# Tag Info

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I don't really like the whole wave-particle duality business because it obscures the more startling truth about particles: they aren't sometimes waves and sometimes particles, and they also don't transform into waves sometimes before reforming as particles, they are something completely different. It's like the story of the blind men and the elephant: a ...

9

The waves of quantum mechanics are probability waves. The solutions of quantum mechanical equations are the wave functions and the square of the wave function gives the probability of finding the particle at $(x,y,z,t)$. That is why the solutions for the electrons in the field of a nucleus are not orbits, but orbitals, i.e. probability distributions. The ...

7

It seems that you have misundrstood the wave-particle duality. What happens in the double slit experiment is that the electrons impact at the screen as they were particles. But they also interfere, just as waves. So you can see a wave-particle behaviour. But it doesn't say that the electron is destroyed, becomes a wave and then a particle again (as you ...

3

The wave-particle duality thing becomes important when you are dealing in a microscopic scale where quantum mechanics becomes relevant and you have to discard your ordinary notion of particle and wave. So don't expect to relate "particle" or/and "wave" notion that you usually get from picturing a marble or water wave from classical world surrounding you. ...

3

This is a fundamentally pointless question because negative mass doesn't exist (or so we think!), but I'll answer anyway because the answer is so unexpected. Suppose we take our two massive bodies: Then the gravitational force between them is repulsive because: $$F = \frac{G m_1 m_2}{r^2}$$ and $m_1$ and $m_2$ have different signs. But let's ...

3

The atmosphere is more or less at equilibrium with regards to being compressed by gravity, so there is no atmospheric heating caused by gravity (otherwise the atmosphere would gradually get closer and closer to the Earth, which clearly is not the case). Heating of the atmosphere is almost entirely due to radiant energy from the sun, along with terrestrial ...

3

The gravitational self-energy of any arbitrarily-shaped object with density function $\rho(\mathbf{r})$ and corresponding gravitational potential $V(\mathbf{r})$ is $$E=\frac{1}{2}\left\langle V,\rho\right\rangle=\frac{1}{2}\left\langle \nabla^{-2}\rho,\rho\right\rangle=\iiint\nabla^{-2}\rho(\mathbf{r})\rho(\mathbf{r})d\mathbf{r}.$$ Under a scale-parameter ...

3

Pretty much no. The problem is that you are not (pardon me :-) ) a rigid body, so you're going to feel a certain amount of force from the wind regardless of what sort of weights you're carrying. What can help is walking with your feet farther apart, which gives you a more stable base to work from, and to learn to turn your body sideways to the wind as ...

2

It can be seen to follow from a more general statement, namely: "if a parameter in the theory is such that the symmetry gets enhanced when it vanishes, then at every order in perturbation theory the corrections to this parameter will be proportional to its bare value". This is because perturbation theory respects the symmetry of the classical theory. If ...

2

Non-conservation of charge in Majorana terms The Dirac mass term is $m\bar\psi \psi$ where one field-factor $\bar\psi$ is complex conjugated (aside from other transpositions included in the Dirac conjugation) and the other is not. So one may assign a fermion number $1$ to $\psi$ which means that $\bar\psi$ automatically carries $-1$ and in the product, the ...

2

This going to be a rather approximate answer because it involves lots of estimated quantities like the current density of matter and the value of the cosmological constant. The second Friedmann equation tells us: $$\frac{\ddot{a}}{a} = \frac{-4\pi G}{3}\left( \rho + \frac{3p}{c^2} \right) + \frac{\Lambda c^2}{3}$$ It's conventional to take $a = 1$ at ...

2

If You know density $rho_r$ at some temperature $T_r$, there is a following formula for density: $rho=rho_r[1+b(T-T_r)]$, where $rho$ is the density at temperature $T$ and $b$ is called coefficient of cubical expansion, evaluated at reference temperature and density ($rho_r$ and $T_r$). It is valid for liquids.

