5:52 AM
So apparently asymptotically safe gravity correctly predicted the higgs mass
Around 12:50

6:03 AM
Yes, here is the 2009 preprint: arxiv.org/abs/0912.0208

@NiharKarve I can't believe that this didn't make headlines
Instead we hear about stringy landscape

7:02 AM
@MoreAnonymous see:
8

This paper was published in Phys Lett B in 2009, and predicted the Higgs mass to be 126 GeV based on the asymptotic safety of gravity. Is this prediction taken seriously by the theory community, or is it considered only to be a lucky guess? EDIT: Thanks to Heidar for this fantastic compilation ...

Betteridge's law applies :-)

@JohnRennie For a few moments in my life I actually knew what I wanted to do a Phd in
:P

7:34 AM

In an unusual reversal of events, I first found this paper - I don't remember how I came to that - that seemed to make strange claims (and is also weirdly disparaging of the people allegedly making mistakes), and then found an 8 year old question here that asked about the same confusions I had so I had somewhere to write down my thoughts

I am unable to arrive at the equation at the bottom starting from 5.7. Starting from eq 5.7, I expanded the mu index which would give four terms. In total, there would be five terms inside: four gamma terms and m. Assuming we're working with four components, I can see four differential equations.

The last equation looks like the four individual differential equations have been summed up to zero. But isn't this a weaker condition than requiring each differential equation to equate to zero?

@Yashas Even if it's a weaker condition, it's still true, isn't it?

@ACuriousMind Yeah but it might admit more solutions than what the original Dirac equation would have?

So? I'm not sure why the author writes that equation down, but I don't see a claim there that it's equivalent to (5.7), so what's the problem?

7:44 AM
@ACuriousMind Oh, I am not sure why I thought it's supposed to be equivalent but now I don't see why that summed form was even mentioned.

I think it's probable the author just make a mistake :P

8:07 AM
@ACuriousMind maybe you learned about it here :p

@bolbteppa Huh, I had completely forgotten that, but no, I didn't find this in the depths of my download folder, I stumbled across it a few weeks ago

I vaguely remember a discussion about this point in the book the other answer cites and that book is decades old so at the very least it's a known question mark

but good to see that me from one year ago agrees with me today :P

@Yashas the thing at the bottom is just (5.7) with the matrix indices written out?

@bolbteppa $\psi_k$ makes it look like they are summing over all components?
each component of $\psi$ should have it's own differential equation, right?
oops, I think I understood

8:17 AM
$\psi$ is a column vector, with components $\psi_1,...,\psi_4$, so the equation is just usual matrix multiplication equation $M_{jk} \psi_k = 0$ for the matrix $M_{jk} = (i \gamma^{\mu} \partial_{\mu} - m)_{jk}$

@bolbteppa Yup, got it.
So I need to write the last equation four times by varying j from 1 to 4 to get all four equations.

Yes, of you can write the whole out as a big $4 \times 4$ matrix, where the entries of the matrix $\gamma^{\mu}$ are in general just $\gamma^{\mu}_{ik}$ so it's ugly, but if you have a standard explicit representation for the $\gamma^{\mu}$'s it simplifies in appearance
@MoreAnonymous there's so much to say about what's going on in that video before even clicking on it

9:16 AM
Talking about some issues with that 'Higg's mass' paper

0

Let's say i have written an answer to a question. I feel like i have done a good job of answering it and covered all the points that needed to be covered. Now, if i see some another answer that i feel brings up some point, that i realise i should have covered in my answer as well. Is it okay to e...

2 hours later…
11:01 AM
How to prove this: $\gamma^\nu\partial_\nu\psi = -im\psi$

That looks like a fairly trivial re-arrangement of the Dirac equation @Yashas

Now I forget again, why is it that if $(\gamma \cdot p)^2 = - p^2 = 0$ then the rank of $(\gamma \cdot p)$ is half the max $n$. Obviously if $A^2 = 0$ then $\mathrm{im}(A) \subset \mathrm{ker}(A)$ so $2\mathrm{im}(A) = \mathrm{im}(A) +\mathrm{im}(A) \leq \mathrm{im}(A) + \mathrm{ker}(A) = n$ so $\mathrm{im}(A) \leq n/2$, but the equality...

