The h Bar

Secret

6:00 PM

They are

kT cannot be negative, so is E-E0+0.5

barrycarter

@Secret Then you're fine. Any other restrictions on B and K?

Secret

nope

barrycarter

@JohnRennie Right, but Bob can still assign "Carol 1 hour after I left, in her time frame" a spacetime coordinate, even though it's not one he can see while accelerating.

John Rennie

Yes

Secret

I also breifly read about the hypergeometric function, it seems to act like some kind of array (as in, it contains the raw materials needed to build nearly any function)

John Rennie

6:01 PM

And that event has a physical significance

barrycarter

@Secret For mathematicians, it's more about selecting objects from bins.

Slereah

OH WAIT

barrycarter

@JohnRennie The significance that Bob can never see it until he stops accelerating?

Slereah

I guess perturbation theory can't tell me the stress energy tensor at the scattering event

barrycarter

Did I say "bins"? John is rubbing off on me.

John Rennie

6:02 PM

@barrycarter Yes

Secret

Indeed, $_pF_q$ looks more like a cabinet to me and the Pochhammer symbols are like the cabinets to be choose from

Slereah

Since the states are only defined for $t \approx \pm \infty$

(times $(1+i\varepsilon)$)

barrycarter

@JohnRennie OK, now that you've totally screwed up my worldview on constant acceleration, want to take a jab at simultaneity?

Slereah

So I guess that is indeed a thing for non-perturbative methods

John Rennie

@barrycarter I'll have a go, but simultaneity is generally not a useful concept in relativity

barrycarter

6:04 PM

Falling and rising factorials

In mathematics, the Pochhammer symbol introduced by Leo August Pochhammer is the notation (x)n, where n is a non-negative integer. Depending on the context the Pochhammer symbol may represent either the rising factorial or the falling factorial as defined below. Care needs to be taken to check which interpretation is being used in any particular article. Pochhammer himself actually used (x)n with yet another meaning, namely to denote the binomial coefficient . In this article the Pochhammer symbol (x)n is used to represent the falling factorial (sometimes called the "descending factorial", "falling...

@Secret There is quite a relation between them, yes.

@JohnRennie Then how come it gets mentioned like a zillion times on this site?

John Rennie

@barrycarter it doesn't get mentioned by us GR experts (I use the term expert advisedly :-) it get's mentioned by newbies who think it matters.

barrycarter

"You forgot about the relativity of simultaneity" <- oft-heard quote

John Rennie

@barrycarter right, that phrase means simultaneity has no absolute meaning

barrycarter

@JohnRennie Hmmm... I would really prefer to solve relativity problems with that cool metric you gave me, but even you said that it was a first step, not a tool to solve problems.

John Rennie

You can do everything, that is everything starting from the metric ,though as you found with acceleration you may have to walk a long, long way from that starting point.

barrycarter

6:07 PM

@JohnRennie Yes, me need shortcuts. Actually, maybe I can make you answer the question I really need an answer to.

John Rennie

Go on then

barrycarter

@JohnRennie Instead of asking all these subquestions to try to figure it out myself.

@JohnRennie [drum roll]

John Rennie

The audience hold their breath

barrycarter

@JohnRennie Bob starts $s_0$ away from Carol (in Carol's frame), traveling at $v_0$ and accelerating constantly at $a$. Give me formula for how Carol and Bob see each other's clocks, distance, and velocity.

skill patrol

::gasp::

John Rennie

6:09 PM

The equations in Carol's frame you know, because they are just the relativistic rocket equations.

barrycarter

@JohnRennie I've been try to solve this by discrete approximations, having Bob drop beacons, etc.

@JohnRennie Ah, but not exactly, because Bob doesn't start at 0 velocity.

@JohnRennie The distance I'm pretty sure is just a simple addition.

John Rennie

@barrycarter yes but that's just fiddling with the origin (in Carol's frame)

barrycarter

@JohnRennie The distance, yes. The velocity, not so much.

IE, the non zero start velocity is nontrivial, I think

John Rennie

Well choose an inertial frame in which Bob starts from rest, derive the equations in that frame and Lorentz transform to get back to Carol's frame. Messy but straightforward.

barrycarter

Unless... I create a third observer that's moving at $v_0$ and not accelerating...

