The h Bar

imbAF

12:16 AM

regarding the electromagn. tensor , what is the difference between these 2 notations $F^{\mu \nu}$ and $F^{\mu}_{\nu}$

imbAF

12:50 AM

Anyone can help me with a proof of symmetry for the electromagn. energy stress tensor ?

Nihar Karve

3:35 AM

@imbAF Different transformation properties? I'm not sure what kind of answer you're looking for here. Do you know the meaning of upper and lower indices?

@imbAF Have you heard of the Belinfante procedure?

imbAF

9:15 AM

no I havent

I am trying to prove that the electromag. stress energy tensor is symmetric. The following was done $F^{\mu}_{\lambda}F^{\lambda \nu}=F^{\mu \delta}g_{\delta \lambda}F^{\lambda \nu}=(-F^{\delta \mu})(-F^{\nu \lambda})g_{\delta \lambda}$

As we can see here, we displaced the metric tensor at the right side

can you do that in tensor algebra?

moving tensors without caring? Isn't that a problem? Doesn't it cause any problem?

ACuriousMind

@imbAF what do you mean by "moving tensors without caring"?

imbAF

it's in the formula i habe

gave

$g$ was in between

ACuriousMind

all of $F^{\delta\mu}, F^{\nu\lambda}, g_{\delta\lambda}$ are just numbers

imbAF

then it was moved to the right

ACuriousMind

why wouldn't they commute?

imbAF

9:27 AM

numbers do

but since they are part of matricies (in our case)

we know that for 2 matricies A B =/= B A

ACuriousMind

just because they are "part of matrices" doesn't mean they're somehow not numbers

if you write matrix multiplication as $(A\cdot B)_{ik} = \sum_j A_{ij}B_{jk}$, then you can equally well write $\sum_j B_{jk}A_{ij}$, there's no difference

imbAF

Yes, but here you have numbers multiplying too : A,B matrices than : $A \cdot B$=/= $B \cdot A$, or not?

aha ok

ACuriousMind

note that the latter is not $(B\cdot A)_{ik}$, it's still $(A\cdot B)_{ik}$

imbAF

I didn't get this last part

ACuriousMind

I'm saying $(A\cdot B)_{ik} = \sum_j A_{ij}B_{jk} = \sum_j B_{jk}A_{ij}$

changing the order of multiplication of the components does not change anything on the level of the matrices as long as you leave the indices as they are

(that's one reason physicists love index notation so much :P)

imbAF

9:33 AM

I see

I was considering always that matrix multiplication is not commutative

and I thought, how can we move the components so freely around

@ACuriousMind one additional question

In einstein notation, we say that the double index represents a sum. So if we have $F^{\mu \mu}$ and $F^{\mu}_{\mu}$, here the first one only is a way of representing an element of the 2nd order tensor (matrix) while the 2nd one is a sum over the elements of the main diagonal. But in the einstein convention, the first term also has a repeated index. Which means, we should sum over it. But that doesn't make sense.

So is it really enough to say that we need repeated indicies, and not repeated indices + one should be an superscript and the other a subscript (idk what is the term used in this case, i forgot it) , to indicate summation?

ACuriousMind

@imbAF $F^{\mu\mu}$ is simply not an allowed term with summation convention

if you have a double index and it's not one upper, one lower, then something has gone wrong

imbAF

wait

$\Theta^{\mu}_{\mu}=g_{\mu \nu} \Theta^{\mu \nu}$

here you have the same index, but one up and one down

I got it

My mistake

Slereah

10:54 AM

Is there a gauge theory with an $SL(4)$ gauge

The closest I can find is unimodular gravity but it's not a direct reduction

ACuriousMind

nice gauge theories have compact gauge groups :P

Slereah

What about naughty ones

$SO(3,1)$ isn't compact

ACuriousMind

sure, and GR sucks :P

John Rennie

It kind of does for r < 3r_s

ACuriousMind

11:12 AM

::groan::

satan 29

@JohnRennie is that a general relativity joke too specific for me to understand?

