elfmotat

Yes, [math]k \cdot x = k_\mu x^\mu[/math]. The two lines look the same, but the solutions of [math]\varphi (x)[/math] are not the same because the field equations contain additional terms in interacting theories.

If you solve for the creation operator in terms of the field in free theories, you get: [math]a^{\dagger} (\mathbf{k}) = -i \int d^3 x \, e^{-ik \cdot x} \, \overleftrightarrow{\partial_0} \, \varphi (x)[/math] where [math]f \overleftrightarrow{\partial_0} g = f \partial_0 g - g \partial_0 f[/math]. In interacting theories this becomes the definition of creation (and annihilation) operators. But with this definition, in interacting theories they obviously become time-dependent. So we write: [math]a^{\dagger} (\mathbf{k},t) = -i \int d^3 x \, e^{-ik \cdot x} \, \overleftrightarrow{\partial_0} \, \varphi (x)[/math]. When doing scattering we assume that in the limit [math]t \to \pm \infty[/math] these operators become the operators of the free theory.

Fermions cannot and do not "act like bosons." If it acted like a boson then we would call it a boson. I already provided you with a test: check and see if the Pauli exclusion principle holds. Particles and virtual particles come from the same fields, so if virtual particles were "little universes" then particles would be too. And if these little universes differed from each other in a way that could be tested by experiment, then the Pauli exclusion principle would not hold. But it does hold, obviously, so they cannot differ in any way that can be verified by experiment. Virtual particles themselves are not directly observable. If you observe a particle, then by definition that particle is not virtual.

You are proposing that virtual particles are not only composite, but also incredibly complicated. Therefore they are not identical particles. Then fermions shouldn't obey anti-commutation rules (i.e. the Pauli exclusion principle), and all of quantum field theory breaks down. So no, this makes no sense.

There appears to be nonzero cosmological constant, which implies by the Friedmann equations that it takes the form of a constant energy density (called "dark energy") that fills up space. Positive cosmological constant implies a negative pressure, which drives expansion. We have no idea what "dark energy" actually is, or why it is there. Yes, but coordinates don't have any physical meaning. The physical distances between points in space increases over time.

The creation and annihilation operators become functions of time when interactions are allowed, and we assume that in the far past and the far future there are no interactions taking place. The reason the Feynman rules work is due to the LSZ reduction formula. A lot of intro texts don't ever mention LSZ, or even wave-packets in general, which can cause confusion. See for example here: http://isites.harvard.edu/fs/docs/icb.topic474176.files/LSZ.pdf . Also, you can make nice pretty bras and kets with \langle and \rangle. For example, "\langle p | x \rangle" renders as: [math]\langle p | x \rangle[/math].

Klauber addresses this point in section 8.9 of his book (which I highly recommend by the way). Particles are indeed typically wave-packets, which are a bit harder to deal with than momentum eigenstates. The way most textbooks deal with this is to make an approximation. Interactions only occur where the interaction part of the Hamiltonian is nonzero. For example, take it to be the QED interaction: [math]\mathcal{H}_{int} = -e \bar{\psi} A^{\mu} \gamma_{\mu} \psi[/math]. In order for the interaction to be significant, the overlap of [math]\psi (x)[/math], [math]\bar{\psi} (x)[/math], and [math]A^{\mu} (x)[/math] has to be significant. Typically this overlap only occurs in a localized region of space, so an approximation can be made: we take the fields to be evenly distributed sinusoids (momentum eigenstates) inside some box of volume V, while outside of the box the fields are all zero. He also provides the following (very helpful) illustration:

The horizon is called a Rindler horizon. What it means is that if you place a charge and a radiation detector inside a uniformly accelerating box, the detector would not detect any radiation. Inertial observers will be able to detect that radiation. (Hypothetically, of course. As I've said, these effects are far too small to be measurable.)

Awesome. I've never seen any articles on this before.

We were discussing how best to respond to crackpottery, which is something I think is relevant to the thread. You implied that I believe the appropriate way to respond to legitimate questions is to "frighten" the asker, and that I'm intolerant of the mentally handicapped. Now you're admonishing me for being "defensive" after I clarified my positions. That's not exactly fair.

Which response(s) do you find defensive? The one in the thread you linked to? I'm not sure why you're interpreting it that way, because I definitely didn't feel like I was being defensive when I wrote it.

I actually just responded to that post. I told you: if someone asks a question, I'll respond (respectfully) if I know the answer. I responded both honestly and respectfully. Did I immediately outright deny energy conservation violation? Yes. Was I a jerk about it? No. I think you're mistaking my policy of being honest with intentionally being confrontational and dismissive, which I don't do.

