116,084 reputation
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bio website motls.blogspot.com
location Czech Republic
age 41
visits member for 4 years, 7 months
seen 3 hours ago

Hi, I am a string theorist and a publicist.


Apr
30
answered Effect of the tail of the cat in the falling cat problem
Apr
30
comment How to interpret this vertical circular motion problem?
Let me just say a different example with exactly the same interpretation you claim to have used. "Calculate the velocity needed by a rocket to escape from Earth's gravitational pull." Solution (yours): "One needs at least 10 kilometers per hour for the rocket to get to Cape Canaveral in time, otherwise the U.S. Congress would cancel the funding." Would you give some credits for that?
Apr
30
comment How to interpret this vertical circular motion problem?
Even if one agreed that the problem "find the angular velocity that achieves something" could possibly mean that one should calculate the angular velocity at a completely different moment than the moment when "something" is achieved, which is ludicrous, your solution would still be wrong. If one asked about the minimum angular velocity, the minimum angular velocity is at the top because $\sqrt{g/l} < \sqrt{5g/l}$. Your attitude to this disagreement is truly unscientific. You're only looking for evidence of one type, the favorable to you, which is scientifically dishonest. Abusing this website.
Apr
30
comment How to interpret this vertical circular motion problem?
I don't claim that you are confused about the formulation. My claim is that you are confused about the right solution i.e. about elementary physics itself. The question is totally unambiguous. It asks you about the angular velocity that prevents the water from falling out. The first step in the solution is to realize that the water may only fall out at the top of the circle. I am confident that most kids in the kindergarten would solve this first step correctly but you have not so you deserve 0 points. The problem doesn't have to give you a hint that the top is the issue; it can be derived.
Apr
30
revised How to interpret this vertical circular motion problem?
deleted 7 characters in body
Apr
30
revised How to interpret this vertical circular motion problem?
added 263 characters in body
Apr
30
answered How to interpret this vertical circular motion problem?
Apr
30
comment Looking for a way to simplify a physics formula
Dear Samantha, the answer may be obtained by generalizing the recipe of Prof Chelsea that she applied to the velocity: youtube.com/watch?v=Qhm7-LEBznk
Apr
30
comment Concept of a point particle in quantum mechanics
The term "pointlike" means pretty much exactly the opposite, doesn't it? It means that the internal scale of the particle is zero. In practice in such cases, we only know that the length scale of such particle's inner guts is shorter than a certain threshold we may access, for example $10^{-19}$ meters we may access by the LHC now. But in the language of QFT, these particles are exactly point-like. What it exactly means is described by the maths of QFT.
Apr
29
comment Kerr-Newman black holes and infinite charge
Sorry, if you want to bring some charge, you're talking about Kerr-Nerman solutions, not just Kerr solutions, right?
Apr
28
answered Concept of a point particle in quantum mechanics
Apr
28
comment Fractional statistics
The configurations in which the particles overlap is singular (e.g. because the interaction energies etc. blow up) so a priori, one isn't allowed to assume that things are smooth around this point. Indeed, for the existence of nontrivial anyons, the point is singular. The burden of proof is on the opposite side than you suggest. If you wanted to prove that there are no anyons because the paths may be deformed - through the coincident positions - you would have to show that it's OK to get through this singular point. In trying to prove so, you would fail because it's not legitimate.
Apr
27
comment Classical limit of a quantum system
Dear @Sys, you may also look at the classical trajectories. The force or acceleration is equal to $a \delta'(x)$ here. So the velocity will get a bump proportional to $a\delta$ and the position will get a step function. So I guess that classically, such a potential will make the particle jump by a fixed distance once it reaches the point. Good that such a thing doesn't occur in classical physics. I still think that for a fixed $a$ and a classical limit of a high kinetic energy, the jump will be very short again. It's plausible that for a finite $a$, the jump in $x$ will be just $\sim\lambda$.
Apr
27
comment Classical limit of a quantum system
Dear @Sys, it's a virtue and necessity, not a bug, that the delta-function is infinite at $x=0$. If it were finite at a single point (i.e. interval of length zero), like in your example, it would have no impact on the particle because zero times finite is zero. So your potential as you wrote it is physically identical to $V=\infty$ for $|x|<d$ and $0$ otherwise which is just a well with the standing wave energy eigenstates. The finite modification of $V$ at one point, by $a$, plays no role at all. A potential with $a\delta(x)$ in it would be another problem.
Apr
26
comment Hydrogen radial wave function infinity at $r=0$
Good way to phrase the priorities. ;-)
Apr
26
comment How does $F = \frac{ \Delta (mv)}{ \Delta t}$ equal $( m \frac { \Delta v}{ \Delta t} ) + ( v \frac { \Delta m}{ \Delta t} )$?
Yup, your comment sounds totally fine and is a way to learn Calculus in Newton's way. But the $\Delta$ symbols are meant for people who don't have to know Calculus and derivatives. You may still want to imagine that all these things are finite, just small, and the very small terms are neglected. But what they mean is really $dt$, $dm$, $dv$ etc. in the calculus-infinitesimal sense while these people also implicitly say "don't ask me about calculus, it is not my goal to explain it now, instead, I want to explain some physics you would quite properly formulate only if you knew calculus".
Apr
26
comment Which new insight did $E=mc^2$ give us?
Physics is about learning right facts about the physical world and about correct equations relating various observable quantities. Physics is about $E=mc^2$. Choosing one particular insight out of many related facts and calling it "essence" isn't physics, it's a rhetorical exercise and emotional self-brainwashing. Are you looking for some cheap slogans, or are you trying to understand physics? The physical importance of $E=mc^2$ is $E=mc^2$, a relation between two quantities (namely energy and mass) that were previously thought to be unrelated. 1 may and should try to understand what it means
Apr
26
comment Which new insight did $E=mc^2$ give us?
Dear @Gerenuk, yes, the total mass $m$ in $E=mc^2$ is also the mass that appears in Newton's formula for the gravitational force. It's one of the equivalent ways to look at the mass. Another equivalent one is the inertial mass determining the resistance towards acceleration. The inertial and gravitational masses are equal (in sensible units) due to the equivalence principle. It's also true that kinetic energy may be used to produce new particles. All these things are aspects of $E=mc^2$. I don't understand what you mean by the question "is that the essence".
Apr
26
comment Are there any non magnetic materials that attract to each other as if they were magnetic?
All materials attract gravitationally but the attraction is very weak. Charged materials may attract electrically. However, the magnetic force is the only other long-range force and if two materials can attract each other magnetically, then I would call them "magnetic materials". So the answer is probably No. ;-) The term "magnetic materials" or "nonmagnetic materials" isn't really good physics. We distinguish diamagnetic, paramagnetic, ferromagnetic, antiferromagnetic, and other materials...
Apr
26
comment Hydrogen radial wave function infinity at $r=0$
Dear @Jerry, you were a minute faster but shorter ;-). I think that $\sin^2\theta$ should be just $\sin\theta$.