# Tag Info

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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 ...

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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 = ... 7 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. ... 5 Although the question is about Minkowski space, I think it may be helpful to consider the more general case first. It's not possible to distinguish past and future simply based on the metric. In fact, there are spacetimes that are not even time-orientable. This is similar to the way in which a Mobius strip is not an orientable surface. So the same ... 5 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 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 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. 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)$$The Lorentz norm of ... 3 The time contour really has nothing to do with renormalization. Rather it is something you choose at the outset for the purpose of the calculation you want to do. With any choice of time contour the renormalization theory is pretty much the same. What renormalization does (understood in terms of Kadanoff/Wilsonian renormalization group) is generate higher ... 3 Yes it is. The volume form on any (pseudo-)Riemannian manifold (M,g) of dimension n, where g is the metric, is given in local coordinates (x^1, \dots, x^n)$$ \sqrt{|\det (g_{\mu\nu})|}dx^1\wedge \cdots \wedge dx^n $$where \det(g_{\mu\nu}) is the determinant of the metric in these coordinates. In cartesian coordinates, the determinant of the ... 3 As others mentioned, special relativity (by definition really) doesn't have anything to do with curved surfaces! Special relativity has a particular metric (minkowski metric) which has no curvature. If your interested in manifolds (particularly integration on them, since integration in minkowski space is pretty trivial) and things like that, you really ... 3 What you've missed is that the distance along the x' axis is not the same as the distance along the x axis. The locus of events that are 1 unit of proper distance from the origin is a hyperbola. This can be used to calibrate the x' axis. See calibration hyperbola. 2 Your questions 1 and 2 can be answered if you consider one more geometrical aspect of the group of the one-dimensional Lorentz bosts and that is the orbit of a single point. In other words, the set obtained by transforming one fixed point, say, (0, 1), by Lorentz boosts of all possible \beta's. If you would apply several transformations successively, this ... 2 Since a worldline along the time axis on Minkowski diagram is at rest, it is more intuitive to measure angles from that axis instead, as then 'slope' is (space)/(time), i.e., a velocity. Then we have the trigonometric relationship:$$\frac{v}{c} = \tanh\alpha$$where Minkowski spacetime follows hyperbolic trigonometry because of the sign difference in the ... 2 Why do you stop your largest angle with ten 9s after the decimal point? If you added more of them, then you'd get a smaller bound for the velocity. And you keep adding 9s ad infinitum and you'll "eventually" reach 89.\bar{9}=90. So eventually, you'll see that the velocity could be arbitrarily small. This just means that the worldline can be vertical... and ... 2 There isn't an official standard name for opposite null lines. Note that opposite null lines are not a coordinate-independent geometric (invariant) notion, and hence it is not a very useful concept. If two null lines happen to lie on opposite sides of the light-cone in one reference frame, then they may not lie on opposite sides of the light-cone wrt. a ... 2 Maybe you are getting confused with the notation. Notice that$$ \partial_\mu N^\mu \equiv \partial_0 N^0 +\nabla\cdot\vec{N},$$where \nabla\cdot\vec{N} \equiv \partial_iN^i. Now, another way o writting this expression is:$$\partial_\mu N^\mu = \eta^{\mu\nu}\partial_\mu N_\nu=\partial_0 N_0 - \partial_i N_i. Vectors are tensor with only one ...

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The geometry of special relativity is called Lorentzian geometry, or in full: the "pseudo-Riemannian geometry of Minkowsk spacetime". This is also the Cartan geometry of the Lorentz group inside the Poincaré group. See on the nLab at Lorentzian geometry for further pointers. See the References there for introductions and surveys.

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We assume that we are working in units where the speed of light $c=1$. Given a vector with time and space components $A=(A^t, A^x)$, the Minkowski square of the vector is \begin{align} A^2 = -(A^t)^2+(A^x)^2 \end{align} We can classify the vector $A$ according to the sign of its square length. In particular, in the given signature (with the minus sign ...

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If the object is standing still his world line is vertical in minkowski diagram. It doesnen't matter if he is observing a black hole or an apple... EDIT 1: After taking a closer look at your image I saw that the cone you drew is colored yellow. For the sake of simplicity lets consider a 1D example (we have spatial dimension "$x$" and time timension $ct$) ...

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I'm not sure what you mean by "as a light cone". An point-object's path through space-time is represented not by a cone, but a curve. I can probably help you on your first question. First, pick a position for your person at time zero. Plot this point on your space-time diagram. Then, choose a (short) time later. Plot the position for your person. (Did the ...

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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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