# Tag Info

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No, the equation reflects the fact that energy is conserved. If a particle has an energy, E, then in the absence of any interactions the energy will remain E. The distribution of E between KE and PE will vary- that's suggested by the left hand side of the equation which shows a potential varying over space.

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I use the signature convention $(+,-)$. Any spacetime-displacement vector can be written as $$\tilde T=a\hat t+ \tilde s,$$ where $\hat t$ is unit and future-timelike and $\tilde s\cdot\hat t=0$ (that is, $\tilde s$ is purely spacelike to the observer with 4-velocity $\hat t$. Assume $a\geq 0$. Since $a$ is the "temporal displacement" and $\tilde s$ is ...

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It may not be immediately apparent, but the answer does have to do with the energy-time principle. But it's a good question, because it exposes some implicit assumptions that are often made when working out solutions to the Schrodinger Equation. The infinite potential well problem is well-known to students in introductory quantum mechanics, since when you ...

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My advice is to remember at all times that quantum mechanics is just a set of rules for building mathematical models of reality, so some of its predictions may be features of the model rather than features of reality. The QM model of a particle in an infinite potential well predicts a set of exact allowed energy values, corresponding to a set of ...

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The fundamental energy-momentum relation in special relativity is $E^2=p^2+m^2$ (let’s ignore factors of $c$, since they will factor out in the end — that is, let’s work in natural units). Taking the derivative with respect to $p$ on both sides gives $$2E\frac{\partial E}{\partial p}=2p,$$ since $m$ is a constant. Thus, $$\frac{\partial E}{\partial p}=\... 1 We have E=\gamma mc^2 (relativistic total energy) and p=\gamma mv, where \gamma=1/\sqrt{1-v^2/c^2}. Therefore, we can write:$$E=\frac{pc^2}{v} \space,$$differentiating both sides WRT p implies:$$\frac{dE}{dp}=c^2\left(\frac{v-p\frac{dv}{dp}}{v^2}\right)=\frac{c^2}{v^2}\left(v-\frac{p}{\frac{dp}{dv}}\right) \space. \tag{*}Moreover, we have: ... 4 pushing requires energy In fact, pushing does not require energy. You can lean a ladder against a wall and without consuming energy it will tirelessly push against the wall until the wall crumbles or the ladder is moved away. If a human being wants to push on the wall with their arm then they will find that human arms are inefficient machines and in this ... 0 I'm considering a local inertial frame, at a point \mathcal{P} of D dimensional spacetime, in which there's an electric field (any configuration). The calculations below are valid even in General Relativity, and for any electric field in spacetime of D dimensions (I don't care about axes aligned to the field!). I'm using c = 1 and metric signature ... 0 Here is an elementary argument for 3+1 dimensions, which I'll then generalize to d+1 dimensions, although I'm unsure if I got the latter right, and would appreciate comments. In the case of a purely electric (or purely magnetic) field, we expect the pressure and tension to depend only on the field at that point -- not on the field in some neighborhood, or ... 1 First thing to note is the equation \text{energy density} = |\text{tension along the $x$ direction}| $$(in units with c=1) would be valid for any dimension, not just 4. The reason is simple: for purely magnetic (or electric) field along the x axis, the field strength tensor is invariant under boosts along the direction of the field. So the stress ... 0 Force is dependent of distance r, so thing move like this$$\newcommand{\newln}{\\&\quad\quad{}} \begin{align}&\int^{r_b}_{r_a}\mathbf{\vec{F}}\cdot d\mathbf{\vec{r}}=-(U_a-U_b) \newln \Rightarrow \int^{r}_{\infty}\mathbf{\vec{F}}\cdot d\mathbf{\vec{r}}=-(U_r-U_\infty) \newln \Rightarrow \int^{r}_{\infty}\mathbf{\vec{F}}\cdot d\mathbf{\vec{r}} =-...

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The force between two point charges is an example of an inverse square law where the force is proportional to the reciprocal of the separation squared, $F\propto \frac {1}{r^2}$, as illustrated below. As the separation increases so does the force decrease. The area under a force against separation graph is the work done and as long as the area ...

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Heat is energy transfer between the two bodies due solely to temperature difference between them. Once thermal equilibrium is reached between the two bodies ( they both reach the same final temperature in the calorimeter) heat transfer between the two becomes zero. The energy that transferred results in changes in internal energy of the bodies. The energy ...

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Loosely speaking, at great distance the force required gets infinitely tiny, so the two infinities largely cancel out.

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Work isn’t $Fd$ when the force changes with position. It’s an integral, and the integral is finite even over an infinite distance because the force goes to zero sufficiently rapidly at large distances. The integral is a standard homework problem so I am not going to write it or evaluate it.

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Your equation involving $\gamma m$ is useless for photons because $\gamma$ is infinite and $m$ is zero. That product is indeterminate. Your professor is correct.

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The conservation of momentum approach is the correct one as the "collision" between the conveyor belt and the sand is "inelestic" for the following reason. Just before hitting the belt the horizontal speed of the sand is zero. Some time after hitting the belt the sand has the same horizontal speed as the belt. In between kinetic frictional forces ...

