The h Bar

Sir Cumference

01:40

oh hey winterbash started

2

John Rennie

05:27

Hats! :-)

3

Slereah

09:35

morning

The Last Airbender

09:58

2

Q: Work done while compressing an ideal gas (The physical significance of $\int dPdV$)

Today in our chemistry class we derived the pressure-volume work done on an ideal gas. Our assumption was that $$P_{ext}=P_{int}+dP$$ so that all the time the system remains (approximately) in equilibrium with the surrounding and the process occurs very slowly (its a reversible process). Now $$W_...

yh05

@JohnRennie I understand why getting a measurement from experiment that is different from what is expected from a scientific law does not necessarily disprove the law.

I'm curious how the laws are tested experimentally to higher degrees of precision in general. And since laws are commonly expressed as mathematical equation with various variables, how do we repeat the experiments to cover as much possible values of the variables as possible? Is it impossible to test the law under all possible values for the variables (there would be infinitely many tests)?

Slereah

It is impossible, yes.

There are infinitely many different physical models we can come up with to fit experimental data

To narrow it down we use a variety of heuristics

Which are, if not truthier, at least more practical to use

John Rennie

@yh05 in principle getting a measurement from experiment that is different from what is expected from a scientific law does disprove the law.

In practice we always worry that there might be some experimental error in our measurement.

Slereah

10:14

Well that's part of the whole bit of scientific law that nobody ever teaches in class

The ~correspondence rules~

Also the whole business of predictability and epistemology and all that nonsense

John Rennie

Unless we're lucky and we get a measurement that is blatantly different usually something gets disproved gradually by multiple measurements that disagree.

Slereah

In practice we don't have enough informations about the initial conditions of our experiment, nor the means to compute it

We don't know the position and momentum of every particle in the universe

So we only get a degree of confidence

We just have a few informations on what we hope is relevent

But in real experiments, 1) we only have limited control and knowledge of the experimental setup (ie we can't prevent the temperature from varying somewhat) and 2) we don't necessarily know everything relevent to the measurements

A good example is that whole experimental error of the LHC

For a while at the LHC there was a weird measurement discrepency happening every day around the same time

Turned out that it was caused by a train passing nearby

John Rennie

Re your second question, there are usually key features of a theory and we test those first. If our experiments agree with the key features then experimenters may or may not test the more subtle predictions or they may lose interest.

Slereah

There's also the whole LIGO business being disturbed by trucks a few miles away~

Hm

I can't find that train for the LHC business again

what was it

radioprotection.unblog.fr/2010/09/11/les-tgv-perturbent-le-cern

Ah there it is

This is probably the best thing to show if someone ever wonders why it's a good idea to wait a while before interpreting experimental results

yh05

10:34

Where can I read about how scientific laws are tested?

Slereah

Depends how you want to read about it

skillpatrol

Scientific law

Scientific laws or laws of science are statements, based on repeated experiments or observations, that describe or predict a range of natural phenomena. The term law has diverse usage in many cases (approximate, accurate, broad, or narrow) across all fields of natural science (physics, chemistry, biology, geology, astronomy, etc.). Laws are developed from data and can be further developed through mathematics; in all cases they are directly or indirectly based on empirical evidence. It is generally understood that they implicitly reflect, though they do not explicitly assert, causal relationships...

yh05

I read it. I'm looking for a detailed description of how scientific law is tested.

skillpatrol

which law?

Slereah

10:50

I don't really know any standard texts on the topic but look into epistemology and whatever else

It's a mix of rather abstract philosophy and practical concerns sort of deal rly

yh05

I can't find any relevant one with google search. =(

Slereah

The ideal case is that you have an experimental setup. You input all the initial conditions that you are using for that experiment, your theory gives you what you should get in return. You perform the experiment, and the results from the experiment should match up with what you predicted.

this is rarely the case due to a variety of reasons

For instance :

1) Measurement errors : the way you measure both the initial conditions and the final ones is unlikely to be perfect, due to
1a) Systematic errors : Errors that always happen during the measurement (for a start, no measuring apparatus has infinite precision, the influence of the measuring apparatus on the value itself, etc)
1b) Random errors : Due to our lack of perfect knowledge, we can only get fairly broad measurements (ie : temperature) and this is likely to change in a random manner during the experiment

skillpatrol

@yh05 have you seen this?

