The h Bar

barrycarter

4:00 PM

@Gustavo6046 Hello!

John Rennie

@barrycarter the centre of the wheel

barrycarter

@JohnRennie OK, the torso now.

John Rennie

It's true!

Gustavo6046

Isn't here some kind of feed that displays the latest questions posted over Physics over the course of a minute?

barrycarter

@JohnRennie Yes, but trivially so!

Gustavo6046

4:01 PM

I didn't see one scrolling up-page.

barrycarter

chat.stackexchange.com/rooms/info/71/the-h-bar?tab=feeds

@Gustavo6046 I see one at the top of this page, not sure if its RSS.

John Rennie

@Gustavo6046 right now I'm seeing a popup for the question "Pulley inertia momentum of inertia". Is that what you meant?

Gustavo6046

what I meant is, a feed that shows the latest questions on the chat.

barrycarter

@JohnRennie I once answered a question re how close would the moon have to be in order to have an epicircular orbit instead of the current one which doesn't cross its path.

Gustavo6046

Like the one in the 2nd Monitor.

barrycarter

4:02 PM

@Gustavo6046 A chat feed would be fantastic.

Hmmm, you can feed the starred chat items.

John Rennie

@Gustavo6046 I don't know of anything like that

ACuriousMind

Uh...there is a feed in the upper left corner?

And there's the bot that posts meta questions directly into the chat

Gustavo6046

@ACuriousMind not for questions... at least I don't see one. And I meant non-meta questions.

barrycarter

@ACuriousMind I think he wants it the other way around. Chat questions, not site questions.

Gustavo6046

Also, this is so far the best place I could find for my modelling-related questions.

ACuriousMind

4:04 PM

@Gustavo6046 It pops up at irregular intervals, not sure what determines those

barrycarter

@Gustavo6046 Oh, you mean site questions? I see one at top, it's URL is chat.stackexchange.com/rooms/info/71/the-h-bar?tab=feeds

ACuriousMind

I think we purposefully didn't configure the bot to post them to chat because no one wants to see all the homework questions :P

barrycarter

I think there are some magic settings where you can turn it off.

John Rennie

@barrycarter the moon is tidally locked - no points in/on it move in an epicycle type motion

barrycarter

@JohnRennie No, I meant if you look at the moon's orbit from the Sun or other "stationary" point. It looks nearly circular. If the moon were a lot closer it would intersect its own path.

Gustavo6046

4:05 PM

Is there another place to post my modelling questions?

@barrycarter Thanks

barrycarter

@Gustavo6046 Fashion modelling?

Gustavo6046

no

John Rennie

@barrycarter aha, got you. That's quite a fun question for a rainy afternoon.

Gustavo6046

I mean, models like Wavefront .obj

barrycarter

I was teasing :) Could you be more specific?

Gustavo6046

4:06 PM

and stuff like the Unreal model file format specification.

barrycarter

@JohnRennie It never rains here, but yes.. I answered it, but I don't remember the answer.

Gustavo6046

I just posted a related question. But... I'm not sure this is the right place.

barrycarter

@Gustavo6046 You could try [stackoverflow.SE]

Gustavo6046

Stack Overflow is a code site!

barrycarter

What's the magic code word for stackoverflow?

John Rennie

4:07 PM

There is a game development SE somewhere ...

Gustavo6046

John Rennie: oh of course!

barrycarter

@Gustavo6046 There's also a graphics artists site.

Gustavo6046

I'm already logged to it

but I had no idea it could have someting to do to modelling,

barrycarter

@Gustavo6046 Stack Overflow will answer file format stuff.

John Rennie

however I doubt the sort of topics you suggest would be suitable here.

Gustavo6046

4:08 PM

This is the question:

I'm voting to close this question as off-topic because it is not about physics but about computer graphics file formats. — Sebastian Riese 54 secs ago

barrycarter

Ever since the great breakup, there are many sites for each purpose.

Gustavo6046

ops

what I mean is the question, not the comment in it :P

barrycarter

@Gustavo6046 It's about imaginary physics, which I personally love, but...

Gustavo6046

could at least move it?

0

Q: How do I convert between polygonal texture coordinates and vertex texture coordinates?

