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10:00 PM
My understanding of QFT is very very shallow. But if you were to have a hypothetical empty universe, would you still get peturbations like virtual particles?
Well
I guess I need to fix that question
 
which is nothing but pop-science
I guess you have in mind the heuristic picture of phantom-like particles popping in and out of the vacuum
 
No
not really
 
there is nothing in the mathematics of QFT that suggest that such a picture is in any sense accurate
I guess you have in mind the heuristic picture of phantom-like particles popping in and out of the vacuum
 
I have more of an image of electrons getting closer
and a disturbance being made in the middle
with that being the virtual particle
thats a bit like a regular particle
but according to HUP has a few wonky properties since it's not around for long
 
Anonymous
Any idea how to prove $x^5/5!-x^7/7!+x^9/9!....>0$ for all positive $x$ ?
 
10:04 PM
they are a formal concept, with a very precise meaning
"virtual particles" are not to be regarded as something particle-like
 
Wait, they're not? I thought they're just short lived particles that borrow their energy according to HUP
with their lifetime being inversely proportional to their energies
 
o
 
the HUP has nothing to do here
 
Oh
How do you estimate the lifetime of a virtual particle then?
 
10:05 PM
particles cannot borrow energy from anywhere
@Phase again, a virtual particle is not to be regarded as something particle-like
 
Guess I got led astray by a wonky teacher
 
in technical terms, a virtual particle is a contraction of fields in the Dyson expansion of the $S$ matrix
it has nothing to do with actual particles
 
Tbf they probably didn't study this far
 
people use the words "virtual particles" because it makes communication more efficient
but it is a very misleading name
let us introduce a better name
something like "virtual contraction"
 
I mean
 
10:08 PM
now you see all those questions are meaningless
 
That might work if I had any idea of any of that
I've barely gotten through Shankars Principles
I assume from that though that it's just mathematical formalism
 
of course, I dont expect you to understand the technical definition
 
rather than physical things?
 
exactly, just maths
it has no counterpart as far as the physical world is concerned
a lot of mathematical concepts have a corresponding reality associated, at least in principle
"virtual particles" is not one of them
@Blue $x^5/5!-x^7/7!+x^9/9!....>x^5/5!>0$, right?
bc it is decreasing and alternating
 
Wait so
What about the Casimir effect?
Doesn't that have a virtual particle effect?
 
10:13 PM
what about it?
what do you mean by a virtual particle effect?
 
An old tutor said to me that Virtual particles provided the momentum for pushing the plates closer together
if you take two conductive, non charged plates and put them a short distance apart
 
that is the hand-wavy explanation
it doesn't stand up to inspection
 
@Phase The virtual particle myth is really pervasive, but the fact of the matter is that "virtual particles" only exist in the perturbative expansion into Feynman diagrams, and are absent in other approaches. Every effect you can "explain" with virtual particles you can also derive without ever mentioning them.
 
Oh
im just gonna stop taking what irl physics people tell me as fact
that was probably dumb of me
 
and, to be fair, in the typical calculation of the Casimir effect you dont even go beyond tree level, so you dont really have virtual particles
 
10:17 PM
@Phase It's not your fault - as I said, the virtual particle myth is really pervasive, and even many people who should know better continue to perpetuate it, for reasons I cannot fathom
 
Once I finish a QM textbook like Shankar's, what path should i take so that I can eventually start actually learning QFT?
 
@ACuriousMind for the same reason physicists perpetuate the myth of dy/dx being a ratio
 
Anonymous
@AccidentalFourierTransform How can you say that it is decreasing? Derivative?
 
@Blue yes
 
Anonymous
@0celóñe7 Derivative is giving me a series again. I can't prove $f'<0$....
 
10:19 PM
Sum the series.
 
$$E^2 = p^2c^2 + m_0^2c^4$$ [link](https://physics.stackexchange.com/a/352083/149711)
$E$ is energy.
$p$ is pressure.
$c$ is light speed.
$m$ is mass?
 
@0celóñe7 The problem is that the virtual particles idea creates much more confusion than it alleviates. The ratio thingy just works almost all of the time - it's wrong, but it doesn't lead you astray more often than it works.
 
