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3:06 AM
@EmilioPisanty It's bumpin'
3:19 AM
@JohnRennie hi are yiu there
@DanielSank can yiu answer my question
@yuvrajsingh I have no idea.
What is your question?
4:16 AM
@yuvrajsingh morning :-)
5:05 AM
@JohnRennie morning sir
@yuvrajsingh morning :-)
@JohnRennie i have question
Basically it is about electrostatic interaction energy
5:07 AM
@JohnRennie what is it
Basically i was solving a question of two hollow spherical shell
Containing charge $Q$ and $-2Q$
Of inner radius a and outer radius b
OK ... ?
@JohnRennie two conducting spherical shell of radius R and 2R given cgarge Q and 2Q respect.inner shell is provided with a switch which can ground the inner shell
Switch s is intially open and energy stored in the system is u1 after switch closed energy stored is u2 ,find u1/u2
What i did potential of 2Q at Q ,,but my friend told told me that i aslo need to find interaction energy of thus i do not know what is this
I need to work for a few minutes ...
@JohnRennie what is this interaction energy
@JohnRennie are you there
5:33 AM
@yuvrajsingh hi, sorry, something came up at work.
@yuvrajsingh shall we move to the problem solving room? People get annoyed if we do do problem solving stuff in this room.
@JohnRennie yes sir
2 hours later…
7:19 AM
@NovaliumCompany morning :-)
@JohnRennie Have a spare minute :P?
7:22 AM
Well, I have a solution of charged TiO2 particles and water. I plan to remove as much water as possible (I'll set it out on the sun for a few hours), but I was wondering should I put surfactant (soap) into it, because I don't want the TiO2 particles to evaporate with the water. Also, if micelles form around the TiO2 particles, it will be much better later on for the black dye not to cover the TiO2.
Are you going to redisperse these particles in oil?
baby oil yep
What do you intend the surfactant to do once you've redispersed the particles in the oil?
Keep the black dye (that I'm going to add to the mixture of baby oil and TiO2 particles) from covering the white TiO2 particles.
It's not obvious why a surfactant would do that.
7:26 AM
It forms micelle around the TiO2 particles, no :D?
Have you tried just dispersing some TiO2 in oil and adding dye to see what happens? If you leave the TiO2 to settle it should be obvious whether you get a white layer on the bottom or not.
I haven't tried. I guess I'll do that first. Thanks. Also, I managed to generate 360V from a really crappy circuit I designed (which barely works) but still. Would it do anything on a $1 cm^3$ container?
@NovaliumCompany yes, it should still move the particles, just not as fast as 1500V would.
@JohnRennie Alright, thanks for the help
7:45 AM
hard times
@enumaris wut :D?
I forgot how to explain the confusion between a probability distribution given by some ensemble (e.g. canonical ensemble has $p\propto e^{-E/kT}$) and the fact that in stat-mech in equilibrium every microstate is equiprobable...
You would think I would remember how to derive this...but my thermo is too rusty
8:13 AM
well hopefully my follow up message will clear things up a bit lol
8:38 AM
Duplication of another person's qualia experiment:
Consider the following diagram:
Let's say we share a lamp which emit some color
To both of us, the color matches the top left circle
The above image is seen from my mac
I then "transcribe" those colors as close as I can from the Mac display to my PC's display and get completely different RGB values despite they are adjusted to be identical as seen from my eyes
I then post this resulting image of the adjusted colors into the chat.
The person, still viewing the same computer as he is, then said the PC screen's reproduction of the bottom left color as seen from his computer is blue
I can then go to his computer, transcribe the color that is displayed into my PC and adjust it so that they look identical from my eyes
Now the other person, when looking at my PC, despite seeing completely different RGB values, recognise the same color he saw in his PC (as otherwise there will be inconsistency in perception)
Thus, without asking the other guy to copy that color to my PC and hence asking him to adjust it until the color from the two screen look identical to him, I have effectively duplicated the qualia of his color into my PC
So, while none of us have idea what kind of "blue" we each are seeing, we can reproduce this same experience across all computers using the above scheme, thus effectively duplicated another person's qualia
I tried making $1 cm^3$ plexiglass container by cutting and gluing the plexiglass and insulating it but the result is really crappy and I was wondering if I can order for someone to make them for me, maybe 3D printing agency donno?
