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6

Yes - you can even propel spaceships with it - Solar Sail Although Solar radiation pressure at the Earth is around 9E-6N/m2 while the thrust from a Saturn V rocket is 34 MN, so you would need a solar sail something like 2000Km on a side to get the same acceleration


6

That's because the relation $p=\gamma mv$ doesn't hold universally. As you just showed yourself, using this relation for a photon would lead to a contradiction because the energy of a photon isn't zero. A heuristic way of seeing why this relation won't hold for a photon is by recognizing that $$p=\gamma mv =m\frac{d x}{d\tau}$$ but a photon doesn't ...


5

You can understand the expression by attempting the limit for $v\rightarrow c$ and $m\rightarrow0$. Notice that $\gamma\rightarrow\infty$ when $v\rightarrow c$. Therefore $m v \gamma$ is an undetermination of the form $0\cdot\infty$. From your expressions, you cannot say that $E=0$. The limit does not exist, and this implies that this expression is not valid ...


3

by putting $p=\gamma mv$ and then get a value for $m$ (which will be 0 for a photon) and therefore rendering the equation to $E=0$ First, let's write this out in full (in 1D) $$p = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}} $$ Then, solve for $m$ $$m = p\frac{\sqrt{1 - \frac{v^2}{c^2}}}{v}$$ Now, holding $p$ constant, see that the limit of $m$ as $v ...


3

1) Yes indeed, the absence of a $\Delta$ in the second expression is just a typo. 2) The last expression is derived assuming that mass is a constant. If it helps, just set the mass equal to 4, or something. If we want to know how the quantity $ 4v $ changes, we really only need to know how the quantity $v$ changes. Suppose $v$ changes from $v_1$ to $v_2$. ...


2

If there is oscillation, the equilibrium is stable in at least one dimension. The situation where momentum keeps increasing could apply to a driven mass oscillating about a stable equilibrium. It would also apply to a non-driven mass at a saddle point. Consider that the mass begins oscillating approximately, but not exactly, along the concave up curve in ...


1

I'll provide an answer in non-relativistic quantum mechanics. The short answer is that momentum and mass commute, so a particle can have a well-defined momentum and mass simultaneously. But really, mass isn't considered an operator in quantum mechanics; it's a parameter, a number. So for some system, it is presumed that the mass is known always. There's no ...


1

I can give you one example. In a semiconductor reverse-biased p-n junction, a potential barrier exists that prevents electrons from crossing the junction. There is an energetically-forbidden region in the vicinity of the junction. The wave functions of electron states in both the valence and conduction bands are real exponential in this region. ...


1

In the dynamical systems jargon, this is the usual Lyapunov stability. If additionally the system reaches the equilibrium state, then the equilibrium is called asymptotically stable. For the formal definition, see this link: http://www.scholarpedia.org/article/Stability#Definitions:_Stability_of_an_Equilibrium It is also worth to remark that a equilibrium ...


1

Yes. It should be: $$\frac{dp}{dt}=\frac{d(mv)}{dt}$$ I'm using $d$ instead of $\Delta$ because I am thinking about the limit where the changes in $p$ and $t$ are very small. Then these are called infinitessimal changes, and denoted by a $d$. Usually, when one considers simple problems in Newtonian mechanics, what one does is study a given object with a ...


1

According to this article (the choice of article has no significance other than it came up first in my Google search): Remington's 12-gauge 2 3/4-inch Premier Magnum turkey load has 1 1/2 ounces of shot and a 1260 fps muzzle velocity Converting these figures to metric, $m_{\text{shot}} = 0.0425$ kg and $v_{\text{shot}} = 384$ m/sec, so the momentum of ...


1

The current density formula you consider is not right, because $\frac{d(\psi\psi^*)}{dt}$ is always 0. The current density is defined as $\bf{j}=(i\hbar/2m)(\psi\nabla\psi^*-\psi^*\nabla\psi)$. In your one dimensional case change $\nabla$ to $\partial/\partial x$, you will get your answer. A more physical definition of current density operator is this: ...


1

Well that's a result using differentiation and derivation. Have you studied calculus? If not, there is a simple way to look at it. $$\frac{\Delta p}{\Delta t} = \frac{\Delta (mv)}{\Delta t}$$ (Yes there should be a $\Delta t$ in the denominator, too.) Now, what does $\Delta(mv)$ mean? It represents the change in the quantity $mv$. For current situations, ...


1

You are right, there is a $\Delta$ missing in front of the $t$. $\Delta v = v_2 - v_1$. If the mass is not changing, then $\Delta (mv) = mv_2 - mv_1 = m(v_2 - v_1) = m\Delta v$. Hope that helps. The equation that includes $\frac{\Delta m}{\Delta t}$ is not Newton's second law. The second law is valid only for systems of constant mass. An equation like ...



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