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123
123
6:40 AM
Hi All...
Hello @JohnRennie Sir
 
John Rennie
John Rennie
@123 Hi :-)
 
123
123
@JohnRennie Nice to see you sir...
 
John Rennie
John Rennie
And you :-)
 
123
123
I have a question... confusion to tell my students.
also my own questions.
 
John Rennie
John Rennie
OK ... ?
 
123
123
6:42 AM
Moment/Torque is depend on reference point let say O , perpendicular distance to the line of action of force.
1) This reference point can be inside the body
2) Outside the body
3) and can be fixed in both above cases
My confusion is that what is the benefits of having choosing freely reference point in torque formula.
Because when we change reference point for the same rotating object it gives us different value every time. It does not have unique answer as in force.
 
John Rennie
John Rennie
All the angular properties like torque and angular acceleration can be computed about any point, but some point are more useful than others.
 
123
123
@JohnRennie Great i need these answer which clear the topic. Thank you
 
John Rennie
John Rennie
The angular version of Newton's second law τ = Iα is obeyed no matter where you take your reference point, but in most cases taking the point anywhere except for the centre of rotation is not very useful.
 
123
123
Is it also possible we fixed a rigid body at some reference axis within the body/outside the body and body is rotating about its reference axis. and we calculate torque about any other reference point.
 
John Rennie
John Rennie
If you calculate your torques, angular accelerations, etc about the axis of rotation then everything works out nice and simply, so of course this is what we tell students to do. You can use any axis you want, but then the calculation will get messy.
 
123
123
6:50 AM
@JohnRennie Great , these answer help me to read the book in more intuitive and understandable way. here CENTER OF ROTATION means reference axis at which body is rotating it could be fixed/free?
 
John Rennie
John Rennie
@123 yes
 
123
123
But i read angular acceleration, rotational inertia are tensor quantities, Why these are tensor quantities?
 
John Rennie
John Rennie
Is angular acceleration a tensor? I would have thought not. I'll have to Google it.
 
123
123
Rotational Inertia is tensor.. I don't know angular acceleration is tensor or not.
I thought because rotational inertia is tensor the angular acceleration should also be tensor because $\tau = I \alpha$
 
John Rennie
John Rennie
The moment of inertia is indeed a tensor. But as far as I know angular velocity and angular acceleration are vectors, or more precisely pseudovectors though as far as students are concerned we can just take them to be vectors.
 
123
123
6:58 AM
@JohnRennie Okay... It means in torque formula. tensor (rotational inertia) is multiplying with pseudovector (angular acceleration). The output torque is vector.
 
John Rennie
John Rennie
Yes. In general a rank 2 tensor is something that relates two vectors.
In this case we have τ = I α where α and τ are vectors, and I is the rank 2 tensor that relates them.
You can easily see that I cannot be a simple number because the moment of inertia changes with the axis i.e. it is different in different directions.
 
123
123
@JohnRennie Great...
 
John Rennie
John Rennie
But this isn't something students need to worry about until they get to university. Although tensors are basically simple they can be confusing when you first study them. I would not mention tensors to your students unless you want to confuse them!
 
123
123
Pls tell me in solving physical problem, is it important to select carefully right choice of reference system (rectangular, polar, circular, cylindrical etc..) also their right choice of choosing a position of origin and set a orientation of reference system.
I read it in klepenner & kolenkow. If we carefully meet the above condition it will simplify our mathematics and solution.
 
John Rennie
John Rennie
Yes, in general choosing the right way to formulate a problem will make it simpler. Knowing the right way is something you can only build up with experience.
 
123
123
7:07 AM
@JohnRennie You are right. I think tensor as matrix , how matrix behave when it multiply with vector or another matrix. Is this the right way of thinking tensors? Because it give me intuition of tensor and behavior of tensor on other quantities.
 
John Rennie
John Rennie
Tensors and vectors are frequently written as matrices, and this is a nice way to represent them because then we just use the usual rules for matrix algebra.
 
123
123
@JohnRennie Yes but i only saw vectors as 1xn or nx1 matrix, and tensors as nxn matrix. Is this true in general?
 
