IMHO, the notation $\int_a^b\mathrm{d}x\,f(x)$ is much cleaner than $\int_a^b f(x)\,\mathrm{d}x$, because the integration variable ($x$) and its associated integral range $(\int_a^b$) are kept ...

Just to add to the otherwise excellent answer by Triveth. He still leaves the origin of elliptical galaxies unexplained, i.e. how do the stars attain the spheroidal shape? After all, most stars form ...

In $4\times10^9$yr, M31 and MW (Milky Way) will have merged to form an elliptical galaxy. The internal spiral structures of either progenitor and their bars will be destroyed in the process, leaving a ...

You're confusing buoyancy with surface tension. The former applies to (macroscopic) objects which are (at least partly) immersed, while the latter applies to inter-fluid surfaces. Thus, a ...

If $$\mathcal{L} = \boldsymbol{a}\dot{\boldsymbol{q}} + \tfrac{1}{2}\dot{\boldsymbol{q}}^t\mathsf{\boldsymbol{T}}\dot{\boldsymbol{q}} - U(\boldsymbol{q})$$ with some constant vector $\boldsymbol{a}$ ...

Short answer: They don't. Contact binaries are a possible evolutionary phase of close binary systems, where both stars fill and/or overflow their Roche limit, i.e. 'kiss' each other at the L$_1$ ...

This is the pressure-gradient term integrated over all volume, converted to a surface integral and using Gauss' theorem. Note that physicists prefer the differential form of such equations (see also ...

Well, this depends obviously how far they are away. If their observable universe and ours overlap, then objects in that overlap will be observable by us and them. However, we will see them from ...

A plane wave is obviously a solution to the Schrödinger equation, but it cannot be normalised and hence cannot represent a particle. The condition for $\psi(\boldsymbol{x},t)$ to be normalisable, i....

The relation $\Delta v_x = \Omega h$ correctly expresses the difference in the inertial velocity between the top and bottom of the building/tower; this is not the cause of the problem. The naive ...

If $m$ < $n$, then your $F$ is a function from $m$-dimensional flat space to $n$ dimensional flat space, which maps $\mathbb{R}^m$ onto a $m$ dimensional surface in $\mathbb{R}^n$. If the motion ...

The smoke is a mixture of air and tiny 'dust' particles, which are much larger than the air molecules and undergo Brownian motion. Watching the dust particles under a microscope in a smoke cell (where ...

The appropriate definition of symmetry uses infinitesimal quantities, not just small quantities. Thus, in terms of your question, the Lagrangian is symmetric if $dL/d\epsilon=0$ at $\epsilon=0$. In ...

In an elastic collision the masses of both objects, the total kinetic energy, and the total linear momentum are conserved. The kinetic energy has contributions from the motions of the objects as well ...

Liouville's theorem holds for all Hamiltonian systems. If your definition of an asymptotically stable point $\boldsymbol{x}^*$ means that trajectories from points $\boldsymbol{x}$ in some ...

The problem is that the $x^3$ term also contributes to the first order (in $J_r$) correction to $H$ and we must go to second-order perturbation theory. Using the Deprit perturbation series (...

This is simply because for (most) structures the damping is weak, in the sense that $c^2\ll4km$, when the damping time $2m/c$ (the time over which the damping term takes energy out of the system) is ...

Your logic is correct, including the choice of tangential velocity as x = r*cos(phi); y = r*sin(phi); vx=-v*sin(phi); vy= v*cos(phi); Where you (tried to) set v in centrifugal balance. However, ...

Let's make a simplified ideal model of the situation: ignore heat losses to environment (evaporation, surface conduction) and gains from other sources (pumps running hot). Also, assume that the flow ...

Sciences use mathematics only as a tool. In almost all such applications, mathematical problems (such as pointwise vs uniform convergence) are not inherent to the scientific problem at hand, but arise ...

When neglecting air drag, the vertical and horizontal motions decouple and we have the equations \begin{align} \ddot{x} &= 0,\\ \ddot{y} &=-g. \end{align} The initial conditions (at $t=0$) are ...

I think your considerations are led by the daily experience that what you see actually happens as you see it. However, this is an illusion. All you perceive are photons, but they can (1) take their ...

As to your first question: the meaning is that the particle is not a free particle after all. This is the problem of a particle in a one-dimensional infinitely deep square potential well: $\Phi=0$ for ...

The post you're referred to explains reasonably well why spiral galaxies are discs. So why are stars not disc like. The simple answer is: a disc-like object cannot be a star. But what is a star? A ...

Only in certain simple situation can both gauges be satisfied. $V=0$ is obviously one of them. For a general electric and magnetic field configuration, however, only one gauge can be satisfied.

No, this makes no sense. The 'speed' that you would obtain (and which other answers provide) is not actually a speed. It is merely an angular speed (that of the Earth's spin) multiplied with a ...

In a coordinate system rotating at constant angular rate $\omega$, neither energy nor angular momentum are conserved and one has coriolis and centrifugal forces. The bead is forced outwards by the ...