The h Bar

Slereah

12:34 AM

onlinelibrary.wiley.com/doi/10.1002/sapm196039164/abstract

anyone happens to have access to this article?

Slereah

12:46 AM

jstor.org/stable/43634504?seq=1#page_scan_tab_contents

also this one mb

0celo7

@Slereah seems to be a European thing

@Slereah I do

0celo7

1:38 AM

@ACuriousMind wtf the exercises in Milnor are so hard

and they use multiple results from throughout the text

skill patrol

2:57 AM

Hi @dmckee

0celo7

3:37 AM

@ACuriousMind

"The introduction of numbers as coordinates...is an act of violence..." - H. Weyl

skill patrol

4:02 AM

spoken as a true number theorist

DanielSank

5:01 AM

@0celo7 :)

John Rennie

@Obliv This is a cluttered desk!

google.co.uk/search?q=alan+guth%27s+desk&tbm=isch

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

8:26 AM

My work always tried to unite the
Truth with the Beautiful, but when I
had to choose one or the other, I usually chose the Beautiful.

Hermann Weyl

Slereah

So apparently there is a wavefunction for bipolar coordinates

It is either in the form of an infinite series or an integral

John Rennie

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Weyl came up with a truly elegant version of general relativity, but failed to spot that if you moved an object through a closed loop the object's size could change. Beautiful, but ... erm ... not true! :-)

But then he did invent gauge theory so we can forgive him :-)

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:-)

here is the source of the above quote @JohnRennie

@yuggib o/

How's the sushi?

user116211

8:43 AM

Hallo people!

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Yo, wazzup?

user116211

Came from shopping after buying snacks for enjoying Euro Cup ;P

Slereah

@JohnRennie Is that like conformal GR or something

John Rennie

@Slereah weylmann.com/weyltheory.pdf

Slereah

have you even checked if it was true

walk in a little circle

check if you get taller

John Rennie

8:51 AM

Hi @ally

Come to discuss relativity?

user128101

Hi John, thanks for your kind invitation, would it be more appropriate to open a specific chatroom?

Slereah

nah no need

John Rennie

No, just chat here. Someone else might be interested.

Slereah

Things are pretty slow on sunday

user116211

@Slereah ??

John Rennie

8:53 AM

Sunday? Have you been messing with those CTCs again?

@ally shall I start, or do you want to kick off the discussion by asking a specific question?

Slereah

Oh right

When you got no job every day is sunday :p

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Spain better not win again @MAFIA36790

user116211

@Slereah ah!

user116211

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ I'm going with Germany.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Good choice.

user128101

8:55 AM

@JohnRennie, as you wish. If I got it right, your argument was: gold has increased mass because every observer agrees on its value, whereas different observers have different opinions on the mass of the particle at LHC. Is that correct?

John Rennie

@ally No. All observers will agree on the invariant mass of protons at the LHC.

user116211

@Slereah start a blog... like Lumo; you'd be busy.

John Rennie

Different observers will disagree about the proton momentum, but when you take that moemmtnum and calculate the amss from it all observers will get the same value of $m$.

user128101

@JohnRennie, I am referring to the mass increase due to KE

John Rennie

The mass doesn't increase with velocity.

user128101

8:57 AM

@JohnRennie, yes you said that many times, I don't get it since you did not refute my arguments about objective clues that mass has increased, shal I repeat them?

John Rennie

Go on then ...

user128101

the speed of the p is certified by clock, mass is certified by Lorentz formula, mass increase is certified by increased radius

John Rennie

If we do measurements at the LHC we can measure the particle momentum and we can measure the velocity. Yes?

@ally: Hello?

user128101

At LHC they measure the time to make a rvolution, measure and controll the magnitude of B so we know for certain that speed is .999999991 c B is 8.22T, KE = 7 TeV,right? Now considering the invariant mass the radius shoul be 7460 time smaller, and that is confirmed by Lorentz formula. So we

user128101

know for certain that the effective mass is 7460 times greater, right?

