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seen Apr 15 at 19:18

Feb
4
comment When can one write $a=v \cdot dv/dx$?
This implies that a function must be invertible in order to have a well defined spatial derivative, but that isn't true. Many functions must restrict their inverses to a particular domain, while still having well defined derivatives outside that domain (e.g., sine and cosine). Even if the velocity cannot be expressed analytically as a function of position, the spatial derivative may still be well defined, and thus the OPs original integral may still be valid, even if no clean expression for v(x) can be found.
Feb
3
comment Do wide-angle videos make the first-person view seem slower than perceived in real life?
@AndrewCheong The transform that attempts to achieve what you are looking for is the stereographic transform. However, since few fisheye lenses use this mapping, you may find that the transform from equisolid-angle to stereographic projections remove too much scene or diminish image quality to an unacceptable level near the edges.
Feb
3
comment Do wide-angle videos make the first-person view seem slower than perceived in real life?
@AlistairAdams I've amended the answer to address the difference in edge effects between normal and fisheye lenses.
Feb
3
revised Do wide-angle videos make the first-person view seem slower than perceived in real life?
Typo fix.
Sep
30
awarded  Explainer
Sep
7
awarded  Yearling
Sep
7
awarded  Yearling
Apr
22
comment A man running on the treadmill
@Ben Crowell: The trouble is that, in the presence of certain fields, there are multiple ways to define frames in which Newton's first and second laws hold. It is impossible to distinguish a field-accelerated frame from an inertial frame without making reference to some notion of a non-accelerated reference frame, and this usually comes down to pointing to the rest of the system and assuming that 1) that part is holding still, while 2) this part is accelerating due to some field. It is possible, although non-trivial, to construct a field which reverses the roles. Hence the mystery.
Mar
26
comment Why is $\vec j\cdot \vec e$ the joule dissipation?
There is no microscopic interpretation. Even the simple $\vec{j}=\sigma \vec{e}$ doesn't have a microscopic interpretation. Quantities such as resistivity and conductivity are inherently statistical. They depend on collisions to create entropic increases, i.e. thermalizing the kinetic energy that charged particles gain from being accelerated by electromagnetic fields. Joule's expression is an improvement over Ohm's law in that it describes dissipation at a point in space, but when the volume being described has too few particles in it, the description is no longer valid.
Mar
25
answered Why is $\vec j\cdot \vec e$ the joule dissipation?
Mar
6
comment A man running on the treadmill
I wasn't aware that it had this name (I think I read it in one of Feynman's texts), but that sounds like exactly what I'm talking about.
Mar
5
answered A man running on the treadmill
Mar
3
answered How can the nucleus of an atom be in an excited state?
Mar
3
comment What does it mean for something to be a ket?
Ask your bra if he knows anything about quantum mechanics. If he does, buy him a beer, and ask him what the wavefunction is.
Mar
3
comment Electric potential at a point outside a charged sphere
@Rafael: nicely edited. I wish there were a way to upvote an edit.
Mar
2
revised Ionized Depletion Region, Why aren't those charged being excited?
added 4 characters in body
Mar
2
comment Electric potential at a point outside a charged sphere
When asking questions on physics.stackexchange, you can use symbolic math. For example, your last equation can be written \$V=Q/(4 \pi \epsilon_0 r)\$, which will produce the output $V=Q/(4 \pi \epsilon_0 r)$.
Mar
2
answered Electric potential at a point outside a charged sphere
Mar
2
answered Condition for closed orbit
Mar
1
comment The field of uniformly charged ball (without Gauss theorem)
What is your justification for rewriting $|\mathbf{r}_0-\mathbf{r}|^3 as (r^2+r_0^2)^{\frac{3}{2}}$?