Tag Info

New answers tagged

-5

Some people have very long explanations of why the universe can be created out of nothing "because the universe always does this". As a counter argument: There is no proof that the universe always does this. It is only theoretical. If you believe that the whole universe can be created out of nothing, it means you could create energy and matter out of ...


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 $$ ...


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 ...


0

Indeed. It is due to the law of conservation of angular momentum. The angular momentum of the rotating element within the motor will exactly cancel that of the rest of the motor, thereby giving zero net angular momentum, as with the initial conditions.


0

The casing will spin in the opposite direction. That is the principle of reaction wheels.


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.


20

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 ...


0

The way you distinguish between different interactions in the universe is by looking at the mediator bosons thereof, which appear when scattering two particles under this or that interaction. Take two electrons as example and scatter them up: if you identify that photons take part to the process then it is an electromagnetic interaction; if you find out ...


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 ...


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 ...


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. ...


0

You can't just acquire velocity (and hence momentum) without an equal and opposite momentum going to something else. So you are wrong that there are things you can do to acquire velocity. Unless you want to leave parts of yourself behind or are capable of stealing momentum from, say, an electromagnetic field. However, if you are an extended body in a ...


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 ...


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, ...


0

Here it explains what I think you are asking: We work with a formulation of Noether-symmetry analysis which uses the properties of infinitesimal point transformations in the space-time variables to establish the association between symmetries and conservation laws of a dynamical system. Here symmetries are expressed in the form of generators. We ...


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

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 ...


0

It follows from $L$ being a function $\propto\dot{x}^2$. With this at hand, you are left with two choices: $\left(\nabla_{\dot{x}}L\right)'\sim\left(\dot{x}\right)'=0$ implies $\dot{x}=\rm const.$ $L=0$ implies $\dot{x}=0=\rm const.$ Either way, you get that the velocity is constant in time (for this particular, free-particle case).


0

I believe this is discussed in Nonlinear Dynamics and Chaos by Strogatz. I don't remember the details, but it's worth looking at his discussion of an energy function.


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 ...


0

You assert that total electromagnetic momentum is small in a small region. Even when this is so, it is not right either morally or mathematically to then throw away a term like $\frac{d}{dt} \vec{P}_{em}$ because something can have a very large time derivative even if it is quite small. A stiff spring oscillating with a small amplitude can have large ...


0

Great question. I didn't check all the math, but I see a problem. In the first graphic, you described a collision process in a single reference frame. Your picture is lovely. In the second graphic there is a problem. In the rest frame of the 2kg ball, the 2kg ball sees a 4kg ball approaching it. During the collision, momentum is transferred from the 4kg ...


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

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.


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 ...


0

Since the unfractured disk is assumed (it is so assumed, right?) to be of uniform density, it has a center of mass at the geometric center of the disk. Then, since the center of mass of two distinct masses lies on a line connecting the two, the CMs of the two fragments must lie on a diameter of the disk. Not only that, if the radial distance of the CM of ...


0

I think what you would do here is find the momentum. Momentum = Mass x Velocity. so lets say the objects speed is 10 mph, and the weight is 15 lbs. When the object in motion hits the resting object, it will deliver 150 lbs of force. If the object in motion is going 30 mph and weighs 2000 lbs (lets say a car) then the force it will deliver upon impact with ...


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 ...


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.


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 ...



Top 50 recent answers are included