...I'm back. And I passed that fucking PDEs exam! You guys. I'm nearly finished with this degree. But before I do. Can anyone help with some...uhhh...probably comes under the heading of graph theory, though is mostly just manipulation of summations. I can't grasp it, conceptually. Question: Solution: The crucial step I don't understand is how the third expression in the solution equates to the fourth. Though additionally if anyone has any resources for improving understanding of summation notation, that'd be aces. I find it unintuitive. Edit: 10 mins of scribbling & staring into space later...I've verified that both expressions are accurate, but I'm still stuck on how to manipulate the summations algebraically. Man, this field is so cool. I wish I understood it.

Ignoring the summation over j_{l-1} for now, we have sum_{j1=1}^{n}...sum_{jl-2=1}^{n}A_{ij1}A_{j1j2}...A_{jl-2jl-1} This is a product of matrices. In particular, let's consider only the first two factors and summing over j_{1}. We have sum_{j1=1}^{n} A_{ij1}A_{j1j2} = A_{i1}A_{1j2} + A_{i2}A_{2j2} + ... + A_{in}A_{nj2} which is the (i,j_{2}) entry in A^{2}. So similarly for the first three factors: sum_{j1=1}^{n}sum_{j2=1}^{n} A_{ij1}A_{j1j2}A_{j2j3} = sum_{j2=1}^{n} (A^{2})_{ij2}A_{j2j3} = (A^{3})_{ij3} And so on for the rest of the chain. Since there are l-1 factors and the outer indices are i from the first factor and l-1 from the last factor, we end up with (A^{l-1})_{ijl-1} i.e. the (i, j_{l-1}) entry of A^{l-1} as our end result. Combinatorics and graph theory is basically all induction. Just start with something being 1, and then 2, and then 3, and then try to convince yourself that going from 2 to 3 is like going from 3 to 4, and therefore from n to n+1.

I have no idea what's going on in the first 2 equations but going from the third to the fourth seems to be just taking the j_{l-1} summation and moving it all the way outside, bracketing the whole thing between the first summation sign and the k_{jl-1}, which you can do because k_{jl-1} only depends on j_{l-1} so it can be moved outside all the middle sums. The string of sums inside the brackets ends in summing over j_{l-2}. Does that help? EDIT: ninja'd, also lol I was wondering why they didn't use induction but turns out I just didn't realize it because they went from l to l-1 instead of the other way around.

Matrix indices are kind of a mess to try to read, which is why I got really into Einstein notation when I first found out about it, and then into Cvitanovic Birdtracks diagrams. Anything to make the bookkeeping easier.

Cvitanovic Birdtrack diagrams are basically like Feynman diagrams with all the physics stripped off. A matrix becomes a little box with a wire going in and a wire going out, and you get the product of two matrices by just connecting the outgoing wire of one matrix to the ingoing wire of the other, leaving you with a thing with a wire in and a wire out, which you can then put in a box. We can also talk about higher order tensors as boxes with more wires going in or out. Einstein notation shows you what indices to match by having them use the same symbol; birdtracks show you what indices to match by connecting the corresponding wires. The benefit of the birdtracks system is that you then have all the power of graph topology to tell you when two different-looking things are really the same. This is really handy for things like Lie group computations, since birdtrack diagrams tend to be invariant/equivariant naturally, rather than needing that to be imposed or proven. Birdtracks evolved from Penrose graphical notation which is mostly for general relativity; birdtracks are a bit more general in scope since they cover more Lie groups, and are less rigid in structure, but they don't really have good notation for covariant derivatives, or any type of derivatives for that matter.

Maths babes! Physics babes! I'm back to ask for help again. I've been utterly floundering on my "Electromagnetism & Special Relativity" module, and I forgot that I knew some people who know all about this. So. In advance: I'm sorry, all my questions are going to be hard to follow because I don't understand what I don't understand. So please bear with me. Could someone explain to me in small words the η^{µν} tensor for transforming co-/contravarient vectors? For the purposes of my course it's defined as And for context, this is the relevant problem that I can't solve: I'll be back with a hundred more questions, probably. Sorry about that. Your help is much appreciated.

@EulersBidentity so you know how the Einstein Summation Convention is basically "generalizing matrix multiplication to any sort of tensor"? Well, any index in a tensor can be either covariant (index is below) or contravariant (index is above). And the way to go from one to the other is to multiply by a special tensor, which, in the case of special relativity, is the Minkowski tensor, η, which has the special property that η_{μν} is the same as η^{μν}. (this is not the case in general relativity, where you use a different, more generalized tensor) How do you do the thing? One index at a time. Suppose you want to raise the index "b" of the tensor F_{ab}. How do you do it? you multiply F by η^{bβ}. The result is, F_{ab} η^{bβ} = F_{a}^{β} (you can think of it like "the b's cancel out") To raise both indices, you need to do it twice. To lower an index it's the same thing, only with the lower-index version of η ...am I being clear or sort of confusing?

