11
votes
Why are Coulomb forces on two charges the same?
Simple, partial answer
One way to explain this is that there are two contributions to the force, that get multiplied together. A way to express this is:
$$
F = q E
$$
where $q$ is the charge that the ...
- 45.1k
7
votes
Accepted
Why does a current loop obey Newton's third but a charged particle doesn't?
In general, a moving set of charges will create time-dependent electric and magnetic fields, and so the Poynting vector will in general be time-dependent. But the Poynting vector is (proportional to) ...
- 46.3k
5
votes
Why are Coulomb forces on two charges the same?
You can also think of it from the conservation of momentum perspective. For a closed system (say, a universe with only two charges), the net momentum change has to be zero.
Therefor $$dP = P_{\rm ...
- 947
3
votes
Two cars travelling towards each other with the same velocity and different mass. They crash. Which one has the largest acceleration?
The force they exert on each other will be equal and opposite during the collision (Newton's third law of motion). Let's call the magnitude of this force $F$, which is the same for both cars.
Then ...
- 105
2
votes
Does the momentum operator applied to a position state vanish?
$i\hbar\,\partial/\partial x$ is the momentum operator in the position representation. It is not the momentum operator that acts in the Hilbert space of abstract kets like $|p\rangle$, so your eq. $(2)...
- 1,007
2
votes
What went wrong in the following calculation of $\langle p'|[x,p]|p'\rangle$?
The issue you have (as mentioned in comments and @J. Murray) is that none of the things you are evaluating are defined. We instead calculate $\langle p'| [ X , P ] | p \rangle$. This is manipulated as
...
- 24k
2
votes
Accepted
What went wrong in the following calculation of $\langle p'|[x,p]|p'\rangle$?
The wrong step was in assuming that those expressions are well-defined in the first place.
The "generalized" bras/kets $|p\rangle$ and $|x\rangle$ are not true members of the Hilbert space, ...
- 66k
2
votes
Two cars travelling towards each other with the same velocity and different mass. They crash. Which one has the largest acceleration?
As the comments have pointed out , the cars dont have the same initial velocity
one is $50 km/hr$ and other is $-50km/hr$
Let us assume the velocity of small car is positive so
and the final velocity ...
- 620
2
votes
Accepted
Expectation of momentum in the bound state
$\renewcommand{\Re}{\operatorname{Re}}\renewcommand{\Im}{\operatorname{Im}}$Another way to phrase the derivation of @Latrace and @Gonenc, maybe more physically, is to recognize that the Wronskian ...
- 5,413
1
vote
Accepted
Are $E_i A_j + E_j A_i$ components of a tensor?
From your remark about the electric field not being a tensor itself, I presume you are referring to Lorentz tensors (rather than tensors in three-dimensional space). The quantities $E_{i}A_{j}$ are ...
Buzz♦
- 15.4k
1
vote
Question about sign convention in electron-hole crystal momentum
It is good to keep in mind that, ultimately, a hole is just a vacant electron state in a valence band. It merely provides a convenient (but optional) language for describing conduction due to valence ...
- 11k
1
vote
Accepted
Questions related to conservation of momentum
Since the mass is thrown out with a sideways velocity relative to the car, it still has a velocity $v$ in the direction of the car immediately after the throw. This means that the momentum ...
- 46.3k
1
vote
The Expectation Value of Momentum Operator
It depends on your definition of bounded state. If a bounded state is just a smooth, rapidly vanishing for $|x|\to \infty$, wavefunction, the the claim is generally false.
Indeed, consider a smooth ...
- 67.8k
1
vote
Accepted
Do $p_\mu$ and $\gamma^\mu$ commute?
but I thought $\gamma^\mu p_\mu \not=p_\mu\gamma^\mu$, where is my logic flawed?
Your logic is flawed right at the beginning--right from your premise: "I thought $\gamma^\mu p_\mu \not=p_\mu\...
- 15.7k
1
vote
Accepted
Derivation of Quantum Momentum Operator in Griffiths
The integration-by-parts for 1.30 & 1.31 are explained in this other post: Integration by parts to derive 𝑑⟨𝑥⟩/𝑑𝑡.
Equation 1.25 is actually exactly as it seems - the product rule results in ...
- 21
1
vote
How to evaluate the action of a fractional differential momentum operator?
δ(x) is very confusing, so, as per comment, I'll use f(x) instead. Moreover, since there is only one independent variable, I'll simplify notation to $\partial= \frac{\partial}{\partial x}$. You then ...
- 58.8k
1
vote
Replace sum or subtraction of momentum vectors with an equivalent vector
The resultant vector does not have the same angle as the vectors being summed. Equation (1) should be
$$m_eu_e\cos\alpha = m_1u_1\cos\beta + m_2u_2\cos\beta$$
and similarly for equation (2).
- 23.9k
1
vote
Time-evolution of momentum eigenstate under harmonic oscillator Hamiltonian
I'm not sure what you are really after (smells like an attempt to compute the Mehler kernel unconventionally), so I'll assume you are seeking the evolution matrix elements in momentum rep, $\langle p'|...
- 58.8k
1
vote
Motion/momentum of a wave packet
If you are familiar with the Hamilton Jacobi formulation of classical mechanics, this relationship is natural. $S$ is the action variable. The wavepacket assumption is to obtain Hamilton’s equations ...
- 8,661
1
vote
Why is angular momentum and energy conserved in a lever but momentum is not? Why the apparently unequal treatment?
Why is angular momentum conserved in general?
You may know that conservation laws come from symmetries of the Lagrangian.
The Lagrangian is symmetric under space translation? The system will conserve ...
- 2,829
1
vote
Fundamental "definition" of momentum
When we say that massless particles have "momentum", are we even talking about the same thing as when we say that particles with mass have momentum?
The motion of the electric field in an EM ...
- 11
1
vote
Is the additional term in the canonical momentum exactly equal to the momentum of the electromagnetic field?
The canonical momentum of a particle in an electromagnetic field is given by $$\textbf{P}=m\textbf{v}+q\textbf{A}$$ Is the term $q \textbf{A}$ equal to the momentum of the electromagnetic field (which ...
- 35.3k
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