Noether's theorem vs. Heisenberg uncertainty principle

In continuation of another question about Noether's theorem I wonder whether there exists some kind of relationship between this theorem and the Heisenberg uncertainty principle.

Because both the principle and the theorem relate energy with time, momentum with space, direction with angular momentum. When this is a general fact then e.g. electrical charge and electrostatic potential(*) should be partners in an uncertainty relationship too. Are they?

I feel that these results look so basic and general that I hope that a pure physical reasoning (without math or only with a minimal amout of math) exists.

Also compare this question where again momentum and space are connected, this time through a Fourier transform.

(*) i.e. electric potential and magnetic vector potential combined.

Answer

Noether theorem is as valid in CM(*) as in QM(**). It deals with conservation laws and symmetries. In CM the variables are certain, in QM they may be uncertain.

HUP belongs to QM and gives a limitation on canonically conjugated variable uncertainties in a given state.

If some variable in QM is uncertain, it does not mean its expectation value is not conserved. A superposition of free motions states $e^{ipr}$ is also a free motion state although the momentum, for example, may be uncertain. The dynamics of the momentum expectation value is determined with an external force, like in CM (see the Ehrenfest's equations). No external force, no variation of the expectation value <p(t)>.

So I do not see any relationship between HUP and Noether.

(*) Classical mechanics (**) Quantum mechanics

quantum mechanics - Is a dice roll deterministic?

This should be understood as distinct from the question of is it possible to predict the outcome of a roll, which seems to be an issue related to intractability and observation?

What I'm really asking is, if quantum indeterminacy is a factor of more than just observational limitations and intractability, where outcomes may be independent of of the prior state of the system, does it affect outcomes in the world of classical mechanics?

My interest is related to certain combinatorial problems, whether random number generation needs to be integrated, and if so, how it might be treated differently from uncertainty arising out of imperfect or incomplete information or intractability.

riddle - Songs & Artists from Haikus

The goal is to determine the song title and artist from these haikus. All the artists will have received mainstream success in some form.

Haiku #1

"A possessor of

marionette expertise."
- Alloy Performer

Haiku #2

"Degradation of
one's overall attitude."
- One Extra Digit

Haiku #3

"A compilation

of objects, short in stature."
- Eyes Closing Often

Haiku #4

"Within an insect's
abode, a transformation."
- Superb Resonance

Haiku #5

"Rock's adversary

opening layers of skin."
- Chi Town Neighbourhood

The haiku itself holds no significance over the answer.

--

Some information about myself (to be used as hints, if you wish):

I'm from

North America.

Specifically

Canada.

I was born during

the mid 80's.

My favourite types of music are

some form of Rock 'n' Roll.

Answer

Haiku 1

Master of Puppets, Metallica

Haiku 2

Drag You Down, Finger 11

Haiku 3

All the Small Things, Blink 182

Haiku 4

Change (In The House Of Flies), Deftones

Haiku 5

Papercut, Linkin Park

Answers for 1,2,3 and 5 were previously provided and confirmed by OP.
Adding answer here for the 4th Haiku.

angular momentum - How does the mass loss in a binary system affect the semi-major axis of the orbit?

It is common that during the stellar evolution one of the stars in a binary system would transfer mass into the the other, resulting in the increase of mass of one star and decrease of mass of the other. Or, if one of the stars happens to evolve through the red giant phase while the other is still on the main sequence, the loss of the mass should be really significant and consequently the orbit should also be changed dramatically.

I have a hard time visualize how the semi-major axis is changed as a result of mass loss. According to the Kepler's third law the semi-major axis a is proportional to $(M_1+M_2)^{1/3}$ but I am not sure whether the period can be fixed or not.

Answer

The behaviour depends on whether the mass is gained by the companion (conservative mass loss) or lost from the system altogether (unconservative mass loss).

For conservative mass loss, one conserves angular momentum and the total energy and requires that $M_1+ M_2 =M$, which is constant. In this case we have $$ M_1 M_2 a^{1/2} = {\rm constant},$$ so $$\frac{a_f}{a_i} =\left(\frac{M_{1,i}M_{2,i}}{M_{1,f}M_{2,f}}\right)^2$$ and what happens to $a$ depends on whether the more massive or less massive star loses mass, since the product $M_1 M_2$ is maximised when the two are equal. Thus if the mass ratio becomes more similar (more massive star loses mass) then $a$ increases, and vice versa.