I'm not sure this is the right place to ask this question, but the 3D Graphics (or whatever it's called) site on SE isn't ready already, so please move there once it's done. :) Wavefront .OBJ models take texture coordinates per polygon, which I even think is right. However, according to the Unre...

coordinate-systems

barrycarter

I'd still take a shot at posting it on stackoverflow.com

John Rennie

4:10 PM

@Gustavo6046 If you click the flag link under your question you can leave a message for the moderators asking if they will migrate it.

Gustavo6046

ok, thanks.

barrycarter

@JohnRennie Oh, you can flag a question for migration?

John Rennie

One of the flag options is to leave a message for the moderators. There's no specific "migrate" flag option, but obviously the message for the moderators can say anything.

barrycarter

Cool. I always used comments to suggest migration.

Good to know I can bug the mods directly.

John Rennie

@barrycarter presumably you just require the Moon's orbital speed around the Earth to be greater than the Earth's orbital speed round the Sun. That way there will be a period of retrograde motion when the Moon is on the side of the earth far from the Sun.

barrycarter

4:17 PM

@JohnRennie Where were you when I needed you? :P

@JohnRennie You could refine that slightly by noting that R^3/T^2 is a constant for a given star system.

John Rennie

Kepler's nth law - can't remember the value of n

barrycarter

6

Q: Moon's orbit around the Sun

The Earth revolves around the Sun and the Moon revolves around the Earth. Out of curiosity I started thinking about the orbit of the Moon around the Sun and expected (assumed) it to be as follows: But on Wikipedia and some other sites I found out that the orbit is actually like this: I have...

orbit moon sun earth natural-satellites

user507974

guys, is there a general name for the family of techniques where when its difficult to resolve small changes you do some kind periodic signal injection and it provides fine scale phase information that helps increase precision (i think my description may be a bit off)

barrycarter

That's the limiting case (I'm surprised it let me put that in)

@user507974 You mean like convolution or something?

user507974

@barrycarter I guess convolution is kind of a analagous to what happens mathematically but I guess I'm talking more in experimental physics here

barrycarter

4:22 PM

@user507974 All physics is math, but OK. You want to inject this into the observed data, right? Or into an experiment itself?

user507974

yea, kind of if I understood it right

barrycarter

That was a 2-choice question, not a yes-no question :)

You want tweak data or the experiment itself?

user507974

experiment signal (which is observed data technically)

barrycarter

@user507974 So that really is math. Any reason you don't want to apply mathematical techniques to it?

user507974

not really, just wanted to know if there was any names it went by but convolution probably is a good description of it

barrycarter

4:26 PM

@user507974 WP it and see what happens, you might find other words/terms for it

user507974

WP?

barrycarter

In physics, wherever there is a linear system with a "superposition principle", a convolution operation makes an appearance. For instance, in spectroscopy line broadening due to the Doppler effect on its own gives a Gaussian spectral line shape and collision broadening alone gives a Lorentzian line shape. When both effects are operative, the line shape is a convolution of Gaussian and Lorentzian, a Voigt function.

Wikipedia

Convolution

In mathematics and, in particular, functional analysis, convolution is a mathematical operation on two functions f and g, producing a third function that is typically viewed as a modified version of one of the original functions, giving the integral of the pointwise multiplication of the two functions as a function of the amount that one of the original functions is translated. Convolution is similar to cross-correlation. It has applications that include probability, statistics, computer vision, natural language processing, image and signal processing, engineering, and differential equations. The...

Voigt profile

In spectroscopy, the Voigt profile (named after Woldemar Voigt) is a line profile resulting from the convolution of two broadening mechanisms, one of which alone would produce a Gaussian profile (usually, as a result of the Doppler broadening), and the other would produce a Lorentzian profile. Voigt profiles are common in many branches of spectroscopy and diffraction. Due to the computational expense of the convolution operation, the Voigt profile is often approximated using a pseudo-Voigt profile. All normalized line profiles can be considered to be probability distributions. The Gaussian profile...

Gosh, I love doing that.

Secret

This
$$\int_{-\infty}^{\infty}\sqrt{K-\sqrt{B}t_3}e^{-t_3^2}dt_3$$
Is what happens when I convolve a gaussian and a maxwell boltzmann to fit some data

barrycarter

@Secret Do you just need the answer, or do you want to find a method to solve it?

Secret

That is because the 'constants' are actually
$$K=E-E_0+0.5$$
$$B=kT$$
and E is our variable

barrycarter

4:30 PM

@Secret By using Mathematica :)

@Secret Question though. A convolution generally results in a function. What you have there will result in a number. Is that what you want?

