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1

Why can't we simply assign to every point of the Bloch sphere a phase $e^{i\phi}$? This is the idea of a section of a fibre bundle. You are considering in this case a base space $S^{2}$ with fibre $S^{1}$. Locally the fibre bundle looks like $S^{2}×S^{1}$. However you want to consider a fibration such that the global space is not the trivial product ...


0

There is a mathematical convention for direction angles $\alpha$ in an $x$-$y$ diagram: counter-clockwise from positive horizontal axis (your "east" direction). Since $\tan \alpha = y/x$ is a periodic function with period $\pi$, $\arctan (y/x)$ has two solutions in the range $[0, 2\pi]$, while your calculator may give you just one solution, mostly in the ...


0

Typically in motion problems angles tend to be measured from the horizontal. For example, in projectile motion problems, the launch angle of the projectile is nearly always measured from the horizontal, and when doing force inclined-plane problems, the same is usually true. However, there are situations where that is not always true. Sometimes in statics ...


2

There is no hard and fast convention. What is important is that the direction you state is unambiguous. You might state a given angle as $20^\circ$ South of West, or $200^\circ$ counterclockwise from East. Neither is "more correct", but note that neither is ambiguous, either.


2

Here is your picture, with a couple of additional angles and segments drawn: Do you see it now?


2

Perhaps looking at it like this will help clarify things: You have two triangles, one with angle $\alpha$ and one with angle $\alpha + \mathrm{d}\alpha$. The sides opposite those angles differ by $\mathrm{d}x$, but the hypotenuses are essentially the same, both equal to $r$. Write an equation expressing the fact that the triangles' opposite sides differ ...


2

We have $$\tan \alpha=\frac{a-x}{b}\implies\sec^2 \alpha \,d\alpha=-\frac{dx}{b}\implies\frac{r^2}{b^2}d\alpha=-\frac{dx}{b}\implies rd\alpha=-\cos\alpha\,dx,$$ which is the desired relation with a minus sign, that must be present, since an increase of $x$ decreases $\alpha$.


1

Yes it takes integration. Take a small section of the rod ${\rm d}x$ with density $\rho(x)$ and cross sectional area $A(x)$. The mass of the rod is $$ m = \int \limits_0^\ell \rho(x) A(x) {\rm d}x $$ The center of mass $x_C$ is defined as $$ x_C = \frac{1}{m} \int \limits_0^\ell x \rho(x) A(x) {\rm d}x $$


2

The volume of the shape drawn there is $dS\,v\,dt\cos(\theta)$. $dS$ is not a cross section, it is at an angle to the axis $v$ is aligned with.


0

I don't understand... Those two cylinders have the same volume $S.L$: So all particles exactly on (or beyond) surface $S$ at point $A$ and time $t$ will be exactly on (or beyond) surface $S$ shifted at point $B$ by time $(t+dt)$ assuming they all move at constant velocity $\vec{v}$. I don't see how the orientation of the surface changes anything. I can't ...


1

You can't simply multiply $dS$ with $\vec{v}$ to obtain the volume. If we assume a coordinate system $x$ perpendicular to $dS$, and $y$ in the plane of $dS$. The volume is defined as: $$V=\int^{s_{begin}}_{s_{end}} \: S \vec{dx}_{\perp S}$$ However, if we want to express it as a function of $v dt$ we get: $$V=\int^{t_{begin}}_{t_{end}} \: S \: \vec{(v ...


0

The way I see it, with larger $\theta$ (as long as it's not $\pi/2$) it is still cylinder of volume $dS.v.dt$ that will pass through the surface $dS$


2

Look at large $\theta$ : fewer particles per unit time will reach the surface because their perpendicular velocity is much less. When $\theta = \pi/2 $ zero particles cross the surface.


