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The special theory of relativity describes the motion and dynamics of objects moving at significant fractions of the speed of light.

1
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Yes, the frequency does change. This is a component of the relativistic Doppler effect. The effect comes from two sources: first, there is the classical (nonrelativistic) Doppler effect, which comes …
answered Mar 1 '13 by David Z
3
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Are you asking about the formula for the ellipse, or the formula for how angles in the sky are distorted? In the former case: it's easy to work out the shape and size of the ellipse by considering Lo …
answered Mar 13 '11 by David Z
0
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As Stephen has answered, the answer to the first question is trivially yes. Albert and Rick perceive themselves to be at rest, so it would be nonsensical if one of them couldn't accelerate further. R …
answered Jan 7 '12 by David Z
2
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he says a 4 vector is defined as the tangent vector to this curve That is not true in general. A four-vector is not always defined as the tangent vector to a curve. In the book they are computing …
answered Nov 1 '11 by David Z
1
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since the back end of the train is moving towards one light pulse while the front end of the train is moving away from the other This is not true in the reference frame of an observer on the trai …
answered Aug 15 '15 by David Z
2
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$E=\pm(p_{x}^2+m^2)^{1/2}$ That is true, though. The relativistic energy-momentum relation, $E^2 = m^2 + p^2$. So you can use it. In fact, if you didn't already know about rapidity, this would b …
answered May 2 '15 by David Z
9
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The modern answer I know a body's momentum is equal to the product of its velocity and mass. No, it's not. At least, not in general. $p=mv$ is an approximate formula that works well for massive …
answered Apr 26 '15 by David Z
13
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If the light is bouncing (off the mirrors) in the same direction as B's spaceship is moving, A would see exactly what B does: a single beam of light bouncing off each mirror alternately, retracing …
answered Sep 6 '11 by David Z
1
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Look at where you expand $c\mathrm{d}t'$ into $\gamma(v)[1 - \beta u_x]c\mathrm{d}t$. The last step of that is wrong, and you can tell because the units don't match up in the final expression: $\beta …
answered Nov 28 '12 by David Z
3
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It's true that the CM frame of a set of objects is distinguished with respect to that set of objects. But that doesn't qualify it as a preferred rest frame, as far as special (or, locally, general) re …
answered May 30 '12 by David Z
14
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The advantage of this particular definition is that differences in pseudorapidity are invariant under boosts along the $z$ axis. Specifically, consider a Lorentz transformation corresponding to a boos …
answered Jul 14 '11 by David Z
13
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My understanding of special relativity is that it is fundamentally based on the constancy of the speed of electromagnetic radiation Actually, no, that's true. Special relativity is (one could say …
answered Sep 2 '12 by David Z
0
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You're right that, if the result of a calculation depends on the units you use, there's something wrong with it. In fact, even more than that, you can't even get any result out of $p^2 = \frac{h^2c^2} …
answered May 28 '12 by David Z
1
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This statement: $x_1 = \gamma_1(x' + v_1t') \neq \gamma_2(x'' + v_2t'') = x_2$ Note that it is not allowed to assume that e.g. $x'$ and $x''$ are transformations of $x, t$. is a direct cont …
answered Sep 4 '12 by David Z
5
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It's a matter of terminology. You can define an object's mass to be proportional to its energy in the reference frame you observe it from, which includes kinetic energy in the mass, or you can define …
answered Jul 27 '14 by David Z

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