@JohnRennie Are you prepared to revisit our friends Bob and Carol?

John Rennie

@barrycarter oh all right then :-)

barrycarter

@Secret But neither have a dependency on $t_3$?

@JohnRennie OK, Bob accelerates away from Carol at a slow rate... long-term, what does Bob see?

Secret

(sorry accidentally overwrite the previous messsage by mistake: (In principle I need the answer, but I want to learn how to compute these in general (e.g. by hand and by recognising special functions)

@barrycarter, yes E has no dependence on $t_3$

barrycarter

@Secret The fact that you're integrating from -Inf to +Inf makes things easier. You're not trying to find the anti-derivative in general right?

John Rennie

@barrycarter Bob observes a spacetime geometry described by the Rindler metric.

barrycarter

4:34 PM

@JohnRennie OK, in English please?

John Rennie

The Rindler metric is hard to explain in words, but look at it this way ...

Secret

I only need the $\infty$, $-\infty$ rsult because I am just evaluating the convolution and hoping something in closed form pop out since our QR least square fit algorithm is still not working

John Rennie

Suppose Carol gives Bob a head start of $x$ metres then shines a light ray at him.

barrycarter

@JohnRennie I'm trying to get you to say something you said before that made no sense to me, but OK.

Secret

thus unable to check whether the curve has fitted correctly numerically

John Rennie

4:36 PM

Nothing can move faster than light light, so the light ray must eventually catch up with Bob. Right?

barrycarter

@Secret Have you considered a power series approximation?

@JohnRennie Correct.

@JohnRennie In fact, if Bob remained at a constant velocity, this would be the thought experiment to derive the formulas for SR.

@JohnRennie For Bob, the light ray must arrive in x/c time (assuming bob's standing still at the start)

John Rennie

@barrycarter But no! Bob's velocity tends asymptotically to $c$, and if Bob has a head start of $c^2/a$ or greater the light ray will never reach Bob. More precisely the distance between Bob and the light ray tends asymptotically to zero as time goes to infinity.

Secret

@barrycarter I could try that. It wasn't the first thing that pop into my mind since I tend to use approximate expressions last, trying to retain an expression as symbolic as possible if it can

barrycarter

@JohnRennie We might be talking about different types of acceleration. I'm talking about Bob accelerating with reference to his current frame (I can explain that more if that makes no sense)

@Secret I agree it's really ugly. I'd pop it into WA or something and see what you get. The answer find help you find the solution.

Secret

PS earlier in MSE, it was said the integral can be computed. However I don't understand how because I never came across bessel K functions before nor that exp(cx^(2/3)+x)^2 integral
http://math.stackexchange.com/questions/1752956/how-to-integrate-int-infty-infty-sqrta-bxe-x2dx-and-why-integrat#comment3577142_1752956

John Rennie

4:40 PM

@barrycarter Me too. The $a$ is the constant acceleration Bob measures in his rest frame i.e. he feels as if he is in a gravitational field producing a force on him of $F=ma$, where the $m$ is Bob's mass.

In Carol's frame she observes Bob to be accelerating at $a/\gamma^3$.

barrycarter

@JohnRennie OK, so the reason the light ray never reaches Bob is because Bob is accelerating and so light no longer has to travel at c towards him?

Secret

John Rennie

@barrycarter so far I'm viewing everything from Carol's frame, and in Carol's frame Bob's velocity tends asymptotically to $c$.

Secret

@barrycarter I got the following
It seemed to suggest the solution is in terms of besselK(x^2) but I am not really sure especially I don't know how they are computed

barrycarter

@JohnRennie OK, that makes sense...

John Rennie

4:43 PM

But his velocity approaches $c$ fast enough that the light ray cannot reach him IIF he has enough of a head start. The head start required is $c^2/a$ metres.

Ok now let's switch to Bob's frame.

barrycarter

@JohnRennie But his approach to the speed of light is asymptotic. You're saying it's asymptotic fast enough that the light ray never catches up.

John Duffield

@barrycarter : my pleasure Barry. Methinks I'll write a new one. One that draws upon my experience and also talks about the trouble with physics.

John Rennie

@barrycarter Yes, exactly.

barrycarter

@JohnRennie I am having trouble accepting this. Can you provide a rough reason? I can run some math on it.

@Secret Integrate[Sqrt[K - Sqrt[B*t]]*Exp[-t^2],t] yields no result on Mathematica.

