Talk:Lagrangian/Archive 1

Archive 1

Old comments

• Can someone please explain where $\partial _{\mu }$ came from? -- Taral 17:38, 17 Jun 2004 (UTC)

$\partial _{\mu }$ (shorthand for ${\frac {\partial }{\partial x^{\mu }}}$) crops up here instead of ${\frac {d}{dt}}$ that you see in, for example action, because the function that is being varied, $\ \phi (x)$ is a function of 'n' variables $x^{\mu }$ ($\mu$ = 1,2,3 ... n) as opposed to q(t) which is a function of 1 variable t. -- Amar 10:03, Jun 18, 2004 (UTC)

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• Lagrangian, Lagrangian mechanics and Action (physics) all contain duplicate material! Phys 19:00, 24 Jul 2004 (UTC)

Agree. Will try and attempt a merge sometime soon. [[User:AmarChandra|Amar | Talk]] 09:14, Jul 28, 2004 (UTC)

If no one objects, I'll add the content of this page to Lagrangian mechanics and set up a redirect here. - mako 30 June 2005 00:19 (UTC)

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• Can someone explain why is the spelling "Lagrangian" preferred here to "Lagrangean"? I would expect the pattern "Shakespeare -> Shakespearean".

I would suspect the difference is time period. Shakespeare -> Shakespearean was probably a valid English formation for a significant period of time. Lagrange lived more recently by ~200 years, when Lagrange -> Lagrangian was probably favored.

Personally, I had never even considered "Lagrangean" as an option, but apparently it is sometimes used: Google search turns up 1:27 for Lagrangean:Lagrangian. --Laura Scudder 16:54, 3 Apr 2005 (UTC)

Isn't it because Lagrangean is pronounced with a hard g, so to keep the soft j sound it's spelled Lagrangian? Fephisto 17:19, 30 December 2006 (UTC)

Well, usually when one refers to a function or object, one uses (-ian), e.g. Lagrangian or Hessian or Jacobian or Wronskian or Hamiltonian. So you guys are probably right that we should talk about Lagrangean mechanics, just as we talk about Euclidean geometry, but then there is also Riemannian geometry and Minkowskian geometry, and having to talk about a Lagrangian when dealing with Lagrangean mechanics would just be awkward. This is just the way physicists mangle words.

Just a note, the term Lagrangian is much newer than the object itself. In older literature it is just called Lagrange's function. :) --Lionelbrits (talk) 18:17, 27 November 2007 (UTC)

Lagrange was originally known as Giuseppe Lodovico Lagrangia, under which name he did his work on this stuff. That's why it's the Lagrangian. I think this should be clarified on the wiki page. Susilehtola (talk) 08:26, 25 January 2014 (UTC)

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I tried to Edit the article by using a less confusing method of representing Kinetic and potential energy but the sub scripts k and p do not show as they should. Could this be solved in any way. I think rather than assigning variables T and V for kinetic and potential energy respectively; It is rather better to use E(k) or E(p) respectively. Any consideration is appreciated.

27 June 2008 —Preceding unsigned comment added by 99.247.108.206 (talk) 16:35, 27 June 2008 (UTC)

To 99.247.108.206: See Help:Displaying a formula for information on how to make Tex do what you want. I think that we should stick with T and V which are standard notation rather than $E_{k}$ and $E_{p}$ which would actually be confusing to many people who are used to the standard. JRSpriggs (talk) 19:28, 27 June 2008 (UTC)

Functional

Is the Lagrangian itself a functional? It doesn't seem to be by the definition given. It's the expression for the action that's a functional, mapping a set of functions onto R. It's true that the coordinates that are the domain of the Lagrangian can be expressed as funtions of some parameter, but I'm not sure if that really makes the Lagrangian a functional; it certainly seems to be a differant case from the action integral.

