Hi All..

Hello @JohnRennie Sir.

@123 hi :-)

What is the difference between line integral W = F . dl, and W = F . dr

None. In both cases we are integrating wrt the displacement. In one case you are calling the displacement dl and in the other you are calling it dr.

@JohnRennie Okay... dr is the position vector or displacement vector?

When you integrate F.dl you are integrating along the path taken by the object.

Also in order to find Work Done do we need to separate this equation in components?

Actually doing the integral can mean you need to break all the vectors into components.

It depends on the exact details of the calculation.

@JohnRennie It means in W = F.dl , the path of the particle is given. In W = F.dr path is not given

If r is the position vector, then dr is the change in the position vector, and the change in the position vector is the displacement of the object. So F.dr does give the work dW.

In case of projectile trajectory using energy equation. I found symon mechanics separate the components. Because there is no force in the direction of x-axis. So dot product in work does not justify x-axis.

@JohnRennie if dr is position vector. What exactly is dl , is arc length of the actual path. How to compute dl.

sorry dr change in position vector (displacement vector)

You are just calling the position vector r in one case and l in the other. It's the same vector called by different names.

@JohnRennie Thanks. I have cleared almost many thing related to KE, PE and work done. But there are few things I wanted to understand. Pls help

What if I have object at height h1 and I moved it to height h2 under uniform gravitational field. There is PE difference. Why we need PE difference in that cases what special information of knowing PE difference rather in term of forces.

@JohnRennie means. Because forces, mass, velocity, are important to us not work, PE, KE . But in order to find force, path, velocity we need work, KE, PE.

What if at height h1 we have PE isn't there any useful parameter in term of force which defines property of space at height h1 or h2 rather knowing PE

I'm busy at the moment. Hopefully I won't be too long ...

Hope I explain my point well. I don't think so. :D

@JohnRennie OooOkay... Sir

@JohnRennie Sir, When you are available pls look at a glance of my question.

@123 I can't tell what you are asking ...

@JohnRennie means my question is not clear?

Yes

Let me try again.

For example in uniform gravitational field. If an object at height $h_1$ has $PE_1 = mgh_1$ when it is moved to $h_2$ PE becomes $PE_2 = mgh_2$. The difference in PE is $\Delta PE$ . Which is conservative field.

@JohnRennie My point is that PE , KE and Work done are not the parameters we want. The useful parameters for us are force, velocity, path, position. but PE, KE, W are parameters easy to calculate and by know these we can calculate F, v, path etc...

OK ...

If PE is parameter at height h is mgh, which is not the parameter we needed, in term of force don't we have information property of space at that point rather knowing PE.

We do know the force because it's F = -dU/dh

Aaaaaaahhhhhh... May be i am close to asking

@JohnRennie Yes... But it has PE function. At h1 we have mgh1, at h2 we have mgh2. The value of PE changing as we change position. What is changing in term of force at h1 and h2.

What about Force at h1 and Force at h2 and difference in force at different height. as in the case of PE. What is the harm of knowing difference in force directly rather difference in PE.

"The second line says something like that the space of all 1-dimensional supersymmetric euclidean field theories (≃ supersymmetric quantum mechanics) “at grade n” is homotopy equivalent to the n-th space in the Ω-spectrum representing periodic K-theory."

Why must you do this, Urs Schreiber

I'm still a bit unsure about something we're doing in SUSY. If we consider massive reps of the SUSY algebra, and we "go to a frame where $P^\mu=(m,0,0,0)$" as we discussed a few days ago, we are considering a subset of the vectors in the representation whose eigenvalues wrt the momentum generators are those in the bracket. This is fine, but if we then find that the Lie bracket $\{Q,Q\}$ reduces to something similar to the Clifford algebra, doesn't this only hold

when we're considering states "in the rest frame"? Or does the ability to obtain any other "moving" state by the action of the Lorentz generators mean that this new Clifford algebra looking bracket holds on all states?

Because only in that case could I believe that it's actually a useful thing to consider :/

Then again you are contracting the $\mu$ index on $P^\mu$ with a gamma matrix, so maybe this does then hold for all states

hmm I maybe have just typed myself to the answer again lol

it's fine, rubber ducking isn't only for programmers ;)

:D

Hey everybody

Is this calculation not wrong?

I think the result should be ${F_y}_p=\frac{T_0}{a}\left(\psi_{p+1}+\psi_{p-1}-2\psi_p\right)$

This is from the PDF lecture notes of my course, and my teacher also showed this in online course as well. I'm confused.

looks like a typo with the brackets to me, yes

I thought i was mistaken, when my teacher kept talking over it without noticing the mistake

it's often surprisingly hard to notice such errors in stuff you've typed yourself

I wouldn't worry too much about it

Yes, you are right. Anyways, thanks for checking it.

