightening. The approach via potentials exposed by Dan is easier to understand, though why forces are represented by potential is basically an axiom, you cannot prove it but it's easier to swallow for me. Commented Dec 9, 2018 at 19:57

Suppose you have a system whose kinematics are completely characterized by a set of kinematic variables $$\left(q^a: 0 ≤ a < m\right)$$ for configuration and their time rates of change $$\left(v^a: 0 ≤ a < m\right)$$, where $$v^a = dq^a/dt$$. Suppose its dynamics is completely characterized by a set of dynamic variables $$\left(p_a: 0 ≤ a < m\right)$$ for momentum, and their time rates of change $$\left(f_a: 0 ≤ a < m\right)$$, with $$f_a = dp_a/dt$$.

Suppose further that the system's constitution can be described by specifying the dynamical variables $$p_a = p_a(q,v,t)$$ and $$f_a = f_a(q,v,t)$$ as functions of the kinematic variables and the time, such that The Principle Of Reciprocity holds: $$\frac{∂p_a}{∂v^b} = \frac{∂p_b}{∂v^a},\quad \frac{∂p_a}{∂q^b} = \frac{∂f_b}{∂v^a},\quad \frac{∂f_a}{∂v^b} = \frac{∂p_b}{∂q^a},\quad \frac{∂f_a}{∂q^b} = \frac{∂f_b}{∂q^a}.$$

As examples, you can cite any number of cases: gravitational dynamics, coupled harmonic oscillators, pendulums, conservative systems, etc., and merely calling out of The Principle Of Reciprocity in each case - showing that it's true in all of those cases - should, all by itself, be a major "Aha!" moment ... because that's actually where everything is coming from.

Suppose, now, that the dynamics have translational symmetry with respect to one of the configuration variables $$q^0$$, with $$m > 0$$. In infinitsimal form, this symmetry can be described as: $$δq^0 = ε,\quad δv^0 = \frac{d}{dt}δq^0 = \frac{dε}{dt} = 0,$$ with $$δq^a = 0,\quad δv^a = 0\quad (0 < a < m),\quad δt = 0.$$ Then, applying the symmetry to the dynamical variables and imposing invariance, we get: $$0 = δp_a = ε\frac{∂p_a}{∂q^0},\quad 0 = δf_a = ε\frac{∂f_a}{∂q^0},$$ from which it follows $$\frac{∂p_a}{∂q^0} = 0,\quad \frac{∂f_a}{∂q^0} = 0,$$ i.e. that all the dynamical variables are independent of $$q^0$$. By The Principle Of Reciprocity, it then follows that: $$\frac{∂f_0}{∂q^a} = \frac{∂f_a}{∂q^0} = 0,\quad \frac{∂f_0}{∂v^a} = \frac{∂p_a}{∂q^0} = 0,$$ so that $$f_0$$ is independent of $$(q,v)$$ and reduces to a function $$f_0 = f_0(t)$$ of $$t$$ alone. Since $$\frac{dp_0}{dt} = f_0(t),$$ then $$\bar{p_0} = p_0 - \int f_0(t) dt,\quad \frac{d\bar{p_0}}{dt} = \bar{f_0} = 0,$$ so, $$\bar{p_0}$$ is conserved. Replace $$\left(p_0,f_0\right)$$ by $$\left(\bar{p_0},\bar{f_0}\right)$$ and pretend that this was $$\left(p_0,f_0\right)$$ all along. The Principle Of Reciprocity is unaffected by the change, and the dynamic law continues to hold: $$d\bar{p_0}/dt = \bar{f_0} = 0$$. So, you have conservation of momentum for $$\bar{p_0}$$, which is now $$p_0$$.

