Terry Tao 最小作用量原理理解为微分方程的另一种 formulation 解运动微分方程，相当于最小化一个泛函，即作用量 解方程2x-3=0相当于最小化一个函数x^2-3x。T 正比于dx/dt的平方, V 正比于 x^{2} . 这

物理学中“作用量”这一概念该如何理解?

最小作用量原理是个很神奇的东西, 可以导出各种理论. 但"作用量"本身是个什么东西呢? 在国际单位制下它具有角动量的量纲, 或者是能量X时间, 这有什么特别含义么?

以及, 计算作用量时被积分的那个拉格朗日函数代表了什么? 一个物理系统的拉格朗日函数是如何获取的? 是有一定规则, 还是根据经验拼凑修补呢? 就我接触过的知识看来, Lagrangian都是大神们先知先觉就给写出来了, 或者通过某些神奇的技巧就给推出来了, 比如经典电磁场, 还有量子场论里的标量场矢量场等等. 就没有一个通用规则说什么东西的存在应该对应哪些项的.

当然以上概念作为抽象的理论要素直接接受也不是不可以, 就像能量这个概念其实也是挺抽象挺不直观的, 只是日常生活常用, 大家习以为常了而已. 在此只是想了解下大家对此有啥自己的理解, 相互交流下.

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匿名用户

杜纸钱、何史提、周怡帆 等人赞同

• 类比：
  解方程相当于最小化一个函数。
  解运动微分方程，相当于最小化一个泛函，即作用量。
• 作用量的量纲，较有意义的类比，是与普朗克常数的量纲相同。有一种作用量的形式是（Maupertuis' principle）被积分项对应的是相空间中的体积元。
• 很多微分方程（不仅是物理学中的）都可以用变分法表示（可理解为最小化某个作用量），但是并非所有微分方程都可以。具体的适用条件还在研究，但已经有很多结果。涉及较深数学，请自己看这里的讨论。

因此请将最小作用量原理理解为微分方程的另一种 formulation，
并在相空间和不确定关系的基础上理解其量纲。

编辑于 2013-04-30 2 条评论 感谢 分享 收藏 • 没有帮助 • 举报

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Tao Chen、匿名用户、csxiao 等人赞同

1. 理论力学给了我们很多线索，在1D单个质点的情况下，L中常出现的是动能与势能的差，即L=T-V。通过Legendre变换， 得到H=T+V，正好是个守恒量，物理意义是能量。比如对于谐振子：
T 正比于dx/dt的平方, V 正比于 . 这里没有写准确的表达式，因为你会发现这其实已经从形式上给出了很重要的东西。
2. L应该是在洛仑兹变换下的标量，这个要求很重要，尤其对于场论。
3. 从计算的角度讲，如果研究场，它的泛函极值对应的微分方程应该是经典的场方程。如果研究是单粒子方程，则是经典的动力学方程。
举个例子：
经典的电磁场的协变形式，即张量表式形式，及其有利于我们凑这个L的形式。
我们有：
还有：
经典场方程：
这些量都是在处理经典的真空Maxwell方程时得到的结果。而且我们发现A的四维形式完美的把三维的A和电势结合在了一起。
然后作类比： ，前者明显比后者完整，是连续场中相比1D粒子情况下必须出现的东西。主要是将位置的梯度包含了进去，因为描述的是场，是物理量的分布。有人会问，为什么不包含更高阶的情况，实际上是有人作尝试的。
光这样当然不行，上面是dx/dt的平方，而且L必须是标量，于是自然想到张量的缩并可以，于是有：
, 还不如直接用这个了：. 好，算是动能项吧。 看看1D的情况，还有,好吧，直接用A缩并就可以的！于是又有了....诸如此类的类比和猜想。最后别忘了加上：


不论对于标量场，矢量场还是Dirac场，都会频繁的出现对场的作用，都是类比于1D动能。而场自身的平方，或是缩并，都类比于1D下谐振子的势能，

L的形式很多时候都是基于类比，对称性的要求构造出来的，到底正确与否还得看和实验的比较。

算是起个抛砖引玉的作用吧，并不那么严谨。比如场论中矢量场的构造还得兼顾量子化时的需要，比如约束条件等等。那里会有几套量子化方案。。。严格的说来矢量场量子化是约束体系的量子化问题。。。。

编辑于 2013-05-02 6 条评论 感谢 分享 收藏 • 没有帮助 • 举报

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前面回答的都不错。
我就说一下拉格朗日量是如何构造的。
经典情形下，我们只需要用量纲分析即可，但是要记住，我们在经典场的情形下是可以不需要考虑相互作用项，分别说
标量自旋0场，克莱因戈登方案，引入原因是量子化薛定谔方程方案，四维动量。实际上的原因很简单啊，拉格朗日量的量纲在4维时是4，标量场\phi是1，\partial是1，所以我们需要(\partial\phi)^2。
自旋1/2场，狄拉克方案，量纲分析法，\bar\psi量纲3/2\partial是1\psi是3/2。
矢量自旋1场，麦克斯韦方案，量纲分析同自旋0场。