Slereah

A somewhat common experimental error that I have heard about is

The scientist himself

I have heard of a few experiments in which the source of error was the scientist approaching the setup to read the results

because the scientist's body emitted heat, reflected light/sound/neutrons, etc

those are hard things to predict

yh05

11:18

It is not possible to test a law under all possible values for the variables present in the law. How do we tackle this problem? Can I have a detailed explanation, please?

John Rennie

11:46

@yh05 there isn't a single detailed explanation. It depends on what the law is. If you can come up with a specific example then we could discuss how it has been tested, but there isn't going to be an answer that applies to all laws.

Novalium Company

12:57

omg

let's make an AI where it will decode what a person on a keyboard is typing by just recording the audio where it will have learned with first listening and looking at what the person is typing (supervised, labeled data)

#hacking

#bless

Slereah

that's been done already

Slereah

13:40

Aaron Stevens

@Slereah Then I would point out that a negative second derivative is not sufficient to lower the badness.

Slereah

You can decrease the rate of rising of the badness

skillpatrol

a president did come up with a new proof of the Pythagorean theorem

Slereah

13:57

also

"U.S. President Richard Nixon, when campaigning for a second term in office announced that the rate of increase of inflation was decreasing, which has been noted as "the first time a sitting president used the third derivative to advance his case for reelection.""

JMac

@Slereah I assume he didn't explicitly talk about "derivatives" and instead talked about them as rates of change.

Aaron Stevens

@Slereah Right, but I would much rather the first derivative actually become negative

skillpatrol

that was an explicit demo of a jerk using the jerk to advance his case for reelection...

Aaron Stevens

63

Q: Ball hits curve of same curvature

I was doing some physics problems for homework and, while procrastinating, I came up with a theoretical scenario that I couldn't figure out the result of. The following is from a side view and in a frictionless, ideal environment: A ball moves toward a wall at a constant velocity. At the...

homework-and-exercises newtonian-mechanics forces classical-mechanics

come on, one more vote

JMac

@AaronStevens Saw it on HNQ the other day and made sure to cast mine.

Aaron Stevens

14:04

@JMac It is a great physics question. It's just the way it has been asked makes it off topic.

JMac

@AaronStevens Besides not showing any analysis, the diagrams uses comic sans. It deserves a custom close reason for that.

PM 2Ring

@skillpatrol Garfield's proof is ok, but it's not a major breakthrough. It's essentially an older proof cut in half. ;) But I will admit I thought it was cute when I first encountered it, several decades ago.

Paul Painlevé

Paul Painlevé (French: [pɔl pɛ̃ləve]; 5 December 1863 – 29 October 1933) was a French mathematician and statesman. He served twice as Prime Minister of the Third Republic: 12 September – 13 November 1917 and 17 April – 22 November 1925. His entry into politics came in 1906 after a professorship at the Sorbonne that began in 1892. His first term as prime minister lasted only nine weeks but dealt with weighty issues, such as the Russian Revolution, the American entry into the war, the failure of the Nivelle Offensive, quelling the French Army Mutinies and relations with the British. In the 1920s...

skillpatrol

did you catch my Nixon comment?

11 mins ago, by Slereah

"U.S. President Richard Nixon, when campaigning for a second term in office announced that the rate of increase of inflation was decreasing, which has been noted as "the first time a sitting president used the third derivative to advance his case for reelection.""

7 mins ago, by skillpatrol

that was an explicit demo of a jerk using the jerk to advance his case for reelection...

