right now, assume both 乙肝,丙肝 are treatable, then it is not 路径依赖;
if 丙肝 is not treatable, and you get it, then your life, luck are all 路径依赖: you get 乙肝, you are ok; you get 丙肝, you are done this life.
热统计物理学,时间有方向,路径依赖,初试点或条件很重要,生死攸关
but there is this theoretical argument, that if we make
旋转变换_百度百科
baike.baidu.com/view/2028050.htm轉為繁體網頁
缺少字詞: isotropic
chap07邻域运算_百度文库
wenku.baidu.com/view/101a3f8583d049649b665809轉為繁體網頁
Laplace算子的旋转不变性的证明| NARUTOACM
www.narutoacm.com/archives/laplace-rotation-invariant/轉為繁體網頁
am going to give an answer that's probably more difficult than necessary. Classicaly, two frames are inertial if observer in each frame can agree on Newton's Laws (law of inertia, f=ma, action/reaction). Turns out that they must be moving at constant speed relative to each other. If you are in an accelerated frame, then newton's laws won't hold. The first one, for example, would break down, because object would have acceleration without force (there is an ambient force due to the fact that the observer is accelerating). The problem is how do we jump from one frame to another so that measurements in each frame agree. By simple assumption of homogeneity (no preferred origin) and isotropic (no preferred direction) of space one can prove that there are three kinds of transformation possible: one with a speed minimum, one with a speed maximum (Lorentz), and one with neither (Galileon). The first kind turns out to violate causality. And the third kinds turns out to be invalid because some other law of physics is not invariant under it (Maxwell's Eqn). So one conclude that the second one to be correct and the third one to be a limiting case of it (limit as the speed limit goes to infinity). And for all law (included EM) to be correct, this speed limit must be the speed at which EM disturbance travel in vaccum, i.e. the speed of light. Arriving at this conclusion requires 5 assumptions: 1) all Law of physics is invariant when you go from one inertial frame to another. 2) the transformation from one frame to another is linear 3) spacetime is homogeneous 4) spacetime is isotropic 5) Causality is preserved. I like this derivation better because the speed limit turn out to be a result, not an assumption. And the 5 assumptions above are more "obvious", "trivial", and "intuitive". In short, inertial frame are frames that moves at constant speed with respect to each other. And by assuming the 5 assumptions above, and using some group theory and linear algebra. One arrive at the Lorentz transformation which has stand the tests of experiments
Reference https://www.physicsforums.com/threads/what-is-an-inertial-reference-frame.11148/
Maxwell's equations - Wikipedia, the free encyclopedia
en.wikipedia.org/wiki/Maxwell's_equations
Wikipedia
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[PDF]EM Waves, Wave Propagation in Linear/Homogeneous ...
web.hep.uiuc.edu/home/serrede/P436/Lecture.../P436_Lect_06.pdf
[PDF]Chapter 1 Electromagnetic Theory ² Maxwell's Equation
www.colorado.edu/...fa05/Chapter1.pdf
University of Colorado Boulder
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[PDF]Lesson 9 - NICADD
nicadd.niu.edu/~piot/phys_630/Lesson9.pdf
[PDF]Electromagnetic Waves in Homogeneous Media - MIT
web.mit.edu/.../Lec02_Waves-v11....
Massachusetts Institute of Technology
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如图所示,坐标空间中有场强为E的匀强电场和磁感应强度为B的匀强磁场,y轴为两种场的分界面,图中虚线为磁场区域的右边界.现有一质量为m,电荷量为-q的带电粒子从电场中坐标位置(-L,0)处,以初速度v0沿x轴正方向开始运动,且已知L=
m
v
2
0
qE
.
试求:要使带电粒子能穿越磁场区域而不再返回电场中,磁场的宽度d应满足的条件.
v |
2
0
|
答案:解:带电粒子在电场中做类平抛运动,设粒子进入磁场时的速度大小为v,速度方向与y轴的夹角为θ,如图所示,则:
vy=
?
=v0
故
v=
=
v0
cosθ=
=
,θ=45°;
粒子在磁场中做匀速圆周运动的半径为:R=
要使带电粒子能穿越磁场区域,磁场的宽度应满足的条件为:d<(1+cosθ)R
即:d<
答:要使带电粒子能穿越磁场区域而不再返回电场中,磁场的宽度d应满足的条件为d<
.
vy=
qE |
m |
L |
v0 |
故
v=
√
|
|
√
| 2 |
cosθ=
vy |
v |
| ||
2 |
粒子在磁场中做匀速圆周运动的半径为:R=
mv |
qB |
要使带电粒子能穿越磁场区域,磁场的宽度应满足的条件为:d<(1+cosθ)R
即:d<
(1+
| ||
qB |
答:要使带电粒子能穿越磁场区域而不再返回电场中,磁场的宽度d应满足的条件为d<
(1+
| ||
qB |
点评:本题考查带电粒子在匀强磁场中的运动,要掌握住半径公式、周期公式,画出粒子的运动轨迹后,几何关系就比较明显了.
分析:粒子在电场中受到的电场力的方向向上,粒子做类平抛运动,水平方向做的是匀速运动,竖直方向做的是匀加速直线运动,从而可以求得带电粒子运动到Y轴上时的速度;当磁场的运动的轨迹恰好与磁场的右边沿相切时,此时的磁场的宽度最大,根据粒子的运动的轨迹可以求得磁场的宽度最大值.
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