Every physics education hands you two rules early and never quite justifies them.
The first is conservation of energy: in an isolated system energy is never created or destroyed, only moved between forms — a pendulum trading height for speed, a bomb converting mass into heat and light. The second is conservation of momentum: with no external forces, the total momentum before a collision equals the total momentum after.
Both are true. Both are useful. Neither is explained. They arrive as facts about the world, on the same footing as "the sky is blue" — something you check, not something you understand.
They are not facts about the world. They are consequences of something simpler, and the connection is one of the most beautiful results in physics. Here it is in one sentence:
Energy is conserved because the laws of physics do not change from one moment to the next. Momentum is conserved because they do not change from one place to another.
That is Noether's theorem. This post derives it — not by analogy, not by assertion, but by an argument you can follow with nothing beyond first-year calculus. By the end you will have derived both conservation laws from scratch, checked them numerically, seen exactly where they fail, and understood why the result quietly runs most of modern physics.
There is no hand-waving in what follows and no step marked "it can be shown". The one worked example is carried all the way through with real numbers, and the two conservation laws are then measured on it to confirm the derivation did what it claimed.
The elementary route taken here — three points, two segments, one clever ordering — is due to a 2004 paper by Hanc, Tuleja and Hancová, cited at the end.
1. What a symmetry actually is
Ask most people to picture symmetry and they picture objects. An equilateral triangle rotated by \(120^\circ\) looks unchanged. A square needs \(90^\circ\), a pentagon \(72^\circ\). A circle is special: rotate it by any angle at all and nothing happens.
Discrete symmetries admit a fixed list of operations. The circle admits a continuous family — and that difference turns out to be everything.
Hermann Weyl's definition, which Feynman borrows in the Lectures, is the one to keep:
A thing is symmetrical if one can subject it to a certain operation and it appears exactly the same after the operation.
The move that makes this powerful is to stop applying it to objects and start applying it to laws. A law of physics is an equation. Doing something to an equation — shifting the coordinates, rotating the axes, sliding the clock — and finding the equation unchanged is a symmetry in precisely Weyl's sense.
Hold on to the triangle-versus-circle distinction. Noether's theorem has nothing to say about the triangle. It needs a knob you can turn continuously, and turn by an arbitrarily small amount.
2. A law with symmetries
Take Newton's law of gravitation. Two masses, separated by a distance $r$, attract each other with a force
$$F = \frac{Gm_1 m_2}{r^{2}}$$
To do anything concrete we need coordinates, so put the masses in a plane at
$$r^{2} = (x_2-x_1)^{2} + (y_2-y_1)^{2}$$
Everything the force depends on is packed into one number, $r$ — and $r$ is built entirely out of coordinate differences. That is the crack we are about to lever open.
Sliding the whole system sideways
Move both masses by the same amount $a$ in the $x$ direction:
$$x_1 \to x_1 + a, \qquad x_2 \to x_2 + a$$
Recompute the separation. The new horizontal gap is
$$(x_2+a)-(x_1+a) = x_2 - x_1$$
The $a$'s cancel. Not approximately — identically, for any $a$, large or small.
A rigid shift changes both coordinates and therefore changes neither difference. \(r' = r\), so \(F' = F\): a spatial translation symmetry.
Turning it
The same thing happens under rotation, and it is worth seeing twice over — because there are two ways to describe one operation, and the second is the one that generalises.
Two operations, one conclusion. In (a) a genuinely new configuration appears, the same length as the old. In (b) nothing physically moves at all — there is no rotated vector to hunt for. The separation is one object described twice, and although its components change, \(\Delta x^{2}+\Delta y^{2}\) does not. Either way $r$ survives, so $F$ does.
The two symmetries this post is about
Two symmetries of nature matter here, and both are so familiar they are easy to miss.
Space is homogeneous. The laws of physics are the same here as they are a metre away, or on the far side of the Sun. We are hurtling through space right now; in six months Earth will be on the other side of its orbit, the Sun is falling around the galaxy, the galaxy is drifting through the Local Group. The scenery changes constantly. The laws do not.
Time is homogeneous. The laws that governed Newton's apple are the laws that govern the coffee cooling next to you, and they will be the laws a billion years from now. Things change; the rules governing the changes do not.
The two claims that will become momentum and energy conservation: \(x \to x+a\) changes nothing, and \(t \to t+b\) changes nothing.
These sound like philosophical throat-clearing. They are about to do real work.