I'm not confident enough to change this, and there do seem to be multiple definitions of functional, which confuses things. Comments? --Starwed 14:42, 14 December 2005 (UTC)

By the only definition of functional I know, it is indeed the action, not the Lagrangian that is a functional. I'll be bold and see if anyone comes up with another definition. — Laura Scudder ☎ 15:23, 14 December 2005 (UTC)

Vote for new external link

Here's my site full of PDE's with harmonic functions. Someone please put it in the external links if you think it's helpful!

http://www.exampleproblems.com/wiki/index.php/Calculus_of_Variations

Conflict of interest

Like User:Linas, I am concerned about recent edits by the data.optics.arizona.edu anon of this and other articles, edits which present the work of B Roy Frieden as mainstream. In fact, this work is not universally accepted and has been severely criticized (google for sci.physics.research discussion from some years ago, for starters). The conflict of interest is that this anon has confessed to being B. Roy Frieden IRL:

Notable edits by the data.optics.arizona.edu anon include:

1. 150.135.248.180 (talk · contribs)
   1. 20 May 2005 confesses to being Roy Frieden in real life
   2. 6 June 2006: adds cites of his papers to Extreme physical information
   3. 23 May 2006 adds uncritical description of his own work in Lagrangian and uncritically cites his own controversial book
   4. 22 October 2004 attributes uncertainty principle to Cramer-Rao inequality, which is potentially misleading
   5. 21 October 2004 adds uncritical mention of his controversial claim that Maxwell-Boltzmann distribution can be obtained via his "method"
   6. 21 October 2004 adds uncritical mention of his controversial claim that the Klein-Gordon equation can be "derived" via his "method"
2. 150.135.248.126 (talk · contribs)
   1. 9 September 2004 adds uncritical description of his work to Fisher information
   2. 8 September 2004 adds uncritical description of his highly dubious claim that EPI is a general approach to physics to Physical information
   3. 16 August 2004 confesses IRL identity
   4. 13 August 2004 creates uncritical account of his work in new article, Extreme physical information

---CH 21:25, 16 June 2006

This is kind of a meta-question, so skip it if you don't like it. But please don't give me the business just for asking it.

See, I have this "thing" about inconsistency in my universe. Yes, one would think that if they care about truth, that the popes and poobahs and ayatollahs should sit down and talk and decide what's real. "Okay, so just why do you think god is an elephant with six arms, if you would be so very kind enough to explain it please?"

But then again, I EXPECT True Believers of different religions to politely ignore each other while preaching furiously to the people on the street. Why? Because they're presumptuous and ignorant and arrogant and STUPID.

But when the Jedi Priests of my own religion (physics) contradict each other and leave it at that, it gives me the willies. It gives me the creeps. It gives me the heebie-jeebies. It makes me feel like maybe the real truth is going to jump out of a dark alley and assault me, like when it turned out santa claus was actually just another ridiculous lie the grownups tell.

So I'm afraid I must butt in and ask:

Hasn't it been established whether this Frieden guy is right or wrong? Okay, "EPI is a general approach to physics" is an opinion, so the hell with it. You don't have to explain that. But the underlying issue is physics, not opinion. I guess I'm asking So can the Klein-Gordon equation can be derived via his method, or not? It should be trivial (or at least do-able) to determine if the Maxwell-Boltzmann distribution can be obtained his way.

You put "method" in quotes. Is his procedure ambiguous, or not well-defined (like faith healing)? Does his method give the wrong answer (like pre-newtonian gravity)? Does it just happen by coincidence to be correct in a particular but important case (like the Bohr atomic model of hydrogen)? Does it use generally acceptable but not rigorously-proven-legitimate techniques (like renormalization)? Does it deny something generally believed to be true (like creationism)? Does it assert something generally accepted as false (like aether)? Does it posit the existence of new, hypothetical entities (like dark energy)? Does it give only an approximate answer that you guys don't think is close enough to the correct one (like that law the stupid people passed declaring pi equal to exactly three)? Is it obviously wrong (like when my boss fired me for disagreeing that "the Maharishi Yogi is the only scientist who truly understands relativistic physics")? Is it completely incoherent nonsense (like the revolutionary-theory-proving-Einstein-wrong-of-the-day at sci.physics.relativity, or the Scientific Proof of God from Ph.D. particle physicist and crackpot celebrity | George Hammond)?