Aaron Stevens

@JMac On an unrelated note, two users were removed, and someone removed the accepted status on one of my answers, so I have lost 45 reputation points fairly quickly. I think I will make a meta post complaining that someone is out to get me

PM 2Ring

@skillpatrol Yes. I was trying to come up with a good pun involving the word checkers

Aaron Stevens

14:09

@skillpatrol Too bad Nixon was not talking about the third derivative of position

skillpatrol

:D

Aaron Stevens

:)

JMac

@AaronStevens That is definitely a government conspiracy. They are out to get you where it hurts - imaginary internet points.

Aaron Stevens

@JMac I shouldn't expect fairness on a site that punishes critical thinking

JMac

@AaronStevens I think we need a banner on top of the site that reminds users that they will be punished for critical thinking. It would really clear up a lot of confusion.

PM 2Ring

14:12

@AaronStevens Closed.

Aaron Stevens

@PM2Ring Finally... haha

@JMac I have been raising custom flags on posts that use too much critical thinking. For some reason they keep getting declined though

skillpatrol

RIP

JMac

@AaronStevens It's because the moderators and experienced users work together to make sure new users like yourself have a bad time.

Aaron Stevens

@JMac No, they are threatened by my critical thinking and insight

skillpatrol

et al

JMac

14:34

@skillpatrol Facebook and youtube are the two key weapons in fighting the heliocentric enthusiasts who poison our children's minds with things like "outer space" and "science".

skillpatrol

indeed

skillpatrol

15:18

@JMac imho, youtube has more positive applications for the heliocentrics than facebook

JMac

@skillpatrol Youtube has a lot of anti-flat Earth stuff. I'm sure FB has some; but the format doesn't work the same.

skillpatrol

yeah, and twitter is just plain "fake news"

yuvraj singh

@PM2Ring, hi sir.

I gone through the same experience again but this time a different one.

Aaron Stevens

@yuvrajsingh Haven't there been issues of pinging users before?

skillpatrol

15:35

let's not be overly "critical" here :P

Slereah

let's

skillpatrol

's

Slereah

$\mathbb{R}^n$ has an irreducible representation of $O(n)$

Koray

Hi may I ask a question about physics here?

Slereah

But how many reducible representations does it have

Is it also one

John Rennie

15:39

@Koray yes, of course

Slereah

The representation with $n$ trivial groups

Koray

Thank you :) I hope this question makes sense. According to Einstein's equivalence principle, which one has an acceleration: an object free falling to a non-atmosphere planet, or an object staying still on the surface of the planet?

Slereah

The first one travels along a geodesic while the second does not

John Rennie

@Koray you have to define what you mean by acceleration, and you have to do it in an observer independent way i.e. a way that all observers can agree on.

Slereah

The second one is the only one that has an invariant definition of acceleration

(it's the acceleration produced by the electrostatic interaction with the earth)

Koray

15:41

I'm talking about the same acceleration mentioned in the twin-paradox

John Rennie

@Koray typically we use the proper acceleration as this is a well defined quantity that all observers can agree on.

@Koray the twin paradox isn't really about acceleration.

Koray

I'm asking this to understand, if the travelling twin turn using gravitational pull, would there be a difference.

Slereah

@JohnRennie It kinda is

John Rennie

If you're interested in understanding the twin paradox I wrote an answer on the main site specifically to explain exactly what causes the twin paradox (it isn't really a paradox of course).

Koray

Sir I've seen your answer and I is great. But I could not able to understand it. I'll try harder. Thank you very much..

It is great I mean :)

John Rennie

15:44

@Koray it doesn't matter whether the travelling twin turns round by using a rocket or by slingshoting round a massive object.

Koray

Sir thank you very much again, I really appreciate it. Have a good day :)

user8718165

Hello :) @JohnRennie

John Rennie

@user8718165 hi :-)

yuvraj singh

@AaronStevens didn't, t get you in what context u are referring.

PM 2Ring

@yuvrajsingh You had trouble with dacoits again?

yuvraj singh

15:59

@PM2Ring not completely.

Now I am OK, I thought to share with you.

