Studying How to self-study physics past High School Level?

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A teenage physics enthusiast seeks guidance on advancing their knowledge beyond A-level physics. They have studied various textbooks and completed numerous practice problems but are unsure how to proceed. The discussion emphasizes the importance of parallel development in mathematics and physics, particularly the necessity of mastering calculus, including vector calculus, before delving deeper into physics topics like electromagnetism and astronomy. Recommendations include following established university curricula, such as MIT's or Yale's free online courses, and utilizing specific textbooks like Halliday, Resnick, and Krane for foundational physics. Participants suggest exploring hands-on projects related to electromagnetism and astronomy to enhance practical understanding. Additionally, resources for learning mathematics relevant to physics are highlighted, with an emphasis on avoiding overly abstract mathematical texts. Overall, the conversation encourages a structured approach to learning, integrating both theoretical and practical aspects of physics.
  • #91
Mozibur Rahman Ullah said:
I've edited it to remove the offending word!
It doesn't matter whether it's "stage of career" or "stage of education". The fact remains the OP is 13. Responses should be tailored accordingly.
 
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  • #92
Hey, @Mozibur Rahman Ullah - thank you so, so much for putting in all of the time and effort to explain that to me (and sorry for the late response) - it means a lot to know that there are people willing to spend so long helping out a curious 13-year-old girl :D
As for your explanation, I thought it was great! I had to read it through a few times to fully get what was going on, but most of it was really understandable - I got what you were saying, and the only searches I made were on the lines of "what does the word auxiliary mean", or "what is the tensor product" so nothing really that was instrumental to your approach. I enjoyed it a lot - and whilst I may be (very) young, I think you dumbed it down very well :D.
a few things:
Mozibur Rahman Ullah said:
The map T is called the tangent functor and this has nice properties. For example, if M & N are manifolds then T(M×N)=TM×TN
Ok so this is absolutely not the correct thing to ask, especially because I obviously don't really know the appropriate maths but my curiosity is really getting the better of me - why? For me it just doesn't seem immediately intuitive. (also, is the multiplication of manifolds similar to to taking the cross product of vectors? It seems that way... ish)
Mozibur Rahman Ullah said:
t's best to see this as an illustration but I don't know if Physics Forums supports images.
It does!
Mozibur Rahman Ullah said:
Its also worth mentioning that this rule is basically the chain rule we learn to love in calculus, generalised to multivariable calculus and then to manifolds. The way to show this is to work locally, that is by fixing charts and then working out the various expressions.
Yeah, this cleared up a lot. In my mind, it's basically "generalised chain rule" now - which actually makes it very easy to kind of get a gist of what's going on :D
Mozibur Rahman Ullah said:
It's been a long post but I hope this gives you some sense on how the exterior covariant derivative comes about. To be honest, I don't expect you to grasp all this as I've skipped over a lot of detail - for example, what is the wedge product. And don't feel frustrated if it escapes you, as I've already said the exterior covariant derivative is really, really advanced stuff. But I do hope what I've written gives you a feeling for the shape of the terrain you would have to cover to rigourously learn this stuff. If you have any questions on what I've written then feel free to ask.
It's been long - but actually great! I understood where it came from (though I have to agree with @CrysPhys that I have a long way to go till I'm ready to study it in any depth whatsoever) and it was really, really interesting. I think I would have been frustrated if it slipped past me (though most of it didn't... thankfully), because now you've made me curious - and once I'm curious there's no stopping me :) Thank you for the explanation once again - it was really eye-opening!
CrysPhys said:
<<Emphasis added.>> The OP is 13! :rolleyes:
yeah.... 13 but curious! :) (though thanks for the reminder, once again my curiosity gets the better of me sometimes, honestly without you I'd already have confused myself 50 times over via random sources on the internet about differential geometry because @Mozibur Rahman Ullah got me so excited) You were also right about the university courses - I've almost finished the calculus one, and it was very useful - I learnt a lot!
CrysPhys said:
It doesn't matter whether it's "stage of career" or "stage of education". The fact remains the OP is 13. Responses should be tailored accordingly.
I thought it was tailored pretty well! Extremely A bit ahead of what I'm doing right now, but interesting and understandable (even if I in reality know nothing about the subject and probably don't even really understand the exterior covariant derivative at all it just feels like it).
 
