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13

The statement is true for decays, where lifetimes can be measured. It is not true for interactions though. A suicidal electron meeting a positron has a good probability to disappear, together with the positron, into two gamma rays, at low energies. Electron-positron annihilation It is intriguing that this is not true for neutrinos. If an electron ...


8

This is not exactly true. It is believed that net charge is conserved, but there is a weak process called electron capture, where an electron is captured by a nucleus, (usually from an inner "orbital" so there is a spectroscopic signature), a neutrino is emitted and a proton changes to a neutron. So therefore your textbook is wrong!


77

Imagine you are an electron. You have decided you have lived long enough, and wish to decay. What are your options, here? Gell-Mann said that in particle physics, "whatever is not forbidden is mandatory," so if we can identify something you can decay to, you should do that. We'll go to your own rest frame--any decay you can do has to occur in all reference ...


4

In both cases and at all times, the force from the (wall/tire) on the hammer equals the force from the hammer on the (wall/tire) : total momentum must be conserved. However, in the first case, the initial energy is dissipated in the wall (as heat and/or damage), so at the end the hammer is stopped. In the second case the initial energy is stored as ...


2

A collapsing gas cloud is an open system. It loses mass, energy and angular momentum as it collapses. Even if the net angular momentum of the cloud is zero, after the collapse the final planetary disk can have a significant net angular momentum, and the ejected material will have the opposite angular momentum. What can not happen, and that's where your ...


0

Yes. Some of the impulse force is used in work which rotates the rod, and some of the impulse force is used in work which changes the bullet's trajectory. The distribution of impulse force between the rod and the bullet is equal.


0

For the first question: there may be several species of particle in a theory, so charge conservation can be upheld without creating a particle and its antiparticle together. For example, in beta decay, an electron is `created', but no positron, and charge conservation is upheld by a neutron being exchanged for a proton. This partially answers question two ...


2

If we treat the Earth as an isolated system then both its linear and angular momenta will remain constant. To answer your question you need only consider the angular momentum. The angular momentum is given by: $$ L = I\omega $$ Since $L$ is a constant, if the moment of inertia changes from $I_1$ to $I_2$ then we have: $$ I_1\omega_1 = I_2\omega_2 $$ and ...


0

Neither the linear momentum or the angular momentum will change (they are conserved quantities), unless you apply a force or a torque respectively. In this case, the change in radius will likely change the moment of inertia. And as the (constant) angular momentum depends on the product of moment of inertia and angular velocity, then the angular velocity ...


2

From my readings; the key to conservation of momentum appears to be based on defining a closed system to see if any mass crosses the boundaries of the system.


1

Any self-adjoint operator $A$ on some Hilbert space $H$ generates a one-parameter subgroup of the unitary group $U(H)$ by $$t\mapsto U(t):= e^{itA}.$$ Here we shall think that $A$ is the Hamiltonian of some mechanical system, and that $U(t)$ is the flow it generates. In the Heisenberg picture, the time evolution of an observable $F(t)$ is given by ...


1

According to the Wikipedia article "Constant of Motion" A quantity $A$ is conserved if it is not explicitly time-dependent and if its Poisson bracket with the Hamiltonian is zero That is to say, if both $$\frac{\partial A}{\partial t} = 0 $$ $$\{A, H\} = 0 $$ then $$\frac{dA}{dt} = \frac{\partial A}{\partial t} + \{A, H\} = 0, \, \Rightarrow A ...


1

Is the same idea of classical mechanics, but now this quantities can be undetermined if you apply some measurement of the complementary of this quantity. e.g. The total angular momentum is a constant of motion $\vec{L}$. You can measure some component $L_i$ of this angular momentum a lot of times and this yields to the same result (conservation), but if ...


1

A diagram which is first order in $\alpha_\text{EM}$ would have to have one vertex, because $\alpha_\text{EM}\propto g^2$ where $g$ is the factor associated with each vertex (and the amplitude corresponding to the diagram gets squared). There's only one possible vertex in QED, namely the photon-electron-positron vertex, and it's impossible to arrange this in ...


4

I suppose you read this passage in the famous Feynman Lectures. I am fairly certain that what Feynman is referring to (and what you are looking for) is a proof that an electrostatic field is conservative. There are a number of equivalent ways of stating that a vector field is conservative, each of which can be taken as a definition. Let $\vec{F}(x)$ be a ...


2

You cite the example of a (Newtonian) gravitational potential which, naively, seems to depend on the position of the bodies under consideration. However, a little more contemplation is warranted here. Consider the following: When we ask a basic question about the gravitational potential energy of some body relative to Earth, do we have to specify where ...


4

Any energy principle is not being violated since the speed of the photon is never less than $c$ and hence the momentum is unchanging (in the classical sense). Why light travels slower than $c$ in a medium is because of the photons being absorbed and reradiated by atoms in the material. In a sense you can make the analogy of light traveling a longer path in ...


0

This is an inelastic collision (the bullet and wood do not elastically bounce off of each other but instead, 'stick' together) and so, kinetic energy is not conserved in the collision. However, the total momentum just before the collision must equal the total momentum just after the collision. $$p(0-) = m_b v_b = p(0+) = (m_b + m_w)v_{bw}$$ Thus, the ...


