The h Bar

Now suppose you're running alongsdie the left block at the same speed $v$. In that case you see the left block as stationary because you and the block are moving at the same speed. Yes?

user228700

Yup.

10:04 AM

And now the right block appears to be coming towards you ate a speed $-v$:

user228700

Yes...

OK. Now suppose you're running at a speed $v/2$. In that case the motion of the blocks looks like this:

user228700

Riight...

Are you happy with this, because this is the key step in the argument?

user228700

Give me a moment to think about it, please...

user116211

10:07 AM

Well, @JohnRennie, you would love these pics then....

user228700

I mean, I get it in a mathematical sense, but I guess I'm just not able to visualize this actually happening. Never mind, Ik what u're saying...

> I'm just not able to visualize this actually happening

user116211

You do need to get an instinctive feel for this argument.

The point is that all velocities are relative i.e. we measure a velocity relative to some fixed point.

user228700

I do, I do!

10:11 AM

OK, I'll believe you :-)

user116211

user228700

I just...I'm unable to picture myself in a car or something, and seeing the other block move toward me. I hope this inability is OK...(?)

OK, we'll forge ahead and see how it goes.

user228700

As long as I understand what's happening..?

If you look at that last diagram you'll notice it is symmetrical.

user228700

10:12 AM

OK!

user116211

And I love this:

Specifically the total momentum is zero.

user228700

Ah, yes...

That's because the left block has momentum $mv/2$ and the right block $-mv/2$

user116211

user228700

10:13 AM

This makes it a lot easier to understand how in the world net momentum could be zero.

And we know momentum is conserved, so at all points during the collision the momentum must stay at zero.

Now let's put your spring back in (another diagram is pending) ...

user228700

Wait, wait!

user228700

You simplified this by making me move too (which doesn't change anything, yes) but I'm trying to justify momentum adding up to zero in the first picture.

OK. What about it is giving you problems?

user228700

Just that it's not adding up to zero :-P Hang on, why should it have to add up to zero in the first place? Assuming that that first block has always remained moving with that velocity, when we take the system, the momentum won't add up to zero, because it didn't start out with zero momentum!

10:21 AM

Remember that I said velocities are always measured relative to some point?

user228700

Yeah...

Suppose you choose to measure all velocities realtive to you, which is a perfectly good way to measure velocity. Then your velocity is always zero. Yes?

user228700

@JohnRennie Yeah, but...OK?

Aha, since you deleted the post I'm guessing you're getting it. The reference point can be moving.

Suppose you're standing still by the side of a road and you watch a car pass you at 10 m/s, what speed is the car doing?

user228700

See, this is why I deleted it :-P I didn't want to give u the impression that I was under the impression that the reference point can't be moving.

10:24 AM

Back to my question: Suppose you're standing still by the side of a road and you watch a car pass you at 10 m/s, what speed is the car doing?

Just got back from my Chem exam, I think I absolutely rocked it :)

user228700

Oh no :-( Sir, I assure you! I'm very comfortable with kinematics!

user228700

I understand that the reference point can be moving too!

@Kaumudi OK so the velocity of an object can be different if we take different reference points. Yes?

user228700

...which is why I deleted that message. I was gonna continue on that line but seeing as u were going to continue, I deleted it...

user228700

10:26 AM

@JohnRennie Yes, yes!

user116211

@BernardMeurer \o/

and that means the momentum of an object will be different if measured from diffreent reference points.

user228700

Yes.

user116211

@BernardMeurer, Now that the paper went well, prepare for the Uprising against that madman.

So if we go back to our blocks (assume right is positive) then in the first image the total momentum is $mv$.

user228700

10:27 AM

Yes...

In the second image the total momentum is $-mv$

@MAFIA36790 Lol, he actually put a wrong question on the exam

And in the last image it is zero.

user228700

Yep.

Which god bless he did, because that meant b) was impossible, and I didn't know how to do b) :p

user116211

10:27 AM

@BernardMeurer WOW! Madman he is indeed.

user228700

Okay, I see ur point.

That doesn't mean momentum is changing and not being conserved.

It's just that we've shifted our reference point.

user228700

U mean that momentum is relative too.

@MAFIA36790 He wanted to know if you hit W or Cs with a 500nm radiation which one would have photoemission

@Kaumudi Exactly!

user116211

10:28 AM

@BernardMeurer, Anyways, now the exam is over, start reading a good book now.