Slereah

GR doesn't need unitarity

Slereah

11:28 AM

SL(4) gauge would give rise to some scalar potential and I tried looking into scalar gravities, but couldn't find anything about that

Nobody seems to have tried a gauge version of Nordstrom theory

John Rennie

11:46 AM

@satan29 :-)

imbAF

12:21 PM

the metric tensor $g_{\mu \nu}$ is a diagonal matrix with the first element a pozitive 1 and the rest neg. 1

shouldn't $g^{\mu \nu}$ be the opposite?

With the first diagonal element a neg. 1 and the rest pos. 1?

ACuriousMind

@imbAF why should it be?

what is your definition of $g^{\mu\nu}$ in terms of $g_{\mu\nu}$?

just write it down and carefully compute it

imbAF

I know that both represent a matrix with diagonal elements 1, the first being pos. and the rest neg

but

Metric tensor

In the mathematical field of differential geometry, one definition of a metric tensor is a type of function which takes as input a pair of tangent vectors v and w at a point of a surface (or higher dimensional differentiable manifold) and produces a real number scalar g(v, w) in a way that generalizes many of the familiar properties of the dot product of vectors in Euclidean space. In the same way as a dot product, metric tensors are used to define the length of and angle between tangent vectors. Through integration, the metric tensor allows one to define and compute the length of curves on the...

Here you see the reverse

ACuriousMind

where?

imbAF

and if they are the same, what is then the difference? does one help lower an index while the other to raise an index?

I highlighted it

Metric tensor

In the mathematical field of differential geometry, one definition of a metric tensor is a type of function which takes as input a pair of tangent vectors v and w at a point of a surface (or higher dimensional differentiable manifold) and produces a real number scalar g(v, w) in a way that generalizes many of the familiar properties of the dot product of vectors in Euclidean space. In the same way as a dot product, metric tensors are used to define the length of and angle between tangent vectors. Through integration, the metric tensor allows one to define and compute the length of curves on the...

Lorentzian metrics from relativity

ACuriousMind

I can see that the URL is weird, but whatever you're doing doesn't do anything for me except open the article at its beginning

imbAF

12:26 PM

Lorentzian metrics from relativity in this section, in the end

ACuriousMind

why don't you just link the section you're referring to, the table of contents is a bunch of links for a reason

imbAF

Like this? en.wikipedia.org/wiki/…

ACuriousMind

that worked

I don't see anything about $g_{\mu\nu}$ and $g^{\mu\nu}$ being different there

imbAF

the matrices are different

ACuriousMind

I just see the statement that whether the metric is 1,-1,-1,-1 or -1,1,1,1 depends on a choice

@imbAF Are we reading the same thing? I don't see any statement that $g_{\mu\nu}$ is different from $g^{\mu\nu}$, I just see a statement that $g$ can be one matrix or the other.

imbAF

12:31 PM

I took it in alignment to what we learned briefly about the contra and covariant notations of a four vector

ACuriousMind

maybe you shouldn't interpret things into the article it's not actually saying :P

imbAF

How could I know lol, I am just blindly walking in a bridge

that;s how it feels to do tensor algebra with no tensor algebra classes

ACuriousMind

it's just saying that whether you choose to have 1,-1,-1,-1 or -1,1,1,1 depends on your convention, you can do relativity with both, you just need to be careful that the physical meaning of positive/negative spacetime interval reverses

welcome to sign conventions, one of the most annoying eternal curses of theoretical physics :P

imbAF

lmao so it's not only me lol

One thing for example, I am leaving the rest and focusing on this part of my notation

$(-F^{\delta \mu})(-F^{\nu \lambda})g_{\lambda \delta}=F^{\delta \mu}F^{\nu \delta}$. Could this also be written as : $F^{\nu \lambda}F^{\mu}_{\lambda}$. I think no, the reason being that in the first term $\delta$ is in the left and not the right. We could do that by using the antisymmetri of it, which would give me a minus in my formula. But I am not sure. Basically I am asking why did the metric tensor "collaborated" with the 2nd term and not the first?