4

The rest mass of the system is conserved, it's just that the rest mass of the system isn't the sum of the masses of the parts. The rest mass of a system is just the length of the total energy-momentum vector. And that vector is conserved, so the length is conserved. The sum of the rest masses of the parts is not conserved. But that simply isn't the rest ...


3

Here is my comment in more details For any system or single elementary particle mass $M$ is defined as $$ M = \sqrt{E^2 - \textbf{P}^2} $$ where $E$ is total energy and $\textbf{P}$ is total momentum. For an elementary particle (like electron) mass is always conserved. For a system $M$ is conserved as long as $EdE - \textbf{P}d\textbf{P} = 0$, in ...


2

What does conservation of mass mean in classical mechanics? Weight and mass are the same , we know the mass by weighing it, and if we add 1 kilo of sugar to another kilo of sugar, we will have two kilos of sugar. That is what is meant classically that the mass is conserved. Dissolving a kilo of sugar to a kilo of water will give you two kilos of sirop. The ...


3

For the rest mass we have $$m^2=p^2=p^{\mu}p_{\mu}$$ where $p^{\mu}=(E,\vec{p})$. It is Lorentz invariant, which means the rest mass of the particle is always the same no matter in which frame the observer is in. While for relativistic mass, it's simply equivalent to the total relativistic energy, which is always conserved. Note the difference between ...


7

Mass, or more correctly, rest mass is not conserved in special relativity. Particles are able to be created and annihilated in special relativity, for instance, an electron and a positron can interact to produce two photons: $$e^++e^-\rightarrow 2\gamma $$ Here mass is clearly not conserved, because both the electron and positron are massive but photons are ...


3

Yes. The so called rest mass $m_0$ is the magnitude of energy-momentum four vector and is always conserved. It is not just conserved but is also invariant since it is the magnitude of a four vector. Perhaps you are confused between rest mass and relativistic mass. The relativistic mass is the total Energy divided by $c^2$. $m\equiv \gamma m_0 $ This ...


1

Gauss-Ostrogradsky theorem does not actually "care" for a "type" of space or functions that you've got, so it has the same form in curved space and in flat space: $$ \int_D d^4x \frac{\partial}{\partial x^\mu} Anything = \oint_{\partial D} d\sigma_\mu Anything $$ where $d^4 x$ and $d\sigma_\mu$ are constructed from differentials of coordinates the same ...


0

When two object orbit each other, the shape of their orbit is independent of their relative mass. In fact they will each have the same shape of orbit, but scaled by the mass of the other. So if you have two objects of mass $m$ and $2m$ respectively, then the former will have an orbit (circle or ellipse) that is exactly twice as big as that of the other. ...


3

To say that the orbit becomes more circular the greater the Sun's mass is not true. Instead, the eccentricity (i.e. how much the shape of an orbit varies from being circular) is governed by a couple of factors. If you have a planet orbiting about the Sun with a mass much less than that of the Sun, and you know the following for an instantaneous point in the ...


3

Disclaimer: Let us here avoid the discussion of how to assign a stress-energy-momentum (SEM) pseudo-tensor $t^{\mu\nu}$ to the gravitational field. The word pseudo here refers to the fact that $t^{\mu\nu}$ is not a tensor wrt. general coordinate transformations; only a rigid subgroup thereof. In other words, the pseudo-tensor ...


4

No. The shape of the orbit, i.e. how elliptical it is, does not depend on the relative masses of the two bodies. All objects in the solar system orbit around the centre of mass of the solar system. For obvious reasons, namely that the Sun contain far and away most of the mass of the solar system, the centre of mass of the solar system is quite close to the ...


2

Yes, objects with mass all attract to each other and move each other of course, except that the star doesnt change it's theoretical orbital shape depending on it's mass, it probably just experiences small tidal forces that aren't much bigger than it's own centrifugal forces. The oscillation of the star position is a complex dynamic based on it's surrounding ...


3

Let $\xi^\alpha$ be a Killing vector of a metric $g_{\mu\nu}$, i.e. it satisfies $$ \nabla_\mu \xi_\nu + \nabla_\nu \xi_\mu = g_{\mu\alpha} \partial_\nu \xi^\alpha + g_{\nu\alpha} \partial_\mu \xi^\alpha + \xi^\alpha \partial_\alpha g_{\mu\nu} $$ Then the quantity $$ Q = \xi^\alpha u_\alpha $$ is conserved along any geodesic. To see this, we can compute $$ ...


2

But why is it the component of the dual vector $p_\phi$ that is constant rather than $p^\phi$? From the bottom of page 189: The geodesic equation can thus, in complete generality, be written $$m \frac{dp_\beta}{d\tau} = \frac{1}{2}g_{\nu \alpha,\beta}\;p^\nu p^\alpha$$ We therefore have the following important result: if all of the ...


