# Tag Info

0

Assuming you're talking about a hamiltonian for a two-dimensional quantum system, any self-adjoint (aka hermitian) $2\times2$ matrix with complex entries will do. Any such matrix must equal its adjoint (conjugate-transpose), and this restricts the matrix to have the following form: $$H = \begin{pmatrix} a & c+id\\ c-id & b \end{pmatrix}$$ ...

2

As far as we can tell (up to energy scales we've measured so far), spacetime is a nice and smooth manifold. It might happen that the smoothness is approximate and spacetime is discrete at a much more microscopic scale, or it could turn out that spacetime is smooth all the way through. Short answer: We don't know. About the notion of energy quantization: ...

-1

I have a quantitative answer which is a thought experiment avoiding all but the simplest equations. An object going from velocity v=0 to v=1 needs to be pushed or pulled in some way. In my explanation I will use the same method to push the object from v=0 to v=1 then from v=1 to v=2, then v=2 to v=3, etc. I will show how the energy of movement embodied in ...

9

A general photon isn't too dangerous. Most photons that we encounter have the power to heat our bodies and not much else. The heat we absorb from photons daily isn't that much, so this is rarely a problem. Now, an interesting thing about photons is that two photons of a lower energy do not make a single photon of higher energy (frequency). So a million ...

16

I have a somewhat non-physics answer for you. If you allow me to broaden your question a bit to "why doesn't light kill or otherwise make all life on Earth impossible" the answer is that the Earth is in what we call "the habitable zone". If the Sun produced so much light or light at such high energies that it would kill you, it also would heat the planet ...

9

This question is more interesting than I thought at first. I like it. There are several different parts to an answer to this question; I'll just contribute a couple that have something in common: our bodies (and everything else, it has nothing to do with bodies) also emit photons about as fast as they absorb them. On the macroscopic/thermal scale, we have ...

1

I suppose $f$ is just an arbitrary scalar function on the manifold. I'm not well-versed with the concept of Ricci flow, so I'll try to give a simple operational answer. I also don't understand what exactly you're looking for. The Ricci scalar $R$ roughly represents the amount of energy stored in spacetime (as curvature). The dilaton is a scalar field which ...

43

Individual photons are very small and don't have much energy. If you put a lot of them together in one place you can hurt somebody - by simply supplying enough power to melt an object (ask any spy on a table underneath a laser beam). There is another very odd feature of photons. Although lots of them can provide a lot of energy and heat an object, it takes ...

3

You have to take into account the differentials. The actual equation is $$f_\text{MB}(\mathbf{v})\,\text{d}v_x\text{d}v_y\text{d}v_z = n\left(\frac{m}{2\pi k_BT}\right)^{3/2}e^{-mv^2/2k_BT}\,\text{d}v_x\text{d}v_y\text{d}v_z.$$ Changing to spherical coordinates, we get $$\text{d}v_x\text{d}v_y\text{d}v_z = ... 0 The distribution function allows to find some kind of probability by summing of f_vdv if it is the distribution over velocities or f_EdE if it concerns energies. One can be transformed to other as:$$f_v(v) dv=f_v(E) v dE,$$since dv=vdE if E is a function of v Taking into account E=\frac{mv^2}{2}, one gets:$$f_v(v)dv=f_v(E) ...

1

This kind of exponential decay toward "equilibrium" can be derived when one looks at a Markov process. In this case, if we call $S_t$ the state of the system at time $t$ and $S_{t+1}$ the state at time $t+1$, one has for the evolution: $S_{t+1} = T S_t$ where $T$ is called the transition matrix. This implies that $S_t = T^t S_0$. The idea is then to ...

0

This form of $dE/d\tau$ is valid only when the system is not too far from equilibrium and linear response assumption is valid. The fact that $dE/d\tau$ depends on the difference $E - E(0)$ alone is a consequence of assuming a linear response.

5

The work in the first law is exactly the usual work $W=\int Fdx\rightarrow\int PdV$. For point particles, this is enough to completely specify the behavior of the system using Newton's first law, or energy methods. However, for macroscopic objects, the motion of the internal components (in thermodynamics these would be particles) have some additional degrees ...

2

The $W$ term in the first law expression exclusively refers to the mechanical work done by a system and all other things , all other possible exchanges of energy are clubbed together in $Q$. Suppose I am the system under consideration , and I apply a force on a block and that does some mechanical work (that is the point of application moves a distance) ...

5

It is just easier, i.e. less expensive, to build and maintain them that way. There exist alternative designs that are more efficient but also more difficult (= more expensive) to build, put up and maintain. You can check those out via this link.

2

Energy and momentum has to be conserved. That the electron / photon has to have enough energy for the excitation is obvious. What is interesting is what happens when they have too much energy. For radiative transitions between bound states the orbital anuglar momentum has to change by 1. This means that the photon has to be absorbed which in turns means ...

-5

As we know that opposite charges attract each other, we consider gravitation as negative and matter as positive.. Thus there is a attraction between the two... That's y we say that real particles are always positive as they r nthng but matter/energy.. However, inside the black hole the gravitational field is so strong that even real positive particles ...

