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1:30 AM
Hello everyone
I am an undergraduate physics student looking to become an experimental physicist, and I am looking for some advice as to how to do well in the laboratory.
In a week I will begin a course called "Advanced experimental physics", where my group and I will have to do complex experimental physics (at least compared to the usual basic things done in the undergraduate levels). We will be monitored by my university's top researchers in the field, so we would like to make a good impression.
 
 
3 hours later…
4:32 AM
@NickHeumann Without knowing the details of your area or experiment... and I am a computational person who has had to do a lot of head scratching and questioning of experimental people... I would want to see that you A) understand the importance of boundary/initial conditions in order to generate repeatable and reliable measurements, and B) understand and want to quantify the assumptions and uncertainties that go into every measurement made
If A and B are done, and you can clearly communicate your setup and results, then anybody in the world should be able to replicate your results -- either experimentally or computationally (ideally -- simulations probably have a long way to go to get things right, but the better you can answer A and B, the fewer excuses we have for being wrong)
Since this is a lab course, I would want to see some thought into the design of experiment also. How much depends on how the course is set up. But, if you want to measure X, you should have put some thought into all the things that influence X and how you can control and measure each of those things
 
 
1 hour later…
5:39 AM
8 hours ago, by ACuriousMind
How could you ever measure "infinity"?
The problem of physical infinity is that given finite resources, there is really no way to distinguish it from some humongous quantities
You have to somehow stumble upon the axioms that justify it in the first place, of which give contradictions if the object is finite to prove it is a candidate of physical infinity
Infinity thus behave like a counterfactual concept. Lacking a framework to model them, they are literally only describable as something that does not satisfy the behaviour of finite objects
 
5:55 AM
@ACuriousMind Very long ruler
 
On physical infinity
Suppose I have a box which has volume 3 cm3 and inside, there is another box which measured to have volume 3 cm3
This will be an example of something that has a self injection that is not a surjection
We can go further and test whether the space is eulidean by checking how parallel lines behave near the boxes. If the space is euclidean, then it is more plausible that the box behave like infinity at least for objects of size less than 3cm3
 
 
2 hours later…
8:15 AM
I am trying to find the expression which is similar to the hamiltonian but it seems that I am missing something here,

$\hat{\phi}(x) = \int \frac{d^3k}{(2 \pi)^3} \, \frac{1}{ \sqrt{2 E_{k}}} \left( \hat{a}_{k} + \hat{a}^{\dagger}_{-k} \right) e^{i k \cdot x}$
$\hat{\pi}(x) = \int \frac{d^3k}{(2 \pi)^3} \, \sqrt{\frac{E_{k}}{2}} \left( \hat{a}_{k} - \hat{a}^{\dagger}_{-k} \right) e^{i k \cdot x}$

$\hat{P}^{i} = -\int d^3x \, \hat{\pi}(x) \, \partial_{i} \hat{\phi}(x)$

what I got is
$P = \frac{1}{2} \int \frac{d^3p}{(2\pi)^3} \, p^i \left( \hat{a}_{p}^{\dagger} \hat{a}_{p} + \hat{a}_{p}
How to get rid of the second, third, and the forth terms?
 
Hm
I think I solved the first 3 chapters of Peskin in some notebook
Lemme see if I still have it
 
I appreciate that!
 
does this help
It's something about how the integral behaves under change of sign of $p$
Actually
I think I might have asked on PSE about that?
 
In the Hamiltonian case, I found that

$\hat{a}_{p}^{\dagger} \hat{a}_{p} \text{must equal} \hat{a}_{-p}^{\dagger} \hat{a}_{-p}$
but I didn't understand how it works.
Now, I check the notes
 
8:32 AM
My notes tend to be a bit shorthand :p
 
It's fine.
just here ...
 
