The h Bar

Hari Prasad

19:16

@barrycarter what happened?

barrycarter

@HariPrasad What do you mean?

Hari Prasad

@barrycarter Help with relativity?

barrycarter

@HariPrasad Oh yes, can you help?

Hari Prasad

@barrycarter I'll try

what do you what to know?

barrycarter

@HariPrasad OK. I'm 10 light years from Earth and see Earth's clock reading 2000. Since I know that I'm 10 light years away, I assign that event x=10 and t=-10

Does that make sense so far?

Hari Prasad

19:18

@barrycarter Alright

go on.

barrycarter

@HariPrasad Then I instantly accelerate to 0.8c towards Earth. The Earth is now 6 light years away. I'm still seeing Earth's clock read 2000, but I now assign it t=-6 since I'm 4 light years closer. Good so far?

Hari Prasad

@barrycarter fine

barrycarter

@HariPrasad OK, so the same event went from 10,-10 in one frame to 6,-6 in another. Is that correct?

Hari Prasad

@barrycarter yes

barrycarter

@HariPrasad However, when I apply the Lorentz transform for 0.8c, I can't seem to get this answer.

ACuriousMind

19:22

@barrycarter What event exactly did you assign those numbers to?

barrycarter

@ACuriousMind Earth at the year 2000.

There is a Lorentz transform that converts 10,-10 to 6,-6, but it's not for 0.8c.

The gamma for 0.8c is 0.6, so this is very close to being correct, I think.

(or 1 divided by 0.6, whichever convention you prefer)

ACuriousMind

@barrycarter Okay. Why do you think your arbitrary assignment of 10,-10 should transform into 6,-6? Where you put the zero of time is arbitrary, what transforms meaningfully are time and space differences.

barrycarter

@ACuriousMind Thinking...

ACuriousMind

To expect the 10 light years transfrom into 6 lightyears is reasonable - that is the spatial difference between you and the Earth. But the assignment of -10 is arbitrary, it's not a time difference between two specific events.

barrycarter

I assign t=0 to when the Earth's year 2000 light rays hit my ship.

IE, the time between when the rays left Earth and the rays reached me.

ACuriousMind

19:29

Okay, that works. Now, why should that difference transform into -6?

barrycarter

Because I was seeing Earth at the year 2000, and I'm still seeing Earth at the year 2000. Since I'm now 4 light years closer, I must also assign Earth time to be 4 years later.

Note that any Lorentz transform will convert 10,-10 to x,-x for some value of x.

ACuriousMind

Wait a moment. What do you actually get when transforming 10,-10 by 0.8c?

Hari Prasad

@barrycarter I have something to say. The gamma for 0.8c is 1.6 not 0.6 as you stated earlier

barrycarter

@ACuriousMind 30,-30 if I use +.8 and 3.333, -3.3333 if you use -.8

@HariPrasad Sqrt of 1-v^2 ?

(or it's inverse, whichever you prefer)

ACuriousMind

@barrycarter Gamma is the inverse of that.

barrycarter

19:32

@ACuriousMind I think 3.333, -3.333 is the "more correct" answer, it just doesn't agree with what I think should happen

@HariPrasad OK, so it's 1.66666, (I see your point)

ACuriousMind

How do you get your "6 light years", then?

barrycarter

@ACuriousMind Lorentz contraction no?

physics.stackexchange.com/questions/246165/… (John Rennie's answer is what I'm quasi-working with)

Hari Prasad

@barrycarter Lorentz contraction no. for 0.8c is 0.59

barrycarter

@HariPrasad Well, 0.6, right?

ACuriousMind

@barrycarter Remember, Lorentz contraction is what happens to meter sticks, not distances.

The crucial point in the derivation of length contraction is that both end of the object that is being contracted travel at the same speed

barrycarter

19:36

@ACuriousMind Ummm, suppose I dropped a marker before accelerating. The marker was 10 ly away from Earth pre-acceleration and 6 ly away after acceleration, right?

