The h Bar

Slereah

17:10

Hm

This lattice intro gives the matrix for a scalar field as $$\sum_{x,y} \varphi(x) (\delta_{x+a\hat{\mu},y} + \delta_{x-a\hat{\mu},y} - 2 \delta_{xy})\varphi(y)$$

But that would give $$\sum \varphi(x) \varphi(x+a\hat \mu) + \varphi(x) \varphi(x - a\hat\mu) - 2\varphi(x)^2$$

When the actual derivation gives me

$$\sum \varphi(x+a\hat \mu)^2 + \varphi(x)^2 - 2\varphi(x) \varphi(x + a\hat\mu) $$

Is there a transformation of one into the other I am not seeing

Oh wait

Or are they using a mix of the two lattice derivatives, maybe

$\frac{\phi(x+a) - \phi(x)}{a}$ and $\frac{\phi(x) - \phi(x-a)}{a}$

No that wouldn't work either

ACuriousMind

@Slereah It works

They use a mix because the convergence behaviour of the mix is better than using the same derivative twice

Slereah

But that would give me a term in $\phi(x+a) \phi(x-a)$

Is there a trick to know

Like variable change or something

ACuriousMind

I don't see your issue: $\nabla_B\nabla_F \phi_x = a^{-1}\nabla_B\left(\phi_{x+a} - \phi_x\right) = a^{-2} \left(\phi_{x+a} - \phi_{x+a-a} - \phi_x + \phi_{x-a}\right)$.

Which is $a^{-2}\left(\phi_{x+a} +\phi_{x-a} - 2 \phi_x\right)$.

There's no trick involved.

Slereah

17:26

But the lagrangian is $\partial \phi \partial \phi$

ACuriousMind

Which is equivalent to $\phi\Delta\phi$ upon integration by parts

Slereah

Oh right

Well see, that was a trick

Hm

What's the equivalent of integration by part in a summation

ACuriousMind

You integrate by parts before you discretize

Slereah

Well yes, but STILL

I want to know :V

Though I guess it might be a trivial thing to show

ACuriousMind

@Slereah Unsurprisingly, it's summation by parts :P

Slereah

17:29

heheh

Well, it checks out!

also

How reasonable is it to mix two different derivatives for that

From what I recall of non-relativistic path integrals, taking the integrals at different points on the lattice was a pretty important decision

ACuriousMind

@Slereah Very. Upon taking the continuum limit you have convergence of order $a^2$ instead of just $a$ that you would get by using the same twice

@Slereah I don't know what's that supposed to mean

Slereah

Well if you had a function of $x$ in the action

Say $\int dt A(x(t))$

The summation was $\sum \varepsilon A(x_i + \lambda \Delta x_i)$

The lambda gives you the type of stochastic integral this is in the continuum limit

Of course this isn't quite the same here, since it's a derivative and all, but still I wondered

Then again the derivatives of $x$ didn't have any mention of this, so probably safe

ACuriousMind

@Slereah Why would that be a "stochastic integral"? You're not taking a path integral here anywhere, you're just writing down the discrete action.

Slereah

Well as you recall, path integrals are done over fractal curves

And those cannot be integrated with Riemann integrals

ACuriousMind

You're not doing a path integral here, you're just writing down the action!

Slereah

17:42

So the integral for the action has to be a fancy integral

Well yes, but EVENTUALLY it will be in a path integral :p

If you choose the wrong lambda, this is equivalent to a wrong operator ordering

Though I think this doesn't matter for the definition of the derivative here

This would only be a problem for something like $f(\varphi(x))$ here I think

Not quite sure if there's an equivalent for QFT of that?

Never seen anything regarding this

ACuriousMind

I don't think the lattice version cares at all about that. The measure is really just $\prod_x \mathrm{d}\phi_x$, there's no fancy Wiener measure or somesuch involved.

Slereah

What happens in the continuum limit, though

ACuriousMind

@Slereah Since we don't have a rigorous definition of the path integral in 4D, that's an unanswerable question.

Slereah

What about in 2D :p

I guess that might be an issue for like

Sine gordon

ACuriousMind

@Slereah Taking the continuum limit of lattice theories is a horribly difficult thing

Slereah

17:51

Even for trivial examples?