He actually said, and I quote, on the question "Only one of these will experience the photoelectric effect, which one will it be?"

user228700

That we don't start out with zero momentum in every frame of reference, but this doesn't mean that it's not ultimately conserved, yeah?

the answer was: none

@MAFIA36790 Is Chang good? It's on my syllabus

@Kaumudi Correct. For any given reference point momentum is always conserved, but if we switch to a different reference point the momentum may change.

user116211

@BernardMeurer I would say Atkins.

user228700

10:30 AM

Aah, this is a huge (seemingly obvious, as it may be to less dumb people) revelation!

@MAFIA36790 People gave you shit on the chat the other day for recommending Atkins though

@JohnRennie Thoughts on this?

user116211

@BernardMeurer I would still say the book is good.

@JohnRennie You have an amazing voice

@Kaumudi this principle is known as Galilean relativity.

Galilean invariance

Galilean invariance or Galilean relativity states that the laws of motion are the same in all inertial frames. Galileo Galilei first described this principle in 1632 in his Dialogue Concerning the Two Chief World Systems using the example of a ship travelling at constant velocity, without rocking, on a smooth sea; any observer doing experiments below the deck would not be able to tell whether the ship was moving or stationary. == Formulation == Specifically, the term Galilean invariance today usually refers to this principle as applied to Newtonian mechanics, that is, Newton's laws hold in all...

user228700

@BernardMeurer Whoa, where is that coming from? Also, did u check out that vid. on MO theory I recommend..?

user228700

10:32 AM

@JohnRennie Ah, OK...

@Kaumudi I did yeah, not all of it though b/c time, but a good portion of it. Thanks for that :)

@Kaumudi I'm a bit surprised your physics teachers didn't cover this. Or maybe going by what you've said about them in the past I shouldn't be surprised :-)

user228700

Should've been acquainted with this before. Sadly, I wasn't. Okay, shall we continue and add that wretched spring?

user228700

@JohnRennie I don't have any teachers...except u guys. I study from home.

OK let's put in the spring ...

user228700

10:34 AM

@BernardMeurer Cool. Glad it helped :-)

I've drawn this at the moment the two blocks touch the spring, so the spring is uncompressed i.e. at it's natural length.

user228700

Yeah, OK...

And the problem was to find the minimum length of the spring, or something like that?

user228700

*Maximum compression.

Same difference :-)

Can you now see how to do this, or shall I continue ranting?

user228700

10:36 AM

Yeah, sorry :-P I'm turning blind.

user228700

Erm...can u continue, please? I tried before and I wasn't able...I'm sorry :/

user228700

OK, hang on, trying again..

OK. We are working using a reference point where the total momentum is zero. And that means the velocities of the two blocks must always be equal and opposite. OK so far?

user228700

Just one second...I'm trying (again) to see if I can figure it out on my own...

user228700

So for how long will these two blocks keep moving?

user228700

10:42 AM

I mean, as soon as the springs starts experiencing tension, it pushes the blocks away, decreasing their velocity...and that K.E gets stored in the spring, yeah?

user228700

And the spring comes under tension the moment the two blocks even start moving. ie. At t=0

user228700

(t being time)

You don't need to worry about time. You can just use conservation of energy.

user228700

Yeah, no, I just added that for good measure...

user228700

OK, so when the two blocks stop moving, the spring attains max. compression, yeah?!

10:45 AM

@Kaumudi yes. That's supposed to be the lower diagram in my picture above i.e. spring length $\ell$

user228700

ie. All the kinetic energy of the two blocks is transferred to the spring.

user228700

@JohnRennie Ah, OK...

Ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

user228700

Wait, what happened?

user228700

:-o

10:46 AM

yo @mods

user228700

I'm scared :/

Q: What parts of a physics undergraduate curriculum have been discovered since 1966?

shouldn't this one be community wiki?

7

What parts of an undergraduate curriculum in fundamental physics have been discovered since, say, 1966? (I'm choosing this because it's 50 years ago). Physics textbooks from before 1966 seem to contain everything in fundamental physics that is taught to undergraduates nowadays. I would like to d...

user228700

@JohnRennie: Omg, what happened?! Did I say something wrong..?

@JohnRennie You know that some of us have become allergic to that picture, right?

@Kaumudi no, that was just me reacting to you spotting the way to do the problem. It wasn't a scream!

user228700

10:48 AM

Ohh. Phew. I was so scared :-P

user116211

@EmilioPisanty Reminds me of a MAN.

@EmilioPisanty why the rubber sheet image is associated with an article on galilean relativity escapes me. Oh well ...

@EmilioPisanty Johnrennie has no control of that. That pics just pops up whenever a gravity related wikpiedia article is posted

user116211

Welcome to The h Bar @NatalieHogg.

@Kaumudi: back to the spring?

