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14

I would say that it is a result of time reversal symmetry. If you consider the projectile at the apex of its trajectory then all that changes under time reversal is the direction of the horizontal component of motion. This means that the trajectory of the particle to get to that point and its trajectory after that point should be identical apart from a ...


10

If the theory is invariant under translations in space, then linear momentum is conserved by Noether's theorem. If the theory is quantum, conservation holds only on the level of the expectation values (because that's the only meaningful level where you can talk about momentum as a number that's conserved in time), but it still holds. There is no way out. ...


6

I would consider that since acceleration is a constant vector pointing downward, that the time the projectiles downward component takes to accelerate from V(initial) to 0 would be the same as the time it takes to accelerate the object from 0 to V(final)


4

A lot of things have to hold to get that symmetry. You have to neglect air resistance. You either have to throw it straight up, or the ground over there has to be at the the same altitude as the ground over here. You have to through it slowly enough that it comes back down (watch out for escape velocity) But if you have that, then the simplest explanation ...


4

I think its because both halves of a projectile's trajectory are symmetric in every aspect. The projectile going from its apex position to the ground is just the time reversed version of the projectile going from its initial position to the apex position.


4

Comments to the question (v2): Noether's (first) Theorem is really not about Lie groups but only about Lie algebras, i.e., one just needs $n$ infinitesimal symmetries to deduce $n$ conservation laws. If one is only interested in getting the $n$ conservation laws one by one (and not so much interested in the fact that the $n$ conservation laws often ...


3

This is a very good question. First of all, you are absolutely correct that, for a single electron, invariance under $\phi\to-\phi$ means invariance under $y\to-y$. This is obvious just from looking at the coordinates $x=\rho\cos(\phi)$ and $y=\rho\sin(\phi)$. It is clear that $\phi\to-\phi$ means the exact same thing as $y\to-y$. The answer is a little ...


3

One could say that this is an experimental observation; after one could envisage, hypothetically, where this is not the case. This is not hypothetical once you take air resistance into account. One could say that the curve that the stone describes is a parabola; and the two halves are symmetric around the perpendicular line through its apex. But ...


3

Comments to the question (v2): First of all, recall the notion of an (off-shell) quasi-symmetry. It means that the action $S[\phi]$ changes by a boundary integral under the transformation of the fields $\phi$ and spacetime point $x$, cf. e.g. this and this Phys.SE posts. Since the action $S=\int_R \!d^nx~{\cal L}$ is an extensive variable, it is clear ...


2

(Non-kinematics math attempt but just some principles) It is a partial observation in that It hits the ground with same speed. Angle by which it hits the ground is the same (maybe a direction change) It takes equal time to reach to the peak and then hit the ground They are equally strange coincidences. Which are more fundamental? Consider the following ...


2

the parity operation is $P: (t,x) \rightarrow (t',x') = (t,-x)$ lets see how this effects this term... $\epsilon^{\mu \nu \lambda} a_\mu \partial_\nu a_\lambda = \epsilon^{\mu 0 \lambda} a_\mu \partial_0 a_\lambda + \epsilon^{\mu j \lambda} a_\mu \partial_j a_\lambda$ $=\epsilon^{\mu 0 \lambda} a_\mu \partial'_0 a_\lambda - \epsilon^{\mu j \lambda} a_\mu ...


2

Assuming no quantum gravity, $\eta^{\mu\nu}$ is a constant and can be pulled out of the derivative and what remains looks like a $\delta^k_{\mu}$ or $\delta^k_{\nu}$-type expression (in the sense of a Kronecker $\delta$), pulling the $k$ into the $\partial^\mu$ or $\partial^\nu$ respectively. If you are confused about where the minus sign comes from, I ...


2

Noether theorem tells you that if you can find a (one parameter) group of infinitesimal transformations $\alpha$ and $\beta$ such that: \begin{equation} t'=t+\alpha\epsilon \end{equation} \begin{equation} q'^\mu=q^\mu+\beta^\mu\epsilon \end{equation} and your lagrangian is invariant under this group of transformations, then the quantity \begin{equation} ...