ACuriousMind

In general, if you don't use the right prescriptions, you'll mostly end up with the trivial theory

Slereah

Like free fields

Or maybe fields coupled with classical potentials

ACuriousMind

@Slereah And...what would you look at in the free case to determine whether what your limiting process spits out is the "correct" thing?

Slereah

Regular QFT, I suppose

ACuriousMind

No, specifically, which object do you look at?

You have no coupling and no interesting observables

Slereah

17:54

Well currently I'm trying to compute just a transition amplitude

So I'll check it against the Heisenberg formalism for that amplitude

Danu

Speaking of coupling constants, I heard an interesting statement in Math. Gauge Theory

ACuriousMind

@Slereah What non-trivial transition amplitudes are there in a free theory?

Slereah

Well this relates to a question I had a few months back

How do I do a path integral between two boundary conditions

Danu

That the coupling constants are the prefactors of $-B_{\mathfrak g}$ (the Killing form)

Slereah

And my example was from one coherent state to another

I'm trying to do it via lattice methods

ACuriousMind

17:56

@Danu What?

Danu

@ACuriousMind So on a simple Lie algebra of a compact Lie group, there is a unique Ad-invariant inner product

ACuriousMind

@Slereah I'm not sure that is a well-defined question.

Danu

given by a scalar multiple of $-B_{\mathfrak g}$

Slereah

Isn't it?

ACuriousMind

@Danu I know what a Killing form is

Slereah

17:57

I mean

It's basically how you derive path integrals from Heisenberg

Danu

@ACuriousMind A bilinear form on any Lie algebra---in the case the algebra is simple it is non-degenerate and I think compactness guarantees it's negative semi-definite

Slereah

It's just that people use almost exclusively the one for the propagator

Where boundary conditions don't matter

Danu

@ACuriousMind Oh.

How did I misread that as don't know?!

ACuriousMind

@Danu Yes, it does by Bonnet-Myers, as the Killing form is the negative Ricci tensor on the group for the connection given by the Lie bracket.

@Danu I don't know. The part of your statement that doesn't make sense to me is that the coupling constant has anything to do with the Lie algebra

The coupling constant is the $g$ in $S_\text{YM} = \frac{1}{g^2}\int F\wedge{\star} F$, no?

Danu

Yeah.

ACuriousMind

18:01

So, I can choose that $g$ however I like, no?

Danu

We're doing pure math for now, so the connection to actual Lagrangians is not cleared up yet

ACuriousMind

It's just a parameter of the theory

Danu

@ACuriousMind Yes, and the same for the multiples of $-B_{\mathfrak g}$

They're parameters

Slereah

Hopefully the method will just be the same as doing the path integral for a point particle from $x_1$ to $x_2$, except it will be from field configuration at x $\varphi_1$ to $\varphi_2$

Hopefully it will not be too infinite

Danu

My point was that it's surprising to see how coupling constants are formalized

ACuriousMind

18:03

@Slereah You integrate over a region in field space. Usually, the integral is over all of field space. What exactly do you mean by "integrating between two boundary conditions"?

Slereah

Well

At $t=t_1$, the field (on the hypersurface blablabla) is in the configuration $\varphi(t_1,\vec{x}) = \phi_1(\vec x)$

And similarly for $t_2$

Danu

I'll get back to you once I find out exactly how it translates to Lagrangians @ACuriousMind

AccidentalFourierTransform

@Slereah I wonder if that can be done with the Faddeev-Poppov method

Slereah

It's a rather simple thing to do but it seems to never be done in path integrals

(simple with Hilbert vectors, anyway)

Gonna try to do it with coherent states because hopefully those aren't too terrible

ACuriousMind

@Slereah So, you want to integrate over the set of fields that obey those boundary conditions?

Slereah

18:06

yes

ACuriousMind

Well...you'd need to know how to compute a path integral in the first place for that. Usually you just use its formal properties, you don't actually evaluate it

Slereah

I am highly unusual

B)

ACuriousMind

Also, I have a slight feeling that the space of fields obeying a certain boundary condition might be a zero measure set in the cases where the measure is actually well-defined.

