Tag Info

New answers tagged

0

I'm going to assume you have some familiarity with linear algebra, as the math becomes much less tedious than trying to do three-dimensional trigonometry with the $x$, $y$, $z$-coordinates directly. You are looking for a function describing the line of the horizon. Since it is a circle and thus a one-dimensional object, I'm going to call it $\vec{h}(t)$, ...


0

Any, the Hamiltonian whos potential nergy is a function of $r$ only has rotational invariance. Since the components of $\mathbf{L}$ are generators of rotation, it can be shown that $[\mathbf{L},V]=[L^2,V]=0$


0

Angular momentum is defined as, $$ L=r\times p\equiv r\times\frac{\hbar}{i}\nabla $$ From this it follows that $$ \left[L_x,\,x\right]=\left[L_x,\,p_x\right]=0 $$ where $$ L_x\sim yp_z-zp_y $$ Noting that $$ \left[A,BC\right]=\left[A,B\right]C+B\left[A,C\right] $$ the expected results can be determined.


1

Yes, for the hydrogen atom Hamiltonian, $L^2$ commutes with the Hamiltonian because it commutes with $r^2$.


0

This properly belongs on math.se, but to properly derive these you need to remember that we can write a vector in terms of basis vectors. The vector $\vec{A}$ is unchanged, but it is just expressed as a different linear combination: $$\vec{A} = A_x \hat {x} + A_y \hat{y} = A_r \hat{r} + A_\theta\hat{\theta} $$. Because you can write $\hat{r}$ as a linear ...


2

Note that you haven't actually found the general solution in spherical coordinates... What you have there is a solution known as a spherical wave, which describes a set of spherically symmetric wave fronts that diverges from (or converges towards) the origin $r=0$. However, in general, a wave could also be a function of the angles $\theta$ and $\phi$, which ...


6

The two solutions are different because they have different boundary conditions. In the first case, the equation is indeed $$ \frac{\partial^2u}{\partial t^2} = c^2 \nabla^2 u = c^2 \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right) u. $$ Here though we specify $u(t,x=x_0)$ to be some value ...


1

Looks like you wish to express matrix addition with matrix multiplication. This is done with adding extra (service) dimension, which components are always 1 for vectors. $\begin{bmatrix} a_1& a_2& a_3 & 1\\ \end{bmatrix} \times \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ b_{41} & ...


1

The rotation matrices you used only apply to rotations around cartesian axes (x,y,x). You cannot generalize it to spherical coordinates in the way you did. See page 4 here for the rotation matrix in spherical coordinates.


3

There appears to be a typo in the coordinate transformation for $r$: it should be $$ r = \rho\,\exp(\tau/\alpha). $$ With this in mind, your reasoning is correct; if you work out the partial derivatives, you'll find that $g_{\tau\tau} = 1$, and you can work out $g_{\rho\rho}$ in a similar way. P.S. also note that $t = \tau - ...


2

My guess would be that $\mathbb E^n$ denotes Euclidean space. In addition to having geometric structure (angles and distances) and motions (rotations, translations, reflections) - not all of it terribly useful in the 1-dimensional case - it is an affine space. Affine spaces have no notion of distinguished origin or zero point. We can use a vector space like ...


0

I don't see the difference between your "coordinate time" and just time. Yes, time is relative. Since time is relative, it is incorrect to find a reason (or cause). A cause is, by definition, some local event in the past, which affects some other event (consequence) in the future. Time relativity is not an event. It is not local, and is not in the past. ...


0

At the research of the answer, I would recommend to consider the example of an astronaut going very far in a lifetime (example: moving near speed of light, he reaches a star 5000 light years away in only 50 years). I would ask the reversal of "Why is coordinate time frame dependent?": why proper time is passing slower than the observer's dilated time? It ...


1

Tilting an object in space changes its apparent dimension (think of trying to get furniture through a door: the width of an object depends on its orientation). Objects in relative motion are tilted in space and time (or rather, spacetime), and different observers will see things unfold under different perspectives. Personally, I find relativity of ...


0

One reason for the difference is time dilation--a given coordinate clock at rest in one frame will be running slow as measured by coordinate clocks at rest in another frame. It can be demonstrated that this follows logically from the two postulates of special relativity, using the "light-clock" thought-experiment detailed here. The second postulate says that ...


0

The ambiguity is resolved by choosing the coordinates first. So: 1)Set up a cartesian system with x,y,z coordinates. 2)Pick the group you would like to study. Let's say $P4_32_12$ from page 1151 of this document: http://mcl1.ncifcrf.gov/dauter_pubs/284.pdf that DavePhD recommended. 3)You can then see that the asymmetric unit is given by $$0\leq ...


0

Here are some examples, this should get you started: http://math.mit.edu/~mckernan/Teaching/12-13/Autumn/18.02/l_16.pdf


5

Let's look at your last line: Why not simply postulate a global coordinate system and let the coordinate mapping from $M$ to $\mathbb{R}^n$ and the resulting metric tensor describe the geometry? P.S. By Rn I mean the set of all n-tuples with the usual definitions of sum and scalar multiple, not Euclidean n-space. $\mathbb{R}^n$ has a particular, simple ...


2

First of all, the metric tensor is one additional piece of structure one inserts on a smooth manifold to measure lenghts and angles. The metric is indeed not present in all applications of Differential Geometry to Physics (see e.g. Lagrangian Mechanics). In that case, it is important to know also how to deal with manifolds without metric tensors. Now, about ...



Top 50 recent answers are included