@Charlie Thanks.

Is it that the $\gamma$'s are irreducible or something

11:26 AM
Or we just say it has the maximum possible rank because physics

I searched "What is a tensor" in the internet and found this interesting answer from William Oliver (the first answer).
https://www.quora.com/What-is-a-tensor
It is interesting to me because It goes sort of contrary to almost all "what is a tensor?" type videos in youtube.

Why is there no energy term in the exponential? Isn't Dirac equation time-dependent and we should be plugging in the complete wavefunction?

@bolbteppa you need some additional information about $p$ to conclude that $\gamma\cdot p$ has rank 2, since e.g. for $p=0$ you'd just have rank 0.
@Yashas there is an energy term there, $p$ and $x$ are 4-vectors.

@Prithubiswas not sure about that answer at all, e.g. it looks like he's saying that people are wrong to call a vector a tensor because a vector is not a linear map and it's just that people get confused because a linear functional is a tensor, but a linear functional is just a vector in the dual (vector) space...

@Prithubiswas that answer just looks as if it is entirely unaware of the fact that the space of multilinear maps, i.e. tensors, is itself a vector space and hence tensors are certainly not only a "generalization" of vectors but really also just vectors :P
the viewpoint that tensors are multilinear maps is sometimes a better way to understand some things than viewing them as vectors in the space of tensors, but both viewpoints are fully equivalent, it's silly to claim one of these is the "true" way to define tensors

11:44 AM
@ACuriousMind "The space of multilinear maps" isn't that the tensor product of a vector space and a dual vector space according to wiki?

that's the space of bilinear maps
but yes
in general a $(p,q)$-tensor on a vector space $V$ is an element of $V^{\otimes p} \otimes (V^\ast)^{\otimes q}$.

Assuming $p \neq 0$ it's supposed to be so obvious that $(\gamma \cdot p)$ has half the max rank if $p^2 = 0$ that no extra explanation is ever needed apparently :\

There's a few standard ways of introducing tensors that are technically equivalent but all highlighting different properties and every source thinks their definition is the most "simple and intuitive". But what you end up with is a huge, garbled mess of overlapping definitions and re-phrasings that make any attempt to learn tensors from multiple sources about 100x harder than it should be

@bolbteppa in that case you have at least 2 non-zero components of $p$ (and you can in fact always transform into the frame where it has exactly 2 non-zero components), so w.l.o.g. you have $p_0 \gamma^0 + p_i \gamma^i$ for fixed $i$ (no summ. conv.). If you look at an explicit representation of the $\gamma^\mu$ it's easy to see this sum always has rank 2

I dont really get how physicists define tensors , (probably because I am in hi school).But here is how I think of tensors ... a (p,q) tensor is a special kind of multilinear map of the form VxVx..VxVxVx..xV* > R , where in the cross product , we have V p times and V* q times.(I guess this is kind a abstract definition)

11:49 AM
It's not a cross product it's a tensor product
$V\times V\neq V\otimes V$

@Charlie I did not use tensor product , it is cross product (check the wiki)

The tensor product can be constructed as a quotient space of the cartesian product, but it is certainly not equivalent to $V\times V$

I spell out the different but equivalent definitions of the tensor product in this answer of mine

ah I see what you've written now, sure. Cartesian product might be better wording
The quotient space definition is probably the nicest imo

also tensor products are a while another level of abstract concept.It is used to construct a space from multiple vector spaces and tensors are thought to be a member of it (But again , there are different definitions of tensor and different people use/buildup different definitions for convenience)
@Charlie That brings me to the wormhole of abstract algebra xd.
@Charlie also I used the cartesian product for multilinear maps and tensor products (circle with cross) to construct quotient spaces.

11:58 AM
The magnitude of the electric force (or Coulomb force) between two electrically charged particles is equal to $F_{12}(r)=k\frac{|q_1q_2|}{r^2_{12}}\vec{r}_{12}$
Shouldn't by saying magnitude there is no $\vec{r}$?
And if this is Coulomb force then shouldn't the vector be divided by it's magnitude in order to be unit vector?
I am sure I am not wrong.