ACuriousMind

$F^{\delta\mu}F^{\nu\delta}$ is wrong, one of the $\delta$s should be lower

whether you call that index $\delta$ or $\lambda$ doesn't matter

and why are you still worried about the order of the terms? didn't we establish that everything here is a number being summed over and so they commute?

imbAF

12:42 PM

oh fck

I wrote that last part wrong

and Now I have to do it again

order

Ok let me first write that thing correctly

@ACuriousMind We have this :$(-F^{\delta \mu})(-F^{\nu \lambda})g_{\lambda \delta}=F^{\delta \mu}F^{\nu}_{\ \ \delta}$

ACuriousMind

or ${F_\lambda}^\mu F^{\nu\lambda}$, depending on whether you carry out the sum over $\lambda$ or over $\delta$

imbAF

Ok

so what you did here was considering the first and the 3rd term correct?

ACuriousMind

I would describe it as "I carried out the sum over $\delta$"

imbAF

Better

you carried the sum over deltas

ACuriousMind

that it's the first or third terms is immaterial, as we established, the order doesn't matter

imbAF

12:48 PM

now I asked about the order, and you told me that we discussed about it, but you misunderstood me

what i mean by order

was the ordering the the index

let me elaborate what I am trying to say

ACuriousMind

I mean, the order of the indices does matter for $F$ somewhat, but the sum over $\delta$ (or $\lambda$) doesn't really care what position the two indices being summed over are

one has to be upper, the other has to be lower, that's all that matters there

imbAF

apparently, it doesn't

but are these the same

ACuriousMind

you are perhaps confused because in $(A\cdot B)_{ik} = \sum_j A_{ij}B_{jk}$ one instance of the index is always the second and the other the first, and you inferred that that should be a general rule

but it isn't - if I write $\sum_j A_{ji}B_{jk}$ that's a perfectly fine expression, it's just the expression for $(A^T\cdot B)_{ik}$ instead

imbAF

to much info to fast to process

one second

ACuriousMind

chat messages have no expiration date, you can take as long as you want to read what I wrote ;)

imbAF

12:53 PM

I will try to write what I am trying to find out, I think it has to do with what your wrote right now

but i cannot be certain

since I simply cannot

first of all are these the same $F^{\mu}_{\ \ \nu}$ and $F_{\mu}^{\ \ \nu}$

no wrong

writing

now it's as I wanted it to be

ACuriousMind

no, the indices differ in being upper/lower, so these are different objects

you have ${F^\mu}_\nu = g^{\mu\rho}g_{\nu\sigma}{F_\rho}^\sigma \neq {F_\mu}^\nu$

imbAF

and what about $F^{\mu}_{\ \ \nu}$ and $F^{\ \ \mu}_{\nu}$ ?

ACuriousMind

they differ by a $-$ since $F$ is anti-symmetric

imbAF

but why? we don't have index change

$\mu$ is still up while $\nu$ is still down

ACuriousMind

so?

imbAF

1:06 PM

well

ACuriousMind

anti-symmetry means $F_{\mu\nu} = -F_{\nu\mu}$

imbAF

I know antisymmetry is if you swap them ,no?

exactly

ACuriousMind

This straightforwardly extends to ${F^\mu}_\nu = -{F_\nu}^\mu$ and $F^{\mu\nu} = -F^{\nu\mu}$

imbAF

shouldn't it be $F^{\mu}_{\ \ \nu}=-F^{\nu}_{\ \ \mu}$ ?

