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6

Your book may be treating things a little backwards from the way they are usually done. The usual way is to define the momentum four-vector as the combination $(E/c, \vec{p})$, where $E$ is already known to be the total energy (the thing that reduces to $mc^2 + \frac{1}{2}mv^2$ for $v\ll c$) and then go on to show that it satisfies the properties expected of ...


5

Microscopically, i.e. in the quantum theory the scattering with radiation is a collision of particles with photons such as $$ e^- + \gamma \to e^- + \gamma$$ The momentum vectors of the particles above are $$ \vec p_1+\vec p_2= \vec p_3 + \vec p_4$$ where the identity holds due to momentum conservation. But in general $\vec p_1\neq \vec p_3$ and $\vec ...


4

It Depends on Your Model How closely do you want to model reality? The truth is, most collisions are some mixture of inelastic and elastic. (That is, momentum is transferred, but not always "cleanly," some of that energy gets transferred into deforming the objects.) You can see this sort of thing if you watch slow-motion videos of things striking other ...


4

If the total kinetic energy before the collision equals the total kinetic energy after the collision, the collision is elastic. Otherwise, it isn't elastic. given the mass, the velocity, and the 'angle' the two objects are going two be when they collide - how can I know if I need to compute an elastic or an inelastic collision? The mass, velocity ...


3

There is not such thing as a "partially elastic" collision. Classical collisions between particles can be separated into two categories: elastic and inelastic. Elastic collisions are defined as collisions in which no energy leaves the system (i.e. $E_i = E_f$). All other collisions are inelastic, as some energy is lost ($E_i > E_f$). A perfectly inelastic ...


3

I understand that the inner product of two 4-vectors is conserved under the Lorentz transformations Yes, $p_1.p_2$ is a Lorentz invariant So that the absolute value of the four momentum is the same in any reference frame. It is not correct to speak about the "absolute value" of a (quadri)vector. Which is conserved in a Lorentz transformation ...


2

There is, effectively, only gravitation and friction acting on the pack of gum. However, the friction is not that strong (it is mostly independent of the velocity of the book, and dynamic friction is weaker than static friction) and it doesn't have that much time to act. Hence it doesn't affect the momentum of the gum noticeably. This is very related to the ...


2

$V_e=V-v_e $ Why does this occur? This is just saying that the exhaust velocity is measured with respect to the engine. If the rocket is moving forward ($V$), then the observed exhaust velocity ($V_e$) with respect to the ground (or other specified frame) is reduced by the engine's velocity. And since we are concerned with the changes in velocity with ...


2

Normally you solve an elastic collision with just momentum and energy conservation, because you really don't know what happens at impact. The formulas are given in this answer. For equal masses in one dimension the velocities are exchanged. It turns out your interaction time and acceleration multiply to get the correct velocity change, but since you ...


2

In special relativity, if you add two velocities, you have to use the formula $$v = (v_1+v_2)\left(1+\frac{v_1v_2}{c^2}\right)^{-1} \text{ .}$$ So you cannot simply add two velocities together. Usually, velocity is not a good variable to work with in special relativity. It's much easier to use four-momentum conservation, which is simply given by $$p = p_1 ...


2

I don't understand why the time component of the 4-vector [ $m~\gamma(|\vec v|)~(c, \vec v)$ ] is being denoted as $E/c$. So the underlying question is two-fold: Why is "energy" considered the time component of some 4-vector at all?, and Why this specific time component expression, among time components of all different 4-vectors imaginable? (Where ...


2

You can write kinetic energy as $$K=\frac{1}{2}mv^2=\frac{mv^2}{2}=\frac{(mv)^2}{2m}.$$ You can take the next few steps of rewriting this in terms of momentum $p=mv$.


1

To parameterize the degree of inelasticity you use the "coefficient of restitution" which is 1 for elastic processes and 0 for inelastic processes. This is described by $$ \text{coef. of restitution} = c_R = \frac{\text{final relative speed}}{\text{initial relative speed}} = \frac{v_2 - v_1}{u_1 - u_2} \,. \tag{*} $$ This also tells you how to compute the ...