John Rennie

@barrycarter there's a proof in:

0

A: What is the proper way to explain the twin paradox?

Appendix - why the Rindler metric? After reading my answer you could be forgiven for feel a bit cheated because it all depends on my claim that the accelerating twin observes a spacetime described by the Rindler metric not the Minkowski metric and I did kind of pull this out of the air. It isn’...

Secret

4:46 PM

WA is weird when it came to integrating symbolic expressions. Mathematica can handle that problem but my mathematica has expired and it will need a few weeks before it arrive

John Rennie

See the last section of the answer I've linked. Search for Proof that you can outrun light if you can't see it.

barrycarter

@JohnRennie OK, I am still in disbelief, let me check my own numbers.

@Secret I'm running the infinity integral now.

Secret

ok

John Rennie

@barrycarter you're about to discover why the Rindler metric is so weird :-)

barrycarter

@JohnRennie I'm going to cheat, hang on

@JohnRennie v = c th(aT/c) = at / sqrt[1 + (at/c)2] from the relativistic rocket means Carol sees Bob's speed as at / sqrt[1 + (at/c)2]?

John Rennie

4:50 PM

@barrycarter err yes, I think that's the correct equation though I would need to check to be sure.

barrycarter

@JohnRennie But it's not the tanh() one right?

John Rennie

@barrycarter Correct

barrycarter

@JohnRennie I'm not happy about this, hang on while I prove what you proved in the question/answer.

John Rennie

The tanh equation is the one that has Bob's time as the argument

barrycarter

@JohnRennie I'll accept for the moment that you're right, and that Carol's light beam will NEVER catch Bob, at least not in Carol's frame. Now, Bob's frame?

John Rennie

4:52 PM

I'm a bit surprised you're working with the equation for the velocity. I would use the equation for distance and subtract $ct$.

barrycarter

That would actually work better, good point.

I'm running @Secret's integration and don't want to disturb Mathematica, but will do that in a sec.

John Rennie

@barrycarter OK. If light emitted from a distance $c^2/a$ away from Bob can never reach him that means there is an event horizon at that distance. Because an event horizon is defined as a surface from which light can never escape.

barrycarter

@JohnRennie OK, but I'm more interested in what Bob actually sees.

John Rennie

Just to clarify, are we taking about seeing or observing ?

barrycarter

@Secret $
\frac{K \left(\left(\sqrt[4]{-1}-i\right) B \Gamma \left(\frac{3}{8}\right)
\,
_3F_3\left(\frac{1}{8},\frac{3}{8},\frac{7}{8};\frac{1}{2},\frac{3}{4},\frac
{5}{4};-\frac{K^4}{B^2}\right)+K \left(\left(\sqrt[4]{-1}+i\right) K \Gamma
\left(\frac{7}{8}\right) \,
_3F_3\left(\frac{5}{8},\frac{7}{8},\frac{11}{8};\frac{5}{4},\frac{3}{2},\frac{7}{4};-\frac{K^4}{B^2}\right)-2 \left((-1)^{3/4}+i\right) \sqrt{B} \Gamma
\left(\frac{9}{8}\right) \,
_3F_3\left(\frac{3}{8},\frac{5}{8},\frac{9}{8};\frac{3}{4},\frac{5}{4},\frac

@JohnRennie Let's do both. Seeing = what's entering his eyeballs, observing = he accounts for light speed travel time?

Secret

4:57 PM

OMG I think I am seeing some kind of hypergeometric functions here

In terms of my problem, I am not sure if matlab has hypergeometric functions

In terms of curiosity, I sometimes really wonder how Mathematica compute all these crazy integrals

barrycarter

@Secret Yes, it's the HypergeometricPFQ function.

John Rennie

@barrycarter I can't do seeing in Rindler coordinates because light doesn't travel at a constant velocity and calculating the trajectory of a light ray becomes a formidable problem. The best I can do is observing i.e. assigning events to points in spacetime.

barrycarter

@JohnRennie So, Bob can say where/when something happened, but not when he'll actually SEE it happen?

Secret

http://au.mathworks.com/help/symbolic/hypergeom.html
Hmm, it seems we are good, let's see how this thing works...

John Rennie

It's possible to calculate when Bob will actually SEE something. What I'm saying is only that the calculation is too hard for me.

So I have to restrict myself to the observing side of things.

barrycarter

5:01 PM

@JohnRennie OK, close enough

John Rennie

Good. The reason I went on about the light ray catching Bob, or failing to, is that Bob's spacetime contains an event horizon at a distance $c^2/a$ from him.