Apparently it integrates to zero because it's an even function
According to this gentleman
 
delta func. is even but P ...
it is an infinite number
 
Well you know
It's one of those QFT thing
Sweeping infinities under the rug
 
the same case as in the H
:D
 
8:40 AM
Because you're being very naughty and multiplying distributions together
Even though that is verbotten
3
Q: Schwartz impossibility result

MickGI was wondering what made it impossible to define a product of distributions. Googling, I found two questions, one of which stated the following impossibility result: There is no associative algebra over $\mathbb{R}$ containing $\mathcal{D}'$ as a vector subspace and the constant function 1 a...

v. naughty
 
So there is a contradiction between this $\delta_0 = 0$ and $\delta_0 $ = inf. number
 
I think this is correct here
Like the energy is infinite due to silliness, but I think the momentum is properly zero, because otherwise it wouldn't be rotationally invariant
Lemme try to think of a proper argument for it
Man those notes were rushed
It probably makes sense by considering it before integration
ie it's something like $$\int dx dy \delta(x - y) \delta(y)$$
I think
 
8:56 AM
@Slereah $\int dx dy \delta(x - y) \delta(y) = \int dx \delta(0) = \infty$ ?
 
Something like that
I forget exactly where the $\delta(0)$ thing comes from
I think it makes sense that it's zero if you keep the integral as $$\int \frac{d^3x}{2}\frac{d^3p}{(2\pi)^3}\frac{d^3k}{(2\pi)^3}$$
Well
It makes sense for a physicist anyway
A mathematician would surely frown
 
it comes from the commutator a_p a_p+
and the integral over k
sorry without mentioning "and the integral over k"
 
Well it's $$\int d^3x d^3 p d^3 k p^i e^{i(p-k)x} \delta(p-k)$$
Or something
 
@PM2Ring or @JohnRennie - Does it matter how much surfactant (soap) I put in the solution of charged TiO2 and water? I'm planning to cut a piece of a solid soap that's used in the shower and bath.
 
Hm
is there a decent handwave to say that this is zero
 
9:09 AM
@NovaliumCompany if you're using water you shouldn't need any surfactant.
 
@Slereah So, I want to clarify what I have understood. to get rig of delta, either we interpret it ( $\delta(0)$) as inf. number ignore it as usual, or, we perform the integral over $p$ (product of distributions with $\delta(0)$ to give zero. Am I right?
 
I'm not sure yet
Let me think it over
see the thing is
That the energy is infinite doesn't matter too much
Since, unless considering GR, we can only measure differences in energy
and you can renormalize the Lagrangian anyway
Without changing the dynamics
But having the momentum not be exactly zero is weird, because the vacuum is supposed to be invariant under rotation
So I think this should be zero in a stronger sense
this isn't a super rigorous argument because QFT nonsense but it's just a gut feeling
Wait
\begin{eqnarray}
\int d^3x d^3 p d^3 k p^i e^{i(p-k)x} \delta(p-k) &=& \int d^3x d^3 p p^i e^{i(p-p)x} \\
&=& \int d^3x d^3 p p^i
\end{eqnarray}
Hm
Still not great
But if you're integrating $p^i$ it's just a linear integral
$$\int dp\ p = \lim_{x \to \infty}[p^2/2]^x_{-x}$$
Which is indeed zero
And the other integrals are just a constant which you can set to zero
I think that's a decent argument
 
9:30 AM
I couldn't understand where this $x\delta (x)=0.$ come from?
 
Well a dumb argument for it is that $\delta(x) = 0$ outside of its support $\text{supp}(\delta) = \{ 0 \}$, and $x = 0$ on that support
So it is zero everywhere
 
ok
So this is the argument of what we discussed earlier?!
$p\delta (p=0)=0$
without performing the integral
@Slereah once again, the claimed result $\delta(0) = 0$ is wrong from math.stackexchange.com/questions/1357877/…, right?
 
Oh $\delta(0)$ isn't a well-defined quantity
but yes it is very much not $0$
 
But I couldn't recognize what this post is talking about?
 
