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7

Perhaps the simplest and most intuitive approach is to regularize the hard wall potential $$V_0(x)~=~\left\{ \begin{array}{rcl} 0 &\text{for}& x<0 \cr\cr \infty &\text{for}& x>0\end{array}\right. $$ as $$ \lim_{\varepsilon \to 0^+} V_{\varepsilon}(x) ~=~V_0(x).$$ For instance, one could choose the regularized potential as $$ ...


6

If you collide two ideal billiard balls, then that would be what you would call a perfectly elastic collision. If you have a large dense collection of billiard balls, and you slam a new one into the collection, then there are a whole lot of elastic collisions, transfering energy and momentum in many ways that you would be hard-pressed to calculate exactly. ...


4

A simple version of this is bremsstrahlung, i.e. an electron that decelerates and produces electromagnetic radiation / photons. By your reasoning the energy of the electron should only be able to go into other electrons: maybe it should radiate other electrons, maybe a single electron shouldn't lose energy as it travels. But the electron can transfer some ...


4

Collisions can be elastic or inelastic. Elastic collisions are collisions where the incoming and outgoing kinetic energies are the same and only the angles change The same holds true classically, example: billiard balls ; and in the elementary particle framework. example: An electron hitting an electron has a probability of scattering elastically. ...


3

In a typical introductory physics class, the terms elastic and inelastic refer to whether the macroscopic kinetic energy is the same before and after the collision. By this I mean the large-scale motion of the objects that you typically consider in collisions, like carts or balls, not the detailed particle motion/potential. You are correct that the total ...


2

Earth is really, really big (in comparison to that projectile). In order for an object to completely penetrate it, it would need to have enough force to go through 12,742 kilometers of solid and liquid. It would need either extreme mass or extreme speed. in the case of extreme mass, at a certain point, the object wouldn't go straight through earth as much ...


2

Energy-momentum conservation is a stronger statement than the statement* that the inner product $p_\mu p'^\mu$ is conserved. It states that the sums are conserved individually/coordinatewise - $P_1+P_2=P_1'+P_2'$. As I see it, the conservation of the inner product is a statement about change in reference frames, whereas the conservation of energy and ...


1

It seems like you know what the answer is, but you just don't know how to prove it. You are right though. To make things simpler, just view things in the center of mass frame. Then the total momentum is zero, and, like you said, the total angular momentum is just the sum of the orbital momentum of each planet plus the sum of the spin angular momentum of ...


1

I think you have too many parameters, and not all of the necessary ones To simplify your thinking: Change to a frame of reference in which one billiard ball is initially at rest at the origin, and the second is moving at velocity $V$ from right to left along the straight line $$y=k, \,k>=0$$ A collision will take place if and only if $k<2\sigma$. ...


1

Using $e^-e^-$ or $e^+e^+$ means that the final states need to be charged and have lepton number of two. This produces a different set of potential final products then $e^-e^+$. One such example would be $e^- e ^- \rightarrow \mu ^-\mu^- $. While this may be an interesting collision for some new theory, such interactions can only produce a very particular ...


1

When a type of quantum (elementary) particle is absent it just means that the field of that particle is in the quantum "vacuum state". But "vacuum state" does not mean absence of everything concerning that field. The vacuum state has various physical properties in spite of its name: It is nothing but a possible state or quantum configuration of the field, ...


1

As your read suggestion suggested, you can approximate physical reality with "no collisions are simultaneous". The reason is that the physical world is full of indeterminacies (AKA errors) due to thermal fluctuations and many other sources. What this means is that, even if the strict mathematical solutions that you will find by assuming simultaneous ...



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