No, I don't think that's correct.

The coordinate acceleration scales as a' = a/γ³ and the relativistic mass scales as m' = γm

So the product is going to scale as 1/γ²

@JohnRennie I've seen many references like this one that indicate F_x = F'_x

sciencebits.com/Transformation-Forces-Relativity

Havent you, like, spent at least a year on this? If not, months? Can't you see that the correct thing to do is to give up on dead end ways to do physics? The actual answer is that if you want to do modern physics, you should stop insisting upon pre-modern ways to look at physics.

users.physics.ox.ac.uk/~Steane/teaching/rel_A.pdf

No I was thinking about these things a while ago, but had to divert my attention to other activities

now I'm revisiting it again

The transformation law of forces in relativity is a straightforward question that any physicist should be able to understand

Well, nobody can stop you from being haunted by a ghost of your own making. But at least we have helped.

I'm just trying to get a good grasp on relativistic momentum

Because the traditional approach it is taught is inadequate

I am not sure what "traditional approach" you are talking about, but there are excellent teaching materials out there, though drowned out by the mediocre.

@Kenshin disclaimer a) i havent done this kind of SR specifically in like over a decade and b) i haven't looked that closely, but if you're referring to the last line on that webpage, arent they saying v=0, i.e. there is no relative motion between the two frames?

@qwerty there is definitely something being zeroed out. The one I linked is much more trustworthy and had a lot of other extra terms.

The traditional approach I believe is Lewis and Tolman

It is definitely a lot more ugly than is typically argued for.

@qwerty, it is the last line, but provided F_y and F_z go to zero, then we get F_x = F'_x regardless of the x direction velocity

@JohnRennie Oh, you might be interested to note that the notes directly critiqued your PoV, in a way that I also completely agree.

@naturallyInconsistent, I've seen the approach here: G. N. Lewis and R. C. Tolman, “The principle of relativity and non-Newtonian mechanics,” Philos. Mag. 18, 510–523 (1909). but this is what I consider to be inadquate and seems to be the approach propagated in most modern textbooks

@naturallyInconsistent the only other appproach I've seen is here: arxiv.org/pdf/physics/0402024 but it also has its own isues

So that's why I'm revisiting it from first principles

and have discovered a paradox

that noone seems to be able to resolve

@Kenshin Oh no, that is light clock and there is also barely any algebra in the momentum side of the discussion. This is incredibly woeful by modern standards. This is nowhere near the state of the art as should be mediocrely done in modern textbooks either

@naturallyInconsistent Where do I find the "notes"? I can't see them on the page that was linked?

@Relativisticcucumber Can't be. I'm a currently living the condensed matter dream

Or should I say nightmare?

@JohnRennie do you see the Oxford Andrew Steane's notes Rel_A.pdf that miao miao just linked above? That's a big tome

@naturallyInconsistent I'm sure there are better treatments and I've seen many, but none the less now that I've found the paradox I need to resolve it

Ah, your link.

But the paradox relies on first understanding that Fx = F'x

@naturallyInconsistent What page?

@Kenshin this is just wrong.

I've seen the proof of this, but there's no point in me continuing with the paradox unless you guys agree that Fx = F'x

you don't agree that Fx = F'x?

The closer I get to freedom from these stupid exams, the more I feel that will to study physics again. Yay :D

it can be proven

*subject to force in y and z directions being 0 of course

@HerrFeinmann yayy!

If I survive superconducitivity I'm back

This is the right moment to listen to Sinmas songs

@JohnRennie software page 22, labelled page 49, Section 3.5.5 Equation (3.62). Yes, I'm this level of precise when teaching students too, you're welcome

@Kenshin The pdf I linked showed a different result.

no it didn't

@HerrFeinmann yay~

I looked at that pdf already

@naturallyInconsistent I'm this level of precise when talking to myself lmao

I realized how much I like to write bibliography

@Kenshin what if I provide the page number? Will you admit to being wrong?