1

Perfectly elastic and perfectly inelastic collisions are just limiting cases on a scale of how much kinetic energy is retained. As noted in @Nathan's answer, if you work in the center-of-mass frame, a perfectly inelastic collision results in 0% of the kinetic energy retained, while perfectly elastic collisions have 100% of kinetic energy retained. So, you ...


1

The Heisenberg inequalities reads : $$\Delta x \Delta p_x \geq \frac{\hbar}{2},$$ where $$\hbar = \frac{h}{2\pi}. $$ Therefore, for a free particle, your add the expressions $$ p_x=mv_x$$ so you get : $$\Delta x \Delta v_x \geq \frac {\hbar}{2m}$$ you have the uncertainty on the velocity : $\Delta v_x \geq \frac{\hbar}{2m\Delta x}$ the Heisenberg ...


1

Try showing that the kinetic energy of a particle of mass $m$ and momentum $\vec { p } $ can be written as $$K=\frac{p^2}{2m}$$ The solution to your problem should be clear from there. This expression for kinetic energy is actually a pretty handy formula to have memorized.


1

Here's a parallel answer to LuboŇ°'s but purely classical. Start by noting that the momentum vector of a plane wave with wavelength $\lambda$ is: $$ \vec{p} = \frac{2\pi}{\lambda} $$ In some elastic scattering experiment, e.g. X-ray or some other diffraction measurement, we have something like: where $\vec{p}_{in}$ is the momentum of the incoming wave ...


1

Recall the Heisenberg principle which tells us that a we cannot know both position and momentum at the same time $$\Delta x \Delta p \geq \hbar/2$$ For a particle, we have actually at most $\Delta x = L$, so that we cannot in principle have a sharp momentum state. Fortunately, there are eigenstates which are eigenstates of $\hat{P}^2$, that is, states ...


1

Taking the standard $[0,L]$ problem, eigenfunction and energy eigenvalues are: $$ \varphi_n=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}, \qquad E_n=\frac{\hbar^2\pi^2n^2}{2mL^2}. $$ This means that stationary 1D box systems (e.g. insulated ones) only admit states with a discrete set of possible energies, as above. Now, that as far energy is concerned. What about ...


1

In general, the elasticity of a collision is dependent on the properties of the colliding objects. In a perfectly elastic collision, no kinetic energy is dissipated, which means the collision creates no heat, no sound, etc. In a perfectly inelastic collision, the maximum possible amount of kinetic energy is dissipated as heat, sound, etc. This corresponds ...


1

theoretically that might be possible but practically it is IMPOSSIBLE you just cant apply same force on every atom of every single molecule of the car. and you should also see at the general formula of momentum in which force is inversely proportional to the time taken and directly proportional to the change in momentum according to your fictional scenario ...


1

A wall is not just standing on the ground, its contact with the ground is made fast by foundations. Pushing against it is pushing against a system wall + ground. Since it is in addition a very rigid solid, it will transmit all of your push to the ground. You can probably be convinced that the resistance of this system to motion is much greater than yours. ...


1

The continuity equation (without sources) is usually written as follows $$\partial_t \rho + \nabla \cdot \mathbf{j} = 0$$ If you identify $\rho$ as the mass density, integrate over some volume $V$ and use the divergence theorem you get the result that you mention in your question. Namely, the change in mass in $V$ equals the amount of mass flowing through ...


1

It hadn't. The metric of the space is changing, as they are described by the Friedman-equations. There is no explosion-like thing, it is a common misconception.


1

Well, it doesn't need to have any form before you choose a representation for the operators. In general, any Hermitian operator $Q$ satisfying $[Q,\psi]={\cal F}$, is the generator of the transformation $\psi \rightarrow\psi + i\epsilon{\cal F} $. Let's try to figure this out. Consider the action of the unitary operator $U(\epsilon)=e^{i\epsilon{\cal Q}}$ ...


1

Using Fourier analysis, and setting $\hbar$ to 1 (I leave it to you to reintroduce it consistently using dimensional analysis), we have $$ f(x) =\int \tilde f(k) e^{ikx} dk\\ xf(x) = \int i \frac{\partial \tilde f(k)}{\partial k} e^{ikx} dk $$ where we used integration by part. Applying the commutation relation holds $$ i (\frac{\partial( p_x[\tilde ...



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