ACuriousMind

that's not even a valid equation

single indices are not allowed have different upper/lower position on different sides of an equals sign, that's another sign that something went wrong

You can really just write this out in terms of the anti-symmetry of $F_{\mu\nu}$: ${F^\mu}_\nu = g^{\mu\rho}F_{\rho\nu} = -g^{\mu\rho}F_{\nu\rho} = -{F_\nu}^\mu$

especially if you're still getting used to index notation, you shouldn't guess what happens, you should actually try to derive it

imbAF

1:12 PM

I do

but apparently

order matters

of the indices

what I mean by that

Charlie

I never actually took the time to learn Fourier transforms so I'm doing a bit now, why does the integration over $p$ disappear in 2.21 when we insert the Fourier transformation?

of $\phi(x)$

imbAF

As you pointed out this was wrong :$F^{\mu}_{\ \ \nu}=-F^{\nu}_{\ \ \mu}$ where $\mu$ is more left then $\nu$ and by simply swapping you will get an expression where $\nu$ is more in the left then $\mu$. While the correct that you wrote ${F^\mu}_\nu = -{F_\nu}^\mu$ $\mu$ is still superscript and $\nu$ is still subscript but in the term with the neg. sign $\nu$ is more in the left then $\mu$

Charlie

Order of indices matters insofar as $X_{\mu\nu}\neq X_{\nu\mu}$ in general, look at the second to last thing ACM wrote, that is the full set of steps that derive the relationship

imbAF

And ofc it makes sense

Yes I did

Charlie

Are you still confused about something then?

imbAF

1:23 PM

yes, one more additional thing

ACuriousMind

@Charlie you're pulling the derivative "under the integral sign"

imbAF

maybe it's a rule and I am not aware of it

ACuriousMind

the neat property of Fourier transformations is $\partial_x \mathscr{F}(f(p)) = \mathscr{F}(pf(p))$

i.e. the derivative w.r.t. $x$ becomes multiplication by $p$

Charlie

I see where the $|\vec p|^2$ comes from, but the integral $\int \mathrm d^3p e^{i\vec p\cdot \vec x}$ isn't present in 2.21 and I'm not sure why

I'm probably missing the most obscenely obvious thing lol

@imbAF what's still confusing?

imbAF

I am writing it atm

Charlie

1:26 PM

ok lol

as you were

imbAF

it might be a bit length or badly asked, but I am trying my best for it to be understandable, articulation ain't my forte in this part of physics

Charlie

lengthy and poorly articulated describes half the text written about physics in human history, i'm sure you'll be fine

ACuriousMind

@Charlie I find it more helpful to think about this in a way that doesn't even write down any integrals: If you write $\phi(p,t)$ as the Fourier transform of $\phi(x,t)$, then $(\partial_t^2 + (p^2 + m^2)) \phi(p,t) = (\partial_t^2 + (p^2 + m^2))\mathscr{F}(\phi(x,t)) = \mathscr{F}((\partial_t^2 + (\partial_x^2 + m^2))\phi(x,t)) = \mathscr{F}(0) = 0$, where the second-to-last equality is just the original x-space equation, hence (2.21) is true.

Charlie

AH

I was looking at it going "if the integral is zero the integrand isn't necessarily zero", but that IS true in the case of the function argument of the Fourier expansion right

I see, thank you!

I will go back to existing in a world where I'm not stumped by things written on page <30 of Peskin and Schroeder lol

we'll see how long that lasts

ACuriousMind

yeah, it's a bit tricky because this isn't an "the integrand is zero because the integral is zero" argument, but the way your screenshot is written makes it look like it should be one

note that I'm not plugging the $\phi(x,t) = ...$ equation into the x-space equation (which is what that text would at first glance seem to imply I should be doing), I'm starting with the l.h.s. of the p-space equation

you do this carefully two or three times and after that you just get used to "Fourier transform the equation" and never think about it again

imbAF

1:37 PM

Ok here it is, what I basically wanted to know

sorry for the lengthy thing

And the main thing I wanted to ask , here : $F^{\delta \mu}F^{\nu \lambda}g_{\lambda \delta}=F^{\delta \mu}F^{\nu}_{\ \ \delta}$ we are summing over the $\lambda$s and you can clearly see how the indices are ordered when summing over $\lambda$ ($\nu \lambda \lambda \delta$ ) the $\lambda$'s come one after the other with no other index in between . What I wanted to know, and @ACuriousMind confirmed it was that this can be also written as

$F^{\delta \mu}F^{\nu \lambda}g_{\lambda \delta}=F^{\delta \mu}g_{\lambda \delta}F^{\nu \lambda}={F_\lambda}^\mu F^{\nu \lambda}$ but here if we sum over delta we see that between the two $\delta$ indices there are other terms and the ordering is as ($\delta \mu \lambda \delta$) the $\delta$'s don't come one after the other as above, there are other indices in between.

