# Tag Info

## New answers tagged conservation-laws

0

Different vacua will necessarily yield different answers to this question. Also, you would need to actually know what the energy value (vacuum expectatation value) of the vacua we are in actually is. Evidently, we don't. The fact that the Higgs mechanism does not violate conservation of energy in this vacua does not mean that it cannot violate it, nor ...

2

I've made a small illustration depicting the key idea. If this is in coherence with what you've asked, we could summarize some important points about the case. Total energy of the Earth and Bar Magnet system is given by the equation: $KE + PE = \frac{1}{2}mv^{2} + \frac{GMm}{R}$ While PE is there for both Earth and magnet system (combined), KE is ...

2

This should go on forever, and current should keep appearing across the load resistance. This is a contradiction. Since there is current through (not across) the load resistance, there is work being done on the load: $p = i^2R$. Let's be clear on this: the coil-load system does no work on the pendulum, the pendulum does work on the coil-load ...

1

The force on the pendulum only applies when the pendulum is in the vicinity of the coil. At that moment the harmonic motion of the pendulum is distorted. It 's amplitude is lessened and with it the upward motion. So the kinetic energy of the pendulum is converted into gravitational energy and electric energy. But the gravitational energy is less than without ...

10

As the magnet approaches the solenoid, a current is induced. The current generates a magnetic field. The field repels the magnet, slowing it's approach. The amplitude of the oscillations diminish. If there was no resistance, this would work in reverse as the magnet receded from the solenoid. The magnetic field would accelerate the magnet. The magnet would ...

2

Actually pendulum wont oscillate forever.Its energy turns into heat in resistor.In other word it's domain would be something like this figure.

1

The short answer is that there is a induced force on the magnet. This induced force will make the pendulum loose energy in the same proportions as there is electrical energy being generated. A good experiment to show this effect is by having a small bar magnet and a copper pipe Or solenoid. When you let a small bar magnet drop from a certain height it will ...

2

While I agree with the caveats made by dmckee in his comments, there is an obvious interpretation of stopping power as the change in momentum caused by the projectile. The mass and velocity of the projectile are $m$ and $v$ respectively, and the mass of the target is $M$. Since the target is stationary the initial momentum is just $mv$. Assuming the ...

0

Yes the skater does increase the angular momentum by doing work; pulling her arms in. You do work on a swing (sitting up and down) to increase your angular momentum likewise.

1

A proton has a positive charge so by charge conservation it is not possible to reduce a proton to uncharged radiation particles such as photons (assuming that is what you mean by "pure-energy") Because of gauge invariance charge conservation is likely to hold good in all future physics, but we can't be totally sure of that. It is possible that some charged ...

0

The simple answer to the main question is, yes. There are two ways to annihilate matter without using anti-matter. One is called fission, and the other is called fusion. Although only some of the matter is converted into energy in either of these processes, the efficiency of the "annihilation" is not in the main question. If 100% annihilation is required, ...

0

The forces act on BOTH the bodies involved, not on the same one! That's why the statement of Newton's third law is: The third law states that all forces exist in pairs: if one object A exerts a force FA on a second object B, then B simultaneously exerts a force FB on A, and the two forces are equal and opposite: FA = −FB Source:Wikipedia. So it isn't ...

1

First of all, mathematical definitions of force and momentum aren't really very intuitive or common-sensical. Just ask Aristotle for his common sense laws of forces! The fact that momentum is conserved in closed systems is a highly non-trivial fact, as is the Third Law. The reason that these laws exist at all is because you can't really 'see' or' feel' ...

1

Alternatively, and qualitatively, think about the components of velocity (in the x y directions) have changed. Along the x axis, velocity has reduced, so the re has been a force in the -x direction. In the y axis, velocity has changed sign, so there must have been a force in the -y direction. Hence the total force is down and to the left, ie quadrant ...

1

The total impulse is the change in momentum (note that this is a vector equation): $$\vec{I} = \vec{p}_{final} - \vec{p}_{initial}$$ You know the momentum before and after the collision so you can calculate the total impulse, both magnitude and direction. Impulse if force times time, so the direction of the force will be the same as the direction of the ...

1

This post has some information about impulse that you might find useful. Homework Question involving Momentum You will not find conservation of momentum useful here. True, the total momentum of object + wall is unchanged by the collision. But the momentum of the object does change. Since $\Delta P = J = F_{avg} \Delta t$, the direction of $F_{avg}$ and ...