Slereah

Well probably

ACuriousMind

Since, in the rigorous path integral of QM, you have to choose the conditional Wiener measure that is already adapted to the boundary conditions

Slereah

18:09

I mean there are infinitely many configurations with those

But still

The same applies to the path integral of a point particle

ACuriousMind

I.e. you're not using a measure on the space of paths and restrict to certain subsets, you're defining a new measure for every integral with a different boundary condition

Slereah

Yet it still spits a result

Not sure it will translate well to QFT but I am curious

ACuriousMind

So one should expect the same for the path integral - you can't just say "I want to integrate over just these fields", you have to define your integral from scratch if you change the boundary conditions.

Slereah

Well hopefully the lattice calculation will converge nicely :p

ACuriousMind

I mean, the Faddeev-Popov trick of just writing a $\delta$ function works for physicists, but the actual thing to do is to quotient out the gauge orbits out of the space of all fields and then integrate over that quotient.

@Danu I have a suspicion it has to do with physicists usually using a scaled version of the field, you get the usual physics action from $\frac{1}{g^2}\int F\wedge{\star} F$ by rescaling the field (or rather the Lie algebra generators) with $g$, which of course also rescales the Killing form.

Danu

18:15

@ACuriousMind Yeah, I realize that

ACuriousMind

Together with all the factors of $\mathrm{i}$ floating around this makes translating between different conventions horrible :P

Slereah

Hm

I am wondering what would be the best idea for boundary conditions

Maybe something with a simple Fourier transform

Just some plane waves

Light travels at $c$ in all frames

And light has no rest frame

barrycarter

hmmm?

Of course, the time dilation and Lorentz contraction cancel each other out, as they must.

Slereah

There are tons of procedures in QFT that are done only in one formalism every time and it annoys me that it's so rare that people try to prove it for all of them

barrycarter

Hail @JohnRennie

John Rennie

18:23

@Danu the amount of time I spend in the chat room is directly proportional to the amount of work I'm trying to avoid doing :-)

2

barrycarter

@JohnRennie Can two observers in different frames who are a non-0 distance apart meaningfully synchronize clocks?

John Rennie

Actually it's not directly proportional - there's quite a higher power law involved!

ACuriousMind

It might even be exponential!

barrycarter

1/(total work - work i am avoiding)

John Rennie

@barrycarter Yes, see:

Einstein synchronisation

Einstein synchronisation (or Poincaré–Einstein synchronisation) is a convention for synchronising clocks at different places by means of signal exchanges. This synchronisation method was used already by telegraphers in the middle 19th century, but was popularized by Henri Poincaré and Albert Einstein who applied it to light signals and recognized its fundamental role in relativity theory. Its principal value is for clocks within a single inertial frame. == Einstein == According to Albert Einstein's prescription from 1905, a light signal is sent at time from clock 1 to clock 2 and immediately back...

but in general the clocks won't stay synchronised

barrycarter

18:26

@JohnRennie I only need single instant synchronization for my evil purposes.

John Duffield

@AccidentalFourierTransform : I would venture to suggest that some people are more concerned about downvoting good questions and answers.

Slereah

shoo shoo

Loong

2

barrycarter

@JohnRennie OK. Points E and A are 8 ly apart. Ship S travels at 0.8 c past point A. S and E want to synchronize to t=0 the instant that happens. Possible?

Must be a log graph.

There also appears to be a delta t shift.

John Rennie

@barrycarter In principle yes. S and A can synchonise because they are at the same point. A and E can synchronise (eventually) because they are in ther same inertial frame.

barrycarter

18:35

@JohnRennie OK. So S and A synch easily. For E, because it's 8 ly away, when they first see S pass A (at their time t=8), they declare the synch time to be t = what?

John Rennie

@barrycarter I feel you're probing for a signifance that isn't there. E knows how far they are from A and therefore how long a signal takes to get from A to E.

barrycarter

@JohnRennie So when E first sees this event, he simply declares t = 8 ?

John Rennie

Well E certainly could do that. Whether there is any point in doing so depends on what your ulterior motive is ...

barrycarter

@JohnRennie The point is to make t = 0 for all 3 observers when S passes A.

@JohnRennie The problem: E has no way of knowing when S passes A until 8 years after it happens.

