# Tag Info

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You are making confusion between operators and their representations on the position basis. Your hypotheses 1) and 2) are wrong (more ill-interpreted): as you can easily notice the left hand sides contain an element in the Hilbert space whereas the right hand sides contain numbers (functions evaluated in a point $x$, therefore a number). To answer your ...

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$E = pc$ is only true for massless particles. For massive particles you have the mass-shell relation: $E^2 = m^2c^4+p^2c^2$ After you use $E=T+mc^2$ and you can find $p$

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To see why the exhaust speed is important, let's do a calculation. Let's start with a rocket of mass $m$ going at speed $u$. (We measure all speeds with respect to some inertial reference frame.) Now, suppose it exhausts a tiny amount of propellant of mass $\delta m$ and the propellant is traveling at speed $u_P$. After it exhausts that fuel, the rocket ...

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Imagine the rocket before and after throwing a small ("infinitessimal") amount of fuel out its exhaust. You apply the momentum conservation notion by equating the increase in the rocket's forwards momentum with the momentum of the fuel thrown backwards. The easiest inertial frame to do one's analysis in is that of rocket immediately before the increment ...

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It you are working with second quantization (quantum field theory), then what you wrote under "1." and "2." is incorrect. Your physical degrees of freedom are not coordinates $x$, but rather field variables $\phi$. So, $$\left| 0 \right> \sim \exp \left\{ -\frac{1}{2} \int d^3 x \phi(x)^2 \right\}$$ This object should be called wave functional ...

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Due to the amount of momentum (and energy) transferred to you. Assuming a completely inelastic process (and an idealized one dimensional setup) you get from the conservation of momentum ($v$ is the initial velocty, $p$ the initial momentum, $v'$ is the final velocity, $M$ is the mass of the object hitting you, $m$ is your mass): $$p = M v = (M + m) v'$$ $$... 1 In quantum mechanics observables are represented by (some classes) of self-adjoint operators on some Hilbert space. Saying that you can precisely measure a quantity given by the operator A means that your state can be one of the eigenstates of that operator. Likewise, if you want to precisely measure two quantities A, B together your state needs to be an ... 0 Am I missing something here? Well, you (among others) seem to be missing that to measure "momentum" is defined through the application of the gradient of the translation operator \nabla \hat T_{\mathbf r}[~] := \frac{d}{d \mathbf r_{\mathcal S} }[~] to what's given through observational data (e.g. concerning a particular object A under ... 1 Am I missing something here? Yes. What you're missing is "the mass of a body is a measure of its energy-content". Read Einstein's original paper, and take note of this: "If a body gives off the energy L in the form of radiation, its mass diminishes by L/c²". Next, imagine your body is a massless photon in a gedanken mirror-box. It isn't actually at rest ... 6 They're both saying the same thing: the relativistic momentum is given by$$ \mathbf{p}=\gamma(v)\,m\mathbf{v} $$The confusion, it seems, is that you are using Feynman's m=\gamma m_0 as equivalent to the m in Resnick & Halliday's text; the actual correlation is Feynman's m_0 to Resnick's m--both of these terms are the (invariant) rest mass. ... 1 The way to define momentum in a special relativistic context is the following: Start with the trajectory of the particle parametrized by its proper time x^{\alpha} (\tau); define the four-momentum by p^\alpha = m \frac{\mathrm{d}x^{\alpha}}{\mathrm{d}\tau}, where m is the mass of the particle (note that I'm only using one mass, not distinguishing ... 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 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 Imagine this situation: at time t=0, we have a infinite long straight wire with current zero, and a charged particle q with zero velocity. at time t=T, we make the current to be I, thus we have a  \mathbf{B} field, and  \mathbf{A} field. during this process,  \mathbf{A} is build up from zero to some value, therefore we have induced electric field ... 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 The p in the Fourier transform of the (free real scalar) field$$ \phi(x) = \int \left(a(\vec p)\exp(-\mathrm{i}px) + a^\dagger(\vec p)\exp(\mathrm{i}px)\right)\frac{\mathrm{d}^3p}{2p^0} $$is a number, it is essentially a change in coordinates on \mathbb{R}^n like every Fourier transform. The canonically conjugate momentum \pi(x) := \frac{\partial ... 1 Static magnetic fields by themselves have no momentum, you need an electric field and a magnetic field to have momentum. Also, the momentum comes from the total fields. So even if you think of two magnetic fields \vec B_1 and \vec B_2 and two electric fields \vec E_1 and \vec E_2 the momentum density is \epsilon_0\left(\vec E_1+\vec ... 1 I use this arrangement in my introductory classes when time allows, but not for momentum. I call it the "work-energy mini-lab". (It's a mini-lab because I don't expect a detailed write-up and guide the class through some of the harder analysis.) Some things to note. By measuring how far the mass hanger has to drop you get the distance over which the ... 1 You need to keep the amount of time the same as well. The change in momentum of a system is equal to impulse delivered to it, which is just the time-integral of the force:$$ \Delta \vec{p} = \vec{J} = \int_{t_1}^{t_2} \vec{F} \, dt = \vec{F} \Delta t \text{ (if $\vec{F}$ is constant with respect to $t$.)} $$So just having \vec{F} constant isn't enough; ... 0 Great question. I didn't check all the math, but I see a problem. In the first graphic, you described a collision process in a single reference frame. Your picture is lovely. In the second graphic there is a problem. In the rest frame of the 2kg ball, the 2kg ball sees a 4kg ball approaching it. During the collision, momentum is transferred from the 4kg ... 1 I didn't redo your calculations and I assume that they are correct, which actually doesn't play any role in what I'll describe now. Notice that in the second scenario the 2kg ball will inevitably start to move. By keeping it still you change the reference frame one more time, which invalidates the use of conservation laws. You cannot use the conservation of ... 4 If the initial momentum is mv, then when the particle bounces off the wall, it is going in the opposite direction, so its new momentum is -mv. The difference is 2mv because mv - (-mv) = 2mv. By the way, this is not really how it works; molecules will speed up or slow down when they hit the wall. If the molecule is moving slowly, it's likely to speed ... 2 Imaginary wavevectors are possible and, as ptomato's answer correctly points out betoken evanescence. I'd like to add a few words to his answer that might help clear up your confusion. Imaginary wavenumbers always betoken Evanescence. Sometimes the vague term "nearfield" is used to connote something not propagating. Evanescence is NOT dissipative; this is ... 2 Yes, indeed the out-of-plane component of the wave vector of a surface plasmon is imaginary. A purely imaginary wave vector means the wave does not radiate in that direction, but instead is evanescent. (That's what you get if you plug in a purely imaginary k_z = -i\alpha into the formula$$ E(z, t) = e^{i (k_z z - \omega t)} = e^{-\alpha z} e^{-i\omega ...