I don't want to see a definition of four force

but the transformation law from one reference frame to the next

yes please show the page number

@Kenshin software page 40, labelled page 67, Section 4.1.1 Equation (4.7). And yes, I'm expecting an apology

actually page 67 I was just looking at that now

and it confirms what I'm saying

That F_x = F'_x

equation 4.6

This is the last day of Christmas holidays, so Sinmas song for you all :P

equation 4.6 when you put in what energy is, you get F_x = F'_X

@Kenshin That clearly had a $\vec u\cdot\vec v$ in the denominator, and this can very well not be zero.

please apologise

I am not going to apologise because I made it clear that the transformation law is not so simple as you wrote. You didn't specify enough to have the offending term become zero.

I did, I said that force is in the direction of velocity

i.e F_y and F_z are 0

when that is the case

we get F_x = F'_x

this is exactly what is shown at the bottom here: sciencebits.com/Transformation-Forces-Relativity

and also consistent with the paragraph I pasted

@Kenshin Hm, unless this is just a sarcastic quip, you shouldn't ask users to apologize - even more so someone who is trying to help you.

This is what you said. It is insufficient for the denominator to go away if you said just these.

@HerrFeinmann you missed something

@HerrFeinmann only because he asked me to apologies first when he thought Iw as wrong

Ah

Sorry

@naturallyInconsistent that is sufficient

@HerrFeinmann no, dream! hehe

"forces and velocities in y and z directions are zero" is sufficient

@Kenshin No, it is not sufficient. The particle can be moving in the x direction too. That's literally how $\vec u\cdot\vec v\neq0$

Well, that holds for anyone who asked for apologies first :P

@HerrFeinmann you should have read the chat log for the particular situation before making a comment like that

@naturallyInconsistent I never said the particle had to be stationary in the x-direction

the particle can be moving in the x-direction and F_x = F'_x still applies

@Kenshin in which case then the correction term literally applies, and your expression is just straight up wrong.

No, let's start with equation 4.6 and we can prove mathematically I'm right

first I want you to write dE/dt in terms of F_x, F_y and F_z for me

No, that's on you, not on me.

dE/dt = (v_x / c) F_x + (v_y / c) F_y + (v_z / c) F_

I'm not interested in a dead-end way to do modern physics. Whatever "paradox" it is you have discovered, it is not of fundamental importance to modern physics, even though, yes, maybe it would be a cute curiousity to ponder.

I'd rather help you understand energy-momentum in SR. That is far more productive a conversation.

I understand that well already

I'm trying to understand forces in relativity, because that's where the paradox arises

essentially when you plug in dE/dt expressed as the force components

you can factor out F_x

and the factor cancels the denominator

If that is the case, I'm not sure why you cannot resolve the paradox yourself.

the factor is preserved for f__y and F_z

but disappears for F_x

so tthat's why when F_y and F_z are 0

you get F_x = F_x' as your text suggests when it says

it doesn't require v_x to be 0

@naturallyInconsistent maybe I could eventually solve it but I thought people here would have done the work previously

since it's a very basic relativity concept

forces are fundamental to Newtonian mechanics so surely how they apply in relativity is very basic for all people here

@naturallyInconsistent it may be that there's a fundamental problem with the way relativistic momentum is defined that leads to the paradox

@Kenshin yes, the odds are that someone would have sorted out that before. However, you have some unstated assumptions that are just not realistic: that people would go out of their way to work with something known to be tedious and a dead-end even though it was important to Newtonian mechanics. Yes, there are some basic treatments, e.g. constant forces, say, that quite many textbooks cover, but it is just out of the way of the standard curriculum, for great reasons

But if we want to advance science beyond the current state, it may mean we need to understand things better that are outside of the current curriculum

@Kenshin that cannot be the case, because it is momentum and energy that are fundamental and are the assumptions from which the rest of the tapestry of modern physics is built upon. To reject them would be to have to reconstruct most of the Standard Model. You'd better have amazing evidence before attempting that

It's possible that forces are more fundemental, but that it was easier for people to work with energies and momentums mathematically, and that's why the standard model is framed in those terms

it may be that working with forces will give the same results for most calculations as the current standard model

@Kenshin That is true, but you dont seem to have a thing that would provide a credible challenge to the status quo, nor have to demonstrated an understanding of the standard fare as it is. I mean, maybe if you could have stated your problem, we could at least judge?

well first we need to accept that F_x = F'_x

once you agree (obviously assuming forces in y and z direciton are zero), then I can proceed to the next bit of the paradox

but if you don't agree, of course the result won't look paradoxical to you

@Kenshin no, forces are the first in line for the guillotine when the quantum revolution came. There is no future physics theory in which it is fundamental to anything.

