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5:12 AM
I read not to ask to ask, just ask, so... uhh.. hi.
I'm wondering about d = (1/2)vt
I don't really understand what it means for the area to equal the distance in said Cartesian chart, but that aside doesn't that lead to d = (1/2)d?
d = (1/2)vt
v = d/t
d = (1/2)(d/t)t
d = (1/2)d
What am I missing?
@here
@JohnRennie @DavidZ @JMac @peterh-ReinstateMonica
 
@RubelliteYakṣī hi :-)
 
Are you thinking about constant acceleration?
 
I thought it was instantaneous velocity
I'm trying to understand the ^2 in E=mc^2, which I worked back to d = (1/2)vt. So, looking at the area of the graph I understand why 1/2, but then I started thinking, v is simple distance over time, so how could distance - 1/2 distance
 
If an object starts from rest and accelerates with a constant acceleration, a, then the distance travelled in a time t is s = ½ a t^2.
But we can also write this using the final velocity v, where v = at.
That gives us s = ½vt
 
5:22 AM
but this isn't about acceleration, it's about speed. That's d/t/t vs d/t, right?
 
Is an affine parameter a parameter which parameterizes a geodesic?
 
Sorry if this sounds basic. I haven't had physics in 20 years
 
BTW this is nothing to do with the ^2 in Einstein's equation
 
Yes, but I was working backward from those
E(v^2/2c^2)+KE_2 = KE_1
Perhaps I'm overcomplicating, just taking the equation from that image: d=(1/2)vt, how does that not mean d=d/2?
 
Do you know basic calculus i.e. simple differentiation and integration?
 
5:32 AM
I remember having done limits
Supposing I was asking this before Newton invented calculus...
Is it really d_1 & d_2? I'm just trying to get a handle on what the real world representations are rather than symbol manipulation
@JohnRennie My understanding of the ^2 in Einstein's equation is that all things are already moving at c (speed of causality) through time and we are squaring it because of relative velocity, right?
 
No.
 
Oh, in that case I have 2 questions, then
 
There are various semi proofs, which are really plausibility arguments that proofs.
 
To really prove it you need to know about special relativity e.g. you need to understand what four-vectors are and their properties.
 
5:38 AM
length, width, height, time?
 
Yes
 
The problem is that if I starting throwing terms like energy-momentum four vector it's going to sound like voodoo.
It isn't of course, it's just that all disciplines have their specialised terminology and it sounds like black magic unless you're familiar with the area.
This applies to bricklaying or ironmongery as much as it does to physics!
 
Okay, then, instead of trying to show me the derivation, why not just say what the symbols mean? It'a all aymbolic logic, but the symbols stand in for real world phenomena. So, in the E=mc^2 equation, E is energy, the ability to do work, yes?
 
I'm afraid I need to work now for an hour or so, but I'll be around later if you want to pick this up.
 
5:43 AM
sure...
The reason I sought out a chat room was because I figured I'd forget to check back in on a forum, tho
IOW, I may or may not remember to return
I really thought the d=(1/2)vt thing would be super simple for people in here to explain
Have a good day
👋
 
6:26 AM
@RubelliteYakṣī well yes it is. If you travel for a time t at a velocity v then the distance travelled is s = vt. In this case you are accelerating with constant acceleration a for a time t so the initial velocity is zero and the final velocity is v. That means the average velocity is ½v.
Then distance travelled is the time multiplied by the average velocity so it's s = ½vt.
But this still has nothing to do with Einstein's equation.
 
Ohhhh!
I see. Thanks
It has more than nothing to do with it
KE=Fd
I was wondering about one thing, looked into it, then wondered about another thing
@JohnRennie So, as stated, the distance s on the left side is not the same s as v=s/t, because that is the average s/t, right?
 