9:01 AM
More precisely:
The two images are the same images on two computer screens
I cannot illustrate obviously how the colors look identical in those two screens, but they do despite their obviously different RGB values
That the red boxes exists provide a channel to transfer any color from the top computer to the bottom one. And if any of the color matches some objective light source seen by another person, it means without getting the other person's help, one can duplicate the experience from one computer to another
Thus the objective difference in the computer display colors that are otherwise identical to one observer's point of view provide a way to capture some of their qualia
9:19 AM
@DanielSank unfortunately the singer/accordionist/songwriter just passed away =(
10:09 AM
@NovaliumCompany I wonder if you could make a cell by sandwiching some form or washer or O-ring between two sheets of perspex.
i.e. have a piece of perspex as the base. Put an O-ring on it. Put another piece of perspex on top and clamp the two pieces of perspex together. Then you have a disc shaped shell in between the two sheets of perspex.
Drill a hole, or holes, in the top sheet so you can put stuff into the cell you've created.
10:22 AM
@JohnRennie Interesting idea
But how am I going to insulate it?
and where would I get transparent O-ring from?
(I was thinking of 3D printing but I'm not sure, maybe I can make them myself)
@NovaliumCompany Suppose you leave out the dye to begin with, then if your cell works one side will go white and the other will go clear. Yes?
@JohnRennie Yep
So why not put your electrode either side of the cell i.e. the particles would move horizontally?
Obviously this would be no use as a screen, but you could see if your particles were moving.
That is what I'm going to do
But how am I going to insulate it?
and where would I get transparent O-ring from?
Let me draw a diagram ...
10:32 AM
sorry :D
That's a side view.
Drill holes in the top plate for you to fill the cell then stick the electrodes into it.
Can I ask, or you are drawing a diagram?
10:52 AM
@NovaliumCompany Why are you using hot glue? As the Wikipedia article mentions, the usual glue for Perspex is cyanoacrylate, aka super-glue or krazy glue, which is very chemically compatible with Perspex.
@PM2Ring I didn't see what the Wikipedia articles mentioned :P
Can I use my solder to melt the edges xDD?
@Secret I use my i1 Display Pro for that.
@NovaliumCompany In theory, yes. But it's tricky to get the temperature right, unless you've had a lot of practice.
10:59 AM
My only problem is insulating whatever the container is. What's the best way to do that?
Melting with solder, using superglue?
In ancient times, before the invention of super-glue, people would usually join Perspex using the solvents mentioned on Wikipedia, but it's not good to breathe those fumes.
So I guess I'll go buy some superglue
where do I get the O-ring from?
@NovaliumCompany Sorry, I don't understand the question. But if melting is involved, you'd do it with the soldering iron, not with hot solder.
@PM2Ring My problem is making the container waterproof. (solution can't get in nor out)
@NovaliumCompany It should be waterproof if you glue it correctly.
11:05 AM
Yep, that's why I'mma buy some superglue, but where do I get the o-ring from?
@NovaliumCompany No idea. Your electronics parts shop may have flexible washers that are large enough.
Thanks, last question. If I use conducting plates to make the electric field and they are outside the container, would it still work?
How are you going to see through the conductive plates?
I'm not going to, it's just a really rough prototype. Later I'll probably buy ITO
So how will you be able to tell whether it's working or not if you can't see anything?
11:17 AM
I can see from the sides, where the conducting plates aren't. (I'm making a cubical container)
Oh. I thought with this talk of O-rings that you'd changed the design of the cell, to be like John's diagram.
Not yet, maybe, but I'll try first with the cubical one that I already made
12:07 PM
@JMac Brilliant.com uses it
@AaronStevens Oh man that's awful. Like super awful. They even italicized it so that it looks even more like the velocity terms; just in case the weird notation doesn't confuse you on it's own.
@PM2Ring @JohnRennie What can I use as conducting plates? Will aluminum foil do it?
12:25 PM
@JMac Yeah.... Not so brilliant in my opinion
1 hour later…
1:52 PM
@PM2Ring @JohnRennie I pretty much evaporated almost all of the water out of the TiO2 and now the TiO2 looks quite solid. The charge has not been lost right?
2:37 PM
@NovaliumCompany The charge has been lost. The TiO2 particles get a charge because the Ti-O-H groups at their surface ionise into TiO- and H+. Bet when you remove the water they reform neutral TiOH.
3:08 PM
@JohnRennie :-(
@JohnRennie What can I do?
@NovaliumCompany In a non-aqueous solution you need a different way to create the charge as sodium hydroxide won't work.
I don't know how the charge is created in oil.