John Rennie
John Rennie
@123 Yes
 
123
123
or vectors can be more than 1xn or nx1
 
John Rennie
John Rennie
Vectors are written as n x 1 matrices
Then in matrix form an equation like τ = I α becomes:
(n x 1) = (n x n) (n x 1)
 
123
123
7:13 AM
@JohnRennie Thanks science and math can be understand intuitively. I have heavily disagreement. Why teachers don't teach these subjects like that. We need teachers like you and curius mind
Is there only two types of motion 1) Translation 2) Rotation
@JohnRennie Why books add 3) Oscillatory Motion as third type
 
John Rennie
John Rennie
Do they?
 
123
123
@JohnRennie Yes in our 10th and 12 standard books they wrote 3 types of motion.
How can we understand oscillatory/ vibratory motion in term of translation or rotation? Under which type of motion it fall.
 
user178758
user178758
7:34 AM
How does the textbook present them?
 
123
123
@user178758 Textbook present oscillatory motion as third type of type.
 
user178758
user178758
Does the book compare and contrast the similarities and differences?
 
123
123
@user178758 No.. If there are only two types of motion. Oscillation must fall any one of them or both.
Motion
In physics, motion is the phenomenon in which an object changes its position. Motion is mathematically described in terms of displacement, distance, velocity, acceleration, speed, and time. The motion of a body is observed by attaching a frame of reference to an observer and measuring the change in position of the body relative to that frame with change in time. The branch of physics describing the motion of objects without reference to its cause is kinematics; the branch studying forces and their effect on motion is dynamics. If an object is not changing relatively to a given frame of reference...
In this type wiki article they oscillation as separate type.
Also circular motion as separate type and projectile motion also
 
user178758
user178758
that's the standard way of presenting them
 
123
123
@user178758 Are you asking or telling me?
 
user178758
user178758
7:44 AM
I am stating a fact :-)
 
123
123
@user178758 Ooookay.. We were read in 10th standard only three type of motion.
 
user178758
user178758
try reading from your textbook to your students
it will help them to learn
 
123
123
Okay
 
 
2 hours later…
123
123
9:28 AM
Why constant "k" is used in newton's second law of motion F = k ma . What is the purpose of it. When constant is needed physically or mathematically in this case or generally.
 
B. Brekke
B. Brekke
@123 Where did you see this? I have never seen Newton's second law written like this
 
123
123
@B.Brekke I have seen in many books also teachers explain it in different ways . I don't understand who is correct.
Principle of mechanics synge in chapter 1. also in physics book.
 
B. Brekke
B. Brekke
9:44 AM
Are you confusing it with Hooke's law?
 
123
123
some teachers said we create two new physical quantities here FORCE and MASS that's why it is needed. POM synge book said by selection proper unit system we can set k = 1. book relate it to unit system.
 
B. Brekke
B. Brekke
@123 Okay, in that case the $k$ constant is used to relate the unit of force with the unit of mass. If you use Newtons for force and kilograms for mass, as we often do, you set $k=1$. However, if you use Newtons for force and grams for mass, you need a conversion factor $k$.
I prefer to express my variables in the SI system, and then use Newton's second law. In that way, I can always use $k=1$.
Note that the units of acceleration also matters
 
user178758
user178758
10:09 AM
@123 we don't "create" physical quantities; we measure physical quantities with standardized units, right?
 
Slereah
Slereah
10:27 AM
user image
2
 
Loong
Loong
10:46 AM
 
 
3 hours later…
DIRAC1930
DIRAC1930
1:44 PM
Do non rel. spinors transform under $SL(2,\mathbb{C})$?
 
Slereah
Slereah
1:55 PM
SU(2) afaik
 
DIRAC1930
DIRAC1930
2:06 PM
Okay so if I were to construct an action with a potential term that transforms as a scalar under that group, would I just have $\epsilon_{\alpha \beta}\psi^\alpha \psi^\beta$?
There must be a Hermitian adjoint there somewhere right?
 
Manas Dogra
Manas Dogra
2:58 PM
While defining operator products of energy momentum tensor and fields in OPE, we take them to be radially ordered(time ordered). Becker,Becker,Schwarz book on string theory says that this is done so that the product have convergent series expansions...Why would time ordering imply convergent series expansion?
 
DIRAC1930
DIRAC1930
3:15 PM
Maybe it can't be what I suggested because it isn't real
 
DIRAC1930
DIRAC1930
3:31 PM
What about $\epsilon_{\alpha \beta} \psi^\alpha \overline{\psi}^\beta$ where $\overline{\psi}^\beta = (\epsilon \psi^\dagger)^\beta$?
@ACuriousMind What do you think?
 