You get the distance by integrating the velocity.
There are two special cases:
- constant velocity: in this case s = vt where v is that constant velocity
- constant acceleration: in this case s = ½vt where v is the final velocity
 
okay...
You're more... explaining the underlying principles of the equation than answering my question, though.
In the case of constant velocity s=st/t
 
I'm not sure what your question is ...
 
in the case of constant acceleration s=?
My guess is s_final = (s_avg t )/ 2t
 
6:41 AM
If you start from rest and accelerate at constant acceleration a for a time t then s is the distance travelled. It isn't obvious what you mean by s_avg. The distance travelled is just the distance travelled. What would the "average distance travelled" be?
 
v=s/t
I'm asking you to reformulate the equation using the component parts of v
so I can understand specifically how s=s/2 is wrong
 
s/t is the average velocity. If v is the final velocity, i.e. the velocity at time t, then the average velocity is given by v_av = ½v.
And the distance travelled in the time t is s = v_av t = ½vt
 
Okay, but in
s = ½vt, that v is made up of parts, yes?
 
I'm not sure what v is made up of parts means. v is just the velocity at the time t.
 
OMG
What is velocity?
It is the distance traveled in an amount of time
 
6:46 AM
Velocity is ds/dt.
 
here the d is calc stuff?
 
It isn't simply distance divided by time.
@RubelliteYakṣī yes
 
okay, but if I start driving my car and you have a stop watch, and you record my speed at the moment you turn off the stop watch, that is my final velocity, yes?
Assume I'm driving in a straight line. I've gone a certain distance in a certain amount of time
 
When you say I record your speed, how am I doing this? With a radar gun?
 
Why are we taking derivatives instead of just using the odometer and stopwatch?
Let's say a computer is connected to the speedometer and the stopwatch
 
6:49 AM
I think you are mixing up average velocity with instantaneous velocity
 
There is a final velocity the moment the stop watch is stopped
Then you take that final velocity, add it to my starting velocity and divide by 2, yes?
thus the 1/2
now, the starting distance was 0. just as the starting velocity was (and for the same reason)
and the starting time was 0 as well
so the average velocity should be 1/2 the final distance as compared to the final time, yes?
since v=s/t
 
@RubelliteYakṣī no
 
no to which?
 
Average velocity is just distance travelled divided by time taken.
v_av = s/t
v = ds/dt
 
Above you wrote, "That means the average velocity is ½v."
Now you are saying it's not?
 
6:58 AM
It is always true that:
v_av = s/t
v = ds/dt
In the special case of constant acceleration it is also true that v_av = v/2
 
I'm sure you are very skilled at manipulating symbols, but you aren't so great at putting the meanings of those symbols into words. I don't understand how this can be, since the point of physics is to describe the real world—as opposed to pure mathematics
Thanks anyway
 
7:58 AM
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2 hours later…
10:11 AM
Hi everyone. I have a question about the microcausality condition in QFT, i.e. the commutator of any two operators vanishes for space-like spacetime intervals. Is this a relation holding in any Lorentz invariant interacting QFT or there could be a priori a violation of it?
In other words, under what circumstances the microcausality condition holds for interacting fields? Weinberg states that is a sufficient condition to have a Lorentz invariant S-matrix and he comments explicitly that it is not necessary. What other kind of Lorentz invariant theories he is talking about?
 
 
1 hour later…
11:14 AM
I have mixed feelings about this answer. On the one hand, it correctly says that all electrons are the same. OTOH, it tells the OP to "Imagine [electrons] as small rotating balls". I don't think that's a good idea, but is it bad enough to warrant a downvote?
 
 
3 hours later…
2:26 PM
Are deSitter space and antideSitter space both vacuum manifolds?
 
Jim
2:50 PM
@CaptainBohemian IIRC, yes
 
@Jim Ok. Thank you.
 