3. The principle of stationary action
To convert symmetry into conservation we need one more ingredient, and it is the strangest idea in classical mechanics.
Throw a ball straight up. It rises, slows, stops, falls. Plot its height against time and you get a parabola. Everyone knows this. But why that path? Newton says: because \(F = ma\), integrated. There is a completely different answer that turns out to be deeper.
Throughout this post I will use one concrete example, in units where \(m = 1\) and \(g = 2\), so the flight lasts exactly 2 seconds and peaks at a height of 1:
$$x(t) = 2t - t^{2}, \qquad v(t) = 2 - 2t$$
Time runs horizontally, height vertically. This picture — a spacetime diagram — is where the rest of the argument happens. Note the two endpoints: they stay nailed down from here on.
At every instant compute the kinetic energy $T$ and the potential energy $V$, and take their difference. That combination is called the Lagrangian:
$$L = T - V = \tfrac{1}{2}mv^{2} - V(x)$$
The difference, not the sum. This trips people up constantly, so it is worth a picture.
The sum \(T+V\) is the flat green line — that is the conserved thing, and we have not earned it yet. The difference \(T-V\) is the dashed line, and that is the thing you integrate.
Now accumulate that difference along the whole trajectory. The result is the action:
$$S = \int_{t_1}^{t_2} Ldt = \int_{t_1}^{t_2} \bigl(T - V\bigr)dt$$
For our ball, \(L = 4t^{2} - 8t + 2\), and the action of the true path is exactly
$$S = \int_0^2 \bigl(4t^{2}-8t+2\bigr)dt = -\tfrac{4}{3}$$
Deform the path. Keep the two endpoints pinned, but let the middle bulge upward or sag downward, and compute the action of each deformed path.
Top: the true path and two deformations. Bottom: the running total \(\int_0^t Ldt'\) along each. All three start at zero; the true path finishes lowest, at \(-4/3\), while both deformations finish at \(-0.89\). The action is not a property of an instant — it is a property of an entire history.
Every alternative path has a larger action than the real one. Of all the paths connecting two fixed events, nature takes the one that makes the action stationary.
"Stationary", not "least"
That word matters, and here is where I want to be more precise than the usual telling. Let the deformation be
$$x_\varepsilon(t) = 2t - t^{2} + \varepsilon \sin\left(\frac{\pi t}{2}\right)$$
The sine vanishes at \(t=0\) and \(t=2\), so the endpoints stay put for any \(\varepsilon\). This integral is worth doing rather than waving at, because the interesting thing is which term disappears.
The velocity picks up a matching term,
$$\int_0^2 \tfrac{1}{2}v_\varepsilon^{2}dt = \frac{4}{3} + \frac{8\varepsilon}{\pi} + \frac{\pi^{2}}{8}\varepsilon^{2}, \qquad \int_0^2 Vdt = \frac{8}{3} + \frac{8\varepsilon}{\pi}$$
Both carry the same linear term, \(8\varepsilon/\pi\). Subtracting one from the other to form
$$S(\varepsilon) = -\frac{4}{3} + \frac{\pi^{2}}{8}\varepsilon^{2}$$
Look at what is missing. There is no term linear in \(\varepsilon\). It did not get small — it cancelled identically, and it would have cancelled for any variation vanishing at the endpoints, not just this one.
A first-order change in the path produces only a second-order change in the action. That flatness at \(\varepsilon = 0\) is the entire content of the principle.
So the precise statement is: the true path is the one for which \(\delta S = 0\) to first order, under any variation that holds the endpoints fixed. Sometimes that stationary point is a minimum (as here). Sometimes it is a saddle. "Least action" is a historical name, not a theorem.
*Formally: if
Hold on to this one fact: moving the interior of a true path, with its endpoints held fixed, does not change the action. Everything below is that fact, deployed cleverly.
But how does the ball know? The action principle has an uncomfortable smell of teleology about it. The ball at \(t=0\) appears to be surveying every possible future, computing an integral over each, and selecting the winner. That cannot be what is happening, and it isn't.
The resolution is quantum mechanical, and it is Feynman's. In quantum mechanics the particle really does take every path — each contributing an amplitude \(e^{iS/\hbar}\), a unit arrow whose angle is the path's action measured in units of \(\hbar\). Now ask what survives when you add up all those arrows. For a path where the action is changing as you deform it, neighbouring paths have wildly different phases, their arrows point every which way, and they cancel. But near a stationary path the action is flat by definition — so neighbouring paths share almost the same phase, their arrows line up, and they add.