Anything that happens farther away from Bon than $c^2/a$ (in his coordinates) is behind an event horizon and can never affect him.

barrycarter

@JohnRennie OK... I'm trying to trick you into saying something weird, and then point out that Bob can stop accelerating and then we're back in SR?

@JohnRennie OK, fair enough. So what is he observing in terms of Carol's clocks and meter sticks?

I realize he can never actually see those things as long as he continues accelerating.

John Rennie

@barrycarter Correct. Switch back to Carol's frame for a moment ... if Bob stops accelerating it means that all those light rays that were chasing him can now catch him. So the event horizon has disappeared.

barrycarter

@JohnRennie Right, but I'm hoping to show that they do in such a way that things are weird.

John Rennie

Starting at time zero Bob observes carol accelerating away from him at an initial acceleration of $-a$ ...

barrycarter

5:06 PM

@JohnRennie Right. Or observes, since he can never see light from Carol again.

@JohnRennie The question we're doing now is a little different from my initial setup, but I'm guessing they will relate?

John Rennie

So for early times Carol's velocity increases linearly with time. However Carol's acceleration slows, then reverses i.e. Bob observes Carol to start slowing down again.

barrycarter

@JohnRennie That's where I'm having the issue. When Bob is "near the speed of light" compared to Carol, why isn't Carol receding at light speed (observation wise)?

John Rennie

As Carol approaches the event horizon at $-c^2/a$ Bob observes Carol's velocity to tend asymptotically to zero. So Bob observes Carol to take an infinite time to reach the horizon.

It is in fact very similar to when someone falls into a black hole.

barrycarter

@JohnRennie Well, wait. The light beam Carol shone can reach the initial point where Bob started, right?

John Rennie

ah my phone's going ...

barrycarter

5:12 PM

You're on your phone?

John Rennie

yes, sorry ...

barrycarter

Hmmm, interesting how your phone went just as I asked the tough question ;)

John Rennie

Ok the call is ended though they might ring back. It's someone who is paying me to code for them, so I'm afraid they get priority :-)

You still there?

barrycarter

@JohnRennie Yup :)

@JohnRennie I thought you meant your battery was dying.

John Rennie

Ok, suppose at time zero carol starts sending pulses of light towards Bob, one a second.

barrycarter

5:19 PM

@JohnRennie OK, as Carol's velocity decreases, does it decrease fast enough that her distance (to Bob) asymptotes to a point?

John Rennie

does it decrease fast enough that her distance (to Bob) asymptotes to a point Eh?

barrycarter

@JohnRennie You said Carol's velocity decreases from Bob's view point, correct?

John Rennie

Yes, but it doesn't change sign. Carol doesn't speed away then reverse and come back

barrycarter

@JohnRennie Correct, but if the velocity goes to 0 fast enough, its integral, the distance, may also asymptotically approach a single value.

John Rennie

@barrycarter aha, yes, exactly. Carol's distance from Bob (in Bob's coordinates) tends asymptotically to $c^2/a$.

Which you may remember is the distance to the event horizon.

barrycarter

5:22 PM

OK, wow. But, if Bob stops accelerating, Carol's velocity jumps back up to .9999 c or whatever?

John Rennie

So Bob observes Carol to approach the horizon but slow as she approaches it and take an infinite time to reach it.

@barrycarter yes

barrycarter

@JohnRennie OK that was the part that confused me. Even if Bob is accelerating at 1m/s^2, his observations change drastically if he stops accelerating.

John Rennie

@barrycarter yes

barrycarter

@JohnRennie This makes me unhappy, but OK.

John Rennie

But at low accelerations the horizon is so far away i.e. $c^2/a$ is so big, that it wouldn't be noticable.

barrycarter

5:24 PM

@JohnRennie Right, but suppose we don't give Bob a head start... same thing would happen, right?

John Rennie

Suppose Carol starts firing light rays at Bob one a second

The first ray released at time zero hits Bob instantly because Bob and Carol are at the same point.

barrycarter

Right.

John Rennie

The second light ray takes longer to reach Bob, obviously, the third longer still and so on.

barrycarter

If Bob is accelerating slowly, many of the first light rays will hit Bob, OK.

You're saying, at some point, the light rays will never reach Bob?

John Rennie

By the nth light ray Bob has a big enough head start that the nth light ray cannot reach him.

