ajb

By positive definite you exclude the possibility of zero as an eigenvalue, as for positive semidefinite you include zero as an eigenvalue. Is that right DH?

For my undergraduate studies we had something like 4 hours of lab work a week. Then there were extended projects in the 3rd and 4th years which could involve a lot more lab work. Much more emphasis was placed on theoretical work. I found it difficult to get many of the experiments working satisfactorily. I am sure that is why I went into theory/maths instead of experimental work.

Sorry Albers, been away for christmas. I am no expert in building solutions. I assume that you can indeed have very different metrics in different regions, but you will have to match them in a smooth way. I beleive that this matching can be difficult.

I suggest if you want to learn some physics Halliday Resnick and Walker is a good place to start. It will teach you just about everything a first year undergraduate in physics should know.

My first thought is that the topics you list are very wide and that it is impossible to get more than a superficial knowledge of all of them. Of course, I think that everyone should have some knowledge of all aspects of modern science, but the reality is that there is so much out there and there exists many different approaches to science. I am very guilty of not knowing much biology or experimental physics. My honest advice would be to get a real understanding of a few select topics and just get aware of some of the others. As to what topics, that is up to you and will be hugely influenced by who you talk to. Either way, good luck.

The antisymmetric part of [math]g^{ab}[/math] (note upper indexes) is a bi-vector and as such could be used to define a bracket on functions on the manifold. If this bi-vector satisfies the condition [math][g,g]=0[/math] where [math] [,][/math] is the Schouten-Nienhuis bracket (a generalisation of the Lie bracket), then we have a Poisson structure, i.e. a Poisson bracket on functions. This bracket would now also satisfy the the Jacobi identity. This is still classical. To make it quantum you would want to deform the Poisson algebra or use Dirac's canonical quantisation. What is true is that the commutator of quantum mechanics can be formulated as a classical Poisson bracket if you take the manifold to the the Hilbert space of states. (Not sure if the infinite dimensions makes this difficult in practice). Everything would look like classical mechanics. Now, if you were to further insist that the antisymmetric part of the metric were non-degenerate, i.e. [math]t^{a}g_{ab}t^{b}= 0 \implies t^{a} , t^{b} = 0[/math] for all vectors [math]t[/math], then you would have a symplectic structure. (We assume non-degeneracy for the symmetric part). So simply, I don't think you can view it as the quantum commutator, but it is the "classical origin" of it.

The exterior derivative is a derivation on differential forms. Locally you can always represent differential forms as [math]\omega = \omega_{0}(x) + \omega_{\mu} dx^{\mu} + \frac{1}{2!}\omega_{\mu \nu} dx^{\mu} dx^{\nu} \cdots [/math] where you think of the [math]dx^{\mu}[/math] as formally being Grassmann odd. (they are functions on a particular supermanifold). Thus we have the antisymmetry properties of differential forms; [math]dx^{\mu}dx^{\nu} = - dx^{\nu} dx^{\mu}[/math] etc. The exterior derivative (aka de Rham differential) is the homological vector field [math]d[/math] which in local coordinates looks like [math]d = dx^{\nu} \frac{\partial}{\partial x}[/math] As it is homological we have [math]d^{2} = \frac{1}{2}[d,d] = 0[/math] So you define the action of the exterior derivative on a differential form as [math]d[\omega] = d\omega[/math]

You could have an antisymmetric part in general, although the definition of a metric is that it is symmetric. An antisymmetric part would be more like a symplectic form. As you know [math]F_{ab}[/math] is antisymmetric and you can use this as part of a symplectic form. I see no reason why you could not consider the object [math]\hat{g}_{ab} = g_{ab} \pm \Omega_{ab}[/math]. Where [math] \Omega[/math] is the symplectic two form. (You may want some factors in there etc... ). I would suggest starting with a symplectic form as it is non-degenerate. Maybe it has been done already? What you could do is start from a manifold [math]M[/math] and build the symplectic manifold [math](T^{*}M, \Omega )[/math]. In local Darboux coordinates we have [math]\Omega = dx^{a}\wedge dp_{a}[/math]. Then in these local coordinates you can build [math]\hat{g}_{ab}[/math] and then start asking about it properties. One natural question is if there exists a connection which preserves this. It would be a "mix" of the Levi-Civita and Fedosov connections some how. (If it exists). Another extension is super Riemannian geometry. The metric for the odd part is antisymmetric (really it is all supersymmetric).

