« first day (4345 days earlier)      last day (591 days later) » 

1:54 AM
0
Q: How to make my question on-topic

Kman3I posted the following question several hours ago: Electric field inside sphere - Griffiths Problem 2.14. In doing so, I showed my attempt in detail and provided my thought process behind it. Within two hours (and downvotes within minutes) -- and after only one answer was posted that I felt did n...

 
 
5 hours later…
7:21 AM
@fqq and trivial it was, thanks a lot for the help fqq! saved me some troubles
 
7:45 AM
In Special Relativity, if I have two linearly independent lightlike vectors is there a canonical way to construct a pair of orthogonal timelike and spacelike vectors?
For example, from u and v of lightcone coordinates we can make t and x easily...But that's simply because u and v are of that form...Is there a method for any arbitrary lightlike vectors?
 
7:56 AM
Two null vectors are orthogonal only iff they're proportional, so if $u$ and $v$ are independent, $$\langle u+v, u+v \rangle = 2\langle u, v \rangle \neq 0$$
Therefore it has either a timelike or spacelike norm
and $u - v$ has the norm of the opposite sign
After that, you can construct a pair of orthogonal timelike and spacelike vector by the Gram Schmidt process
 
8:27 AM
@Slereah But won't this step change the spacelikeness or timelikeness?
 
8:40 AM
There is a specific version of Gram-Schmidt to build an orthonormal basis in Minkowski space
Not really specific but since the process uses the inner product in it, if you change the inner product it adapts
it just will not work with null vectors since you have to divide by the norm
If you have a set of linearly independent non-null vectors $v$, then you can build an orthonormal basis as
\begin{eqnarray}
u_0 &=& v_0\\
u_1 &=& v_1 - \frac{\langle u_0, v_1 \rangle}{\langle u_0, u_0 \rangle} u_0\\
u_2 &=& v_2 - \frac{\langle u_0, v_2 \rangle}{\langle u_0, u_0 \rangle} u_0 - \frac{\langle u_1, v_2 \rangle}{\langle u_1, u_1 \rangle} u_1\\
u_3 &=& v_3 - \frac{\langle u_0, v_3 \rangle}{\langle u_0, u_0 \rangle} u_0 - \frac{\langle u_1, v_3 \rangle}{\langle u_1, u_1 \rangle} u_1 - \frac{\langle u_2, v_3 \rangle}{\langle u_2, u_2 \rangle} u_2\\
\end{eqnarray}
and then normalize it with \begin{equation}
e_i = \frac{u_i}{\| u_i \|}
\end{equation}
 
 
1 hour later…
9:56 AM
@JohnRennie Good afternoon Sir. Sir, may I ask a question ?
 
@An_Elephant Hi, yes you're welcome to ask :-)
 
Unfortunately my question is perhaps homework related. But thanks very much. Consider a very very long straight wire carrying a current. A circular loop of some radius and at a finite distance from the wire is kept . The wire and the loop are in the same plane.
Then, what is the magnetic force on the loop?
We need not to find the actual expression for force
But only tell the dependency of force on the radius and distance.
The answer key says that force is proportional to ( radius/distance)^2 . But isn't the force zero due to symmetry?
 
The loop is in the plane of the wire, so one side of the loop is closer to the wire than the other side. Yes?
I can draw a diagram if it will help ...
 
No sir
Wait a minute. I am sending original diagram.
 
OK ...
 
10:07 AM
 
Yes, that's the diagram I was going to draw.
 
Sorry sir.
Now sir tell.
 
The point on the loop nearest the wire is at a distance d-a and the point on the loop farthest from the loop is at a distance d+a. Yes?
 
Yes sir.
 
And the field of the wire is B = μ₀I/2𝜋R where R is the distance from the wire. Yes?
 
10:12 AM
Yes sir. And field is going into the paper
 
Yes. Now consider a small length d𝓁 of the loop at the point nearest the wire. The force on this element is F = B i d𝓁 where i is the current flowing in the loop and B = μ₀I/2𝜋(d-a). Yes?
 
Yes sir . And this force is towards the wire.
 
And now do the same calculation for the force on the opposite side of the loop. Now we get:
F = μ₀I/2𝜋(d+a) i d𝓁
and the force is in the opposite direction. Yes?
 
Yes sir
 
So if we combine these two elements there is a net force:
Fnet = μ₀I/2𝜋(d-a) i d𝓁 - μ₀I/2𝜋(d+a) i d𝓁
towards the wire.
Yes?
 