So for all times $n$ and later Carol is behind the horizon

Anthony

5:27 PM

Is there a way to derive the quantization of light as $hf$, or is it simply assumed?

I assume we can solve the schrodinger equation to get it, I just don't think I've seen it anywhere.

barrycarter

@JohnRennie So we're in the same situation. Bob observes Carol approach a single point, her velocity slowing to 0?

John Rennie

@barrycarter Yes

barrycarter

@JohnRennie And if he stops this extremely gentle acceleration, suddenly, she's gone from 0 to near c in a fraction of a second?

John Rennie

@barrycarter yes. I know it seems odd, but you can justify it by examining what happens to light rays fired from Carol to Bob in Carol's rest frame.

All that's really happening is a change in whether the light rays can reach Bob or not.

barrycarter

@JohnRennie I think maybe that's part of my unhappiness with SR. We ignore light travel time. In this example, we're having Bob observe events that he can never ever see.

John Rennie

5:31 PM

What it means is that Bob's coordinates (while he is accelerating) form a chart that does not cover all of spacetime.

In fact this is a fantastically important observation because this is a common occurrence in general relativity

barrycarter

@JohnRennie Oh! You're saying that, while Bob is accelerating, the + - infinity on his x coordinate are finitely far away in Carol's frame?

John Rennie

@barrycarter other way round. An infinite Carol-Bob distance in Carol's coordinates maps to a distance $c^2/a$ in Bob's coordinates.

barrycarter

@JohnRennie Wait. Bob can still assign coordinates to anything in the universe, right?

John Rennie

Actually I think I'm talking rubbish here.

Bob's coordinates do extend to $\pm\infty$ but part of the spacetime is behind the horizon.

barrycarter

@JohnRennie Can I ask you about simultaneity next?

John Rennie

5:34 PM

Quick change of subject there! Are we done with Bob and Carol?

barrycarter

@JohnRennie Meaning, if Bob assigns coordinates greater than c^2/a, there is nothing in the universe that can have those coordinates?

@JohnRennie Oh, I thought you'd given up on them based on your rubbish (English: trash) comment.

John Rennie

Remember that the Rindler coordinates are static so they assume Bob has been accelerating for an infinite time and will continue to accelerate for an infinite time.

barrycarter

@JohnRennie OK, but as Bob accelerates, there is some finite limit to the x values he can assign to events in spacetime?

A limit that decreases asymptotically to c^2/a ?

John Rennie

Yes, nothing that is, or ever has been, causally linked to Bob can ever get farther away from him than $c^2/a$.

barrycarter

@JohnRennie Oh, but things he has never seen can get further away?

John Rennie

5:39 PM

Well Bob's coordinate system extends past $x = -c^2/a$. Bob can assign coordinates to things farther away but he can never know what exists for $x < -c^2/a$.

barrycarter

Do you need to see (not just observe) a spacetime event to assign it coordinates?

Anthony

I'll ask one more time- is the quantization of light postulated independently from the schrodinger equation, or can we use the schrodinger equation to show it?

barrycarter

@Anthony I don't think anyone's ignoring you, we just don't know the answer :)

John Rennie

@Anthony hi Anthony, I'm afraid we're in the middle of an argument about relativity and there doesn't seem to be anyone else around. Have you searched the site? I'm sure quantisation of light has been asked about before.

Anthony

@JohnRennie I'll look some more, I've been fishing around. It's cool, I thought there was someone else in here, sorry.

John Rennie

5:41 PM

@barrycarter the question is whether something is causally linked to Bob. That is, can an event have any effect on Bob.

If an event has no causal link to Bob then how can Bob ever know that event exists?

barrycarter

@JohnRennie Right, so if Bob observed a star 10 light years away, the star 10 years ago is linked?

John Rennie

Yes. Photons, gravity waves, cosmic rays, whatever emitted by the star could reach Bob and have some effect on him.

barrycarter

@JohnRennie That was my question. Does relativity allow Bob to assign coordinates to events he has never seen and will never see? Perhaps through an intermediary observer (or does that violate causality too?)

John Rennie

@barrycarter if/when you read John Duffield's book you'll realise this is John's issue. Events that are behind a horizon cannot have any effect on an observer outside the horizon. Not ever. Not via an intermediary or in any other way.

So what does it mean to say those events exist?

barrycarter

@JohnRennie Events can exist in spacetime even if a given observer can't observe them, right?