When you're drawing a Feynman diagram, the combinatoric factor shouldn't depend on which order you connect the legs in, should it? If I'm drawing a Feynman diagram for scalar field theory at second order, I'm getting different combinatoric factors depending on which order I connect the pieces in

As in, if I draw myself two four point vertices next to each other, if my first connection goes from one to the other, there's 4 equivalent ways of doing that, if my second connection is again from one to the other there's 3 equivalent ways of doing that, then the last two are fixed

I'm not sure what you mean - what's the case where the combinatorial factor is different from the 12 you get here?

I tried to illustrate it here, in the first picture the first line I draw between the two vertices has 4 options (the four legs on the other vertex), the second one has three options (the remaining 3/4 legs on the other vertex) and the last two are closed on themselves and have no freedom. In the second diagram I draw the exact same thing but doing the connections in another order

the first connection has 3 options, the other 3 legs on the same vertex, the second has 4 options, the 4 legs on the other vertex, and the third connection has also three options, the remaining 3 legs on the other vertex, then the last connection is fixed and closes the right hand vertex on itself

And unless I'm literally just not able to count, even though I get the exact same diagram, it seems like the number of equivalent ways of arriving at that diagram is different, depending on the order in which I make the connections

oh, unless it doesn't count if you're connecting the vertex to itself

no that wouldn't make sense

argh

I think this somewhat heuristic "ways to connect the legs" is really a confusing way to think about the symmetry factor for Feynman diagrams.

hmm

Abdelmalek explains here in point 2 a more formal way to think about this - it's really about symmetries of the graph

For your diagram, there's one symmetry for each of the loops, one symmetry swapping the two lines connecting the two vertices, and one symmetry switching the two vertices with each other entirely, so I'd say this diagram has symmetry factor $2^4 = 16$

The answer I'm given says it is 12 :C

thinking about the diagram as a bunch of "half-edges" might seem weird at first, but I think viewing the factor as a property of the finished diagram rather than something you figure out via the "drawing process" is rather more satisfying

@Charlie ah, now look again at the 4 symmetries I enumerated - they are correct, but my $2^4$ is only correct if each of these is an independent element of order 2

I'm not sure what order means here

or rather, let's do this formally as the answer I linked suggests: We have 8 half-edges: $\{a_1, b_1, c_1, d_1, a_2, b_2, c_2, d_2\}$, which form edges $(a_1, a_2), (b_1, b_2), (c_1, d_1), (c_2, d_2)$ and vertices $(a_1, b_1, c_1, d_1), (a_2, b_2, c_2, d_2)$

I'm already lost, i think I don't know enough about symmetry groups in general to do anything formally, I'm going to just go through some online resources for it, seems like something that makes sense as soon as you get it once

I don't think you need to know much about groups, you just need to know how to count

my counting skills are apparently not what I thought they were :P

my "element of order 2" just means that the symmetry gives the original thing back when you apply it twice, i.e. its "square" is the identity operation

oh right

So, symmetries are some sort of permutation of the half-edges that preserve both the edges and the vertices. e.g. the permutation that just exchanges $b_1\leftrightarrow b_2$ is not a symmetry because it doesn't preserve the 4-tuples that form the two vertices

but the permutation that exchanges $a_1\leftrightarrow a_2, b_1\leftrightarrow b_2, c_1\leftrightarrow c_2, d_1\leftrightarrow d_2$ does that, and it also perserves all the edges. When you do this twice, you get back the original thing, so this contributes 2 to the symmetry factor

the only other permutations that preserve the vertices are those that send half-edges to half-edges with the same index (I chose the names that way)

so, of those there are three preserving the edges $a_1\leftrightarrow b_1,a_2\leftrightarrow b_2$ and $c_1\leftrightarrow d_1$ and $c_2\leftrightarrow d_2$

It naively sure looks like it should be 16

...which are all order 2 too, so this gives me again my 16. are you sure the 12 is really correct @Charlie? 12 would suggest there's an element of order 3 hiding here somewhere but I really don't see it.

if your answer key is wrong, that would explain why you're confused about how to do this exercise :P

P&S give some similar examples, if you go by those examples you'd (naively) get 16

I suspect whoever designed this exercise tried to compute the symmetry factor the same way Charlie did, but never noticed the second way of counting so just wrote 12 as the answer :P

