What is energy?

Pyramid9 · 2ヶ月前

Mass and energy are two sides of the same coin, they are both properties of matter. Matter is stuff, and we add labels to stuff - mass, energy, momentum.. and describe how they relate to each other.

Imagine you have money, but you also have cabbages. I mean money as the abstract notion of value that does't actually requires a dollar bill. The mass of the cabbages has an intrinsic monetary value, but even if you have an equation that relates the money you can get by selling cabbages to the mass of the cabbages, that doesn't mean that money and cabbages are the same thing.

E = mc² is exactly the same thing. It's an exchange rate between how much energy you have in a mass "m" of cabbages - cabbages are your matter, btw. Before Einstein, you would say that E = 1/(2m) p² , where p is momentum - which means that you only factor how fast cabbages are moving in order to know its price. What Einstein did was to correct this equation by adding an mc² term to the price of cabbages.

SalmanShaheen · 4ヶ月前

whereas before industrialization everybody had a (shitty) job farming

Are you serious? Are you implying that in post-colonial Brazil where slavery was legal or just abolished, before the words civil rights even made sense, everyone had a shitty job and thus industrialization ruined that magic wonderland of exploitation? Today even the jobless in Brazil earn comparatively more than someone on a minijob in Germany.

This surplus in state money is not possible without hue productivity which leads to more revenues and thus more taxes - and by not forcing the population into mindless meaningless jobs in their early 20s and directing government resources towards education and reinsertion, you can provide conditions for an "efficientization" of the means of production without sending everyone into poverty. Allowing the market to be efficient doesn't automatically mean negating a social state - which is exactly the premise of nordic social democracies: we mostly embrace the free market, but we use the excess wealth in order to provide a good quality of life and education for everyone.

Anyway, these 4 examples have nothing to do with each other, they developed under completely different institutions, and sit now in different economic situations. Spain was thriving until before the financial crisis an Brazil is in an upward path; Syria was a Theocracy while China, well, it's difficult to quantify. From these, only Syria is doing worse than it did 100 years ago, but it's not hard given the circumstances, right? They do have one thing in common, though, neither had institutions that allowed for industrialization to happen until very recently. The fact that it happened in Britain in the 18th-19th century was no coincidence, as post-revolutionary Britain was the only place that allowed the constructive destruction to happen.

Interestingly enough, it could have happened a few centuries earlier, and one example is how Elizabeth I treated one invention (at the time presented by Willian Lee): "Thou aimest high, Master Lee. Consider thou what the invention could do to my poor subjects. It would assuredly bring to them ruin by depriving them of employment, thus making them beggars."

SkiFreeOrDie · 4ヶ月前

Because it can happen. It's a shitty answer but that's the reality. On the case of the neutron, this process is allowed thanks to the exchange of a W boson which changes a down quark into an up quark, like this, and since the neutron is more massive than the end proucts, it is kinematically possible.

This diagram has a mathematical interpretation: once you use the Feynman rules, it gives you the amplitude - which tells you the probability that for some time interval, this process will happen. This tell you what a lifetime is: the expected time you have to wait in order to see a neutron decaying.

So at the end, nothing really causes the neutron to decay, it only decays because it can.

SalmanShaheen · 4ヶ月前

It's been that way all through history. Why do you think it's different this time? Innovation comes, some people lose their jobs, 10 minutes later new jobs are created 100-fold. The thing is that the jobs that are created are primarily hidden, so in the beginning it all looks catastrophic. Do you think you'd be living in such a nice society if every former telephone operator and mail delivery person received a compensation when you go online? Heck, computers already took so many jobs, and every factory has some degree of automation, but we're not complaining about it, are we?

At this age, a lot of people are getting paid to make reddit and facebook posts. Could you envision that 15 years ago when probably some publicists were crying over competition because of the internet?

iPlaySkullgirls · 4ヶ月前

That design is very different, in his design it would stop in the stable position once the magnetic fields are aligned. The reason why his design works is because he took off the coating in only half of the wire, which means that once the magnetic fields are aligned there is no current, so only the unstable alignment remains.

