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32

We know through experimental observation. That is the beginning and end of the subject of physics, at least the part of it the tells it apart from, say mathematics. Conservation of momentum is simply an inductively reasoned hypothesis to summarize certain patterns in experimental data. You are alluding to the conservation of momentum's being "explained" ...


12

The answer by WetSavanna... is complete but I want to particularly address the part I hope it is clear that I'm not trying to suggest that I don't trust these laws to be true but rather that I'd like to know how we know they are true. Physics theories are mathematical models that fit current observations/data and are predictive of new ones. Prediction ...


6

Update 1: 1) Note added in proof: The photon stress-energy densities obtained below more or less heuristically are identical to those obtained in more rigorous approaches from the electromagnetic stress-energy density tensor. 2) The physical reason why the stress-energy argument retrieves the detailed balance result in the OP, but is inequivalent to simply ...


5

There is no doubt the Newton's third law holds in this situation. The source of confusion might the fact that you are neglecting the time interval of the collision so it is better to approach this problem by the momentum principle. The impulse of a force $F$ is given by $$I\equiv\int_{t_1}^{t_2} Fdt=\Delta p,$$ where $\Delta p$ is the momentum change due to ...


4

A eigenstate of a crystal hamiltonian can be written as a Bloch function in space representation $$ \psi(\mathbf{r}) = e^{i\mathbf{k}\mathbf{r}} u_\mathbf{k}(\mathbf{r}) $$ $u$ is periodic with respect to the unit cell. The momentum is now given by $$ \langle \psi|\hat{\mathbf{p}} |\psi\rangle = -i\hbar \int e^{-i\mathbf{k}\mathbf{r}}u_\mathbf{k}^*(\mathbf{...


4

When exploring deep questions in physics, like you are, it is important to remember that nature appears to obey the laws of physics. Empirical studies such as science cannot prove ontologically that nature does obey the laws of physics, or obey laws at all. For an extreme test case, consider the concept of idealism. In idealism, one claims that there is ...


4

We know or reasonably assume that momentum and energy are conserved because of two reasons mainly: Mathematical plausibility: If we assume that nature follows mathematical descriptions, then e.g. Noether's theorem makes it a necessity that momentum is conserved. Otherwise, something with the mathematical description would be wrong. And so far, in the vast ...


4

$\hat{X} = i\partial_{k_x}$ in momentum space is wrong. The left hand side is an operator, the right hand side is the representation of an operator when acting on functions, taken with respect to the scalar product with the basis $|\ p\rangle$. In particular we have $$ \langle x\,|\,\hat{x}\,|\,\phi\rangle = x\phi(x) $$ and $$ \langle p\,|\,\hat{x}\,|\,\...


3

The forces acting on the fluid are changing its momentum. There are forces acting on the stationary portions of the control volume boundary, and there are also forces being applied to force fluid into, and acting to prevent it from flowing out of the control volume. The total rate of change of momentum in the control volume at any instant of time is equal ...


3

Car collision "damage" usually goes with the energy in the zero momentum frame. In both cases that is (since in the zero momentum frame, the two cases are equivalent, assuming the masses of the cars are equal): $$E_1 = 2 \times \frac{1}{2} m v_{rel}^2 = m \left( 30 \frac{km}{h} \right)^2$$ Therefore a priori there is no difference between the two situations....


3

TL;DR: The physics of hitting things are not as easy as exerting a constant force on something. What I am trying to say with that is that Newton's law of course applies, but it would be more obvious to see it if you were just pushing/leaning against the wall with your weight. Then I'd say the two walls probably feel roughly the same. So what is different ...


3

Although this is almost an engineering question, there are some simple physical constraints that mean the answer is an emphatic no. At staging, significant change in momentum / trajectory of the payload-bearing stages can only arise from high impulse transfer between the payload bearing and jettisoned stage; this statement is a reformulation of conservation ...


3

As indicated in the answer to your previous question, since as $E(=30\:\mathrm{eV})$ is higher than $V_2(=20\:\mathrm{eV})$, that particle is not bound, it's not an eigenstate of the system's Schrödinger equation (it's a scattered state). Its wave function for $x \to +\infty$ would something like: $$\psi=c_1e^{-ikx}+c_2e^{+ikx}$$ ... where both complex parts ...


3

If you shoot light out the backside of your craft then indeed you would move towards the front. But then you'd use up some energy in order to go because the light that leaves would carry energy as well as momentum. The energy of your craft would decrease and thus the mass of your craft would go down. You'd arrive at your destination with a less massive craft ...


2

I would call this quantity "first moment of mass" or just "moment of mass". Have a look at this wikipedia article to read about the general concept of moments in physics. As pointed out in the comments to your question, the first moment of mass is closely related to the center of mass (CoM). For a collection of particles with masses $m_i$ and positions $\...