  • #93
Mozibur Rahman Ullah said:
Great, I'm glad you feel its worth learning. But I'd hold your horses about learning this stuff seriously at this stage of your education. This is advanced stuff and the exterior covariant derivative is really advanced stuff! Typically, you will learn vector analysis in undergraduate courses and differential geometry in graduate courses and the exterior covariant derivative is most likely taught at a second course at this level. Certainly I wasn't introduced to it in my masters course on theoretical physics. Nor does the textbook I recommended, Lee's Smooth Manifolds explains this. I first learnt about it in Baez & Muniain's Gauge Fields, Knots and Gravity which I have already recommended. It takes the physics approach to differential geometry and so avoids all the very careful constructions in math whilst still being relatively rigorous. The chapter you should look at is chapter II - 3, Curvature and the Yang-Mills equation, pg.250. But I urge you to study the text systematically, or at least go through it trying to understand the concepts that they introduce to get a sense of the terrain. It pays to go through the text step by step. But I also think at this stage of your education you should be getting the fundamentals of physics down - that is really, really important - and not be distracted by advanced and really advanced stuff. I promise you they will still be waiting for you later on in your career. It's a problem that I've had in my own education, so I feel your frustrations. I explained what I did in my previous post as motivation that this stuff is worth learning and not as an indication that you should be learning this stuff at this stage of your education.

Nevertheless, I've written an exposition below which I hope will give you a sense of the terrain without getting bogged down in detail. I hope it helps without further confusing you and leading you astray!

First, to really get grips with the exterior covariant derivative you need to be comfortable with manifolds, tangent bundles and vector bundles as well as their spaces of sections - basically tangent fields and vector fields, as well as the wedge product (a generalisation of the cross product which works for all dimensions, the cross product only works in 3d) and tensor product and of course differential forms and the exterior derivative, oh as well as understanding that tangent fields can also be interpreted as certain differential operators. This is a long list of mathematical technology, so I've written an exposition that gives you a birds-eye view of some of the neccessary concepts without getting into the detail that's needed to really grasp these ideas and make use of them.


The first thing to learn is manifolds. This is basically a formalisation of one aspect of the notion of covariance. In fact Einstein said:


"So there is nothing for it but to regard all imaginable systems of coordinates, on principle, as equally suitable for description of nature"


Manifolds are spaces that are equipped with charts, that is coordinate systems. They are also equipped with transition functions that map between these different charts. This sounds complicated, but it's not so bad as we learn to manipulate manifolds. For example, if you had two manifolds ##M## & ##N## then you can multiply them ##M \times N## and you can take their disjoint sum ##M \sqcup N##. For example if we took the real line is a manifold as well as the circle. If we multiply them we get the infinite cylinder and if we add them we just get the line 'next' to the circle. So we have a calculus of manifolds. It's also important to distinguish between manifolds by themselves and manifolds that are embedded in Euclidean space. For example, if you imagine a sphere - by the way, mathematicians call the surface of a ball, a sphere and the interior of a ball, is just called a ball, then you will probably imagine a sphere in space. This makes it easy to see what the tangent planes on a sphere are. This is a sphere embedded in space. When a mathematician refers to a sphere, it is a sphere all by itself, sort of hanging in the void. This makes it difficult to understand what the tangent planes are as there's nowhere we can take the tangent plane. There's only the void surrounding the sphere. However, there is a way of introducing the tangent planes. This relies on maths but a quick way of seeing this, is simply to invoke a theorem that all manifolds are embeddable in a Euclidean space of high enough dimension and then just take the tangent planes as usual. The manifold with all the tangent spaces attached is called the tangent bundle over that manifold.


Now if ##M## is a manifold - for example a sphere - then ##TM## is the usual notation for the tangent bundle over ##M##. The map ##T## is called the tangent functor and this has nice properties. For example, if ##M## & ##N## are manifolds then ##T(M \times N) = TM \times TN## and obviously ##T(M \sqcup N) = TM \sqcup TN##. This is already important. For example, the tangent space to the infinite cylinder can now be easily calculated. It is ##T(S^1 \times \mathbb{R}) = T(S^1) \times T(\mathbb{R})##.


It also turns out that if we have maps ##f: M \rightarrow N## and another map ##g: N \rightarrow P##, then ##T(g\circ f) = Tg \circ Tf##. To understand this intuitively, it best to see the situation geometrically. When we map one manifold to another, then we also map the tangent planes on the first manifold to the second manifold. Then if we have a composition of maps then the tangent maps also compose. It's best to see this as an illustration but I don't know if Physics Forums supports images. Its also worth mentioning that this rule is basically the chain rule we learn to love in calculus, generalised to multivariable calculus and then to manifolds. The way to show this is to work locally, that is by fixing charts and then working out the various expressions.