4

We (physicists) believe the reason is this: known symmetries and conservation laws. For example, the mutual annihilation of a proton and positron would remove $2\,e$ charge units from the Universe. This violates the conservation of charge principle, which can be seen to arise from the application of Noether's theorem to the global gauge symmetry of the ...


3

One has to realize that Kepler's laws are a mere approximation. The motion of planets around the sun is a two-body problem. In case of such two-body problems, both the bodies revolve around the center of mass. But it turns out that Sun is much much heavier than the planets. So the center of mass of the system is very close to the Sun and Hence it is a good ...


0

Actually here $\rho$ and ${\bf v}$ are function of $(t,\vec{x}) \in \mathbb R \times \mathbb R^3$, as it is usual in the so-called Eulerian description of a continuous body. There is no reference to the curves describing the evolutions of the particles of the fluid $\vec{x}_{\vec{x_0}}= \vec{x}_{\vec{x_0}}(t)$ as instead, it is the standard in the so-called ...


11

Absolutely. If Compton scattering occurred in first order in $e$, the only contributing diagram would be the obvious one. Say we're in a frame with the electron initially at rest and an incoming photon in the $z$ direction. Then the electron 4-momentum is $$p^\mu_{\text{in}} = (m,0,0,0)$$ while the photon 4-momentum is $$k^\mu_{\text{in}} = ...


2

For two objects to remain in a stable circular orbit, the force acting on them must be equal to the centripetal force corresponding to their rotation. $$F=\frac{mv^2}{r}$$ or in terms of angular velocity $$F=m\omega^2r$$ where $r$ is the radius of orbit in this case. As the gravitational force acting on the two stars is the same. $F$ is equal in both ...


0

From the chain rule we have, $\frac{d \rho}{dt} = \frac{\partial \rho}{\partial t} + \frac{\partial \rho}{\partial x}\frac{dx}{dt} + \frac{\partial \rho}{\partial y}\frac{dy}{dt} + \frac{\partial \rho}{\partial z}\frac{dz}{dt}$ $\therefore \frac{d \rho}{dt} = \frac{\partial \rho}{\partial t} + \frac{\partial \rho}{\partial x} v_x + \frac{\partial ...


3

I will be assuming that the system has some angular momentum about the center of mass initially. If the system has no angular momentum then both the stars would accelerate towards each other and end up colliding. The problem is a two body problem. In such cases, both the stars would revolve around the center of mass. If the distance between two stars is ...


2

Yes, we expect all astrophysical black holes to have nonzero rotation. If nothing else, a rotating black hole that absorbs even a single particle with net angular momentum will then have nonzero angular momentum.


1

Gyroscopes depend on the conservation of angular momentum. Orientation and navigation gyroscopes are finely balanced/symmetrized so that gravitational fields will not exert external torque and modify the angular momentum. As the container which holds the gyroscope moves, a gimbal mount allows the gyroscope to maintain a constant rotational axis ...


2

Why is angular momentum conserved when a planet revolves about sun in an elliptical orbit? Why is linear momentum not conserved in this case? $$\rm \text{no external }\color{red}{torque}\to\color{red}{angular}\text{ momentum conserved}\\ \text{no external }\color{red}{force}\to\color{red}{linear} \text{ momentum conserved}\\$$ There is no external ...


1

Both are conserved if you consider the whole system: If the earth looses linear momentum, the sun will gain it and vice versa. Subsystems may violate conservation laws (e.g by transfering energy/momentum). This is called local violation. But globally conservation laws will always hold. The question why they hold globally in the first place, can be answered ...


1

Of course we can. First off, when discussing conversation laws, continuity typically refers to the conservation of mass specifically: $$\frac{D \rho}{D t} = 0 $$ in a Lagrangian frame or: $$\frac{\partial \rho}{\partial t} + \frac{\partial \rho u_i}{\partial x_i} = 0$$ Now, for momentum. This equation holds for both fluids and solids and is an expression ...


2

I'll make an example, to make things clear. Take a two body system, in which the particles are seperated by a constant distance $d$ and have mass $m_1 = m_2 = m$. This is a holonomic constraint, since $$ | \vec{r}_1 - \vec{r}_2 | = d $$ with the particle-positions $\vec{r}_1$ and $\vec{r}_2$. This system is therefore reduced to 5 degrees of freedom (6 minus ...


6

I think this should help clear things up. Suppose you take a rod at rest and apply a force $F$ perpendicular to the rod at a distance $r$ away from its center of mass for a short time $\delta t$ - short enough that the orientation of the rod does not change much during the time the force is applied. The rod's linear momentum will become $$F\delta t$$ (from ...


1

Acceleration of the center of mass is always $F/m$, so if force and mass are the same, the center of mass will accelerate the same way, regardless of the point where the force acts. After the same time of experiencing the same force, the body in the rotating case has greater kinetic energy than in the non-rotating case. This is due to greater work done by ...


2

Let's dive right into an example -- let's say you are simulating a fluid. First, you need to pick your reference frame. Are you going to simulate a fixed domain in space and have your fluid move through it, meaning you have a grid and at each point on the grid you store and solve for the fluid properties (Eulerian frame)? Or will you track each discrete ...



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