Or what?

Note that I am not asking for a description of his specific "theory". Not only am I almost certainly too dumb to understand it, but IMHO (with emphasis on the "H"), my question should be answerable without hiding behind jargon, even though the jargon is more conclusive and concise.

I know that physics whackos like to quote Einstein. But nevertheless, he did say "if you can't explain it to your grandmother, you don't understand it", and Feynman was legendary for explaining stuff using words like "stuff". I myself have explained why three-dimensional acceleration is really four-dimensional rotation in a way that any smart high-school kid could understand. The interval metric and Riemann spheres, too.

My concern here is the charge of "conflict of interest". That concept does NOT belong in physics. If he's wrong, then exclude him for that. But it's cowardly to use an irrelevance to hustle an unpopular opinion out the side door. I couldn't care less if Jim Maxwell had taken out a patent on the light bulb before publishing his book, because his equations comprise an astoundingly elegant, simple, and beautiful explanation that I believe is even more important to revealing the deep nature of reality than Newton's.

And the hell with Wikipedia rules about "conflict of interest". Wikipedia "rules" are applied when crooked, dishonest Wikipedia admins want them applied, and ignored when they get in the way of the club's mandatory opinion. We wouldn't delete Aldrin's contribution to an article about the decision to make Armstrong the first man on the moon instead of him, even though he feels very strongly about it.

SO, I ask again: If this Frieden guy is an optics physicist for the University of Arizona who published in a legitimate journal, why can't his work be cited here?

Or is there no more convincing reason than "conflict of interest"?

Curiosity is gonna kill me someday,

HelviticaBold 04:17, 10 February 2012 (UTC)

To Helvitica Bold: Eh, what? If there is a question in all that rambling, I missed it. Please ask it short and clear. JRSpriggs (talk) 12:10, 10 February 2012 (UTC)

Inaccessible

This article is not very accessible because the equations are not explained in words, such as the equation that shows that S does not change (which is an assumption based on conservation of energy I suppose). Also the different forms of the differential are not clear. I haven't been able to find the meaning of reverse wiggle d for example, which closes me out of a proper understanding. As I recall this is a post grad level d symbol, so it is not reasonable to expect a person to know it. --Centroyd 02:41, 22 July 2007 (UTC)

The action does not change because you can show that ${\frac {\delta }{\delta q_{i}}}S\left(q_{i})\right)$ leads to the Euler-Lagrange equation, which is zero. That's where the zero comes from. It's all covered in action (physics). The $\delta$ symbol is usually used to indicate a small but non-differential change. It's probably used here to remind us that we're changing a functional, which requires we jump through some extra hoops. — Laura Scudder ☎ 14:47, 22 July 2007 (UTC)

The $\delta$ indicates that the differential is a virtual displacement differential. A force acting through a virtual displacement does virtual work. Virtual displacements are not functions of time (time is taken to be held constant when considering virtual displacements). When (1) the only net applied forces to a system are defined by scalar potentials (i.e. they are conservative), and when (2) the inertial forces are used to to convert a dynamic system into a static system, and when (3) the generalized coordinates are chosen to match the internal degrees of freedom of the system, then the principle of virtual work for applied forces says that the constraints do no virtual work (because they act normal to the paths of the virtual displacements), that the difference of the inertial and applied forces also does no virtual work because the system is in static equilibrium, and that the system operates in accordance with Lagrange's equation. Sorry this isn't more detailed, but this is just a talk page. The reference I got this information from is: Torby, Bruce (1984). "Energy Methods". Advanced Dynamics for Engineers. HRW Series in Mechanical Engineering. United States of America: CBS College Publishing. ISBN 0-03-063366-4. ChrisChiasson 20:25, 22 September 2007 (UTC)

