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19

In this case, gravity is still an external force. In a zero-g environment, the mouse would also begin to move around the inside of the wheel, opposite the rotation it causes in the wheel, which would keep the angular momentum at zero. This would happen because the only way for the mouse to exert a force on the wheel and rotate it is for it to push itself in ...


7

I think you've confused global and local charge conservation. The quote you included actually skips parts of the video, and I think the video may actually have been edited as well, so it is not obvious from the text alone what Feynman is referring to when he says "the second form of charge conservation...". He's actually referring to the first sentence. ...


6

These types of theories that physicists such as Krauss espouse of a "Universe Coming From Nothing" are quite flawed, as by no means are they talking about nothing! Further, the concepts of particles, mass, and energy are not even well-defined when talking about the universe in general. I wrote a paper on this (excuse the shameless self-promotion), it can be ...


3

The statement is really about the transformation between inertial co-ordinates and co-ordinates fixed to the body. This is expressed by: $$D_t = d_t + \omega(t)\times\tag{1}$$ where $D_t$ is the "total" derivative, i.e. the time derivative in the inertial frame, $d_t$ is the time derivative in the frame fixed to the body. Since there there are no torques ...


3

I'm not sure whether everything's clear from your description and from Geoff's Answer. If you haven't already worked it out, the essential problem that Feynman and Einstein are getting at is the relativity of simultaneity. In the second situation, which Geoff's answer explains well, a relatively moving observer will not see the disappearance of charge and ...


2

The angular momentum of the earth && mouse wheel system does not change. When the earth pulls on the mouse, the mouse pulls on the earth, so no net moment is seen any arbitrary point in the universe.


2

I don't have that text, but I can find the table of contents on the internet. Somewhere in that text (most likely chapter 5 on non-inertial reference systems), there should be a derivation that for any vector quantity $\boldsymbol q$, the time derivative of that vector in an inertial frame and a rotating frame that share the same origin are related by $$ ...


2

In a collision it's often the case that it's hard to measure exactly how long the collision lasts and exactly how the force between the objects changes during the collision. Squishy objects like nerf balls will collide relatively slowly while hard objects like billard balls will have a short collision time. However there is a well defined quantity called ...


2

1) Yes, the charge is truly and exactly conserved. What it is confusing you, I think, is that the current for a scalar field explicitly depends on 4-potential $A$, whereas that for a spin-1/2 does not. This is obviously related to the number of derivatives in the Lagrangian kinetic term and, likewise, to the number of derivatives in the current. It can help ...


2

If $\vec{F}$ is a conservative force field, then it satisfies the property $$ \tag{1} \vec{\nabla} \times \vec{F} = 0, $$ and it can be written as $$ \tag{2} \vec{F} = \vec{\nabla}V, $$ for a scalar function $V$ (which corresponds to potential function in physics). Note that, when you put $(2)$ into $(1)$ it becomes a "curl of a gradient" and is ...


2

You cannot change your linear or angular momentum in open space at all. You need something to transmit it to. if you swing your legs your body will rotate in the opposite direction while you swing, and stop when you stop swinging. If you are out of fuel there is no way to accelerate. Only by releasing mass you could change momentum, as Bender well shows you, ...


2

Noether's theorem states that there exists a conservation law for every continuous (in fact, differentiable) symmetry. Reflection is a discrete symmetry, so the theorem is not applicable here. But, in quantum mechanics, you have the parity operator $P$, that reflects the coordinates $$P\psi(\vec{r}) = \psi(-\vec{r})$$ Since $P^2 = I$, the operator $P$ has ...


1

An elastic collision is defined as one which conserves energy. When you jump against a wall, most of your kinetic energy is dissipated as heat into your tissue as your legs and muscles absorb the impact. Therefore, energy is not conserved so by definition this is an inelastic collision.


1

Never seen that before, so I just tried it. Cool. I believe that the membrane between the yolk and the white is elastic, so when you first, gently, give the egg a little angular momentum, you are only spinning the white. As the yolk catches up the effective moment of inertia drops, and conservation of momentum therefor implies a higher angular velocity.


1

I didn't redo your calculations and I assume that they are correct, which actually doesn't play any role in what I'll describe now. Notice that in the second scenario the 2kg ball will inevitably start to move. By keeping it still you change the reference frame one more time, which invalidates the use of conservation laws. You cannot use the conservation of ...


1

(After possibly introducing more variables) then OP is essentially considering an autonomous system of $n$ coupled 1st order ODEs $$\tag{1} \frac{d\vec{z}(t)}{dt}~=\vec{f}(\vec{z}(t)), \qquad f: U \to \mathbb{R}^n , \qquad U\subseteq \mathbb{R}^n, $$ i.e. without explicit time dependence, so that the system (1) possesses time translation symmetry. OP is ...


1

You're correct. To find the equations of motion, we have: \begin{align*}c_i&=\frac{\partial L(v^2)}{\partial v_i}\\ &=L'(v^2) 2 v_i \end{align*} so that $L'(v^2) v_i$ is constant for all of time. Firstly, you could imagine a world in which all paths ${\bf x}(t)$ are valid mechanical paths. Then the Galilean transform of a valid mechanical path ...


1

I think I can remember the derivation for a conservative force field in classical mechanics, wich is a somewhat stronger assumption than pure time-translation invariance. Let $\vec{F}$ be a conservative force field, that is $$ \nabla \times \vec{F} = 0 $$ or alternatively $$ \phi := -\int_\gamma \vec{F} \cdot d\vec{a} $$ does not depend on the path $\gamma$ ...


1

The usual approach: write down conservation of angular momentum, linear momentum, and energy. Assume the impact is elastic and infinitely short duration. In that time the spring didn't move and the third particle didn't come into the equation. That means the problem can be reduced to two simpler problems: two particles that hit elastically (after collision ...



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