DIRAC1930
DIRAC1930
4:00 PM
Does anyone know what to do?
 
 
1 hour later…
DIRAC1930
DIRAC1930
5:23 PM
Sorry, I forgot to mention that $\psi^\alpha$ are non relativistic quantum fields
 
DIRAC1930
DIRAC1930
5:57 PM
I don't know. What I wrote doesn't seem to be invariant under $SU(2)$
But that would mean that $\psi_1 \psi_1^* + \psi_2 \psi_2^*$ isn't which would mean that I've made a mistake I think
 
 
2 hours later…
ACuriousMind
ACuriousMind
7:29 PM
@DIRAC1930 just use the normal spinor terms with the $\gamma$s and everything, you just have that the $\gamma$ are simply Pauli matrices for 3d
 
DIRAC1930
DIRAC1930
How exactly do we construct invariant terms? We have the Hermitian inner product from QM but also the transformation of spinor fields classically
We want a term whose expectation value is invariant too?
right
 
ACuriousMind
ACuriousMind
I'm not sure what you mean, exactly
if the operator is invariant (classically), then the expectation value is, too
the representations on the fields and on the space of states need to be compatible, otherwise QFT doesn't work
 
DIRAC1930
DIRAC1930
7:45 PM
Don't quantum fields transform like $U \Psi U^\dagger$ or something though?
Oh but that does something like $U \Psi U^\dagger = D \Psi(\dots)$
where $D$ is a representation of the group
right?
 
ACuriousMind
ACuriousMind
yes, that's one of the Wightman axioms and what I mean by "compatible"
 
DIRAC1930
DIRAC1930
So for $SU(2)$ it is correct to say that $\epsilon_{\alpha \beta}\psi^\alpha \psi^\beta$ is an invariant
But where does $\psi^\dagger$ come in
 
ACuriousMind
ACuriousMind
it's invariant, that doesn't mean it's real
you want your Lagrangian to be real
 
DIRAC1930
DIRAC1930
But if I write $\epsilon_{\alpha \beta} \psi^\alpha {\psi^\dagger}{}^\beta$ it's not invariant correct?
I need the $\psi^\dagger{}^\beta$ to be something else
 
ACuriousMind
ACuriousMind
you probably shouldn't write the index $\beta$ upper on the $\dagger$
 
DIRAC1930
DIRAC1930
7:50 PM
Okay sorry
 
ACuriousMind
ACuriousMind
because conjugating changes how it transforms
 
DIRAC1930
DIRAC1930
$\epsilon_{\alpha \beta}\psi^\alpha \psi^*{}^\beta$
Does that make sense?
 
ACuriousMind
ACuriousMind
where, exactly, is the difference between $\ast$ and $\dagger$ here?
 
DIRAC1930
DIRAC1930
Well dagger comes in when you write the above as $\psi \epsilon \psi^\dagger$ in matrix notation
$*$ is just complex conjugate
 
ACuriousMind
ACuriousMind
sure, but we aren't in matrix notation
you wrote down components
 
DIRAC1930
DIRAC1930
7:52 PM
Yes that was my mistake
 
ACuriousMind
ACuriousMind
what is the difference between the components of $\psi^\ast$ and $\psi^\dagger$ that you think makes this valid where it wasn't with $\dagger$?
my point about the index position was that conjugating the vector changes the rep from the "normal" to the conjugate one
and usually in this notation, the upper/lower distinction should be precisely that between a rep and its conjugate
otherwise you'll have to explain to me what the upper/lower mean in your notation
 
DIRAC1930
DIRAC1930
So does the thing transform like $\epsilon_{\alpha \beta} (U \psi)^\alpha (U \psi)^*{}^\beta$?
 
ACuriousMind
ACuriousMind
why is the $\epsilon$ not transforming at all?
 
DIRAC1930
DIRAC1930
I don't know what I'm doing thats why lol
I don't know how to transform it
 
ACuriousMind
ACuriousMind
well, but why are you writing the indices on it like that/why are you trying to construct an invariant with it?
 