John Rennie

9:06 AM

No, because the Lorentz equation for the force on the particle is a non-relativistic approximation and doesn't apply to relativistic particles.

user128101

The fact the some observers may not agree is irrelevant, but I have also shown that even a moving observer will check that

John Rennie

See for example:

lecture-notes.co.uk/susskind/special-relativity/lecture-7/…

and a quick Google will find you many simiar articles.

user128101

I thought that indeed Lorentz was particularly made for relativistic particle and speeds

user128101

I was referring to the 1/sqrt(1-v^2/c^2) are you referring to the same equation

John Rennie

The Lorentz equation describes the force on a charged particle:

Lorentz force

In physics, particularly in electromagnetism, the Lorentz force is the combination of electric and magnetic force on a point charge due to electromagnetic fields. If a particle of charge q moves with velocity v in the presence of an electric field E and a magnetic field B, then it will experience a force F = q E + q v × B {\displaystyle \mathbf {F} =q\mathbf {E} +q\mathbf {v} \times \mathbf {B} } (in SI units)...

So in this case you're (mis) applying it to calculate how particles curve in the magnetic fields at the LHC.

user128101

9:09 AM

I was referring to the mass formula, that tells us that at that speed mass becomes 7460 greater

John Rennie

OK, that's the equation m' = m/sqrt(1 - v^2/c^2) Yes?

Are you starting by assuming that that equation is true?

user128101

yes, I was also referring to the measurement of the radius by a moving observer, can we stick to these to data?

user128101

Lorentz tells us that mass is greater, and in the lab frame we confirm the radius is as many times greater, right?

John Rennie

The trouble is that you're starting by assuming what you're trying to prove.
What I'm saying is that the parameter m' in that equation is not the mass, or more precisely not a mass in any physically meaningful way.

user128101

So, lets assume the mass increase is real

ACuriousMind

9:13 AM

@0celo7 Weyl was awesome

user128101

No, i do not follow you. I am not trying to prove anything

Slereah

IIRC Einstein said that he had no idea what Weyl's papers were about

John Rennie

If you have a particle moving in a circle there must be a force towards the centre to cause the circular motion. Yes?

user128101

Please, try to follow my sequence, else I am lost: I am applying the usual formulae, everything says mas is 7460 greater!,

user128101

Is that questionable?

ACuriousMind

9:15 AM

@ally yes:

23

Q: Why is there a controversy on whether mass increases with speed?

Some people say that mass increases with speed, some people say that the mass of an object is independent of its speed. I understand how some (though not many) things in physics are a matter of interpretation based on one's definitions. But I can't get my head around how both can be 'true' is an...

user128101

@ACuriousMind, it's already difficult for me to discuss with one

John Rennie

What I'm trying to do is lay out the arguments in a simple logical order to pinpoint where we are disagreeing. It will take me about five minutes. At the moment I can't follow your argument because you haven't stated it clearly.

Why not take the five minutes to follow my argument and see where it leads?

skill patrol

^

user128101

NO, because you have already been so kind to answr many of my questions, and chatting with me.I need you to patiently to follow my logic step by sted and say when it breaks down

John Rennie

OK. What's your step 1?

skill patrol

9:19 AM

Try it, you might like it :-)

user128101

Once more, At LCH the measure Ke, speed, B field can you doubt that?

John Rennie

OK, so far so good. Yes I can measure momentum, energy, velocity, the strength of the magnetic field and the radius of the circular motion. This far we agree.

user128101

At that Ke and speed lorentz formula tells us what is effective mass, 7460 invariant masses, right?

user128101

In our frame, the radius of revolution confirms that invariant mass hass hugely increased, right?

John Rennie

@ally Aha! No. That's where you're wrong.

user128101

9:22 AM

Ok shoot

John Rennie

Well this is what I was trying to explain before.

Suppose we have a proton with mass m moving in a circle of radius r at velocity v, then the acceleration towards the centre is a = v^2/r. Yes?

user128101

If effective mass were theinvariant mass what would radius be?

user128101

forget for a second about SR,

user128101

what would radius be

John Rennie

@ally that's what I'm about to do. Are you happy with my statement about the acceleration above?

ACuriousMind

9:25 AM

If you ask John to follow your explanation step by step to see where it is wrong it would behoove you to in turn follow his explanation of why it is wrong step by step

user128101

I am not considering acceleration for now, let's stick to the radius

John Rennie

I'm about to calculate the radius, but you need the acceleration to do it. That's why I'm starting with the acceleration. Are you happy for me to proceed?

user128101

what is the radius of a p at .9 c?

user128101

OK, proceed

John Rennie

OK, just to reprise, the acceleration is related to the radius and particle velocity by:
a = v^2/r
This is just basic mechanics.

Now let's proceed using Newtonian mechanics not SR.

user128101

9:28 AM

in a B field it becomes a = vB

user128101

right?