Anonymous
$x^4/4!-x^6/6!+x^8/8!-....$
 
Anonymous
How?
 
@EnderLook $p$ is momentum.
 
10:20 PM
That's a logarithm (maybe)
 
ups
 
Put it into wolfram and see what it gives
 
Anonymous
Somehow relate it to $e^{-x}$
 
Anonymous
As Balarka claims
 
Anonymous
Lemme see
 
Anonymous
 
Anonymous
cos is definitely not positive for all real $x$
 
Anonymous
nor negative
 
50/50
 
but $\cos(x)-1+\frac{x^2}{2}$ is positive everywhere
 
10:26 PM
so is mod(cos(x))
 
Anonymous
@AccidentalFourierTransform Proof?
 
not that that's relevant
 
Anonymous
Don't say graph :P
 
$\cos(x)-1+\frac{x^2}{2}\ge \frac{x^4}{4!}$
 
?
How bout this?
 
Anonymous
10:31 PM
wooo..that works
 
Shameless method but i guess as long as its the right result
 
Anonymous
How did you get it though?
 
I just figured that since cos(x) can be expressed as a convergent series, I could just put cos(x) - (1 - x2/2!) into Wolfram to get what you wrote the sequence to be
and wolfram took it from there
Pretty sure you can do that sort of stuff with Convergent series
haven't studied it formally tho
 
Anonymous
Hmm...I could have just related it to $\cosh(x)$...meh
 
oh no
i should have been in bed by now
now im hungry again
#FirstWorldProblems
 
10:37 PM
It doesn't?
huh
 
yes it do
im off to bed anyway
buh bye
 
afaik for any absolutely convergent series stuff like what I did above should work
cya
 
@Phase that's what AFT said
 
o
I didnt read it
 
Those imaginary exponentials combine to cosine
 
10:38 PM
Well yeah
but he was asking in terms of e
 
Anonymous
But then...that's another struggle
 
Anonymous
With imaginary terms
 
Anonymous
hmmm
 
Anonymous
2
A: Proving $x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} \space \space \forall x \in \Bbb R$ using Taylor's expansion

Jack D'AurizioFor any $x>0$ we have $\sin(x)<x$, hence by applying $\int_{0}^{t}(\ldots)\,dx$ to both sides we get $1-\cos t < \frac{t^2}{2}$. By applying $\int_{0}^{x}(\ldots)\,dt$ to both sides we get $x-\sin x<\frac{x^3}{6}$, which can be rearranged as $\sin(x)>x-\frac{x^3}{6}$. By performing the same trick...

 
Anonymous
Aaaaahhhhhhh....I didn't notice this
 
Anonymous
10:39 PM
wow
 
Anonymous
@Blue: $$ \frac{x^5}{5!}-\frac{x^7}{7!}+\ldots = \int_{0}^{x}\int_{0}^{a}(u-\sin u)\,du\,da $$ is the integral of a positive function. Just the same argument in disguise. — Jack D'Aurizio 29 mins ago
 
@ιllιlılMostafaıllııllıl not my field so I can't comment
But it looks like they're getting Javier García de Abajo's affiliation wrong, he's at icfo
 
Anonymous
11:13 PM
It seems the best way is to differentiate till you get something like $x-\sin(x)$...then go reverse from there to find the graph
 
Anonymous
Jack's method is also a good alternative (which is more of a trick)
 
11:46 PM
@Blue what are you even trying to do?
 
Anonymous
11:58 PM
@0celóñe7 Prove $(x^5/5!-x^7/7!+x^9/9!-...)>0$. I guess it is best to take $F(x)=\sin(x)-x-x^3/3!$ try to plot it's graph by differentiating. $F'(x)=\cos(x)-1-x^2/2$. $F''(x)=-\sin(x)-x$. $F''(0)=0$. $F''(x)<0$ for positive $x$....use these facts to plot the graph for $F(x)$
 
can this be mod-deleted? it's absolutely incomprehensible: physics.stackexchange.com/questions/351864/…
(or, you know, can anyone with enough rep vote to delete?)
 
@Blue that's a sine function
 
Anonymous
@0celóñe7 Yes. So?
 

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