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The analogy is not well justified. In chemical systems, temperature can be introduced as $$T = \left(\frac{\partial U}{\partial S}\right)_{V,N}.$$ The constant specifiers are important. You can't introduce additional constraints like $U = \text{const}\cdot N$. And if energy was just constantly proportional to $N$, independent (explicitly) of $S$ and $V$, the ...

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The mean photon energy in a photon gas at temperature $T$ is $$\bar{E_\gamma}=\frac{3\zeta(4)}{\zeta(3)}k_BT\approx 2.70\,k_BT$$ where $\zeta(x)$ is the Riemann zeta function and $k_B$ is the Boltzmann constant. The derivation of this from Planck's Law is a standard homework exercise, so I won't give it. For $T=5500\text{ K}$, $\bar{E_\gamma}=1.28\text{ ... 0 No, not unless the wall moves. If the force is not applied through a distance, no work is done. 0 Need to address a few of your statements before answering. Work is just change in the energy of an object. Work can cause a change in the energy of an object, but it is not the energy change of an object. The change in energy of an object is the sum of the changes in its internal (microscopic) and external (macroscopic) energy. Work is a mechanism for ... 1 Realising that there is a connection between two properties does not automatically mean that this connection must be direct proportionality. Couldn't the connection be, say, quadratic? And that is actually the case. Work$W$done equals kinetic energy$K$gained (if we start at$v=0$): $$\require{cancel}W=K=\frac12 mv^2\qquad\text{ so } \qquad W \propto v^2\... 0 If you apply a force F to something initially at rest, and keep the force going over a distance D, the force accelerates the thing to a speed V where VV=2FD/M. In other words the velocity of the body doesn't rise in proportion to FD, but the velocity squared does. If you rearrange the last expression you get mVV/2=FD. I.e. the kinetic energy is the work ... 0 The units of F.ds are those of energy, to start with. But the "why" is a matter of observations, experimant.: Work is closely related to energy. The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work done on the body by the resultant force acting on that body. Conversely, a ... 0 About work: Let me begin with a little bit of history. James Prescott Joule did experiments along the following lines. He would have a calorimeter, with paddles inside so that moving the paddles would churn the water inside, and the friction of that churning would raise the temperature of the water. One of the ways of driving the paddles would be to have a ... 0 This is a rare case where it may help to get more technical. Work is the dot product of the force vector and the displacement vector, and it can be defined as follows$$\vec{F}\cdot \vec{d} = |F||d|cos(\theta)$$Intuitively, it wouldn't make sense to say that an upward force is causing a box to move horizontally across a room (mathematically that would mean ... 0 Let me not answer this question but only provide you with a hint. You are already close to what you are looking for. I don't think that your approach is flawed. There are just a few gaps that you have overlooked. You are right in concluding that the work W should depend on the mass m and the velocity \vec{v}. However, you know that W is a scalar. ... 0 First of all, your total energy should include the kinetic energy as well,$$E_{tot} = U + T$$where U is any potential energy (gravitational, electromagnetic...). This is sort of the definition of "total energy". The Virial Theorem relates the mean kinetic energy of a system of particles with the mean of it's potential energy. This is an important ... 1 I've learned that for an ideal gas, internal energy is solely a function of temperature (because we ignore the potential energy from attractions). That is correct. The molecules of an ideal gas are considered to be point masses with no intermolecular (van der Waal) forces. Consequently, the point masses of an ideal gas do not interact with each other.... 1 This answer is meant to complement exp ikx's answer. I'm just trying to get you to see these expressions in various ways. Let \bf \hat u be the 4-velocity of the observer. Let \bf p be the 4-momentum of the particle. With the (-,+,+,+) convention, \bf \hat u\cdot \hat u=-1 and {\bf p\cdot p}=-m^2 So, think of \bf -p \cdot \hat u as the ... 2 What does E=−p⋅u stand for in this context? (for massive particle) Is it really energy? Let c=1 and signature metric be (-,+,+,+). In a momentarily comoving reference frame, \mathbf u=(u^0,u^1,u^1,u^3)=(1,0,0,0) and hence$$-\mathbf p\cdot\mathbf u=-p^0u^0\eta_{00}=p^0=E$$It is a four-vector expression and is preserved under Lorentz ... 1 1) Conservation of momentum in the absence of external forces is not valid in non-inertial frames. This is because Newton's second law, in which \vec{F} stands for the net external physical (not including pseudo forces) force, is valid only in inertial frames. Next, you can solve this by considering the (block + wedge) as a combined system in the ground (... 3 Yes, massless gravitons would have energy and momentum, just like massless photons and massless gluons. Matter radiates gravitons when it accelerates in certain ways, such as two stars orbiting around each other. The energy radiated in the gravitons typically simply reduces the sum of the kinetic energy and the gravitational potential energy of the ... 1 The answer is to resubstitute U(S,N). First, for the caloric equation:$$ \left(\frac{\partial U}{\partial S}\right)_V = U_0 \left(\frac{V_0}{V}\right)^{2/3} \text{e}^{(S-S_0)/C_V} \frac{1}{C_V} = \frac{U}{C_V} \stackrel{!}=T \quad\to\quad U=C_VT $$and for the ideal gas law:$$ \left(\frac{\partial U}{\partial V}\right)_S = U_0 V_0^{2/3}(-2/3) \frac{... 