3:11 PM
@apt45 AFAIK we simply do not know whether all relativistic interacting theories fulfill microcausality
 
Jim
@PM2Ring try leaving a comment on the answer. If it doesn't improve, then downvote
 
Crucially it also depends on what your definition of a "quantum field theory" is
E.g. microcausality is one of the Wightman axioms, so axiomatically you don't have a QFT if you don't have microcausality in that approach
 
3:24 PM
@ACuriousMind thank you. Let's put aside axiomatic QFT. What if microcausality is not satisfied in a local relativistic QFT but still the S-matrix is Lorentz invariant? Do we have counterexamples?
This paper arxiv.org/abs/0709.1483 is claiming that "For a Lorentz-invariant theory microcausality can be proven in a variety of ways"
I am not aware of any of this variety of ways they are talking about, since they refer to the vev of the commutator.
 
I don't really know anything specific there, I'm afraid
 
Could it be possible, in principle, to have a relativistic QFT, compute the spectral density of the two-point function and find that the field commutator doesn't vanish for space-like intervals? Or the computation would be automatically consistent with the microcausality requirement?
@ACuriousMind you don't need to know anything about the paper. It's just a statement about the fact that microcausality can be proven non-perturbatevely (and they cite the textbook of Weinberg)
No sorry, I think the spectral density has nothing to do with it. Still, I am wondering whether an explicit computation at one-loop of the two-point function could reveal a violation of microcausality.
 
vzn
3:42 PM
4
Q: Is microcausality a statement about locality?

WillAs far as I understand it locality is the rejection of action-at-a-distance. By this I mean that in a given frame of reference at a given instant of time (in that reference frame), two physical objects can only interact if they are in physical contact. Lorentz invariance requires that this is the...

 
3:55 PM
@vzn thanks for the link. @ACuriousMind you wrote "Only for very few and low-dimensional theories, e.g. scalar fields in 2D, it is proven that a theory with arbitrary polynomial interactions obeys the Wightman axioms". Do you have a reference about this fact of scalar theories in 2d?
 
@apt45 The classic is Glimm and Jaffe's work, e.g. in their book Quantum Physics. They construct the 2d theory as a Euclidean theory obeying the OS axioms, which via OS reconstruction means there is a corresponding Wightman theory. Note That I was talking there about the full list of Wightman axioms - it can well be that microcausality alone without the rest is much easier to establish
 
@ACuriousMind yeah, sure. Thanks.
 
vzn
honestly am having trouble parsing the defn of "microcausality". this pg en.wiktionary.org/wiki/microcausality defines it as "The spacelike local commutativity or anticommutativity of fields." in a sense, A or Not A. but everything satisfies A or Not A...
 
Whether commutativity or anti-commutativity is the "correct" way to express causality depends on whether or not you have a fermionic or a bosonic field. See physics.stackexchange.com/q/134577/50583 for a formal derivation of why fermions must anticommute
 
4:12 PM
@ACuriousMind therefore when we talk about Local field theories this is equivalent of talking about commuting fields at space-like intervals? So when the commutator is not zero we really talk about a non-local theory?
Or local QFT is more referred to local Lagrangians?
 
What exactly "local" means depends on the context
some people use "causality" and "locality" almost interchangably, for others they have nothing to do with each other
 
Yeah sure, this is the root of confusion about this topic. So, if I deal with a Lagrangian where the interactions are polynomials in fields and their derivatives, I might have in principle a violation of microcausality of the underlying quantum-theory?
 
Perhaps? I haven't really thought all that much about that condition - as I said, I only think that we can't prove it in full generality, I don't know any counterexamples
 
Ok thanks!
 
4:28 PM
In 1+1D non-relativistic quantum mechanics, the systems cannot have degeneracy. Does this also hold in 1+1D QFT also?
 
@expikx You have to be more precise in what you mean by that, and the answer is likely "we don't really know" in any case. In general, we do not understand interacting QFTs or their bound states very well and have to use some QM approximation (physics.stackexchange.com/q/217575/50583). Additionally, much of what you think of as a bound state in ordinary QM becomes a resonance in QFT (physics.stackexchange.com/a/12571/50583).
 
5:01 PM
A Bar walks into Einstein.
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ooops wrong frame of reference.
 
 
1 hour later…
6:13 PM
Hahahaha omg
 
 
2 hours later…
8:04 PM
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11:40 PM
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