In algebra, you think of tensors as multilinear maps. In geometry, it is more useful to think of tensors as objects who's components transform in a certain way under diffeomorphisms (or maybe some subgroup of the diffeomorphism group). There is no need for me to be any more specific here as a quick google search will show you the details. The point here is that the components of the tensor transform as to compensate for how the basis changes leaving the entire object invariant. (From this point of view tensors are just scalars on a larger space!) I think this fact is not stressed enough in physics texts.

Tensors, tensor densities and connections are simply fundamental in physics and differential geometry. As the Lorentz transformations are in fact linear infinitesimal diffeomorphisms in hindsight it is not surprising that tensor objects under Lorentz transformations are tensors in the general sense.

I don't know the book by Adler, Bazin, and Schiffer. It seems quite old, but then by the 1960's most of the framework of general relativity was established. It is possible that notation has changed slightly. Wald I would also recommend. I think that the books by Nakahara and Bertlmann give the best introduction to differential geometry as needed in physics.

Have a look at the notes by Carroll. They are a good introduction, but are not really general enough if you want a good understanding of differential geometry. There are some notes via my website that may also be useful, but I know there are some typos. Really the only way to learn differential geometry is to do differential geometry. Most physics books on general relativity are poor and lack the mathematical background needed.

The angle between two vectors remains the same after a conformal transformation. You have used the metric to define what you mean by [math]\cdot[/math] and [math]||[/math] This is all very standard stuff Albers, as you often talk about aspects of general relativity I thought you would be quite familiar with this.

True, but in quantum mechanics time is not a Hermitian operator. It is not an observable in the usual sense. It is a lot more involved as to how you can derive this relation.

The inner product is really a map that takes a vector and a dual vector and spits out a number (or whatever your field is). Generally, you can think of it as a generalisation of the dot product. You can use the inner product to define what you mean by the angle between vectors. I don't think this concept is always useful. A conformal transformation on (pseudo-)Riemannian manifold [math](M,g)[/math] is a diffeomorphism which preserves the metric up to a scale. More precisely, Let [math]f :M \rightarrow M[/math] be a diffeomorphism. Then [math]f[/math] is said to be a conformal transformation iff [math]f^{*}g_{f(p)} = e^{2 \sigma}g_{p}[/math] where [math]\sigma[/math] is a function of [math]M[/math] Now, conformal transformation on "stretch" they don't "twist". That is they do not preserve lengths but preserve angles (as defined via the inner product/metric).

Even more loosely, everything travels in 4-d space-time. They only thing anybody agrees on is the length of any two journeys in space-time that start and end at the same points. Thus, for this to be true any observers on different paths must experience different "amounts" of space and time in order for the total to be the same. This is the origin of Lorentz invariance and all the other weird effects.

The responses so far are correct. There is no rest frame for the photon and so it makes little sense to try to talk about a proper time. You can heuristically think of it as photons experiencing "no time", but I don't think that is a precise statement. It is more of an analogy by letting the speed tend to c in the formula for time dilation. As the Lorentz group in non-compact, you cannot consider a Lorentz transformation actually at c.

Differentiation and calculus in general must be the most widely applied section of mathematics. I think it has applications in just about all branches/applications of mathematics. So what is it used for? As has been stated, differentiation gives you the infinitesimal change in something with respect to some variable.

Yes, thank you for the correction.

You should look up noncommutative geometry, the Moyle plain and deformation quantisation. There are many papers on quantum field theory on a noncommutative spacetime in various contexts.

Marsden is well known. His book 1981 book Calculus Unlimited is available on his website as a PDF. Good luck

On a vector space the only thing you can do in general is say that they are equal. This is because we have a well defined zero-vector, 0. The vectors a,b are equal iff a-b = 0 This is then understood as the components being equal. (you should then show that it does define an equivalence) Generally there is no product or inner product between vectors (or more correctly for inner product between vectors and dual vectors). As the magnitude of a vector requires a inner product (or a metric) generally we do not have the notion of a magnitude.

This was my observation also! For [math]e^{-t^{2}}[/math] you will need the error function, which is related to the normal distribution.

This is the n-body problem and it's stability. This is an old question in classical mechanics. The n-body problem is in general non-integrable and may be chaotic!

Why not try it and find out for yourself?