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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 ...

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.

12

I don't like these definitions of momentum. They may try to give an intuitive explanation but some will still find it hard to understand why these definitions correspond specifically to $m\vec{v}$ in Newtonian mechanics. The truth is that momentum is just a vector that we discovered that it is conserved when there is no external force on the system. It is ...

7

The answer to this is true for all definitions in physics. You are asking why angular momentum is defined as $\mathbf L=\mathbf r\times\mathbf p$ instead of $\mathbf L=\hat r\times\mathbf p$. The answer is because the former is very useful. In other words, not only is it useful to know the orientation of the momentum $\mathbf p$ relative to the origin, it is ...

7

I believe the answer is actually very simple. To be "in motion" simply means to have $\vec{v} \neq 0$. Nothing more, and nothing less. (Thus, as a consequence being in motion implies that speed $\neq 0$) In physics we don't usually talk about objects having "more" or "less" motion. It's more of a yes/no thing. I'm not convinced that the truck has "more ...

7

$d\vec p$ is not actually a vector, but a differential volume element. It is a bit of sloppy notation because sometimes $d\vec p$ means the vector $(dp_x,dp_y,dp_z)^T$ and sometimes it is the volume element $dp_xdp_ydp_z$. You can also use $d^3p$ to denote a volume element but both are used often. A more proper way to derive $m^3$ is using the Jacobian. ...

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 ...

6

As pointed out in the comments, your argument cannot be applied to systems with more than 2 interacting bodies. For $N$ bodies, you will arrive at $$\sum_i^N\sum_{j\neq i}^N\mathbf F_{ij}=0$$ which does not specify $\mathbf F_{ij}=-\mathbf F_{ji}$ for all $i,j$ for $N>2$. Therefore, Newton's third law is not obtained here. It must be an additional ...

6

The first thing you need to understand: you are applying the creation of physics definitions backwards. You are asking, "why isn't work given by this equation?", but this question doesn't make sense if you think about it. It is not the case in physics where we think, "Hmm... I want to define something called "work". What should it's equation be?" This doesn'...

5

The piston which is referenced in basic thermodynamics is not a real, dynamical object with mass or inertia. It's an idealization which serves only as the "handle" by which the volume of the system can be changed. To understand the model, consider an ideal gas with fixed $N$. The state of the system is a point in $(P,V,T)$ space which lies on a surface $... 5 Note that in many applications, there is a simplification: the Hamiltonian is quadratic in the momenta, i.e. the path integral over the momenta is Gaussian, which can be performed exactly. On the other hand, if the Hamiltonian has higher-orders of momenta, then the path integral over the momenta can generically not be performed exactly, and we have to rely ... 4 It is not really correct to say that those two equations disagree. An indeterminate form does not give a specific value, it could have any value. So it doesn’t disagree with any specific answer. When you see an indeterminate form it is an indication that you need to use a different formula. Often, if you look at the derivation for the equation you will ... 4 Your equation (2) is an expression for a particle with mass; it does not apply to photons. 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 No. There would be no direction for the vacuum expectation value of the momentum since all directions in a vacuum are equivalent. 3 Angular momentum is the moment of momentum just as torque is the moment of force and velocity is the moment of rotation. All the moment of xxx quantities are evaluated using the full radius vector since the moment arm (minimum perpendicular distance) to the line where the quantity is acting though is needed. $$\matrix{ \text{(moment of rotation)} = \... 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 ... 2 \newcommand{ket}[1]{\left|#1\right>}Marcus has given the mathematical answer using the commutation rules of ladder operators a_p and a_q^\dagger. I want to give a bit the intuition behind the operator P^\mu and how you can almost guess the answer. Note that this intuition is solidified by the calculation that Marcus gives and in no way replaces it.... 2 Use the canonical commutation relation [a_p, a^{\dagger}_q] = (2\pi)^3\delta^3(p-q): We have that a_{p_2}p^{\mu}a^{\dagger}_pa_pa^{\dagger}_{p_1} =p^{\mu}a_{p_2}a^{\dagger}_pa^{\dagger}_{p_1}a_p + p^{\mu}a_{p_2}a^{\dagger}_p[a_p, a^{\dagger}_{p_1}] The first term is ignored, because when considering <0|p^{\mu}a_{p_2}a^{\dagger}_pa^{\dagger}_{p_1}... 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 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 ... 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 Here's an intuitive way, I believe, to understand it. First off, we need to have some definitions, to understand exactly what we're talking about - in particular, we need to know what exactly we mean by a "closed system" or even a "system" in general, here. This is an important and crucial part of any deductive reasoning (and missing it is the source of ... 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 Okay, so, for a free particle, with the hamiltonian$$ H = - \frac{\hbar^2}{2m} \Delta $$living in a cavity of volume V (for any shape of the cavity) number of energy eigenstates with energy less than E has the following asymptotic$$ N(E) \sim \frac{V}{(2\pi \hbar)^d} \omega(\sqrt{2mE}),\, E \gg \frac{\hbar^2}{2ma^2}$$Here$\omega(R)$is a volume of$...

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

The total momentum at any point will be zero but as there are minute fluctuations in electric and magnetic fields, it can be stated that vacuum has momentum. Do check out the existence of virtual particles. This might help you understand deeper about properties of vacuum

1

The initial work, 𝑓𝑑𝑥=𝛿𝑄, at the very beginning will give the piston momentum, after initially being at rest. Thus, no further heat should be required to expand the gas after an initial infinitesimal flow of heat, unless there are friction forces that will bring the piston to rest. What you are describing would not be considered a reversible ...

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