3:27 PM
What if I leave some water with the TiO2 and put that in the baby oil?
3:38 PM
Hmm, I have a really crappy idea about something. If 15V are required for a really small cell, and I need 1500V for my big cell, can I just hook many small conducting plates on both sides of my big cell, each with ~15V across them? You get the idea.
3:52 PM
Good news! I found containers that I can buy that will do the job perfectly! mart-ina.com/index.php?route=product/…
2 hours later…
6:18 PM
Could someone explain this invariance?

\Lambda_{\frac{1}{2}}^{-1} \gamma^{\mu} \Lambda_{\frac{1}{2}}=\Lambda_{\nu}^{\mu} \gamma^{\nu}
7:07 PM
@Slereah Are you busy?
1 hour later…
8:17 PM
@Student404Mus explain it in what sense
@Slereah The author said that the gamma matrices are invariant under simultaneous transformations of vector and spinor indices
I couldn't see the invarian here!
That is certainly true, yes
for example...
$\gamma^\mu$ maps two spinors to a vector
ie $\bar \psi \gamma^\mu \psi$ is a vector
$x^{\mu} \rightarrow \Lambda^{\mu}_{\nu} x^{\nu}$
8:20 PM
So there exists a vector $X^\mu = \bar \psi \gamma^\mu \psi$
Therefore, if you rotate your spinors, $\psi \to \Lambda \psi$
The resulting object is $\bar \psi \Lambda_{1/2}^{-1} \gamma^\mu \Lambda_{1/2} \psi$
But this object is also a vector
Can someone help me this? A rod AB travels along a parabola with A at origin and B at (2,2). B moves with constant speed up along the parabola. Equation of parabola: $y = x^2/2$. What is the angular acceleration of the rod, acceleration of end point A?
And as usual, $X^\mu$ behaves in the usual way under rotation
ie $X^\mu \to \Lambda X^\mu$
therefore the two must be equal
Could someone give me a hint? I would be really grateful.
@PolarBear If you're doing angular acceleration, you need an angle. A convenient choice in this case is the angle relative to the $x$-axis.
Do you know how to express the angle between the x-axis, and the line segment AB? (A being the origin is another convenient feature here)
@Slereah I understood.
8:30 PM
@Semiclassical Yes, I do.
Okay. How?
@Slereah The above statement does not imply the invariance of gamma matrice under rotations (Lorentz trans.)?
Well it's not invariant
it's covariant
i.e. gamma's maps two spinors to a vector
@Semiclassical I just realized I don't know
8:31 PM
The invariant quantity would be $$g_{\mu\nu}(\bar{\psi} \gamma^\mu \psi)(\bar{\psi} \gamma^\nu \psi)$$
Is it allowed to say gamma's are not Lorentz invariants?
Well fairly obviously they are not Lorentz invariant, since $\gamma \to \Lambda \gamma$
But they are covariant
okay. consider the plane with the xy-axes. pick a point (x,y) (in the first quadrant for convenience) and draw the line segment to the origin
Instantaneous angle at any point in time?
Invariant would be $\gamma \to \gamma$
8:33 PM
Always the invariant objects would be their multiplication twice?
but physicists tend to be a bit vague with terms and sometimes you get invariant $\approx$ covariant
This is a sin
you can form that into a right triangle with base x and height y. How do you get the angle from that?
@Slereah the sinvariance of physics
@Student404Mus Scalars are invariant
@Semiclassical 45 degrees, at that point
Really? For any point I could possibly pick in the first quadrant?
8:34 PM
Okay wait, instantaneous right
ie the value at a point does not change depending on the coordinate system
Oh, you mean general
We have to trace the path of B
Right. So you've got $y=x\tan \theta$ for instance
Another fuzzy trick would be the differential operator $\gamma^{\mu} \partial_{\mu}$
8:35 PM
@Semiclassical Yes
So, $\theta = \tan^{-1}(x/2)$?
Hmm. Yes, that will do nicely.
Note, though, that $x$ is itself a function of $t$.
So, differentiate it twice?
and input the value?
In principle, yes. The trickiness here is that you need to figure out what $x(t)$ is.
@Slereah under Lorentz transformation, the Dirac equation would be

\left[i \gamma^{\mu} \partial_{\mu}-m\right] \psi(x) \rightarrow\left[i \gamma^{\mu}\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \Lambda_{\frac{1}{2}} \psi\left(\Lambda^{-1} x\right)
And because A is at origin, that'll directly give the answer for angular accleration?