2 When work is done, the energy is stored either in form of potential or kinetic. Just to be clear, it is an inclusive "or". When work is done all the energy may be stored as potential energy, or it may be stored partly as potential and partly as kinetic energy, or it may be stored completely as kinetic energy, as illustrated in the following 3 scenarios ... 1 It requires same amount of energy to generate mg upwards but in the first case, no energy is stored while in the second, some energy is being stored. Here is the key mistake. The energy required is not the same in the two cases. In the first case no energy is required, this matches with no energy being stored. In the second case energy is required, this ... 0 Collision of the balls in billiards is under the influence of a wide variety of factors. This includes spin, static friction, rolling friction, air resistance, angular velocity of the balls etc. Hence the expected observations given below will not be observed in practise. From the information you provided, the collision would be elastic in nature. This ... 1 Problem Statement The first case (work done by forces) considers the net work done. The second case (work done by pressure) considers only the work done by external forces. The analogy seems not to be exact. Why not? Background Forces and Work Let's expand the definition of net work. The net is the sum of external AND internal. The definition of work ... 1 I think the key point here is that a gas system needs to exert a permanent force on its surroundings to maintain its equilibrium at rest. Changing of system perspective in your example, the person pushing the block already has to overcome the 9N force acting as a resistive force to set the mass in motion, so the total work done by this person is not 1$x$,$x$... 2 I have a friend who could probably do a better job of explaining this, but I’ll give it a try. The work here is called boundary work, or p dV work, and applies to a closed system typically represented by a gas in a cylinder fitted with a piston. The piston is normally considered massless so no work is done on the piston itself. So your analogy of two ... 0 Thank you @dmckee for the comment. We have to write the radial functions as the sum of the regular and singular Bessel function solutions, i.e. $$R(\rho) = Aj_l(\rho) + B\eta_l(\rho)$$ where the$\eta_l(\rho)$is the spherical Neumann functions, defined as $$\eta_l(\rho) = -(-\rho)^l\bigg(\frac{1}{\rho}\frac{d}{d\rho}\bigg)^l\frac{\cos\rho}{\rho}$$ ... 0 I don't think it can be modeled that way. It is too complicated due to all the different molecule sizes. And disturbances. But I think a statistical model would do. Heat is a Electromagnetic wave causing the electrons to vibrate the whole molecule. The vibration is what we feel as heat. Sound causes molecules to vibrate by vibrating the molecules. There is ... 2 Its the mass of the rotor relative to the mass of the load. No matter what a motor must drive its own rotor. Lets say that rotor weighs 1 kilogram. If you were to drive a load that also weighs a kilogram, then 50% of the energy is driving the load and 50% driving the rotor, so it cannot be more than 50% efficient. If you increase to 2 kilograms as a load, ... 0 So we are given: $$\psi(x)=\begin{cases} Ax, & 0< x<\frac{a}{4} \\[1em] A(\frac{a}{2}-x), & \frac{a}{4}<x<\frac{3a}{4} \\[1em] A(x-a), & \frac{3a}{4}<x<a \end{cases}$$ And we have to calculate the most probable energy. I'm assuming this is the initial wavefunction,$\Psi(x, \ 0)$of the system. In order ... -1 All of the other answers are great, and I highly recommend reading them. However, I think something is missing if you don't attempt to get an intuitive understanding with geometry: This triangle shows that the equation$E^2=(mc^2)^2+(pc)^2$can be represented via a sort of reverse Pythagorean theorem. The special case of$E=mc^2$can be found by setting$p\$ ...

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A device that does this is called a synthesizer. It can be programmed to emulate many types of sounds fairly well, but complicated sounds like the human voice are harder to synthesize.

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Here's how to visualise it. 1) When the puck hits the slide, the puck will rise up the slide and the slide will recoil. 2) As the puck rises it will lose KE to PE and also share KE with the recoiling slide. 3) At the top of its rise the puck will be stationary with respect to the slide. At that point the puck and slide will be a combined mass of 4m. ...

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First, you should understand what "conservation" means in a physics sense. Conservation of a quantity means it is neither created nor destroyed in a time-limited process (we're not talking about eternity past or future). It is merely moved around to other systems/objects or (for energy) changes forms. The means that if a quantity (like momentum or total ...

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The mistake you are making is saying the second equation is due to "conservation of energy". The equation only addresses conservation of kinetic energy. Kinetic energy is not conserved. But total energy is conserved. The kinetic energy that is lost is dissipated primarily as heat because the collision is inelastic. Hope this helps.

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What you are talking about is an inelastic collision, which is defined by loss of kinetic energy (converted into e.g. heat or sound, as you had before the edit). Check this wikipedia page for more info. On a side note, if you have two expressions that tell you different things, this means that at least one of them is wrong. You don't just choose one of them ...

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