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12:16 AM
regarding the electromagn. tensor , what is the difference between these 2 notations $F^{\mu \nu}$ and $F^{\mu}_{\nu}$
 
12:50 AM
Anyone can help me with a proof of symmetry for the electromagn. energy stress tensor ?
 
 
3 hours later…
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?
 
 
6 hours later…
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?
 
@imbAF what do you mean by "moving tensors without caring"?
 
it's in the formula i habe
gave
$g$ was in between
 
all of $F^{\delta\mu}, F^{\nu\lambda}, g_{\delta\lambda}$ are just numbers
 
then it was moved to the right
 
why wouldn't they commute?
 
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
 
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
 
Yes, but here you have numbers multiplying too : A,B matrices than : $A \cdot B$=/= $B \cdot A$, or not?
aha ok
 
note that the latter is not $(B\cdot A)_{ik}$, it's still $(A\cdot B)_{ik}$
 
I didn't get this last part
 
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)
 
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?
 
@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
 
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
 
 
1 hour later…
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
 
nice gauge theories have compact gauge groups :P
 
What about naughty ones
$SO(3,1)$ isn't compact
 
sure, and GR sucks :P
 
It kind of does for r < 3r_s
 
11:12 AM
::groan::
 
@JohnRennie is that a general relativity joke too specific for me to understand?
 
GR doesn't need unitarity
 
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
 
11:46 AM
@satan29 :-)
 
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?
 
@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
 
I know that both represent a matrix with diagonal elements 1, the first being pos. and the rest neg
but
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
 
where?
 
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
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
 
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
 
12:26 PM
Lorentzian metrics from relativity in this section, in the end
 
why don't you just link the section you're referring to, the table of contents is a bunch of links for a reason
 
that worked
I don't see anything about $g_{\mu\nu}$ and $g^{\mu\nu}$ being different there
 
the matrices are different
 
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.
 
12:31 PM
I took it in alignment to what we learned briefly about the contra and covariant notations of a four vector
 
maybe you shouldn't interpret things into the article it's not actually saying :P
 
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
 
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
 
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?
 
$F^{\delta\mu}F^{\nu\delta}$ 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?
 
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 :$(-F^{\delta \mu})(-F^{\nu \lambda})g_{\lambda \delta}=F^{\delta \mu}F^{\nu}_{\ \ \delta}$
 
or ${F_\lambda}^\mu F^{\nu\lambda}$, depending on whether you carry out the sum over $\lambda$ or over $\delta$
 
Ok
so what you did here was considering the first and the 3rd term correct?
 
I would describe it as "I carried out the sum over $\delta$"
 
Better
you carried the sum over deltas
 
that it's the first or third terms is immaterial, as we established, the order doesn't matter
 
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
 
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
 
apparently, it doesn't
but are these the same
 
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
 
to much info to fast to process
one second
 
chat messages have no expiration date, you can take as long as you want to read what I wrote ;)
 
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
 
no, the indices differ in being upper/lower, so these are different objects
you have ${F^\mu}_\nu = g^{\mu\rho}g_{\nu\sigma}{F_\rho}^\sigma \neq {F_\mu}^\nu$
 
and what about $F^{\mu}_{\ \ \nu}$ and $F^{\ \ \mu}_{\nu}$ ?
 
they differ by a $-$ since $F$ is anti-symmetric
 
but why? we don't have index change
$\mu$ is still up while $\nu$ is still down
 
1:06 PM
well
 
anti-symmetry means $F_{\mu\nu} = -F_{\nu\mu}$
 
I know antisymmetry is if you swap them ,no?
exactly
 
This straightforwardly extends to ${F^\mu}_\nu = -{F_\nu}^\mu$ and $F^{\mu\nu} = -F^{\nu\mu}$
 
shouldn't it be $F^{\mu}_{\ \ \nu}=-F^{\nu}_{\ \ \mu}$ ?
 
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
 
1:12 PM
I do
but apparently
order matters
of the indices
what I mean by that
 
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 $\phi(x)$
 
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 $\mu$
 
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
 
And ofc it makes sense
Yes I did
 
Are you still confused about something then?
 
1:23 PM
yes, one more additional thing
 
@Charlie you're pulling the derivative "under the integral sign"
 
maybe it's a rule and I am not aware of it
 
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 $p$
 
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?
 
I am writing it atm
 
1:26 PM
ok lol
as you were
 
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
 
lengthy and poorly articulated describes half the text written about physics in human history, i'm sure you'll be fine
 
@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.
 
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
 
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
 
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.
 
0
Q: Why some questions are now off-topic which were on-topic some days ago?

Billy IstiakUsually 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?

 
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.
 
huh
that's embarrassing, but I don't see how it could not always have been misspelled
3
 
Hey all! Isn't the derivative of an auto parallel curve been 0 (in general relativity)?
 
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
 
@ACuriousMind Can we exclude possibility of hacking?
:P
 
rob
105
Q: How can we get rid of misspelled and unused (or "zombie") tags?