It definitely says $4\cdot 3\int dy_1dy_2\Delta_{y_1y_2}^2\Delta_{y_1y_1}\Delta_{y_2y_2}$

where we've factored out the interaction factor and the propagator between the two external points that goes with that diagram

Unfortunately through all of my struggles to get a consistent answer myself at no point did I find 16 lol

ah

I managed to get 12, 18 and 36

we're talking about different symmetry factors -.-

oh no

@Charlie there's an $\frac{1}{n!}$ in front of that expression somewhere, right?

or something similar

yeah we've factored out $(1/4!)^2$ which comes from the factor in the interaction term

the overall factor when that $4\cdot 3$ hits that fraction will be $\frac{1}{16}$ if what I'm doing is consistent

This has some examples apparently including this one and it gives 16

I don't see that, at second order we'll have $\sim \frac{1}{4\cdot 3\cdot 2}\cdot\frac{1}{4\cdot3\cdot 2}\cdot 4\cdot 3=\frac{1}{48}$

@bolbteppa yeah, the problem is that Charlie's "combinatorial factor" is different from the overall "symmetry factor"

@Charlie this I would take as the indication that the correct number for what you're asked to compute is actually the 36

yeah I had noticed there was some confusion in a few places in answers I read about which exact number people were talking about

oh

It's a fact that the symmetry factor - as in, the overall prefactor the value of this diagram will be multiplied with - of a Feynman graph with automorphism group $G$ (in the sense I explained above) is $\frac{1}{\lvert G \rvert}$

but you're computing a different number, that will give this factor only after being multiplied by whatever $\frac{1}{n!^k}$-like expressions you can "pull out" of the sum of diagrams

both of these computations are "valid" and should yield the same factor when compared knowing how the two numbers they compute are related. So either we've missed an additional symmetry of order 3 above, or the correct solution to your exercise is 36, not 12.

Can anyone recommend a lecture series on statistical physics for a mathematician youtube.com/watch?v=D1RzvXDXyqA&ab_channel=Stanford seems good but wonder if anyone else has anything up their sleeves.

ok, ty for your help, I apparently need to do a lot more on this

This answer might have what you want @Monty? physics.stackexchange.com/questions/202885/…

@Charlie don't let this get to you, actually computing Feynman diagrams by hand like this is hard work that's susceptible to a lot of silly but hard to catch errors

core great find cheers :)

and it's also not really explained all that well in most texts because while everyone "should have seen" this type of computation it's not actually something anyone does as part of their daily work I think - we have computers for that nowadays

Yeah I am a bit upset to find that after going to the trouble of learning how to derive correlation functions through both Wick's theorem and the functional integral, the one thing I can't get involves nothing more than counting :P

@Monty what about this and then this

I just know my exam is going to explicitly say "include combinatoric factors" and I'm going to end up weeping onto my exam script

@bolbteppa much appreciated

@Charlie just guess a random power of 2

you never know, it might be right :P

Joking aside it may legitimately come to that, they said you can't use calculators but they didn't say anything about dice

::sounds of rolling dice:: "Critical Hit! Take that, (FP) ghosts!"

P&S say you barely ever have to calculate one of these beyond second order so their examples and the ones I just gave and any more you've seen are probably enough :p

sure, just memorize the symmetry factors of all diagrams up to order 2, easy!

Because this exam is open-book I will mostly likely just end up writing down all of them on a piece of paper so I don't accidentally have a stroke trying to calculate them mid-exam

actually a valid tactic

@Charlie open book but no calculators?

Interesting policy

what would you need a calculator for in a theoretical physics exam? :P

Well, they say you can't do a lot of things, because it's an at-home open book exam (bc university has been shut all year) it's mostly just an honour code

that being said they told us that last year (same procedure) they zeroed like 150 people's exam marks because they were caught colluding

in one instance two students accidentally submitted the same exam script

@ACuriousMind well they're the ones who explicitly disallowed them :p

@Charlie unbelievable

@Charlie lol

(un)fortunately I didn't actually know anyone on my course last year, they were all 3rd year students and I was an outsider joining the class so I couldn't have colluded even if I wanted to

I feel like "collusion" is a far too fancy word for two people just handing in a copy of the same solution :P

and tbh there isn't that much of an advantage having the exam be open book, they just don't ask definition questions and if you have to waste time looking through lecture notes to get an answer you've kind of wasted time you don't have so you have to learn everything anyway

@ACuriousMind @Slereah sorry for another question. How is the signature the same in Eddington Finkelstein coordinates. There is one negative and 2 positive term in the metric