ACuriousMind

In this case, you don't have such a distance - the distance is between you and the earth, which are now moving at unequal speeds.

barrycarter

@ACuriousMind Yes, you showed me a fascinating example of that earlier, but, in this case, I'm pretty sure the distance does contract. That's how the "faster than light" relativity rocket works.

ACuriousMind

What does "the distance contracts" even mean? How are you measuring it?

barrycarter

@ACuriousMind That's a good question, but I think we're getting a little metaphysical. Isn't it generally accepted that we can measure time and distance in any frame?

Presumably we bounce light beams off something and measure the time they take to return.

ACuriousMind

@barrycarter It's not metaphysical, saying how you measure things in relativity is crucial.

barrycarter

19:39

@ACuriousMind OK, but I'm saying we do have some definition of distance already, right?

ACuriousMind

@barrycarter Ah, but light takes time to travel. In this case, you will travel a significant distance until any beam you bounce off the earth returns to you

@barrycarter Well, locally, distance is what rulers measure.

Just like time is what clocks measure

The issue is how you do non-local measurement, such as this "distance between me and the earth" thing you're trying to do here

barrycarter

@ACuriousMind Hmmm... I think you've brought up a good point, but I think the point has been settled. You can measure distance and time between two events somehow.

ACuriousMind

You claim your Lorentz transform "should" give 6 lyrs, but you have thus far failed to demonstrate that. It's that thing with the Lorentz contraction all over again: You're getting an appearent contradiction because you expect the transfrom to do something that it doesn't.

barrycarter

@ACuriousMind You've pointed out the somehow might be quite interesting, but if we can accept a method exists, we can continue.

@ACuriousMind OK, good point. Let me rephrase: the event is: Earth at the year 2000 is the start of the event, my spaceship at 2010 is the end of the event. The light beam traveling between the two is the event. Is that OK so far?

ACuriousMind

No, because you're again using "event" to mean something other that "point in spacetime" :P

barrycarter

19:43

Sorry.

Hari Prasad

@ACuriousMind What does "Proper distance between two events" mean in GR?

barrycarter

I meant, that's the distan... yes

I meant that's two events, because we need two events to use the Lorentz transform.

ACuriousMind

@HariPrasad I would call the invariant interval between two spacelike separated events that.

barrycarter

But I see a problem already.

OK, but the problem doesn't fix my situation.

@ACuriousMind Let me tighten this up a bit. Event I: the Earth at the year 2000, Event II: a planet at the year 2010 (the planet from which I'm about to accelerate).

ACuriousMind

So, let's carefully state again what your situation is. You have the event "Light is emitted at earth at New Year's Eve in 2000", and assign it (10,-10), while you sit at (0,0). This is in a frame where you and the earth are at relative rest.

barrycarter

19:45

Yes, I'm good with that.

ACuriousMind

You then pass to the frame where something is at your location, but travelling at 0.8c. For simplicity that "something" is also you for the purposes of this discussion.

barrycarter

Yes, or we can use the planet I just invented :)

The planet is at rest w Earth, I am moving.

ACuriousMind

Okay

barrycarter

So... are you good with the original coordinates being (10,-10)?

ACuriousMind

Yes, that's fine.

barrycarter

19:48

Do you agree that ANY Lorentz transformation of 10,-10 will give x,-x for some value of x?

ACuriousMind

Yes, the factor is $\gamma(1-v)$.

barrycarter

Which actually gives the answer I want.

ACuriousMind

(Any Lorentz boost, btw, there are more Lorentz transformations than boosts ;) )

barrycarter

BUT there's a problem.

Right, but this specific case we are essentially measuring light travel time which must remain constant.

To compute x', I can't just multiply by gamma.

Divide by gamma.

Wait, why gamma(1-v) and not gamma(v)?