Physics Meta

0

Q: Why some questions are now off-topic which were on-topic some days ago?

Usually some questions were on-topic at the beginning of the community. But now they are strictly considered off-topic. resource-recommendations is one of them. Do we consider them off-topic because the community is growning faster and faster?

imbAF

Therefore I assumed (because no one said anything in the lecture whether it's possible or no) that you cannot lower or raise indicies unless we are like in the first situation. (where the indices we are concerned with are directly one other the other with no other indices in between).

So I thought of doing this :$F^{\delta \mu}F^{\nu \lambda}g_{\lambda \delta}=F^{\delta \mu}g_{\lambda \delta}F^{\nu \lambda}=-F^{\mu \delta}g_{\lambda \delta}F^{\nu \lambda}=-F^{\mu \delta}g_{\delta \lambda}F^{\nu \lambda}=-F^{\mu}_{\ \ \lambda}F^{\nu \lambda}=(-F^{\mu}_{\ \ \lambda})(-F^{\lambda \nu})=F^{\mu}_{\ \ \lambda}F^{\lambda \nu}$ which is not what @ACuriousMind wrote, his answer was ${F_\lambda}^\mu F^{\nu \lambda}$

rob

Y’all .. has the meta tag for resource recommendations always been misspelled? It doesn’t have a “d” in it.

physics.meta.stackexchange.com/questions/tagged/…

ACuriousMind

huh

that's embarrassing, but I don't see how it could not always have been misspelled

3

More Anonymous

Hey all! Isn't the derivative of an auto parallel curve been 0 (in general relativity)?

ACuriousMind

1:43 PM

you can't change how a tag is spelled, you can only create new tags and then synonymize afaik, and there's no correctly-spelled synonym

More Anonymous

@ACuriousMind Can we exclude possibility of hacking?

:P

rob

105

Q: How can we get rid of misspelled and unused (or "zombie") tags?

During the re-tagging of questions, tags sometimes become orphaned from existing questions. Are these zombie tags ever removed from the tags list? What if a tag is misspelled and needs to be removed? How do we get rid of it? Return to FAQ Index

support faq tags tag-pruning

It looks like the way to fix it is to re-tag and then propose a synonym. There are only 45 questions using the misspelled tag.

Though here is ancient advice that moderators can rename tags.

ACuriousMind

2:05 PM

@rob I don't see anything in the UI that would allow me to do that, do you?

rob

I’ve found it (with some help from Teacher’s Lounge): it’s the “merge” tool. But because I thought creating a synonym might be the right approach, now I need two people to downvote my proposed correctly-spelled synonym before we can merge.

Apparently.

ACuriousMind

alright, I've downvoted your obviously wrong proposal :)

rob

hey, go easy on me, I’ve only been a moderator for five years, I’m still learning the ropes

aha! I found the even-more-secret UI, and the misspelling is gone.

ACuriousMind

hurray for mod power

imbAF

2:43 PM

is this correct $\partial_{\mu}g^{\mu \nu}=0$

?

More Anonymous

@imbAF No its not coordinate invariant

imbAF

but it contains only constants, or that has no importance ?

More Anonymous

if it were the covaraint derivative instead of partial derivative then its a different story

imbAF

what you mean with covariant derivative?

$\partial_{\mu}= (\frac 1 c \frac{\partial}{\partial t},\nabla)$ ?