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The equation is just the kinetic and rest energy, it does not include potential energy. But potential energy in relativity is not the proper concept. The linked question has some useful answers, but I think your true question is about how to learn to do things relativistically that you used to do non relativistically. And since the other answer so far takes ...

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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 ...

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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 ...

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In classical mechanics there is no distinction between free and bound as far as this relation is concerned. In relativistic quantum mechanics (i.e. QFT), a particle that satisfies this relations is said to be "on-shell" or a physical observable asymptotically free particle. It is certainly not satisfied for virtual particles, but they are as their name ...

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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 ...

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So, I suppose that $Φ(k)$ is the probability density of the momentum. Is this true? Almost. $\Phi(k)$ is the probability amplitude for the momentum of the particle. The probability density is obtained as usual by squaring the amplitude, giving $|\Phi(k)|^2$. For a free particle, all values of momentum are always allowed, which enables the superposition ...

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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.

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$N.s$ is the unit of Momentum and Impulse. Let's consider, what the quantities itself are so that you might be able to correlate them with their units. Speaking colloquially, Momentum is a measure of strength and a measure of how difficult it is to stop an object, and Impulse is the measure of how much the force $changes$ the momentum of that object. ...

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The law states this: Two objects masses and with velocities $m_1$, $m_2$, $\vec{v}_1$ and $\vec{v}_2$ collide with contact normal $\vec{n}$. The final velocities are $\vec{v}_1^\star$ and $\vec{v}_2^\star$ such that the coefficient of restitution $\epsilon$ is defined by \vec{n}\cdot \left( \vec{v}_2^\star - \vec{v}_1^\star \right) = -\epsilon \;\left( ...

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A force is applied for a certain time.

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