@JohnRennie Hmmm, wish I'd thought of that earlier.

John Rennie

6:12 PM

But why would you do that? Sticking needles into your eyeballs is more fun and probably more productive.

barrycarter

@JohnRennie Didn't you just suggest that?

John Rennie

I suggested how you could have Bob start at $v \ne 0$, but I still think it's a pointless complication

barrycarter

@JohnRennie Right, but if Bob DOES start at non-zero v, I still need to solve the problem. Oh great, now I suppose you want the question that inspired the question I asked you.

1

Q: Special Relativity as Applied to an Interstellar Starship?

I'm once again confused by the twin paradox. Let's say I am on an interstellar starship flying at 0.6c from a star 30 light years away from Earth to a star 50 light years away from Earth and 40 light years from each other. How would you estimate the difference between the time on the ship when yo...

It's basically a multi-stage problem. Accelerate, coast, decelerate.

And the other thing I want to do is add in "eyeball times", when we determine what each observer actually sees, and, finally "mirror times"-- how long would it take to exchange a message with the other observer.

John Rennie

Solving that problem is easy and doesn't require all the stuff I've been blabbing about for the last two hours.

barrycarter

@JohnRennie Please, be my guest!

I've been struggling with it for weeks.

John Rennie

6:17 PM

Work in the Earth frame. Ignore Bob's fame as we don't need to work in it.

barrycarter

OK, but we can get to it later, right?

github.com/barrycarter/bcapps/blob/master/STACK/…

That may or may not help. I derived formulas for each leg of the journey.

John Rennie

The trajectory of the first stage, the acceleration is given by the relativistic rocket equations. So we know $s(t)$, $v(t)$ etc. Yes?

$t$ is Earth time.

barrycarter

@JohnRennie Yes, good so far.

John Rennie

The elapsed time on Bob's clock is just the length of the trajectory calculated by using the (flat space) metric to integrate along the curve.

So we now what time Bob's clock shows for any point on the curve.

barrycarter

@JohnRennie Wait... that part confuses me.

@JohnRennie Are you using the magic metric we talked about before?

John Rennie

6:21 PM

Hmm, I wonder how far back we need to go with this.

the metric I'm talking about is the Minkowski metric diag(-1,1,1,1).

barrycarter

@JohnRennie The one you showed me earlier, dx^2 - dt^2 in this case, right?

John Rennie

Yes

barrycarter

OK, so you believe that, in this case, it IS ok to use that metric to solve this problem?

John Rennie

Yes. Carol's frame is inertial and in flat spacetime so her spacetime geometry is described by the Minkowski metric.

barrycarter

@JohnRennie Right, but dx^2 - dt^2 should be the same for both, right?

John Rennie

6:23 PM

That's why we are going to work in Carol's frame, because it's simple.

barrycarter

@JohnRennie OK. The equations for the accel leg are easy, as you point out. Pretty much also for the coast leg. The decel leg is where it gets ugly.

dmckee

^ The very key to doing relativistic calculations.

John Rennie

@barrycarter See this is where the low cunning of us physicists comes to the fore.

The deceleration step is the time reversal of the acceleration step.

barrycarter

@JohnRennie Wait.

@JohnRennie I just saw it.

@JohnRennie Symmetry.

dmckee

@JohnRennie I was about to type the same thing.

barrycarter

6:26 PM

@JohnRennie Same equations just backwards in time with a distance increase.

John Rennie

@dmckee Like I said, the low cunning of the experienced physicist :-)

dmckee

low cunning is what keeps us employed.

John Rennie

@barrycarter careful, you're entering the twilight zone

It had never occurred to me that corrupting the purity of mathematicians could be such fun ;-)

5

barrycarter

@JohnRennie Actually, that's what I ended up doing it turns out. I guess my big unhappiness is that the formulas are insanely ugly and there should be a more general formula.

John Rennie

Insanely ugly? That's a bit harsh ...

barrycarter

6:28 PM

OK, fairly ugly?

John Rennie

Well it's usually the case that differential equations are much prettier than their solutions. That's the universe for you.

barrycarter

Right. But I mean: the relativity rocket equations aren't too bad. But once you start time reversing and time shifting them and stuff, they get seriously ugly. If I could figure out a one time formula for s0, v0, a (with v0 possibly being negative), I could plug and chug.