More Anonymous

Covariant derivative

In mathematics, the covariant derivative is a way of specifying a derivative along tangent vectors of a manifold. Alternatively, the covariant derivative is a way of introducing and working with a connection on a manifold by means of a differential operator, to be contrasted with the approach given by a principal connection on the frame bundle – see affine connection. In the special case of a manifold isometrically embedded into a higher-dimensional Euclidean space, the covariant derivative can be viewed as the orthogonal projection of the Euclidean directional derivative onto the manifold's tangent...

Also are you doing general or special relativity?

And i gotta bounce in 8 minutes

btw

imbAF

2:51 PM

special

More Anonymous

Ah ... then dont worry about what I said

In that case your correct

the partial derivative is 0

imbAF

I am simply trying to find out what comes when we do $\partial_{\mu}\Theta^{\mu \nu}$

\ok

ok

thx

ACuriousMind

3:05 PM

nothing meaningful, you can't just apply $\partial_\mu$ to stuff and expect the result to be a proper scalar/tensor

imbAF

yes, I thought so

but I wasn't sure

I simply thought that since the metric is 2nd order tensor with constants, applying the covariant derivative would give zero or is simply ignored

ACuriousMind

but you're not applying the covariant derivative

you're just applying the derivative

the result of $\partial_\mu g^{\mu\nu}$ is different depending on your coordinate system

e.g. it is zero in special relativity for Cartesian coordinates, but it is non-zero in spherical coordinates!

imbAF

in special or general relativity

ahaa

well I wasn't aware of it, but know I know

tho that raises more questions. as to why you get another result depending on the coordinate system

ACuriousMind

because $\partial_\mu$ is "bad", you can't treat the derivative like you'd treat a vector $A_\mu$

imbAF

what you mean by "bad" and treat ?

you mean I can't treat it as a vector?

a four vector*

ACuriousMind

3:10 PM

the reason we like to write stuff like $A_\mu T^{\mu\nu}$ for a vector $A$ and a tensor $T$ is that this expression holds in all coordinate systems

it's a vector itself, $V^\nu = A_\mu T^{\mu\nu}$

but if you use $\partial_\mu$, then $\partial_\mu T^{\mu\nu}$ doesn't transform like a vector with free index $\nu$ under coordinate transformations

imbAF

This is a bit advanced for what I understand of special/ gen. relativity atm

ACuriousMind

the reason is that coordinate transformations do something like $T^{\mu\nu}\mapsto (f'^\mu)_\rho (f'^\nu)_\sigma T^{\rho\sigma}$ and if there's a $\partial$ in front then you have to use the product rule

which gives you something very different from what you get when there's just a vector $A_\mu$, there's no product rule for multiplication

imbAF

but

I will have to copy paste all of this and come back later :P

Right now I am more concerned with the rules in tensor algebra, for example

ACuriousMind

@imbAF I mean, all I'm saying is that $\partial_\mu g^{\mu\nu}$ can be zero in one coordinate system and not zero in another

imbAF

yes i understood that

but as I told you, I need to know the why to that,

tho it can wait for a later moment

ACuriousMind

3:15 PM

this isn't how a "proper" tensorial expression behaves - usually, when you have $T^{\mu\nu\sigma} = 0$ in one coordinate system then that means $T$ is zero in every frame

so if you want do learn "tensor algebra", you should stay away from expressions that involve $\partial_\mu$!

imbAF

I mean

Slereah

What if I want a differential algebra!

imbAF

I am trying to do this, or rather understand this example in our class $\partial_{\mu}\Theta^{\mu \nu}$

which includes that partial thing

and

here I am confused : $$\partial_{\mu}{(\frac 1 4 g^{\mu \nu}F_{\lambda \tau}F^{\lambda \tau})}=\frac 1 2F_{\lambda \tau} \partial^{\nu}F^{\lambda \tau} $$ . I understand that the partial "interacts" with the metric tensor and raises it's index

ACuriousMind

@Slereah then you have questionable taste ;P

imbAF

but I don't understand why it acts only in one of them

+ somehow you have a 2 term coming from somewhere

ACuriousMind

3:18 PM

@imbAF anti-symmetry strikes again

imbAF

where?