@ACuriousMind saw you post.Great answer.This is exactly what I had in my mind.

may be may be $\vec{r}$ is already unit vector

12:13 PM
@ACuriousMind your fluent transistions to corrdinate definition to tensor product definition was cool.I kind of wish they taught this in textbooks.

I'm sure I didn't invent this :P

12:47 PM
@ACuriousMind Thanks. Why is 'u' assumed to be dependent only on three-momentum?

@Yashas since given three components and the mass of an on-shell momentum you can compute the fourth, you can write any function of on-shell 4-momenta for a fixed mass as functions of the 3-momenta if you like

@ACuriousMind what about the dependence on position?

what dependence on position?

@ACuriousMind why u(p) but not u(p, x)?

it's an ansatz, you just don't need it

1:53 PM
Hello World
Can a regular comtupar Engynar become one of the best qutnam Comtuparing Expert of the world?
Let's see

2:12 PM
One of the guys in the video posted above
From the article
"...only finally released a paper this year after years promoting the theory in public forums while questioning the legitimacy of peer review, lamenting the need to provide evidence, and otherwise dismissing critics or skeptics hesitant to accept his theory with open arms"
and all this is just from *one* of the people in this 'alternative physics' video posted above...

eh, Smolin and Hossenfelder are fine I suppose
this is a field day for Lubos, he'll be writing up a stinker on this
all his favourites in one video

Is Weinstein generally considered a nutcase in the community? It seems so strange that someone with a PhD in mathematics could be so detatched

I mean it's STEM
We're all insane

But we're not all "I have a theory of everything and everyone else is wrong" insane :P

Well some are

2:27 PM
Which in itself isn't all that uncommon, but from someone with actual qualifications is a pretty small subgroup surely
from Harvard no less

expensive schools aren't particularly better at making you not insane
just look at the US government

What's bizarre is that his theory is criticised pretty heavily for having physical and mathematical gaps. Which is understandable for someone who's flicked through a few Wikipedia entries and formulated a theory of everything, but not really for someone who's gone to the trouble of getting a PhD

It happens

I was under the impression that a PhD is a relatively hard qualification to fudge :P

Penrose's a pretty heavy GR hitter but he ended up having some pretty weird theories

2:32 PM
unlike a degree which I can say first hand you can get away with quite minimal work

A PhD is just a fancy degree

While deriving Dirac equation from relativistic dispersion relation and the linearity condition, we can show that the matrix parameters can be any matrix with dimensions greater than four and even. The textbook I am reading didn't explain why it went ahead with just four. What about the other possible dimensions?

Any multiple of 4 for a start I think?

Well you have to make a choice at some point

You can just make diagonal matrices of matrices
with the same properties
I do recall @RyanUnger posting an article once
about the proof of the number of dimensions of the gamma matrices

2:35 PM
With four-component vectors, the free particle solution admits two E >0 solutions corresponding to spin up and down, and two E < 0 solutions corresponding to spin up and down of the antiparticle.

I was wondering how someone would interpret the solutions if they started with six components or 8 components or whatever.

Well for a start, you can have, say, a system of multiple spinors
you could interpret it that way
There are also equations for particles of different spins, although I don't know if they obey the same algebra

@Charlie if anything it is a stark example of why this stuff rarely happens outside of the 'system', at the very least there's zero impetus forcing one to do anything beyond self-motivation, absolutely nothing stopping someone from saying things like 'I'm unable to find my old notes from years ago where I wrote this... I cannot even remember the standard theory' similar to that which you'll find in this 'masterpiece'
@Yashas the idea is, you're trying to find the simplest matrices which will work, it wont work in 3 dimensions because you need 4 matrices and the 3 Pauli matrices form a basis for 3D space so you can't get 4, thus you try 4 dimensions for the matrices.