I also agree that the motivation and derivation of the Euler-Lagrange equations could have been described less tersely, with resulting considerable increase in accessibility. Another way to understand them (less physical, more mathematical) is that they represent a necessary condition that a lagrangian L must satisty if, among all possible integrands that have the same boundary values and are suitably smooth (i.e. have enough continuous derivatives to justify taking the derivative under the integral, etc) it is one with minimum action. This of course is not saying why L should minimize the action (this was a deep observation of Hamilton, as I understand it) or how one determines what L actually is for a given situation, but so be it. In analogy with the conditions for an extremum of a function (df/dx = 0 at the extremum), one can apply a technique called the calculus of variations to the functional S (S is a "function of a function") in order to find conditions that an extremum of S must satisfy. Continuing the analogy, the calculus condition df/dx = 0 is an equation which possibly can be solved to obtain an x that is an extreme point of f(x). The condition that the derivative of S with respect to each of the coordinates qi must be zero, provides the analogous condition for a function L to minimize S(L), and leads to the Euler-Lagrange equations, a set of differential equations that a function which is an extremum of the action must satisfy. Again, continuing the analogy, minimizing S does not give L, just a set of equations that L would satisfy if it exists. For a function, df/dx = 0 does not guarantee that x leads to an extremum of f; f(x) could be a maximum, a minimum, or neither, but if the derivative is 0 it is called a stationary point of f. Similarly, L is called a stationary point of S, as satisfaction of the Euler-Lagrange equation is a necessary but not sufficient condition for an extremum. Markssss (talk) 13:36, 10 January 2011 (UTC)

Brackets for functional dependence

Since this seems to be the subject of some edits and reverts, perhaps the opposing parties could offer their comments here. I will say that, as an observer, the brackets for functional notation look dumb (that's just my opinion - no offense, hopefully). ChrisChiasson (talk) 14:35, 22 November 2007 (UTC)

Clarity is paramount in communication. It is more important to get the message across without mistakes than to adhere to the most common practice. While it is more usual to use parenthesis to indicate functional dependence than to use brackets, the use of parenthesis runs the risk that the reader will confuse it with the use of parenthesis to indicate grouping — the precedence order of operations. Also there is the risk that the reader will think that the two items are to be multiplied rather than that one is a function of the other.

In advanced books, these problems are often avoided by omitting mention of the functional dependence since the reader is assumed to know that the quantities are dependent variables. However, in Wikipedia, I think that we should not make such an assumption. JRSpriggs (talk) 01:10, 24 November 2007 (UTC)

Yeah, this is the notation I've always seen when clarity on functional dependence is needed. Yes, it looks silly, but it is the most transparent. — Laura Scudder ☎ 13:34, 24 November 2007 (UTC)

This seems to be a complete dumb informalism. See the page of Action (physics). The lagrangean is somethimes writeen using [] and sometimes ().

This ilogical formalism (informalism) that is valid for some examples and invalid for others. If the use of () is more suitable for the reader to read, it can be used without misunderstandment. There is no formalism like ( use parenthesss for it and brackets for that.)

I can use without misunderstanding (1+(2+(3+(4))) prefered by {1+[2+(4)]}

Util it is formalized in a proper way, it is more logical to be used in a way to please the reader. —Preceding unsigned comment added by 201.8.101.218 (talk • contribs) 22:01, 30 November 2007 (UTC)

I agree that clarity is paramount, but:

• the notation is not even consistent throughout the article, let alone with the most common convention;
• square brackets can be used to show grouping and multiplication, too, e.g. $54[(x-1)^{2}+(x+1)^{2}]^{4}$;
• someone unaware that $f(x)$ can mean something different than f times x will find most of the article incomprehensible anyway, and using $f[x]$ wouldn't help them understand it any better; on the other hand, somebody who does know functional notation might be confused by the square brackets (as I was);
• if it's not clear which the arguments of a function are, it is better to explicitly state them (see Euler–Lagrange equation#Statement) than using square brackets and hoping that they magically clarify it all;
• when seeing square brackets used instead of round ones without any obvious reason (i.e. without any nested subexpression surrounded by round brackets), I would suspect that they are being used with a special meaning (for example, see Nearest integer function; or maybe an integer index, as in Discrete-time Fourier transform). -- A r m y 1 9 8 7 ! ! !  11:34, 25 September 2008 (UTC)