量子化之后我们要考虑的东西就很多了，还要考虑重整化
举个栗子，
考虑标量自旋0的相互作用\phi^4，量纲分析\phi是1，4个\phi是4。同时满足重整化要求的量纲分析。

所以写拉格朗日量的第一步都是量纲分析。后来做出的理论直接应用量纲分析的听起来不大一样的是超对称量子力学的拉格朗日量。然后用了数学推广到2d 超对称sigma模型。

后面如何写拉格朗日量的方法就多了，比如在2维的情况下，我们将系统束缚在一个以标量\phi为半径的圆上，我们就可以加一项\lambda(1-\phi^2)这里量纲分析是2维下，拉格朗日量量纲为2，标量自旋0场\phi永远是1。
再比如用凯勒流形作为限制写拉格朗日量，得到sigma model。
等等等等，我们第一步写拉格朗日量就完成了。
第二部是考虑（经典）对称性。（我其实不太喜欢经典对称性这个词，经典对称性只要说出来就意味着量子化后是不对称的，否则就直接叫对称性了，而量子化后不对称说明自然界不存在这种对称性，也就是说经典对称性是不存在的。。。）对称性可以再剔出一部分不可能的拉格朗日量。
接下来就要考虑量子化了！这个就是麻烦太多的事情，可以将拉格朗日量进行筛选。

编辑于 2014-05-01 3 条评论 感谢 分享 收藏 • 没有帮助 • 举报

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Physics is not a finished logical system. Rather, at any moment it spans a great confusion of ideas, some that survive like folk epics from the heroic periods of thee past, and others that arise like uptopian novels from our dim premonitions of a future grant synthesis. ——Weinberg, Gravitation and Cosmology

个人看来，在经典物理里，拉式量就是物理理论的妈妈。给定了一个系统的拉式量，就决定了这个系


http://mathoverflow.net/questions/101395/which-differential-equations-allow-for-a-variational-formulation

Which differential equations allow for a variational formulation?

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Many ODE's and PDE's arising in nature have a variational formulation. An example of what I mean is the following. Classical motions are solutions q(t)  to Lagrange's equation ddt ∂L(q,q ˙ )∂q ˙  =∂L(q,q ˙ )∂q ,  and these are critical points of the functional I(q)=∫L(q,q ˙ )dt.  Of course one needs to be precise with what considers a solution to both equations. This amounts to specifying regularity and a domain of the functional. This example is an ODE, but many PDE examples are possible as well (for example electromagnetism, or more exotic physical theories). Once one knows a variational description of the problem, many more methods are available to solve the problem. Now I do not expect that any PDE or ODE can be viewed (even formally) as a critical point of a suitable action functional. This is because this whole set up reminds me of De Rham cohomology: "which one-forms (the differential equations) are exact (that is, the d  of a functional)?". The last sentence is not correct, but the analogy maybe is? Anyway, my question is: Are there any criteria to determine if a given differential equation admits a variational formulation? fa.functional-analysis ca.analysis-and-odes ap.analysis-of-pdes share|improve this question asked Jul 5 '12 at 11:58 Thomas Rot 6031415

1   I'm sure that the answer is yes and that I have read it once. But where ? –  Denis Serre Jul 5 '12 at 13:33 3   Ricci flow was not known to be a gradient flow until Perelman introduced his F-functional. I guess the setting was more complicated here, because there is a big gauge group of diffeomorphisms, but thats an example of where people were not sure if it could be given a variational formulation until it was done so. –  Otis Chodosh Jul 5 '12 at 15:02 6   expanding a bit on the comment of Otis: It was of course know that the Ricci flow is not the gradient flow in a strict sense of a diffeo-invariant functional (reason: gradients of diffeo-invariant functionals are always divergence free, the Ricci tensor is not divergence free), so the amazing and surprising discovery of Perelman was that the Ricci flow actually is a gradient flow up to diffeomorphisms. –  Robert Haslhofer Jul 5 '12 at 16:05 Differential equations with a Hamiltonian formulation, automatically have a Lagrangian formulation (formally, at least). Of course, this simply moves the difficulty over to the question of when one can determine that an equation has a Hamiltonian formulation, but in practice, having a conserved integral of motion tends to be a useful clue in this regard... –  Terry Tao Jul 5 '12 at 19:49 Regarding Ricci flow: in retrospect, the existence of gradient shrinking Ricci solitons, and the absence of periodic-up-to-diffeomorphisms "breather" type solutions, was an indication that Ricci flow had a gradient-flow-up-to-diffeomorphism interpretation. One could also build upon this observation to try to guess the functional that gives this gradient flow, by writing down the integrals that vanish for gradient shrinking solitons, and then trying to express those integrals as a gradient-up-to-diffeomorphisms of something, though this is still far from an easy task... –  Terry Tao Jul 5 '12 at 19:53  | show 1 more comment