Obliv

what are the operands/symbols here? $a\equiv b ~mod~n$ the $\equiv$ means congruence, what does mod n mean?

barrycarter

Plus I could come up with "new" formulas for what the eyeballs see and mirror bouncing time.

@Obliv a=b mod n means that b-a is a multiple of n

@Obliv it's really say a =mod n= b

@Obliv The mod n is part of the congruence.

Obliv

are those equal signs $\equiv$ or =?

barrycarter

6:31 PM

@Obliv They're what you typed, i'm too lazy, but I've seen a = b (mod n) too

dmckee

@JohnRennie Kind of the converse of

abstrusegoose.com/105

2

John Rennie

@dmckee :-)

Secret

@JohnRennie Based on what we know about relativistic kinemetics, is it correct to say that given two relativisitc objects A and B moving pass each other and that B is at rest wrt lab frame while A is not. then, in the lab frame A is seen to have some mass and momentum E but in A's rest frame, this translates into A perceiving itself to have a higher rest mass?

John Rennie

It's a shame that abstruse goose gave up.

barrycarter

@JohnRennie Plus, I guess I sort of like deriving some of the formulas myself using SR and derivatives.

@Secret I'm not even sure if the great @JohnRennie can sort that question out.

John Rennie

6:34 PM

@barrycarter given that the trajectory is in three chunks there isn't going to be a nice clean equation for the whole journey.

dmckee

@Secret Absolutely not. A will always measure it's mass as $m_A$.

John Rennie

It's always going to be a messy stitching together of different equations.

barrycarter

@JohnRennie No, but if I could come up with reasonably good formula for s0, v0, and a, I could apply it pieceweise.

@JohnRennie It seems a shame to solve for this special case, when a general solution would be so much more useful.

John Rennie

The general equation is $a' = a/\gamma^3$. Have fun integrating that one.

barrycarter

@JohnRennie I have the POWER of Mathematica :P

John Rennie

6:36 PM

$a'$ is $d^2x/dt^2$ in Carol's frame.

barrycarter

You meant 2, I hop e:P

John Rennie

Off you go then. Make $a$ an arbitrary function of $t'$ or if you prefer $t$.

barrycarter

@JohnRennie Hmm.. that might be one way to go, though I'd still need to figure out constants My thought experiment was having Bob accelerate in discrete intervals.

@JohnRennie Now I suspect treachery. Is that function known to be non-integrable?

dmckee

@barrycarter Well what John proposes is just the limit of that with short intervals, after all.

barrycarter

(and before anyone says it, I mean closed-form non-integrable, of course)

John Rennie

6:38 PM

Well it's obviously integrable for constant $a$ because the results are the relativistic rocket equations.

dmckee

It's just that it can get unreasonably hard very, very quickly.

Secret

@dmckee I see, looks like I mixed up relativistic mass and rest mass again

John Rennie

But start making $a$ variable and it's going to get messy.

barrycarter

@JohnRennie True, although they have special initial conditions that might make or break the integrability.

Obliv

what is meant by "differing by an integral multiple of n"? Isn't it integer multiple of n?

dmckee

6:39 PM

@Secret This is why we keep telling you there is only one mass and it is a Lorentz scalar.

barrycarter

@Obliv It means 17 = 5 (mod 12), because 5 and 17 are 12 apart.

dmckee

It really is easier to do most problems in that frame of mind.

barrycarter

@Obliv Also, 29 = 17 = 5 = -7 (mod 12)

Secret

agreed

John Rennie

@Obliv See:

Modular arithmetic

In mathematics, modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value—the modulus (plural moduli). The modern approach to modular arithmetic was developed by Carl Friedrich Gauss in his book Disquisitiones Arithmeticae, published in 1801. A familiar use of modular arithmetic is in the 12-hour clock, in which the day is divided into two 12-hour periods. If the time is 7:00 now, then 8 hours later it will be 3:00. Usual addition would suggest that the later time should be 7 + 8 = 15, but this is not the answer because clock time "wraps...

