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2:26 AM
0
Q: Is there a way to retrieve one's own edit of a different participant's PSE question, if the edit was rejected within hours of being sent?

EdouardI spent a few hours this morning, attempting an edit of a PSE question (similar to many bearing on those aspects of space that are often described as representing its expansion) that I'd attempted to relate to two cosmological models: Penrose's CCC model and Poplawski's torsion-based model. Alt...

 
 
3 hours later…
5:42 AM
Hi this is a very elementary question. How can I write the Time ordered product of three operators? T[H(t_1)H(t_2)H(t_3)]? I don't really understand the notations given on wikipedia. Thanks!
 
6:34 AM
@NamanAgarwal assuming $H$ is bosonic, then the time-ordering operator simply reorders the operators such that the associated time decreases from left to right
for instance, if $t_3 > t_1 > t_2$, then $\mathcal T\{H(t_1)H(t_2)H(t_3)\} = H(t_3)H(t_1)H(t_2)$
 
6:52 AM
how does one go about writing equation of the motion w.r.t to time?
 
@napstablook By using conservation of string length
 
I tried conserving angular momentum and I got v as a function of theta
but I couldn't convert this further to v as a function of time
 
@napstablook Angular momentum isn't conserved. Can you think of something else that is?
@napstablook The force from the string is not directed towards O
So how can angular momentum be conserved about O?
 
ok how does one go about using conservation of string length?
 
Hint: The force from the string is always perpendicular to the velocity of the object
 
6:58 AM
work energy theorem use?
 
@napstablook Yeah, energy is conserved, not angular momentum
It's a common mistake to conserve angular momentum, for example physics.stackexchange.com/questions/640606/…
 
@VincentThacker I see so velocity of the particle should remain constant no? why have you written $vsin\theta$ is constant?
 
@napstablook Yeah, that's right. The post I linked is the 3-dimensional version of the problem. I was just using it as an example.
 
ok I have been using the proper conservation but I relate change of $\theta$ by equate torque = $I\alpha$ and and tension force since velocity is constant
@VincentThacker thanks I got it
 
@napstablook you're welcome
 
 
2 hours later…
fqq
8:57 AM
@MoreAnonymous not really, it's been commonly used in English since the middle ages, just slightly newer than the plural they. Then some grammarian invented a "rule" that generic he should be used, and other Victorian grammarians started policing everyone else's language with this and other random rules.
@NamanAgarwal if you want a single expression without if clauses you can use Heaviside theta functions
 
9:10 AM
@fqq and now a few amateur grammarians started policing everyone else again.
That's why I am no longer a mod on this hostile site working for free for a criminal company.
 
9:48 AM
@Charlie I know it's one of my stock phrases, but I didn't realize I use it that often :P
 
0
Q: Why my watched tags aren't yellow?

Mauro GilibertiThe questions tagged with my watched tags used to be yellow, and I found that super useful. Why are they not yellow now? Is this a problem on my end, or a change of the site?

 
@FadedGiant You were a mod?
 
10:14 AM
@fqq
Frankly i need the expression for the time ordered product interms of the the step function explicitly.
i can write that for when T is applied over two operators. But for three or more I can't see how i can generalize the expression
@NiharKarve
 
fqq
just replace every "if $t_i>t_j$ with $\theta(t_i-t_j)$
@NiharKarve e.g. this would be $\theta(t_3-t_1)\theta(t_1-t_2)$ etc
 
let me write it explicitly and i will get back to you to get my solution checked if that is okay.
would take 5mins mostly
is this correct? @fqq
 
10:41 AM
@NamanAgarwal yes
 
 
1 hour later…
11:43 AM
Anybody who can assist my friend in resizing his self declaration image
 
12:21 PM
I'm afraid figuring out how pagesetting works in LaTeX is an unsolved problem of physics
3
 
 
3 hours later…
3:36 PM
@Slereah ( in some context to the discussion that day)- An accelerating charge radiates. If the charge is at rest wrt an inertial frame and I accelerate wrt the charge will I see the charge radiating. Or otherwise if a charge is accelerating wrt an inertial frame and I accelerate with the charge at the same rate will I see the charge not radiating. (I can’t connect this with Electrodynamics I have studied
 
Oh man that's an old debate IIRC
I forget if there's a resolution to it
Feynman had a whole paper about it I think?
 
if you're talking about special relativity, then acceleration is independent of observer, so there's no actual conflict there
so presumably you are talking about GR?
 
Depends what you mean by SR
 
@NiharKarve we had a very long discussion regarding whether general covraince is vacuous or is there a principle of general relativity which would imply all the frames are equivalent.
 
If you do special relativity in accelerated frames, then that's about the same issue
 
3:43 PM
I am trying to come with something which should say that indeed all frames are equivalent
@NiharKarve GR or accelerated frames in SR
 
@Shashaank All inertial frames are equivalent
 
@Shashaank this Wikipedia page might be relevant
 
@Slereah is my conclusion that just covariance under a group of transformations is not sufficient for having a relativity principle corresponding to transformations of that group. In nutshell I am saying that just because SR (or NM) is invariant(covariant) under Lorentz transformation (galelian) doesn’t necessarily imply a principle relativity of inertial frames. I know books sate that
 
It depends what you mean by "principle of relativity" I guess?
 