John Rennie

Newton's second law tells us F = ma, so the force on the proton must be:

F =ma = mv^2/r

Yes?

user128101

yes , but v= p/B , right?

user128101

Imean r= p/B

John Rennie

OK, and as you say the Lorentz force law for a proton ina magnetic field B tells us:

F = evB

where e is the proton charge.

user128101

9:32 AM

yes, I calculated that at .9c the radius of a proton should be 1,4 cm, at c a little larger

John Rennie

For circular motion the two forces must be equal so we equate them to get:

evB = mv^/r

and we can use this to calculate r. I haven't done the caculation, but if you say the answer is 1.4cm I'm happy to believe you.

So we have a problem because that obviously isn't the radius of the LHC. Yes?

user128101

Yes, if we consider that massis actually 7460 times bigger, we get the actual radius, 4 km

John Rennie

@ally correct, but that isn't the reason we got the wrong answer.

user128101

what is then?

John Rennie

We got the wrong answer because Newton's second law, F=ma, and the Lorentz force law, F=evB, are non-relativistic approximations and gve the wrong answers if we apply them to relativistic objects.

user128101

9:35 AM

it cannnot be a coincidence the figeres are too exact

John Rennie

As it happens, the error turns out to be a factor of sqrt(1-v^2/c^2)

user128101

I am not considering newton at all, why do you bring it in? F=evB is a valid formula

user128101

thats why I said, consider v =.1 c or any other non relativistic speed

John Rennie

F=evB doesn't apply to relativistic objects.

See lecture-notes.co.uk/susskind/special-relativity/lecture-7/… for the details

user128101

does it apply for v = .14 c?

ACuriousMind

9:39 AM

(Aside: I'm not sure what you calculated, but I get 34.27 cm for the proton. )

John Rennie

It is exact only in the limit of v = 0

It is a pretty good approximation up to a few percent of $c$

There is no exact cutoff when it ceases to apply, the error just gets gradually larger.

skill patrol

(Aside: @ACuriousMind the Brits spell "behoove" with one less o :p )

John Rennie

But look: the point is that we can absorb that factor of sqrt(1-v^2/c^2) if we say the mass changes by that factor

ACuriousMind

@skillpatrol You might have noticed by now that my spelling is neither fully American nor fully British :P

John Rennie

And that's why the concept of relativistic mass was devised.

It was to absorb that annoying factor.

skill patrol

9:41 AM

Indeed @ACuriousMind :D

John Rennie

But it's a physically meaningless procedure. It's just saying "hey my equation only works if I change the mass"

Instead of "hey I'm not using the correct equation"

user128101

but if we apply the evB formula, results are in accordance with the lorentz mass formula

user128101

Ok, but this is completely new to me, I thought your main argument was that the mass increase had to be rejected because not all observers agree

user128101

That is a weak argument, isn't it?

John Rennie

I can explain what I meant by that, but that's another few minutes. Shall I go ahead?

Secret

9:44 AM

https://en.wikipedia.org/wiki/Ununoctium

If there's a way to perturb the energy states in all spontaneous decay routes of these superheavy elements externally so as to inhibit their radioactivity, it would be great, as right now they are simply way too unstable to study for more than mere moments

should read up what controls the energetic barrier of particle decays...

user128101

You are welcome, your advice is precious

John Rennie

Even in regular mechanics some things are observer dependent. So for example you and I will disagree about the kinetic energy of an object if you are moving relative to me or vice versa. Yes?

But some things we will agree on. For example we will agree on the mass of the moving particle (Newtonian mechanics remember). Yes?

user128101

No, John, that is a fallacy. Observers are partially blind, you cannot trust them

user128101

You can't trust anyone, actually. So differents conclusions mean nothing

John Rennie

I'm assuming we have ideal observers who don't make mistakes. If you reject that notion it's going to be hard to make any progress.

user128101

9:48 AM

I am honoured to talk to you, give me the benefit of dout and follow me step by step

John Rennie

OK, go ahead.

user128101

you are flying in a plane, at 3000 km/h all windows shut, no intruments. What is your speed you feel you are still? you open a window, you change your mind, you have intruments, you get a more perecise indo, do you follow me?

John Rennie

Yes, I think so.

And what that shows is that speed is a somewhat ill defined parameter.

user128101

you have no instruments, but you have an accelerometer you used during lift off , everything is blind, do you know your speed? the fact that you change your mind does if affect your actual speed?