0

It is simply a matter of definition. It is defined in a way such that in infinite distance the potential energy is 0, therefore as you get closer, the potential energy is expressed in a form of kinetic energy and the amount of potential energy "available" decreases. Just definition.

0

First things first: the total mechanical energy is always kinetic energy plus potential energy. So if your answer sheet actually said $KE - PE$, it's wrong. But what I suspect it really said is that the potential energy is negative, so the formula you wind up with is $$\underbrace{\frac{1}{2}mv^2}_{KE} \underbrace{- \frac{Gm_1 m_2}{r}}_{PE}$$ Now, the ...

1

The other answers tackle the statistical/thermodynamic aspect. I will tackle the "falling apple " aspect. Why does the apple fall? From this observation onwards nature was modeled mathematically as interactions between masses, in this case, charges in the electromagnetic case etc. The observations of gravitational interactions led to a mathematical model ...

3

Large systems with many degrees of freedom (e.g. a ball consisting of many molecules) tend to settle into low energy states. This is a direct consequence of two fundamental laws, the first and second laws of thermodynamics: energy conservation and entropy increase. A system with many degrees of freedom can be in many different microscopic states (think ...

1

I will address such sample systems as a point (or a small metal ball) rolling or bouncing on some hard surface with hills and pits, and an atom which can be either in excited or in basic state. I. If we consider an ideal closed system, then the enegly is conserved. But real systems do not (exactly) behave this way. For a macroscopic mechanic motion we can ...

1

The probability of finding a system in a state with energy $E$ is $P(E) = \exp(-\beta E)/Z$, where $\beta = (kT)^{-1}$, $k$ being the Boltzmann constant and $T$ being the absolute temperature. $Z$ in the formula for $P(E)$ is the canonical partition function. For our purpose, we can consider it as a factor introduced to ensure that $0 \le P(E) \le 1$. The ...

0

The most important thing about feeling is that its relative and depends on the temperature of sensor point on skin. That's the reason we need thermometer to measure temperature with which everyone could agree. Energy transfer is involved here, but it has very little to do with Amanda's feeling. Amanda's tongue and mouth was at high temperature due to ...

0

One way is to find out the internal change energy of the system and infer the heat transfer to the system from that and the work done: $$\delta Q_\text{to}=dU-\delta W_\text{on}.$$ If you have a handle on the system's entropy, on the other hand, then you can use the Gibbs relation, $$\delta Q_\text{to}=TdS,$$to find the heat delivered. In general, though, ...

3

You are talking about relativity and gravity together so the question can only be answered in the context of general relativity, but concepts like gravitational potential energy and gravitational force acting over a distance are Newtonian and do not really carry over to general relativity. However, the gravitational field does contribute to total energy and ...

3

Depends on what you're doing. General relativity handles it for you, in the sense that the Einstein field equation links geometry to the non-gravitational stress-energy tensor. That general relativity is non-linear can be interpreted in part as gravity itself contributing to gravity, but it's generally not even possible to localize gravitational energy in a ...

0

If you have a path on $p-V$ diagram that is parametrized by some parameter $x$, so that $p=p(x),\,V=V(x)$, then: $$dU=\delta Q+p(x)dV(x)$$ here $Q$ is the total heat received by the system (it is negative if system releases heat). I write $\delta Q$ to indicate that $Q$ is not a function of state, and $\delta Q$ is not a full differential. Assume now that ...

0

AB-isothermal. $\Delta W=Q$ ; area under the curve depicts the work done , ie. heat intake. $\Delta W=n\Bbb RT\ln\dfrac{V_2}{V_1}$ BC-isobaric. Can't be calculate directly from curve. otherwise use ,$Q=\Delta W + \Delta U$ where $\Delta W=P(\Delta V)$ and $\Delta U=\dfrac f2 p\Delta V$ CD-isochoric. $Q=\Delta U=\dfrac f2 V\Delta p$ . DA-adiabatic . ...

2

You have a few different but related questions so I will try to explain them in a simple, no-math way. If a radio tunes to a specific frequency, where does the excess energy go? Almost every object that has radio waves (electromagnetic waves) around it absorbs some of the radio energy. When the radio waves hit the electrons in the atoms and transfers ...

1

When radio waves hit the antenna it creates an electric potential difference between the antenna and ground. An electric current flows from antenna to ground, through the radio receiver. The radio receiver is able to extract information (the signal) from this current and amplifies it. Virtually all the electromagnetic energy collected by the antenna flows to ...

1

The answer is the energy goes into the gravitational field. If you take the simplest case of a spatially flat homogeneous cosmology with no cosmological constant then the equation for energy in an expanding volume $V(t) = a(t)^3$ is $E = Mc^2 + \frac{P}{a} - \frac{3a}{\kappa} (\frac{da}{dt})^2 = 0$ $M$ is the fixed mass of cold matter in the volume, ...

3

The first law of thermodynamics says "the increase in internal energy of a body is equal to the heat supplied to the body minus work done by the body". Assuming there is no heat flow (for simplicity), this says "the increase in internal energy of a body is equal to the work done on the body". Since you are doing work on the gas, the internal energy ...