Ruiner · 8ヶ月前

The terminology exists because of the "evolution operator", which is the exponential integral of the Hamiltonian. This operator obeys the condition that U(t,t0)* = U(t0,t), so U*U is essentially going forth in time and then back again. Since it is unitary, U*U= Id, which is to say that evolving forth in time and then back again leaves the system unchanged.

When you do perturbative quantum field theory, the evolution operator is replaced what we call the "S" matrix. Instead of evolving the system, the S matrix is essentially telling you "the probability that state A will evolve to state B", and you build this object perturbatively - as a sum of individual processes represented by feynamn diagrams. In this case, the "operator unitarity" of the S matrix translates into the "sum of probabilities equals one" unitarity that is usually referred to in the high energy physics literature.

Ruiner · 8ヶ月前

Suppose that you possess a very peculiar power: you are able to look at a cup of tea and track and the positions and momenta of the molecules in this cup of tea. Forget quantum mechanics, at this level they all interact in some boring way and with your supercomputer you are able to track the evolution of the system. Also, someone just gave you a cup of tea after adding sugar and stirring for a while, and in order to test your superpower, asked you the following question: "what did I do first: poured water in the cup and then sugar or the opposite?"

The laws that govern the system are "unitary", which means that if you know every position and velocity now, you can in principle change the direction of time and know what was happening 10 minutes ago. But if you look at the system after it has reached maximum entropy, you'll find out that it seems completely insensitive to the details of the initial condition of the cup of tea: the system has reached equilibrium, and the only parameter that describes it is temperature.

So where has this information gone? Well, the information must for surely be there, since the laws of physics for sure have not been broken by a cup of tea - even if physics has a strong bias for coffee. It's just trickier to get hold of this information, and this apparent "information loss" is at the origin of why entropy increases.

Unfortunately we mortals do not have access to your superpower. It's just not possible to know every position and velocity of 10³⁰ particles. What we can do is the following approximation: we measure the velocity of some particles that are flying around and devise a probability function to find a particle with a certain velocity "v". What we find is that once a system has reached equilibrium, this probability is the so-called "Maxwell–Boltzmann distribution".

Before reaching equilibrium, however, this function can be anything you decide to prepare: this is what happens once you pour the tea in the cup, you have a situation which is far from equilibrium. We can describe the time evolution of the system by writing a dynamical law for the evolution of this probability distribution: this is called the Boltzmann equation, and it tells you how the average velocity of particles evolve when they interact among each other.

Solving Boltzmann equation is not an easy story, but it turns out that it has one peculiarity: given a nice enough interaction, the "fixed point" of the evolution is the "Maxwell–Boltzmann distribution", which is to say that thermal equilibrium will always be reached, no matter where you started from!

But where has the information gone? Well, the information is not encoded in the average velocities, it is instead hidden in correlations! In order to understand why this is important, just think about the opening of a billiard game: right after the impact, the balls fly in average with the same velocity distribution, but how they fly around is extremely dependent on the details of how you aim your strike. This information can be extracted from by looking at how they are correlated with each other: i.e: "the probability that ball 1 has velocity v1 knowing that ball n has velocity vn". There are almost infinitely many configurations which look almost the same as each other if you don't care about the details! On the other hand, the probability that after striking only a few balls will fly with almost all of the energy while the other stand still is very small.

By starting from the initial knowledge of all the velocities of every particle and sticking the simplified description of the distribution function of velocities of particles, we are ignoring all the correlations among the particles in the system. It just turns out that there are a huge number configurations which are very different at the microscopic level - once we take into account these correlations - but look exactly the same from the point of view of the distribution function. (just think about the billiard opening again). Equilibrium is reached not because of a force, but just because since everyone is interacting, it's very very unlikely that they won't equilibrate. And entropy is at the end just a measure of this coarse-graining. It tells you, in the statistical physics language, how many "microstates" exist for a single macrostate.

