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It refers solely to trees, and Vallombrosa is the place "where th' Etrurian shades / high overarch't imbow'r."
这个比喻指的就是树木，地点是瓦隆布罗萨,“在瓦隆布罗萨，厄特鲁利亚的光辉逐渐暗淡“
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So if we want to talk about the volume of that, we just talk about the surface area, which is 4 pi r squared, and we multiply that by the thickness d r.
如果我们要讨论它的体积，我们要用的是表面面积，也就是４πｒ的平方，乘以厚度ｄｒ
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So instead of v bar, we write p v bar minus b, equal r t.
现在考虑，这些气体分子之间。
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And so, you know from your Newtonian mechanics, as you were learning in 8.01, the dynamic force here mv^2/r is mv squared over r.
在8。01节对牛顿动力学系统的学习中，我们可以知道这里的运动受力,就是。
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This is the initial part of the simile that describes the leaves: "thick as Autumnal Leaves that strow the Brooks / in Vallombrosa, where th' Etrurian shades / high overarch't imbow'r."
这是描写落叶的比喻的开始部分：,“他们人数众多，稠密得像秋天的繁叶,/在瓦隆布罗萨，厄特鲁利亚的光辉逐渐暗淡“
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And in fact, an example of a methodological attempt was done by Henry Gleitman at University of Pennsylvania, who built a tickle machine, which was this box with these two giant hands that went "r-r-r-r."
实际上,宾夕法尼亚大学的亨利·葛雷曼,曾经进行过研究方法上的尝试,他发明了一台挠痒痒的机器,这是一个装有两只大手的箱子,双手会不停的去挠痒痒
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You have to discount something that's one period in the future by dividing it by 1+r.
你必须把未来一段时间内的资金通过,除以1+r来贴现
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When you vary time a little bit and ask, "How does R change?"
当你把时间改变一个微元,然后问,"位矢 R 会怎样变化"
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We're saying the probability of from the nucleus in some very thin shell that we describe by d r.
某一非常薄的壳层dr内,一个原子的概率,你想一个壳层时。
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And orbiting around this is a lone electron out at some distance r.
有一个单电子,在环原子核的轨道上运行。
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And now the force, in its most general term / is q1q2 over 4 pi epsilon zero, which is the conversion factor r squared.
库仑力的最基本形式,就是，其中r是一个变量。
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q1*q2/ That's simply q1, q2 over 4 pi epsilon zero R.
那只是简单的。
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r So the circumference is 2 pi r.
周长=2π
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So, what we can do to actually get a probability instead of a probability density that we're talking about is to take the wave function squared, which we know is probability density, and multiply it by the volume of that very, very thin spherical shell that we're talking about at distance r.
我们能得到一个概率,而不是概率密度的方法,就是取波函数的平方，也就是概率密度，然后把它乘以一个在ｒ处的,非常非常小的,壳层体积。
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If you get them backwards, logr you will integrate one over r and will get log r.
如果你逆推的话，对1/r积分得到。
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And we wrote something that looks, the energy is equal to minus the Madelung constant times Avogadro's number, 0R0 q1 q2 over 4 pi epsilon zero R zero.
我们写下了，晶格能等于负的马德隆常数,乘以阿伏伽德罗常数，乘以q1q2除以4πε
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dE/dr=0 We take dE by dr equals zero at r equals r naught.
当r等于r圈时。
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And we can find r naught by looking for the minimum.
我们可以通过寻找最小值得到r圈。
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And this equilibrium spacing is denoted r sub zero or r naught.
这个平衡间距,用r下标0或者r圈表示。
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When you get one over R, you get a gradual fall.
当你用1除以R，你得到逐渐的下降。
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Generally, if it pays c dollars for every period, the present value is c/r.
推广开来,如果在每段时期内支付c美元,那么它的现值就是c/r
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Then, I was talking it over with another graduate student and he said, well if you want to find out why don't you ask J.R. Hicks?
随后,我和另一位研究生讨到了此问题,他说,如果你想知道,干嘛不直接去问J·R·希克斯
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So I have a distance here of 2 r naught.
因为有一个2r圈的距离。
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So the square root of n squared r e over r h.
这里的n值是什么呢？
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I'm going to write the particular case of R and take derivatives.
我会写出 R 的一个特例,然后对其求导
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If it's going in a circle, you will say from now on, that it, indeed, has an acceleration, even though no one's stepping on the accelerator, of amount v^2 over R.
如果它在一个圆周上运动,你会说从现在起它其实有加速度,即使没有人去踩油门,加速度大小为 v^2 / R
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Energy goes as one over r.
能量和1/r成正比。
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and a radius R and that's where your particle is.
半径为 R,这就是质点的位置
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That's what I meant. This is called R.
这就是我说的 R
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The size of that is v^2 over R.
它的大小是 v^2 / R
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