Obliv

6:40 PM

vzn linked me that yesterday I understand the concept I just didn't know you could say an "integral" multiple

Secret

Meanwhile, crazy function is STILL calculating

barrycarter

@Obliv That's syntactic sugar. Just say "multiple"

Obliv

well now I know what the % operator in CS means mathematically I think

John Rennie

I keep seeing really interesting questions flashing past while I'm stuck here trying to teach physics to a mathematician ...

3

barrycarter

@JohnRennie Ooooooooh!

@JohnRennie Right up to the neck now, cowboy.

John Duffield

6:44 PM

@JohnRennie : what issue? If you're accelerating through space you might think there's this big black Rindler horizon behind you, but everybody else will later assure that what was behind you was space and stars as normal. You couldn't see them, that's all.

barrycarter

@Obliv That is the mod operator, yes.

Obliv

@barrycarter is this right? $a \equiv b$ if $a~ mod~ n = b~ mod~ n$?

manshu

If we have an object in hemispherical bowl at the horizontal position and we are to find the velocity at the lowermost position, then do we take into account the rotational energy too?

John Duffield

@JohnRennie : we can see things 46 billion light years away. When cosmologists say 13.8 billion light years they're referring to light travel time, ignoring the way space has expanded whilst the light has been in transit.

barrycarter

@Obliv Yes, that's correct. When we say a (mod n) we mean the number b such that a = b (mod n) and b is between 0 and n-1

The event horizon goes away once Bob stops accelerating though. It's not gone forever.

John Rennie

6:47 PM

@manshu it depends on the question. The question will usually say something like the ball rolls without slipping

barrycarter

@Obliv So saying a = b (mod n) and a (mod n) have slightly different meanings.

manshu

@JohnRennie it says that the mass slips

John Rennie

@manshu do they specify a friction coefficient between the ball and bowl, or do they say it is frictionless?

manshu

It's frictionless

John Rennie

If it's frictionless there is no torque on the ball so it cannot start rotating. In that case just consider translational kinetic energy.

manshu

6:51 PM

But the object is revolving around the center of the hemisphere. Then why shouldn't we consider rotational motion?

John Rennie

@barrycarter I know it's not of interest to you personnally, but understand the event horizon in Rindler coordinates is key to understanding the event horizon of a black hole. They are very closely analogous.

The difference is of course that you can't just turn off the acceleration with a black hole.

barrycarter

@JohnRennie I'm sick to death of Mr Rindler... wasn't he a dragon in the old AdventureLand game?

John Rennie

@barrycarter pass

barrycarter

I'll find a formula... I'll find one I tell you! Fools! I'll show you all!

manshu

@barrycarter Be nice :p

John Rennie

6:54 PM

The thing with relativity (both flavours) is that with enough work there's a point where everything clicks and it becomes intuitive.

Learning all these pesky geometries is an essential part of the process

Even though it can be tedious at times.

barrycarter

@JohnRennie Oh yeah, someone once said relativity was a geometric theory, though I can't remember who.

John Rennie

Obviously someone with a remarkably deep understanding of the area

skill patrol

ACM

barrycarter

@skillpatrol It's @JohnRennie's go to phrase.

Except I don't really believe in geometry. Not enough numbers. That's why they invented trig.

skill patrol

where do you think trig came from?

barrycarter

6:56 PM

@skillpatrol It was a virgin birth.

skill patrol

no motivation

John Rennie

Algebraic geometry? Proof if any were needed that mathematicians can take a simple idea and turn it into something so complicated that even other algebraic geometers can't understand them.

barrycarter

Geometry is what happens when you give mathematicians too much paper.

@JohnRennie Rotations are closer to linear algebra, though.

@JohnRennie And you can prove many (all?) of the triangle identities using trigonometry.

@JohnRennie And it's theoretically possible that trig will yield an identity that could not be discovered by geometry alone.

skill patrol

have you looked at The Elements?

Obliv

when it says there are precisely $n$ distinct equivalence classes mod $n$, namely $\bar{0}, \bar{1}, ... , \overline{n-1}$ and these residue classes partition the integers $\mathbb{Z}$ how can it partition $\mathbb{Z}$ when the union of these congruence classes do not equal the entire set $\mathbb{Z}$?

John Rennie

6:59 PM

@barrycarter Sorry, I dozed off for a moment there. What were you saying?

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