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Grad student: jlackm.wordpress.com


Aug
30
comment How was the formula for kinetic energy found, and who found it?
@user17574 There are much much simpler proofs, all of these conserved quantities, momentum, energy, angular momentum could be proved rigorously just from Newton's laws. Also considering the Lagrangian involves the kinetic energy, I don't think this helps.
Jul
18
comment The Hole Argument
Well I'm glad you see what I mean. So then are the two spacetimes distinguishable? I mean, can we tell which one we "live" in? I think part of my confusion arises from the fact that the coordinates don't seem to be defined before finding for the metric.
Jul
18
comment The Hole Argument
But they only mention the coordinate transformation thing later, on page two they seem to be literally considering different metrics on the same manifold. They even write $G(r)$ and $G'(r)$, and since they are using the same letter I can only assume that $r$ represents the same point on the manifold, resulting in two different metrics. Which part of this isn't right?
Jul
18
comment The Hole Argument
I don't understand that, why do you call them the same thing if they are two different things? You seem to have two different metrics on the same manifold, hence should be able to use the same points.
Jul
18
comment The Hole Argument
So take two points $p,p'$ on $M$. Using $g, g'$ you measure two different distances between them, but since they are isometric there is no observable difference in the physics?
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 $g'(p)\ne g(p)$.
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?
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?
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.
Jun
8
comment Is length/distance a vector?
I didn't interpret the OP's question in the formal sense of a vector space, that isn't usually what physics people mean when they use the term vector, otherwise anything could be a vector. They usually mean vectors as in velocity vectors, formally tangents to curves or whatever. It makes sense to write the position of a particle at $(0,1)$ as $\hat{y}$ in cartesian coordinates, but how would you do so in polar coordinates? In polar coordinates the same position is written $(1,\frac{\pi}{2})$, but I don't think identifying that with $\hat{r}+\frac{\pi}{2}\hat{\theta}$ works.
Jun
8
comment Is length/distance a vector?
I disagree that position is a vector, position is specified by coordinates, but it's not a vector; it's not a quantity with a magnitude or direction...the displacement vector is a different thing and is a vector. Just thinking in terms of differential geometry, coordinates and vectors are not the same thing. @Oaoa What you are saying would be right in a different context, but not in this one.
Jun
3
comment What exactly is implied by Einstein's insight in this scene from the NOVA series “$E=mc^2$ Einstein's Big Idea?”
I've always wondered about the clock tower thing...was Einstein just really lucky with his reasoning that if he were traveling near the speed of light, the clock would slow down? because I don't see how this implies that light is the speed limit...I mean as long as light has a finite speed, this effect would be observed. It seems like faulty reasoning that luckily led him to the right conclusion, unless I am misunderstanding.
May
30
comment Momentum of particle in a box
One thing I've noticed, is that the "de broglie wavelength" isn't even a possible wavelength, interestingly enough (at least for n=1).
May
30
comment Momentum of particle in a box
Ok, only thing is it seems like after you make a measurement of momentum, and the wavefunction collapses to $\int_{-\epsilon}^{\epsilon}dp\,\psi(p)e^{-ipx}$ or whatever, it won't solve the boundary conditions, which seems problematic? One of the justifications I saw online for the statement that there are only discrete momentum values is the de broglie wavelength, with the wavelength being the one corresponding to the argument of $\sin$. This make me wonder is the de broglie wavelength is really good for anything, other than if you know the particle is in approximately a momentum eigenstate.