The determinant of the metric doesn't change sign as you cross the horizon in the ingoing Eddington-Finkelstein coordinates, David Tong talks about this worry on page 240 of his GR notes that you can find here: damtp.cam.ac.uk/user/tong/gr/six.pdf

@Charlie yes I have seen that page but doesn’t he focus on that the determinant is not 0 anywhere. Isn’t the signature defined by the difference of the positive and the negative signs along the the diagonal eleme

elements

To my understanding the argument is that since the determinant of a matrix is the product of it's eigenvalues, to flip the signature of the metric you would need to flip the sign of the determinant. Since the determinant is a continuous function you would therefore need to continuously pass the determinant through zero in order for the metric to change signature and since the metric is non-singular at any points other than the origin the signature does not change as you cross the event horizon

@Shashaank it's only the difference of the positive and negative signs of the diagonal elements when the metric is diagonal

the metric is not diagonal in EF coordinates, so you can't tell anything from the signs of the individual terms in there

(and Charlie's argument for why the signature cannot change as long as the metric is a continuous function is correct)

Thanks@Charlie and @ACuriousMind. I get it. @ACuriousMind and @Charlie but is there any other way of figuring out the signature of the metric if it is not diagonal rather than depending on the determinant or is the only other way is to calculate the eigenvalues in that basis and check their signs..

Eigenvalues are basis independent

calculating the eigenvalues of a 4x4 matrix is typically not hard, I'm not sure why you're looking for another way

since it's just determining the roots of a polynomial of degree 4, you can just let a computer do it

e.g. in the Schwarzschild coordinates it's very obvious what the signature is, you don't even need to compute anything

Metric isn’t a Linear operator that it would have eigenvalues in that sense and as far as I know the concept of eigenvalues and eigenvectors holds for linear operators only...so I had presumed that by eigenvalues of metric one means eigenvalues in a particular basis

@Charlie

@Shashaank the metric is a linear operator on the tangent spaces at each point

Isn’t it a bilinear form

and one of the conditions for it to actually be a metric is that the signature of that operator on each tangent space is the same - otherwise there would be at least one tangent space on which it was degenerate, but one of the properties of a proper pseudo-Riemannian metric is that it is everywhere non-degenerate

@Shashaank a bilinear form has an equivalent expression as a linear operator

If you can write it as a matrix is it not automatically a linear transformation? I thought that was a thing

@ACuriousMind @Charlie my understanding was that only (1,1) tensors represent linear transformations and the metric is a bilinear form. From my linear algebra course, I was of the understanding that eigenvalues and eigenvectors are not defined for bilinear forms

@Shashaank a bilinear form $g$ has an associated symmetric matrix $g(e_i,e_j)$ for $e_i$ some basis of the space

I thought Sylvester's law of inertia was specifically about eigenvalues of quadratic forms, but I don't know all that much about that it

it doesn't matter that the form is "conceptually" not a linear operator, the eigenvalues of this matrix are the same as the "eigenvalues" of the quadratic form

@ACuriousMind So why in linear algebra we didn’t have eigenvalues of the bilinear form and of inner product. Why did we restrict eigenvalues to linear operators only there

I don't know what you mean by that

"we" certainly have eigenvalues of quadratic forms, they are the $\lambda_i$ in its diagonalization, see en.wikipedia.org/wiki/Quadratic_form#Real_quadratic_forms

Isn't the metric a linear operator if it provides a linear isomorphism between tangent and cotangent bundle fibres?

@Charlie yes, but eigenvalue is a concept for a linear operator from a space to itself

you can't talk about the "eigenvalue of the map $V\to V^\ast$"

you can only talk about eigenvalues of maps $V\to V$

not even if $V^*\cong V$?

@Charlie well, you have to fix such an iso then

finite-dimensional vector spaces are always isomorphic to their duals, but not canonically so

I always hesitate when I see canonical used in math vs physics because it seems like we use those terms differently to the purists

if you have an iso $f : V\to V^\ast$, you can of course talk about the eigenvalues of $g\circ f^{-1} : V\to V$ for any $g : V\to V^\ast$

but if you take the operator $g$ itself as that iso, you get the identity, which doesn't really tell you anything :P

@Charlie I have no idea what physicists mean but in mathematics a "There is X, but no canonical one" always means that there is more than one thing that has the properties of X and there's nothing telling you which one is "best" to choose

one can make this notion precise in category theory and the idea of natural transformations, but the above is really the "spirit" of it

ah ok

in this case, there are infinitely many isomorphisms between a vector space and its dual - choose any bases $v_i$ and $v'_i$ of the spaces and send $v_i\mapsto v'_i$