Unconservative mass loss is more complicated and depends on how the mass leaves the system. The most simple case is spherically symmetric mass loss that escapes without interacting with the other star and leaving with the specific angular momentum of the mass-losing star. In this case $$\frac{a_f}{a_i} = \frac{M_i}{M_f},$$ where $M$ is the total system mass. Since $M_f < M_i$ then the separation always widens.

electrostatics - Applying Kirchoff voltage law to a short circuit

If you consider an ideal wire with no resistance that shorts an ideal battery, the only voltage drop that exists is the emf of the battery, with nothing to balance it.

Obviously in the real world such a scenario is impossible, for the wire will have some resistance, but in this ideal example, is KVL not violated?

Answer

No, it doesn't break. All it means is that all the voltage is on the ideal wire. According to Ohm's law there will be infinite current to account for the zero resistance. That's what happens in a real circuit - if you short a battery the current is very high.

everyday life - Why do corn flakes clump together in milk when stirred?

The Cheerios Effect is the name the scientific community has given to this phenomenon. When cornflakes or oats are mixed with milk and stirred, they tend to clump together. Why does this happen? Can physics explain this?

Answer

Assuming your question is essentially: "Why do the flakes actually 'clump' together ?", the answer is one that involves probability.

Suppose you have only two wet corn-flakes on the surface of milk isolated from each other. Say somehow they start moving toward each other. As they come together, they stick. Now, once stuck, the flakes will not seperate from each other on their own. (probably because they have attained a state of lower energy). In other words, the probability of them seperating is literally zero.

Now let's say there are many such flakes which are isolated from each other on the surface. Let's start strirring the milk in a random fashion and not necessarily in a circular manner. As you stir randomly, the flakes start moving in random directions on the surface of milk. Some may move toward each other, and some opposite. But those which move towards each other are by fate destined to stick and once stuck they don't move apart. Therefore, there are many probabilities that the flakes may come together, but there is zero probability that they come apart. That is, the flakes which are stuck remain stuck even though there may be other unstuck flakes randomly moving about on the surface. In time, even those flakes will stick to the ones which are already stuck. Thus, the flakes finally 'clump' together.

Hope it helped !

spacetime - Interval preserving transformations are linear in special relativity

In almost all proofs I've seen of the Lorentz transformations one starts on the assumption that the required transformations are linear. I'm wondering if there is a way to prove the linearity:

Prove that any spacetime transformation $\left(y^0,y^1,y^2,y^3\right)\leftrightarrow \left(x^0,x^1,x^2,x^3\right)$ that preserves intervals, that is, such that

$$\left(dy^0\right)^2-\left(dy^1\right)^2-\left(dy^2\right)^2-\left(dy^3\right)^2=\left(dx^0\right)^2-\left(dx^1\right)^2-\left(dx^2\right)^2-\left(dx^3\right)^2$$

is linear (assuming that the origins of both coordinates coincide). That is, show that $\frac{\partial y^i}{\partial x^j}=L_j^i$ is constant throughout spacetime (that is, show that $\frac{\partial L_j^i}{\partial x^k}=0$).

Thus far all I've been able to prove is that $g_{ij}L_p^iL_q^j=g_{pq}$ (where $g_{ij}$ is the metric tensor of special relativity) and that $\frac{\partial L_j^i}{\partial x^k}=\frac{\partial L_k^i}{\partial x^j}$. Any further ideas?

Answer

In hindsight, here is a short proof.

The metric $g_{\mu\nu}$ is the flat constant metric $\eta_{\mu\nu}$ in both coordinate systems. Therefore, the corresponding (uniquely defined) Levi-Civita Christoffel symbols

$$ \Gamma^{\lambda}_{\mu\nu}~=~0$$

are zero in both coordinate systems. It is well-known that the Christoffel symbol does not transform as a tensor under a local coordinate transformation $x^{\mu} \to y^{\rho}=y^{\rho}(x)$, but rather with an inhomogeneous term, which is built from the second derivative of the coordinate transformation,

$$\frac{\partial y^{\tau}}{\partial x^{\lambda}} \Gamma^{(x)\lambda}_{\mu\nu} ~=~\frac{\partial y^{\rho}}{\partial x^{\mu}}\, \frac{\partial y^{\sigma}}{\partial x^{\nu}}\, \Gamma^{(y)\tau}_{\rho\sigma}+ \frac{\partial^2 y^{\tau}}{\partial x^{\mu} \partial x^{\nu}}. $$

Hence all the second derivatives are zero,

$$ \frac{\partial^2 y^{\tau}}{\partial x^{\mu} \partial x^{\nu}}~=~0, $$

i.e. the transformation $x^{\mu} \to y^{\rho}=y^{\rho}(x)$ is affine.