ACuriousMind

in getting you the 2

use the product rule as you probably intended to, then use anti-symmetry to reduce to the two resulting terms into a single one

imbAF

ok, but the metric tensor does what i described correct?

and then it's gone right?

ACuriousMind

you can just do the sum over $\mu$ and get $\partial^\nu$, yes

imbAF

it looks as if it ignores the first term

but that's not the case obviously

ACuriousMind

3:21 PM

stop guessing

just do the calculation

use the product rule, and stare at the result until you see why it's equal to the result on your r.h.s.

imbAF

btw is it correct to call it a term? $F_{\lambda \tau}$ or is there a more appropriate name for it?

ok will do

rb3652

3:40 PM

Hey folks, quick question: is a dissociation curve and melting curve the same thing?

Looking at this phase diagram, for instance:

I'm very new to this so I'd just appreciate a quick heads-up.

When I search for dissociation curve, I get results about oxygen and hemoglobin, which is not what I'm looking for

imbAF

I can't do it. I have to somehow using antisymmetri show that $(\partial^{\nu}F_{\lambda \tau})F^{\lambda \tau}=F_{\lambda \tau} (\partial^{\nu}F^{{\lambda \tau}})$ that way I can get a 2, which then reduces 1/4 to 1/2 , but the antisymmetries are $F^{\mu \nu}=-F^{\nu \mu}$ and $F^{\mu}_{\ \ \nu}=-F_{\nu}^{\ \ \mu}$. I used them and I didn't get anything useful

rb3652

Or is the dissociation curve literally the conditions at which a molecule will "dissociate" (separate)?

ACuriousMind

@imbAF I'm sorry, it's not antisymmetry (but try raising/lowering indices instead)

imbAF

yes i tried to do that

i multiplied the first F term with 2 metric tensors

basically

for the first term only

$\partial^{\nu}(F^{\rho \sigma}g_{\rho \lambda}g_{\sigma \tau})F_{\rho \sigma}g^{\rho \lambda}g^{\sigma \tau}$. The partial doesn't interact with the metric tensors so I get delta's of the form $\delta^{\rho}_{\ \ \rho}$ $\delta^{\sigma}_{\ \ \sigma}$

but then I will get 16, so a big mistake

imbAF

4:01 PM

not $\delta$\s but rather $g^{\rho}_{\ \ \rho}$ $g^{\sigma}_{\ \ \sigma}$, and each gives a 4, so 16 in total

ACuriousMind

4:23 PM

@imbAF You have an index more than twice there, that's another one of those things that can tell you you've done something wrong

when you choose $\rho,\sigma$ for the raising/lowering on the first $F$, you should choose two different indices for the second $F$

imbAF

and why is that?

is there any intuition to it

or just a result of spending time with such problems

and you know tricks now, on how to solve them?

ACuriousMind

it's just how raising/lowering works: When you write something like $A_\mu = A^\nu g_{\mu\nu}$ the index you choose (in this example $\nu$) needs to be an index you're not already using since you're introducing a new summation

the rules for "good" expressions are: No index more than twice, indices that occur twice must occur once as upper and once as lower, and on both sides of an equals sign you need the same number of free indices (i.e. indices that occur only once) and these free indices need to be in the same upper/lower position

imbAF

but that is what I did in my calculations, the only thing is that i took the same symbols for two different terms

ACuriousMind

the expression you wrote down has four $\rho$s

imbAF

hold on

ACuriousMind

4:29 PM

and that's led you to get some ${g^\rho}_\rho$ that aren't correct

imbAF

I see two tho, at $g^{\rho}_{\rho}$

where are the 4 ?