Physics Meta

16:19

0

Q: On Winter Bash 2019 page

Winter Bash is on! I am very excited about hats though I don't own one yet :-( I was just hanging around clicking on random cards on the Winter Bash page when I came across this after I clicked on the "Where in the World" card. Shouldn't it be hats instead of has? I mean is that really a ty...

discussion bug

Secret

16:44

Abstraction (computer science)

In software engineering and computer science, abstraction is: the process of removing physical, spatial, or temporal details or attributes in the study of objects or systems in order to focus attention on details of higher importance, it is also very similar in nature to the process of generalization; the creation of abstract concept-objects which are created by mirroring common features or attributes from various non-abstract objects or systems of study — the result of the process of abstraction.Abstraction, in general, is a fundamental concept to computer science and software development. The...

For a universe to be simulatable, it is actually demanding something very strict:

It is saying that every thing can be reduced into instructions that can be combined in a well defined manner

Abhas Kumar Sinha

@Secret Pls, I hate that, This was a chapter in our OOP C++ class. I almost got bored This is only usefull if you are more towards abstraction kind of programming.

skillpatrol

perhaps, he is :-)

Secret

well, considering that I just spent the last whole week reading and going through the JDK 8 section on classes, I think that is where I am heading

Abhas Kumar Sinha

@Secret :)

Secret

In fact, my whole way of comprehending this world relies on something very similar to that

Abhas Kumar Sinha

16:48

@Secret You must have a deep understanding of Assembly to know it's real purpose.

skillpatrol

^

Secret

The reason why I can talk about highly subjective things like the impression of some given art had on me or other people, or a mystical experience, or can classify up to 50 different kinds of nothingness is because they are reproducible enough that I can treat them as data types

even though there is no way to speak of an ineffable, they can be identified when they are experienced again

and that is more than enough to use them as data

because many ineffable, are actually quite specific in how they behave

As for assembly language, well I once touch on that back in my high school, but I don't think I really knew the details of it including how each keyword is converted into a binary string

Abhas Kumar Sinha

@Secret very ezy

PM 2Ring

17:02

@Secret Sure, if anything requires uncomputable numbers you aren't going to simulate it with a system that can only produce computable numbers. The simulation hypothesis is fun to think about, but IMHO it's a bit silly. Why would anyone waste vast amounts of computing power simulating a world as large & as detailed as this one appears to be... unless they're cheating, and it doesn't really contain as much information as it appears to.

user8718165

@PhysicsMeta I deleted this from here and posted it on the Main Meta site.

Secret

well, what I am hypothesising is that if our universe is really a simulation, it is build on a very strange programming language that already contain any level of infinities, and so unless for inconsistency reasons that are imposed by some kind of incompleteness theorem, what we think are uncomputable numbers are actually computed relatively straightforward in that alien programming language. Thus the philosophically interesting thing bout the simulation hypothesis, IMO, is it makes you

think about what really makes programming and computation, programming and computation

So unlike most people, the reason I am interested in simulation hypothesis is not because of existential stuff about creators, but because it makes me think really hard on the question: What really is a programmable thing

skillpatrol

17:29

thinking really hard is hard work

JMac

18:16

@PM2Ring The issue is that I think it would be very hard for us to try to understand the motivations of anything capable of simulating our universe. Although it may seem incomprehensible to us to put that amount of effort into a simulation, we also don't really have any idea how whatever creates the simulation would even perceive it. Maybe we're a side effect of a 5-D toaster.

Aaron Stevens

18:27

You are So Mean @Aaron — Saikat Das 54 mins ago

Lol

Kyle Kanos

simulation hypothesis is probably bogus. there's troubles enough with It peeps these days working on a single computer, but tying an ever-increasing network of billions of minds together without a single hiccup?

though i'm sure SH advocates would chalk it up to "well it's a really advanced computer"

i believe there also issues with deterministic evolution of systems

JMac

@AaronStevens Every one of that users posts have basically frustrated me since I've noticed them a few days ago. Especially their formatting. The overuse of bold and italics is frustrating enough, but the overuse of emojis (overuse being any number > 0) is what really gets me. That and they straight up reposted an accepted answer to the same question and just added bold.