8:38 PM
the speed along the parabola may be uniform, but the direction is changing throughout
so the x-velocity is not changing uniformly
The Dirac equation is not strictly speaking invariant, since the resulting object is a spinor
but since it's equal to zero it is in effect invariant
@Semiclassical Yeah
So, how to find acceleration of A after that?
But if you write it properly, the actual equation is $$[i\gamma^\mu_{A\dot{A}} \partial_\mu - m I_{A\dot{A}}] \psi^A = 0$$
Naturally, $\gamma^{\mu} \rightarrow \Lambda^{\mu}_{\nu} \gamma^{\nu} $
Therefore the resulting object is actually a spinor $\psi_{\dot{A}}$
8:40 PM
As I said, the issue at this juncture is to find $x(t)$.
Properly the Lorentz invariant equation is $$\psi^{\dot{A}} [i\gamma^\mu_{A\dot{A}} \partial_\mu - m I_{A\dot{A}}] \psi^A = 0$$
@Semiclassical Oh sorry, thinking
it needs to be such that the speed is uniform. That means that $v(t)=v_x^2+v_y^2=v_0^2$.
I get $v_x = \sqrt{\dfrac{0.09}{x^2+1}}$
where are you getting that from?
8:43 PM
It's given that the velocity of B is $0.3$ m/s*
ah, okay
one weird thing about this: If $y=x^2/2$, then the units don't make sense
it'd be fine if you had $y=ax^2/2$ with $a=1$m, for instance
Hmmm. So, I put the value of $x$ in that relation I obtained?
no. that'll be x at that time, not x(t)
you've got a ways to go, unfortunately
So yeah, sorry, need to integrate and find
Is there a smaller way to do this by taking different coordinate system?
yeah. what you've got is $dx/dt=v_0/\sqrt{1+x^2}$
I sorta doubt it, since at the end of the day you want angular acceleration
8:47 PM
"The point of these construction is not the vector spaces themselves (for if you have seen one complex, two-dimensional vector space, you have seen them all)"
that ODE is separable, though, so it's not so bad
Yes, next to angular acceleration of A
I say that, but now I look at the solution. I can get $t$ as a function of $x$ easily enough
I have one doubt though, if A has some acceleration, and we're finding the angular acceleration of the rod wrt origin, where A sits, won't that be the angular accn wrt A
yes? angular acceleration is always with respect to some axis
8:50 PM
Oh yes ^^"
I feel I have forgotten almost all of high school physics after 2 months of vacation
before we go further, it'll be useful to take a step back
we've got $\theta=\tan^{-1}(x/2)$ where $x$ is a function of $t$
to make this not so horrible, let's take derivatives before we substitute x(t)
Ah, that will be better
it's still a bit painful
but it'll help
8:53 PM
image of problem: https://ibb.co/XX2rmkz
$FR/2 - FaR= I_c \alpha$ with $I_c$ is momentum of inertia of cilinder.
$FR/2 - FaR= I_c \alpha = I_c a_c/R$ why can I write ending equality?
why $I_c \alpha = I_c a_c/R$? I have my center on the center of mass ..

$a_c$ is acceleration of center of mass
$R$ is radius of cylinder
@Semiclassical I tried to solve it. It's weird.
yeah, this is why I said to differentiate first :P
what you'll end up with is theta''(t) expressed in terms of x(t), x'(t), x''(t)
ewww, who puts a variable after the squareroot like that
@Slereah if you could see this transformation in details:
\begin{aligned}\left[i \gamma^{\mu} \partial_{\mu}-m\right] \psi(x) & \rightarrow\left[i \gamma^{\mu}\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \Lambda_{\frac{1}{2}} \psi\left(\Lambda^{-1} x\right) \\ &=\Lambda_{\frac{1}{2}} \Lambda_{\frac{1}{2}}^{-1}\left[i \gamma^{\mu}\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \Lambda_{\frac{1}{2}} \psi\left(\Lambda^{-1} x\right) \end{aligned}

I disappointed why he factored (lambda's_1/2) instead of
that's just ugly bruh
9:00 PM
and at that point in the calculation, you're free to evaluate at the $t$ for which $(x,y)=(2,2)$
by the fact that gamma obey this law of transformation
so if you manage to compute x(t=T) (easy), x''(t=T) (harder, but you've got the expression for it) and x''(t=T) (hardest, and requires another differentiation first)
then you can plug those into $\theta''(t=T)$ to get your answer.
your harder and hardest are the same
do u mean x'(t=T) for the first one
@Slereah huh?
