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Jul
18
revised The Hole Argument
added 76 characters in body
Jul
18
revised The Hole Argument
added 799 characters in body
Jul
18
comment The Hole Argument
@ACuriousMind I didn't just change the coordinate system, I changed the metric also, ie. if $p$ is a point on the manifold, then we have $g'(p)\ne g(p)$.
Jul
18
comment The Hole Argument
But if $p$ is a point on the manifold, then we have $g'(p)\ne g(p)$.
Jul
18
revised The Hole Argument
added 799 characters in body
Jul
18
asked The Hole Argument
Jul
2
awarded  Curious
Jun
25
comment How do we know that $F = ma$, not $F = k \cdot ma$
@JánLalinský I think the notion may be problematic if you don't define it...how do you measure it without knowing what to measure? In any case, I was in fact talking about inertial mass, and I think it does get at what the second law is about. Inertia is a bodies resistance to movement...this makes that statement precise.
Jun
24
comment How do we know that $F = ma$, not $F = k \cdot ma$
@JánLalinský Well you haven't proposed an alternate definition of mass...would you prefer it if I said inertial mass?
Mar
18
answered How do we know that $F = ma$, not $F = k \cdot ma$
Mar
7
accepted Momentum of particle in a box
Nov
10
answered Motivation for form of Lagrangian
Oct
28
comment Does GR imply a fundamental difference between gravitational and non-gravitational acceleration?
I think you have it wrong. An accelerometer in free fall would measure $0$ acceleration. The statement that you can't tell the difference between being in free fall and at rest in a vacuum is (approximately) true. You can clearly tell the difference between free fall and upward acceleration in an elevator. In one if you drop a ball it will fall beneath you, in the other it won't. And if you mean downward acceleration of the elevator, then that is exactly free fall so the statement is vacuous.
Oct
21
comment Diffeomorphism Invariance of General Relativity
@twistor59 Do you mean that if $\varphi:M\to M$ is a diffeomorphism, and $\phi$ solves $\eta^{\mu\nu}\nabla_\mu\nabla_\nu\phi=0$, then $\varphi^*\phi$ won't solve $\varphi^*\eta^{\mu\nu}\nabla_\mu\nabla_\nu\varphi^*\phi=0$ where the covariant derivative are now with respect to $\varphi^*\eta$?
Oct
21
comment Diffeomorphism Invariance of General Relativity
Thanks, though it's a bit technical. Do you have a simple example illustrating the difference between a theory that is diffeomorphism invariant and one that is not?
Oct
20
comment Diffeomorphism Invariance of General Relativity
@twistor59 Ok, but $\eta^{\mu\nu}\nabla_\mu\nabla_\nu\phi=0$ isn't a tensor equation, if you do the change of coordinates you get $g^{\mu\nu}\nabla_\mu\nabla_\nu\phi=0$. I still don't see what the problem is, ex: you can solve the wave equation in cartesian or polar coordinates, you just have to do the proper change of variables. The Euler-Lagrange equations are the same in any coordinate system, so what problems do active diffeomorphisms cause?
Oct
20
asked Diffeomorphism Invariance of General Relativity
Jul
31
awarded  Yearling
Jun
8
comment Is Biot-Savart law obtained empirically or can it be derived?
But how I've seen it done, Maxwell's equations are derived from the biot-savart law, which would make this circular.
Jun
8
comment Is length/distance a vector?
I'm not saying it's not standard terminology, I just don't think they are the same kind of vector, and it can be confusing to bag them together...It is standard to call angular momentum a vector, but it's clearly not the same kind as a velocity vector. If you change the place of the origin, the position vector changes, a velocity vector won't.