@Kenshin I wont have to agree or disagree, to at least give your problem a fair shot.

ok let's suppose tis' the case

Now consider two particles of equal mass, each exerting a symmetric force on each other

Now consider the proper frame of particle 1

That particle will have a force say F exerted upon it for some time interval dt

Now consider particle 2's proper frame

in particle 2's proper frame, it is a symmetric scenario, and it will also see a force of F (in opposite direction) acting on particle 2

ok, but let's go back to particle 1's proper frame again

but this time let's consider the force acting on particle 2 in particle 1's frame

Here is where we invoke F = F'

we know that the force on particle2 in particle 2's frame is F, so then it is F' in particle 1's frame, and F' = F, so the forces remain equal and opposite in particle 1s frame

with me so far?

Just continue~

ok now in each of the particles' proper frames, the interval of force is dt

however this means in particle 1's frame, particle 2 appears to interact over a duration dt' instead

So now in particle 1's frame the story is as follows: It sees a force F on itself applying for duration dt, and it sees a force F (Equal and opposite) applying to particle 2 but applying for a duration instead of dt'

Ok now we know that F = dp/dt

so the change in momentum dp is equal to F * short duration force is applied

So the change in momentum in this scenario for particle 1 is Fdt

and the change in momentum for particle 2 is Fdt'

and thus the total change in momentum is not zero

and momentum doesn't appear to be conserved

so that's the paradox

Ok, so actually it seems like your problem is closer to relativity of simultaneity than anything else. Do you have a determined functional form of your force law, or any other specifics that you want to assert?

no I was hoping to work with a generic force F, only specifying that F applies for some duration dt

in some particles proper frame

and by symmetry does the same in the other particles' proper frame

that's it

@naturallyInconsistent Mind, I'm not taking sides. Just saying that it could lead to unpleasant/heated conversations

But anyways, it's not my place to tell users what to do, so I myself crossed a line

I accept that in one particle's frame, the interaction of the other particle appears to be at a different time, but it should still be over a duration of dt' with a force F, so the total change of momentum over the interaction should be Fdt' even if not simultanous

@Kenshin Well, I don't see why you are transforming the dt when you are dealing with this problem. In particle 1's frame of reference, you can have particle 1 feeling +F dt and particle 2 feeling -F dt and so momentum is conserved. By symmetry, you can see that particle 2 also sees that momentum is conserved

But if the duration of F is being applied over dt in particle 2's proper frame, shouldn't F be being applied only over a duration of dt' in particle 1's frame?

@ACuriousMind I finally edited my answer to make it not imprecise, I hope :')

@HerrFeinmann well, again, you should read the chat log as to why that conversation flowed into that particular direction before making a judgement. Like, do you really expect miao miao to be taking random stances without first checking what would be an appropriate response?

@Kenshin That's part of the issue. If you are discussing forces, you would have to be extremely clear what it is that is being mentioned in this respect, and one tends to have to have specific force laws before realising that what I'm talking about is the actually real situation. Whereas if you pretend to be general and leave the thing vague, then you are extremely likely to drop into misconceptions like your problem, transforming something that should not have been transformed.

I'd even agree with you that naïvely, one should think that one would be needing to transform the time interval too

but it happens to be wrong here.

@naturallyInconsistent Not judging and my previous message was actually an apology for that :D

@HerrFeinmann that may not have been clear so if you could please apologies for that as well

@naturallyInconsistent yes I think that's the resolution, that the force transformation law F'_x = F_x is on the basis that dp is calculated by multiplying dt to both

i.e. I guess the short duration of interaction is already accounted for in the derivation of F'_x = F_x

I dont know how to reply to the two of you except with YAY

maybe with judgement eyes lol

thanks guys laterz

i'm glad this one had a happy ending

xD

yay~

yay

Still, I think, we should all work on a better "discussion culture"

tobias and herr f are like two of the nicest ppl regularly on chat. i don't think i've ever seen you two quarrel over anything, even physics

hehe, thanks :) I don't know if this is true for me, but at least I think I am trying to stay calm