Folks; This notation is seriously silly and confusing. Use parantheses!!! If you do not understand functional dependence via parantheses, then you need to start gathering knowledge and background a lot earlier in the process than this article. Square brackets to me generally mean cyclic permutations, ala the Levi-Civita tensor (symbol). Symbology is important, but it shouldn't be confusing. LAncienne (talk) 21:59, 26 June 2009 (UTC)

Lagrangian Mechanics

This article says "This article is about Lagrange mechanics" -- then why is there a separate article called Lagrangian Mechanics? Why isn't that article linked here, or at the Lagrangian (disambiguation) page? PenguiN42 (talk) 19:45, 1 February 2008 (UTC)

This article does link to that one — in the sentence "Under conditions that are given in Lagrangian mechanics, ..." in the lead. If you want to put an additional link on the disambiguation page, be my guest. JRSpriggs (talk) 07:03, 3 February 2008 (UTC)

Lagrangian for an idealized fluid

What is the lagrangian for the idealized fluid, which leads to the stress-energy tensor given in the article:

$T^{\alpha \beta }\,=(\rho +{p \over c^{2}})u^{\alpha }u^{\beta }+pg^{\alpha \beta }$

--Michael C. Price ^talk 06:09, 30 July 2008 (UTC)

Square bracket

Why are square brackets used to delimit function arguments in parts of this page? If nobody objects, I'm going to change them with round brackets. -- A r m y 1 9 8 7 ! ! !  15:02, 24 September 2008 (UTC)

I object. See my reason above at Talk:Lagrangian#Brackets for functional dependence. JRSpriggs (talk) 06:21, 25 September 2008 (UTC)

As I said in the comments on [[Talk:Lagrangian#Brackets for functional dependence] This notation is seriously silly and confusing. Use parantheses!!! Please change it. If readers do not understand functional dependence via parantheses, then they need to start gathering knowledge and background a lot earlier in the process than this article. Square brackets to me generally mean cyclic permutations, ala the Levi-Civita tensor (symbol). Symbology is important, but it shouldn't be confusing. This notation is confusing (and annoying as well). LAncienne (talk) 15:27, 27 June 2009 (UTC)

what exactly is the lagrangian??

Is the lagrangian always K.E-PE????? I dont think so??? so WHAT REALLY is the physical significance of the lagrangian??? Superficially it appears to be KE-PE,but the lagrangian for a charged particle in a vector potential is 1/2MV.V +qV.A-qW. where V=velocity

     A=vector potential 
     W=Scalar potential.

If the hamiltonion for the above lagrangian is calculated it turns out to be 1/2MV.V+qW. So what does the term eV.A represent??? It appears to be the interaction energy. But why does it dissapaer in the hamiltonion. —Preceding unsigned comment added by 59.95.4.155 (talk) 13:56, 5 October 2008 (UTC)

The $qV\cdot A$ term is not in the Hamiltonian as it is an additional momentum (This can be seen by taking derivatives of the Lagrangian with respect to velocity), when you construct the Hamiltonian from the momentum, velocity and Lagrangian the terms cancel. Gobaskof (talk) 13:33, 3 May 2010 (UTC)

Just as total energy is the sum of kinetic energy and potential energy, so total (i.e. canonical) momentum is the sum of kinetic momentum and potential momentum. Kinetic energy can be calculated from mass and kinetic momentum. So one subtracts potential momentum from canonical momentum, p, to get kinetic momentum; then calculates kinetic energy; then adds potential energy to it to get total energy, i.e. the Hamiltonian.