Yay, tensor indices. They're the worst. Okay, first, think of electrical plugs. You have plugs and you have sockets, and you can stick a plug into an socket but you aren't allowed to stick a plug into another plug, and you can't stick an socket into another socket , and it's all terribly homophobic but there isn't much we can do about that. Now, when I say "stick a plug into an outlet", I mean "take an upper index and a lower index and contract", which in turn means sum over all possible ways to set the two indices equal to each other. So if we have η^{μν}, which has two sockets, and a vector s_{ξ}, which has a single plug, we can plug s into η to get η^{μν}s_{ν} = η^{μ0}s_{0} + η^{μ1}s_{1} + η^{μ2}s_{2} + η^{μ3}s_{3} But if we had another object T^{ξ} with one outlet, we can't stick that into η^{μν}. Let's look at η^{μν}s_{ν} again. It still has one socket free. So we can think of η^{μν} as a way to take something that has a single plug and produce something that has a single socket. Conversely, if we had something like θ_{στ} with two plugs, we could use it to turn something with a single socket into something with a single plug. I'm starting to regret having made the homophobia quip, because there are all sorts of implications here. Aaaanyway, how do we get something like θ_{στ}? Well, now we have to think about relativity. Relativity states that absolute displacements are invariant and hence physically valid. Hence η^{μν}, which is our mathematical description of the (square of the) absolute displacement, is physically valid. What else is physically valid? There is an object with one plug and one socket that is also physically valid, namely the Kronecker delta δ_{σ}^{μ}, which takes on the value 1 if the value of σ equals the value of μ and is 0 otherwise. In other words, T^{σ}δ_{σ}^{μ} = T^{μ}. The Kronecker delta is physically valid because it doesn't do anything. We also have an object, sometimes called the Levi-Civita (I'm just gonna say LC from now on) tensor, ϵ^{τμνσ} with four sockets. We'll talk about this one later. We can take two objects, regardless of plug and socket type, and just put them next to each other, which is just tensor multiplication. We take T^{σ} and U^{μ}_{ν} and stick them together to get T^{σ}U^{μ}_{ν}, which has a total of two sockets and one plug. This is physically valid. This includes multiplication by scalars or scalar-valued functions, because scalars are just things with no plugs and no sockets. Addition of two things that have the same number of sockets and the same number of plugs is physically valid; you just have to use the same index labels. And finally, contraction, i.e. sticking plugs into sockets is physically valid. So now we can start building new, physically valid things. In particular, although we can't "build" it in the usual sense, we can use η^{μν}, δ_{σ}^{μ} and contraction to build an object with two plugs, by saying that θ_{σμ}η^{μν} = δ_{σ}^{ν}. This is sixteen linear equations in sixteen unknowns, so you can solve for the values of θ_{σμ}, but you're probably better off just guessing because most of the answers are going to be 0 anyway. Oftentimes you'll see θ_{σμ} written with a η instead of θ, because physicists live in a haze of confusing notation and sketchy assumptions about coordinate systems. I'm gonna keep writing θ. Okay, speaking of things with plugs or sockets, we're now given this object F^{μν} with two sockets, and we're asked to produce F_{μν}, by which they mean turn our object with two sockets into an object with two plugs via physically valid means. Well, we know how to turn one socket into one plug, using θ_{σμ} and contraction. So now we just do it twice, once to each of the sockets of F^{μν}. Now back to LC. This is our hypervolume object, in that if you take four vectors and stick them into the four sockets of LC, you get a number that is equal to the signed hypervolume of the hyperparallelipiped formed by those four vectors. Alternatively it's the determinant: you turn the entries of the four vectors into the columns of a 4x4 matrix and take the determinant and you get ϵ^{τμνσ}t_{τ}u_{μ}w_{ν}s_{σ}. If you unravel that a bit, you find that ϵ^{0123} = 1, that ϵ^{τμνσ} is 0 if the values of any two indices are equal, and that swapping the values of any two indices gives you a minus sign. So ϵ^{0023} = 0, and ϵ^{1023} = -1, and ϵ^{1203} = 1, and ϵ^{1230} = -1, and so on. Finally, and most dangerous, is ∂_{μ}. Unlike all of our other objects so far, this isn't just a bunch of numbers or scalar-valued functions. This is a bunch of derivatives. When we say ∂_{μ}F_{νσ}, we mean ∂F_{νσ}/∂x^{μ}, and that's different from F_{νσ}∂_{μ}. This doesn't mean anything particularly exciting at the moment, but it will become exciting if you move to general relativity, because it turns out that it's harder to make ∂_{μ} generally invariant. But in terms of sticking plugs into sockets, ∂_{μ} is just another object with a plug. So, recap: upper indices are sockets, lower indices are plugs, things need to be relativistically invariant, and you can use invariant things with multiple plugs or sockets to turn sockets into plugs or plugs into sockets. Also, for the sake of everyone's sanity you should probably be setting c to 1. [EDIT] So it turns out that v and ν look even more similar in this font than I had initially thought, so I changed all of the 'v's into 's's, except for one which I changed into a 'w' and changed the 'w' next to it into an 's'. As always, a mathematician is running out of letters.

