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35

To maintain lepton number as a conserved quantity. Consider, in detail, what's going on in a beta decay (well, I'm going to ignore the nuclear context). The reaction is then $$ n \longrightarrow p^+ + e^- + \nu \,,$$ where you should take the symbol $\nu$ to mean some neutrino (without prejudice about matter-type or anti-matter-type for the moment). There ...


21

TL;DR: every time you use momentum conservation. One way to see this is to take a close look at Newton's cradle: Image is published under GNU Free Documentation License You can start with Newton's second law: $$\mathbf{F}=m\mathbf{a}=m\frac{d\mathbf{V}}{dt}$$ By calculating the scalar product with the velocity vector on both sides of the equation we ...


13

The reason why one often thinks that all the familiar phenomena can be explained just on the basis of the second and the first law of Newton is that it is not clearly emphasized (mainly in school textbooks) that Newton's second law for a system of particles can take the form of $F_\textrm{external}=\dfrac{\mathrm d}{\mathrm dt}(p_\textrm{system})$ only when ...


4

We shall consider the man to be at the middle of the box initially. If the man starts walking left, the box shifts towards right (assuming no friction) such that the centre of mass of the man box combined system remains on the tip. If the man reaches the end of the box and starts jumping, the box begins to oscillate about the centre of mass and lf cases are ...


4

The 'rule' that "matter can not be [created or] destroyed" simply isn't true. Matter can be created and destroyed under the right circumstances, it's just that those circumstances are not met for macroscopic quantities of matter in places where humans live. This means that the conservation of matter can be taken as true in almost all of chemistry and large ...


3

All of your claims are essentially true. The angular momentum of light, in both its orbital and spin varieties, is indeed angular momentum that can be transferred to matter to make it spin and give it the garden variety of mechanical angular momentum. This is well explained in the relevant Wikipedia section, with good references for experiments that show it. ...


3

Since $[\hat{H},\hat{H}]=0$, you also have $[\hat{H},\hat{U}(\tau)]=0$. So, if we consider Schrodinger's equation, we have (neglecting factors of $\hbar$ for simplicity) $$ i\frac{d}{dt}|\psi(t)\rangle = \hat{H}|\psi(t)\rangle $$ Multiplying both sides by $\hat{U}(\tau)$ gives $$ i\hat{U}(\tau)\frac{d}{dt}|\psi(t)\rangle = \hat{U} (\tau)\hat{H}|\psi(t)\...


3

Regarding total momentum conservation, the point is that in non-inertial reference frames inertial forces are present acting on every physical object. Momentum conservation is valid in the absence of external forces. However, if these forces are directed along a fixed axis, say $e_x$, or are always linear combinations of a pair of orthogonal unit vectors, ...


3

The corresponding symmetry group is the Lorentz group and yes we can use Noether to derive conserved quantities: Invariance under translations $\rightarrow$ momentum conservation Invariance under rotations $\rightarrow$ spin and angular momentum conservation Invariance under boost $\rightarrow$ some strange, not really useful, conserved quantity


2

Answer posted by Lubos Motl in the comments; I reproduce most of it here. This answer was posted in order to remove this question from the "unanswered" list. Some (sketches of) answers to your questions, one by one: Physical states have to be invariant under gauge symmetries, so all of them are singlets and there are no nontrivial representations, (and 3....


2

The short answer to your question: Yes, they can. But in the particular example you are considering, they don't. As mentioned by Jahan, it is the gravity that gives the system (man+box) a net non-zero momentum. A rather interesting point to worry about in this scenario is 'who gives the system a net non-zero angular momentum?' Since gravity acts through the ...


1

You have proved with your analysis that $v_2$ cannot be zero. $m_2$ must be moving with some non-zero velocity, either in the same direction as $m_1$, in which case $v_2$ must be smaller than $v_1$ (or $m_1$ will never catch up to $m_2$) or $m_2$ must be moving in the opposite direction to $m_1$. With the information given, there are 4 unknowns: $v_1, v_2, ...


1

force is change in impulse over time, so saying that impulse is lower than static friction does not make sense, in the same way that saying that some speed is larger than some acceleration, they are different things. Impulse from a moving object is transferred to the one at rest through a force, that results in an exchange in momentum. During the ...


1

It is a little different in General Relativity. Let's start with Special Relativity and all the 3 forces of the Standard Model in physics. Then we will talk about gravity and the universe. In The Standard Model spacetime is Minkowski, meaning flat in all 4 dimensions. If it is that way clearly any direction and position is equivalent. That's called ...


1

Whilst the motion you intend is not altogether clear, the motor will indeed move as you say. But the attached rigid bodies also move, such that the center of mass of the whole system is stationary (or, more precisely, its state of motion unperturbed by the system's internal motions). Work out the path of the center of mass in your system to check this ...


1

The net force isn't zero. There is an external force, namely gravity. If you did this experiment in outer space, where there truly was no external force, you wouldn't see the box topple over.



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