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15

I will first describe the naive correspondence that is assumed in usual literature and then I will say why it's wrong (addressing your last question about hidden assumptions) :) The postulate of relativity would be completely empty if the inertial frames weren't somehow specified. So here there is already hidden an implicit assumption that we are talking ...


9

First, set the units so that the speed of light is equal to one, so that the path of light rays in space-time are at 45 euclidean degrees. Note that a moving observer has a space-time path which is tilted relative to the t-axis, and the t-axis describes the path of a stationary observer (what I really mean is, get comfortable with space-time pictures). Then ...


8

The ordinary twistor space is parameterized by $(\lambda^\alpha,\mu_{\dot\alpha})$. Here, the $\alpha$ is a 2-valued $SL(2,C)$ spinor index of one chirality and the dotted index is its complex conjugate, the index of the opposite chirality. At the level of spinors, vectors are equivalent to "spintensors" with one undotted and one dotted index. $$ V_\mu = ...


5

Let's build up to this. Suppose you know about x and y coordinates in the Euclidean plane, but to you they're just arbitrary labels for points, like zip codes or phone numbers. Then suppose someone tells you that observers can view the plane from different directions, but the laws of geometry stay the same. You now know that x and y aren't really separate. ...


4

One picture is worth a 1000 words here... The important point is that we make a snapshot of the moving object in a time coordinate which is not its proper time. Indeed, if we drew just one image of the moving object in a cut given by $t' = \rm const.$, we would get a projection on the $x$ axis longer than $l_0$. But we must use $t = \rm const.$ instead if ...


4

I would like to add some further points to the answers above on the Twistor Space <--> Spacetime correspondence. The Twistor space T is a four complex dimensional space with elements described by $(Z^0,Z^1,Z^2,Z^3)$ or $Z^{\alpha}=(\omega^A,\pi_{A'})$ in spinor terms. The incidence relation between Minkowski points and Twistors is given (in spinor ...


4

1) As rotations in Euclidean geometry moves points along circular arcs, boosts in relativity move points along hyperbolic arcs. $\Delta s$ is only invariant in the sense that any given point lies on only one hyperbola (only one level curve of $x^2 - c^2 t^2$) which has a unique $\Delta s$ with respect to the origin. In other words, comparing between two ...


3

Start with (iii) $ T^\mu{}_\mu = g_{\mu\nu}T^{\mu\nu}$ I don't think this can be correct because both indices appear twice. What's wrong with $ g_{\mu\nu}T^{\mu\nu}$? Both indices are contracted. Explicitly it means $$ \sum_{\mu=0}^3\sum_{\nu=0}^3 g_{\mu\nu}T^{\mu\nu}$$ which is a perfectly good scalar. $g^\mu{}_\mu =2$ here I summed all the ...


3

To expand on Qmechanic's point, we can demonstrate how null directions get moved around even in flat space by Lorentz transformations: Given a Lorentz vector $X^a$ you can construct a 2x2 Hermitian complex matrix $$X^{AA'} = \frac{1}{\sqrt{2}}\left(\begin{array}{cc}X^0+X^3 & X^1+iX^2 \\ X^1-iX^2 & X^0-X^3\end{array}\right)$$ The Lorentz norm of ...


3

As others mentioned, special relativity (by definition really) doesn't have anything to do with curved surfaces! Special relativity has a particular metric (minkowski metric) which has no curvature. If your interested in manifolds (particularly integration on them, since integration in minkowski space is pretty trivial) and things like that, you really ...


3

Given Luboš answer, a good place to learn about this stuff is straight from the horse's mouth:Spinors and Space-Time: Volume 1, Two-Spinor Calculus and Relativistic Fields and Spinors and space-time: Spinor and twistor methods in space-time geometry.


3

What you've missed is that the distance along the $x'$ axis is not the same as the distance along the $x$ axis. The locus of events that are 1 unit of proper distance from the origin is a hyperbola. This can be used to calibrate the $x'$ axis. See calibration hyperbola.


2

Your questions 1 and 2 can be answered if you consider one more geometrical aspect of the group of the one-dimensional Lorentz bosts and that is the orbit of a single point. In other words, the set obtained by transforming one fixed point, say, (0, 1), by Lorentz boosts of all possible $\beta$'s. If you would apply several transformations successively, this ...


2

Since a worldline along the time axis on Minkowski diagram is at rest, it is more intuitive to measure angles from that axis instead, as then 'slope' is (space)/(time), i.e., a velocity. Then we have the trigonometric relationship: $$\frac{v}{c} = \tanh\alpha$$ where Minkowski spacetime follows hyperbolic trigonometry because of the sign difference in the ...


2

Why do you stop your largest angle with ten 9s after the decimal point? If you added more of them, then you'd get a smaller bound for the velocity. And you keep adding 9s ad infinitum and you'll "eventually" reach $89.\bar{9}=90$. So eventually, you'll see that the velocity could be arbitrarily small. This just means that the worldline can be vertical... and ...


2

There isn't an official standard name for opposite null lines. Note that opposite null lines are not a coordinate-independent geometric (invariant) notion, and hence it is not a very useful concept. If two null lines happen to lie on opposite sides of the light-cone in one reference frame, then they may not lie on opposite sides of the light-cone wrt. a ...


1

We want to find the (coordinate) elapsed time between two events. Let those two events be the origin and $ct=1, x=0$. Clearly, these two events are co-located in the unprimed coordinate system and so, the unprimed coordinate elapsed time is equal to the proper elapsed time between these two events. To find the coordinate elapsed time in the primed ...



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