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  CERN's Shrinkage Dance Time marches on at the usual rate in your own reference frame no matter how fast you go. It’s the other guys who have timing issues from your point of view. But it's not that their clocks run any slower or faster than yours. It's just that they have a different perception of simultaneity (because of the vx/c^2 term in the time transform) so they want to start or stop counting at the wrong time. Events in Einstein's philosophy can only be simultaneous in one reference frame so one or the other gets a head start in any other context. You're going to get a different result if you start counting before I depart for example. That's time dilation in a nutshell. There's also a scaling factor, but that's really just a calibration exercise to ensure that you're comparing apples to apples. You're also supposed to maintain your geometry in your own reference frame at cruising speed no matter what you’re made of so the other guys hav...
The Missing Ingredient Is the rest frame for a Lorentz transformation truly dealer’s choice? The question boggled my mind for the longest time because it seemed perfectly obvious to me. Why did Hafele and Keating start with a non-rotating Earth for example? Because that context has no angular momentum. The cosmic rest frame has no net linear momentum so the same logic should apply. The universe has no net momentum so the cosmic rest frame is the best context for distinguishing between hitting the gas and hitting the brakes. But momentum doesn’t factor into the equations. Time dilation is agnostic to the direction of motion and proper time intervals are invariant so who cares about the cosmic rest frame? We can transform events into the cosmic rest frame before calculating proper time intervals: Elapsed cosmic time from launch to reunion is Tb  =  t / sqrt(1- Vb ^2/c^2) Elapsed traveller time from launch to waypoint is To  = ( Tb /2 )  * sqrt(1- Vo ^2/c^2) Elapsed tr...
Derive the Lorentz Transformation The fastest way is to generalize Galilean relativity like this: X = A(x - vt) T = B(t - Cvx) where the constants A, B, and C are to be determined. The reverse transformation is: x = (X/A + vT/B) / (1 - Cv^2) t = (T/B + CvX/A) / (1 - Cv^2) These equations have to work if x is swapped with X and t is swapped with T so: A = B = 1 / sqrt(1-Cv^2) The constant C puts an upper limit on the velocity v so it should be expressed in terms of a speed limit: C = 1 / c^2 The value of ‘c’ is pretty much arbitrary because it’s really just a unit conversion factor. Any non-zero real number will do the trick. You could set it equal to one with no loss of generality. You just have to measure space and time in the same units. The last step, which was Einstein’s deep insight, is to notice that the same is true of Maxwell’s equations. i.e. The phase speed of an electromagnetic wave in the absence of any external fields is also unity if you measure space and time in the same...

The Brass Tacks of the Ehrenfest Paradox

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Coordinate Transforms for Rotating Reference Frames   Back Story Wikipedia explains the conundrum in the Ehrenfest paradox thusly. Imagine a disk of radius  R  rotating with constant angular velocity  . The reference frame is fixed to the stationary centre of the disk. Then the magnitude of the relative velocity of any point in the circumference of the disk is  . So the circumference will undergo  Lorentz contraction  by a factor of  . However, since the radius is perpendicular to the direction of motion, it will not undergo any contraction. So {\displaystyle {\frac {\mathrm {circumference} }{\mathrm {diameter} }}={\frac {2\pi R{\sqrt {1-(\omega R)^{2}/c^{2}}}}{2R}}=\pi {\sqrt {1-(\omega R)^{2}/c^{2}}}.} This is paradoxical, since in accordance with  Euclidean geometry , it should be exactly equal to  π . (Image credit: By Geek3 - Own work, CC BY 3.0,  https://commons.wikimedia.org/w/index.php?curid=24038062 ) Conventional Wisd...

The Brass Tacks of the Spaceship Paradox

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  Does the Thread Break? Back Story Bell's Spaceship Paradox is a lesson in formation flying at near light speed. Here's the challenge: Point two spaceships in the same direction and line them up, one behind the other Tether them with a delicate thread and pull it taut Try to get the whole assembly up to speed without breaking the thread On the one hand, the entire assembly is subject to Lorentz contraction so everything, including the thread, should survive. On the other hand, each component of the assembly is subject to Lorentz contraction so the thread should break as the spaceships shrink. It's a conundrum. Here are the worldlines for the ends of the thread from the two perspectives of interest: Rest Length CERN insists that you maintain your geometry in your own reference frame no matter how fast you go or what you’re made of. Otherwise it would be a dead give-away for detecting your own motion at cruising speed with the curtains closed. But the worldline x=L+vt transf...

The Brass Tacks of the Ladder Paradox

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Lorentz Contraction and Perceptions of Simultaneity Back Story In the Ladder Paradox (aka. Pole-Barn Paradox), we are given a ladder (or pole), which is too long to fit inside our barn when both are at rest. A smart aleck physicist, who knows a thing or two about Lorentz contraction, claims he can make it fit (if only for an instant) by throwing it through the barn at a sufficiently high speed. We presume the ladder is perfectly rigid so it won't be contorted by the force of acceleration. That would be cheating. In fact, let's put rockets on both ends just to be sure. Here we go: Hmmm. What went wrong? We followed the recipe, but the ladder didn't shrink. In fact, just the opposite. In its moving reference frame, the ladder actually thinks it has gotten longer (by a factor of γ .)   Our smart aleck friend is undaunted though because, he says, the rest length of the ladder must be measured from the moving frame of reference. Its prelaunch length is therefore shorter so it wi...

The Brass Tacks of Black Holes

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Twin Paradox with Spacetime Curvature None Shall Pass Here's what the Twin Paradox looks like in the vicinity of a Schwarzschild black hole if both twins go walkabout on radial trajectories with equal and opposite velocities. Ages are quoted in terms of bystander time, not observer time (aka. coordinate time.) Note that even the light ray, which emanates from the origin, diverges off towards the end of time without ever reaching the horizon. Escape Plan The takeaway is, anything can escape from the grips of a black hole if you are willing to wait long enough. It's just that the longer an object spends on an inbound trajectory, the longer it takes to emerge after it turns around. However, a black hole is bad news if your personal integrity relies on travellers (or QM wave functions) reuniting at the launch pad at the same time. Terminology As an aside, I was once chastised by a GR expert for describing a black hole as a region in space with an unusually high concentration of mas...