DIRAC1930
DIRAC1930
7:58 PM
Because I want to see what the potential term properly looks like in non-rel QFT
 
ACuriousMind
ACuriousMind
I'm not really sure why you're trying to use an $\epsilon$ at all, really
what "potential term"?
 
DIRAC1930
DIRAC1930
$\epsilon$ is the invariant tensor of $SU(2)$
It's used to raise and lower indicies
In the Lagrangian
 
ACuriousMind
ACuriousMind
I mean, your $SU(2)$ representation has an inner product (it's unitary), no?
just take the inner product of $\psi$ with itself
that's invariant and it's a $\psi^2$ "potential term"
square this to get arbitrary even terms
this is why I don't like index notation :P
 
DIRAC1930
DIRAC1930
So the only inner product I should care about is the one invented by Dirac or whoever?
The Hermitian inner product of QM
 
ACuriousMind
ACuriousMind
no, the representation on the target space of fields isn't "quantum"
it's of course isomorphic to representations you might use in QM, but we're talking about the classical field here
if you can't construct a classically useful Lagrangian you can't even begin to do QFT, after all
 
DIRAC1930
DIRAC1930
8:05 PM
In non rel QFT, classical fields aren't really mentioned at all hence my confusions
So the way to do things is canonical quantization there too?
 
ACuriousMind
ACuriousMind
Weinberg would disagree, but sure :P
 
DIRAC1930
DIRAC1930
But $\epsilon_{\alpha \beta} \psi^\alpha \psi^\beta$ is invariant but it's not the correct one used
 
ACuriousMind
ACuriousMind
again, it might be invariant, but I'm pretty sure it isn't real
 
DIRAC1930
DIRAC1930
Before I move on to making it real, $\epsilon$ doesn't transform right?
 
ACuriousMind
ACuriousMind
I don't understand how it's supposed to not transform and yet has two free indices
what do the indices mean if it doesn't transform?
 
DIRAC1930
DIRAC1930
8:13 PM
I thought that it just does the contraction
 
ACuriousMind
ACuriousMind
it might be such that it is the same after transformation (just like the Minkowski metric is invariant under Lorentz transformations), but that's different from not transforming at all
yeah, it's like a metric
but a metric is not an invariant as such - it's not a scalar
 
DIRAC1930
DIRAC1930
My confusion is that when finding the invariant tensor, we assume that it transforms into itself
 
ACuriousMind
ACuriousMind
yeah, think of it exactly like the Minkowski metric in SR
 
DIRAC1930
DIRAC1930
So if I had a random tensor of numbers $A_{\alpha \beta}$, it's impossible to say how it transforms?
 
ACuriousMind
ACuriousMind
???
if you write down the indices, that tells you how it transforms!"
the whole point of upper/lower index notation is that the indices tell you how many transformations/inverse transformations to add if you do a transformation
 
DIRAC1930
DIRAC1930
8:17 PM
$U_{\alpha}{}^n U_{\gamma}{}^m A_{nm}$
SO it transforms like that
 
ACuriousMind
ACuriousMind
yes, exactly
just because that's equal to the original $\epsilon$ in the case of the specific $\epsilon$ doesn't mean it doesn't transform
and that's what "eats" the two transformations from the two $\psi$ and would make your term invariant
 
DIRAC1930
DIRAC1930
Okay thanks
Now all we need to do is find something that a) transforms as a scalar under $SU(2)$, and b) that is real
 
DIRAC1930
DIRAC1930
8:35 PM
Ah I can't do it
But if I could, would I then just replace the fields with field operators and be done?
 
 
2 hours later…
Slereah
Slereah
10:53 PM
The classical example of an SU(2) scalar is the Higgs field
Although I guess that one is a complex field
 
DIRAC1930
DIRAC1930
11:24 PM
I'm a bit confused by the inner product
$\epsilon_{\alpha \beta} \psi^\alpha \psi^\beta = \psi^\alpha \psi_\alpha$
However
$\epsilon_{\alpha \beta} \psi^\alpha \psi^\beta = -\epsilon_{ \beta \alpha} \psi^\alpha \psi^\beta = -\psi_\beta \psi^\beta$
because $\epsilon = -\epsilon^T$. I don't get why they aren't the same
 
imbAF
imbAF
what do we mean when we say a point moving in phase space?
 
DIRAC1930
DIRAC1930
11:58 PM
Do I have to take into account that spinor fields are anticommuting?
 

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