John Rennie

You'll have to define what you mean by actual speed

user128101

9:54 AM

yeah, your speed relative to earth is well defined, same is the fuel you burned, the friction you met, the Ke you possess, the speed. The fact that you realize or not does not matter, right?

John Rennie

Your speed relative to the ground is well defined, yes.

user128101

But if windows are blind, you think you are not moving, if you have instruments you slowly change the accuracy of your measurements, but that does not matter

John Rennie

When you're sat in the plane with all the windows shut all you know is your speed relative to the plane, which is zero. So it isn't the same as the speed relative to the ground.

user128101

Yes, your judgement is wrong, you think you have no Ke, but you are wrong. The same applies to any relative observer. The ignorance of actual data does not affect it, so any conlusion by different observers is justified and mean nothin

John Rennie

No, that statement is wrong, because there is no correct value for your KE.

Your KE relative to the ground has some value while your KE relative to the plane is zero.

user128101

10:00 AM

You agreed above that speed relative to ground is well defined

user128101

Now you might go to next step and argue that that speed is not absolute,right?

John Rennie

Yes, it's a well defined quantity that is different to your speed relative to the plane.

@ally Aha! Yes, that's my point. Speed is not absolute!

Speed is an observer dependent quantity.

user128101

No, I foresaw that, try to reflect now, since this is a widespread fallacy

ACuriousMind

...what

skill patrol

:-/

John Rennie

10:02 AM

You're about to disagree with every physicist, but go ahead ...

user128101

That is not physics, it is plain, basic logic

John Rennie

Everything is physics ...

skill patrol

Explain it then logically.

user128101

Now the man on the plane thinks he has no Ke , but he is wrong, since he has same Ke of the plane but he simply ignores it,right?

ACuriousMind

@JohnRennie Some mathematicians work hard to do things we can't possibly use in physics. We get them all eventually ;)

skill patrol

10:04 AM

Please :-)

user128101

That is not math, it's the basis of any physical coherent world

John Rennie

The man on the plane isn't wrong. KE is an observer dependent quantity. Relative to the plane the man does indeed have no KE.

While relative to the ground he has a KE > 0

Both statements are true

user128101

Sure, but he does have Ke, sameas plane, the pilot knows it , he doesn't, that is the difference

ACuriousMind

You're always omitting relative to what they "have ke"

user128101

That is understood by now, follow me.

John Rennie

10:06 AM

The man in the sealed plane knows his KE relative to the plane is zero. He doesn't know his KE relative to the ground because he has no way of measuring it.

ACuriousMind

"having ke" is not an invariant property of a thing, the plane has a different kinetic energy with respect to the earth than with respect to the sun

skill patrol

relativity

user128101

Sure he is not stupid, he is ignorant, nevertheless we know he is moving at 3000 km wrt earh

user128101

right?

John Rennie

OK, and that allows us to calculate his KE relative to the earth

user128101

10:08 AM

Physics, science is all about knowlwdge

user128101

right?

Secret

Interesting to brought this out (because I always thought this for energy only occur in relativistic conditions). But how is conservation of energy be understood?

for now, let's consider newtonian mechanics
Suppose I and a lump of ball move with the same velocity v wrt the ground, thus the ball is stationary wrt me. Therefore the ball has no KE wrt me

Now in the ground frame, the ball has a KE of $\frac{1}{2}mv^2$

So if I initially started in the ground frame, and then accelerated myself into the moving frame, I will see the ball's KE slowly go to zero wrt me. But then where does the ene…

John Rennie

@ally I can't say "yes right" or "no wrong" because the statement "science is all about knowlwdge" doesn't mean anything

ACuriousMind

@Secret I answered that very question long ago.

user128101

as you wish, science is abouth anything?

John Rennie

10:10 AM

@Secret it would be cool to defer this until we've sorted ally's issue

Secret

ok

John Rennie

Science starts from a set of assumptions and uses equations to make predictions.

skill patrol

Knowledge is relative to the knower!

John Rennie

So if we have an object with a velocity relative to us of v we know the KE = 1/2 mv^2

user128101

So , the man in the plane , if heis a man of science, or at least of logic, should conclude that he has same Ke/speed of the plane , and that,alas he doesn't know what that is

John Rennie

10:12 AM

The point is that KE is different for different observers, just like velocity is

user128101

Let's conclude this case and then move to two observers

John Rennie

The man in the plane knows his KE relative to the plane is zero

@ally OK, if that will simplify the argument

user128101

...and if he is wise should only that he doesn't know what is the Ke of the plane and try to find out, not to conclude that Ke/speed is relative

John Rennie

But KE is relative

How can it not be? We've agreed speed is relative and KE = 1/2 mv^2

user128101

that's is a pre-judice we are trying to refute

John Rennie

10:14 AM

So if the v in that equation is relative that means KE must be relative as well

ACuriousMind

@ally You're not refuting anything. For instance, why are you using the speed relative to the earth and not the speed relative to the sun to compute "the Ke"?

user128101

that formula only raltes Ke to speed, could we please refer only to one of your choice?