2

Comments to the question (v5): If an action functional $S$ is invariant under a Lie algebra $L$ of symmetries, the corresponding Noether currents & charges do not always form a representation of the Lie algebra $L$. There could be (classical) anomalies. In some cases such (classical) anomalies appear as central extensions, cf. e.g. Ref. 1-3 and this ...


2

Suppose $g$ is the generator of a certain symmetry (i.e. the generating function of an infinitesimal canonical transformation) and you are interested to know how the observable $f$ changes after the "action" of $g$. In the Hamiltonian formalism the change is found to be $$\delta f \approx \epsilon\{f,g\}$$ which can be related to the time evolution of an ...


1

You are right that that the symmetry breaking breaks all three symmetries of $SU(2)$. Thus the $SU(2)$ generators give you three goldstone bosons in the theory with broken symmetry. However, we have not yet considered all of the symmetries of the original theory. We know that the full symmetry group has six generators and that five of them must be broken. ...


1

Hint: The fusion rule Clebsch-Gordan-like coefficients $c_{ij}{}^k=c_{ijk}$ are related to the 3-point function $\langle \phi_i(x) \phi_j(y)\phi_k(z)\rangle$ of 3 primary fields, which in turn is totally symmetric.


1

The answer is that the term symbol refers to the entire, multielectronic, molecule. You are indeed correct that a single electron in a $\Sigma$ state must be symmetric under reflection about a plane that contains the internuclear axis. If you have multiple electrons, however, you can still have a global antisymmetry under such reflections, and get their ...


1

How atomic orbitals merge into crystal band structure is, well, complicated to capture in simple models. As seen in, say, Ashcroft and Mermin (chapter 28), the energy surfaces for Si have symmetry along the <100> directions. In contrast, the surface for Ge have symmetry along the <111> directions, with the band minimum at the zone edge. A comparison ...


1

JakobH's comment as an answer: The electroweak gauge group $\mathrm{SU}(2)_L \times \mathrm{U}(1)_Y$ is broken into the electromagnetic $\mathrm{U}(1)_\text{em}$ by the Higgs field acquiring a non-zero vacuum expectation value, granting masses to the quark and $W^\pm,Z$ bosons. Thus, at the scale where up- and down-type quarks have very different masses, we ...


1

Except that they don't have the same eigenvalue. To understand it in terms of differential operators; divide the first equation by $1/r^2$, this gives you an eigenvalue equation for the total hamiltonian operator $H = -h^2\nabla^2/2m + V$, whose eigenvalue is clearly $E$. Then we divide $\nabla^2 = \hat{R} + 1/r^2 \hat{\Omega}$ in your notation (into radial ...


1

Asking "Why" in physics often leads out of physics and into philosophy. Why is there light? Because God said so. Physics only answers questions about how the universe behaves and how to describe that behavior. If a why question can be answered with physics, the answer is "because it follows from a law of physics." A law is just a mathematical description of ...


1

When one throws a stone.... Your arm is capable of propelling an object at up to around 150km/h (and that's with some practice). At that speed the many factors like air resistance are negligible. Let’s load a 16-inch shell into a gun (you will find several on the ISS Iowa), aim it 45 degrees up and press the button. The shell is going to go up at ...


1

Comments to the question (v1): Concerning the notion of on-shell and off-shell, see also Wikipedia and this Phys.SE post. In the context of an action formulation, on-shell means that the Euler-Lagrange (EL) equations $$\tag{1} \frac{\delta S}{\delta\phi^{\alpha}(x)}~\approx~0$$ are satisfied. It seems that the potential confusing point is the notion of an ...


1

Conservation of momentum! Force × Time = Impulse = Δ Momentum Since the average force is the same going up and down, and since the momentum change is the same going up and down as well, the time during which the force is applied must also be the same.


1

In the modern crystallography there is a notation of aperiodic crystals (or quasicrystals). They are crystals with normal basis $\mathbf a,\mathbf b,\mathbf c$ and a set of propagation (or wave) $\mathbf k$-vectors that are incommensurate with the metric $\mathbf a,\mathbf b,\mathbf c$. The atomic positions (or/and occupancies) are modulated according to ...



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