58

The core of the Noether theorem in all contexts where it arises is surprisingly elementary! From a very general point of view, one considers the following structure. (i) A set of "states" $x\in \Omega$, (ii) A one-parameter group of transformations of the states $\phi_u : \Omega\to \Omega$, where $u\in \mathbb{R}$. These transformations are ...


57

You would have much more mass than 100 kg after the wood was burned. As it turns out, wood is made of cellulose and lignin. Both are cross-linked glucose polymers, so a good approximation of what you would get is given by the chemical reaction of burning glucose: $$\rm C_6H_{12}O_6 + 6O_2 \to 6CO_2 + 6 H_2O$$ This means that 6 oxygen molecules combine with ...


55

Electrons can only be created and destroyed in processes that keep electric charge constant. There are three Standard Model interactions involving the electron: $\rm W^-\to e^-\bar{\nu}_e$ and $\rm \gamma\text{ (or Z)}\to e^-e^+$.$^1$ the first case, the W boson has the same charge as the electron, so no charge is created or destroyed. In the other cases, a ...


41

The two results that you got are absolutely acceptable both mathematically and experimentally and either of them can happen if the experiment is done (but it's not random) practically with all situations similar to the one posed in the question. Why isn't it random ? Because the result depends (with certainty) on the arrangement of the piece of mass $4\; kg$...


36

What a lovely conundrum! I'm afraid that the solution is a rather simple. You didn't specify in your maths whether the 4 kg mass was ejected in the direction of the initial velocity of the 5 kg mass and the 1 kg in the opposite direction, or vice versa. The different pairs of solutions correspond to the two different cases. It's not your fault that you didn'...


26

The low viscosity of the coffee means that you can rotate the cup without significantly moving the liquid it contains: there's simply not enough friction to 'drag' the liquid by the cup's wall. It would be a different picture with a viscous liquid like oil or runny honey. It's useful to remind us what Newtonian viscosity $\mu$ really is. (Source) For $\mu \...


25

The conservation of momentum can be derived from the invariance of the Lagrangian under spatial translations. This follows from Noether’s theorem, so it applies for special relativity. Furthermore you can use Noether’s theorem to derive the form of relativistic momentum. Regarding determining if something is a law of physics: that can only be done ...


23

A lot of debris has probably fallen back to earth. To stay in orbit you need enough angular momentum to overcome attraction. But if the collision happened at an angle a portion of the debris could have enough angular momentum to sustain orbit. Here is a nice video of how the collision could have happened. Here are some snapshots from the video in case the ...


22

If you consider the two magnets together as a system, there is no external force acting on it. Since there is no external force acting on it, the centre of mass of the system remains unchanged. As a result, no matter how the magnets move, they will do so in such a fashion as to keep the centre of mass constant. From this, it should be easy to see that the ...


21

Momentum is conserved only if there is no net external force on the system. Consider the snowball and the tree as the system. In your case, the earth provides an external force on the tree, so the momentum of the snowball/tree system is not conserved. If the tree is "suspended" (not attached to the ground) momentum would be conserved, but the ...


21

It is quite simple. It is that you are entirely overlooking the fact that the marble is interacting with the track and with whatever is supporting the track and ultimately with the Earth. The combined momentum of the the marble, the track and the Earth is conserved throughout. If you repeated the experiment with the marble on a light track which was on a ...


20

Depends on what your sack manages to capture This was a thing that finally helped kill the phlogiston theory of fire (that burning something means releasing the phlogiston enclosed in it): most things got lighter by burning them, but some got heavier! This could only be explained by having some materials contain negative-mass phlogiston, which pretty much ...


18

Try and think of a Noether charge besides a momenta (linear, angular, Hamiltonian). It's rather hard to do in point particle mechanics because there really aren't any we talk about (it's easier in field theory to come up with things that aren't just momentum). So, let me pose the following question, which I think will make the point I want to get at: without ...


17

Yes, the "artificial gravity" can last for a very long time (the other answers address the caveats to "forever"), but we are not getting anything for free: the person at the edge of this rotating spaceship has a force continuously applied to them, but as long as they stay in place (with respect to the ship) this force does not do any work,...


16

Of course the answers like 'it is because of low viscosity' are good, but it is also nice to overcomplicate this problem. You are not rotating the cup The cookie does rotate/move in some sense but only slightly. The reason that the cookie is not rotating a lot is that you are actually not rotating the cup. Instead, you are giving the cup a short twist, which ...


