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Feb
20
comment Variational Principle to find Energy Eigenfunctions
$a=H\psi\,,b=\psi\,.$ Note that $\langle H\psi,H\psi\rangle=\|H\psi\|^2\,.$ So we have $|\langle H\psi,\psi\rangle|\le\|H\psi\|\|\psi\|=\|H\psi\|\,.$
Feb
20
comment Variational Principle to find Energy Eigenfunctions
The Cauchy-Schwarz inequality says that $|\langle a,b\rangle|\le\|a\|\|b\|$ with equality iff $a=\lambda b$ where $\lambda$ is a scalar.
Feb
2
comment Every Relativistic Field Satifies the Klein-Gordon Equation?
Ok thanks. One more question: Is there a sense in which quantum fields satisfy the Klein-Gordon equation? I mean since they are operators. I've read that particles satisfy the Klein-Gordon equation, something about their mass being on-shell, but it's not really clear to me how to go from a field to a particle, some confusion resulting by the presence of real particles and virtual particles. Though I know that light (or electromagnetic waves) satisfies the Klein-Gordon equation with $m=0\,.$
Feb
1
comment Every Relativistic Field Satifies the Klein-Gordon Equation?
Without forces? The analogous classical mechanics Lagrangian is $\frac{\dot{x}^2}{2}-kx^2\,,$ equivalently $F=kx\,,$ so doesn't the term involving $m$ result in a force of some kind? Is the definition of noninteracting "linear"?
Feb
1
comment Every Relativistic Field Satifies the Klein-Gordon Equation?
What's the definition of a free field? Is it defined to be a field with that Lagrangian density? This is how I've seen it defined. Which would make it a tautology...
Jan
3
comment Is conservation of energy a set of principles that is inevitable in any 'possible world'?
You can look up the variational complex, or look at Peter Olver's book "Applicatons of Lie Groups to Differential Equations". It's similar to how you determine if a function is the gradient of a scalar actually. I've written a bit about it, but it's not the best exposition.
Dec
29
comment Is conservation of energy a set of principles that is inevitable in any 'possible world'?
This is something that I see frequently, but I find it a bit disingenuous as presented. Conservation of energy doesn't merely follow from the fact that the laws of physics should have time symmetry, it must further be imposed that the laws of physics come from an action principle. This is highly nontrivial and one needs to know the laws in order to actually verify that this is true. However given Newton's laws it is easily checked that energy is conserved and one needs not invoke Noether's theorem.
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?