Emilio Pisanty

@JohnRennie how'd you get a glasses-with-number-on-top hat?

> post a message in chat within ±12 hours of the UTC New Year’s begin that gets starred

ACuriousMind

That's the most frivolous use for a time machine I can think of

2

Kyle Kanos

@EmilioPisanty so everyone's gonna star something in chat?

Who can jump higher than the highest mountain?

Emilio Pisanty

18:43

Perhaps we can make a room for people to go get their stars?

Kyle Kanos

Answer: anyone, mountains can't jump

ACuriousMind

@EmilioPisanty More seriously, see meta.stackexchange.com/q/339893/263383

Emilio Pisanty

@KyleKanos nice starbait =)

Kyle Kanos

@EmilioPisanty actually, i was going to type that anyway. reading my son's 104 Story Treehouse book & it's littered with terrible jokes

Emilio Pisanty

speaking of creating new chatrooms

0

Q: The create-chatroom page incorrectly specifies the conversations as licensed under cc-wiki

The create a new chat-room page is probably one that doesn't get a lot of traffic, and this is starting to show: The text reads: All conversations here are public, and they are logged and licensed under the same cc-wiki license as The Stack Exchange Network. and yes, that is a link t...

bug chat licensing

that's an interesting quirk

Aaron Stevens

18:54

@JMac well, at least they are not mean like I am

Ultradark

19:10

I invented a space that is isometric to the Minkowski plane. What are some other examples of spaces that are isometric to the Minkowski plane?

Kyle Kanos

19:43

not enough views on this: math.stackexchange.com/questions/3469153/… ?

Ultradark

who downvoted?

Kyle Kanos

no idea, votes are anonymous

Ultradark

@KyleKanos do you know other examples of spaces that are isometric to the minkowski plane?

Kyle Kanos

not even sure i have rep enough to downvote on math.se

@Ultradark nope.

Physics Meta

0

Q: Callout is slightly wrong "Highly active..."

Minor issue .. I asked this question: http://physics.stackexchange.com/questions/223610 Currently the question has the callout: "Highly active question. You have enough reputation to answer this question. The reputation requirement helps protect this question from spam and non-answer activi...

discussion

ACuriousMind

19:48

@Ultradark You can get an infinity of examples of spaces isometric to Minkowski by taking any diffeomorphism from $\mathbb{R}^2$ to some target and defining a metric on the target by demanding that the diffeomorphism be an isometry.

Ultradark

@ACuriousMind what do you think of the question?

-1

Q: Prove that this space is a Pseudo-Euclidean space and isometric to the Minkowski plane

Consider the following matrix which transforms points in $\Bbb R \cap (0,1)^2,$ according to a parameter ${\mathfrak S}.$ $$ \mathfrak T_{\mathfrak S} = \begin{bmatrix} (e^{e^{\mathfrak S}})^{-1} & 0 \\\ 0 & (e^{e^{\mathfrak{ -S}}})^{-1} \end{bmatrix}$$ I want to prove that this space is a P...

differential-geometry euclidean-geometry transformation rotations

this one

ACuriousMind

It doesn't make any sense to me.

Ultradark

What doesn't make sense, so I can improve it?

the matrix acts on points in the unit square, and the flow is an image of hyperbolic rotation

ACuriousMind

@Ultradark Everything, really. What is this matrix, and how does it "transform" the points? Are you just applying it as a linear map? You suddenly talk about "this space" but you didn't define any space (if you mean the image of the linear map, say so and declare what metric you're using on it), nor have you explained why you think it should be isometric to Minkowski space.

Ultradark

I guess I'll delete it, it's not a very good question

the matrix transforms the points by letting the parameter change

ACuriousMind

20:08

I don't know what you mean by that. Mappings are easiest to understand if you write them in the standard notation $(0,1)^2 \to ?, (x,y)\mapsto ?$.