@qwerty Seems that you didn't watch through the end credits

hehe

@qwerty I'm not really nice, but I rarely lose my cool. So long as the opposing part keeps the discussion rational and doesn't let emotions slide into the discussion, I have no reason to "raise my voice"

@TobiasFünke i've never had a message removed or been told off by acm but i think my frustration shows sometimes. actually i get frustrated when people throw shade even when it's not directed at me so maybe i'm commiting a meta-sin :p

@HerrFeinmann lol

I see

Most of the times in the chat I do not possess the necessary knowledge to argue with 100% certainty but I can guarantee that if someone attacks me on something I'm sure about, I'll make a job out of totally annihilating that person :P

lol

I don't know if I'm lawful neutral or lawful evil

@HerrFeinmann stab stab stab

@TobiasFünke i don't know what tone to read this in xD

lol

3rd tone~

in neutral one :p

@qwerty Imagine a chill Tobias, reading a newspaper while having breakfast and then raising his head with a "mhh", followed by "I see" and then back to the newspaper

"quiet tobias judgement/sarcasm tone" ? :P hehe

@HerrFeinmann haha. yes

@TobiasFünke there iz no neutral, centrist scum!!!

For me it's just Tobias collecting information

lol

@naturallyInconsistent haha

@HerrFeinmann pssst. don't tell everyone!

The thing that frustrates me the most in conversations is when people get emotionally invested or take opposition as an offense. The greatest offense someone can give me is to refuse to analyze the reasons of a fight :P

we haven't seen PM 2ring on chat in a while... he also seemed extremely nice

oh and slereah never gets into quarrels either

@qwerty slereah gets transcendental

It's funny that I was about to type it before reading nI's reply

Bolbteppa hasn't written for a long time either

i think i started visiting hbar regularly around september last year and i don't think i've met bolbteppa

@Relativisticcucumber not even as a theorem :P

>_>

🙈🙉🙊

lol

:D

@qwerty then you haven't been in a Landau gate

What happened to RR? He logged yesterday

@HerrFeinmann They've been here every now and then

and even currently is :d

@HerrFeinmann they got suspended for a while

but i dont think he got suspended these past few days

@qwerty Oh so suspension lets you log but not write

A chat suspension only blocks you from the chatrooms.

I mean log in the hBar.

then, yes

sorta like read-only

Mew!

I got 3 hours of sleep

But i made the train so W?

Cool

Time to do more lagrangian problems

Hru @handan_toddler

@Allie M I A O ~

Meow is always good

👍👍

Physics is kinda

cool

I like how thousands of extremely smart people over hundreds of years have paved the way for us

On the shoulders of giants.

I had to rush out the door to make the train and i forgot a winter cowt

Coat

This walk is going to suck

Maybe i should make it a run

Sounds... healthy?

Mew

I hope i get into a good grad program

Hi everyone. In this answer, I do not understand the following sentence: "$E(m,v)$, if it is invariant, must be proportional to the mass, because you can smack two clay balls side by side and get twice the heating, so $$ E(m,v) = m E(v)$$." Can anyone shed some light on this?

@Bml What exactly do you find unclear?

@ACuriousMind How is the equation $mE(v)=E(m,v)$ a consequence of the invariance?

@Bml It's not a consequence of the invariance alone but of the other argument in the sentence - that you get double the energy if you smack two balls in succession into the wall (it makes no sense that the second ball would deliver a different amount). But whether you do that in "succession" or just with one lump double the mass shouldn't matter either, so $E$ must be linear in $m$ since we can linearly combine masses and expect the energy to just add up.

@ACuriousMind "you get double the energy if you smack two balls in succession into the wall (it makes no sense that the second ball would deliver a different amount)": I still do not understand why this is obvious. The energy of a ball is $E(m,v)$, that is, it is a function of its mass and velocity, and if this is equal to $mE(v)$, it means that the dependence must be linear as you say.

However, I cannot understand why this is evident. Could you write down some equations that would make me understand better?

@Bml There are no other equations. It's part of Ron's derivation here that you know heat is additive (either from thermodynamics or empirically via thermometers), i.e. doing the same kind of heat-producing thing twice yields double the heat.

