1,035 reputation
310
bio website folk.ntnu.no/jabirali
location Norway
age 24
visits member for 2 years, 11 months
seen yesterday

I'm currently a student of Applied Physics and Mathematics, and enjoy tinkering with computers in my spare time.


Mar
23
comment Isn't all light polarised?
I think it might be easier to think of unpolarized light as randomly polarized light. When the individual photons in a beam have random polarizations, then the net polarization of the beam is zero, and so we call the light unpolarized.
Mar
21
comment How to solve the inverse square law equation of motion
How would you eliminate $t$? Time only enters into your equations in the form of time derivatives...
Mar
19
comment What's the difference between NMR and EPR?
@Sparkler, the Wikipedia links you provided state that the frequency is similar to ... (60–1000 MHz) about NMR, and the great majority of EPR measurements are made with microwaves in the 9000–10000 MHz (9–10 GHz) region. So unless I've missed something, it seems to be one to two orders of magnitude in difference?
Mar
14
comment Negative pressure, tension, and energy conditions
@Steeven: I think it is easy to imagine negative relative pressure, as in less than the pressure of the atmosphere; but it is harder to think about negative absolute pressure, in the sense less pressure than vacuum, i.e. dark energy...
Mar
13
comment Is it not impossible to see a single atom using visible light?
@jameslarge, thanks, I've updated my post. [I already knew that we could produce "harder" X-rays than 0.1 nm; but I agree with you that my phrasing was misleading.]
Mar
13
revised Is it not impossible to see a single atom using visible light?
added 3 characters in body
Mar
13
answered Is it not impossible to see a single atom using visible light?
Mar
12
comment Production of electric field
What kind of answer do you seek when you ask for a reason that this happens..?
Mar
7
awarded  Nice Answer
Mar
4
comment Why doesn't $ds^2 = 0$ imply two distinct points $p$ and $p'$ on a manifold are the same point?
As for a measure of length on the interval: you can still use the Euclidean length $L = \sqrt{x^2 + y^2 + z^2}$ to measure distances in relativity as well; but different observers will in general not agree on its value.
Mar
4
comment Why doesn't $ds^2 = 0$ imply two distinct points $p$ and $p'$ on a manifold are the same point?
Massive particles travel exclusively on time-like intervals, but yeah, sounds like you got it :). I think I found another way to visualize what's going on. Say that you're at the origin, and shoot a photon towards point $x$. At exactly the same time (in your frame of reference), you start running towards point $x$ as well. The spacetime distance $s = \sqrt{c^2t^2-x^2}$ will then be some kind of measure of how far ahead of you the photon is, at the time $t$ when you arrive at point $x$.
Mar
4
comment Why doesn't $ds^2 = 0$ imply two distinct points $p$ and $p'$ on a manifold are the same point?
@StanShunpike, what I mean is that according to special relativity, a massless particle can only move from $p$ to $p'$ if $(p'-p)^2=0$, a massive particle can only move from $p$ to $p'$ if $(p'-p)^2>0$, and no particle can move from $p$ to $p'$ if $(p'-p)^2<0$. So the spacetime distance $\text{d}s^2$ between the two points determines which kinds of particles are allowed to move from $p$ to $p'$ and not.
Mar
4
comment Is it possible to make an electromagnet w 2 like ends?
If you want, you can take two bar magnets and push their south poles very hard towards each other. Assuming that you're strong enough to overcome the magnetic repulsion between them, you then end up with an object that has one north pole at each end, but a double south pole on the middle. (It might help trying to draw the field lines of this configuration). The case with electromagnets should behave the same as two bar magnets pushed together like this; they repel each other violently, but yes, they do have one north pole sticking out at each end, and a double south pole at the interface.
Mar
4
answered Why doesn't $ds^2 = 0$ imply two distinct points $p$ and $p'$ on a manifold are the same point?
Feb
28
comment Why is the relativistic Lorentz factor defined this way?
You can derive it from the assumption that the speed of light is the same for all observers; wikipedia has an accessible derivation.
Feb
27
comment Superposition in classical Mechanics
In Newtonian mechanics we have $F=m\ddot{x}$, so you tend to get linear differential equations if the forces are linear functions of $x$ and its time derivatives. Typical examples of this in mechanics, are systems that include ideal ropes (which mediate a constant force), linearized gravity (vertical force $F=-mg$), ideal springs (linear force $F=-kx$), and linearized kinetic friction (as in $F = -c\dot{x}$). The example in your figure, seems to include exactly ideal ropes and linearized gravity.
Feb
26
answered Superposition in classical Mechanics
Feb
26
answered Is my calcualtion from the mass of a photon right?
Feb
25
comment Can we derive most fundamental laws from the Action Principle?
However, in quantum mechanics, a particle doesn't take the path of least action, but it takes all paths between points in spacetime, and each path is weighted by a phase factor $\exp(iS/\hbar)$, where $S$ is the action. This is also known as the path-integral formulation of quantum mechanics and quantum field theory, if you wish to read more about it. The path of stationary action in classical mechanics, is then simply the path with the maximal constructive interference in quantum mechanics.
Feb
25
comment Can we derive most fundamental laws from the Action Principle?
Actually, the principle of stationary action only holds exactly in classical mechanics; so you can use it to derive e.g. Einstein's field equations in general relativity, or Maxwell's equations in classical electromagnetism. Mathematically, the most convenient way to do this, is to take whatever is to the right of the integral sign in the action, and substituting it into the Euler-Lagrange field equations.