John Rennie

@ally I'm not sure what you mean by refer only to one of your choice

user128101

one thing at a time, John,Earth we can touch and measure, Sunis far away

John Rennie

OK

skill patrol

10:16 AM

You can not refute what you observe relative to yourself

user128101

refer only to KE or speed,it's the same, right?

John Rennie

KE and speed are linked by KE = 1/2 mv^2, so if you know v you can find KE and the other way round.

So if that's what you mean by the same then I agree

user128101

That's why it is uslessly redundant to quote themboth

John Rennie

OK, but I still don't see your point

user128101

it is going to be a little difficult, but , try

user128101

10:19 AM

we concluded that man on the plane can only conclude he has sameKe of the plane, he doesn't know what is it,but that Ke (wrt earth) is not relative to an observer

John Rennie

There's a problem with that statement

user128101

if you think that any ignorant other observer can question that Ke you should prove that

John Rennie

I'm guessing you mean to say the man in the plane doesn't know his KE relative to the Earth. Yes?

user128101

ANY observer, John

John Rennie

You mean the man in the plane doesn't know his KE relative to any observer except an observer sitting next to him in the plane. Yes?

user128101

10:23 AM

the Ke of the plane is a hard, objective scientific fact /datum, nobody can question, It can be greater, if earth is moving, not less, by any means

John Rennie

The KE of the plane relative to the earth is something we can all agree on, yes.

skill patrol

Do you agree that an observation depends on the observer?

user128101

Any observer who draws a different conclusion is simply ignorant

John Rennie

@ally well, I'd say wrong rather than ignorant, but yes I think I agree.

user128101

ignorant , literally, ignoring some vital info

skill patrol

10:26 AM

Why is he ignoring it?

John Rennie

But what is special about the KE relative to the Earth?

skill patrol

An observer does not ignore vital info

user128101

If you agree so far, John, most of the job is done. If you understan this very simple,basic irrefutable logic , you cancarry it as many steps forward as you wish

user128101

Ke relative to earth is easily checkable, tht's why we chose it.

John Rennie

I agree that all observers will agree about the KE relative to the Earth. But I don't see how that gets us any further forward.

user128101

10:30 AM

It's very simple , John, Eart is our universe of discourse, if earth wer the Universe we would have the absolute, since it is not you have to advance your 'knowledge'

John Rennie

@ally I don't know what that means. The Earth is an arbitrary reference frame that happens to be convenient.

We tend to measure velocities relative to the Earth because it suits us.

But we can measure velocities and KE relative to anything we choose.

user128101

That makes no difference, John, aslongas you canmanage. Now,first, If you use the same frame for two or more observers, you reach same conclusion: what different observer say carries no weight, their evaluation can be distorted in all ways

John Rennie

You say If you use the same frame for two or more observers, you reach same conclusion and yes I agree.

And if the two observers use different frames then they will get different results i.e. reach different conclusions. Yes?

user128101

GOOD, it was not so bad, after all. WE better star from earth, it is easier, much easier. Now , frames , according to basic logic should be gradually enlarged, not invented. Those are the steps I was suggesting

John Rennie

Just to make sure we agree, are you agreeing that:
>if the two observers use different frames then they will get different results i.e. reach different conclusions

skill patrol

10:36 AM

^

user128101

That is too wague to comment, make a concrete example with the plane

user128101

but the point is: who cares if they reach different conclusions as long as it is not speed of light?

John Rennie

The man on the plane thinks he is stationary so his KE is zero. The man on the ground thinks the plane is moving at v so he thinks the KE = 1/2 mv^2. So different conclusions. Yes?

ACuriousMind

@ally What does that mean? Why would "basic logic" dictate that frames should be "gradually enlarged"? Why would we "better start from Earth"? It's simply convenient for us to use Earth, but nothing privileges Earth over any other reference object except us living on it.

user128101

we agreed the man on the plane is ignorant, why recall it?