Now the covariant derivative ##D## is a replacement for the directional derivative in differential geometry. The usual notion of a directional derivative doesn't work as it relies on the embedding of the manifold in Euclidean space. However, that theorem of embedding the manifold in a Euclidean space of high enough dimension comes in useful again. It turns out that the covariant derivative is the orthogonal projection of the directional derivative onto the tangent space. I say tangent space rather than tangent plane because whilst a 2d manifold has a 2d tangent space - aka a tangent plane - a n-dimensional manifold has a n-dimensional tangent space. Usually, when the covariant derivative is introduced it is done intrinsically but I find this geometric way of defining it more intuitive.


We need the covariant derivative because the exterior covariant derivative combines the exterior derivative and the covariant derivative. I haven't introduced the exterior derivative but we can at least see how the exterior covariant derivative is defined on Baez & Muniain, pg.250


We are going to go for an inductive definition. This is because we have differential 0-forms, 1-forms, 2-forms all the way upto ##n##-forms where ##n## is the dimesion of the manifolds. A differential k-form is said to have order ##k##. The space of k-forms on a manifold ##M## is denoted by ##\Omega^k(M)##. It will turn out that the exterior derivative of a k-form will result in a form of one degree higher, aka a k+1-form. In symbols:


##d: \Omega^k(M) \rightarrow \Omega^{k+1}(M)##


We really ought to have a superscript ##k## on the ##d## but this is just usually left implicit to avoid notational clutter.


Now, its important to realise 0-forms on a manifold are precisely real valued functions on the manifold. Then the exterior derivative ##d## of a differential 0- form - aka a function ##f## - is defined by:


##df(v) := v(f)##


Here ##v## is a vector field, really a tangent field. Now I've already said above that tangent fields can be reinterpreted as a certain differential operator. So ##v(f)## is saying ##v## is deriving ##f##.


Now the exterior covariant derivative is defined by introducing an auxilary covariant derivative ##D## and this must live on some vector bundle ##E##. We haven't introduced vector bundles before. They are a generalisation of tangent bundles. And just as tangent bundles have a covariant derivative, so do vector bundles. Now instead of having k-forms we have instead E-valued k-forms. The space of all E-valued k-forms is written as ##\Omega^k(M, E)##. Then the exterior covariant derivative maps E-valued k-forms to E-valued k+1-forms. Or in symbols:


##d^D : \Omega^k(M, E) \rightarrow \Omega^{k+1}(M, E)##


Here again the ##d^D## ought to have a superscript ##k##, but its left implicit. Now to run the inductive definition of ##d^D## we need to find out what E-valued 0-forms are. These are just vector fields of E. We denote them by latin letters, so in the following formula for ##d^D##, by the letter s.


##(d^Ds)(v) = D_vs##


Thus we see in the zeroth step of the inductive definition we see we have replaced the derivative ##v## acting on ##f## by ##D_v## on ##s##. Here ##f## is a 0-form, aka a function, and ##s## is an E-valued 0-form, aka a vector field in ##E##.


We then inductively define ##d## for higher differential forms via a variation of the Liebniz rule for the derivative of products (Baez & Muniain, pg.63) :


##d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^k \alpha \wedge d\beta##


Here ##k## is the order of the differential form ##\alpha##. The symbol ##\wedge## is the wedge or exterior product - it's the generalisation of the cross product to any dimension because the usual cross product just works in 3d only. The corresponding inductive definition for the exterior covariant derivative is on Baez & Muniain, pg.250. It is also a variation on the Liebniz product rule:


##d^D(s \otimes \alpha) = d^Ds \wedge \alpha + s \otimes d\alpha##


Here ##s## is an E-valued 0-form and ##\alpha## is an E-valued k-form. Here the symbol ##\otimes## is the tensor product.


It's been a long post but I hope this gives you some sense on how the exterior covariant derivative comes about. To be honest, I don't expect you to grasp all this as I've skipped over a lot of detail - for example, what is the wedge product. And don't feel frustrated if it escapes you, as I've already said the exterior covariant derivative is really, really advanced stuff. But I do hope what I've written gives you a feeling for the shape of the terrain you would have to cover to rigourously learn this stuff. If you have any questions on what I've written then feel free to ask.
Nitpick : ## T(M \times N) ##~##T(M) \oplus T(N)##.
 
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  • #94
TensorCalculus said:
>thank you so, so much for putting in all of the time and effort to explain that to me (and sorry for the late response) - it means a lot to know that there are people willing to spend so long helping out a curious 13-year-old girl :D

Ah, I thought you were a 13 old year boy! A gender assumption on my part! I'm glad that what I wrote was taken in by you and understood. Richard Feynman once said that you only understand a concept if you can explain it to a student - so I guess I can say I understand the exterior covariant derivative! ;-).
TensorCalculus said:
>what is the tensor product?