2

I have asked a professor about this and he gave me the answer. After replacing $\mathbf{k}$ by $-i\nabla$ in $H(h\mathbf{k})$, we are actually getting a new Hamiltonian that acts on envelop of wave functions. To make this answer relatively complete, I will briefly introduce the main steps focusing on only one band. Suppose the band we are interested in ...


2

Let's consider the frame in which initially both the masses are at rest to be the frame $O$. In frame $O,$ momentum conservation is trivially followed because of the symmetry of the problem. For the energy conservation, we require that $M = m \sqrt{1-v^2}$, where $m$ is the initial rest mass of each of the particles and $M$ is the final rest mass of each of ...


2

If there is no gravity in your simulation, your thoughts are on the right lines. The amount of centripetal force (=thrust) required to maintain a space-craft of mass m in circular orbit of radius r at speed v is $F=mv^2/r$. Thrust has to be directed towards the centre of the circle and has to be maintained constantly - so unlike orbiting a planet using ...


2

@ChesterMiller: answer is good, but I would just simplify it. What you say is that the total momentum is the momentum of the contents of the control volume plus the sum of that which traverses the boundaries. If you take the change in momentum per unit time, you have momentum flux, which equals force.


1

The more fundamental thing to understand are the conservation laws, particularly the conservation of momentum and of energy. The availability of energy gives rise to force. When you move your fist towards an object at a particular velocity you contain within it a kinetic energy and momentum. Materials are held together with binding forces that, at the ...


1

A simpler version of your construction: A man runs at 3/4 times the speed of light on top of a train that drives at 3/4 times the speed of light. Is the man now running at 1.5 times the speed of light? No, he isn't. If he was, there would be no issue with relativity in the first place. In fact he is never reaching the speed of light. It is ...


1

Thanks to Ján Lalinský for getting me to double check my faulty memory. My assumption of $\langle\chi_n|[H_0,z]|\chi_m\rangle = 0$ was wrong. Here is the solution after double checking my work. $$\langle\chi_n|[H_0,z]|\chi_m\rangle=\langle\chi_n|H_0\,\,z|\chi_m\rangle-\langle\chi_n|z\,\,H_0|\chi_m\rangle$$ $$\langle\chi_n|[H_0,z]|\chi_m\rangle=\langle\...


1

You must look at all forms of energy. Just before the explosion, the projectile has gravitational potential energy GPE, kinetic energy KE, and also chemical potential energy CPE stored in the dynamite. Just after the explosion the 3 fragments all have the same GPE as before. The CPE has disappeared in the explosion. As Jim says, we must assume that it ...


1

Let's first address the general question--do "waves" of any kind have (or can they have) inertia? I suppose here by "inertia" you mean "resistance to changes in velocity." This is certainly true of waves in, say, water--you've certainly felt resistance to your hand if you sweep it through water to make a wave; the destructive force of a tsunami is a more ...


1

When the man throws the ball, both the ball and the man get equal momentum in the opposite directions. Since the collusion is elastic, i.e: no loss in energy, the ball rebounds with momentum of the same magnitude but in the opposite direction. At this point, both the ball and the man have momentum in the same direction with equal mangitude. When the man ...


1

When a man in frictionless surface throws the ball in forward direction, by conservation of linear momentum he gets pushed back (exactly the case in space where astronaut throws something back to move foreward).Here,when man throws the ball, the momentum of ball and man are exactly equal and their velocities are in opposite direction. But you need to note ...


1

I don't know what numbers you are using, or how much precision you are expecting, but the problem could be your assumption that the earth - sun system is an isolated system. First there is earth's moon that (if not taken into account) might induce errors. Then there is the effect of other planets, especially jupiter. All of these must be considered if you ...


1

You have proved with your analysis that $v_2$ cannot be zero. $m_2$ must be moving with some non-zero velocity, either in the same direction as $m_1$, in which case $v_2$ must be smaller than $v_1$ (or $m_1$ will never catch up to $m_2$) or $m_2$ must be moving in the opposite direction to $m_1$. With the information given, there are 4 unknowns: $v_1, v_2, ...


1

Momentum ($p$) is "really" $mv$, even for light and EM fields. This can be proven by the use of $E = mc^2$. The momentum for a photon (EM) is $p = mv$. Where the mass is given by $m = E/c^2$ and $v = c$. Substituting these into the equation, one obtains, $p = E/c$. Although this equation "looks" different from $p = mv$, because it was derived using the ...


1

The velocity you calculate by linear momentum is the velocity of centre of mass of the body under study. Angular momentum should be conserved at a point where the impulsive forces act.



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