10:48 AM

Emilo: it's the problem of the pics generating algorithm

user228700

@JohnRennie Why? I thought we were done..?

@JohnRennie actually.... yeah, what the hell?

it's not even in that article

oh, no, wait

it's in the footer

@Kaumudi well yes, I guess we are. Initially the energy is all kinetic so it's $\tfrac{1}{2}m(v/2)^2 + \tfrac{1}{2}m(v/2)^2$. Yes?

user228700

Yeah, no, Ik how to work out the math!

And when the blocks have stopped moving it's all potential energy $\tfrac{1}{2}k(\ell_0 - \ell)^2$

10:50 AM

user228700

What was troubling me was I didn't know what was even happening!

inside a collapsible section

right under.... the man himself.

user228700

And now I know! Although I still dunno how it works when I'm not working in the frame of reference of the COM :/

So we spent a while getting used to the idea of using a different reference frame, but once we were happy about that the answer just jumped out!

user228700

@JohnRennie Yes! :-)

10:52 AM

@Kaumudi all we did to jump to the COM frame was subtract $v/2$ from all velocities. Yes?

user228700

Yep.

So the velocity of the left block changed from $v$ to $v/2$ and the right block changed from zero yto $-v/2$.

So if we want to switch back to the stationary frame we just add $v/2$ to all velocities.

user228700

Yeah...

You can do all your caculations in the COM frame, where it's easy because it's symmetric ...

user228700

Yeah...

10:54 AM

then when you've finished just add $v/2$ to all velocities to get back to the lab frame.

user228700

Yes, OK...

user228700

Omg sir. All my life, I've been struggling with is :-P

In pretty much all of kinematcs we operate in the COM frame if we can.

Even calculating particle collisions in the LHC is done by working in the COM frame.

If you want to read more see:

Center-of-momentum frame

In physics, the center-of-momentum frame (zero-momentum frame, or COM frame) of a system is the unique (up to velocity but not origin) inertial frame in which the total momentum of the system vanishes. The center of momentum of a system is not a location (but a collection of relative momenta/velocities). Thus "center of momentum" means "center-of-momentum frame" and is a short form of this phrase. A special case of the center-of-momentum frame is the center-of-mass frame: an inertial frame in which the center of mass (which is a physical point) remains at the origin. In all COM frames, the center...

user228700

Wow.

user228700

Just. Thanks so much sir :-D

user228700

10:57 AM

SO MUCH.

@BernardMeurer: sorry, I didn't mean to ignore you. I was caught up in kinematics.

user228700

This is what my brain feels like:

@BernardMeurer: Congrats on the exam! :-)

user228700

^My brain :-P

10:59 AM

@BernardMeurer it sounds very English, doesn't it? :-)

user228700

@JohnRennie: Thank you thank you thank you!!

The only thing is that to get used to this you need to do lots of worked examples using the COM frame. Although what I've said may make sense now, tomorrow doubts may creep back in!

And things can get more complicated. For example suppose the masses are different for the two blocks.

David Z

@EmilioPisanty no (as far as I'm concerned)

That means the velocities in the COM frame won't be the same, but we'll always have: $$m_1v_1 + m_2v_2 = 0$$

So the velocities are always proportional to each other at all times during the collision.

@JohnRennie It does, but you sound so young man! How do you do it?

@Kaumudi Thanks :)

@JohnRennie No worries, kinematics comes first :)

11:04 AM

@BernardMeurer I think it's just because my voice is quite high. Actually I'd prefer to have a manly deep voice but oh well ... :-)

@JohnRennie Nah man you sound nice, you could work in phone sex and pretend you were a young stallion in the peak of his youth

Hmmm, I wonder if that pays better than being a computer nerd? Sadly I suspect not :-)

@JohnRennie OH, This reminds me of something important I need to ask you about!

Phone sex?

Now I'm worried :-)

My university has published two openings for people to work as server admins for the uni servers. They're Linux servers. What do you think?

I.e. Thoughts on your own job

11:07 AM

Go for it. You'll learn tons of useful stuff.

(The openings are for first or second year students of CEng and IEng)

Some of server admin stuff is routine and boring, but there's loads of cool stuff as well.

Yeah, I was interested because networking is kind of black magic for me right now

Absolutely go for it. DO IT NOW!

OKAY

11:08 AM

What does the job pay?

I'll send them my CV

They didn't say, I guess I'll find out during the interview or something

Doesn't matter, it's worth it for the experience alone.

Yeah, that's my thinking

You'll be picking up hugely marketable skills, and even if you go on to do something else in IT knowledge of how servers work will be very useful.