13

The definition of an antiparticle is dependent on having the opposite quantum numbers of the particle so that they can annihilate, i.e. the sum of the conserved quantum numbers are zero. Thus the answer by @mpv is adequate. The implication of your question is then: is baryon number conservation a strict law or an emergent law that may be violated at some ...

2

I just started here so I don't have the rep. to comment and I don't have the time for a full answer, but the black hole idea mentioned in the comments above is a fine answer. See, for example, http://arxiv.org/abs/0908.1803v1 and How would a black hole power plant work?

16

I am assuming that by "energy" you mean photons. So you want to transform protons into photons. It is not possible. It would violate several conservation laws - mainly the charge conservation (protons are positively charged), but also baryon number conservation. The antiparticle is necessary to cancel these quantum charges to make the transition possible.

1

If you substitute the decomposition in, you get: $$\partial_t \rho^0 + \partial_t \rho^{E1} + \nabla \cdot (\rho^0 \mathbf{u}^E) + \nabla \cdot (\rho^{E1} \mathbf{u}^E) = 0$$ Typically the decomposition used assumes that $\rho^0$ is constant in time and that $\rho^{E1}$ is random in time, such that it's mean value is 0. Therefore, $\partial_t \rho^0 = ... 2 Elementary particles differ in flavour from their antiparticles, so conservation laws do, indeed, restrict whether particles or antiparticles can be produced in certain processes. (Compare, e.g., the photon, which has zero for all its flavour quantum numbers, and is identical to its antiparticle.) For example, when a neutron decays, the result is a proton, ... 0 OP wrote (v3): Working out the conserved quantity, we get that the$z$-component of angular momentum$ L_z = m x(t)\dot{y}(t) - m y(t)\dot{x}(t)$is conserved for any path$(x(t),y(t),z(t))$. Well, it should be stressed that the conservation law in Noether's theorem is an so-called on-shell conservation law. It does not hold for any curved (off-shell) ... 1 When you're asking practically important questions about the mechanical strength of materials, size matters for many reasons. Let's talk about dislocations, although the same discussion also applies to other defects and impurities. The density and motion of dislocations are critical factors in determining the mechanical strength of materials. A large ... 1 Define$F(u):= \int_0^u f(s) ds$, so the equations for the field$u(t,\vec{x})$can be re-written as $$\frac{\partial^2 u}{\partial t^2}-\Delta_{\vec x} u + \frac{dF}{du}=0\::$$ If defining $${\cal L}:= \frac{1}{2}(-\partial_t u\partial_t u + \nabla u \cdot \nabla u) + F(u)\:.$$ this Lagrangian density leads to your field equations. Moreover, as you can ... 1 A conservation law typically has the form$\frac{\partial}{\partial t} (\text{Volume Density of a quantity}) + \mathrm{div}\,(\text{Flux per unit area of that quantity}) = 0$, in the local / differential version. The integral version is more familiar:$\frac{\partial}{\partial t} (\text{Quantity inside volume}) = - \text{Total flux of quantity through ...

2

The limit to which you refer is known as the thermodynamic limit in statistical mechanics. It consists in taking the limit of infinite particles ($N\rightarrow \infty$) and infinite volume ($V\rightarrow \infty$) while keeping a finite density $N/V$. In a solid, both electrons and atomic nuclei contribute to the thermodynamical and elastic quantities, such ...

2

A "kosher" way to do this employs test functions. Consider a test function $\phi:\mathbb R^4\to \mathbb R$. Notice that \begin{align} \int_{\mathbb R^4} d^4 x\, \partial_\mu j^\mu(x) \phi(x) &= ec\int_{-\infty}^{\infty} ds\,u^\mu(s)\int_{\mathbb R^4} d^4x\,\partial_\mu\delta^4(x - X(s))\phi(x) \\ &= -ec\int_{-\infty}^{\infty} ds\, ...

1

Is there a rigorous derivation of the limits for continuum properties in solid mechanics Continuum properties stop holding whenever we reach the quantum mechanical regime, i.e. where atoms and molecules become distinct and follow quantum mechanical solutions rather than classical collective ones. What spatial scale separates the two regimes ...

