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Feb
1
comment How does Loop Quantum Gravity describe particles?
@Nick You seem to have a fundamental misunderstanding about what LQG is and what it describes. It's an attempt to model gravity quantum mechanically. It has nothing at all to say about electrons or atoms - those aren't what the theory seeks to describe.
Jan
9
comment How is complex permittivity measured?
Rolled back edit after incorrect and unnecessary addition.
Dec
24
comment What exactly does the Kretschmann scalar implies and how does it work?
When you test whether or not a singularity is removable, you're basically just testing whether or not scalars make any sense there. The easiest way to do this is probably with the volume element. Or, more simply, the metric determinant. If det(g) is zero or infinite then volumes don't make sense, so the singularity is not removable.
Dec
23
comment What exactly does the Kretschmann scalar implies and how does it work?
Wouldn't it be easier to determine if a metric has a removable singularity by whether or not the volume element $\sqrt{det(g_{\mu \nu})} d^4 x$ makes sense? If it blows up or becomes zero at the singularity in question then it is not removable.
Dec
23
comment Black hole and relativity
They would appear to asymptotically slow down as they approached the event horizon, and the light coming from them would be asymptotically red-shifted until they were no longer visible. Note that this isn't what 'actually' happens to the person who's falling, as they would experience nothing unusual when they approached and passed the event horizon (besides some potentially very strong tidal forces if the BH is small enough). This is just what would appear to happen due to the BH's effect on light.
Sep
27
comment $\hbar \rightarrow 0$ in quantum mechanics
@UnlimitedDreamer We do the same sort of thing when we take $c \to \infty$ to get from relativistic equations to Newtonian equations. $c$ has a definite value, but by pretending it is infinite our equations become Newtonian. In the same way, when we pretend Planck's constant is zero our equations go from quantum to classical. Edit: Just realized this is an old question. Not sure why it popped up in my recent questions page.
Mar
22
comment Why does the electric field escape a black hole?
A field isn't a 'thing' that can be sucked into a BH. A field is a mathematical object that has value(s) at every point in space. There is no reason to ad hoc assume that the EM field outside of a BH is zero.
Jan
22
comment Does potential energy always equal kinetic energy?
@StanShunpike $T=V$ under special circumstances. What's important is that $T+V=constant$ over time. There may be some points on a particle's path where $T=V$, but it certainly won't be true everywhere.
Jan
22
comment Does potential energy always equal kinetic energy?
But $T \neq V$ in general. However $dT/dt = -dV/dt$ in general.
Jan
22
comment Confusion about what the Euler-Lagrange equation says
I'm not sure I understand what you're asking. Given a Lagrangian, the E-L equations tell you what the equations of motion are.
Dec
30
comment Can one of Newton's Laws of motion be derived from other Newton's Laws of motion?
@Timaeus x(t)=t^3 does not have a(t)=0. It has a(t)=x''(t)=6t. The only moment where a=0 is at t=0. It is implicit in my answer (and I have now explicitly said it) that I'm setting a(t)=0 for all t, not just at a single instant. a(t)=0 does in fact imply v(t)=const.
Dec
30
comment Can one of Newton's Laws of motion be derived from other Newton's Laws of motion?
@Timaeus Why the strawman? I clearly never said "a=0 at a single instant is enough to tell you v=const." When I say "set F=0" I'm implicitly doing this for all time, and that is indeed enough to tell you that v=const. You're arguing against a strange strawman that I clearly never stated nor implied in any way.
Dec
29
comment Can one of Newton's Laws of motion be derived from other Newton's Laws of motion?
@Timaeus Here in the real world we need only consider accelerations. Realistic equations of motion are always first or second-order.
Dec
23
comment What are the relative limitations of the Schrödinger, Pauli, and Dirac Equations?
Both the Dirac equation and the Klein-Gordon equation fail to make much sense if you attempt to use them as single-particle wave equations. They both require the "correct" interpretation, which is in the context of quantum field theory. When you mix relativity and QM you're inevitably led to many-particle systems, where particles can be created and destroyed.
Dec
12
comment What's the Cause of Quantum Entanglement?
@Sofia Just so you know, you said "the-function" instead of "wave-function."
Dec
5
comment Can Schwarzschild black holes evaporate?
The Schwarzschild metric is static by definition. If the metric changes over time then it's not the Schwarzschild metric. Simple as that.
Dec
5
comment Find time-parametrization given path and speed of a particle
Hint: the speed is the magnitude of $r'(t)$.
Dec
1
comment The virtual particles are only a fictive tool in equations? DO they exist or DON'T? And if they exist, why do we call them VIRTUAL?
@PeterShor I see what you're saying, that anything we interact with must be a bit off-shell, but I think you're carrying the interpretation of Feynman graphs too far. They are just approximation tools. Particles aren't "really" in momentum eigenstates either. If I was going to describe what's "really" going on I wouldn't use perturbation theory or virtual particles to begin with.
Dec
1
comment The virtual particles are only a fictive tool in equations? DO they exist or DON'T? And if they exist, why do we call them VIRTUAL?
@PeterShor An external line doesn't necessitate observation, as you seem to imply. That's just bad logic. External lines correspond to particles which can be observed, but they don't need to be.
Dec
1
comment The virtual particles are only a fictive tool in equations? DO they exist or DON'T? And if they exist, why do we call them VIRTUAL?
@PeterShor It seems like you're going out of your way to make it sound absurd. Virtual particles are internal lines by definition. Internal lines are not observed by definition. Therefore, virtual particles are not observed by definition. It's not complicated, and it's not circular. Also notice that I never said, "particles are virtual because they're not observed." That simply does not follow from the logic.