Ultradark

it's a rotation matrix

math.stackexchange.com/questions/3469153/… Made some edits, hopefully it's better now

ACuriousMind

It's not a rotation matrix (except for when the parameter is zero, then it's the trivial rotation).

Ultradark

how is it not a rotation matrix?

ACuriousMind

Rotation matrices by definition are of the form $\begin{pmatrix} \cos(\phi) & \sin(\phi) \\ -\sin(\phi) & \cos(\phi) \end{pmatrix}$, this one is not.

Ultradark

that's for circular rotation

ACuriousMind

20:15

hyperbolic rotations are the same form for $\cosh, \sinh$ instead of $\cos,\sin$, which this one also is not.

Ultradark

you can express a hyperbolic rotation in the form of $$\begin{pmatrix} e^s & 0 \\ 0 & e^{-s} \end{pmatrix}$$

ACuriousMind

Ah, right, sorry. So what you're saying is that you're looking at the map $(0,1)^2\to (0,e^s)\times (0,e^{-s}), (x,y) = (e^sx, e^{-s}y)$.

Ultradark

@ACuriousMind well I was looking at the family of level set curves that the points flow across

I'm not sure if these curves are geodesics

@ACuriousMind so this matrix $$\begin{pmatrix} e^s & 0 \\ 0 & e^{-s} \end{pmatrix}$$ acts on points to make them flow along rectangular hyperbolas. And this matrix $$\mathfrak T_{\mathfrak S} = \begin{bmatrix} (e^{e^{\mathfrak S}})^{-1} & 0 \\\ 0 & (e^{e^{\mathfrak{ -S}}})^{-1} \end{bmatrix}$$

acts on points to make them flow along a function known as $f(x)=\exp(\frac{1}{\log(x)})$ for $x,f\in(0,1)$

So I'm pretty sure this latter matrix is mapping the former matrix to another space, and hence the latter matrix is an image of hyperbolic rotation, which intrinsically must be the same as hyperbolic rotation

and following this line of thought and with some help from the people in the math chat, we came to the conclusion that the space is Pseudo-Euclidean and isometric to the Minkowski plane

Aaron Stevens

21:31

Should questions that belong on the Main stack exchange meta be closed on Physics Meta?

ACuriousMind

@AaronStevens No - after all, meta is about "the software that powers Physics Stack Exchange". I know that SE monitors at least bug on all child metas (even if they don't act on it)

Aaron Stevens

@ACuriousMind Oh ok. It is just that when I have asked questions on Physics Meta before that is more about the general SE site, I am told it belongs there and not on Physics Meta

Or maybe I am conflating that with the cases where the question has already been asked on the main meta site

ACuriousMind

Well, it would be better on main meta, but it's not wrong to post it to ours.

Aaron Stevens

Sounds good, thanks

ACuriousMind

We generally try to be rather lenient about what goes on meta, since it is the first (and sometimes only) place for support users go to.

Muhammed Ç. TUFAN

21:49

Hello

May I ask you a question?

ACuriousMind

As the room description (upper right corner) says: Don't ask about asking, just ask! :)

Muhammed Ç. TUFAN

Oh, thanks :)

Kyle Kanos

Also, asking if you can ask a question is asking a question

Muhammed Ç. TUFAN

Do you know about the center of mass frame?

ACuriousMind

Yes

That was an easy question ;)

Aaron Stevens

21:55

@ACuriousMind Teach me more!

ACuriousMind

Now that's harder!

Muhammed Ç. TUFAN

Sorry, I'm back

Now consider we have 2 objects that have mass M1 and M2

The first object has V1 vector velocity

And the second object has V2=0

You can imagine this collision

But after the collision the objects seperate

On the top, it shows the before collision and on the bottom it shows after the collision.

Is everything OK until here?

The collision is elastic by the way.