During my PhD I worked as a programming advisor i.e. people come to me for help getting their programs to work. Having to deal with all sorts of different stuff is a great learning experience.

BTW I just finished fixing another laptop, this one for my niece.

* Be interested in networking
* Liking GNU/Linux and being interested in learning new things
* Knowing what the 169.254.0.0/16 address range is
* Knowing what STP, DHCP an VRRP are
* Knowing something about TCP, 802.11 and SIP

These are their skills list

I actually hope I learn something about Linux, it's been a while :p

user228700

11:12 AM

@JohnRennie Aha, I just did a problem like this and got it right :-D

I haven't been doing a lot with it lately, so learning curve has been stationary

My guess is that as a noob the work they give you will be a bit routine to start with, but if you show an interest I'm sure they'll be happy to let you get into the cooler areas.

@Kaumudi :: Applause :: :-)

user228700

:-D

@JohnRennie I did the coolest thing the other day, made me feel like a super hero. Middle of the night, someone calls me from Brazil

@Kaumudi it is a very useful skill to have in your toolbox

@BernardMeurer Yes ...

11:14 AM

It's a guy from my Makerspace saying one of the computers that run Arch had gone haystack and PostgreSQL was murdering baby ~~pandas~~ cpu's and it was insane

So at 3.A.M I SSH'd into a computer in another continent and fixed it in under 30 minutes

Felt like the God of Linux

:-)

That's kind of what I do, though all the servers I work on are in the UK and they all run Windows.

I normally use RDP thorugh the Internet, though sometimes through a VPN instead.

I do RDP through SSH with X11 forwarding, it works okay

Anyway, I'm off to start making lunch. I feel strangely happy now. This chat can be a rewarding experience when we're not preoccupied with arguing about politics and flagging each other :-)

user116211

@JohnRennie You missed everything yesterday.

@JohnRennie Flagged

11:25 AM

@MAFIA36790 I read the chat log. Not pretty reading. If I had my way all discussion of politics and religion would be banned from this room with a 24 hour ban for all offenders!

@BernardMeurer :: John reaches for the kickmute button ::

@JohnRennie You just say that because you're a pesky atheist who votes for trump

::hides::

:p

user116211

@BernardMeurer flagged as extremely offensive.

user116211

ARTofCode is coming....

@MAFIA36790 Flagged as unreasonable flagging

No... Not him...

You (collectively) actually managed to get the room frozen last night!!

Though I think that was an over-reaction by MetaEd.

11:28 AM

@JohnRennie That was stupid though, that meta dude should've just banned the people fighting

Like, I was actually talking about Electronegativity with Mafia when he did that

To be fair, freezing the room is a standard way for mods to deal with a fight. It gives them a few minutes to work out who the offenders are so they can issue approriate suspensions.

But I agree that in this case it was unwarranted. There wasn't really a fight, just some erm spirited debate :-)

user116211

@JohnRennie I said to mute the chatroom for a while for someone was flagging a lot.

@JohnRennie Two people are fighting? Ban both of them for 5 minutes, figure out what's happening, let the other continue their things

user116211

A LOT ;P

Muting the room when it's not like a generalized chaos feels like shooting the hostages or something

11:31 AM

But @0celo7 if you're listening, you have seriously, seriously got to rethink your flagging strategy. You are, as we aging hippies say, cruising for a (moderatorial) bruising.

@0celo7 He's telling you to flag twice as much

@BernardMeurer well that would quickly solve some flagging problems, but it's not an optimum solution :-)

@JohnRennie It's optimal for the degree of enjoyment I'll get from it :D

Anyway I'm off. See you all later. Bernard, let us know if you get that job. It sounds absolutely perfect.

user116211

@JohnRennie Have a healthy lunch :)

11:35 AM

@JohnRennie Have a good lunch man! I'll let you know, thanks :)

user116211

yeh @Bernard, apply for it; this will provide a better experience than that madman's chem horror show.

@MAFIA36790 I think a colonoscopy is better than my Chem class

user116211

11:59 AM

@BernardMeurer yeh, it very much seems so.

Did you have your test yet?

user116211

2 hours ago, by Bernard Meurer

Just got back from my Chem exam, I think I absolutely rocked it :)

Cool

user116211

What about your QM exam?

@JohnRennie Dunno what that means.

12:04 PM

in Mathematics, 4 mins ago, by Secret

Hey guys I have a elementary calculus problem:
Suppose I want to show that the arbitrary constants of an integration does not matter:
$$\int uv' dx=u(v + C_1)-\int u'(v+C_2) dx$$
$$=uv + uC_1-\int u'v+ u'C_2 dx$$
$$=uv + uC_1-\int u'v dx- (u + C_3)C_2$$
$$=uv-\int u'v dx + u(C_1-C_2) - C_3C_2$$
But I got an extra arbitrary constant that is multiplied to u. Is there a reason why $C_1=C_2$?

user116211

@Secret You are playing with by parts?

