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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.

0
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We assume that the two sets of coordinates are related by transformations that preserve the metric, namely all the matrices $\Lambda$ such that $g' = \Lambda^{T}g\Lambda = g$. This follows in turn fro …
answered Jul 7 '16 by gented
1
vote
...the Relativistic Mass is defined as... is wrong. There is no relativistic mass. The mass is $m_0$ according to what its meaning is (namely, as coefficient appearing in the Lagrangian in corr …
answered Sep 25 '15 by gented
2
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2answers
An observation more than a question. Take any function $f$ (additional hypotheses may follow) and evaluate it on any two points $x_1, x_2\in\mathcal{D}_f$. Define then the sum of these two points as …
asked Jul 7 '15 by gented
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Mass is not entrapped energy. In special relativity one sees that the energy of a particle can be written as sum of two contributions: one corresponding to the "true" kinetic term (i.e. the velocity …
answered Jun 1 by gented
-1
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> Why is it the velocity that must be the same in both frames and not the distance between two objects? It is an experimental fact that the speed of light is the same in any reference …
answered May 27 by gented
1
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I am finding a difficulty understanding how a tensorial tool (mathematical tool) is the one that describes an idea so physical like the continuous matter. How can those two ideas relate? That see …
answered Oct 10 '15 by gented
2
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Let $\square = \eta^{\mu\nu}\partial_{\mu}\partial_{\nu}$ and let $\Lambda$ be any operator leaving the metric invariant, i. e. $\Lambda\eta\Lambda^{-1}=\eta$. From the above it follows that the comp …
answered Dec 11 '15 by gented
3
votes
I wonder why light does not have infinite speed. It does not have infinite speed because the experiment of Michelson and Morley has proven that it has the same constant speed in any reference fra …
answered Oct 1 '15 by gented
1
vote
The expression that you want to get automatically follows from the definition of the electromagnetic tensor as $F_{\mu\nu} = \partial_{\mu}A_{\nu}- \partial_{\nu}A_{\mu}$. In particular one has $$ \pa …
answered Nov 7 '16 by gented
9
votes
It is an experimental fact that light moves at the same speed in every reference frame, no matter the underlying theory: see the experiment of Michelson and Morley. Every kinematic and dynamical quant …
answered Jul 7 '15 by gented
2
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Comment on the answer of @Michael: The short answer is that you need antiparticles is false. In Quantum Field Theory you have perfectly working solutions also without antiparticles, i. e. for r …
answered Jun 26 '15 by gented
1
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The classical limit in special relativity is to be intended as expansions around $\beta=0$ of any function thereof, stopping at the lowest powers in $\beta$. For example one has: $$ \gamma := \frac{1} …
answered Mar 19 by gented
0
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In general one has $$ K_{\alpha\beta} = \frac{1}{2}K_{\alpha\beta} + \frac{1}{2}K_{\alpha\beta} + \frac{1}{2}K_{\beta\alpha} - \frac{1}{2}K_{\beta\alpha} = \frac{1}{2}K_{\left\{\alpha, \beta\right\ …
answered Aug 1 '16 by gented
1
vote
Start from $d^4p = dp_{\mu}\,dp^{\mu}$ which is manifestly Lorentz invariant. In order to obtain the actual measure to integrate against you have to pair this up with the mass-shell condition for the …
answered Jun 27 '15 by gented
1
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There are two different questions that are unrelated to each other. The first one is how the 4-velocity is defined. By definition, velocities are tangent flows on a differential manifold, therefore d …
answered Sep 20 '15 by gented

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