15

Yes, while the air you are sucking is moving (say left) into your mouth, due to conservation of momentum you would move slowly to the right. The force would be due to the difference in air pressure in front and behind you. The trouble is that to maintain a constant force to the right you'd have to maintain the 'suck' which isn't possible for a person to do. ...


14

Theories of the moon's formation are hypothetical and a subject of active research. While the impact theory is the most popular at the moment, it's not a done deal. So the true answer is that we don't know. But proposals for how it might have happened generally suggest a glancing collision, where the second body (Theia) ploughs through the mantle of the ...


14

Relativistic loss of mass is unmeasurable here, but in principle, you’d lose some tiny fraction of the mass by heat transfer to the surroundings. Whether the smoke would weigh more or less than the wood depends on your definition. Oxygen from the air is combining with carbon in the wood to form carbon dioxide. If this counts as smoke, then the smoke weighs ...


14

OP's Lagrangian is indeed constant on-shell, but not necessarily off-shell. In contrast, the principle of stationary action compares various possibly off-shell paths.


14

So the mechanisms which generate and destroy electrons happen to be such that they never violate charge conservation. Let's take pair annihilation for example, an electron and a positron meet and they become two photons. Before, the total charge is zero: the positron has positive charge and the electron has the exact opposite negative charge. Afterward, the ...


13

The connection is simply this... Van der Waals postulated that there were attractive forces between gas molecules even when these weren't in contact. The $a/V^2$ term in his gas equation is a simple way to take account of such forces, without knowing how they vary with separation between molecules, beyond their being short-range. We now know that these ...


13

Noether's first and second theorem only apply to classical theories with an action formulation. The quantum analogs are (generalizations of) the Schwinger-Dyson equations and the Ward-Takahashi identities.


13

I won't bother to reproduce the contents of the paper I want to recommend, but I will try to summarize what you'll find in it. Baez, John C. "Getting to the Bottom of Noether's Theorem." arXiv preprint arXiv:2006.14741 (2020). The main thrust of the paper is to put applications of Noether's theorem in classical mechanics, quantum mechanics, and ...


13

About increase of angular velocity when radius of circumnavigating motion is decreased. The diagram below represents the trajectory of a point mass that is circumnavigating a central point, in the process of being pulled closer to the central point. The dark gray arrow points represents the centripetal force that the point mass is subject to. If the ...


13

You could ask the same exact thing for the Lagrangian of a nonrelativistic particle. It's an exactly analogous situation. $$ L = \frac{1}{2} m \dot x^2 $$ Now, if we differentiate the Lagrangian with respect to time, we get \begin{align} \frac{d}{dt} L &= \frac{1}{2} m \frac{d}{dt} \dot x^2 \\ &= m \dot x \ddot x \\ &\neq 0 \end{align} It's not $...


13

If you model the colliding objects as rigid bodies, then the collision will take zero time. Since the collision has zero duration, there is no issue with the objects having different accelerations. While this is often a reasonable approximation to make to simplify calculations, it doesn't exactly explain really going on here, so we look at what happens if we ...


13

The law still applies, but we have to be careful with the forces. If the book has a mass $m$, we cannot simply say that the book applies a force of $F=mg$. If the table broke, then there was some maximum force it could withstand, which was less than $mg$. The key to remember is that the sum of the forces on an object equals 0 only if that object is not ...


12

I agree you raise a valid point. It is of course the case that material that is thrown up will not enter circular orbit automatically. Compare launching of satellites. The rocket sets its payload up such that the payload will reach an orbital apogee near the intended altitude, and then some more propulsion is needed to circularize the orbit. So the question ...


12

Neutrinos and antineutrinos are indistinguishable by most of their qualities, butt not all. One of the quantities that distinguish them is exactly the lepton number, which make them interact with other particles in a different way. For example, a neutrino can take part in the reaction $$ n + \nu_e \rightarrow p^+ + e^-$$ but antineutrino can't; on the other ...


12

The angular momentum is given by a vector product $\vec{L}=m[\vec{r}\times\vec{v}]$. One o the properties of the vector product is that the result is orthogonal to both factors. I.e. both $\vec{r}$ and $\vec{v}$ should be orthogonal to $\vec{L}$. Or you can say that both $\vec{r}$ and $\vec{v}$ lie in the plane orthogonal to $\vec{L}$. As $\vec{L}$ is ...


Only top voted, non community-wiki answers of a minimum length are eligible