# Tag Info

51

When light is propagating in glass or other medium, it isn't really true, pure light. It is what (you'll learn about this later) we call a quantum superposition of excited matter states and pure photons, and the latter always move at the speed of light $c$. You can think, for a rough mind picture, of light propagating through a medium as somewhat like a ...

48

[5/3 - Extended the answer, made some corrections, and responded to John Duffield's comment] This is actually the paradox that led Einstein to General Relativity. Consider a special case: An electron and positron are at the Earth's surface. Bring them together and they annihilate, creating gamma rays (which is very energetic light). The gamma rays travel up ...

34

Just for completeness, I'll explain how to obtain the time taken for an arbitrary curve. If $h$ is the initial height of the child and $y$ the height once he has started falling. By energy conservation: $$mgh=mgy+\frac{1}{2}mv^2\implies v=\sqrt{2g(h-y)}\tag{1}$$ We know know the speed at any time. Let us denote the horizontal position as $x$. The ...

31

Do black holes violate the first law of thermodynamics? No. See Wikipedia re the first law of thermodynamics: "The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed ...

25

The shape of the slide definitely determines how long it takes to go down it. Consider if the slide was completely vertical. Now, a certain famous [recently deceased :( ] comedian had the astute observational powers to point out that this would, in fact, be a drop, not a slide. Nevertheless, you would quickly reach the bottom. Now imagine if the slide was ...

23

Why does the author say that we would need to know the shape of the slide to find the time taken for the child to reach bottom of the slide? As you've discovered, the speed going down a frictionless slide only depends on the vertical distance. This speed is not the vertical component of velocity. It is the magnitude of the velocity. The vertical ...

19

Your guess at the solution to this paradox is correct. "Pumping energy up" to the space station, regardless of the method you choose, would require an input of at least the amount of energy you would gain in kinetic energy on the way down. This is just a variation on the impossible perpetual motion machine concept. In practice, you would not only not gain ...

16

Work is calculated as force times distance. $$W = Fd$$ The purpose of a simple machine like a screw jack is to lessen the force required. However, the work needed is still the same, so the distance over which you exert the force has to increase. Halving the force requires doubling the distance. In this problem, you want to lift 2000 lbs a distance of 1 ...

16

I believe, the answer is a small but quantifiable, yes, there is a non flat road configuration that would lead to better gas mileage between any two points at the same height. I have numerically solved for such an optimal path. I believe I can give a nice explanation of why that is, but it will take some work, so bear with me. Granted, you can only expect ...

15

If you could take from orbital energy, then it would decrease, until at some point in the future it would zero. Hence, it can't be perpetual.

15

Conservation of energy/mass is the result of a symmetry called time shift symmetry, and if this symmetry is broken energy/mass will no longer be conserved. It is far from obvious that time shift symmetry would be preserved if closed timelike curves were possible, so you can't use conservation of energy as an argument that time travel is impossible.

13

In your working you have assumed that $a=g$ - this is true if the slide is vertical. Slides will have some angle, $\theta$ (e.g. $45^\circ$), which will mean that the acceleration, $a$ is given by $$a = g ~sin \theta$$ Note that $a$ will be less than $g$ because the value of the $sin$ term will be between $0$ and $1$. (except in the case of a vertical ...

12

A classical explanation to supplement Rod's excellent quantum mechanical one: If you make a Huygens construction of wave propagation (I assume you know how to do that) then every point on the wave front is treated as the source of a new wave of the same frequency and phase. How that wave propagates depends on the medium it encounters. So the Huygens ...

12

In step 1 you lower the mass and this generates energy. Let's say you store this energy is a spring, and for the sake of argument let's say the energy stored is 1J. The energy has to come from somewhere, and of course it comes from the rotational energy of the torus so the rotational energy of the torus is now 99J. In step 2 you slow the torus, perhaps by ...

11

We already harvest energy from the Moon. It causes the tides and stress and strain and motion throughout the Earth. As a result, the Moon keeps getting farther away. (And it causes some heating in the Earth). The Moon at one time had a spin that was not locked to the Earth, and the tidal bulges in the Moon's shape caused by the Earth generated heat in the ...