Yes, it is just an accident that the Lagrangian is kinetic energy minus potential energy in most parts of classical physics. That is no longer true in relativity. We do not know what the definitive formula for the Lagrangian is. In practice, it is obtained by working backwards from the Euler-Lagrange equations which are, in part, determined empirically. The ability to find a simple and symmetrical Lagrangian is a valuable heuristic constraint on the equations of motion. JRSpriggs (talk) 03:13, 10 January 2011 (UTC)

I have tried to clarify that the Lagrangian is not always T-V; that is not the definition but only the natural form of the function. However the edit has been reverted both times without reasons.

Capital L and Script L

Is there a difference between the Lagrangian as referred to by "capital L" and "script L"? — Preceding unsigned comment added by 67.6.78.5 (talk) 18:00, 2 December 2011 (UTC)

See Lagrangian#Lagrangians and Lagrangian densities in field theory. We should use $L\,$ for the Lagrangian (action per unit time) and ${\mathcal {L}}\,$ for the Lagrangian density (action per unit volume per unit time). Notice that

$L=\int {{\mathcal {L}}\,dxdydz}\,.$

Unfortunately, we do not always adhere to this distinction. JRSpriggs (talk) 21:05, 2 December 2011 (UTC)

I have a simple question, would the relationship between the Lagrangian and the Lagrangian density in $n-D$ curved space-time be:

$L=\int {\sqrt {\vert g\vert }}{\mathcal {L}}d^{n-1}x$?

Under the Lagrangian#Lagrangians and Lagrangian densities in field theory section I think we should have

$L=\int {\sqrt {\vert g\vert }}{\mathcal {L}}d^{n-1}x$

rather than just

$L=\int {\mathcal {L}}d^{3}x$

Since the first is more general/usable in field theory. Agree? Disagree? — Preceding unsigned comment added by 207.161.187.38 (talk) 19:14, 9 August 2014 (UTC)

Painfully unclear sections

This article is confusing.

• In the subsection Advantages over other methods, it says ϕ(s) are the generalized coordinates. There is no explanation of what s is. Is it a parameter of ϕ?

• In the subsection "Cyclic coordinates" and conservation laws there is no explanation of what q is. It sort of mentions ${\dot {q}}\,\!$ are time derivatives of generalized coordinates. Well - which are the generalized coordinates - q or φ?? Also the section should explicitly state that p is the generalized momenta, then give the formulas (though that is just a minor detail). The formulas look correct, at least that is fine.

• In the section Explanation, it suddenly uses ϕ and s again - WTF is s??? Again - no explanation of what s is. This is a really ironic title because (mystery s aside) none of it is written clearly. It has been tagged for some time (for obvious reasons) and needs to be resolved.

Here it says that φ and s in are "dependent variables". Dependent variables of what? The system? There should be a clear-cut statement that these are the gen. coordinates of the system. if that is what they are, that they are the same variables stated in previous sections.

Furthermore it mentions s = set of n independent variables of the system (gen. coordinates.?) Shouldn't a different letter be used for these variables inside the {} brackets (q rather than s?) It is confusing to denote a set by a letter - then inside the set definition use the exact same letter. Its like using the defining term in the same definition: e.x. "an electric current is a electric current of charge" is bad - "an electric current is a flow of charge" is better. I understand that the set name s is the set of the different s^α, and that α is an index for a specific value of s. But it might be clearer to use another letter for the set, since s alone is previously used as a variable for φ(s) in which s is a value - not a set.

Anyway I am sure others find this confusing, and that its not just me. I would propose to use q for gen. coordinates. - seems to be the most standard symbol. If anyone reading this thinks the article is already clear - it isn't. PLEASE Make the meaning of the symbols more obvious!!!

• Also - I agree with the above comment. Use L for Lagrangian and script L for the volume density of L. As far as I can tell this is also a convention.