ACuriousMind

Oh, that's a multiplication. $\gamma\cdot(1-v)$.

barrycarter

19:52

gamma times (1-vt), right? for x' ?

Er, no. But not as simple as gamma times x (which would totally work here!!!!) or even gamma times 1-v, right?

ACuriousMind

A Lorentz boost is $t\mapsto \gamma(t - vx)$ and $x\mapsto \gamma(x-vt)$. If $x=-t$, then this becomes $t\mapsto \gamma(1+v) t$ and $x\mapsto \gamma(1+v)x$.

so the factor is $\gamma(1+v)$.

Messed up the sign there when doing it my head, sorry :P

barrycarter

OK, hang on, let me see if I can turn on mathjax

OK, one sec.

Yes, OK. In other words, they both get multiplied by the same number.

Obliv

Are there any other examples in math where we use an abstract identity like complex numbers? From what I understand, it's essentially doing this: $A*A = -1$ where we don't know what A is but we develop a set of imaginary numbers $C$ such that $R \in C$.

barrycarter

@ACuriousMind So... what happens?

ACuriousMind

@barrycarter What do you mean. Now that we've laid out the foundation, what is the problem?

barrycarter

20:00

@Obliv Well, the quaternions, obviously, but I think we were talking about those earlier.

@ACuriousMind What's the translation of (10,-10) into the 0.8c reference frame, and why?

ACuriousMind

@Obliv The procedure of adding elements obeying certain equations works for any ring, and is particularly prominent in what one calls "field extensions".

barrycarter

gamma times (1+v) here is 1.6666*(1+.8) which is 3 which doesn't seem to help much?

ACuriousMind

@barrycarter Well, what do you mean by "translation". You'll assign it coordinates (3,-3) in the 0.8c frame.

Again, what is the problem with that?

barrycarter

@ACuriousMind OK, I'm good with (3.333, -3.333) as the result of the transform, but that says the Earth is 3.33 ly away and -3.33 years from what I consider t=0, right?

Correction

The Earth at the year 2000 is 3.333 light years away and -3.333 years behind my t=0, correct?

ACuriousMind

@barrycarter No, that says when the earth emitted the light at the New Year's Eve of 2000, it was 3 lyrs away from you, and that that event happened 3 yrs ago according to your own clock.

barrycarter

20:05

(thinking...)

Obliv

@barrycarter Wait, I can understand that $a+bi$ is an extension of the $R$ by making a plane with the complex numbers. The i is defined as $i = (-1)^\frac{1}{2}$ how are the j and k defined in $a+bi+cj+dk$

barrycarter

@Obliv Instead of saying i = (-1)^(1/2), you can say i is a number that has the property i^2 = -1

@ACuriousMind If what you say is true, doesn't that mean I'm almost at Earth by the time I see the light from 2000?

@Obliv Notice that (-i)^2 is also -1

ACuriousMind

@barrycarter No. You see that light exactly at your (0,0), i.e. right where you are now, beside your planet.

(Recall that light rays are 45° degrees in all frames)

barrycarter

@Obliv One way of thinking about it. I'm taking the real numbers and adding i, j, and k to them to create a new, non-commutative group. Here's how I define the addition and multiplication of the numbers I've added.. (standard rules go here).

Obliv

@barrycarter Yeah because its defined through arithmetic of the real numbers. if $ijk = -1$ for quaternions, does that mean $j$ and $k$ are defined as $j = \frac{-1}{ik}$?

ACuriousMind

20:10

And I dropped a 0. ::sobs::

barrycarter

@Obliv Ummm, I don't think of it as defining numbers in terms of existing numbers. I think of it as adding new numbers with certain properties.

ACuriousMind

@barrycarter: The proper transform of (10,-10) into the 0.8 c frame is (30,-30)

barrycarter

@ACuriousMind OK, that's what @WillO also says, but that doesn't make sense. You can't have 10 light years convert to 30 light years in any time frame. Or 10 years to 30 years.