If you don't accept that, it doesn't work and you need another derivation, e.g. physics.stackexchange.com/a/648635/50583

Anyone can explain how checking for gauge invariance, helps us know whether we have considered all the feynman diagrams for a process or not? In my notes, for the compoton scattering $\gamma +e^- /rightarrow \gamma + e^-$. In order to check the gauge invariance we say that we substitute for the outgoing photon $\epsilon_\mu(k)'$ with $k_\mu'$. And by doing so we get a zero. And we say that this hints that we used all the correct diagrams.

And as such only two diagrams, for the above mentioned process are gauge invariant

But what is the logic here? How exactly gauge invariance tells us anything about the correct nr. of feynman diagrams?

@ACuriousMind OK, but my doubt is another. If we admit that heat is additive, why should it be obvious that the dependence on mass is linear, that is, that $E(m,v) = mE(v)$? Why could it not be $E(m,v) = m^2 E(v)$ or any other dependence on mass? The equations I was asking about are along these lines, because I did not understand what is the obvious reason why the linear dependence on mass is derived from the argument of the two balls.

And isn't gauge invariance related to the invariance of the lagrangian under gauge transformations? And isn't this a default thing, since gauge transforamtions are a subset of the symmetry transformations?

@Allie you will; don't worry :-)

@imbAF The Ward identities say that the scattering amplitudes decouple from the spurious polarizations at each order (if you used a gauge-invariant process to define the orders). By putting in a spurious polarization and checking your amplitude is zero for it, you show that's true. It's not a proof that your amplitude is correct, but if you had gotten a non-zero value, that would have been proof it's wrong.

And since the Ward identities are a consequence of the gauge symmetries, that's a "check of gauge invariance".

In the lecture, nothing about Ward identities and such was mentioned. Is that a common thing?

I guess you're not really checking the Ward identity, you're just checking the unphysical polarization is zero

Scatering amplitudes would be $|\langle f|S|i\rangle|$ ?

Without the absolute value

But yes, I would expect any lecture that discusses QFT and symmetries to discuss Ward-Takahashi identities. It's a standard topic.

@imbAF I don't understand the question. What are you calculating those Feynman disagrams for if not to compute scattering amplitudes?

You're not randomly computing diagrams because you like diagrams, you're doing this because it gives you scattering amplitudes. If your diagram values give zero the corresponding scattering amplitude is zero.

@ACuriousMind I am not arguing anything here. I am trying to understand or learn the terminology

What we are calculating is the cross section

or the differential cross section

And the cross section, as you know, is proportil to the transition probability, which itself is proportional to the transition amplitude

and the transition amplitude is $|\langle f|S|i\rangle|^2$

So, unless I am missing something here, I don't see the scattering amplitude.

If you know what a "differential cross section" is, you must have discussed the relationship between differential cross sections, scattering amplitudes and S-matrix elements. This, however, is completely irrelevant to the question you asked. I'm not sure why you want to immediately derail the conversation.

I don't want to derail

what is spurious polarizations?

@imbAF If you are computing diagrams that involve photons you must have discussed the quantization of the EM field and how only two of the four theoretically possible polarizations are physical, no?

Yes

The unphysical polarizations are also often called "spurious"

So, this process if i am say so, of checking for whether one has considered all the possible diagrams, is possible only for when we have photons involved ?

Since the Ward identities, as you said,say that the scattering amplitudes decouple from the spurious polarizations at each order

@ACuriousMind Does my previous reply clarify what my doubt is?

And spurious polarizations has to do with virtual photos

@imbAF ...how would you put in a polarization of a photon as $\epsilon^\mu = k^\mu$ if there are no photons?

@ACuriousMind I am not arguing this.

@ACuriousMind It's the 0-tuple of polarizations, clearly

Hi besties

From what you told me, I have a simple idea behind why gauge invariance is relevant here

I am asking

indirectly

@Bml If heat is additive, then if you smack $n$ balls of the same mass $m$ into the wall after each other, you get $nE(m,v)$. But this has to be the same heat as if you smacked one big ball of mass $nm$ into the wall, so $E(nm,v) = nE(m,v)$, i.e. $E $ is linear in $m$.

How do they make such a delightful diagral

If a process doesn't involve photos, this process of checking for all the diagrams, from the gauge invariance, cannot be carried in such cases, i.e cases that involve no photons?