I mean that you can’t detect whether the frame you are in is at rest wrt another inertial frame or moving wrt it
Is that ok
 
3:50 PM
Well, if you mean that in a specific way, I guess?
If you have a non-inertial frame, an object under no other forces will experience acceleration
but no such thing under Lorentz transformation
 
@Slereah that because LT don’t allow transformations to accelerated frame
 
yes
 
What I am trying to understand is that just the general covariance of GR under all transformation doesn’t mean that all frames are equivalent in GR
 
Well once again, depends what you mean by equivalent
Changing frames will not change how you express your problem in GR, if you do it independently of coordinates
But in a practical sense, some frames will be more natural than others
 
@ACuriousMind ACM-autoresponse-chatbot.exe needs a driver update by the sounds of it :p
 
3:54 PM
don't get too hung up on vague notions
 
@Slereah That some experiment in one frame detects that it’s motion is being changed wrt another frame
 
Well experiments are independent of frames
 
oh no are we still debating what covariance means
 
But of course, an actual observer will define its own natural frame, if that is what you mean
but you can define the experiment in whatever coordinates you wish, as long as you make the changes to the observer too
 
Yes that’s what I mean
 
3:57 PM
you can change the coordinates all you want, but those changes will not affect statements like "The onboard clock of the observer measures a time of one second" in between two events
It will be a simpler problem to express in some coordinates, but if you do everything properly, everything should end up being the same
 
The point of general covariance is that you shouldn't be able to change the topological properties of spacetime by simply relabelling the points (which is what a coordinate change is)
 
@Slereah yes so I am trying to show that one experiment in one coordinates when done in another coordinate allows us to differentiate between the two frames ( we know that is not possible in SR to differentiate between inertial frames; if we can come up with an example where this is possible in GR we have that there is no principle of general relativity)
 
I mean, it's the same in SR technically :p
If you allow arbitrary coordinates
but then that's just GR with a flat metric
 
Wait what? You can experimentally differentiate between two different inertial frames in SR
The take-home point is that neither of them is preferred in any sense
 
No I mean that you can also can't differentiate between an accelerated and unaccelerated frame in SR, if you allow to describe accelerated frames
 
4:03 PM
@Charlie no you can’t. That’s my point, if we can in GR then we have that accelerated frames are not equivalent
 
It's down to some ambiguity, really
What does SR and GR mean
 
If you measure the degree of time dilation in two different frames, you can definitively say they are different frames
 
@Charlie But time dilation is symmetric because the LTs are symmetric
 
Does SR mean that spacetime is a vector space with a specific symmetry group, or does it mean that the spacetime is flat
 
You can't say that one of them is moving but the other is not, sure, that's what I meant by "neither of them is preferred"
If I have two frames moving wrt. a third at different velocities, I can absolutely experimentally differentiate between them without violating any laws of physics
 
4:05 PM
@Slereah we should be able to differentiate between inertial and non inertial frames in SR. Only inertial frames are equivalent. There will be inertial forces in accelerated frame which won’t be there in inertial ones and with those you could detect that your frame is actually changing its motion
 
@Shashaank Yeah, accelerated frames in SR have an event horizon
 
I think the issue is that you're conflating a coordinate description, which is a mathematical statement, with what the observer is measuring, which is a physical thing
 
@Shashaank What qualifies two coordinate systems as being "equivalent"?
 
Don't confuse coordinates for the actual measurements
 
@Charlie What do you mean by "experimentally differentiate"?
 
4:07 PM
I can perform an experiment to tell which of the two frames I am in
 
@Charlie No, you can't do that for inertial frames in SR
 
@Charlie how. How can you say without looking outside that you can experimentally differentiate between them...can you see whether it is your frame that is moving or the other one or which one is moving with what speed
 
All experiments will behave the exact same way
 
Phew this is more confusing than understanding String theory. Maybe I need to take an appointment with Penrose
 
I mean you can experimentally measure, given two frames moving wrt a third, which of the two frames you are in, just by measuring how fast you are traveling wrt the third
I don't mean you can claim to be stationary or moving wrt anything objective, but that's what I meant by no frames are preferred
 
4:10 PM
How can you measure how fast you are moving without looking
By experiments
 
You can't, you need to be talking wrt another frame. Again, that's what I mean by no frame is preferred
 
@Charlie Yeah then we are on the same page
 
But there is a measurable difference between two frames wrt a third, so in that sense they are not identical
 
 
3 hours later…
7:06 PM
@Shashaank We have a few questions about that, eg
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Q: Does a charged particle accelerating in a gravitational field radiate?

SergioA charged particle undergoing an acceleration radiates photons. Let's consider a charge in a freely falling frame of reference. In such a frame, the local gravitational field is necessarily zero, and the particle does not accelerate or experience any force. Thus, this charge is free in such a f...

 

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