Obliv

@barrycarter >:( but then how can they be related at all to reality? Surely they have some connection to the real numbers in some way if it can describe reality.

barrycarter

@Obliv I'm a mathematician. I have no connection to reality whatsoever, sorry :)

ACuriousMind

20:11

@barrycarter Why not? You just asserted that, but it's obviously the result of the Lorentz transform.

Chris Okyen

I need a quick answer for something off topic from this room

Obliv

ask away

barrycarter

@ACuriousMind Because, after I accelerate, I am clearly 6 ly from Earth, not 30 ly.

Chris Okyen

SE takes too long to wait and for me I don't get them as quickly as many for whatever reason

ACuriousMind

@barrycarter Ah, but that 30 lyrs are the distance of the Earth 30 yrs in the past!

barrycarter

20:12

@Obliv But, if you insist, you can view them as x, y, and z vectors.

Chris Okyen

Obliv it is not about physics

ACuriousMind

In your frame, earth is moving with 0.8c towards you, after all.

barrycarter

@ACuriousMind You mean of my ship 30 years in the past?

ACuriousMind

@barrycarter In the 0.8c frame, the ship doesn't move. That's what "the ship's frame" is all about.

Obliv

@barrycarter What's the point of addressing the complex numbers as an axis of their own when you could just do that with another set of real numbers T_T

barrycarter

20:14

@Obliv Hmmm, you can think of it as another way of doing the same thing.

@Obliv Not everything in mathematics is unique or useful.

Obliv

@ChrisOkyen It's fine, this is a general chat anyway.

barrycarter

@Obliv We mathematicians sort of strive to create things that have no use.

@ChrisOkyen I love the way you want to ask a "quick question", but have now taken 4 sentences without asking :)

@ACuriousMind OK, but I still don't see how we can involve Earth 30 years ago when the problem shouldn't go outside Earth 10 years ago.

@Obliv Plus it simplifies the taking of cross products (not really, but if you don't like matrices, it does)

ACuriousMind

@barrycarter The "X years ago" here are w.r.t. different clocks! The 10 yrs ago are w.r.t. a clock a relative rest w.r.t. earth, the 30 yrs ago are w.r.t. to an onboard clock.

barrycarter

@ACuriousMind OK, it looks like you're agreeing with WillO. I understand what you're saying, but I still think the problem can be solved without going 30 years back in a reference frame I just entered.

ACuriousMind

Incidentally, look where the Earth is now - it travelled 30 yrs with 0.8c, which are 24 lyrs, which means it's now 6 lyrs away from you, which is exactly what length contraction means for the distance between the earth and your planet

Obliv

20:18

@barrycarter back into the textbook I go. I'm sure I'll internalize this abstraction eventually.

ACuriousMind

@barrycarter You're got "going back thirty years". You're just doing some math. Not every step in a calculation has physical significance.

barrycarter

@ACuriousMind No, no, I understand that the numbers work out with that method. I'm just saying there's a better way to get the same answer. @JohnRennie for example only considers the 22.5 year time period from 2000 to 2022.5

@ACuriousMind I totally understand what you and @WillO are saying. The reference frame you enter has always existed, you're just choosing to enter it at a given time.

ACuriousMind

@barrycarter I don't recognize "better" or "worse" ways. Either the reasoning is formally sound or it isn't.

barrycarter

@ACuriousMind However, I think this is a nonstandard treatment of the problem. There should be a treatment that doesn't go outside the scope of 10 light years and 22.5 temporal years.

ACuriousMind

@barrycarter That's an assertion for which I see no reason.

barrycarter

20:21

@ACuriousMind The reasoning gives the right answer, and it's curiously tempting because you can talk about reference frames existing forever even if there's nothing in them, but I believe that is a nonstandard way of looking at things.