ACuriousMind

Everything's fine. Note that you can use MathJax in chat (meta.stackexchange.com/a/220976/263383) to typeset formulae

Muhammed Ç. TUFAN

Yeah, but I'm sorrily I can't see MathJax rendered and I'm using mobile phone :(

sorry*

ACuriousMind

Ah, yeah, on mobile you have to render the LaTeX in your head ;)

Muhammed Ç. TUFAN

22:07

Yeah, I'm good at it but it's hard to write formulas :/

You can imagine the collision by the center of mass frame

The center of mass would have a constant velocity

Now, the question is find the $\theta_1$ by examining the system with center of mass frame.

ACuriousMind

@MuhammedÇ.TUFAN In the center of mass frame, the center of mass would have zero velocity by definition!

Muhammed Ç. TUFAN

My teacher made the wrote all velocities (velocities relative to CM velocity, $\vec{U}_1, \vec{U}_2, \vec{U}_1^{'}, \vec{U}_2^{'}$ in center of mass frame,

Oh, no.

It has a constant velocity.

$\vec{R}_{CM}=\frac{M_1\vec{V}_1+M_2\vec{V}_2}{M_1+M_2}$

The position vector of CM.

ACuriousMind

What is your definition of "center of mass frame"?

Muhammed Ç. TUFAN

$\vec{V}_2=0$ so, the $\vec{R}_{CM}=\frac{M_1}{M_1+M_2}\vec{V}_1$

It's like we sit down at the center of mass of the system and watch everything.

And the total momentum is equal to $\vec{0}$ before and after thr collisions.

Property of the center of mass frame

ACuriousMind

Yeah, and that means that, in the center of mass frame, $R'_{CM} = 0$ and $V'_{CM} = 0$. (Primed now means "as seen in the center of mass frame")

Muhammed Ç. TUFAN

22:18

Mmm

ACuriousMind

And since total momentum vanishes, you have $m_1 V'_1 = - m_2 V'_2$.

Muhammed Ç. TUFAN

The CM has no velocity relative to itself but it has a velocity relative to laboratory frame

ACuriousMind

Yes, that's true

Muhammed Ç. TUFAN

Yeah yeah you are right at the momentum.

ACuriousMind

I'm just saying that, in the CoM frame, the "constant" velocity of the CoM is a very special constant - zero.

Muhammed Ç. TUFAN

22:20

Oh, then I mean the CM has a velocity relative to the laboratory frame.

ACuriousMind

Okay

Muhammed Ç. TUFAN

Alright

As I derived at the top side, the position vector of CM is equal to $\vec{R}_{CM}=\frac{M_1}{M_1+M_2}\vec{V}_1$

Is that OK?

ACuriousMind

The dimensions don't work out - $R$ has dimension of length, but your r.h.s. has dimension of mass times length per time.

Muhammed Ç. TUFAN

Mmm

I'm not sure why it has that dimension but the derivation is true as I know, my teacher and the book that I use is use like that.

ACuriousMind

I'm afraid then your teacher and book are wrong. Equalities need to have the same dimensions on both sides.

Muhammed Ç. TUFAN

22:25

What is the dimension of position vector, I don't know :/

ACuriousMind

It's a position, it has just the dimension of length.

Muhammed Ç. TUFAN

By the way, the R.H.S has dimensions of $\frac{[M]}{[M]}{[L]{[T]}^{-1}=[L]{[T]}^{-1}$ lenght per time as I think

ACuriousMind

Oh, yes, sorry, you're right

Muhammed Ç. TUFAN

Mmm, I'm not really sure why It's not work out

ACuriousMind

for $V_2 = 0$, what you've got there is not the expression for the position of the CoM, but for the velocity (in the lab frame)

Muhammed Ç. TUFAN

22:28

Yeah but I thinj It's ok

When we think about the general definition of position vector of the CoM it has this dimensions

ACuriousMind

Well, you've correctly said that the r.h.s. has dimensions of length per time, i.e. it's a velocity. So the l.h.s. needs to be a velocity, too

Muhammed Ç. TUFAN

Yeah I got you but

Ohhhhhhh

I'm a big dumb haha

I have wrote it wrong!

