timo

Didn't check the math (again ), but when plugging into equation b you probably want to plug in z for z, not for y. If you want to become a physicist (and perhaps also if you want to become an engineer) then learning tex is certainly recommended: pretty much every physicist knows and uses tex, so you can as well pick up that skill early. I'd say that solving systems of linear equations is frequently done in physics, even though I cannot pinpoint a sensible example at the moment. There's two straightforward generalizations of it that are an asset in every physicist's math toolkit (which is why I tend not to realize the special case of systems of linear equations). The first one is linear algebra, which deals with structures called "matrices" that are similar to systems of linear equations. The other one is quite generic: Often, you have a system involving several unknown variables you want to know. What you often do is that you look for a sufficient number of known relations between the variables to be able to solve for their values. That is very similar to the concept of sets of linear equations, but generally the relations do not have to be linear.

There is at least one mistake in the calculations you made (I stopped reading at some point). But the approach is correct.

A photon is an excitation of the electromagnetic field (staying in the particle physics language). Similarly, an electron is an excitation in the electron-field. The proton can, in that image, be considered as a complicated excitation in quite a lot of different fields (since it is not an elementary particle).

Interactions on the quantum field theory level are usually local. In that picture the wave functions should probably overlap for any interaction term to be non-zero.

I don't have the background to understand the scenario you describe (or even judge if you actually properly describe a defined scenario with the very few words). But my advice would be: - "Entropy must never decrease" tends to be overrated a bit (due to it being rated to increadibly important, usually). Strictly speaking that applies to systems with constant energy, and also implies some mathematical subtleties (which you should probably not bother with). - Hence: The scenario you have in mind: Are you sure that energy is conserved? Or does the new structure perhaps have a lower energy than the old one (the notion of "forming bonds" hints at that)? In that case, the energy must go somewhere, e.g. as heat into the surrounding water. That's also where the increase in entropy could happen. That exact mechanism is a random guess of mine, of course. Point is: Is energy conserved in your picture?

Had the spontaneous urge to quote that.

I fully agree that integrals along a curve are overkill. Saying so actually was the intended message of my previous post. But evidently so, that message didn't quite transfer. I blame the scary integral sign (despite the fact that I explicitly performed the integration).

Generally, for a force F along a path P the work done by said force is [math] W = \int_{a}^b \! \vec F \cdot \frac{\partial \vec P}{\partial s}\, ds[/math], where s=a...b is a parameter describing the process along the path (e.g. time). Since the force is constant in this case, this simplifies to [math]W = \vec F \cdot \int_{a}^b \! \frac{\partial \vec P}{\partial s}\, ds = \vec F \cdot ( \vec P(b) - \vec P(a) )[/math]. Reason I am sketching this rather abstract calculation? Firstly, because it is what you asked in the headline. Secondly, because it demonstrates that the force of the wind in this case does not act on the vertical part of the path, only on the horizontal one. Also note that the force of wind in this example seems to be a conservative force.

According to the Big Bang scenario it took something like 13 Billion years, if I remember correctly. Some finite time, at least. Imagine a volume of size 10 units with some constant amount of mass in it. Let the volume shrink by 1 unit per second. After 10 seconds you have a finite amount of mass in zero volume, which people love to call an "infinite density" (whereas I'd rather doubt the validity of the state). It certainly did not take an infinite amount of time to go from finite density to infinite density, but rather an infinitesimally small one. The trick is that it is not the mass that is affected, but the volume that approaches a very special mathematical value - zero. If you'd start out with a non-zero volume of infinite density, then indeed you might need some weird dynamics to ever reach a state of finite density.

A few random comments: - The exact moment of the Big Bang singularity is problematic to describe. Times arbitrarily close to it are a bit better. As Moontanaman indicates, extrapolating the Big Bang scenario back to the singularity is a bit dubious, since we already expect that unaccounted-for physics effects become relevant before that. - It may be a bit inaccurate to say that math breaks down at the singularity. Math breaks down there pretty much the same way as it breaks down when you mutliply a number with zero. This leads to weird effects like 7*0=5*0 despite [math]7*x \neq 5*x[/math] for any other x. But few people say that math breaks down when you multiply by zero. It is the physical assumptions going into the Big Bang scenario that we expect to break down close to the singularity. - The Big Bang scenario is actually not that similar to a black hole. The former assumes mass being spread out evenly everywhere, the latter assumes matter to be concentrated in a tiny sub-region of an otherwise matterless space. In the former case the geometry of space changes with time, in the latter case one usually considers the case where geometry is constant over time.

There is an arbitrarily large amount of things that could be said here. But I think it boils down to the following: If the two alternatives are indistinguishable, irrespective of the experiment performed, then time being considered an illusion or not is just a label.

What do you think are the consequences? In the sense of what do you think is the difference between time being "an illusion" and time not being "an illusion"?

... if the size of the universe is not limited . But that indeed seems to be the current mainstream trend, according to the German Wikipedia.

Or equal to the width of the universe of the universe is cyclic rather than a cavity with hard walls. Or lower if it's not like a torus (in which case plane wave momentum eigenstates may not even be the proper basis for the lowest-energy modes). Or any other value if the "edge" of the universe happens to be well described by neither a box with infinite potential outside nor by periodic boundary conditions. @Sensei: I advice having a look at the units of your presumed calculation. Keeping track of the units is not foolproof (in particular, having the units correct does not guarantee a statement being correct). But checking the units still is the simplest and most helpful tool for sanity-checking. Particularly for people not familiar with physics. In case you do not understand what I say: The Planck Constant does not have units of energy, but of energy times time (which is usually called "action").