ACuriousMind

You can't make up arbitrary restrictions and then complain that the math of relativity doesn't comply with them ;)

@barrycarter There is no "existence" to a frame. It's just a different choice of basis in a vector space.

barrycarter

@ACuriousMind OK, I know it's a lot to ask, but do you know of a textbook that does it that way? I'm sure there's a better way.

ACuriousMind

Attributing "existence" to a frame, now that is metaphysics!

@barrycarter Probably there is one, but I haven't read any textbooks on relativity.

barrycarter

@ACuriousMind But it has to exist for your answer to work. Otherwise, you can't say t=-30

ACuriousMind

@barrycarter The different bases of Minkowski space "exist" in the mathematical sense. But it's not a physical existence, frames are not things that have stuff "in them" or that "begin and cease to exist".

barrycarter

20:23

@ACuriousMind I believe there is a way to solve this problem without referring to reference frames in the past, and I further believe that this is the more standard way of doing things.

There must be a way to go from 10 light years away 10 years ago to 6 light years away 6 years ago without going through 30 years ago. @JohnRennie's diagram seems to show that.

ACuriousMind

"There must be a way"...why?

(I'm not saying there isn't, I'm trying to understand why you are insistent there is)

barrycarter

@ACuriousMind OK, that's a good question. I'm just convinced it can be done without having to refer to empty reference frames.

Obliv

How do you verify this formally: commutativity $w + z = z + w$ for all $w,z \in C$ ? I just said $w = a + bi$ and $z = c + di$ and then $a + bi + c + di = c + di + a + bi$ and I don't really know what to say at this point.

barrycarter

I believe the 3.333, -3.333 holds the answer.

@Obliv Do you have that a+bi = bi+a ?

Obliv

@barrycarter No, how do you even verify that? It seems like its a definition.

barrycarter

20:30

@Obliv Well, when 'i' is defined, were you given that it commutes additively with other numbers?

@Obliv In other words, do you know that r + i = i + r for all real numbers r?

ACuriousMind

@Obliv I am sure you defined the addition of complex numbers as $w+z = (a+c) + \mathrm{i}(b+d)$. Now you use commutativity of the addition of real numbers to write that as $(c+a) + \mathrm{i}(d+b)$ and observe this is exactly the definition of $z+w$. Hence $w+z=z+w$.

Obliv

@barrycarter No, the textbook says to verify, using the familiar properties of the real numbers, the associativity and commutativity of addition and multiplication.

barrycarter

@Obliv Yeah, do what @ACuriousMind says. The definition of addition should work. OK, I'm off for a bit. Maybe I really do need to find a bloody textbook ;)

Obliv

@ACuriousMind Yeah I forgot to look at the definition :x thanks

same I have to head to class now o/ ty for the help @barrycarter @ACuriousMind

barrycarter

20:53

@ACuriousMind OK, I think I figured it out. If we use the year 2000 as t=0, and the year 2010 as t=10, we can do all the conversions without having to use empty reference frames (I think)

(gone for a bit, but will test that "theory" later).

skill patrol

21:45

@ACuriousMind this is an amazing statement :P

Seriously?

ACuriousMind

Where have you been all the times that I've stated that I've read next to no textbooks?

skill patrol

Not here.

Pardon my absence :-)

Master :p

Relativity textbooks are the most numerous in all of physics, so they say.

I think I mentioned that^ before and you disagreed.

Now I know why ;)

42 minutes till 0celo7 is free to return

barrycarter

@skillpatrol Were you quoting @ACuriousMind in saying "this is an amazing statement :P", and, if so, where can I find the original statement?

ACuriousMind

@barrycarter He finds it amazing that I've never read a relativity textbook. You can see what people refer to when you click on the little arrow to the left of the @username.

barrycarter

Oh, got it! I thought that symbol meant he was quoting you.

skill patrol

22:01

There is a small arrow on the left of the message that points to the referrant

barrycarter

@skillpatrol Yes, I now understand!

skill patrol

*referent

Interesting, the OED does not recognize "referent" as a word :-/

At least in the printed version

barrycarter

22:26

Changing the t=0 to 2010 helps, but not enough.