Now, suppose the dynamics has rotational symmetry with respect to two configuration variables $$\left(q^0,q^1\right)$$, with $$m > 1$$ and the symmetry, given in infinitesimal form by $$δq^0 = -ω q^1\quad⇒\quad δv^0 = \frac{d(δq^0)}{dt} = \frac{d(-ω q^1)}{dt} = -ωv^1,\\ δq^1 = +ω q^0\quad⇒\quad δv^1 = \frac{d(δq^1)}{dt} = \frac{d(+ω q^0)}{dt} = +ωv^0,$$ with $$δq^a = 0,\quad δv^a = 0\quad (1 < a < m),\quad δt = 0.$$

Now, we pull a bait and switch. Make the following replacements: $$\left(\bar{q^0},\bar{q^1}\right) = (r,θ),\quad \left(\bar{v^0},\bar{v^1}\right) = \left(\frac{dr}{dt},\frac{dθ}{dt}\right)$$ where $$\left(q^0,q^1\right) = \left(r \cos θ, r \sin θ\right),\quad \left(v^0,v^1\right) = \left(\frac{dr}{dt} \cos θ - r \sin θ \frac{dθ}{dt}, \frac{dr}{dt} \sin θ + r \cos θ \frac{dθ}{dt}\right).$$ Then, the symmetry for $$\left(q^0,q^1,v^0,v^1\right)$$ can be rewritten as $$(δr,δθ) = (0,ω)$$, from which it follows: $$δ\bar{q^0} = 0,\quad δ\bar{q^1} = ω,\quad δ\bar{v^0} = 0,\quad δ\bar{v^1} = ω.$$

Now, define $$\left(\bar{p_0},\bar{p_1}\right) = \left(p_r,r p_θ\right),\quad \left(\bar{f_0},\bar{f_1}\right) = \left(f_r + p_θ\frac{dθ}{dt},r f_θ + p_θ\frac{dr}{dt} - p_r\frac{dθ}{dt}\right),$$ where $$p_r = p_0 \cos θ + p_1 \sin θ,\quad f_r = f_0 \cos θ + f_1 \sin θ,\\ p_θ = p_1 \cos θ - p_0 \sin θ,\quad f_θ = f_1 \cos θ - f_0 \sin θ.$$

This transform ensures that the dynamic law continues to hold $$\frac{d\bar{p_0}}{dt} = \bar{f_0},\quad \frac{d\bar{p_1}}{dt} = \bar{f_1},$$ and that the Principle Of Reciprocity continues to hold, i.e. that $$\frac{∂\bar{p_a}}{∂v^b} = \frac{∂p_b}{∂\bar{v^a}},\quad \frac{∂\bar{p_a}}{∂q^b} = \frac{∂f_b}{∂\bar{v^a}},\quad \frac{∂\bar{f_a}}{∂v^b} = \frac{∂p_b}{∂\bar{q^a}},\quad \frac{∂\bar{f_a}}{∂q^b} = \frac{∂f_b}{∂\bar{q^a}}\quad(0≤a<2≤b and $$\frac{∂\bar{p_0}}{∂\bar{v^1}} = \frac{∂\bar{p_1}}{∂\bar{v^0}},\quad \frac{∂\bar{p_0}}{∂\bar{q^1}} = \frac{∂\bar{f_1}}{∂\bar{v^0}},\quad \frac{∂\bar{f_0}}{∂\bar{v^1}} = \frac{∂\bar{p_1}}{∂\bar{q^0}},\quad \frac{∂\bar{f_0}}{∂\bar{q^1}} = \frac{∂\bar{f_1}}{∂\bar{q^0}}.$$

More directly, it ensures that $$f_0 dq^0 + f_1 dq^1 + p_0 dv^0 + p_1 dv^1 = \bar{f_0} d\bar{q^0} + \bar{f_1} d\bar{q^1} + \bar{p_0} d\bar{v^0} + \bar{p_1} d\bar{v^1}.$$

So, with Reciprocity, we can draw a similar conclusion, this time with respect to $$\bar{q^1}$$ and $$\bar{p_1}$$: everything is independent of $$\bar{q^1} = θ$$, and $$\bar{p_1}$$ is conserved, up to a $$(q,v)$$-independent function of $$t$$. Written out in terms of the original dynamical variables, it is: $$\bar{p_1} = r p_θ = r \left(p_1 \cos θ - p_0 \sin θ\right) = r \cos θ p_1 - r \sin θ p_0 = q^0 p_1 - q^1 p_0.$$