VitaAeterna · 10ヶ月前

This is a more appropriate question to /r/AskScienceDiscussion , as we currently don't have any means to answer this question without too much speculation.

NoahFect · 1年前

Complexity. The equation that you need to solve - the Schrödinger equation - is actually quite complicated once you have many variables. All the electrons interact not only with the nuclei but also among themselves, so it's a very messy thing to attempt to describe something exactly.

In any case, an exact solution in the analytical sense is only in principle possible for the Hyrogen Atom. For more variables only numerical solutions are possible.

felipehez · 1年前

What defines large scale structure are actually tiny quantum fluctuations.

In order to clarify, you need to understand what inflation is. In the very beginning of the universe, it went through a phase of accelerated expansion. This accelerated expansion made the universe extremely flat, in the same way that stretching a sheet makes it uniform. In the end of inflation, the field that carried the energy responsible for the accelerated expansion decayed into particles, and this decay is not completely homogeneous, so in some areas there was more energy concentrated than in others.

So these inhomogeneities is what we can see in the CMB map, which you probably have seen before. Tiny temperature fluctuations of the order of 10^-5 that gave rise to all the structure around us.

pennysmith · 1年前

Before anything, think about these questions:

What happens if the driver is travelling next to a bowling ball and a helium balloon? How will each of them react to the shock?

Suppose that the same person is instead submerged in a stationary tank. All of a sudden a huge hammer hits the tank, but the tank is very solidly attached to the ground so it remains pretty much stationary. What happens then? Where does the energy go?

In order to understand why the first question matters, think about this other example: you're driving in a closed car with a helium balloon floating next to you. Now you suddenly break, will the balloon move forward with you or backwards?

inteusx · 1年前

Standard (in-out) perturbation theory only tells us the probability of a certain asymptotic state at time T = -inf evolve to another state at time T = +inf. You could in principle try to track the time-evolution, but the machinery for that is much more complicated.

samcobra · 1年前

This is not exactly true. Quantum systems also thermalize despite the fact that there is no loss of information.

Classical information also evolves unitarily, and the unitary evolution is given by Liouville equation. The reason why there is effective loss of information is because we disregard correlations: thermodynamics is about 1-particle distribution functions, whereas equilibration moves information into higher order correlation function which we do not care about for all practical purposes.

hikaruzero · 1年前

I think you understood everything already. The definition that they give of \alpha_G - which is here - is actually "some mass" divided by planck mass squared, and in this case the mass was taken to be the electron mass. But when you write the interaction, what you actually write is 1/M_p (h T), where h is the metric and T is the energy momentum tensor. What happens is that one factor of energy "jumps out" of the energy momentum tensor in order to make a dimensionless parameter by joining 1/M_p. So in a sense, the gravitational charge is E/M_p

When you write the EM interaction, what you have in the lagrangian is \alpha_e * A * J, where J is a 4-current and A is the EM vector potential. In this interaction, \alpha_e has dimension 0 - it's really just a number - so it doesn't really depend on energy scales (actually this is a mild lie, but I'll comment on that later).

When you want to describe interactions in QFT, what you do is a perturbative expansion in the interaction coupling constant. The objects that you want to compute are the S-Matrix elements, which can be translated as: what is the probability that n particles with momentum k interact and become some other m particles with momentum k'?

So if you want to compute the "EM interaction between two electrons", you evaluate this object. But that's not all, since higher order terms in the perturbative expansion give you more things to calculate! Whenever you have a "closed loop", you have a divergent integral - and these divergencies are related to contribution from virtual modes at very high energy scales. Dealing with these divergencies is what we call renormalization: the parameter \alpha_e that we write initially is not really what we measure. What we measure is really "the strength at which two electrons interact", but that depends on all these contributions which are actually divergent! So the physical \alpha_e actually has a tricky dependency on the "bare" \alpha_e! In order to solve that, we need to regularize the theory by introducing an arbitrary scale at which we cut-off the divergencies. At the end, by imposing that the physical parameters do not depend on this "regularization scale", we get a finite theory. But as a result, we find out that the couplings "run" with the energy scale at which we perform the experiment. So \alpha_e actually depends logarithmically on the energy - which is the typical thing in a renormalizable theory.