Classical mechanics is what is left over when everything else has cancelled. And this immediately explains the pedantry about "stationary" rather than "least": what makes the arrows align is flatness, not smallness. A maximum or a saddle does the job just as well as a minimum, which is exactly what the classical theorem says and exactly what "least action" would have led you to expect wrongly.
4. The trick: chop the worldline into straight pieces
Here is the move that makes Noether's theorem elementary. It comes from a lovely paper by Hanc, Tuleja and Hancová in the American Journal of Physics, which shows the whole thing can be done with algebra and no calculus of variations at all.
Zoom in on any smooth worldline far enough and it looks straight.
*On a short enough segment the velocity is just the slope,
For a free particle (\(V = 0\)) the action of one segment is therefore
$$S_{\text{seg}} = \tfrac{1}{2}m v^{2}(t_2-t_1) = \frac{m}{2}\frac{(x_2-x_1)^{2}}{t_2-t_1}$$
That formula is the engine. Everything else is bookkeeping.
Now take the smallest interesting piece of a real trajectory: three neighbouring events, making two segments.
Points 1 and 3 are the ends; point 2 is interior, and free to move. The total action is \(S_{\text{tot}} = S_A + S_B\).
$$S_{\text{tot}} = \frac{m}{2}\frac{(x_2-x_1)^{2}}{t_2-t_1} + \frac{m}{2}\frac{(x_3-x_2)^{2}}{t_3-t_2}$$
Interlude: this picture already contains \(F = ma\)
Before extracting conservation laws, it is worth seeing that the same three points give you Newton's second law — which is reassuring, because it means we are not working with a toy.
Put a potential back in and demand only that the action be stationary with respect to the interior point \(x_2\), holding the ends fixed. That is the principle of stationary action applied directly, with no symmetry involved:
$$\frac{\partial S_{\text{tot}}}{\partial x_2} = 0$$
With equal timesteps \(\Delta t\), differentiating the four terms gives
$$m\frac{x_2-x_1}{\Delta t} - m\frac{x_3-x_2}{\Delta t} - \frac{\Delta t}{2}\Bigl[V'_A + V'_B\Bigr] = 0$$
Rearranged, with \(v_A\) and \(v_B\) the velocities of the two segments:
$$m\frac{v_B - v_A}{\Delta t} = -\frac{V'_A + V'_B}{2}$$
The left side is mass times acceleration. The right side is the average force over the two segments, which tends to
$$m a = F$$
Stationary action with respect to the interior point is Newton's second law, in disguise and in discrete form. This is worth holding on to, because it tells you what the \(\Delta S_2 = 0\) step in the next section is really doing: it is not a trick, it is the equation of motion.
5. Momentum, from the homogeneity of space
Step 1: the symmetry
Slide all three events up by the same amount $a$ — a spatial translation.
Both ends of every segment move by the same $a$, so every rise \(x_{i+1}-x_i\) is unchanged, so every velocity is unchanged, so every kinetic energy is unchanged.
For a free particle the action depends on positions only through their differences, so:
$$\Delta S_{\text{tot}} = 0$$
The action is invariant under spatial translation. This holds for any $a$, which means we can take $a$ as small as we like.
Step 2: the trick
Now the clever part. Instead of moving all three events at once, move them one at a time — and choose the order deliberately.
The whole argument, in four panels. Move point 1 (cost: \(\Delta S_1\)), then point 3 (cost: \(\Delta S_3\)), then point 2 (cost: \(\Delta S_2\)). The net effect is the same rigid translation as before.
Since the three sub-moves compose into the full translation,
$$\Delta S_1 + \Delta S_3 + \Delta S_2 = 0 \tag{from the symmetry}$$
Now look hard at the final panel. By the time we move point 2, points 1 and 3 are already sitting at their translated positions. They are the endpoints of the path, and they are fixed. So moving point 2 is exactly a variation of the interior of a true path with its endpoints held fixed.