If your intent is to provoke, then forgive me for not being keen on discussing with you. If your intent is to learn, I already gave a reply. You are correct that a naive and sloppy interpretation of pop-sci books and Wikipedia runs into problems at some point, like in case of "light has zero mass". That's sloppy because I am pretty sure that even pop-sci will say "photons have zero mass". And it is naive to assume that from understanding the words of a sentence one can directly understand the sentence. Contrary to popular misconception statements in physics are strongly context-based. And especially at higher levels of physics knowledge, understanding these contexts and their limitations, and therefore understanding the validity limits of the statements, becomes increasingly important. In your example, if you knew particle physics you'd know how fields are commonly quantized and how particle physics interactions are being calculated: in the basis of momentum eigenstates. Also, even if you knew particle physics you'd have to understand why and how a calculation in terms of momentum eigenstates (the Feynman diagrams you've undoubtedly seen) relates to the real world. The statement about photons being massless has to be understood in this context. On top of that, many terms in physics are not defined very sharply. In your example, I do not know if there is a universal rule whether a superposition of two or more different momentum eigenstates would be called a photon or if only the eigenstate to a momentum is called a photon. I'd call it a photon, maybe Swansont wouldn't. Same goes for a definition of mass, which I only need when I actually need mass (why would I care about a mass of light in your example?). This fuzzyness about terms usually does not become a problem in interaction between scientists, since they (a) are trained to properly define what they mean where necessary situationally, and (b) usually have interactions with the intent to understand each other, not to correct each other. Just because Prof. Hawking doesn't put his statements into context in his pop-sci books (which he is proud of that they contain no math at all) that does not mean that Prof. Hawking does not know about their context. Your asking about things that seem incorrect is an important step towards understanding contexts. Putting forward such questions with the intention to shed light on the rotten building of physics rather than the humble attempt to increase one's understanding is understandable (I've been a young physics student myself at some point, and in hindsight I really don't envy by tutors) but somewhat silly. In fact, picturing a "constructed house of physics" is questionable in the first place.

The invariant mass of a multi-photon system, e.g. two photons, indeed is not necessarily zero. That is well known. For a radially-symmetry elementary excitation of the electromagnetic field, which in a sense you may call a photon, you'd have to define what mass is. Usually, the dispersion relation E(p) refers to cases with a definite momentum. A radially-symmetric "photon" is not a momentum eigenstate (obviously), so you'd have to pick a suitable generalization over the common usage. If you picked m^2 = <E>^2 - <p>^2 (with <.> being the expectation values for the respective measurements) then indeed that kind of photon would have a non-zero "generalized mass". The photons having zero mass refers to single particles in momentum eigenstates, since in every other case mass wouldn't be a property of the particle type. Btw.: What's the point (in the sense of "intention") of your thread, anyways?

[math] -b/i = +bi \neq -bi [/math]. Not to mention that a simple check whether 0+1i equals 0-1i would have sufficed to sanity-check your alleged proof.

Is there any reason to believe that I don't know what I am saying? Is there any reason to assume the reply of other random strangers from the Internet is more reliable than mine?

The electron is generally assumed to be an elementary particle, which I assume is what you mean by "fundamental particle". @Sensei: Interaction with positrons is not an argument against the electron being an elementary particle. It would probably not be counted as one if that was an argument.

That depends on how many significant digits you take into account when actually calculating something. If you calculate on a computer you can probably assume something like 15 digits. In practice, that is pretty indistinguishable from arbitrary precision (your meter scale would have to have an accuracy in the order of one proton diameter for this to matter). If you approximate pi=3 then of course you have less relevant digits.

The graviton would, if it existed. The gluon would, if it existed as a free particle. The gauge bosons of SU(2) and U(1) would if they weren't affected by the Higgs Mechanism. And ironically, being unstable the W and the Z boson probably won't have much of a definable speed unless they move close to the speed of light (because otherwise they decay very quickly, which may well be before they passed a distance that significantly exceeds the size of their wave packet). To be precise, the Higgs boson does not give mass to any particle at all. It's the interaction with the part of the Higgs field that is not the Higgs boson that causes mass terms for elementary particles. It's somewhat debatable to what extend you can define a sensible "speed" for virtual particles. Or call them particles in the first place. Just taking equations from wave dynamics, plugging them on momentum eigenstates and calling that "velocity" and "particle", respectively, appears a bit too shortcut to me. Not only in the particular case of interacting states, where calling the solutions of the non-interacting case "particles" seems particularly dubious to me.

1) Depends on what you call "accelerate". Light can be bent. In a sense, you could call that acceleration. In a sense, you could say that therefore photons can be accelerated. 2) I'd be careful with the statement "like any other elementary particle". Defining an acceleration for an electron may be rather straightforward (just treat it like a classical point-sized particle with charge), whereas defining an acceleration for a gluon sounds way more tricky than for a photon (defining a gluon already sounds pretty tricky to me).

And yet apart from "full accuracy" you can achieve any degree of accuracy you want if you just bother to keep calculating long enough

How is that even possible for a list including "none of the above" ? And what's that rumbling in the Gödel family crypt?

While possibly true, howlingmadpanda and me seemed to have a different interpretation of the statement