45

Much of mass is just binding energy, so in a chemical reaction the electrons rearrange themselves and energy is released and the total mass of the molecules goes down (in an exothermic reaction, for example). The same is true for common nuclear reactions like spontaneous fission of uranium, with the caveat that some important nuclear reactions do involve ...


32

First off, I'd like to point out, because the term comes up often, in strict usage there is no such thing as "pure energy". Energy is not a stuff, as in particles, but a number that is associated with stuff, a quantity, though that quantity can indeed be thought of as acting, given its conservation property, like a sort of intangible "stuff" that you can put ...


28

While there are some good answers, no one has yet mentioned Noether's theorem. Fundamentally, the conservation of momentum and conservation of energy have different characters. Notably conservation of linear momentum is linked with the laws of physics being independent of where something happens and conservation of energy is linked with the laws of physics ...


18

Einstein, in his famous but rarely-read paper "Does inertia of a body depend upon its energy content?", didn't write that matter (or, better, mass) can be converted into energy. The way he stated its result was: $$ \Delta E = \Delta m \cdot c^2, $$ i.e. in the rest frame of any system, variations of mass and variations of energy (of the system) are ...


17

The total energy (internal + kinetic) of hot body isotropically emitting radiation while moving in the vacuum far from other bodies, will decrease, while momentum is conserved.


10

To start with the $m$ in $E=mc^2$ is the relativistic mass, and in particle physics this is out of use it causes confusion. One uses four vectors , where the "length" of the four vector is the invariant mass, uniquely identified with elementary particles, and with systems of elementary particles. Four vectors are good in keeping track of energy, as one of ...


6

Due to the principle of energy conservation, energy is always conserved for an isolated system in an inertial reference frame. Therefore, considering the conservation of all forms of energies (the sum of them), the question does not make much sense. It makes sense if referring to the mechanical energy only, whose conservation is a theorem and not a ...


6

Yeah, it's interesting to contemplate the relation between energy and momentum. Energy can morph from form to form, with kinetic energy being only one of them. Momentum, on the other hand, deals with a subset of all forms of energy: motion. So let's set up a system with the energy oscillating between potential energy and kinetic energy: let's say a ...


4

Go to the gauge and Higgs kinetic & potential terms of the EW model and read off the vertices: The fun should be the actual strengths of the couplings, which I should not spoil by revealing it to you. (To check your work, go to Appendix B of Li & Cheng). Gluons only couple to themselves, among the bosons, so you have cubic and quartic terms for ...


4

If you’re just considering mechanical energy, consider the case of a firework or other exploding object. The total momentum is conserved, but the kinetic energy is increased from the chemical energy of the explosion. More formally, momentum conservation is associated with isotopic space while energy conservation is associated with no changes over time: ...


3

In Newtonian theory, you're dealing with only gravity and EM forces which are conservative and central so there is no example of a closed system where either energy or momentum are not conserved. So energy and linear and angular momentum conservation are always implied by Noether's theorem. Your example is wrong even if you only consider linear momentum for ...


3

You are basically asking about mass energy equivalence, and nuclear reactions. And you would like to know what happens to the (I assume rest) mass and matter in the process. The mass–energy formula also serves to convert units of mass to units of energy (and vice versa), no matter what system of measurement units is used. However, use of this formula in ...


3

If there was a superdense particle of mass M/2 ...... the superdense material falls down so as to keep the centre of mass at rest. Yes it does. Newton's law of gravitation states that : $$\mathbf F_{12}=-\mathbf F_{21}=G \frac {m_1m_2}{r^2} \hat r$$ Therefore the planet of mass $M$ moves towards the superdense particle with acceleration $a$ and the ...


3

Let's go through this step by step. Just to be charitable, let's also assume that you can recover 100 percent of the ball's kinetic energy when it reaches the ceiling, and that the only force acting on the ball as it travels is gravity (in other words, we're ignoring air resistance). Before the ball is initially dropped, it has a gravitational potential ...


3

When the bullet lodges, a large part of the original kinetic energy is converted into heat (and also sound etc), so you can't use conservation of energy to solve the problem. Such collisions are called inelastic collisions. Elastic collisions are when the kinetic energy is conserved. Intuitively, they are when the objects bounce perfectly off each other.


3

Yes, the velocity can vary. This is the power of the principle. In Newtonian mechanics the idea is that if there is a cloud of particles all with forces between them and so all accelerating, nevertheless the total momentum of the cloud is unchanged (assuming no external forces). One way to think of a force is that it is the transfer of momentum between two ...