@JohnRennie OK, let's cruel things up a bit. Bob know that, standing some distance behind Carol, is Dorothy, his ex. However, he accelerates away fast enough that he can never see Dorothy. Can he still assign her (Dorothy) space-time coordinates?

John Rennie

5:45 PM

Right now we can't see farther away than 13.8 gigalight years, because the universe is 13.8 billion years old. Yes?

barrycarter

@JohnRennie Sure...

John Rennie

But there's no reason to suppose the universe just stops at 13.8 gigalight years

We believe there is stuff farther away even though we can't observe it

barrycarter

@JohnRennie Ah, but we have no way of knowing because everything else is accelerating too fast?

John Rennie

Ignore for a moment dark energy (if you don't know what I mean by that just ignore it!)

barrycarter

OK... I think you're going big to answer my question, but OK

John Rennie

5:47 PM

then if we wait we can see more of the universe. When it's 100 billion years old we'll be able to see stuff 100 billion light years away.

barrycarter

OK... not sure that makes sense to me. You're saying stuff spewed out of the Big Bang faster than the speed of light?

John Rennie

No. I'm saying that since light travels at one light year per year in the 13.8 billion years since the Big bang it has only travelled 13.8 billion light years

So if something is 13.9 billion light years away the light from it hasn't had time to reach us yet.

That's all - nothing freaky.

barrycarter

@JohnRennie Oh, and the Big Bang occurred everywhere (by definition?)

John Rennie

Yes. OK I see what concerned you - you were wondering how there could have been something that far away.

barrycarter

@JohnRennie OK, but here's the rub: everything in the universe started out 0 distance from everything else right? Now, some objects have gotten 14 Gly away from each other in less than 14Gy time, is that what you're saying.

John Rennie

5:51 PM

Oh God, why did I start this :-)

I'll discuss the Big bang if you want, but that's a distraction.

barrycarter

@JohnRennie I can go back to the original question if you'd like :)

John Rennie

But look, all I'm saying is that right now there are parts of the universe causally disconnected from us because they are too far away.

barrycarter

@JohnRennie OK, but I'm saying: if every part of the universe touched us at one point, how did they get that far away?

John Rennie

@barrycarter that's a singularly interesting subject :-) But it is a distraction ...

barrycarter

@JohnRennie LOL :) I do appreciate the joke :) OK, so we can pretend a whole large universe just came into existence?

John Rennie

5:54 PM

The point I was trying to make is that causally disconnected isn't the same as doesn't exist

barrycarter

@JohnRennie Ah! I see what you're saying now. So Bob can assign spacetime coordinates to causally disconnected events, even though he can never see them.

John Rennie

There are bits of the universe causally disconnected from us but that doesn't mean we don't believe they exist

@barrycarter Bingo!

barrycarter

@JohnRennie So, we have near-infinite speed changes and assigning coordinates to unviewable events... why the hell not ! :P

John Rennie

The reason I make a big deal about this is that there are people who claim the inside of a black hole doesn't exist because it's causally disconnected.

barrycarter

Actually, they would be unobservable too, since these are events Bob could only speculate on.

John Rennie

5:56 PM

You have a book by one of those people :-)

barrycarter

@JohnRennie Yes, yes :) I will read it with an open mind, but not an empty mind :)

John Rennie

This is the key point ...

barrycarter

@JohnRennie Also, Bob could speculate on an event and assign it spacetime coordinates, and then change his speed to check if that event occurred.

John Rennie

Bob can never observe (or see) Carol reach the horizon at $x = -c^2/a$.

But for Carol the horizon doesn't exist. The distance Carol measures between herself and Bob extends smoothly to infinity as time goes to infinity.

barrycarter

I think I get it. I mean, mapping an infinite space to a finite space is quite easily mathematically.

Secret

5:58 PM

@barrycarter Super crazy function is now inserted into matlab. Awaiting it to finsih computing

John Rennie

@barrycarter yes, in general the maths involved in relativity, even GR, is straightforward. It's the concepts that strain the brain!

barrycarter

However, Carol can still see and observe events that Bob will never see, and, if Bob had some way of knowing about that event, he could assign it a space time coordinate?

@Secret Oh, I actually made that up, sorry ;)

Secret

...

John Rennie

Yes. Carol can release a light ray that will never reach Bob. So by definition Carol is behind the horizon when she releases the light ray.

barrycarter

@Secret Just kidding! But B and K are positive, right?

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