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1

All the other answers that say the "single" integral is simply a shorthand notation are right, but it is well to remember that one can indeed construe the integral as a single integral as a Lebesgue integral (if you do nothing else, look up Lebesgue's very cute little half paragraph summary (on the Wiki page) of his idea in a letter to Paul Montel). If ...


13

It is, in fact, a double integral! The first notation used $$\varPhi_E = \oint_S \vec{E} \cdot \mathrm{d}\vec{A} = \oint_S \vec{E} \cdot \hat{n} \ \mathrm{d}A$$ is simply a more compact notation. It's much easier to write $\mathrm{d} \vec{A}$ instead of, say, $r \ \mathrm{d}r \ \mathrm{d}\theta$ all the time. Furthermore, it's more general, as $\mathrm{d} ...


3

The general formula is indeed a double integral, so the most technically correct way to write it is $$\Phi_E = \iint_S \vec{E}\cdot\mathrm{d}^2\vec{A}$$ But when formulas start to involve four, five, or more integrals, it gets tedious to write them all out all the time, so there's a notational convention in which a multiple integration can be designated by ...


5

It is just a more compact notation. It is implied by the integration element $dA$ that you are integrating over the surface.


5

For doing matrix multiplication, yes, we usually put the dummy indicies together, but this is (assuming one is not working over $\mathbb{H}$) convention really. Don't forget that this is shorthand Einstein convention. When you write $$ \mathbf{A} \cdot \mathbf{B} = A_{ij} B_{jk} = \sum_{j=1}^n A_{ij} B_{jk} $$ you are writing a shorthand version in ...


0

If you are dealing with spacetime indices (i.e. tensors over the spacetime), then symbols like $\delta^{ab}$ or $\delta_{ab}$ don't make sense. If you lower an index of $\delta^a_b$ you will end up with the metric $g_{ab}$, same for raising an index. This is clear from the definition of $\delta$: $$g_{ab}\delta^b_c=g_{ac}$$ and $$g^{ab}\delta_b^c=g^{ac}$$ ...


0

The result is the metric: the effect of the Kronecker delta in your examples is to set $b = c$. The Kronecker delta is really just the identity matrix.


0

You're confusing two things. In tensor calculus, the Kronecker delta should be visualized as basically the identity. What it does is relabel an index. Example: $$g_{ab}\delta^b_c=g_{ac}$$ This has nothing whatsoever to do with the Dirac delta function (it's actually a distribution) in this context. In nonrelativistic quantum theory Dirac delta ...


0

The co/contra distinction only makes sense when talking about vector fields. Even then the difference only becomes apparent when dealing with curved spaces or at least curvilinear coordinate systems The difference comes from how vectors relate back to the undlying space or manifold on which the fields are defined. Contravariant vectors then are what people ...


1

We expect a vector to change in a certain way when we change the scale we use to measure distance. Consider the vector $$\vec{x}=(1, 0, 0)\,\mathrm{m}$$ If we change scale and now measure in centimeters this vector becomes $$\vec{x}=(100, 0 ,0)\,\mathrm{cm}$$ Now consider a vector representing a force: $$ \vec{F}=(1,0,0)\,\mathrm{J/m}$$ where I've chosen ...


2

You are right that $\langle A\rangle_{\psi}$ is a number and not an operator. However, people often write just a number when they actually mean the identity operator times that number. So in the right hand side, $\langle A\rangle_{\psi}$ should actually be $\langle A\rangle_{\psi} \mathbf{1}_H$, where $\mathbf{1}_H$ is the identity operator on your hilbert ...


2

The notion of co- and contravariance depends on context: If you wanted to be as clear as possible, you should actually mention with respect to what the components transform co- or contravariantly. In case of the algebraic dual of finite-dimensional vector spaces, the implied context is a change of basis of the vector space. Then, we can look at how the ...


5

There are two more points that can be made here. Sorry if I repeat someone. In a way you are right that if you have a vector space and its dual there is no intrinsic way to say which space is the original and which is the dual. This is because there is a canonical isomorphism between a vector space and the dual of its dual. In other words if $V$ is a vector ...


2

I) No, it is important to distinguish between covariant and contravariant tensors. OP's link mentions differential geometry. If one has only studied those objects in the context of pseudo-Riemannian manifolds $(M;g)$, which comes equipped with an (invertible) metric $(0,2)$ tensor $g$, then the existence of the musical isomorphism may perhaps unnecessarily ...


0

I will say that the standard definition of vectors and one-forms is not the world's cleanest. A modern definition of vectors would say that a vector space is a mapping from the functions on the space to itself that satisfies the Leibniz rule and is linear (alternately, the vector space is the local linear approximation of the space). Then, the set of ...


22

This is not really an answer to your question, essentially because there isn't (currently) a question in your post, but it is too long for a comment. Your statement that A co-ordinate transformation is linear map from a vector to itself with a change of basis. is muddled and ultimately incorrect. Take some vector space $V$ and two bases $\beta$ and ...



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