John Rennie

@barrycarter true. And your point is?

barrycarter

18:40

@JohnRennie So, when that light arrives, does he say "that light left at t=0, 8 years ago; thus, it is now t=8"?

John Rennie

@barrycarter Well, he could say: I know A set their clock to zero when S passed, that happened 8 years ago, so if I set my clock to +8 years now I'm guaranteed to have my clock synchronised with A.

barrycarter

@JohnRennie OK, that seemed too easy, but that's what I meant.

John Rennie

You're getting hung up what people see again, aren't you :-)

barrycarter

@JohnRennie YES!!! I believe the simultaneity of relativity can be defeated if we look at light travel time.

John Rennie

@barrycarter you mean the fact that simultaneity is frame dependent i.e. the failure of simultaneity in relativity?

barrycarter

18:44

@JohnRennie I'm turning into that other guy. I believe that simultaneity can exist if you can define proper time and proper distance.

Proper distance is easy to define. If I see an object traveling at $v$ at distance $d$, I can find the proper distance.

John Rennie

@barrycarter OK, you're onto a loser but I'd still say go for it as the journey will be fun

barrycarter

@JohnRennie It's always about the journey. However, it seems that you won't join me in the pit of hell?

John Rennie

@barrycarter All this stuff is new and exciting to you, but it's 35 (!!!) years since I studied this. The glamour has worn thin since.

barrycarter

@JohnRennie Well, that's technically not a no, so I shall continue to try to entice you.

John Rennie

I'm always happy to chat about physics - any branch of physics

barrycarter

18:47

@JohnRennie Given any two events in spacetime, can we define both a proper distance and a proper time between them?

John Rennie

Though electrodynamics and fluid mechanics have been known to put me to sleep

ACuriousMind

@barrycarter No, it's either/or

John Rennie

Proper distance and proper time are the same thing. They differ only by a factor of $c$.

barrycarter

@ACuriousMind OK, I don't like that answer. :)

@JohnRennie No, I mean two separate events.

ACuriousMind

@barrycarter Because for space-like separated events, there's always a frame where they are simultaneous, and for time-like separated events, there's always a frame where they happen at the same point. You may not like it, but this is fact.

barrycarter

18:49

@ACuriousMind Yes, but I am waiting for the @JohnRennie to tell me

Slereah

Pretty glad I did my thesis on path integrals

I still have my cheat sheet :p

barrycarter

I will disbelieve as long as I can!

John Rennie

@barrycarter ?? Given two events the proper distance is just the (Minkowski) length of the four vector joining them. The proper time is this length divided by $c$.

David Z

@JohnRennie and by a sign in the metric signature, don't forget that

barrycarter

@ACuriousMind I meant to put a smiley there

@JohnRennie English please. American English if possible.

Slereah

18:50

always useful

John Rennie

@DavidZ nobody cares about signs and factors of $c$. We put them in at the end to get the right answer :-)

barrycarter

@JohnRennie Wait, this is about that metric?

John Rennie

@barrycarter events are points in spacetime, and given two events there is a four-vector going from one to the other.

barrycarter

@JohnRennie And the length of that vector is sqrt(d^2-t^2) ?

John Rennie

If this four vector is $(t, x, y, z)$ the proper length is:

$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $

barrycarter

18:52

For any observer, right?

John Rennie

And the proper time $\tau$ is just $ds^2 = -c^2d\tau^2$

Slereah

Tho what I need now is the formula for vectors and matrices

John Rennie

@barrycarter Yes

barrycarter

@JohnRennie Oh, this is what you were telling me earlier. For any inertial observer, that quantity must remain constant?

John Rennie

@barrycarter Yes

barrycarter

18:53

Shiny!

John Rennie

Careful, you're hearing the siren call of geometry ...

barrycarter

What's vaguely interesting here is that the metric squared is negative if two events are further apart that light can travel between them

John Rennie

Correct. We call this a space-like interval.

barrycarter

Hmmm, but you suggested earlier that we can't use this to solve problems?

John Rennie

@barrycarter Did I?

barrycarter

18:56

@JohnRennie You did. Something about comparing them to the ZF or Peano axioms.

Correction: that we shouldn't use these to solve problems.

John Rennie

I think you're mixing me up with a certain cat.