1

This has a simple closed-form solution. Denoting $m_0,m_1$ as the initial and final person's mass, $v_r$ as the rice speed and $\delta=m_0/m_1$, if the bag is thrown in one single parcel, we have $$\Delta v_1=(\delta-1)v_r$$ By the rocket equation, if the rice is thrown continuously, we have $$\Delta v_2=v_r\text{Log}(\delta).$$ But $$\text{Log}(\delta)\leq ... 1 You need a model for how you throw the rice. The obvious one is that you can expel any mass at the same velocity v relative to you. Letting M be your mass (without the rice), V your velocity in the CM frame, if you throw it as one lump we have momentum conservation. You start with no momentum in the CM frame, so 10v=MV, V=\frac {10v}M. If you ... 0 Given a bag of rice of mass m_b that you can throw with a maximum acceleration \vec{a}_b, by Newton's second law, the most force \vec{F}_b you could exert on the rice is given by$$\vec{F}_b = m_b \vec{a}_b$$By Newton's third law, the reaction force (acting on you) \vec{F}_{you} is given by$$\vec{F}_{you} = - \vec{F}_b = - m_b \vec{a}_b$$Again ... 4 To stop instantly, you would need infinite deceleration. This in turn, requires infinite force, as demonstrable with this equation:$$\vec F=m\vec a$$So when you hit a wall, you do not instantly stop (e.g. the trunk of the car will still move because the car is getting crushed). In a case of a change in momentum, m\vec v, we can use the following equation ... 0 OP, I am new to stackexchange (but a physics veteran) so I am not yet allowed to comment on the post itself, but you should know that the one you chose as the correct answer is only valid for one dimensional curves, and even there it is valid only for a special definition of symmetry that allows for boundary terms (called "quasi-symmetries," as QMechanic ... 0 I recommend you "Symmetries" by Griffiths, or "Symmetry" by Roy McWeeny (it's a Dover book). "Geometry, Topology and Physics" by Nakahara is a good option. 1 The technical answer is No. Surprisingly I think Wikipedia gives the better definition, though I think both authors are trying to say the same thing. Let the action be defined as S[\varphi]=\int d^4x\ \mathcal L(\varphi(x),\partial_\mu\varphi(x)) A differentiable symmetry is a symmetry of the functional that does not change the action ... 4 Yes, provided one uses the correct notions of symmetry for the action and the lagrangian. The setup. We assume throughout that the action can be written as the integral of a local Lagrangian. Namely, let \mathcal C be the configuration space of the system, then for any admissible path q:[t_a, t_b]\to \mathcal C, there exists a local function L of ... 3 First some terminology: In general an infinitesimal transformation of a field theory consists of a so-called horizontal infinitesimal transformation$$ \delta x^i ~=~x^{\prime i}- x^i$$of the base manifold, and a so-called vertical infinitesimal transformation$$ \delta_0\phi^{\alpha}(x)~=~\phi^{\prime \alpha}(x)-\phi^{\alpha}(x) $$of the fields. The ... 0 Well I say charge indeed has momentum... Given F = qE and realising that F = \frac{dP}{dt} It follows \Delta P = qE \Delta t . We define \Delta P to be the charge momentum and E\Delta t = (iota)$$ From this it is clear that the charge momentum (which equal mass momentum) are two different entities, in fact it can be showed that the two exist in ...

1

As Jan noted, the Hamiltonian should have a minus sign: $H=\frac{(p-qA)^2}{2m}$ where $p$ is the canonical momentum, and the expression $p-qA$ is the kinetic momentum $P$. A homogenous magnetic field is an interesting case, because the vector potential in a given gauge does not exhibit translation invariance, but the physical system clearly does. The ...

0

Maybe an example helps. Let $B$ be a constant magnetic field. Then we can take $A=\frac12B×x$. Now $$\frac{(p+qA)^2}{2m}=\frac{p^2}{2m}+\frac{q}{2m}(p⋅A+A⋅p)+\frac{q^2A^2}{2m},$$ and $$p⋅A+A⋅p=l⋅B$$ where $l=x×p$. Thus $$\frac{(p+qA)^2}{2m}=\frac{p^2}{2m}+\frac{q}{2m}l⋅B+\frac{q^2A^2}{2m}.$$ Here we recognise the $l⋅B$-term as the Zeeman term. If we now ...

0

Hamilton's equations state $\dot{P_i} = -\frac{\partial H}{\partial q^i}.$In this case, this is $\dot{P_i} = -\frac{\partial H}{\partial q^i} = -\frac{\vec{P}}{m} \cdot \frac{\partial \vec{A}}{\partial q^i}.$ So the canonical momentum is not conserved.

0

Hint: $P$ is conserved if it is not explicitly time-dependent and if its Poisson bracket with the Hamiltonian is zero. So you just neet to check that: $$\{P,H \}=0$$

Top 50 recent answers are included