My second question is, why the Dirac lagrangian does not equal zero
\mathcal{L}_{\mathrm{Dirac}}=\overline{\psi}\left(i \gamma^{\mu} \partial_{\mu}-m\right) \psi
since it involves the Dirac equation?
9:14 PM
@Student404Mus It's zero on shell
@Semiclassical Oh yeah yesss
it's still a pain, since the derivatives are a bit tedious
Lagrangian mechanics is like super painful to really understand
on teh shell!
but it's manageably so
9:16 PM
Always so many nuances
Using it is fine, it's understanding it that's a pain
@Slereah Do you see that the above transformation is reasonable?
It's 11 PM
there's the old math joke about how the axiom of choice is obviously true, the well-ordering principle is obviously false, and who can say about Zorn's lemma?
I ain't reading Dirac equations
@Semiclassical You can prove finite choice without the axiom of choice
Just be a finitist
A: Relation between the Dirac Algebra and the Lorentz group

bolbteppaIt's pretty annoying that P&S just give you $$S^{\mu \nu} = \frac{i}{4} [\gamma^{\mu},\gamma^{\nu}]$$ from thin air, here is a way to derive it similar to Bjorken-Drell's derivation (who start from the Dirac equation) but from the Clifford algebra directly, assuming that products of the gamma ma...

9:18 PM
i just i feel like there's a similar effect with how appealing one finds hamiltonian/newtonian/lagrangian mechanics
though quite how I'd line them up i'm not sure
Newtonian mechanics is conceptually much simpler. But for a wide class of problems Lagrangian/Hamiltonian mechanics is much easier to implement effectively.
You need to follow what comes before that transformation of the Dirac equation in P&S first
Gitman and Tyutin's book has a pretty good derivation of the Hamiltonian
Without using the Legendre transform
which is nice
@Slereah ??
Basically introducing new coordinates and showing that the resulting variation gives the same EoMs
9:21 PM
The Dirac lagrangian will only be zero for fields $\psi$ which satisfy the Dirac equation
Any answers to my question?
27 mins ago, by ABC
image of problem: https://ibb.co/XX2rmkz
$FR/2 - FaR= I_c \alpha$ with $I_c$ is momentum of inertia of cilinder.
$FR/2 - FaR= I_c \alpha = I_c a_c/R$ why can I write ending equality?
why $I_c \alpha = I_c a_c/R$? I have my center on the center of mass ..

$a_c$ is acceleration of center of mass
\begin{aligned}\left[i \gamma^{\mu} \partial_{\mu}-m\right] \psi(x) & \rightarrow\left[i \gamma^{\mu}\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \Lambda_{\frac{1}{2}} \psi\left(\Lambda^{-1} x\right) \\ &=\Lambda_{\frac{1}{2}} \Lambda_{\frac{1}{2}}^{-1}\left[i \gamma^{\mu}\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \Lambda_{\frac{1}{2}} \psi\left(\Lambda^{-1} x\right) \\
&= \Lambda_{\frac{1}{2}} \left[i (\Lambda_{\frac{1}{2}}^{-1} \gamma^{\mu} \Lambda_{\frac{1}{2}} )\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \psi\left(\Lambda^{-1} x\rig
@Semiclassical What about accn of A?
The term in brackets is where the results of the above post (i.e. the stuff above the derivation in P&S) comes in, Bjorken and Drell derive it more directly/constructively
well, you know $v_y^2+v_x^2=v_0^2$. if you differentiate that you'll get a relation between $v_x,v_y,a_x,a_y$. additionally, from $v_y^2+v_x^2=v_0^2$ you can infer how $v_y$ and $v_x$ are related at the relevant moment.
between those, and your prior formula, you should have enough info to compute a_x,a_y
9:30 PM
are you saying to me?
9:51 PM
Is it necessery to learn GR to study cosmology right
my friend says no
It would certainly be tough otherwise
Nobody seems to talk about it these days
But there is a classical limit to the FLRW model which is sometimes used
But I'd suggest to slap him and tell him to learn his cosmology
I agree with you, I dont know much about that
lol I ll
Newtonian cosmology is of course very interesting but also WRONG
Wrong in many ways
I don't think I'll ever get to read Weinberg's cosmology :(
If you know GR you can try to
10:08 PM
@bolbteppa Actually, I followed P&S till this point but I am saying that when performing any transformation you transform EACH element in the equation, hence, gamma should be transformed as well as I write it

$\begin{aligned}\left[i \gamma^{\mu} \partial_{\mu}-m\right] \psi(x) & \rightarrow\left[i \gamma^{\mu}\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \Lambda_{\frac{1}{2}} \psi\left(\Lambda^{-1} x\right) \\ &=\left[i \Lambda_{\frac{1}{2}} \gamma^{\mu}\Lambda_{\frac{1}{2}}^{-1}\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \Lambda_{\frac{1}{2}} \psi\left(\L
My question is clear, why you exclude gamma from transformation?