I flag things that are seriously inappropriate. Don't know what more there is to say.

@MAFIA36790 Yeah it was an old unsolved question I dug out from a dusty pile of my handwritten notes during lecture

which sounds way too stupid and localised to be asked as a MSE question, thus I planned to kill it in chat since I cannot solve it due to a conceptual problem I don't aware

@0celo7 It was pretty ez though

user116211

@Secret Why did you use two functions $v+ {\rm C_1}$ and $v+{\rm C_2}\,?$

12:10 PM

i am not sure, but isn;t when you integrating something, you always spawn an arbitrary constant, and if you integrate that expression again, you will get a different arbitrary constant?

user116211

@Secret yes; but you have now three functions $u,$ and the two above.

user116211

By parts come from the chain rule.

user116211

You are using three functions.

user116211

Can you connect the links between the three statements?

chain rule in this case has only two (family of) functions, if there are 3 (family of) functions here, then two of them has to be equal?

user116211

12:12 PM

Check it yourself.

I think I have a brain tumor.

Hello guys, please help me out.

user116211

WTH.

@0celo7 Hm? How's AL btw?

12:18 PM

Yesterday I ate kale that tasted like a disgusting pastry, no lie.

@0celo7 Ewh, disgusting

I've linked the problem out there^^. Given I've found out the Xmax, how do I get the radius of curvature?

@BernardMeurer what?

ok it works out now, thanks

@0celo7 What what?

12:19 PM

@BernardMeurer got here at midnight, just woke up

@0celo7 Go eat barbecue

user116211

Eat some Lays.

Take a snap selfie with your soon-to-be father in law

@Secret Could you help me?

@SwapnilDas Don't ping people to help you. We've seen your question, people will answer if they know or feel like it

user116211

12:21 PM

@SwapnilDas, Don't ping someone randomly; if someone is interested, they will ping you.

I see. Thanks. Lots of protest lol.

@Secret You need less messy calculations dude :p

user116211

@bernard, we work synchronously ;)

@MAFIA36790 Indeed, lol

@MAFIA36790 Does 'pinging' mean mentioning?

I'm new, sorry if I sound foolish.

12:23 PM

@SwapnilDas This is a ping

I am pinging @DHMO. Ping.

@BernardMeurer Ok :-)

@BernardMeurer quoi?

@SwapnilDas Familar with the cyclotron formula?, that is what you need here

@DHMO What's your favorite food?

@BernardMeurer water loves polar chemicals

12:24 PM

Oh, Thanks. I had heard of it, not read though. @Secret

@DHMO ¬¬

@BernardMeurer Do I love sodium chloride the most?

user116211

@bernard, ate Cheetos and Lays today simultaneously ;))

user116211

Not a healthy lunch ;P

I love(d) Cheetos :)

12:26 PM

@MAFIA36790 Yuck :p

I only like frozen food

It's been a while since I had something that wasn't frozen

user116211

@BernardMeurer You ate Turkish Delight yogurt? It's an amazing frozen food.

rob

@0celo7 I'd like to talk about this attitude towards flagging. Do you have a minute?

@MAFIA36790 Don't think we have that here, I buy frozen meals because money

user116211

:(

@rob Oh boy, you're in for a frustrative conversation

rob

12:31 PM

@BernardMeurer I plan otherwise :-)

@rob Well, I planned to study computer engineering this semester, yet I have a chemistry course :P

@rob No, sorry.

user116211

@BernardMeurer That's the greatest mystery ever T__T

rob

@0celo7 Fair enough.

Q: Is the simplest form of natural inflation base-2 notation from the Planck scale?

@SwapnilDas how do you find the Xmax?

Danu

12:39 PM

-2

A base-2 model of natural inflation that starts at the Planck scale and goes to the Age of the Universe is here: http://81018.com/chart Abstract: A most-simple, natural process by which natural inflation quietly encapsulates the universe uses base-2 notation from the Planck Time to the Age o...

cosmology astrophysics cosmological-inflation

This post deserves a quick and pain....less? death

Obvious (badly hidden) self-promotion or crackpot theories.

user116211

It should be closed immediately.

user116211

Crackpot...

user116211

Reminds me of Lumo ;P

downvoted (4th downvote casted so far)

Shing

I think it is closed now

removed*

Danu

12:47 PM

Great.