# Tag Info

1

Here is the diagram you are discussing: It seems you are worried by the angular momentum carried by the W+. The W+ is a virtual particle in this reaction. In virtual paths the particle is off mass shell and its mass is unphysical, and angular momentum as a part of a four vector will be a complicated function also having unphysical measure, so ...

1

The W is massive so can be in the spin 0 state (or $s=-1, 0, 1$ in general). The photon is massless so does not have this "longitudinal" polarization. For the massive vector boson, the relevant symmetry group is the little group $SO(3)$, and for the photon it is $SO(2,1)$.

2

Your 1D "geometric" approach obviously fails in higher dimensions, as momentum is a vector quantity whereas you've described it as a scalar. The vectors you describe aren't spatial vectors of course, and so for higher dimensions (and more particles) they would seem to complicate things too much. Anyways, regarding the second part of your question, there is a ...

0

On the internet there is plenty of talk of how the continuity equation applies to conservation of charge, fluid dynamics, and so forth, but I can't find any mention of how it applies to the conservation of energy. Why? Is it because it is problematic to talk about energy current density (j)? The continuity equation is fine for energy, and sometimes ...

0

The fundamental reason why energy is conserved, is invariance of the physical laws by a time translation $t \to t + t_0$, in a Lagrangian formulation. This is a particular case of the Noether theorem, which states, that if a Lagrangian has a continuous symmetry, there is a corresponding conserved quantity. Now, if we look in detail, conservation of a ...

0

Usually, when physicists talk about energy being conserved, they mean Energy being a Noether charge on the fundamental level, c.f. wikipedia. As a very general result, one can derive that the time derivative of Energy is zero $$\frac{d}{dt} E = 0.$$ This result is only true if the Lagrangian description of your system does not explicitly depend on time ...

1

The "center-of-charge" is part of a more general concept that is used quite often in physics: Multipole expansion. The general idea of multipole expansion is the following: If you view a charge (or mass) distribution from a large distance, then most of its internal structure is irrelevant to you. Instead, it suffices to do all calculations based on a few, ...

3

Mass and charge are not so similar for the charge having "center of charge". The notion of "center of mass" appears in many applications when number of bodies move. In this situation, the movement can be splitted into movement of center of mass and individual movements of bodies relative to the center of mass. This occurs because of dual role of mass: it ...

8

Of course you can define such a quantity, but the question is: does it mean anything physically? Contrary to what has been stated in some of the answers/comments, this quantity is not comparable to a "normalized" dipole moment. A dipole is a system of two charges equal in magnitude but opposite in sign. The corresponding dipole moment, which is of great ...

3

There's another way to do this also, more akin to how spacecraft actually do it: Take a weight on a string, hold it up and spin it. You'll turn in the opposite direction. When you stop it you also stop turning. Of course this will produce an off-axis force that will be a real pain to deal with. Real spacecraft do it by means of a set of internal wheels ...

19

For those that are cat-challenged, here's an alternative explanation and demonstration you can try at home! This demonstration was taught to me by my math lecturer. All you will need is: A swivel chair and a heavy object (e.g. a big textbook) Stand on the seat of the chair (watch your balance now) holding the heavy object. Extend your arms forward ...

43

The astronaut can change his or her orientation in the same way that a cat does so whilst falling through the air. After the transformation, the astronaut is still and angular momentum is conserved. There is a rather beautiful way of understanding this rotation as an anholonomy i.e. a nontrivial transformation wrought by the parallel transport of the cat's ...

3

Other answers have pointed out other ways that might be more efficient, but one very simple way to do it is as follows: start with both arms parallel to the body. Then swing them both backward, up over the head, and then back down in front of the body, leaving them back in the starting position. After this manoeuvre, the body will be oriented in a slightly ...

4

Consider the time derivative of the Hamiltonian $$\frac{dH(q,p,t)}{dt}=\frac{\partial H}{\partial q}\dot{q}+\frac{\partial H}{\partial p}\dot{p}+\frac{\partial H}{\partial t}=-\dot{p}\dot{q}+\dot{q}\dot{p}+\frac{\partial H}{\partial t}$$ From this you see that the Hamiltonian is conserved if it does not depend on time,$t$, explicitly. $H$ may or may not be ...

0

There is no inconsistency. $L_z$ is conserved, but $h^s$ is not; hence it is not a paradox that $h^s$ has no action on the trajectory at a certain time and nontrivial action at others. On the other hand, if the trajectory ever does cross the $z$ axis, then $L_z=m(\dot y x-\dot x y)$ will also vanish. Since $L_z$ is conserved, you can conclude that ...

2

If we have some coordinates $q_i$ and some momenta $p_i$, then a generator of a transformation is defined as a function $g(q_i, p_i)$. By definition, this generates the transformation $$q_i \to q_i + \epsilon \frac{\partial g}{\partial p_i}$$ $$p_i \to p_i + \epsilon \frac{\partial g}{\partial q_i}$$ So if we want the generator of translations, we want ...

1

By definition, the tension in the string can only supply a force towards the center of the circular motion, so in one sense the tension is a centripetal force. However, at any moment, the force of gravity can be decomposed into radial and tangential components. It is the sum of the tension and the inward radial component of gravity that supplies the ...

3

Yes, tension only affects the direction of the particle's velocity. This is because it is always perpendicular to the velocity, and because work is actually the dot product of force and displacement: $$W = F \cdot s = |F| \times |s| \times \cos(\theta)$$ , a force perpendicular to the displacement does no work

1

It is easier to start with conservation of momentum, because even though total energy is conserved kinetic energy may not be conserved. For example if you collide two tennis balls covered in glue, they will stick to each other and the final kinetic energy will be zero (total energy is still conserved because the original kinetic energy ends up as heat in the ...

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