When a theory is non-renormalizable, it's impossible to treat these divergent contributions in a sensible manner. The "coupling constants" actually depends way too strongly on the energy, and you have to deal with an infinite number of parameters that need to be renormalized. The intuitive picture is the following: if we start from a fundamental theory and coarse-grain over fine length scales, we get some "effective interactions" that are ignorant on which are the details of the fundamental theory. This new effective theory, although it makes sense when computing processes at low energies, has an intrinsic energy scale after which it stops making sense: so the procedure of "integrating over virtual modes with very high energies" is somewhat ill-defined. We have a natural cut-off in our theory, and that cutoff is given by the scale at which the interactions become strong.

And that brings us to Newton's constant, since it actually gives us the scale at which the gravitational interactions become strong: and that is 1/Sqrt(G), the Planck scale.

hikaruzero · 1年前

The gravitational coupling constant is actually not a dimension-less parameter like the fine-structure-constant. Newton's constant has a dimension of a length-squared, or inverse-mass-squared. In order to get a dimensionless parameter, you need to multiply it by a mass-squared.

In the example of the wiki article, they use the mass of the electron as this mass scale, but in general this scale will be the typical energy of the particles involved in the process. So if you are colliding electrons whose energy is at the order of the Planck mass, \alpha_G will become order 1, while for two rest electrons it's something extremely tiny.

It's important to say that the fact that the coupling constant is an inverse-mass is of extreme importance, because it's what tells you that GR is non-renormalizable.

Levski123 · 1年前

Of course, yes, my bad! Like the harmonic oscillator as well!

TheGulpmaster · 1年前

It doesn't. If the sun collapses into a black-hole, and the mass is fixed, you wouldn't feel a thing.

However, if you try to get really close to the sun, you're gonna hit the surface; whereas if the sun collapses into a black-hole, its size is going to be much much much smaller than the sun, and since gravity increases with your proximity, at the surface of the black-hole you're going to feel a much bigger gravitational pull than at the surface of the sun.

Levski123 · 1年前

Let me give you just a brief overview, and let me know if you can follow:

Schrödinger's equation is something like H f = E f. H is a differential operator and f is your wavefunction.

This is an eigenvalue equation, meaning that I have to look for both the eigenfunctions "f" of H which satisfy this equation and the allowed values for E for which this equation makes sense.

For a given H, there might be an infinite set of E for which this is valid (we call that a continuous spectrum), or just a few values of E which are separated from each other (a discrete spectrum).

When we have bound states - like electrons in the hydrogen atom - the energy levels turn out to be discrete. In order to find out what the energy levels are, you need to actually solve the eigenvalue equation.

In the hydrogen atom, the way to solve the equation is to separate the radial and angular parts, and to work out each one independently. At the end, this leads to the introduction of 3 special numbers which describe the three coordinates in your system (n, l, m). These numbers are actually discrete and characterize fully all the possible solutions of the eigenvalue equation.

Now, each energy state is then described by a different wave-function, and they look like this.

In order to describe transitions, it's a bit more complicated, but essentially you want to introduce a perturbation in your system, and given this perturbation, you want to check the transition amplitude from one orbital to the next.

AltoidNerd · 1年前

There is a lot of things in here.

The reason is that massless integer spin representations of the lorentz group SO(3,1) are actually representations of SE(2). So spin becomes helicity. For an arbitrary number of dimensions, this becomes SE(d-2). So a massless graviton in 4+1 dimensions has extra degrees of freedom, which is pretty cool, since that's the feature that allows Kaluza-Klein type theories to exist.

This is very simple to understand: if you have a beam of light, it only has two polarizations, which span the directions orthonormal to its momentum. A spin-0 state of a spin-1 particle would correspond to its longitudinal component: oscillations parallel to its momentum. So naturally you cannot construct a longitudinal polarization of a massless guy, since for that it would need a frame of reference. And naturally this holds for arbitrary spin.