We know what that costs. Nothing:
$$\Delta S_2 = 0 \tag{from stationary action}$$
There is a subtlety worth naming, because it is the one place the argument could cheat. Why is the *translated* path a true physical path, so that stationary action applies to it at all? Because the laws are translation-invariant: if
Combining the two boxed facts:
$$\boxed{\Delta S_1 + \Delta S_3 = 0}$$
Step 3: reading it as derivatives
What does that condition mean? Moving point 1 by a small amount $a$ changes the action at a rate given by the derivative — same as estimating a change in height from a slope:
$$\Delta S_1 \approx \frac{\partial S_{\text{tot}}}{\partial x_1}a, \qquad \Delta S_3 \approx \frac{\partial S_{\text{tot}}}{\partial x_3}a$$
Substituting and cancelling the common factor $a$:
$$\frac{\partial S_{\text{tot}}}{\partial x_1} + \frac{\partial S_{\text{tot}}}{\partial x_3} = 0$$
Now just differentiate. Only the first term of \(S_{\text{tot}}\) contains \(x_1\):
$$\frac{\partial S_{\text{tot}}}{\partial x_1} = \frac{m}{2}\cdot\frac{2(x_2-x_1)(-1)}{t_2-t_1} = -m\frac{x_2-x_1}{t_2-t_1} = -m v_A = -p_A$$
Only the second term contains \(x_3\):
$$\frac{\partial S_{\text{tot}}}{\partial x_3} = \frac{m}{2}\cdot\frac{2(x_3-x_2)}{t_3-t_2} = +m v_B = +p_B$$
The derivative of the action with respect to an endpoint position is the momentum there.
Put them into the constraint:
$$-p_A + p_B = 0 \qquad \Longrightarrow \qquad \boxed{p_A = p_B}$$
The momentum on segment A equals the momentum on segment B. And nothing in the argument singled out that particular pair — it applies to every adjacent pair along the worldline.
Chain the equalities together and one number is the same at every point of the motion. Momentum is conserved — not as an extra postulate, but as a consequence of the fact that space looks the same everywhere.
An honest caveat, and where the argument earns its keep
A single free particle is the friendliest possible example, and it is friendly enough to be slightly misleading. Look back at the interlude: for a free particle, setting \(\partial S_{\text{tot}}/\partial x_2 = 0\) gives \(p_A = p_B\) directly. For one free particle, "momentum is conserved" and "the equation of motion" are the same statement, so the symmetry argument, while correct, has not yet told you anything the equation of motion did not.
The content shows up the moment you have more than one particle. Take two masses interacting through a potential that depends only on their separation — which is to say, any real interaction:
$$L = \tfrac{1}{2}m_1\dot x_1^{2} + \tfrac{1}{2}m_2\dot x_2^{2} - V(x_1 - x_2)$$
Neither particle's momentum is conserved now; each is being pushed by the other, and both equations of motion are nontrivial. But shift both particles by the same $a$ and the separation \(x_1 - x_2\) is untouched, exactly as in the gravity example we started with. The action is invariant, the three-substep argument runs unchanged, and what comes out the other end is
$$p_1 + p_2 = \text{constant}$$
Total momentum is conserved, even though neither individual momentum is. That statement does not fall out of either equation of motion on its own — it is a property of the pair, and the symmetry is what reveals it. This is the version that does the work in particle physics, where you cannot follow the internal dynamics of a collision but you can still say with total confidence what the momenta must add up to afterwards.
The same reading applies to the ball we have been throwing. Gravity's potential \(V = mgx\) depends on absolute position, so shifting the ball alone is not a symmetry and its momentum is not conserved. Shift the ball and the Earth together, however, and the separation is unchanged — so the total momentum of the ball-plus-Earth system is conserved after all. The Earth simply absorbs the recoil in a way nobody will ever measure. No symmetry is ever really broken; it is usually just that you drew the boundary of your system in the middle of one.
Aside: you have just met Hamilton–Jacobi theory. The relation \(\partial S/\partial x = p\) is not a coincidence of this construction. It is one of the two Hamilton–Jacobi relations, normally arrived at after a great deal more machinery. It also explains why momentum was the quantity that appeared: momentum simply is the thing that answers to a shift in position. Its partner, \(\partial S/\partial t = -E\), is about to show up on cue.
6. Energy, from the homogeneity of time
Same machine, different symmetry. Instead of asking what happens when we shift the trajectory in space, ask what happens when we delay it in time: every event happens later by the same $b$, with the shape of the motion untouched.
This time we allow a potential, since a free particle would make the result trivial. On a short segment the velocity is constant as before, and we approximate the potential by its value at the segment's spatial midpoint.
The midpoint rather than an endpoint: that choice makes the error second order in the segment length instead of first, which is exactly what we need when the segments shrink.