2

I think you're confusion is with notation since unfortunately, two notation often used to denote projected spinors. One notation is to write: $$\psi \equiv \left( \begin{array}{c} \psi _L \\ \psi _R \end{array} \right)$$ In this notation $\psi _L$ and $\psi _R$ are two component Weyl spinors. However, a second notation ...

2

Physics occurring in one spacecraft traveling fast (at a uniform speed) is the same as physics in another spacecraft "standing still". In fact, the whole point is that these words - traveling fast and standing still - are relative. All the things you are describing like relativistic mass are only apparent to observers in other reference frames. The person in ...

2

In principle one has to calculate the pole of correlation functions involving gauge invariant operators like $\text{Tr}F_{\mu\nu}F^{\mu\nu}$. The problem is that due to asymptotic freedom, QCD is not solvable perturbatively at low energies. This is why nonperturbative techniques like lattice QCD are used to calculate such spectra. A key achievement in this ...

1

Electron is accompained by waves, so there still exists electron which has mass. This solves your problem I hope. Look here at what de Broglie says in his Nobel lecture of 1929 (this is an extracted portion): I thus arrived at the following overall concept which guided my studies: for both matter and radiations, light in particular, it is ...

1

Yes it fluctuates but it is a very small fluctuation. Note that unstable particles have a decay rate or width $\Gamma$ that is related to its lifetime $\tau$ by $$\Gamma=\frac{\hbar}{\tau}$$ when you measure the mass/energy of such particles in experiments you always get a Lorentzian or Breit-Wigner distribution like this from which you can measure the ...

1

The simplest setup is for small displacements. Suppose the spring rest lengths are $L_1,L_2,L_3$, the mass has mass $m$, the springs have constants $k_1,k_2,k_3$, the angle is 120 degrees between attachments, and the attachment points are set up so that at rest, the springs are all unstretched. The potential becomes ...

1

If you substitute the decomposition in, you get: $$\partial_t \rho^0 + \partial_t \rho^{E1} + \nabla \cdot (\rho^0 \mathbf{u}^E) + \nabla \cdot (\rho^{E1} \mathbf{u}^E) = 0$$ Typically the decomposition used assumes that $\rho^0$ is constant in time and that $\rho^{E1}$ is random in time, such that it's mean value is 0. Therefore, $\partial_t \rho^0 = ... 1 The problem is- The rocket is not 'fighting' with any force-field, it is 'fighting' with the very nature of space-time. So unless we have something of zero rest mass 'things' will tend to infinity. And yes the thrust will increase but space-time will distort( following Lorentz transformation, no GR effect here) in such a way that reaching 'c' 'tests' our ... 1 Both Newton and Einstein say that all the laws of physics remain same in a frame at rest with respect to the observer and also in a frame moving with constant velocity. Now Maxwell showed that speed of EM wave in vacuum equals 'c' is a law of physics. But this law was inconsistent with the Galilean transformation, hence Lorentz transformation were ... 1 This question is really about history and what was known to the protagonists in your tale and at what time. As a principle, relativity was embraced every bit as fully by Newton and Galileao as it was by Einstein - it's just that Einstein had a few more experimental results he had to gather into relativistic thinking. As in dgh's answer the whole point of ... 1 Atmospheric pressure, at some altitude, is largely due to the weight of the atmosphere above that altitude. The atmosphere also virtually behaves like an ideal gas. If evaporation of liquid water increases the concentration of H2O molecules in the air the density decreases and the air becomes buoyant and rises. As it rises it expands and cools. Rising air ... 1 Amplitude is the maximum displacement at steady state. That is okay for a working definition. So, all you need to do is find out is the maximum value of$|y(t)|$at steady state. There some small algebraic errors in your solution. That said,$y(t) = y_{cf}(t) + y_p(t)$To illustrate my point better, let me use the following form for$y_{cf}(t) ...

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