Oh, hang on, yes.

barrycarter

@JohnRennie You said it was starting too basic.

John Rennie

I said that while everything derives from the invariance of ds^2 this is sometimes too basic a starting point to be useful.

So we start with theorems derived from the invariance of ds^2

barrycarter

@JohnRennie But it seems like it could be really useful.

John Rennie

This was in relation to accelerated motion, where the easiest starting point is considering four-velocity and four-acceleration

But these concepts are derived from the more fundamental concept of Lorentz covariance

barrycarter

18:58

@JohnRennie Ship: I will be traveling at 0.8c a distance of 4.8 ly in 6 years. 4.8^2 - 6^2 = -12.96

manshu

@barrycarter all the best

barrycarter

Wait. It is distance squared minus time squared?

@manshu Greetings.

@manshu That was my ship speaking.

John Rennie

@barrycarter it doesn't really matter. Proper length^2 is x^2 - t^2 but proper time^2 is t^2 - x^2. Choose whichever suits your purposes.

manshu

Have a nice trip with your ship.

barrycarter

@JohnRennie OK, I think I like it positive when the events are "close" in the light sense. So, 12.96 is the metric of this journey, yes?

@manshu The ship is leaving without me. We had a falling out.

manshu

19:01

:D

barrycarter

Meaning, I fell out.

John Rennie

$c^2d\tau^2 = 12.96$

manshu

what is tau here?

John Rennie

$\tau$ is the usual symbol for proper time, just as $s$ is the usual symbol for proper distance

barrycarter

From Earth, the journey takes 10 years over 8 ly, or 10^2-8^2 ... hey, that's not 12.96

manshu

19:04

@JohnRennie OMG!!! I have to re-study all the physics again

John Rennie

Take the spacetiem points in the ship frame that mark the start and end of the journey and calculate $\tau$. Lorentz transform those points into the Earth frame and once again calculate $\tau$. The two values will match.

How can I be sure of this? Because the Lorentz transforms are derived from the invariance of $\tau$.

barrycarter

@JohnRennie OK, but in the ship frame, the distance is 4.8 ly and the time is 6 years, right?

ACuriousMind

@barrycarter You're doing the wrong thing, you don't use the distance traveled or the time it took to compute the spacetime interval. You use the coordinates of the start point and the end point

John Rennie

In the ship frame the Earth starts at (t=-6, x=-4.8) and ends at (0,0)

barrycarter

@JohnRennie OK, fair enough. So that's 36-(4.8)^2 for the metric squared?

ACuriousMind

19:06

Remember our discussion about the Lorentz contraction not matching what you found by Lorentz transforming the points? You're doing the same thing again, expecting the contraction to yield something it is not supposed to yield.

John Rennie

@barrycarter we should probably nip the abuse of terminology in the bud: it's the proper length/time squared not the metric squared.

barrycarter

@JohnRennie OK, but ds^2 right?

John Rennie

The metric is an equation and/or a tensor depending on the context.

@barrycarter Yes

barrycarter

@JohnRennie OK to me a "length" is a metric, but I'll use proper length.

John Rennie

Well, 6^2 - 4.8^2 is -c^2ds^2

barrycarter

19:08

@JohnRennie Yes, but I'm happy to keep c = 1 consistently.

John Rennie

OK, just a factor of -1 then

barrycarter

@JohnRennie The point is... now, how do things look from the Earth frame, and how do we get the same proper length?

John Rennie

(a) you haven't done the Lorentz transforms
(b) I've spent the afternoon in the pub

barrycarter

@ACuriousMind I'm not ignoring you, I'm just trying to follow what John is saying.

John Rennie

(c) I'm not doing the Lorentz transforms for you - I can barely type :-)

barrycarter

19:10

@JohnRennie So it's not correct to say that, in the Earth frame, the ship starts at (8,0) and ends at (0,10)? (in (x,t) format)

John Rennie

@barrycarter Correct, or rather not correct

barrycarter

@JohnRennie Because of that silly simultaneity thing?

ACuriousMind

@barrycarter Okay, but I'm telling you we've essentially had the same discussion before - you're confusing time dilation and Lorentz contraction with the actual result of the Lorentz transformations

barrycarter

@ACuriousMind Actually, I think I might have this... probably not, but who knows.