3:11 PM
Once in a while I get -10 points because the user is banned
WHERE ARE MY INTERNET POINTS
GIVE THEM BACK

more often than not, "user was removed" is actually a self-deletion, not a ban (or rather, deletion by mod - normal suspensions/bans do not remove votes) :P

weird

3:56 PM
Would it be breaking any rules for me to open a new account every day, upvote Slereah's posts once and then delete it 12 hours later so I can slowly drive him to the brink of insanity?

by his own admission he doesn't need to be driven there ;P
2 hours ago, by Slereah
We're all insane
or rather, driving him to the brink would actually increase his sanity

It seems the pursuit of physics instils madness more than I could ever hope to

it's a weird idea
We were supposed to pick berries in the savannah
why are we thinking about the cosmos

Probably something in the berries

To think that one of the first astronaut was a monkey
Poor guy must have freaked out something fierce

4:07 PM
@Slereah poor guy didn't even know he was making history

At least he survived
Not all the space animals did

@Slereah hi
I have been told that you do neural network stuff
I will have to study NN for a project, is it okay with you if I ask a few questions here (not today tho)?

I mean sure, but I'm not an expert
there are better places to ask

just the basic stuff

@satan29 I can try to answer if the question is simple enough. I routinely train deep neural networks for my masters' work.

4:11 PM
oh,cool!
@Yashas are you familiar with the YOLO model?

@satan29 Yes

cool :-) Thats the main thing for my project

@satan29 are you training YOLO based models or using pre-trained models for inference?

Wikipedia says that the Einstein-Hilbert action doesn't always converge if you integrate over all of $R^4$ and if it doesn't you have to integrate over a large compact subspace. What kind of spacetime does this correspond to? Is it only those with singularities?

@Charlie The simplest case is anti de sitter spacetime
(or de sitter, I always forget which is which)

4:18 PM
Ah

It spans $\mathbb{R}^4$ but the curvature is constant
So you're just integrating a constant over $\mathbb{R}^n$

Ah so spacetimes for which the curvature does not vanish at the boundaries
ok that makes sense

This is true of all field theories, but in general people will skirt this by claiming that we're only considering Schwartz function or equivalent
Having homogeneous non-zero fields is a very bad idea in general but sometimes you gotta do it

3 hours later…
6:59 PM
Why do so many things transform like four-vector? I have learned the derivation of Lorentz transformations for position and time but this transformations generalizes to so many things. I don't know why other four-vectors like four-potential or four-current or whatever must transform exactly like momentum four-vector. It's a bit mysterious.
There probably is some deeper mathematical understanding which I mostly lack right now. But encountering so many four-vectors out of nowhere (textbook makes it look like, look here is something, show that it transforms like four-vector), it feels a bit fascinating how it just pops up everywhere.

7:28 PM
It's essentially a consequence of choosing the background manifold of your theory to be Minkowski space. You can formulate theories in $\Bbb R^3$, vectors like $\vec E$ and $\vec B$ are vectors on $\Bbb R^3$, but in special (and general) relativity "time" is just another direction on your manifold and we often find that the 3-vectors we are used to using in non-relativistic theories are actually just 3 of the 4 components needed in the relativistic theory.
Technically when we say something "transforms like a 4-vector" this is a "physics" way of saying that the object you are considering is a vector in $\Bbb R^{1,3}$, for which there exists a representation of the Lorentz group. When we say objects like $\vec E$ and $\vec B$ don't transform like 4-vectors we mean that there does not exist a non-trivial representation of the Lorentz group on $\Bbb R^3$.
@Yashas

2 hours later…
9:26 PM
Physicists
Tell me how spectral theory of rigged Hilbert spaces work

Kreyszig's functional analysis book includes applications to QM, haven't actually read that section but it probably goes into detail about that. Also I'm pretty sure Valter Moretti (a SE user) has written a textbook about the spectral theorem in QM which is pretty rigorous @BalarkaSen

Excellent, thanks! I'll check it out

1 hour later…
10:58 PM
0

Clearly, marking a question as a duplicate of either my own question or of a question that I answered is not against the rules. But, there seems to be a conflict of interest in the context of the "reputation-points economy" because I clearly stand to "profit" if a question is marked a duplicate o...