# Tag Info

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.

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 ...

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 ...

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 ...

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 ...

2

We have long been taught that electric charges are neither created nor destroyed. No, we have not been taught that. We've been taught that electric charge, i.e., the net electric charge, is conserved. Imagine that, within some volume there is some net electric charge Q. Assuming there is no current through the boundary of the volume, we are taught ...

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\, ...

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, ...

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 ...

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?

2

The moon orbits the earth with a near circular trajectory relative to the earth. So add earth's orbital velocity (around the sun) to the moon's orbital velocity (around the earth). This will put the moon into an orbit around the earth, but might make it a bit more eccentric (elliptical). To correct this you can use angular velocity around the sun with ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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, $\partial_t \rho^0 = ... 1 A conservation law typically has the form$\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 ...

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

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

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 ...

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 ...

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