5

This problem was first formulated by Leonhard Euler in 1744 WPlink: "That among all curves of the same length which not only pass through the points A and B, but are also tangent to given straight lines at these points, that curve be determined in which minimizes the value of \begin{equation} \int_A^B \frac{ds}{R^2} \end{equation} It is a problem of ...


0

Let $N(x)$ be the cumulative number of distinctive neighbours in a distance $x$ for a certain point $A$ (all points with a distance of $x$ or less) and $n(x)$ be the distributive number of neighbours (distinctive points with the distance of exactly $x$). $N(x) = n(0) + n(1) + \ldots + n(x)$ $n(0) = 0$ $n(x) = 3*x$ $N(x) = \sum_0^x n(x) = \sum_0^x 3x = ...


0

Here's an image of a hexagonal lattice to help you count http://www-personal.umich.edu/~sunkai/teaching/Winter_2013/hexagonal.png You can image the lattice being infinite in extent outside the image. So pick any point and count its nearest neighbors. You count six. Now pick another point and count them? Do you notice a pattern? You can be sure that this ...


0

Both isotopes are isovalent so electronically they are identical (in essence your not going to get that much difference in bond angle). However in the asymmetric well approximation since deuterium is heavier the $D-O$ bond is lowered down the well i.e. it has a lower ZPE than the $H-O$ bond and is therefore a stronger bond. This means it has a smaller ...


0

According to http://www1.lsbu.ac.uk/water/data.html (and with reference to A. G. Császár, G. Czakó, T. Furtenbacher, J. Tennyson, V. Szalay, S. V. Shirin, N. F. Zobov and O. L. Polyansky, On equilibrium structures of the water molecule, J. Chem. Phys. 122 (2005) 214305), the bond angles H-O-H and D-O-D are pretty much the same - 104.50 and 104.49 deg., ...


1

Comments to the question (v2): One nice property of the 2-norm (as compared to other norms, such as, e.g. the $p$-norm) is that it gives rise to an inner product via a so-called polarization trick. E.g. in the real case the polarization formula has 2 terms: $$ \langle u, v \rangle ~:=~\frac{1}{4} || u+v ||^2 -\frac{1}{4} || u-v ||^2 .$$ There is a similar ...


0

Start by looking at the coin in a frame of reference that is centered on the stationary coin and that is always "looking" at the rotating coin. You will see the coin make one complete revolution. The easiest way to convince yourself of this it to imagine that you have a "mirror image" coin - both of them with the numbers 1-12 (like a clock), but one the ...


0

At the 3:00 position, it would have moved half a circumference. At the 6:00 position, it would have moved 1 complete revolutions, since the top of the moving coin would touch the bottom of the stationary coin. In other words, the top of the moving coin still remains at the top. Therefore, the answer is 2 complete rotations.


1

You just need to think about the relation between L and theta, which to give a clue involves tan.


1

Binocular vision has already been discussed, but it left out an important aspect. A single eye is sensitive to distance. The shape of the lens changes to focus on near/far objects. The reason this is needed is that our pupil has finite size and cannot be modeled as a pinhole. The same physics is going on here as in a lens of a camera focusing on an ...


3

To have depth perception two eyes are needed. Our two eyes are some distance apart which causes the photons from an object to arrive at slightly different angles. The brain then reconstructs the depth field from these differences. Similarly, we can figure out how far nearby stars are by using images made by a telescope at two different times of the year, ...


1

Often in physics, Objects are approximated as spherical. However do any perfectly spherical objects actually exist in nature? Yes, with a couple of qualifiers. For example, a perfectly isolated 4He atom in its ground state is a perfect sphere according to the standard model of particle physics. This follows because the nucleus is in a spin-zero state, ...


4

If we want a sense of localness (or calculus to work), we'd like to be able to obtain the length by adding up the length from pieces of the path (for example using a ruler, or counting paces as we walk along the path between two points). However, even considering just two dimensions we see something interesting for $L_p$. $$\left(|x|^p + |y|^p\right)^{1/p} ...



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