John Rennie

10:38 AM

Are you agreed that is a good example of how observers get different results if they use different frames?

ACuriousMind

And saying that Earth is special because we're living on it is just the age-old "sun rotates around the Earth" argument all over again.

John Rennie

@ACuriousMind chill, I think we may be very nearly there ...

user128101

No, John, it is not a matter of frame, is a matter of 'savvy'. The pilot and the man on earth are on different frames but same exact right conclusion, the passenger does not becaus he is ignorant,

John Rennie

OK answer me this. What does the man on the plane think his mass is? Ignore relativity for now.

user128101

Let's not get too many irons in the fire, let's agree on the basic principle. I hope you, too, can ignore relativity for a while

John Rennie

10:43 AM

I'm asking that question because it is of vital importance to the debate. Suppose the man on the plane weighs 70kg. What does the man on the plane think his mass is?

skill patrol

Passenger = pilot = observer on the plane

user128101

The man on the plane is ignorant, probably doesn't know the difference between mass and weight, can we stick to the example?

John Rennie

Assume mass and weight are the same, and answer the question. What does the man on the plane think his mass is? It isn't a trick question.

user128101

I do not think it's a tricky question, I do not know what tha man thinks besides that he is not moving

John Rennie

Well suppose I said the man on the plane thinks his mass is 70kg. Would you agree with that?

user128101

10:47 AM

I can't agree on people's ungrounded opinions, John, can you? Can we conclude this example, i have to go to lunch , shortly

John Rennie

OK, see you later.

ACuriousMind

@JohnRennie ...you are a very optimistic man :P

John Rennie

@Secret are you still interested?

Secret

Let me just rearrange my thought process because I was destracted by a video...

user128101

I can conclude this example if you have any more considerations, do you agree that even if you consider more observer anywher on earth, in the air, moving in any direction reaching all sortsof conclusions, the Ke of the man in the plane is not affected and remains an objective fact?

John Rennie

10:51 AM

No, I disagree - that is completely wrong.
If you define the KE as the KE measured by an observer at rest on the Earth then I agree everyone will agree on that quantity, but you have omitted that condition from your statement.

user128101

We agreed it is understood. If you want to make an articulate example you think can disprove that, take your time, open a chat room and ping me in the afternoon or any time in the future, we are almost on same time zone. Thanks for your attention

John Rennie

You are welcome

Secret

@JohnRennie ok, using @Acuriousmind's answer, when we switch frames, the time derivative of energy changes, although we knew in all momentarily comoving frame energy is still conversed.

So if in all momentarily comoving frames, energy is still conversed (meaning that the time derivative of energy is zero), where does the energy difference came from before and after the change of the frames from e.g. one that is stationary with a moving object, to one which is stationary wrt the lab frame?

what supplied the extra energy to boost from one frame to another if it is only the frame that is changing (and in every infitesimal step of the process the time derivative of the energy vanishes)?

ACuriousMind

@Secret No, the time derivative of energy doesn't change. It's zero in any fixed inertial frame. Always. The change of energy you are "seeing" is not a consequence of time evolution but of you switching your reference frame.

John Rennie

@Secret Have a read through:

32

Q: How am I able to stand up and walk down the aisle of a flying passenger jet?

The energy of a moving object is $E = mv^2\;.$ That is it increases with velocity squared. I walk at say 3 miles per hour, or lets round that down to 1 meter per second for a slow walk. I weigh less than $100~\mathrm{kg}\;,$ but lets just round that up to $100~\mathrm{ kg}$ for convenience (it i...

newtonian-mechanics energy reference-frames inertial-frames

I think it contains a lot of relevant stuff.

It doesn't directly address your question, but the arguments about relative accelerations and energy changes are relevant.

As for myself, I have a bottle of gin somewhere - a large one. I may be some time.

skill patrol

10:59 AM

I think ally is playing dumb :P

ACuriousMind

@JohnRennie It's the middle of the day!

skill patrol

When backed into a corner he refuses to accept fact.

John Rennie

@ACuriousMind yes, but I feel like I've just gone twelve rounds with Tyson Fury :-)

skill patrol

Tyson

John Rennie

@skillpatrol Ah yes Tyson Fury - a Brit!

skill patrol

11:03 AM

Indeed, fighting July 5

ACuriousMind

@JohnRennie So a large bottle of alcohol is what one should consume after physical exertion? oO

John Rennie

@ACuriousMind it's mental oblivion I am seeking :-)

ACuriousMind

I might take up sports, then...