10:17 AM
Yes sir
 
And this net force is not zero. That's why there is a net force on the loop.
If the magnetic field of the wire was uniform then you would be correct that there would be no net force on the loop, but the field of the wire is not uniform.
 
But sir, consider two length elements mirror image to each other with respect to the horizontal diameter.
Field at them are same but current is opposite so their forces cancel.
Only the parts that you told which are present at extreme contribute to force
Because they don't have mirror images
 
No, the horizontal components of the current are opposite and cancel, but the vertical components are the same so they add.
 
Sorry sir
Now I understand
 
OK :-)
 
10:22 AM
In order to have opposite forces I should not take mirror images but instead diametrically opposite points
 
Yes
 
However, even though the force are opposite but magnitudes are different
 
Because those points are different distances from the wire.
 
Yes sir. Now I realise. The mistake I did is that the mirror images of length elements will cancle out came in my mind because they indeed cancelled but only for the pair situated at extreme top and bottom and this was the only pair equidistant from wire. And I forgot to consider other pairs to confirm :(
Thanks very much sir :)
 
You're welcome :-)
 
 
2 hours later…
12:06 PM
I think I have the Idea
To deal with translation DoF in GR, the trick isn't to consider more general jets for the tangent bundle, but more general jets for the frame bundle
The affine frame bundle has the same associated bundle [the frame bundle] but with a different structure group
And the affine frame bundle does indeed work as a fiber bundle $\mathrm{inv}J^k(\mathbb{R}^n, M)$
I think so anyway
People tend to be pretty vague about affine bundles in general
 
 
2 hours later…
2:28 PM
2
A: Can DART/Dimorphos impact be seen with naked eye?

blobbymcblobbyUpdate: So, impact done. Some images: Shortly before impact: Motion clips from or compiled from ATLAS (Asteroid Terrestrial-impact Last Alert System): A NASA and University of Hawaii project, and 1 meter telescope at South African Astronomical Observatory: Original answer: TL;DR With naked ey...

 
2:40 PM
hi everyone
actually I was reading about atomic clock
I encountered with few questions like why do we use caesium in the atomic clock
like we need a clock which can measure the event something around the speed of light which are so fast that accuracy of a pendulum based clock or normal big bang tower watch
or something,if I have to record electron transition as by like applying high frequency oscillation
why do I need caesium
 
did you try searching the site? :P
66
Q: Why do atomic clocks only use caesium?

PinkiModern atomic clocks only use caesium atoms as oscillators. Why don't we use other atoms for this role?

 
I can use any other metal hit it with as hard generate_ a high frequency oscillation
@ACuriousMind actually more than caesium
I feel when electron transition is what we are measuring then why a particular material or metal
 
I don't understand what you're trying to ask. Did you already read the answers to the question I linked? They both explain that it is not necessary to use caesium but also why it is the most popular choice.
 
@ACuriousMind why we use atomic clock can we come first on that?
like is my logic is correct?
6 mins ago, by Jack Rod
like we need a clock which can measure the event something around the speed of light which are so fast that accuracy of a pendulum based clock or normal big bang tower watch
 
are you asking why an atomic clock can be more precise than a normal mechanical clock?
 
2:49 PM
Yes
 
@Jack The answer linked by ACM says the atomic clock uses "the frequency of the hyperfine transition - a splitting of energy levels caused by the magnetic field of the nucleus"
> The hyperfine structure transition can be used to make a microwave notch filter with very high stability, repeatability and Q factor, which can thus be used as a basis for very precise atomic clocks.
 
@JackRod because we can control microwave oscillations much more precisely than we could ever build a mechanical clock
 
how how do we control in the sense?
@PM2Ring hi sir
 
The early atomic clocks, built in the late 1950s were quite good, but the modern caesium fountain is even better. It uses laser-cooled atoms in free-fall to minimize the error. See en.wikipedia.org/wiki/Atomic_fountain
 
I'm just saying that it's much easier to produce a microwave with a well-defined frequency than to fashion mechanical clock parts to infinite precision (and at some point you just run up against the limits of the materials themselves, the fact that they'll always degrade slightly over time, that you can't really build a mechanical clock that displays nanoseconds, etc.)
 
3:10 PM
> The passive atomic resonator serves as an atomic frequency reference by stabilizing a quartz oscillator in a feedback control circuit. A microwave signal derived from the quartz oscillator drives the atomic resonator and induces transitions between desired atomic states in the resonator.
So a typical caesium clock uses a quartz crystal. The caesium frequency is used to control the quartz frequency.
 

« first day (4345 days earlier)      last day (591 days later) »