@bolbteppa Thanks for the post https://physics.stackexchange.com/questions/381625/relation-between-the-dirac-algebra-and-the-lorentz-group/381638#381638
the derivation of $S^{\mu\nu}$ is now clear
It's not clear, what you've written is wrong, why would the gamma transform
why it wouldn't?!
this is the transformation of Dirac equation under Lorentz transformation under spinor representation
otherwise, I don't know what this transformation is.
wow, this pytorch-transformers library is so nice...
all the major language models are there...and pretrained weights are there too...
and the API even downloads the pretrained weights for you if you don't have it on your machine
10:30 PM
@bolbteppa What is written in B&D

\left(i \hbar \tilde{\gamma}^{\mu} \frac{\partial}{\partial x^{\mu^{\prime}}}-m c\right) \psi^{\prime}\left(x^{\prime}\right)=0
As may be shown by a lengthy algebraic proof, all such $ 4 \times 4 $ matrices $ \tilde{\gamma}^{\mu} $ are equivalent up to a unitary transformation $ U : $ $ \tilde{\gamma}_{\mu}=U^{\dagger} \gamma_{\mu} U \quad U^{\dagger}=U^{-1} $
and so we drop the distinction between $ \check{\gamma}^{\mu} \$ and $ \gamma^{\mu} $"
3.2 of P&S is saying that a field $\phi$ is a scalar field if under Lorentz transformations $x_0 \to x = \Lambda x_0$ we have that $\phi$ transforms into $\phi'$ which is such that $\phi'$ at $x = \Lambda x_0$ is the same as $\phi$ at $x_0$, $\phi'(x) = \phi(x_0)$ but since $x_0 = \Lambda^{-1} x$ it can be written as $\phi'(x) = \phi(x_0) = \phi(\Lambda^{-1} x)$,
i.e. the transformed field $\phi'$ at the point $x$ is the same as the un-transformed field $\phi$ at the un-transformed point $x_0 = \Lambda^{-1} x$.
Thus, if we take $\phi$ at $x$ we can perform a Lorentz transformation on $\phi$ alone (i.e. without changing $x$) by sending $\phi(x)$ to $\phi'(x) = \phi(\Lambda^{-1} x)$. This is useful for Lorentz transforming differential equations at a given point.
A spinor field $\psi(x)$ is a four-component quantity such that the field $\psi'$ at the new point $x = \Lambda x_0$ is a 'transformed' (say 'rotated') version of the field $\psi$ at the un-transformed point $x_0 = \Lambda^{-1} x$, i.e. $\psi'(x) = \Lambda_{1/2} \psi(\Lambda^{-1}x)$, where the $4 \times 4$ transformation matrix $\Lambda_{1/2}$ acting on $\psi$ is a representation of a Lorentz transformation, as described in my post above.
Thus, to Lorentz transform the $\psi$ part of $\psi(x)$ at $x$ without changing $x$ we send $\psi(x) \to \psi'(x) = \Lambda_{1/2} \psi(x_0) = \Lambda_{1/2} \psi(\Lambda^{-1}x)$. Now, if we Lorentz transform a differential equation, like the Dirac equation, because $\partial_{\mu}$ explicitly depends on $x$ we must also Lorentz transform it.
But the only way to 'perform a Lorentz transformation' on the gamma matrices is to perform a unitary transformation on them as described in my post, so they are completely inert under a coordinate transformation, while spinor fields are defined to transform in a certain way under Lorentz transformations, but gamma matrices are just inert matrices.
This is why P&S then insert the identity $I = \Lambda_{1/2} \Lambda_{1/2}^{-1}$ in the second line, to shoehorn in a unitary transformation on the gamma matrices to correct for the transformation of $\partial_{\mu} = (\Lambda^{\nu} \, _{\mu})^{-1} \partial_{\nu}$
11:16 PM
@bolbteppa your help is grateful, thanks!

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