Say you have a ball attached to a hanging spring. You pull the ball down a bit, then let go. What happens? The ball will oscillate up and down for a while, converting potential energy (which you put into the system when you pulled on the ball) into kinetic energy, then back to potential, then back to kinetic, etc. Your system doesn't violate energy conservation in the same way that the ball on the spring doesn't: you put in some initial potential energy (when you lined up north-north), and the system merely converts potential to kinetic, back to potential, back to kinetic, etc. It will continue to oscillate ("oscillate" is probably the wrong word) like that until it has lost too much energy from friction and air-resistance.

Of course not! If someone asks a question I'll respond when I know the answer, and I always do so respectfully. If the question is vague or unclear, I'll ask for clarification. If someone makes statements (i.e. not questions) that I know to be false, I'll correct them. And if an entire post is nonsense, I'll still call it nonsense. What I definitely will not do is self-censor out of concern that the person I'm responding to might happen to have some sort of disability. I think that would be incredibly condescending! Plus, I don't particularly care about the person behind the screen - just what they're saying.

I certainly don't go out of my way to ruin people's fun, but I don't like to coddle anyone either. If somebody says something that's incorrect, I'll point it out. If somebody posts what is clearly a bunch of nonsense, I'll call it nonsense. I don't think people's feelings should be the priority. Honesty is the best policy. Discussions should be about the empirical content of posts - this is a science board after all!

Well now my posts just look silly .

Probably because quarks always appear in bound states which have net integer charge.

Because they have different energy according to blue guy and red guy. (I feel like we're going in circles here.) Say Alice is moving relative to Bob. Bob tosses a ball in the direction Alice is moving. In Alice's rest frame the ball has less energy than in Bob's rest frame. It's the same situation, only with light. It doesn't matter that the light moves at the same speed relative to Bob and Alice, they still report different energies because the energy of a photon isn't related to its velocity! It's only dependent on its frequency, which can be different for two waves that move at the same speed.

The way I interpreted the question, he's wondering whether the neutron is composed of charges that sum to zero, or if it just has zero charge (i.e. if quarks weren't charged).

Frequency of a wave is not the same as speed of a wave. The frequency is how many times it oscillates per second. The speed is how fast it travels. Two waves can have the same speed and different frequency. The fact that energy is not an invariant quantity is not new. This was obvious even in Newtonian physics. For example: if you toss a ball it will have some kinetic energy. But in the ball's rest frame it has zero kinetic energy. Thus energy is not invariant.

I think an easier way would just be to check what section(s) the majority of the threads in the trash can are being moved from.

I feel like the "at least I have ideas" people fail to realize this. Coming up with new ideas is important, but it's something that should wait until someone's well-versed enough in physics to know what constitutes "new ideas." Not that people still in the process of learning can't have good ideas (indeed I'd claim that everyone is still in the process of learning), but it's precisely because they're knowledgeable enough to recognize the gaps in our knowledge that their ideas are "good ideas." In other words, they know what they don't know! Bad ideas are usually the product of people who don't know what they don't know - people blissfully unaware of just how much we do know. As a result, these "bad ideas" often criticize existing theory, or reinvent the wheel, and can usually be trivially falsified. Coloring outside the lines is great, but it's important to know where the lines are in the first place lest us accidentally draw on Mother's carpet. My OP was based on anecdotal evidence (not exactly very convincing). Perhaps my impression that crackpots like to crowd around the physics table is merely an example of selection bias. So the premise of the thread may indeed be false - I don't know how to check that.

That's cute. There's a difference between "getting angry and ready to take it personally" and calling a spade a spade. You intentionally hijacked my thread, and you did it in an obnoxious and purposefully disrespectful manner. Where I'm from they call that trolling. Kindly take your little game elsewhere - the conversation has been sufficiently derailed already.

The neutron consists of three quarks (and a bunch of gluons, but gluons have zero charge): two down quarks and one up quark. The down quark has electric charge -1/3. The up quark has charge +2/3. The net charge of the neutron is therefore: (-1/3) + (-1/3) + (2/3) = 0. The proton is similar. It consists of two up quarks and one down quark, so its net charge is: (2/3) + (2/3) + (-1/3) = 1.

Essentially it's asking for the maximum angle the projectile can be launched from without it hitting the ceiling. But instead of the angle, they want the y-component of the initial velocity. (The two are related!) The initial speed of the projectile is 20 m/s, but what is its initial y-component of velocity?