0celo7

22:40

What a troll

@DavidZ :(

I'm an abused housewife

@DanielSank Nothing. Don't blame the victim.

0celo7

23:00

The feeling of needing 5 minutes to figure out something is nice

If it's instant, there is no satisfaction

If it takes hours, you just get mad

But a nice short bout of thinking feels good

FenderLesPaul

@BernardMeurer is that an option?

for you to attend that is

skill patrol

Welcome back @0celo7

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Welcome back @Slereah

0celo7

23:17

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ two very interesting new posts on my blog since I was last here

I'm improving some proofs in Hawking-Ellis

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Cool

0celo7

with proper notation for one

and their proof of lemma 4.5.2 leaves out a LOT of details

not details that the reader could even reasonably fill in

skill patrol

These guys live on a higher plane :P

0celo7

@skillpatrol or they were lazy :P

skill patrol

Indeed.

FenderLesPaul

23:23

@0celo7 dude

how do we have the same shoes

0celo7

what to prove next

@FenderLesPaul great minds and all that jazz

FenderLesPaul

I like jazz

0celo7

what other shoes do you have

skill patrol

More than blues?

0celo7

let's see

two pairs of gray Nikes, one pair of black Nikes, black Polos, gray Polos, tan Polos, white Jordans

Bernard Meurer

23:24

@FenderLesPaul No, because I haven't been accepted yet

0celo7

the tan shoes I showed you

Bernard Meurer

and to be very honest I don't think I will

0celo7

Nike sandals, Jordan sandals

gray Jordans

FenderLesPaul

@BernardMeurer keep your head up dude you can't know until it comes around

Bernard Meurer

It did make me feel better today when the UPenn guy who interviewed me sent me an email from his personal address saying he was sad I didn't get in and that I should give him a call whenever I'm around the east coast

@FenderLesPaul I only got into one good school so far man :/

0celo7

23:26

@BernardMeurer you're a fuck

FenderLesPaul

one good school is better than none

I would suggest looking into possibilities of taking loans

Bernard Meurer

@FenderLesPaul One good school I can't afford :p

0celo7

I know you got into UCSD and VT

both are good

FenderLesPaul

if you're doing computer science or engineering with the prospect of going into industry

Bernard Meurer

@0celo7 UCSD is the good one

FenderLesPaul

23:26

then loans aren't the worst thing

Bernard Meurer

VT is alright

FenderLesPaul

if you're doing physics or math I would advise against it

0celo7

@BernardMeurer VT is top 15 for engineering

Bernard Meurer

@0celo7 yeah but their projects are just so stupid, at least the undergrad stuff they do in CS/CEng is just silly

0celo7

what?

Bernard Meurer

23:28

@FenderLesPaul I've been considering loans, but even so I can't get a 50k loan every year for 4-5 years

0celo7

what projects

50K a year??

FenderLesPaul

@BernardMeurer you'd have to pay 50k a year for UCSD?

wtf??

Bernard Meurer

@0celo7 Sorry, by projects I mean undergrad 'research'

FenderLesPaul

I have to pay 10k a year for Cornell, what kind of bullshit is that 50k a year

Bernard Meurer

@FenderLesPaul Yep

0celo7

23:28

@BernardMeurer what?

do they not work for grad students?

FenderLesPaul

@BernardMeurer ok I would advise against UCSD

Bernard Meurer

@FenderLesPaul students.ucsd.edu/finances/financial-aid/budgeting/…

FenderLesPaul

if you have to pay 50k a year it's definitely not worth it

they didn't give you any financial aid?

did you apply for aid?

user54412

@FenderLesPaul That's cheaper than average.

FenderLesPaul

or can you not get any because you're international?

Bernard Meurer

23:30

@FenderLesPaul You have scholarships?