0

The short answer is "yes". The energy lost from the photons is taken up by the energy in the gravitational field. Of course energy is a relative concept but if you take the simplest case of a spatially flat homogeneous cosmology with no cosmological constant then the equation for energy in an expanding volume $V(t) = a(t)^3$ is $E = Mc^2 + \frac{P}{a} - ... 0 The only thing that prevents us defining a total conserved energy for the entire universe is that if the universe is infinite then the total energy could be infinite or indeterminate. The statements that say energy is not conserved in general relativity are wrong, irrespective of who says them. You can define energy over any finite volume of space and you ... 1 energy is always positive or 0. And these are just numbers we associate with a body due to its motion according to different set of mathematical rules so that we can study these particles . As such they have no physical meaning . For example momentum is just$m$x$\vec v$. It is just a number we associated with a body by the quantities we defined ourselves ... 6 Imagine that you have just two particles with the same mass and same speed, but going in opposite directions. They have opposite momenta, so the total momentum is zero. But they each have energy, and the total energy is not zero. The reason is because kinetic energy is just$\frac{1}{2} m v^2$. That square means that the kinetic energy can never be ... 1 Use an equal-arm balance beam, with a 250 kg counterweight suspended 5 cm in the air on one end, and the mass to be moved just touching the ground on the other end. Energy storage is in gravitational energy. Losses can be made insignificant. Problem was originally solved by elevator engineers... 2 It sounds like the way you're imagining this is the source of your confusion. Electrons in one atom are not attracted to the electrons in another atom. What actually happens is that it requires less energy for two atoms to come together and share some electrons in a covalent bond. How much can be saved and the configuration of the bonded atoms depends on ... 0 A little trick required here. Perform a substitution first: $$mv \gamma(v) = u ,$$ hence your integral becomes $$\int \frac{du}{dt}dx ,$$ but notice that $$\frac{du}{dt} = \frac{du}{dx}\frac{dx}{dt}dx ,$$ but$\frac{dx}{dt} = v$- the definition of velocity! So the integral ... 0 With your current assumptions you do not have enough equations to solve this problem, since it is two dimensional, which gives you 4 unknown variables:$u_{1}',v_{1}',u_{1}',v_{2}'$where$u$and$v$are the speeds in the respectively$x$and$y$direction, the indexes$1$and$2$indicate which object it is and the apostrophe ($'$) indicates that these ... 0 Here's a formula to help you find final velocities from initial velocities: $$v_1=\frac{u_1(m_1-m_2)+2m_2u_2}{m_1+m_2}$$ $$v_2=\frac{u_2(m_2-m_1)+2m_1u_1}{m_1+m_2}$$ These formulas you get from combining momentum and energy equations. You have to apply both of the above formulas separately in 2 dimensions:$x$and$y$. So you should get ... 0 I found the solution. It was much more easier than I believed on first sight. After a good rest on my couch, I saw easily the big deal: The total energy of the system is get by the kinetic energy of the rotating mass$(1/2)Ix_2^2(t)$, that is positive, and the gravitational potential energy$Mgh$, with$h$obviously pair to$l∗cos(x_1(t))$, that is ... 0 You may recall that one way to solve the problem A mass$m$passes the table top moving upwards with initial velocity$v\$ (in standard gravity). What is the maximum height above the table reached by mass? is to use energy conservation: you equate the maximum gravitational potential energy relative the table top with the initial kinetic energy of mass. ...

1

I apologize "basics foundations of thermodynamics" still does not make a lot of sense to me. Steve B already provided some answer associated to one way to interprete the word "foundation" that is from statistical mechanics. I will kinda play here devil's advocate and assume that you are refering to axiomatic thermodynamics. As far as I am concerned, the ...

0

Yes, it can be proved. But the mathematical rigour only applies to ideal gas, so you might feel disillusioned after all the calculations, considering how much this can be valid in reality. If you insist you want to know, then you may read on. From the first law of thermodynamics, $$\delta Q = \frac{3}2 nRdT + \frac{nRT}V dV,$$ where internal energy ...

1

There's a group I call "thermodynamic purists" who think that thermodynamics is a self-contained system based on semi-mathematical "axioms". I disagree! I think that thermodynamics is fundamentally a consequence of statistical mechanics, and that this is the best way to think about it and understand it. I acknowledge that reasonable people can differ on ...

1

The problem is that neither the potential energy nor the energy necessary to compress the gas is of any importance here. We can set the energy of the compression to zero and still would not gain any energy in this model! So for any real device with losses we certainly can't gain energy. This can best be shown in an experiment by connecting a propeller to a ...

2

Energy is a scalar quantity, a single number, that doesn't change in physical systems where all interactions are inside the system. For a number of particles that only interact with one another at the same location, kinetic energy is this single number that's conserved. When they interact at a distance via the electromagnetic field, the total kinetic energy ...

-1

In regards to wind resistance, it is negligible at running speeds in comparison to its value as air conditioning. The reason bicyclists can cycle for much longer than runners can run is from the much better cooling obtained from greater wind resistance! The increased drag is of course more than made up for by the vastly greater efficiency of cycling.

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