So each segment contributes
$$S_{\text{seg}} = (t_2-t_1)\left[\frac{m}{2}\frac{(x_2-x_1)^{2}}{(t_2-t_1)^{2}} - V\Bigl(\frac{x_1+x_2}{2}\Bigr)\right]$$
and for our two segments,
$$S_{\text{tot}} = \frac{m}{2}\frac{(x_2-x_1)^{2}}{t_2-t_1} - V_A(t_2-t_1) + \frac{m}{2}\frac{(x_3-x_2)^{2}}{t_3-t_2} - V_B(t_3-t_2)$$
where \(V_A\) and \(V_B\) are the midpoint potentials of the two segments.
Step 1: the symmetry
Shift every event later by $b$.
*Every duration survives, because
The action depends on the times only through their differences and on the positions directly — and neither changed. So once again
$$\Delta S_{\text{tot}} = 0$$
Step 2: the same trick
Move the events one at a time, ends first.
Identical choreography to before — only the direction of the arrows has changed. By the last panel the endpoints are pinned, so the final move is an interior variation and \(\Delta S_2 = 0\).
$$\Delta S_1 + \Delta S_3 + \Delta S_2 = 0, \qquad \Delta S_2 = 0 \qquad \Longrightarrow \qquad \Delta S_1 + \Delta S_3 = 0$$
Step 3: differentiate, this time with respect to time
$$\frac{\partial S_{\text{tot}}}{\partial t_1} + \frac{\partial S_{\text{tot}}}{\partial t_3} = 0$$
Only the first two terms contain \(t_1\). Differentiating,
$$\frac{\partial S_{\text{tot}}}{\partial t_1} = \frac{m}{2}\frac{(x_2-x_1)^{2}}{(t_2-t_1)^{2}} + V_A = \tfrac{1}{2}m v_A^{2} + V_A = T_A + V_A = E_A$$
Only the last two contain \(t_3\), and they come with the opposite sign:
$$\frac{\partial S_{\text{tot}}}{\partial t_3} = -\left[\tfrac{1}{2}m v_B^{2} + V_B\right] = -E_B$$
And there is the companion Hamilton–Jacobi relation, \(\partial S/\partial t = -E\), arriving exactly as promised.
Therefore
$$E_A - E_B = 0 \qquad \Longrightarrow \qquad \boxed{E_A = E_B}$$
Energy is conserved, because the laws of physics do not care what time it is.
Notice the sum \(T+V\) appeared here on its own. We did not assume it was interesting; we differentiated the action with respect to a boundary time and it fell out. That is the difference between a definition and a derivation.
7. What actually happened
Both derivations had the identical skeleton. Worth seeing it stripped bare:
Chop the worldline into two short straight segments meeting at an interior point.
Apply the symmetry to the whole picture. The total action is unchanged: \(\Delta S_{\text{tot}} = 0\).
Decompose the move into three sub-moves — the two ends first, the interior point last.
Kill the middle term. By the time the interior point moves, the endpoints are already fixed, so stationary action gives \(\Delta S_2 = 0\).
What survives is a boundary condition: \(\Delta S_1 + \Delta S_3 = 0\), which says the derivative of the action at one end cancels the derivative at the other.
That equality is the conservation law. Whatever quantity the symmetry conjugates to is the same at both ends.
The symmetry supplies step 2. Stationary action supplies step 4. Neither alone gives you anything; together they force a conserved quantity into existence.
It is worth noticing that this is a machine, and machines run in both directions. Read forwards, it says: propose a symmetry, and you have predicted a conserved quantity. Read backwards, it says: measure a conserved quantity you cannot explain, and you have detected a symmetry you did not know the laws possessed. Physics has spent a century running it in the second direction. When a quantity stubbornly refuses to change — baryon number, lepton number, colour charge, isospin — the response is not to record the fact and move on; it is to go looking for the symmetry that must be responsible, and the symmetry usually turns out to be the more fundamental object. That is how a bookkeeping observation becomes a structural claim about the laws of nature.
This is Noether's theorem, in the simplest case it comes in:
Every continuous symmetry of the action yields a conserved quantity.
Emmy Noether proved it in 1918, and in full generality it covers vastly more than particles on a line — it applies to fields filling all of space, and it is the reason symmetry principles are the organising idea of modern physics rather than a decorative one.