I spent basically the first half of my grad school career learning how this stuff works, and the second half learning how to avoid it at all costs.

Really? How come? I actually really like it cause it's all compact and looks nice. Then again I'm only an undergrad so proooobably haven't encountered the worst examples. But like. I'm the kind of person who likes trig identities cause it's all summed up in e^{iθ}=cis(θ) so. I'm a weird kid. EDIT: Wait, I think I misunderstood, you mean you hate having to add up all the terms, right?

I disliked having to figure out what was contracted with what, and the proliferation of indices looked really inelegant to me. Like, suppose I want to write down a tensor with n upper indices and m lower indices without actually fixing m and n. Do I want to have to write T^{a1...an}_{b1...bm}? And if I want to contract that with something else? Or partial symmetrization/antisymmetrization. I mean, I could explicitly contract it with a symmetrizer/antisymmetrizer, I guess, or use that godawful bracket notation, but do I really want to? Also keeping track of tensors with more types of indices than just "covariant" and "contravariant". When you have to deal with things like spinor indices, or adjoint indices, and you need some way to indicating that a tensor has 3 covariant vector indices and 2 contravariant vector indices and 2 left-handed spinor indices and 4 right-handed spinor indices and an adjoint index but you only have "upper" and "lower" types. And even just keeping track what's contracted and what's free requires searching the entire term every time. So instead I started doing all of my computations using birdtracks, which is basically the plug-socket model. Very easy to write down a tensor with an unknown number of indices, because I don't need to explicitly label each index. Very easy to follow contractions because the connection is visual. Very easy to symmetrize or antisymmetrize arbitrary indices because the corresponding cables can be brought together easily. Very easy to make up new types of indices, just by making up new types of plugs and sockets. Very easy to figure out what's uncontracted, because those are just cables that don't lead anywhere. Why write c_{k} = (T^{a1})_{i1}^{i2}(T^{a2})_{i2}^{i3}...(T^{aek})_{iek}^{iek+1}(T^{aek+1})_{iek+1}^{i1}X_{a1}X_{a2}...X_{aek}X_{aek+1} when you can just write:

Or imagine trying to write this: There are 14 tensors there and 16 contractions and two uncontracted indices (marked I and O here) that you would have to search for. If I were being really more formal, there would actually be 17 tensors there and 19 contractions.

I'm realizing now, years after the fact, that in that last diagram there should be a gap in the thin curve leading up to the white circle in the middle, to indicate that the crossing of the two thin curves isn't itself a nontrivial tensor. I think the text explicitly notes which kinds of cables a nontrivial tensor can have so there's really only one reasonable way to interpret the crossing and this particular expression wasn't really of interest to anyone, but just for clarity's sake I should have put in a gap, probably.

These are both great!! I like the plug/socket analogy particularly (despite implications). Since (η^{μν})^{2} is the 4x4 id matrix, are the entries (& position) of F_{μν} the same as the entries of F^{μν}? Or is there a subtlety of indices that I haven't understood? Edit: Sorry about the delay, I'm trying to keep track of all my different modules. This one slid to the back burner but it's my soonest Oh God exam.

There is a subtlety of indices, which you'll figure out if you do the full thing, and it has to do with the fact that you're not actually contracting the two ηs with each other - one of the η's is acting on the columns, the other one is acting on the rows. I think this is one of the times where it's actually for the best to do the full calculation, just to try to get a grasp on what's going on behind the scenes. F_{μν} = η_{bν}η_{μa}F^{ab} = η_{bν} * [ η_{μ0}F^{0b} + η_{μ1}F^{1b} + η_{μ2}F^{2b}+ η_{μ3}F^{3b} ] (remember that each of the terms in the sum is a full-fledged 4-by-4 matrix, it just so happens that most of the terms in each of the 4 matrices are 0, and they don't mess with each other.) Yes, before you ask. It is a pain in the ass and there are better ways to do this, which @Exohedron can probably explain a lot better. But I can't think of them rn

Unfortunately, at this stage in your education every way to do this computation is a pain in the ass; the solution is not to find a better way to do the computation, but to find a way to not have to do the computation (that's the real secret to mathematics). But, alas, you've been given Maxwell's equations in the non-invariant formulation, which means you actually have to worry about components and awful nonsense like that. But yeah, don't try to do things the easy way for the nonce. Do things the hard way, verify that the hard way works, see what the hard way gives you, and then look for an easy way once you've built up a suite of test examples whose solutions you know. I think one of my profs would file this under "calculations you should do at least once in your life, but in the privacy of your own home."

But in the future, the answer is Feynman diagrams, but like, taken seriously instead of just being shorthand.