Actually, I think you will learn more from those books. Anyway, I agree with almost everything mal4mac said. However, I don't think you should risk taking a harder course if you think you won't do well because your GPA matters after all. You can however choose to use CR/NCR so that it doesn't...
I guess the difference is that in math the coordinates (basis) we are picking are not some real physical thing whereas in physics, the coordinates are physical. So (1, 2, 3) is coordinate independent in math but in physics a vector is something attached to a physical space and so it depends on...
I don't quite see the connection with vectors and functions but here is some relevant comments which might help.
From a mathematical point of view v = (1, 2, 3) is indeed a vector as an element of the vector space \mathbb{R}^3 over \mathbb{R} with the usual operations. Furthermore v is...
PHY100H is a bird course (at least if you know any physics at all). I never took it but two of my friends (studying life science) took it and got an easy 4.0. As stated by others, it talks about physics and introduces you to physics in a qualitative way. As far as I can remember, the marks...
Can you explain why the multilinear array definition is coordinate free as this is not obvious to me. If you need to choose coordinates to get the tensor, then the tensor depends on which coordinates you choose. You definitely cannot write one down without choosing coordinates. This definition...
I see my mistake now. I kept corresponding the V's in the multilinear map definition to the V's in the algebraic definition when I should have corresponded the V^*'s to the V's in the algebraic definition
Yes all of them are equivalent but what about tensors as multidimensional arrays where...
Ok, I will explicitly state all the details to be clear. In Calculus of Manifolds, tensors are defined as all multilinear forms with the field being the real numbers,
T: V^n \to \mathbb{R}
The book doesn't go into the dual spaces but including that would look like,
T: V^n \times {V^*}^m...
May be you misunderstood my post. A tensor is a multinear form with the field being the real numbers (at least the way I learned it). You only get a matrix or a multidimensional array representation when you pick coordinates but otherwise they are coordinate free just like linear...
I learned tensors as the multilinear forms T: V^n \to \mathbb{R} (you can also add in the dual spaces) where V is a vector space over \mathbb{R}. So I don't quite see how vectors are tensors (assuming the vector is from the vector space V). A tensor of type (1, 0) would just be a linear...
Interestingly, we used the formula for the "surface area" for the hypersphere in thermal physics! It came up when deriving the multiplicity of a ideal gas with fixed amount of energy (related to probability of each microstate of the gas).
First we just assumed there is one particle. So all...
Its the second case. Since in general there are many topological spaces for one set, X, so the open sets of a topological space of X will depend on which topological space of X you are working with.
I think eigen decomposition is another term for spectral decomposition in the spectral theorem. Although it is stated in a different way than diagonalizing a matrix, the spectral decomposition is related to the Schur decomposition for normal linear transformations.
You will get x as a function of y, say f(y) = x. Then you can try and find the inverse to get y as a function of x, i.e. y = f^{-1}(x). There are a few things that can go wrong. If f is not invertible then that tells you that there is no unique solution y(x) to the differential equation...