MottiDuring 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

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.
 
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.
 
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.
 
hurray for mod power
 
2:43 PM
is this correct $\partial_{\mu}g^{\mu \nu}=0$
?
 
@imbAF No its not coordinate invariant
 
but it contains only constants, or that has no importance ?
 
if it were the covaraint derivative instead of partial derivative then its a different story
 
what you mean with covariant derivative?
$\partial_{\mu}= (\frac 1 c \frac{\partial}{\partial t},\nabla)$ ?
 
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
 
2:51 PM
special
 
Ah ... then dont worry about what I said
In that case your correct
the partial derivative is 0
 
I am simply trying to find out what comes when we do $\partial_{\mu}\Theta^{\mu \nu}$
\ok
ok
thx
 
3:05 PM
nothing meaningful, you can't just apply $\partial_\mu$ to stuff and expect the result to be a proper scalar/tensor
 
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
 
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!
 
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
 
because $\partial_\mu$ is "bad", you can't treat the derivative like you'd treat a vector $A_\mu$
 
what you mean by "bad" and treat ?
you mean I can't treat it as a vector?
a four vector*
 
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
 
This is a bit advanced for what I understand of special/ gen. relativity atm
 
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
 
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
 
@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
 
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
 
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 $\partial_\mu$!
 
I mean
 
What if I want a differential algebra!
 
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
 
@Slereah then you have questionable taste ;P
 
but I don't understand why it acts only in one of them
+ somehow you have a 2 term coming from somewhere
 
3:18 PM
@imbAF anti-symmetry strikes again
 
where?
 
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
 
ok, but the metric tensor does what i described correct?
and then it's gone right?
 
you can just do the sum over $\mu$ and get $\partial^\nu$, yes
 
it looks as if it ignores the first term
but that's not the case obviously
 
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.
 
btw is it correct to call it a term? $F_{\lambda \tau}$ or is there a more appropriate name for it?
ok will do
 
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
 
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
 
Or is the dissociation curve literally the conditions at which a molecule will "dissociate" (separate)?
 
@imbAF I'm sorry, it's not antisymmetry (but try raising/lowering indices instead)
 
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
 
4:01 PM
not $\delta$\s but rather $g^{\rho}_{\ \ \rho}$ $g^{\sigma}_{\ \ \sigma}$, and each gives a 4, so 16 in total
 
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 $F$
 
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?
 
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
 
but that is what I did in my calculations, the only thing is that i took the same symbols for two different terms
 
the expression you wrote down has four $\rho$s
 
hold on
 
4:29 PM
and that's led you to get some ${g^\rho}_\rho$ that aren't correct
 
I see two tho, at $g^{\rho}_{\rho}$
where are the 4 ?
 
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 $\delta^{\rho}_{\ \ \rho}$ $\delta^{\sigma}_{\ \ \sigma}$
 
Yes four, cuz i used the same indices in both terms
and you are saying that is wrong
right?
 
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 $F$
 
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?
 
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
 
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
 
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
 
yes but as you saw
that was the reason why I took double indices
 
it's just a matter of training, if you do these kinds of manipulations often enough they'll become second nature
 
yeah
3rd day :P
 
4:45 PM
Any good reference notes for Komar mass formula?
(hey all)
@ACuriousMind Same can be said about a lot of physics :)
 
indeed
 
@ACuriousMind do you write any blog by any chance. Would be wonderful to read the "must read" books of a curious mind
 
@MoreAnonymous nope
tried a few times, but it turns out I don't have much motivation for writing my thoughts into the void :P
 
@ACuriousMind I'd read it
 
you could do something like a notebook tho, a page a day, or every 2 days
 
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?
 
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
 
@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?
 
5:10 PM
Well you still need it sometimes
For a start if you consider spinor quantum fields in GR
 
@Slereah But isn't that wrong? a classical version of spinors does not exist right?
 
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
 
@Slereah True but GR is a classical limit of a quantum GR theory
 
What I guess you mean is that there's no macroscopic phenomenon where that's necessary
but that's due to statistics theorems
 
@Slereah Yes
 
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
 
Agreed ...
 
there is otherwise no issue taking classical fermions
 
What's the difference between a clathrate and a gas hydrate?
 
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
 
5:54 PM
Is there a name for this identity : $F^{\mu}_{\ \ \lambda}\partial_{\mu}F^{\lambda \nu}=\frac 1 2 F_{\mu \lambda}\partial^{\mu}F^{\lambda \nu} + \frac 1 2F_{\lambda \mu}\partial^{\lambda}F^{\mu \nu}$ ?
 
 
1 hour later…
7:00 PM
@imbAF it's just the anti-symmetry of the field strength tensor along with some raising/lowering of indices
so i doubt there's a name for it
 
While perusing a book, I found a diagram with little flags on it
I wonder if it's about flag varieties
 
clearly it's a flag catalog
 
look at how cute they are
 
7:18 PM
@Semiclassical Yes, I was able to derive it
simple trick apparently
 
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