I remember the maths teacher criticising the use of canonical in "canonical quantisation" when the topic came up lol

no basis choice is better than any other, and there's no additional data we could use to distinguish one of them so we say the isomorphism $V\cong V^\ast$ is non-canonical

yeah that makes sense

this is kinda important because objects that are canonically isomorphic are in some sense "the same": If I have a fixed distinguished iso $f : V\to W$, then any map into/out of $V$ resp. $W$ becomes one into/out of $W$ resp. $V$ by concatenating with $f$ or $f^{-1}$

so any statement about $V$ has a unique way in which it becomes one about $W$ and vice versa - there's no property by which we could really tell them apart

@ACuriousMind if eigenvalues in a sensible concept for bilinear form then so should trace be. But trace of the Minkowski metric is not 2 but 4 ...trace like eigenvalues is concept for linear operator...Is this answer ( particularly) the 2nd paragraph wrong

The trace of the Minkowski metric isn't 4?

@Charlie it is, if your definition of the trace is $A^\mu_\mu$

hmm I guess so

Any ways could you please summarise the correct thing for eigenvalues of a metric ....is it basis independent ....

@Shashaank Okay, let's do this more carefully: On each tangent space, there is a bilinear form $g_p : T_pM\times T_pM \to\mathbb{R}$ and an associated quadratic form.

For each $p$, you can determine the diagonalization this quadratic form possesses by the law of Sylvester as $\lambda_1 x_1^2 + \lambda_2 x_2^2 + \lambda_3 x_3^2 + \lambda_4 x_4^2$.

And at each p what ever basis I choose I will get the same eigenvalues for the bilinear form

it is a fact that these $\lambda_i$ are the same as the eigenvalues of the linear operator defined by the matrix $g_p(e_i, e_j)$. This is not a claim that $g_{\mu\nu}$ would be a linear operator from the tangent spaces to themselves, it is just a fact about the quadratic form in each invidiual tangent space

Ok now I see

Do second rank contravariant tensors or third rank covariant rensors etc also have a concept of eigenvalues or not @ACuriousMind

this isn't a concept of eigenvalues for tensors

it's specifically a concept of eigenvalues of quadratic forms

Ok does the concept of eigenvalues extend to any other object relevant in GR apart from linear operators and bilinear forms or not

no

Ok thanks again

@ACuriousMind sorry just a small thing more. John Rennie told me that in Schwarzschild metric the Schwarzschild coordinates are good even within the horizon and not good just at the horizon....this looks correct after all they characterise points uniquely there. But all the books I have seen say Schwarzschild coordinates are good only outside the horizon...

depends on what you mean by "good"

What does John Rennie mean by good there that the books don’t mean

you'll have to ask that either the books or John :P

it's not exactly a technical term

Hey. Can I ask for undergrad advice here?

Ok yah I see, but then what do you suggest...are they good or not according to your definition of “good”.

@timetraveller69 As the room description says: Don't ask about asking, just ask! If someone can and wants to answer you, they will.

@Shashaank They are well-defined, but you probably don't want to use them because the nature of the $t$ and $r$ coordinates changes at the horizon. I suspect the latter is what your sources mean by "not good".

ACM in the post you made about operators in QFT, you say "If you want to be formal about it, you have to consider quantum fields in the continuum as operator-valued distributions $\phi(x)$ that only yield actual operators when smeared with a test function $f(x)$ as $\phi(f) := \int f(x)\phi(x)\mathrm{d}x$", just out of interest, what functions do you actually integrate the "operator field" with to get the operators we usually use in qft? Or maybe a better way of putting it

when do you actually integrate the field operators like that? Or is this more something formal that doesn't appear often

no physicist in practice does this :P

but note that although we usually compute scattering amplitudes in momentum space as functions schematically like $A(p_1,\dots p_m,q_1,\dots q_n)$, these relate to differential cross sections that only give you finite measureable quantities when integrated over some ranges of momenta

the problem is that on the way there, the physicist considers quantities like $\phi(x)^2$ that doesn't exist for distributions - just like $\delta(x)^2$ doesn't exist

so while one might hope that we're merely "delaying" smearing the field, what actually happens is that nothing of what we do in ordinary QFT is rigorously justified, and this problem of multiplying distributions is closely linked to why renormalization is necessary

note that the usual formalism focuses very much on manipulating states like $\lvert p\rangle$, but you should remember from ordinary QM already that these things don't actually exist as states, but that hasn't ever stopped any physicist from using them