The tensor product is taught badly. One reason is that they don't use geometrical intuition and its usually defined in a very mathematical way which is difficult to understand unless you've done the relevant math - basically the free vector space on a basis and a quotient of a vector space.

The tensor product of two vectors results in a higher dimensional vector. There are actually two ways to think of higher dimensional vectors. The simplest way is that the vector remains the same - an arrow - but it now lives in a higher dimensional space. This isn't the tensor product. The second way of thinking about it is that the arrow itself becomes higher dimensional. This is the tensor product. Thus a 3d arrow is the tensor product of 3 vectors ##u \otimes v \otimes w##. Whereas an ordinary vector has only one way to add, a 3d arrow has three ways to add. That's not so scary when you see it visually - however I haven't figured out how to insert diagrams here yet. A 2d arrow is ##u \otimes v## and is visually represented by a parallelogram in the obvious way. However, there is a proviso - you can internally rescale it without changing the tesor. By internal rescaling I mean you can rescale one side, say ##u##, by a factor of ##a## whilst rescaling the other side, ##v## by ##1/a##. This is because the magnitude of a tensor is its area. For a 3d tensor, which will be a parallelepipid, its magnitude will be its volume. And so on. Another thing to note is that in this geometric picture you can only define tensors upto the dimension of the ambient space. But in fact, it can be defined for arbitrary dimension. This is best shown mathematically.
TensorCalculus said:
>also, is the multiplication of manifolds similar to to taking the cross product of vectors? It seems that way... ish
the multiplication of manifolds follows the Cartesian product of sets. Have you come across this? Basically, a manifold is a set with a smooth structure. Here, the smooth structure is the set of all charts on the manifold. When you multiply two manifolds ##M##, ##N## together you first multiply the two sets together to get ##M \times N## and then the smooth structures on ##M## and ##N## determines a smooth structure on ##M \times N##. We still use the symbol for the Cartesian product ##\times## for the product of manifolds.

I think you also asked why ##T(M \times N) = TM \times TN##. The best thing would be to try a few low dimensional cases to see why its true. For example, the plane which is ##\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}##, the torus ##T^2 = S^1 \times S^1##. Recall here that ##S^n## is the n-dimensional sphere, and so ##S^1## is just the circle. Whilst ##T^2## is the torus which is 2d. And the infinite cylinder ##Cyl^2 = \mathbb{R} \times S^1##.
TensorCalculus said:
>It's been long - but actually great! I understood where it came from @CrysPhys that I have a long way to go till I'm ready to study it in any depth whatsoever) and it was really, really interesting. I think I would have been frustrated if it slipped past me (though most of it didn't... thankfully), because now you've made me curious - and once I'm curious there's no stopping me :) Thank you for the explanation once again - it was really eye-opening!

You're welcome. I think it's fantastic that a thirteen year old is taking so much interest in advanced ideas. You've done remarkably well in taking it in. Well done! If you have any further questions on the material, you're welcome to ask them.
 
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  • #95
Mozibur Rahman Ullah said:
Whereas an ordinary vector has only one way to add, a 3d arrow has three ways to add. That's not so scary when you see it visually - however I haven't figured out how to insert diagrams here yet. A 2d arrow is u⊗v and is visually represented by a parallelogram in the obvious way
Aren't you talking about multivectors here?
 
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  • #96
Hey y'all... we do have a differential geometry forum here... hint hint. :wink:
 
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  • #97
Muu9 said:
Aren't you talking about multivectors here?
Not quite. Usually k-multivectors (or k-polyvectors or k-vectors) are wedge products of k vectors rather than the tensor product of k vectors. But given that there is no specialised term for them, I think it's no great stretch to use the same term for them.
 
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  • #98
jtbell said:
Hey y'all... we do have a differential geometry forum here... hint hint. :wink:
I'm afraid it's a lost cause. :oldbiggrin:
 
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  • #99
jtbell said:
Hey y'all... we do have a differential geometry forum here... hint hint. :wink:
I've noticed. But this thread began as a reply to a request for advice on self-studying physics beyond high school physics and I mentioned differential geometry and it went downhill from there ;-).
 
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  • #100
Classic case of "topic drift." :cool:
 
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  • #101
jtbell said:
Classic case of "topic drift." :cool:
Nice phrase - I'll have to use it somewhere :-).
 