This is, at the end, intimately related to gauge redundancy, since the only lorentz invariant way to write the theory of a spin-1 is a lorentz vector, but a lorentz vector has four (three independend) components, so you need a redundancy to kill one degree of freedom. So it's a pretty deep connection that theories of massless integer spin particles need gauge invariance to exist.

Adding a mass changes the story, since you are already on a different irrep, and this one already has three helicities. In the Higgs mechanism, what you would associate to the longitudinal component of your massive vector field is actually a goldstone boson of the broken fundamental symmetry, so your vector field somewhat borrows the longitudinal polarization from the Higgs.

Harachel · 1年前

It's true, this statement should be always be followed by "with respect to some reference frame." If you're carrying a cup of tea in your spaceship, the cup of tea doesn't care that some guy on the moon is measuring your speed at 0.9999c.

thats_interesting · 1年前

It's very broad. I stayed at LMU, many people went to the Max Planck institutes, and I remember at least a couple who went to Oxford, another to Cambridge and others to Imperial College, Paris and Amsterdam as well. I don't know if there's anyone who ended up in America, that's a less favorite choice probably because of the hassle of taking GRE and paying for applications, compared to the usual application process in Europe which amounts to just saying "Got money?" to a prof.

thats_interesting · 1年前

Yo, I did the TMP, finished a couple of years ago and now on a PhD program.

I don't know about now, but I think it should receive between 30 and 50 students a year. More than half from abroad, I'd say. All the lectures I took, except for one which was a mistake, were great.

The spectrum of subjects is very large, from string theory to pure maths and biophysics. Pretty much all of the professors were easily accessible, and as you can expect, all of them heavily involved in research.

I highly recommend the program, it's easily the best in Germany now, and quality-wise it's not much behind the Part 3. Up to last year according to some survey that they made, everyone who finished the program ended up doing some PhD somewhere in the world, and most leading professors in LMU are very recognized in their fields, so getting a good master thesis advisor is easy.

If your friend wants, write me a PM and I can give my e-mail address and respond to questions with a little bit more detail.

poorasian · 1年前

You might be interested in this: http://en.wikipedia.org/wiki/Quantum_tomography

luxo42 · 1年前

I should add that I work closely to theoretical cosmology, so yes, I'm allowed some self-loathing. It's not pompous to state a fact, and the fact is that this is a terrible career choice and everywhere around me there are 40yr old postdocs fighting each other to death for a position in the middle of nowhere.

Another fact is that scientifically-wise, most topics are already beaten to death and there are few datasets to compare to. There is just so much you can write about solutions to the cosmological constant problem without running in circles, and yet you are supposed to put out two papers a year regardless of how much that will contribute to the field. Same for inflationary models - thanks god for BICEP!

And it's also a fact that in condensed matter you do learn a very much broader framework than working in cosmology. The best physicists I know started in condensed matter, even some of the best cosmologists around also have CM background. CM is where the good ideas in physics come from. Open problems which can actually be tackled are mostly related to questions which are not within the expertise of most chairs. It's tough to spend 3 years learning non-equilibrium QFT when there is no guarantee that this will give you a publication when you can easily play around with some arbitrary model and get a few citations and one place in the hall of honor of the almighty PRD.

So yes, it's by no means an easy field, and it's not a very meritocratic one, and it's very very likely that you will put years of work into some model that just has nothing to do with physics.

ochanihitesh · 1年前

That's actually a good question if one thinks carefully.

Even if you create another universe with exactly the same parameters (particle masses, couplings, etc) that ours do, all the structure we see was created after the inflationary period during an epoch called reheating. The universe was very much homogeneous before, but then quantum fluctuations - which are random! - started creating areas of higher density. This eventually evolved to form what we see today as stars, galaxies, etc etc. So this new universe will not be exactly like ours, it will be slightly different, although qualitatively they will be similar.

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