Well the thing is
In QFT, operators are actually distributions
 
9:43 AM
Could you please in few words give a summary
 
But to obtain the stress energy tensor in QFT, you need to multiply operators
ie $H \approx \pi^2$ and such
But as this post says, it's not mathematically well-defined to multiply two distributions together
This is why we get really weird results
 
Ah hah!
I see
So, mathematically QFT is not acceptable since the indefiniteness of product distributions?
 
@JohnRennie I have this mixture of charged TiO2 particles and water. I'm planning to remove as much water as possible (by increasing the surface area by spreading the solution on a plate and letting it on the sun for a few hours) and I also thought I should put some surfactant so it can form miscelle around the TiO2 particles and they don't evaporate with the water and also later when I put them in baby oil and add black paint, the miscelles should stop the black dye to cover up the TiO2 particles
 
Well there are ways to get around that
But my point is, the way QFT is done, it's a bit non-rigorous
so you have to watch out for some issues
 
@Slereah If we think is this way, even $\delta_0 x = 0$ is wrong?
 
9:51 AM
Well even $\delta(0)$ doesn't make sense
Because distributions aren't defined on individual points of space
The proper way to do it is to have a regularizer
Like $$\delta_\varepsilon(x) = \frac{1}{\varepsilon \sqrt{\pi}} e^{-(\frac{x}{\varepsilon})^2}$$
and then, at the very end of the process, take the limit $\varepsilon \to 0$
but dragging around the regularization through the whole process is a huge pain
 
Always,there should such ansatz exist to solve problems
 
otherwise you end up with this kind of abstract nonsense
 
poooff...
@Slereah Do you have Mathpix?
 
I don't know what that is
 
it is math snipping tool
it allows you to snapshot any figure to latex format
unfortunately this great tool is broken nowadays
since their latest update
@Slereah What kind of tool you use to save time when you write latex in SE?
 
10:03 AM
The proper way to write the Hamiltonian :p
@Student404Mus I use my own two hands
 
This is why you pretend $\delta(0)$ makes sense :p
 
What is the reference of these images?
 
there's also methods for dealing with this in like
Schwarz's QFT book
in the regularization chapter
 
I see
 
10:29 AM
Just when you thought tensors are hard enough in general relativity, computational chemist took it one level higher
I have absolutely no idea what this indices flooding equation is trying to say
and seriously, "globally contravariant" brain shuts down
 
 
2 hours later…
12:53 PM
@Slereah can i ask you a question
 
1:06 PM
Guys, I glued together a container $1 cm^3$ of plexiglass but I'm struggling to insulate it so water doesn't leak. If I cover it whole in hot glue, the transparency gets lost.
 
@yuvrajsingh shoot
 
1:33 PM
@Student404Mus In $\frac{1}{2} \int \frac{d^3p}{(2\pi)^3} \, p^i \left( \hat{a}_{p}^{\dagger} \hat{a}_{p} + \hat{a}_{p} \hat{a}_{-p} - \hat{a}_{-p}^{\dagger} \hat{a}_{-p} - \hat{a}_{-p}^{\dagger} \hat{a}_{p}^{\dagger} + [\hat{a}_{p},\hat{a}_{p}^{\dagger}] \right) $ for the first and third term you have
$$
\frac{1}{2} \int \frac{d^3p}{(2\pi)^3} \, \left( p^i \hat{a}_{p}^{\dagger} \hat{a}_{p} - p^i \hat{a}_{-p}^{\dagger} \hat{a}_{-p} \right) = \frac{1}{2} \int \frac{d^3p}{(2\pi)^3} \, \left( p^i \hat{a}_{p}^{\dagger} \hat{a}_{p} + p^i \hat{a}_{p}^{\dagger} \hat{a}_{p} \right) = \int \fra
 
How to deal with problem of constrained @Slereah
 
To get rid of the $\delta$ you treat it as an infinite constant just as for the Hamiltonian $H$, recall $P^{\mu} = (H,P^i)$ so you're just working out the oscillator expansion of the components of the four-momentum operator
 
1:55 PM
@NovaliumCompany having been in that situation before (albeit with a slightly larger cube) we used a spray glue type deal...i think it was called flex seal? which worked pretty well.
 