FenderLesPaul

@ChrisWhite it's cheaper on an absolute scale but typically the university offers financial aid to drastically lower the tuition

but I guess that's mainly for domestic students

Bernard Meurer

I don't have any aid from the UC system because I'm international

0celo7

How does one define a solid cone in a vector space?

FenderLesPaul

@BernardMeurer scholarships + financial aid

user54412

@FenderLesPaul Only ivies to make themselves look good

Bernard Meurer

23:31

being an international student is a piece of shit, everyone treats you like a bag of cash

user54412

I know plenty of people who can't go to colleges they get into because their middle class families can't afford it.

FenderLesPaul

Damn that blows

0celo7

My family couldn't afford Ivy League.

DanielSank

@ChrisWhite That's hard to believe. They can't go to any college or they can't go to a college they think is worth the money?

FenderLesPaul

I had no idea

user54412

23:32

Also, it's probably worse for international. I don't know if the US government will pay for them, but that's basically the only game in town. US banks are basically prohibited from making college loans anymore.

Bernard Meurer

But yeah, right now I can either:
1. Get into UWaterloo
2. Get into Coimbra or ULisboa in Portugal
3. Go work and move on

Guess I'll wait and see

FenderLesPaul

I have confidence that you'll get into Waterloo

wait are you Canadian?

user54412

@DanielSank They can't go to many colleges. There's always some, but the list of 65k+ schools is a lot more than just ivies nowadays.

FenderLesPaul

or do you live in Brazil?

Bernard Meurer

@FenderLesPaul I'm Brazilian in Brazil

DanielSank

23:33

@ChrisWhite We're talking about Ivies?

Sorry.

FenderLesPaul

So will you have to pay a lot more for Waterloo than a Canadian student?

Bernard Meurer

@FenderLesPaul Yeah, 30k CAD year

0celo7

@FenderLesPaul It is possible for a family to make $300,000+ a year but not have money for college

DanielSank

@ChrisWhite Ohhhhh, I see, you're saying people are academically capable of colleges they can't afford.

FenderLesPaul

@0celo7 that's true I don't disagree

Bernard Meurer

23:33

which is WAY better than the 54K at UCSD or 45k at VA Tech

0celo7

@BernardMeurer So like 15USD a year.

user54412

@DanielSank yes

FenderLesPaul

I wouldn't have been able to afford Cornell without the crazy financial aid and scholarships they gave

and my family is upper middle class

so I can relate

Bernard Meurer

@0celo7 23k USD

0celo7

@FenderLesPaul My FAFSA is fucked up

FenderLesPaul

23:34

ok 23k is definitely better than 50k

and Waterloo is a really good school for engineering

so it wouldn't go to waste

Bernard Meurer

@FenderLesPaul Yeah, Cornell is need blind ~passionate sigh~

They have the IQC

the IQC is the reason I applied there in the first place

FenderLesPaul

Yeah the IQC is pretty dope

I can't wait to visit

Man the tuition in the US really is an issue

Bernard Meurer

I don't know if I can even look at the building while being an undergrad, but anyway, no place better to find out if I do want to go for QC after all

FenderLesPaul

I had to turn down some schools I really wanted to go to because they didn't give me any aid at all

when I was applying for colleges

:/

Like University of Chicago

Bernard Meurer

If I still had economic support from my parents I'd go fro UCSD

I really liked what I saw from the school

FenderLesPaul

23:37

I was actually going to major in math, with interest in algebraic topology, had I gone to UChicago

@0celo7 we could've been compatible!

0celo7

@FenderLesPaul what

user54412

@BernardMeurer Yeah. The problem is China. With the rise of an upper middle class that never existed there (numbering in the millions), many near-broke schools in the US have realized they can charge 70k when parents are desperate to get their kids a US education at all costs, thus allowing them to reduce the cost to locals. Of coarse, one looks bad making rules for Chinese people explicitly, so they charge all international students.