ACuriousMind

39 mins ago, by imbAF

$\partial^{\nu}(F^{\rho \sigma}g_{\rho \lambda}g_{\sigma \tau})F_{\rho \sigma}g^{\rho \lambda}g^{\sigma \tau}$. The partial doesn't interact with the metric tensors so I get delta's of the form $\delta^{\rho}_{\ \ \rho}$ $\delta^{\sigma}_{\ \ \sigma}$

imbAF

Yes four, cuz i used the same indices in both terms

and you are saying that is wrong

right?

ACuriousMind

6 mins ago, by ACuriousMind

when you choose $\rho,\sigma$ for the raising/lowering on the first $F$, you should choose two different indices for the second $F$

imbAF

ok

when you look at an expression that utilizes this kind of notation, do you try and understand what is doing i.e the row is multiplying a column, or you don't concern yourself with that and simply, mechanically without giving it a thought, you use the rules that you know about how to combine indices etc etc?

ACuriousMind

4:38 PM

I just apply the rules. Once you have objects with more than two indices there's no interpretation in terms of matrices anymore, anyway

imbAF

I see

Yeah i got it, the result

but I do think that the rules you listed do come from a concrete example, in the most generla case that we can comprehend from observing matricies

hence why you said take different indices

ACuriousMind

thinking about how it works with matrices is a nice consistency check, sure

but it would be much slower if you actually thought about every step by first "translating" it to the matrix world

imbAF

yes but as you saw

that was the reason why I took double indices

ACuriousMind

it's just a matter of training, if you do these kinds of manipulations often enough they'll become second nature

imbAF

yeah

3rd day :P

More Anonymous

4:45 PM

Any good reference notes for Komar mass formula?

(hey all)

@ACuriousMind Same can be said about a lot of physics :)

ACuriousMind

indeed

More Anonymous

@ACuriousMind do you write any blog by any chance. Would be wonderful to read the "must read" books of a curious mind

ACuriousMind

@MoreAnonymous nope

tried a few times, but it turns out I don't have much motivation for writing my thoughts into the void :P

More Anonymous

@ACuriousMind I'd read it

imbAF

you could do something like a notebook tho, a page a day, or every 2 days

More Anonymous

4:55 PM

Id be an anonymous reader

:P

@ACuriousMind Also I recently had a very naïve thought. For the people who try to introduce spinors in general relativity like don't the have to explain why we never see a classical spinor?

ACuriousMind

what does that have to do with GR? when we do QFT, we introduce a spinor field into the classical field theory that we quantize, too

More Anonymous

@ACuriousMind hmmm ... fair but that the spinor field has no classical counter part ... Over here in GR it seems to have one as GR is a classical theory no?

Slereah

5:10 PM

Well you still need it sometimes

For a start if you consider spinor quantum fields in GR

More Anonymous

@Slereah But isn't that wrong? a classical version of spinors does not exist right?

Slereah

What do you mean by "a classical version of spinors"

We live in a quantum world, technically there's no classical version of anything

there is most certainly a classical formalism for fermions

More Anonymous

@Slereah True but GR is a classical limit of a quantum GR theory

Slereah

What I guess you mean is that there's no macroscopic phenomenon where that's necessary

but that's due to statistics theorems

More Anonymous

@Slereah Yes

Slereah

5:16 PM

it's part of the whole Pauli exclusion thing

You can't have many fermions with the same state

So you can't really have a massive fermion wave like you do with photons

More Anonymous

Agreed ...

Slereah

there is otherwise no issue taking classical fermions

rb3652

What's the difference between a clathrate and a gas hydrate?

imbAF

If i have something like this $F_{\pi \lambda}\partial^{\pi}F^{\lambda \nu}$ am I allowed to rename the letter $\pi$ ?

Since it's sum over it, I should be able to

imbAF

5:54 PM

Is there a name for this identity : $F^{\mu}_{\ \ \lambda}\partial_{\mu}F^{\lambda \nu}=\frac 1 2 F_{\mu \lambda}\partial^{\mu}F^{\lambda \nu} + \frac 1 2F_{\lambda \mu}\partial^{\lambda}F^{\mu \nu}$ ?