The general statement, for the curious: if the action is invariant under \(q \to q + \epsilon K\) and \(t \to t + \epsilon\tau\), the conserved quantity is
$$Q = \frac{\partial L}{\partial \dot q}K - \left(\dot q \frac{\partial L}{\partial \dot q} - L\right)\tau$$
Set \(K=1, \tau=0\) and you get momentum. Set \(K=0, \tau=1\) and you get the second bracket, the energy function — which is where I owe you a correction.
The thing that is conserved is not always \(T+V\). Time-translation symmetry conserves \(h = \dot q\partial L/\partial \dot q - L\), the Jacobi integral. That equals \(T+V\) only when the kinetic energy is a quadratic function of the velocities and any constraints are themselves time-independent. Put a bead on a wire that someone is spinning at a fixed rate and you get a conserved $h$ that is emphatically not the bead's energy — the hand doing the spinning is pumping energy in. The clean statement is that time-translation symmetry conserves $h$; the familiar statement is a common special case.
8. More symmetries, more conservation laws
Once you have the machine, you can feed it anything.
Rotations give angular momentum
Rotate the whole two-segment picture through a small angle instead of sliding it, and run the identical argument.
*Under \(\delta x = -y\delta\theta\), \(\delta y = +x\delta\theta\), the surviving boundary condition says
Boosts give the centre-of-mass theorem — and a wrinkle
Watch the same motion from a frame gliding past at speed $u$: \(x \to x + ut\). Now something new happens. The Lagrangian is not invariant — it picks up an extra piece. But that piece is a total time derivative:
$$L \to L + \frac{d}{dt}\left(mux + \tfrac{1}{2}mu^{2}t\right)$$
A total derivative integrates to boundary terms, so the action changes only at its endpoints — and the argument still runs. This is called a quasi-symmetry, and it is why the symmetry that matters is a symmetry of the action, not of the Lagrangian.
The conserved quantity is \(pt - mx_{\text{cm}}\): the statement that the centre of mass of an isolated system drifts in a straight line at constant speed.
The dictionary
Each continuous symmetry buys exactly one conserved quantity. The last row is deliberately set below the line — see §10.
9. Does it actually work? Let's measure it
Derivations are cheap. Here is the argument's own construction, run numerically: take the thrown ball, chop its worldline into 24 straight segments, and read off exactly the two quantities the derivation builds — the segment momentum
There is a sharp prediction here. Under gravity, \(V = mgx\) depends on where you are — so space-translation symmetry is broken, and momentum should not be conserved. But nothing depends on when — so time-translation symmetry survives, and energy should be conserved exactly.
Left: under gravity the momentum falls linearly — the broken symmetry costs exactly the conservation law it was responsible for — while the energy sits flat to within \(10^{-14}\), machine precision. Right: remove the potential and both symmetries return, along with both conservation laws.
The same system, exhibiting both outcomes at once. That is about as direct a demonstration as you could ask for that the symmetry is doing the work, not the system.
Why programmers should care
This is not only a statement about nature. It is a statement about any discretisation of nature.
Left: two integrators for the same harmonic oscillator. The symplectic method is derived from a discrete action and inherits the discrete time-shift symmetry, so its energy error stays bounded forever; explicit Euler has no such structure and its energy diverges — the two differ by one line of code. Right: the converse. Add friction, and there is no time-translation symmetry of any time-independent action, and no conserved energy.
If you have ever wondered why orbital mechanics and molecular dynamics codes insist on symplectic integrators, this is why. A conservation law is not a bonus fact about a system — it is a receipt for a symmetry. Preserve the symmetry in your discretisation and you keep the law; break it, in physics or in code, and you lose it.
10. Where the theorem bites back
Two places where taking Noether seriously overturns something you were probably taught.
Energy is not conserved in general relativity. The conservation law came from time-translation symmetry. In an expanding universe there is no such symmetry — spacetime itself is different tomorrow — so there is no globally conserved energy. This is not a bookkeeping gap waiting to be filled; the quantity simply does not exist. Light from distant galaxies loses energy as it redshifts and that energy does not go anywhere. Locally, over regions small enough that expansion is negligible, the symmetry is approximately restored and so is the law, which is why nobody notices in a laboratory.