3 Answers 3

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Others give useful references that discuss what is known about the answer, but no statement of the answer itself. The relevant algebraic setting is the variational bicomplex, which is discussed in the works of Anderson and others. In this setting, there are two differentials, the horizontal differential d H   (representing derivatives with respect to independent variables like t  ) and the vertical differential d V   (representing variational derivatives with respect to dependent variables like q(t)  ). Each of these differentials is "de Rham-like" and they anticommute with each other, which explains the cohomological flavor of the answer. A rough statement of the answer is as follows. A Lagrangian L  density gives rise to a set of Euler-Lagrange equations E i =0  as follows: d V L=E i  d V q i −d H θ,  that is, the vertical 1-form E i  d V q i   is vertically exact (up to a horizontally exact term d H θ  ). So, it is necessary for E i =0  to be the Euler-Lagrange system of some Lagrangian that E i  d V q i   is closed up to a horizontally exact term, namely d V (E i  d V q i )=d H θ ′ (=−d H d V θ).  In fact, the same condition is also sufficient, up to obstructions related to the global topology of the manifold where the dependent variables q  take their values. This condition was formulated already classically by Helmholtz. However, the above statement is restrictive in that it answers the question only when E i =0  is already in Euler-Lagrange form. However, there are many transformations that one can apply to the system E i =0  that gives an equivalent system F a =0  . Given only the system F a =0  , is it still possible to decide whether it is equivalent to some system E i =0  in Euler-Lagrange form? This is the hard inverse problem (aka the multiplier problem). The only general result that I'm aware of in that direction is this. If there exists a form ω  of vertical degree 2 and horizontal degree n−1  , where n  is the number of independent variables, such that it is both horizontally and vertically closed modulo the equations F a =0  (namely d V ω=A a F a   and d H ω=B a F a   ), then there exists (again, up to global topological obstructions) a Lagrangian density L  whose Euler-Lagrange equations E i =0  are equivalent to a subsystem of F a =0  . To my knowledge, the above observation first appeared in Henneaux (AnnPhys, 1982) for ODEs and in Bridges, Hydon & Lawson (MathProcCPS, 2010) for PDEs. The calculation demonstrating this observation is given in a bit more detail on this nLab page. (Edit: At risk of shameless self-promotion, I'll also note that I collected these observations in a self-contained paper (arXiv; JMP, 2013).) It reduces the solution of the hard inverse problem to classifying all such forms ω  (corresponding to the so-called characteristic cohomology of the variational bicomplex restricted to F a =0  in the corresponding degree) and checking that there exists a candidate that gives rise to a Lagrangian density whose Euler-Lagrange system E i =0  is equivalent to the full system F a =0  . The calculation of the corresponding characteristic cohomology of the system F a =0  is still non-trivial, but there exist ways of attacking it, which include Vinogradov's C  -spectral sequence mentioned in other responses. share|improve this answer edited Dec 2 '13 at 13:14 answered Jul 6 '12 at 10:16 Igor Khavkine 4,40721231

Nice answer! Since you mentioned the har inverse problem, I'll complete your answer a bit bu mentioning that the problem of determining when a system of paths---one for every tangent direction---correspond to extremals of a variational problem is quite interesting and goes back to Darboux, Hilbert (his fourth problem, for example), Douglas (3 dimensions), and has been worked on by very many people. There is a huge gap between general results (your boxed theorem) and concrete cases like Hilbert's fourth problem. –  alvarezpaiva Jul 6 '12 at 10:49 Thank you very much for this answer. I will ponder on it. –  Thomas Rot Jul 8 '12 at 8:25 add a comment |

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There is a huge amount of literature on this problem. I include some works that seem "classic" and that I've consulted at some point: Tulczyjew: http://www.springerlink.com/content/u9481124734547t6/ 105_419_0">http://www.numdam.org/numdam-bin/fitem?id=BSMF_1977_105_419_0 Takens: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.jdg/1214435235 And the work of Anderson on the variational bicomplex (I think this complex was introduced by I.M. Gelfand circa 1970). There is also quite a bit of good work by Vinogradov and his school (the C-spectral sequence). All this is mostly applications of homological algebra to the theory of PDE's. share|improve this answer answered Jul 5 '12 at 20:43 alvarezpaiva 6,0331237

Thank you for the references, I'll look into it. –  Thomas Rot Jul 8 '12 at 8:26