@yuvrajsingh Constrained what
Hamiltonians?
if you want the standard books on the topic it's like Dirac or Henneaux-Teitelbaum
 
2:15 PM
@bolbteppa I got it. Thank you very much
 
Dirac's probably a better intro
For a start it's not a brick of a book
 
2:43 PM
@heather Can I just wrap it in normal transparent tape? Cuz I need a transparent solution.
 
Why is k = 1 in f = kma ?
 
@HardikSharma the units of force are defined to make $k=1$
 
This was asked on the main site. Let me see if I can find the question.
 
@JohnRennie already did
 
2:46 PM
Ah JMac beat me to it :-)
 
I decided not to even participate :D
 
I knew I had seen it, so I just searched that before trying to answer them
 
When finally someone asks a question I think I can answer, it turns out my brain is still not big enough
There hasn't been a single person in this chatroom which I'm smart enough to help :(
 
That seems like a really bad way to look at things. For one thing, most people at some point in their lives aren't able to help others with physics questions, because they still have to learn. At the same time, to feel confident in helping others you at least need to try at some point, or you'll never feel comfortable. Just don't oversell your knowledge if you're unsure
 
@JMac got it. Thanks a lot.
 
2:51 PM
@JMac Understood
Looking it from the bright side it means I still have a lot to learn.
Why my voltage multiplier Tinkercad circuit doesn't work?
 
 
2 hours later…
4:27 PM
I'm pretty sure this is supposed to output a much higher voltage. What's going on?
 
4:48 PM
Most likely, it's not taking the internal resistance of the devices correctly to charge the caps with the right RC
 
5:05 PM
How GitHub helps you, physicists?
 
5:23 PM
Using the normalization definition of the momentum state
$$
|\mathbf{p}\rangle=\sqrt{2 E_{\mathbf{p}}} a_{\mathbf{p}}^{\dagger}|0\rangle
$$

since (before normalization)

$$
\langle\mathbf{p} | \mathbf{q}\rangle=(2 \pi)^{3} \delta^{(3)}(\mathbf{p}-\mathbf{q})
$$

it follows that

$$ \langle\mathbf{p} | \mathbf{q}\rangle= 2 E_{\mathbf{p}}(2 \pi)^{3} \delta^{(3)}(\mathbf{p}-\mathbf{q}) $$

How someone could prove the result

$$ (1)_{1-\text { particle }}=\int \frac{d^{3} p}{(2 \pi)^{3}}|\mathbf{p}\rangle \frac{1}{2 E_{\mathbf{p}}}\langle\mathbf{p}| $$
The above result is the completeness relation after normalization
 
@Student404Mus You just apply that last expression to a $\lvert q\rangle$ and show that the result is the same $\lvert q\rangle$, i.e. it acts as the identity on that basis.
 
5:39 PM
ah ! I see
 
6:02 PM
@ACuriousMind Had you tried this web application before? snip.mathpix.com
 
nope
 
It help to save time when it comes to writting latex equations
you need only to copy to clipboard the snapshot of the equation you want then paste it in the script editor of latex (in the web application) and it's done
 
7:02 PM
0
Q: Help with book suggestions

Amanda MacaurenniIf I am to ask for book suggestions on a certain topic, where should I ask? Should I ask on Physics Stack Exchange? If so, what tag should I use?

 
 
1 hour later…
9:11 PM
@JohnRennie What do you mean? I want to put surfactant so when the water evaporates the TiO2 particles don't go with it. Also, later on when I mix the TiO2 particles (now with micelles around then) and black dye in baby oil, the black dye doesn't cover the TiO2 particles.
 
 
2 hours later…
10:44 PM
@Student404Mus you can see what your simulation/analysis code was and revert it to what it was before you broke it...if you use git properly. A lot of people also use it for sharing their code or finding other peoples' code
 

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