ACuriousMind

It happens :)

Muhammed Ç. TUFAN

$\vec{R}_{CM}=\frac{M_1\vec{r}_1+M_2\vec{r}_2}{M_1+M_2}$

That is now OK :P

Sorry, forgive me for that mistake :P

ACuriousMind

Yeah, that seems more like it

Muhammed Ç. TUFAN

22:31

Now, I take the derivative of the $\vec{R}_{CM}$

So, it will give me the velocity of the CoM

$\vec{V}_{CM}=\dot{\vec{R}_{CM}}=\frac{M_1\vec{V}_1+M_2\vec{V}_2}{M_1+M_2}$

ACuriousMind

yup

Muhammed Ç. TUFAN

If $\vec{V)_2=\vec{0}$ the velocity of CoM is

$\vec{V}_{CM}=\frac{M_1}{M_1+M_2}\vec{V}_1$

Is that OK?

ACuriousMind

yup

Muhammed Ç. TUFAN

Alright

Now, I will write the velocities relative to CoM frame

$\vec{V}_{1,CM}=\vec{U}_1=\vec{V}_1-\vec{V}_{CM}=\vec{V}_1-\frac{M_1}{M_1+M_2}\vec{V}_1=\vec{V}_1(1-\frac{M_1}{M_1+M_2})=\vec{V}_1\frac{M_2}{M_1+M_2}$

I can write all the other 3 velocities but It will take a long time, you know I use mobile phone :(

Is that OK for you?

ACuriousMind

Yeah, it's fine. Just ask what you're going for and we'll figure it out

Muhammed Ç. TUFAN

22:39

Okay.

Thanks.

Now, you see that $\tan\theta_{1}=\frac{\sin\theta_1}{\cos\theta_1}$

Multiply it with $\frac{{\vec{V}_1}^{'}}{{\vec{V}_1}^{'}}$

so $\tan\theta_1=\frac{{\vec{V}_1}^{'}\sin\theta_1}{{\vec{V}_1}^{'}\cos\theta_1}$

Is everything OK?

${\vec{V}_1}^{'}$ is the velocity of the first object after the collision btw.

ACuriousMind

alrighty

Muhammed Ç. TUFAN

Okay

Now, you see that ${\vec{V}_1}^{'}\cos\theta_1$ is the after collision velocity of the first object on the $x$ dimension.

and the ${\vec{V}_1}^{'}\sin\theta_1$ is on the $y$ dimension

I will make a little transformation here.

${\vec{U}_1}^{'}={\vec{V}_1}^{'}-\vec{V}_{CM}$

${\vec{V}_1}^{'}={\vec{U}_1}^{'}+\vec{V}_{CM}$

Is that OK?

ACuriousMind

yup

Muhammed Ç. TUFAN

Okay, now i will seperate ${\vec{V}_1}^{'}$ into It's components

${{{V}_1}^{'}}_{x}={{U}_1}^{'}\cos\theta+V_{CM}$

${{{V}_1}^{'}}_{y}={{U}_1}^{'}\sin\theta+0$

Is that OK?

ACuriousMind

22:54

Yes - except I want to go back one step - how do you know that $V'_1\cos(\theta_1)$ is the after collision $x$-velocity?

Muhammed Ç. TUFAN

It is showed on the image of the collision. Let me upload it one more time.

ACuriousMind

Common sense from watching billard balls collide heads-on (where $M_1 = M_2$ and $\theta = 0$) tells us that when the angle is zero, the ball 1 will stop and ball 2 will have all the velocity.

Muhammed Ç. TUFAN

ACuriousMind

That picture does not model the case $\theta = 0$ well.

Muhammed Ç. TUFAN

Yeah, I get what you mean. But in this collisions there is no information about the masses, but I don't know why the collision doesn't keep moving on the same dimension :P

ACuriousMind

22:57

Oh, wait, sorry, I got confused - in my degenerate case, $V'_1 = 0$, so everything's fine. Please continue.

Muhammed Ç. TUFAN

I think this is something like a general definition or proof of 2 mass colliding

OK, thanks.

At the last relative velocities do you know what is $\theta$ angle?

(not $\theta_1$)

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