A conservation law is only as good as the boundary you drew. This one is less exotic but catches people far more often. Every "violation" of energy or momentum conservation you will meet in an undergraduate lab is a system whose boundary was drawn through the middle of an interaction — a block sliding to a halt, a ball bouncing lower each time, a circuit warming up. Nothing has broken. The symmetry argument applies to the isolated system, and friction is simply the point at which you stopped tracking degrees of freedom, the energy having gone into molecular motion you declined to model. The genuine failures of energy conservation are not these; they are the general-relativistic ones above, where no enlargement of the system restores the symmetry, because the symmetry is absent from spacetime itself.
Charge conservation is subtler than the dictionary suggests. Electric charge conservation follows from a global phase symmetry, \(\psi \to e^{i\alpha}\psi\) with \(\alpha\) the same everywhere. That is a genuine instance of the theorem proved above. But electromagnetism's defining symmetry is local: \(\alpha\) can be chosen independently at every point. Local symmetries are governed by Noether's second theorem, which yields not new conserved charges but identities among the equations of motion — the reason gauge theories have constraints like Gauss's law rather than extra conservation laws. Two theorems, one paper, routinely conflated.
11. Emmy Noether
The theorem was proved in 1918 by Emmy Noether, in a paper called Invariante Variationsprobleme. Hilbert and Klein had brought her to Göttingen precisely because they were stuck on a conservation-of-energy problem in Einstein's new general relativity, and she was the person who could solve it. She solved it, and in the process reorganised the relationship between symmetry and conservation for the whole of physics.
The faculty refused to grant her a habilitation, on the grounds that she was a woman. Hilbert's response to the senate has survived: "I do not see that the sex of the candidate is an argument against her admission as a Privatdozent. After all, we are a university, not a bathhouse." For years she lectured under Hilbert's name — her courses announced as his, taught by her. She was eventually granted the habilitation in 1919, and never a proper salaried professorship.
In 1933 she was dismissed from Göttingen for being Jewish. She emigrated to the United States, taught at Bryn Mawr and lectured at the Institute for Advanced Study, and died in 1935 at fifty-three, following surgery.
Her contribution to physics is the thing this post is about, but it was arguably a side project. Her deeper legacy is in abstract algebra, where she rebuilt the foundations of ring theory and did as much as anyone to create the axiomatic style that mathematics has used ever since. If you have ever met a Noetherian ring, worked with modules instead of matrices, or been told to think about a structure in terms of its homomorphisms rather than its elements, you have been standing in her house. Her students and collaborators — the "Noether boys" — carried that style into van der Waerden's Moderne Algebra, the textbook that taught the subject to everyone who came after.
There is a detail about the 1918 paper that deserves more attention than it gets. Noether did not set out to write down a general principle about symmetry; she was handed a specific, embarrassing problem. General relativity appeared not to conserve energy in the ordinary way, and nobody — including Einstein, Hilbert and Klein — could say cleanly why. Her answer was that this is not a defect of the theory but a signature of what kind of theory it is: the local symmetry of general relativity puts it under her second theorem, where energy conservation becomes an identity rather than a law. She resolved the puzzle by showing that the question had been the wrong shape. The first theorem, the one this whole post has been about, was in some sense the easier half of the paper.
Einstein wrote to the New York Times after her death:
In the judgment of the most competent living mathematicians, Fräulein Noether was the most significant creative mathematical genius thus far produced since the higher education of women began.
12. Where this goes next
We have derived momentum conservation from the homogeneity of space and energy conservation from the homogeneity of time, using nothing heavier than the chain rule. The same machine produced angular momentum from rotations and the centre-of-mass theorem from boosts.
The obvious next question is the one the dictionary above only gestured at: what symmetry is responsible for the conservation of electric charge? Answering it properly means taking the local/global distinction seriously, and it leads directly into gauge theory and the structure of the Standard Model — where symmetries stop merely constraining the motion and start determining which forces exist at all.
That is the next post.
Notes and sources
The elementary three-point argument is due to J. Hanc, S. Tuleja and M. Hancová, "Symmetries and conservation laws: Consequences of Noether's theorem," American Journal of Physics 72(4), 428–435 (2004), doi:10.1119/1.1591764. A free manuscript version is available here. If you want the argument in more depth than this post gives, start there.
Noether's original paper is Invariante Variationsprobleme, Nachr. d. König. Gesellsch. d. Wiss. zu Göttingen, Math-phys. Klasse (1918), 235–257.
On the distinction between Noether's first and second theorems, and the widespread conflation of the two, see K. Brading and H. Brown, "Noether's theorems and gauge symmetries" (2000).
Every figure here was generated from source, and every number plotted was computed rather than sketched — the closed form.