@imbAF Your question does. The method you are using is putting in $\epsilon^\mu = k^\mu$ as the polarization of your photon(s). You asked me if that can apply to when there are no photons. I don't know how to interpret this in any other way than you asking me if you can do the same when there are no photons, which makes no sense to me.

Consider an entirely different process, with no photons involved whatsoever

can you check for whether you have considered all the diagrams

from gauge invariance ?

Why would there be gauge invariance if there are no photons?

So when you solve the Euler-Lagrange equation, you get the path through phase space that minimizes the Lagrangian (and therefore the path that ultimately gets taken)

Right?

I had no idea that gauge transformations are restricted to EM

@imbAF Well, they're not, but if you have a gauge symmetry then you have some other gauge field with some other particle (e.g. gluons) for which you also have unphysical polarizations and you can do a similar check

@Allie technically it's only a stationary point of the action (the E-L equations do not guarantee a minimum) but otherwise yes

Right

@Slereah once you have grasped the secrets of category theory those of TeX are mere trifles compared to it :P

And when youre solving the Euler-Langrange equation its not w.r.t any specific path, right? You pretty much just get how the system will evolve at any given point in phase space

@ACuriousMind Could you elaborate on this. I don't understand. What I have from my notes is that a trasformation of the type $\psi\rightarrow e^{i\alpha}\psi$ is a gauge transformation global or local

@imbAF I really hope that's not exactly what your notes say. Because the EM Lagrangian is not invariant under that transformation alone.

@ACuriousMind It wasn't specified which Lagrangian is invariant. It was just said that that kind of transformation is a guage one

@imbAF and of course you have encountered a gauge field, it is the 4-potential of electrodynamics, which you quantized to get your photons in the first place!

local if there is spacetime dependency

@ACuriousMind it's the hardest part really

@ACuriousMind We haven't really

@ACuriousMind If $E(nm, v) = n E(m, v)$, I don't see linearity in $m$: if we substitute $n=1$, we have the identity $E(m,v)= E(m,v)$...

@imbAF You really need to start asking questions much much earlier if this is really your level of understanding. You say you are computing diagrams with photons in them and checking "gauge invariance", which means you must have written down the EM Lagrangian in terms of $\psi$ and the 4-potential $A^\mu$.

This Lagrangian is supposed to have gauge invariance (otherwise how does any of what you said make sense?) but it is not invariant merely under the transformation $\psi(x)\mapsto \mathrm{e}^{\mathrm{i}\alpha(x)}\psi(x)$

But one final question I have is the following. When a process involves photos and/or electrons. To compute the matrix element, we average over spin and polarisation of incoming particles and sum over all the possibilities (spin, polarization) of outgoing ones. If this is an accurate statement, why average for the incoming and sum for the outgoing. In both cases, I can argue that we don't know anything about the spins and polarizations

@imbAF You're just assuming that a good model of "not knowing the spins/polarizations" is assuming a uniform distribution over spins/polarizations so you can just take the normal average.

@ACuriousMind You take the average for the incoming ones, but the sum for the outgoing ones.

@Bml The equation holds for every choice of $n$ and $m$.

Why not take the average in both cases? Or sum for that matter

@imbAF Because in the outcome you're just not measuring the spin/polarization

so you sum over all the possibilities you can't distinguish

that's just how probabilities work

@Bml Look, it was $E(nm,v)=nE(m,v)$; you can swap m and n and take $n=1$ at the end.

I know how probability works lol. But I got confused why average for the incoming and sum for the outgoing, when in the lecture and in the book I am reading qft from mandl shaw the argument givne in both cases was that we have no information about spin/polarization

@imbAF And indeed it is in both cases the argument. You just need to think carefully about what "not knowing" means. For the input, that means you need to choose some distribution that models your lack of knowledge, which is generally agreed upon to be the uniform distribution. For the output, that means you don't measure it and so cannot distinguish the cases, meaning you have to add the outcome probabilities for each distinct case you can't tell apart.

It's logical if you think about what each step actually is supposed to model instead of just being confused that a similar-sounding assumption may produce two different formulae.

@naturallyInconsistent OK, so we have $E(m,v) = mE(1,v)$. Do we have to assume $E(1,v) = E(v)$?

