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4

I'll address your issues with definition (1): $E$ is a function of $\vec p$ because $\lvert \lambda_{\vec p}\rangle\sim\lvert \lambda_0\rangle$ where by $\sim$ I mean that they are related by a Lorentz boost. That is, to "construct" these states, you actually first sort out all the states $\lvert \lambda_0\rangle,\lambda_0\in \Lambda$ ($\Lambda$ now denotes ...


3

The boost generators are First of all, note that You've chosen specific representation of Lie algebra generators of Poincare group, which is vector-like matrix representation. There are many representations in general (below I'll write about them). In Your question, You've chosen the matrix representation of Poincare group algebra generators in pseudo-...


3

$\Lambda_{\mu\nu} = {\Lambda_\mu}^\sigma\eta_{\sigma\nu}$. It doesn't "do" anything. $\delta_{\mu\nu}$ and $\delta^{\mu\nu}$ are not tensors, as I explain at length in this answer of mine. The matrix elements of the identity are $\delta_\mu^\nu$, which you could have determined by thinking about the fact that the identity must send vectors $v^\mu$ to other ...


3

That light moves with a fixed speed in vacuum, in all reference systems is an experimental fact. Maxwell's equations fit so well all macroscopic electromagnetic data that the speed of light is fixed is not under question. It is inherent in the construction of the classical theory. Light is made up by a zillion of photons. Photons are elementary particles in ...


2

Since I finally got the first one right, I might as well answer my own question. \begin{align*} [\gamma^{\mu}, S^{\rho\sigma}] &= \dfrac{i}{4}[\gamma^{\mu}, [\gamma^{\rho}, \gamma^{\sigma}]] \\ &= \dfrac{i}{4}\left( \gamma^{\mu}\gamma^{\rho}\gamma^{\sigma} - \gamma^{\mu}\gamma^{\sigma}\gamma^{\rho} - \gamma^{\rho}\gamma^...


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This definition doesn't depend on the metric signature convention. Note that in definition of $\gamma$ the metric doesn't appear anywhere. It is defined purely in terms of "3-vectors" and "3-scalars" measured by particular observer. So it is impossible for metric to appear here explicitly.


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Nothing is problematic with it. As FraSchelle says above—and is also true in the development of many other physics theories, in that they are, over time, purified of the scaffolding that helped construct them*—the original motivation doesn't affect the content of the developed theory. *cf. the top of p. 90 (PDF p. 91) of Stefano Bordoni's When ...


1

The thing about the relativity principle is just that: all non-accelerating frames are equal, in the sense that no inertial frame is more "real" or "accurate" than an other inertial frame. If two frames are moving in a Minkowskian manifold with a constant velocity relative to each other, it doesn't matter which one you choose. No matter what, all the ...


1

There is no way to explain this without explaining relativity first. In Galilean universe (the "classical" physics, which is what most people intuitively assume and think about), light speed cannot be explained. Indeed, the Maxwell equations which is describe how light works, were the first clue that our understanding of space-time was flawed. So the real ...


1

In the special relativity, as explained by Einstein, there are two possibilities. Either the speed of any particle is limited at all or not. Now if the speed is not limited, the Galilean theory pops out. But as we already know, that does not properly explain the transformation of velocities from one to another reference frame accurately when dealing with ...


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There is no frame in which "space" is at rest. Using the length contraction formula $$ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} $$ for the distance to the star is slightly misleading. The distance $L$ in your frame corresponds to two points, $x_0 = 0$ for you and $x_1 = L$ for the star, that you are observing simultaneously, say at time $t_0$. In the star's frame ...


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Suppose, as you said, we have a thin solenoid pointing upward, going through the middle of a loop one light-year wide. Then the three equations $$\nabla \times E = - \frac{\partial B}{\partial t}, \quad \mathcal{E} = -\frac{d\Phi_B}{dt}, \quad B = \mu_0 n I$$ together imply violation of causality, as shown in your question. Maxwell's equations are already ...



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