  • #102
Sorry for my late reply - exam season is taking it's toll, that's for sure.
Mozibur Rahman Ullah said:
Ah, I thought you were a 13 old year boy! A gender assumption on my part!
oh well, it happens all the time :-)
Mozibur Rahman Ullah said:
'm glad that what I wrote was taken in by you and understood. Richard Feynman once said that you only understand a concept if you can explain it to a student - so I guess I can say I understand the exterior covariant derivative! ;-).
yes, you certainly do. I think you explained it very, very well.
Mozibur Rahman Ullah said:
The second way of thinking about it is that the arrow itself becomes higher dimensional. This is the tensor product. Thus a 3d arrow is the tensor product of 3 vectors u⊗v⊗w. Whereas an ordinary vector has only one way to add, a 3d arrow has three ways to add. That's not so scary when you see it visually - however I haven't figured out how to insert diagrams here yet. A 2d arrow is u⊗v and is visually represented by a parallelogram in the obvious way. However, there is a proviso - you can internally rescale it without changing the tesor. By internal rescaling I mean you can rescale one side, say u, by a factor of a whilst rescaling the other side, v by 1/a. This is because the magnitude of a tensor is its area. For a 3d tensor, which will be a parallelepipid, its magnitude will be its volume. And so on. Another thing to note is that in this geometric picture you can only define tensors upto the dimension of the ambient space. But in fact, it can be defined for arbitrary dimension. This is best shown mathematically.
Ah, so it's not nearly similar to what I thought it was. This makes sense though (intuition aside - I've given up on intuitive for this). I won't ask why a 3D tensor is a parallelepiped because despite my curiosity, the proof might be a bit too much for me, but thanks, this was so, so much better than google :)
Mozibur Rahman Ullah said:
the multiplication of manifolds follows the Cartesian product of sets. Have you come across this? Basically, a manifold is a set with a smooth structure. Here, the smooth structure is the set of all charts on the manifold. When you multiply two manifolds M, N together you first multiply the two sets together to get M×N and then the smooth structures on M and N determines a smooth structure on M×N. We still use the symbol for the Cartesian product × for the product of manifolds.
I see. It's all quite standard then, not like the vector product. I wonder how this is computed with actual values.
Mozibur Rahman Ullah said:
I think you also asked why T(M×N)=TM×TN. The best thing would be to try a few low dimensional cases to see why its true. For example, the plane which is R2=R×R, the torus T2=S1×S1. Recall here that Sn is the n-dimensional sphere, and so S1 is just the circle. Whilst T2 is the torus which is 2d. And the infinite cylinder Cyl2=R×S1.
Yes, that makes it a bit more intuitive (no proof though... I guess I'll leave that for later)
Mozibur Rahman Ullah said:
You're welcome. I think it's fantastic that a thirteen year old is taking so much interest in advanced ideas. You've done remarkably well in taking it in. Well done! If you have any further questions on the material, you're welcome to ask them.
Haha - thank you, you flatter me. I do take interest in them, I can't help myself. I think it would be even more impressive if I'd be able to limit myself, and understand that I should look at the basics first, rather than just relying on my curiosity to tell me where to go. After all, I know next to nothing about physics. That's why I came here, and I definitely got what I wanted. A set path I can follow. Physics is truly beautiful, I find it hard to not want to learn as much as I can about it.
jtbell said:
Hey y'all... we do have a differential geometry forum here... hint hint. :wink:
Hint taken - though I do agree that this is a case of topic drift... it's a lost cause, sorry! :D
 
  • #103
TensorCalculus said:
Hello! So I'm a teenage physics enthusiast, who wants to take my knowledge past A-level (or in America I believe this would be high school level) physics.
I've studied multiple textbooks like Young and Freedman's University physics, studied maths from books like mathematical methods for physics and engineering.
I solidified all that by doing lots (like, LOTS) of the practice problems and some Olympiad papers.

I don't really know where to go from here.
I've resorted to surfing the internet, and finding free courses and watching YouTube videos that satisfy my interest, or reading popular science, reading high-school textbooks, and just sitting around.
I'm not entirely sure what books to buy - I fear that if I accidentally skip straight to something too advanced I'll not have strong basics. My teacher told me to just read popular science, but I really enjoy looking at the math behind things and I feel that there are few popular science books (that I've read at least) that satisfy my curiosity.
I do really, really love physics, and want to try and study it further - unfortunately, I'm 13 and have quite a long time till I can study physics in University/College.
Does anyone have any resources they would recommend to me (my interest particularly lies around astronomy and electromagnetism) , or any advice to give?
Schaum's textbooks, just classic.
 
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