John Rennie

It's a good cure for post traumatic stress disorder

Secret

> The change of energy you are "seeing" is not a consequence of time evolution but of you switching your reference frame.

how is the energy supplied so that switching frames will result in different energy "seen".
Since you say the time derivative of energy is always zero, is the picture sort of like there exists many reference frames for an object in motion, and in each reference frame a unique value of energy is assigned to this object. Thus switching reference frames is akin to moving through these stack of reference frames from A to B in parameter(?) space?

->Meanwhile still reading the answer, but I get the argument about energy supplied by muscles to both the plane and the person. What I am wondering ab…

ACuriousMind

11:06 AM

@Secret There is no "energy supplied". Why do you think there needs to be "energy supplied"?

It's simply the case that energy is not a frame invariant.

John Rennie

>It's simply the case that energy is not a frame invariant
...

As I've spent the last two hours trying to explain :-)

skill patrol

Ding, round 2 :P

John Rennie

OK, I give up, what are the conditions for getting that vertical bar on the left?

ACuriousMind

@JohnRennie Writing > text, I think. If you write no space between the > and the rest, it doesn't appear.

> test

John Rennie

> Ah yes

That was easy, thanks :-)

Secret

11:16 AM

Clarification of my question to be posted in a few sec, John, you might be able to help translate it to Acusioudmind because I am not sure how to mke it more linear...

@JohnRennie

typo: Does it mean that as the observer changes from frame A to frame B
typo: Hence energy can change with changing frames but not due to nonzero time derivative, because each frame has a unique value of energy assigned to the object (which value measured wrt other frames) and we are jsut moving between these stack of frames

ACuriousMind

You really need to stop drawing so many pictures and learn how to actually say what you mean instead of other people guessing it.

John Rennie

OK I think I've recovered.

@Secret Let's take a simplified example. You start at rest and the ball with mass $m$ moves at velocity $+v$.

Now at time $t=0$ you start accelerating at a constant acceleration $+a$, and you continue this acceleration until you have matched velocities with the ball.

@Secret Is this a satisfactory scenario to address your question?

Secret

yes, and the question is then, what slows down the ball wrt me given I am not in contact with it (in fact, nothing is actually contacting or exerting a field)

ACuriousMind

I see we need to exorcise the ghost of "causation" again :P

John Rennie

OK initially the KE of the ball is $\tfrac{1}{2}mv^2$, and at the end the KE of the ball is zero, so that implies work was done on the ball. Yes?

Secret

11:24 AM

yes, based on work energy theorem

John Rennie

OK let's switch to your rest frame. In this frame you are stationary and you feel a constant acceleration of $a$. So far so good?

Secret

yup, until the point I match the ball's velocity, where I stopped accelerating

John Rennie

Good. Now the key point is that this is not an inertial frame, and in none inertial frames we generically expect there will be fictitious forces.

Fictitious force

A fictitious force, also called a pseudo force, d'Alembert force or inertial force, is an apparent force that acts on all masses whose motion is described using a non-inertial frame of reference, such as a rotating reference frame. The force F does not arise from any physical interaction between two objects, but rather from the acceleration a of the non-inertial reference frame itself. As stated by Iro: Such an additional force due to nonuniform relative motion of two reference frames is called a pseudo-force. Assuming Newton's second law in the form F = ma, fictitious forces are always p...

In this case, we will find that if we use the frame coordinate $x$ then every object whose position we describe by $x$ and whose velocity is given by $dx/dt$ will experience a fictitious force.

That fictitious force will be F = -ma$, where $a$ is the acceleration I introduced above.

This is the key point, and you have to be happy with this because the rest is just algebra.

skill patrol

Have you seen the movie The Exorcist? @ACuriousMind :P

Secret

So basically in the accelerating frame, everything except me and anything that is "co-accelerating" with me is being exerted by a field of fictious force -ma, thus naturally the ball will be retarded by this fictious force field, because it is not "co-accerlerating" with me?

John Rennie

11:31 AM

@Secret Exactly. And if you work out how long the force acts for, i.e. how long it takes to bring the ball to rest, then work out the distance moved by the ball, then work out force times distance, you will find it comes to 1/2mv^2.

i.e. in this frame the work done by the fictitious force on the ball to bring it to rest is exactly the same as its original kinetic energy

Now ask me about string theory - I'm good at that :-)

Secret

I see, before today's discussion, I always thought fictitious force act locally (i.e. in the free body digram, I only have one arrow instead of a field of them), which is why it is not apparent to me initially

skill patrol

As long as the ball is not a 'savy' observer :D

John Rennie

@Secret and that is why if a student attempts to solve a problem using a fictious force you mark them zero out of ten and tell them to come back when they've done it properly.