John Rennie

@ACuriousMind to be fair, every relativity student does this at first

barrycarter

19:13

For this trip, the proper distance is 10 ly, and the proper time is 6 years?

ACuriousMind

@JohnRennie Yeah, I know, it is an inevitable product of teaching relativity through thought experiments and paradoxes instead of axiomatically as we do with every other established theory-

Slereah

I probably should do the lattice in 2D or I'm gonna go crazy with the evaluation

John Rennie

@barrycarter Proper distance and proper time are related by $ds^2 = -c^2d\tau^2$, so no 10^2 ly \ne -c^2 6^2

@ACuriousMind you're preaching to the converted :-)

barrycarter

(thinking...)

We might be using two different definitions. The proper (non Minkowski) distance between two events is the distance between them to an observer who is stationary with respect to those two points. Yes?

John Rennie

@barrycarter in general no

barrycarter

19:18

@JohnRennie OK, what do you call that quantity? I could've sworn I've heard it called proper distance.

John Rennie

If the two points are simulataneous in the rest frame of the observer then the proper length will simply be the distance between them because it reduces to $s^2 = 0 + \Delta x^2 + \Delta y^2 + \Delta z^2$

Which of course is just Pythagoras' theorem

barrycarter

Oh, that's right, the observer has to see them at the same time.

John Rennie

@barrycarter there's that see word again. In the rest frame of the observer the two points must have the same $t$ coordinate

barrycarter

And if two events are simultaneous in one observer's rest frame, they can not be simultaneous in another observer's rest frame (different velocity), right?

John Rennie

Correct

manshu

19:21

lol

John Rennie

@manshu yes, the irony :-)

barrycarter

Similarly, if two events occur at the same xyz coordinate for one observer, they won't for another (which is actually fairly obvious, even without relativity).

John Rennie

Yes

barrycarter

Even if you account for the velocity, they still won't be the same, correct?

John Rennie

@barrycarter I guess you mean even allowing for a Galilean transformation they still won't be the same. If so, the answer is yes.

Remember that the Lorentz transformation tells us $x' = \gamma(x - vt)$.

barrycarter

19:23

@JohnRennie Yes. OK. Now, can you transform between any two frames by using just the two frames I described above.

@JohnRennie Correct. The gamma is what throws it off.

manshu

@JohnRennie You already mentioned that you can barely type right now.

John Rennie

@manshu Relativity is more fun after a few beers :-)

@barrycarter I'm not sure what that means

barrycarter

@JohnRennie Neither am I .. and I haven't even been drinking. Maybe I should start. OK, let me think...

John Rennie

Any inertial coordinate system can be transformed to any other inertial coordinate system using the Lorentz transformations

manshu

@JohnRennie Einstein himself might have said so in his own lifetime.

barrycarter

19:26

@JohnRennie Yes, but I believe it can be done with just the two frames of simultaneity and origin equivalence (0 distance).

Slereah

Even in 2D damn that's a lot of terms

John Rennie

@barrycarter OK, though I'm still not sure what that means

barrycarter

@JohnRennie OK, for the ship, the journey starts and ends at the same point, right?

John Rennie

@barrycarter same spatial coordinates. Obviously not the same spacetime point.

barrycarter

@JohnRennie Right, but since there is 0 distance (for him) between these two points, 4.8 years is in some sense the "proper time" (in heavy quotes) of the journey?

John Rennie

19:29

The proper time for an observer is exactly the elapsed time recorded by a clock carried by that observer. This is true in accelerated frames as well, and even in GR.

barrycarter

@JohnRennie OK... now I need a frame in which the journey is simultaneous, but I'm pretty sure there's no such frame.

John Rennie

If two points have a time-like separation i.e. you can go from one to the other slower than light, then there is no frame in which the two points are simultaneous.

barrycarter

@JohnRennie OK, I think I'm getting close to this now. E at time 0 and A at time 0 are simultaneous and 8 ly apart, yes?

John Rennie

Yes

barrycarter

@JohnRennie E at time 0 and E at time (anything) are at 0 distance, right?

John Rennie

19:32

0 spatial separation, yes.

barrycarter

@JohnRennie So, could I somehow combine those two transformations to get a movement of both 8 ly and some number of years forward?