2

Yes, this decay is possible, and measured to occur, via diagrams similar to: involving two W bosons. Note that since it involves one more W boson than the $\rm K^0$ to pion decays, the decay is suppressed pretty strongly. There are some subtleties involved here regarding CP violation, depending on whether the original particle is a $\rm K^0_L$ or $\rm K^...


2

The 3D spherically symmetric harmonic oscillator $$ H~=~\frac{p_x^2+p_y^2+p_z^2}{2m}+ \alpha (x^2+y^2+z^2) ~=~ H_x + H_y + H_z $$ is a separable, Liouville integrable, and in fact a maximally superintegrable system with additionally integrals of motion $H_x$, $H_y$, $H_z$, $L_x$, $L_y$ and $L_z$, i.e. nothing exotic like the Laplace-Runge-Lenz vector. (A 3D ...


2

Since Weinberg is, after all, a physics book rather than a book of rigorous math, I'm not convinced that one should attempt to understand "necessary" here in its rigorous logical meaning rather than a colloquial meaning. In any case, whether the Hamiltonian is "necessarily" formed from such fields is an ill-defined question to begin with: Suppose we start ...


2

As dmckee says in the comments, you have found a first order diagram. What are Feynman diagrams? they are the diagrammatic representations of complicated integrals that will give the cross section of an interaction. These integrals are a series expansion , and each term has a diminishing constant allowing to use the first terms to first order, the way you ...


2

For a single isolated particle (with constant mass) conservation of momentum does indeed imply conservation of kinetic energy (and also of mechanical energy, as if it is isolated there is no force applied to it and therefore there is no potential energy). This is easy to prove: Let $\vec{p}=m\vec{v}$ be the momentum and $K=\frac{1}{2}mv^2$ the kinetic ...


2

Example of momentum conservation, while energy is not conserved (sticky collision): Just let two pieces of sticky stuff with opposite momenta (so in total zero momentum), collide frontally. The end result is a piece of this stuff that has also zero momentum but the kinetic energy has conversed into internal heat (energy is alway conserved, just like total ...


2

Momentum is always conserved. Energy is always conserved. Since momentum is always conserved, it's irrelevant what's going on with kinetic energy. If you were considering energy conservation, then you would just want to keep in mind that kinetic energy isn't conserved -- the total energy is. If a system is not isolated, then you can't just count that ...


1

According to the No-hair theorem (or, more generally, conjecture), linear momentum is conserved in a black hole along with angular momentum, mass-energy, and electric charge. Also, nothing in the theory of dark matter contradicts the conservation of momentum as a Noether current due to the translation symmetry of space (see the Noether theorem) plus the ...


1

Conservation of Momentum Momentum of a system is conserved whenever the net external force acting on the system is $0$. Conservation of Energy Energy is conserved whenever the net work done by external forces are zero i.e., there is no exchange of energy from outside. Note that the conservation laws are independent of one another.


1

First of all, what was said in the question and assumed wrong: Constant heat capacity means $$\partial_{\rho}\partial_{T}\hat{u}=\partial_\rho \big(\partial_T \hat{f}(\rho, T) - \hat{f}(\rho, T) - T\partial_T\hat{f}(\rho, T)\big) = \partial_\rho \big(\partial_T f_2(T) - \hat{f}(\rho, T) - T\partial_T f_2(T)\big) = \partial_{\rho} \hat{f}(\rho, T) = \...


1

You indeed have to consider a fixed spatial volume $V$ contained in a set of boundaries $\Sigma$. For steady-state flow and an incompressible fluid the continuity equation degenerates to the given term but there are two fluxes over the surface that need to be considered, one entering the control volume in $S_A$ and one exiting it in $S_B$. As for the ...


1

will the reading on the weighing machine change even for a second due to the gun`s recoil? Yes it will increase briefly. When the gun fires the bullet upward it applies an upward force on the bullet. Per Newton's third law, the bullet applies an equal and opposite downward force on the gun. Since you are holding the gun, that downward force is ...


1

That is a very good question! Before we answer it, please however have in mind that shooting a projectile into air is dangerous and should not be done without proper precautions. The weighting machine reading depends on force acting on it, in this case it is our weight. Finally to properly answer the question, the reasoning is as follows: When we shoot ...


1

Comments & hints: The invariance $\delta L=0$ of the Lagrangian (1) under the infinitesimal transformation $$\delta q^a~=~\epsilon v^a(q), \qquad \delta t~=~0,$$ follows directly from the Killing equation (2). According to Noether's theorem, the corresponding conserved Noether charge $Q=p_av^a$ is the momentum $$p_a~=~\frac{\partial L}{\partial \dot{q}^...


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