FenderLesPaul

you know

0celo7

I'm a dirty engineer

FenderLesPaul

because you like math a lot

0celo7

23:38

Speaking of math

@FenderLesPaul can you look at a proof of mine?

I don't think it's very elegant

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Yup, the percentage of Chinese international students has doubled.

Bernard Meurer

@ChrisWhite Yeah I can see that, I don't know this whole thing left me really 'destimulated' with higher education

FenderLesPaul

@0celo7 ok

0celo7

FenderLesPaul

what is the break point?

0celo7

23:41

@ChrisWhite you're an astro person, and astro people know random things

@FenderLesPaul where $\dot\gamma$ does not exist

FenderLesPaul

ok

user54412

"random things"

0celo7

it should say

"lie in the same half"

FenderLesPaul

yeah I figured

looks good to me

@0celo7 omg we have a heart on snapchat

0celo7

@FenderLesPaul so it makes sense that along a geodesic, $\sigma$ is decreasing

Bernard Meurer

23:42

@ChrisWhite I got me some chocolate coffee beans yesterday btw

user54412

@BernardMeurer :(

0celo7

so along ANY timelike curve, it's decreasing right at the start

user54412

@BernardMeurer :)

Bernard Meurer

@FenderLesPaul You have a heart with @0celo7 on snap and his girlfriend doesn't hahaha

0celo7

@BernardMeurer :(

Bernard Meurer

23:43

@ChrisWhite Yeah, idk what i'll do, it'll be alright though

@ChrisWhite They are pretty damn good

0celo7

the heart disappeared a few days ago :(

:(

@ChrisWhite why, in elastic scattering in QM, is $|\mathbf{p}|$ conserved, but $\mathbf p$ itself isn't

FenderLesPaul

@0celo7 actually is $\sigma(\gamma(t))$ defined by replacing the upper bound of $\sigma(p)$ (which you have as $1$) with $\gamma(t)$?

user54412

@BernardMeurer Don't make decisions until you have all the information. List your options. Weigh the pros and cons. Then think outside the box for more options, and weigh their pros and cons. Take your time, and let chocolate coffee beans be your friend.

FenderLesPaul

@BernardMeurer yaaay

0celo7

@FenderLesPaul mmmm, well, if $\gamma$ is a geodesic, then you replace the $1$ with $t$

Bernard Meurer

23:46

@ChrisWhite Thanks, I'll do just that :)

0celo7

so it's $\sigma(\gamma(t))=\int_0^t \langle\dot\gamma,\dot\gamma\rangle\,\mathrm{d}t'$

FenderLesPaul

ok well $\sigma$ is negative

0celo7

yes

wait a moment

FenderLesPaul

so it makes sense that it's decreasing as $t$ increases

Bernard Meurer

Has anyone here watched a movie from 87 called Brazil (which has nothing to do with Brazil)

0celo7

23:47

$\sigma(\gamma(t))=t\langle\dot\gamma,\dot\gamma\rangle$

because that thing is constant for a geodesic

user54412

@0celo7 I have no idea what the context is, but doesn't conservation of E immediately imply conservation of |p|, where E conservation is the definition of "elastic"? Also, if vector p was conserved, would anything interesting have happened?

0celo7

@ChrisWhite I know why |p| is conserved

I want to know why strict momentum conservation doesn't hold

@ChrisWhite nothing interesting would have happened, no

So why does interesting stuff happen?

FenderLesPaul

@0celo7 $\sigma(\gamma(t)$ is not constant for a geodesic it's $\langle \dot{\gamma},\dot{\gamma}\rangle$ that's constant for a geodesic

whereas $\sigma$ is the integral of that quantity

oh you have a $t$ out front

sorry I missed the $t$

yes that quantity is constant

0celo7

@FenderLesPaul And then it's clear that the derivative of that is always negative

FenderLesPaul

yes

0celo7

23:52

and since the derivative at $q$ agrees with that of the other one

the other one is always negative too!

FenderLesPaul

mmhm

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