Fictitious forces are the root of all evil

Even hardened old physicists like me can get confused if you attempt to work with fictitious forces

skill patrol

I thought you're a physical chemist?

Slereah

he's a soapologist

John Rennie

11:37 AM

We physical chemists have an inferiority complex, so we claim to be physicists when we think we can get away with it.

skill patrol

:-)

John Rennie

@Slereah actually I never did any work on soap, or at least no serious work.

I worked on:

antiperspirants

shampoo

ACuriousMind

So...you infiltrated this SE and gave good answers to an incredible number of physics questions just to compensate an inferiority complex?

John Rennie

toothpaste

Slereah

So what would you say is the best shampoo

John Rennie

11:39 AM

mayonaise (can't remember how to spell that)

tomato ketchup

and so on, but never soap

ACuriousMind

Yeah, but "toothpastologist" doesn't have a nice ring to it ;)

Slereah

What about mayologist

John Rennie

@Slereah is that a serious question i.e. do you want the long answer?

Slereah

Well I have no hair

So guess

ACuriousMind

@Slereah what?

I remember a photo of you eating a reindeer where you had hair

Slereah

11:40 AM

That was a reindeer hat, @ACuriousMind

skill patrol

What's your favorite physical chemistry textbook @JohnRennie

Slereah

The hair was not my own!

ACuriousMind

Oh :D

Secret

@JohnRennie
I also understood that fictious force arises due to the coordinate transformation between frames (i.e. it has a geometric origin).

Now since I prepared that diagram. Is the "more geometric way" to explain that is that since energy is frame dependent (i.e. not lorentz invarient as acuriousmind highlighted), thus we can imagine there is an abstract space which there's a field of reference frames, and in each frame the target object has its frame dependent quantity being assigned a value unique to the frame (but the numbers can look different depending on which other frame you are…

Slereah

Also I do have some hair but usually I keep it shaved

John Rennie

11:41 AM

@skillpatrol Both Moore and Atkins are good in their own ways

I still have my copy of Moore, though my Atkins has gone astray over the years

Slereah

Do you remember IPAC chemistry nomenclature, or do you look it up when you need to

Secret

as a rough analogy, it is similar to how the world look different in red glasses compared to green glasses?

John Rennie

@Slereah in industry nobody uses the IUPAC names.

Slereah

Do you just call it "water" instead of oxidane

John Rennie

For example you might use the chemical sles

Slereah

11:43 AM

You filthy casuals

John Rennie

meaning sodium lauryl ether sulphate

and even the long form isn't a proper IUPAC name :-)

@Secret err, yes, I think so, though I'm not 100% confident I grasped what you mean.

KE is coordinate dependent (are you listening to me ally? :-) so the figure you get is dependent on what coordinates you choose.

The change in KE when you change coordinates has no physical significance

Now if you discovered you'd changed the norm of the energy-momentum four-vector that would be a different matter ;-)

Secret

Ok, let me simplify

Consider the linear accelerated frame scenario:
Suppose we measured everything not frame invarient wrt a frame called the lab frame (it's just any arbitrarily chosen frame to get a reference point)

Suppose we have an abstract space. Inside this space, there's a straight line.

On this straight line, each point is a sequences of frames with increasing v.

We start at the point which correspond to the frame where the ball has v=0.

When we change frames, we move up the stright line, until we reached the frame where $v=v0$

John Rennie

I have to go. See you (all but one of you) later.

user116211

ACuriousMind

@Secret You don't "move" in the "space of frames". Motion happens in spacetime any nowhere else. And the "space of frame" is either the Poincaré group (for inertial frames) or the diffeomorphism group (for arbitrary frames)

user116211

11:53 AM

Does a physics office room look like that o.O

ACuriousMind

"Change of frame" is not something that happens in reality. It's just something that happens in your theoretical description of the situation.

user116211

Okay, this is John Ellis' office at CERN ;/

Secret

@ACuriousMind I see, but is the idea of a "space of frames" a good way to interpret the theoretical/computational(?) process of changing frames?

ACuriousMind

Yes, applying an element of the Poincaré group to everything is the charge of one inertial frame to the other.

Secret