11

First, the energy expectation value of the superposition state you have written down is $$\left(\frac{n_1 + n_2}{2} + \frac{1}{2}\right)\hbar\omega$$ and one might naively conclude that therefore the energy of the state lies in between the energy of its constituents. This naive concept doesn't work, though - the "energy" of a state that is not an energy ...

11

What does this small change means in form of Rotational Kinetic Energy? There's a problem with your calculation: You assumed a constant value for the Earth's moment of inertia. The Moon and Sun raise tides on the Earth itself. These Earth tides result in subtle changes in the Earth's moment of inertia. The signature of these tides can easily be seen in ...

11

Very little of the energy from a rocket engine ever goes to the kinetic energy of the rocket. The only way you get perfect conversion to KE of the rocket is when the propellant is directed in the opposite direction of motion and when the ejection velocity is exactly equal to the speed of the rocket. In that case, the propellant winds up containing 0 kinetic ...

10

Ever since Newton and the use of mathematics in physics, physics can be defined as a discipline where nature is modeled by mathematics. One should have clear in mind what nature means and what mathematics is. Nature we know by measurements and observations. Mathematics is a self consistent discipline with axioms, theorems and statements having absolute ...

8

The mass of a free neutron is 939.566 MeV/c$^2$ (almost 1 GeV/c$^2$, so that's probably where your instructor got the "1" value), and the mass of a free proton is 938.272 MeV/c$^2$. A free neutron will decay into a free proton, free electron ($\beta^-$), and an anti-neutrino, $\bar{\nu}$. The mass of the electron is 0.511 MeV/c$^2$, and of the ...

8

Let's say that the rocket is traveling in the $y$-direction at some velocity $v$, which may or may not be non-zero. A force - in this case, thrust provided by the engine - is applied perpendicular to the direction of motion, in the $x$-direction. This force produces an acceleration, which causes the rocket to move. Therefore, the force is not applied at a 90 ...

7

Exerting a force and providing energy are quite different things. In particular, to provide energy to a body the force needs to perform work, that is, it needs to move the object in the direction that the force acts in. In the case of the Moon, the movement is circular and perpendicular to the gravitational force, so there is no inwards / outwards motion.* ...

7

Even if the laser had perfectly reflecting, i.e. lossless, mirrors at either end of the cavity, and both ends were sealed so no light could escape it would still require a continual power input. That's because excited atoms/molecules can decay by mechanisms that don't involve a photon e.g. collisional de-excitation. The lost energy goes into heating up the ...

7

Update: According to this paper, "On the Interpretation of the Redshift in a Static Gravitational Field", the answer I give below is a common but misleading interpretation. The classical phenomenon of the redshift of light in a static gravitational potential, usually called the gravitational redshift, is described in the literature essentially in ...

6

Virtual particles are not real. Though sounding like a tautology, it is an important one - they are not actual states in the asymptotic Hilbert spaces of a quantum field theory, where particles usually live. They are a name given to internal lines of Feynman diagrams, which, in turn, are mere computational tools in a perturbative approach to QFT. Nothing in ...

6

From the geometry, you can state that in order to move the screw by 1 unit of distance, you have to move the end of the handle by $10\times 20\times 2\pi$ units of distance. Let's call the unit of distance $[L]$ - in this case, an inch might be a good unit but we don't have to be explicit about that. Conservation of energy says that work done on the system ...

6

A battery connected to a capacitor is an RC circuit in the limit $R \to 0$ (i.e., there is no resistor and the resistance of the wire is negligible). One might think that the energy loss is zero in this limit, but this is not the case. For an RC circuit with a battery and an initially (i.e., at $t=0$) uncharged capacitor, we have Q(t) = CV ...

6

It is not true that in all fusion and fission processes the mass of the products is less than the mass of the reactants. This is only valid for exothermic reactions. The change of mass is due to the change in the binding energy of the nucleons (note that the change in binding energy is in the order of 1 MeV, while the mass of the nucleons is around 940 ...

6

Your question asks why the "current quark masses" [see http://pdg.lbl.gov/2011/download/rpp-2010-booklet.pdf at page 21] of the quarks that make up a proton don't add up to the mass of the proton. The problem is that, for the light quarks, the "current quark masses" are very different from the "constituent quark masses" [see wikipedia]. "Constituent quark ...

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