@Bml no, there is a free choice of proportionality constant

Ron's answer just dropped that but as you should well know the conventional choice is $\frac{1}{2}$

@ACuriousMind Ok. I see. I am not saying that I fully understand it. But I will try to read it a couple of times, both answers to my questions. And I hope I have a better understanding

@Madder surely we need some other properties of the $\psi$ and/or $\Phi$ function(s) that is not shown in your image

@ACuriousMind I don't understand. If we solve the answer equation, $E(2v) = 4E(v)$, the solution is $E(v) = k v^2$, and we conventionally place $k= 1/2$. So, $E(v) = 1/2 v^2$, and since $E(m,v) = m E(v)$, then $E(m,v) = 1/2 m v^2$. But since we have seen that $E(m,v) = m E(1,v)$, then I infer that $E(v) = E(1,v)$. No?

@Bml If you define $E(1,v) = E(v)$, sure. You could just as well have make an ansatz $E(1,v) = k_2 E(v)$ and then in the end set the product $k k_2 = 1/2$. It makes no difference.

@ACuriousMind Yes. Also, I have another doubt: why is $E(m,v)$ invariant?

@Bml As the answer says: "I will show that, assuming Galilean invariance, [...]"

It's an assumption, not a derived statement

@ACuriousMind Yes, OK, but what is the intuition behind this assumption? That is, why should one believe that this has foundation in Newtonian Relativity?

To remember the terminology of adjoint modalities where the $p$ modality is denoted by $\Box$ and the $s$ modality is denoted by $\bigcirc$ I just remember that $s$ stands for sircle

@Bml To be honest, I personally find this way of "deriving" energy rather silly. We are no longer bound to Newtonian mechanics and if we use any action formulation of the free particle then you just find $\propto mv^2$ as the conserved quantity associated with time translation invariance by Noether's theorem, none of this weird guessing or thought experimentation necessary.

@Slereah sircle and pquare?

It's the most one can ask for

$p$ is for preceding and $s$ for successive bc $p$ is a comonad where the (co)unit has it precede the object $\Box X \to X$ while the $s$ is a monad with unit $X \to \bigcirc X$

Helpful to remember all that nonsense :p

@ACuriousMind Sure, but I find this topic much more interesting than the usual way in which kinetic energy is presented in Introductory Physics courses: that is, we do the line integral of the force and from there we get an expression that we arbitrarily call "kinetic energy." What I mean is: excluding special relativity and Noether's theorem, how is the invariance of heat from one reference system to another intuitive?

@Bml Stuff you see doesn't burst into flames if you go faster :P

so apparently how hot something is is something of an invariant

seems pretty intuitive to me (though it's perhaps false in special relativity as the topic of relativistic temperature is contentious) :P

What about for a Rindler observer with Rindler radiation tho

@Slereah ...I just said it might be false in relativity :P

If Rindler radiation is the same as the one experienced by an observer stationary on the EH and all that bullshit, I was thinking of the same thing

I'm under the impression that controversy is still not fully resolved (i.e. is temperature not invariant or is the non-invariant definition of temperature wrong)

Only special :p

do we really have to fight over whether or not a Rindler observer is special or general relativity :P

can't I go one day without having to rant about GR

Oh, by the way...

if you say "diffeomorphism" I'm banning you

(not really)

@Bml, v^2 = u^2 + 2as.

So long as the looming threat is present I'm satisfied

Hard to put a lot of cool examples of adjoint modalities and all that before the Hegel stuff in my analysis of the nlab stuff because all the good modalities are either in the Hegelian hierarchy or they're the worst things

@Kenshin ?

Just handwaving here, but even with the current definition of temperature as average kinetic energy, the worst that velocity composition would do is to break isotropy and move some of the energy into a specific component

@Bml that equation is sufficient to understand kinetic energy

Hard to have a nice example when the typical example of adjoint modality is like "a-adic completion / a-torsion appoximation"

Probably you'd have some effect on non 3d systems?

the only basic ones are the even/odd modality and the ceiling/floor modalities

Good modalities but only so much you can get out of them!

I should look at bireflective subcategories since they all define an adjunction

ncatlab.org/nlab/show/bireflective+subcategory#examples

they do look a bit more palatable