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 Apr 22 awarded Yearling Jul 9 comment Pauli Matrices proof As someone already mentioned, your identity only holds for special values of $\phi$. To find out which values of $\phi$ it holds for, note that $\sigma_z^2 = 1$. Combined with the identities you already provided, you can then show that $\exp(2i\sigma_z\phi) = \cos(2\phi) + i\sigma_z \sin(2\phi)$, and check when this equals one. May 2 revised What is the exact gravitational force between two masses including relativistic effects? Fixed latex May 2 suggested approved edit on What is the exact gravitational force between two masses including relativistic effects? Apr 22 comment What is the relation between phase space formulation with Wigner quasi-probability distributions and path integral formulation of quantum mechanics? One formal difference is that the phase space formulation is an extension of classical Hamiltonian mechanics, while the path integral formulation is an extension of classical Lagrangian mechanics. Apr 22 awarded Yearling Mar 23 comment Isn't all light polarised? I think it might be easier to think of unpolarized light as randomly polarized light. When the individual photons in a beam have random polarizations, then the net polarization of the beam is zero, and so we call the light unpolarized. Mar 21 comment How to solve the inverse square law equation of motion How would you eliminate $t$? Time only enters into your equations in the form of time derivatives... Mar 19 comment What's the difference between NMR and EPR? @Sparkler, the Wikipedia links you provided state that the frequency is similar to ... (60–1000 MHz) about NMR, and the great majority of EPR measurements are made with microwaves in the 9000–10000 MHz (9–10 GHz) region. So unless I've missed something, it seems to be one to two orders of magnitude in difference? Mar 14 comment Negative pressure, tension, and energy conditions @Steeven: I think it is easy to imagine negative relative pressure, as in less than the pressure of the atmosphere; but it is harder to think about negative absolute pressure, in the sense less pressure than vacuum, i.e. dark energy... Mar 13 comment Is it not impossible to see a single atom using visible light? @jameslarge, thanks, I've updated my post. [I already knew that we could produce "harder" X-rays than 0.1 nm; but I agree with you that my phrasing was misleading.] Mar 13 revised Is it not impossible to see a single atom using visible light? added 3 characters in body Mar 13 answered Is it not impossible to see a single atom using visible light? Mar 12 comment Production of electric field What kind of answer do you seek when you ask for a reason that this happens..? Mar 7 awarded Nice Answer Mar 4 comment Why doesn't $ds^2 = 0$ imply two distinct points $p$ and $p'$ on a manifold are the same point? As for a measure of length on the interval: you can still use the Euclidean length $L = \sqrt{x^2 + y^2 + z^2}$ to measure distances in relativity as well; but different observers will in general not agree on its value. Mar 4 comment Why doesn't $ds^2 = 0$ imply two distinct points $p$ and $p'$ on a manifold are the same point? Massive particles travel exclusively on time-like intervals, but yeah, sounds like you got it :). I think I found another way to visualize what's going on. Say that you're at the origin, and shoot a photon towards point $x$. At exactly the same time (in your frame of reference), you start running towards point $x$ as well. The spacetime distance $s = \sqrt{c^2t^2-x^2}$ will then be some kind of measure of how far ahead of you the photon is, at the time $t$ when you arrive at point $x$. Mar 4 comment Why doesn't $ds^2 = 0$ imply two distinct points $p$ and $p'$ on a manifold are the same point? @StanShunpike, what I mean is that according to special relativity, a massless particle can only move from $p$ to $p'$ if $(p'-p)^2=0$, a massive particle can only move from $p$ to $p'$ if $(p'-p)^2>0$, and no particle can move from $p$ to $p'$ if $(p'-p)^2<0$. So the spacetime distance $\text{d}s^2$ between the two points determines which kinds of particles are allowed to move from $p$ to $p'$ and not. Mar 4 comment Is it possible to make an electromagnet with two like ends? If you want, you can take two bar magnets and push their south poles very hard towards each other. Assuming that you're strong enough to overcome the magnetic repulsion between them, you then end up with an object that has one north pole at each end, but a double south pole on the middle. (It might help trying to draw the field lines of this configuration). The case with electromagnets should behave the same as two bar magnets pushed together like this; they repel each other violently, but yes, they do have one north pole sticking out at each end, and a double south pole at the interface. Mar 4 answered Why doesn't $ds^2 = 0$ imply two distinct points $p$ and $p'$ on a manifold are the same point?