John Rennie

You mean A at time zero and E at time $t$ ?

barrycarter

@JohnRennie Yes!

John Rennie

Well the proper distance is $s^2 = -t^2 + 8^2$.

barrycarter

So, if the ship takes 10 years to arrive in the Earth frame, the Minkowski length of the event it... 10^2 - 8^2 ?

John Rennie

19:36

If $s^2 > 0$ there is a frame in which the two points have the same time coordinate. If $s^2 < 0$ there is a frame in which the two points have the same spatial coordinates.

barrycarter

@JohnRennie OK, in this case I chose two transforms, one with 0 time and 8 ly distance, the other with 0 distance and 10y time.

John Rennie

If the ship passes A at $(0,0)$ and passes E at $(10,8) then the proper time^2 between the events is $10^2 - 8^2$,

barrycarter

@JohnRennie OK, good. Now, can I use that in some way to find out something about the ship?

manshu

@JohnRennie no..not by pythagoras theorem

it will be plus sign

barrycarter

@JohnRennie Oh, wait, since the journey has 0 distance for the ship, the time must be sqrt(36) or 6, which it is. Shiny!

John Rennie

19:39

@manshu the calculation of the proper time/length uses the Minkowski metric. That's where the minus sign comes from.

barrycarter

@manshu That's why relativity sucks, they can't define even define vector length properly.

John Rennie

I have to drag my sorry carcass off to bed now. I'm at work at six.

barrycarter

@JohnRennie Sleep tight.

OK, I'm fairly sure that I haven't done anything interesting with this metric yet.

manshu

ahh...I really need the mustache like John Rennie.

barrycarter

@manshu It's hard to tell if your av has one, but you could just paint it on.

manshu

19:43

no. it's against my principles

I was

I was an year ago.

barrycarter

I'm trying to figure out something interesting to do with the Minkowski metric.

Danu

Forget about it? :-)

manshu

haha]

barrycarter

@Danu I'm seriously tempted to agree.

ACuriousMind

@JohnRennie You work on Sundays? That sounds like the worst kind of semi-retirement! ;)

barrycarter

19:45

@ACuriousMind I think he once said he works 7 days a week. That damn British pluck.

(and yes, that was rhyming slang)

manshu

That didn't rhyme. You need some time alone with the rhyme.

barrycarter

@manshu That's not how rhyming slang works, berk ;)

Actually I did do it wrong.

manshu

See? I am not berk

barrycarter

Although "porky pies" does have the rhyme in it.

@manshu You're more of a merchant banker.

manshu

ah damn...I am offended again.

barrycarter

19:49

@manshu I was trying to find one you'd understand, Cupid.

manshu

@barrycarter ahhh...I think I now get how you are playing this game

barrycarter

@manshu Your turn, rummy!

Qmechanic

1

Q: Is the standard model a quantized gauge theory?

I have studied some quantum field theory and gauge theory but I am definitely not an expert. I am aware that in quantizing electrodynamics one has to fix a gauge. I have read that for general gauge theory there are additional complications that arise and require things like Faddeev-Popov ansatz o...

Too broad?

manshu

@barrycarter No..it's your, mass goal.

user54412

@ACuriousMind My anecdotal experience is that users will come up with close reasons for those anyway. (I say this as I'm trudging through 50 close reviews...)

barrycarter

19:54

@manshu Nice!

manshu

¯\_(ツ)_/¯

ACuriousMind

@Qmechanic Yes. The first question is a historical one, the second is trivially answered with "Yes.", and the third is a duplicate, as you point out; Neither of these three questions are related in a way that would require them to be asked jointly.

manshu

OMG...I just lost my hand...

barrycarter

@manshu There goes half your sex life

manshu

@Loong Thanks

@barrycarter I have already lived my half life.

after another half, I am expecting the end of the Earth

barrycarter

19:56

@manshu But you will continue to fade exponentially.

manshu

Yeah...Don't remind me that I am just a stick figure.

It hurts

barrycarter

@manshu I think you're a real stick...

@manshu Sticks and stones ARE your bones?

manshu

We differ in sizes and composition. If you are talking about me (the stick figure), then my bones are just sticks

Transcript for