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Mar
22
awarded  Nice Answer
Mar
1
revised Relationship between Quantum superposition and Uncertainty principle
corrected statement based on my initial misinterpretation of the wikipedia quote
Mar
1
answered Relationship between Quantum superposition and Uncertainty principle
Feb
2
revised Redshift due to a static gravitational field and the conservation of energy
added reply to comment
Jan
17
comment Redshift due to a static gravitational field and the conservation of energy
@ThomasElliot I think you're mixing me up with someone else. The total energy is conserved. The square is correct; actually, if you go through the steps to derive the massive particle case, the constant on the left is $\frac{1}{2}(\epsilon^2-1)c^2$. ... Yes, descending photons are blueshifted; I have no idea what anything 'appearing blueshifted but not really blueshifted' would even mean, and I can't clarify that because it's not my statement. Otherwise, the answers in linked question are good: there is energy in the system due to particle position, and it can be exchanged with kinetic energy.
Jan
16
awarded  Yearling
Jan
14
answered Redshift due to a static gravitational field and the conservation of energy
Jan
1
answered Can there be acceleration without velocity?
Oct
28
answered Black holes have conservative potentials. Why don't things “swing” in and out?
Oct
20
comment Can you take the integral of $ d^2x\over dt^2$?
When you pull out $F$ to the front of the integral, you're assuming that it is time-independent. Why?
Aug
25
reviewed Approve How do we see? Where do the photons disappear?
Aug
10
comment General relativity without energy?
Then what's $\nabla$? If it's the covariant derivative as usual, then $\nabla g_{\mu\nu} = 0$ in any spacetime whatsoever in GTR because the connection is Levi-Civita. Maybe it would work better in terms of vanishing Weyl curvature as the additional condition, $C_{mu\nu\sigma\rho} = 0$, because the Weyl curvature is part of the Ricci curvature not determined by the stress-energy of sources.
Aug
3
comment Does acceleration warp space?
What do you mean by "[Mikowski] spacetime no longer looks flat to accelerated observers"? It surely does! The Riemann curvature vanishes in every coordinate chart, which of course includes the Rindler chart adapted to accelerated observers.
Aug
3
awarded  Proofreader
Jul
24
comment Why general relativity over other similar theories?
@ViktorToth The equivalence principle is about gravity. The section you're quoting from is named "Principle of Relativity and Gravitation". It's completely obvious from the get-go that Einstein was seeking a relativistic treatment of gravity. The whole paper is about the implications of gravity to various phenomena, from gravitational time dilation to electromagnetism. I have no idea how one could read it and conclude that Einstein wasn't concerned about reconciling relativity and gravity, and conclude instead that it accelerated frames were the primary motivation.
Jul
24
comment Why general relativity over other similar theories?
@ViktorToth it's rather trivial to treat all frames equally in STR (in the same sense that GTR does): just express your equations of motion in tensorial form. Then no coordinate system, inertial or otherwise, is intrinsically distinguished. The difference in GTR lies in the fact that metric is dynamical and not necessarily flat. ... Saying that Einstein's primary motivation was treatment of accelerated frames is a really bizarre claim. Rather, it was treatment of gravity that obeyed the equivalence principle, a goal presented from the earliest (1907 and 1911) papers on the subject.
Jul
23
comment Why general relativity over other similar theories?
"he was seeking was a generalization of the theory of relativity (later to be known as the special theory) to accelerating frames." -- Just wrong. STR has no absolutely no problems with accelerated frames, and Einstein was actually seeking a relativistic theory of gravity. The oft-repeated claim that one needs GTR to deal with accelerated frames is a very silly myth that has no basis in reality.
Jul
22
answered How can an infinitesimally small object rotate?
Jul
22
comment How can an infinitesimally small object rotate?
@AlecBell The singularity of a spinning black hole is (probably) not pointlike, nor even analogous 'a point in space'. Specifically, the Kerr metric of an isolated rotating black hole has a ring singularity. ... I put in a qualifier of 'probably' because it is is not uncontroversial that the interior of the Kerr metric is a physically realistic description of an isolated rotating black hole. ... Actually, even the Schwarzschild singularity is not really 'a point in space', since it's not a timelike singularity, but that's a different issue.
Jul
22
comment How can an infinitesimally small object rotate?
@AlecBell A region of spacetime separated from the rest of the universe by an event horizon.