--Maschen (talk) 14:08, 3 December 2011 (UTC)

Originally, the Lagrangian was a function of generalized coordinates q_i and their conjugate momenta p_i. These coordinates and momenta were, in their turn, functions of time t. So t was the independent variable and q_i and p_i were dependent variables. Often the q were x, y and z of various particles. So if there were 5 particles, there might be 15 coordinates and 15 momenta. However, in systems where there were constraints, it might turn out that 8 of these coordinates could be calculated from the other 7. In that case, the number of generalized coordinates q would be reduced to 7 and similarly there would be 7 momenta p. That is why they were called "generalized", because they might not be the usual Cartesian coordinates (nor even a subset of them).

This formalism was generalized further to handle field theory. In field theory, the independent variable t was replaced by an event in spacetime (x, y, z, t) or still more generally by a point s on a manifold. And the dependent variables q were replaced by φ the value of a field at that point in spacetime. JRSpriggs (talk) 03:39, 5 December 2011 (UTC)

Yeah...

Thanks very much for your explaination, though i'm sorry I don't interpret that as the answer to the request. As with Lagrangian (see here too), would you or anyone else mind re-writing these sections to make them clearer? Thanks again. --Maschen (talk) 13:35, 5 December 2011 (UTC)

clean up

Main changes:

1. reworded a section on reading conservation from the lagrangian
2. made lead note more specific
3. removed a silly amount of spacing in formulae,
4. used a better equation numbering system in places,
5. re-section slightly and tweak some headings
6. this is another article which uses so much inline LaTeX when we can just use html (except for unclear letters perhaps, like tau, or maths some symbols).
7. removed this pointless statement:

"Equations derived from a Lagrangian will almost automatically be unambiguous and consistent, unlike equations brought together from multiple formulations."

it doesn't add meaning to anything.

F = q(E+v×B) ⇄ ∑_ic_i 09:51, 26 April 2012 (UTC)

Definition of Lagrangian

The definition of the Lagrangian needs to be reviewed - the current defintion is incorrect. It is NOT necessarily the kinetic minus the potential energy; that is merely the "natural form". The Lagrangian of any system is NOT unique; there may be mutiple L functions to describe the same system. The Lagrangian function stems from Hamilton's principle/ principle of least action and it would be better to mention this in the definition. — Preceding unsigned comment added by 110.174.33.20 (talk) 14:13, 3 June 2012 (UTC)

See Goldstein, Poole & Safko, "Classical Mechaincs", 3ed, 2002, Pearson (end of chapter 1.4 "D'Alembert's Principle and Lagrange's Equations) — Preceding unsigned comment added by Austinstkong (talk • contribs) 14:43, 3 June 2012 (UTC)

And a section should be added to discuss this in more detail. — Preceding unsigned comment added by Austinstkong (talk • contribs) 15:03, 3 June 2012 (UTC)

Don't all Lagrangians differ by a linear function of position? — Arthur Rubin (talk) 18:25, 3 June 2012 (UTC)

The question really is what does one mean by "equivalent"? Producing the same motions? Producing the same equations of motion? Giving the same momenta and forces?

For example, if k is a nonzero constant and u(t) is any smooth function of time, then kL+u(t) will produce the same equations of motion as L. But the momenta and forces will change by a factor of k and it will not be invariant under Lorentz boosts if the original was and u is nonconstant. JRSpriggs (talk) 04:03, 4 June 2012 (UTC)

Who introduced the concept of a Lagrangian?

The article states in the lead: "The concept of a Lagrangian was originally introduced in a reformulation of classical mechanics by Irish mathematician William Rowan Hamilton known as Lagrangian mechanics." No source is cited for this statement, and I could not find any that supports this. On the contrary, more than a few sources ascribe this feat to Lagrange himself. Our article on Lagrangian mechanics too states, "It was introduced by the Italian-French mathematician Joseph-Louis Lagrange in 1788" (which is even before Hamilton was born). The article on Hamiltonian mechanics concurs; Lagrangian mechanics is called there a "reformulation of classical mechanics introduced by Joseph Louis Lagrange in 1788".  --Lambiam 15:34, 21 August 2012 (UTC)

Then change the attribution rather than calling for a citation. JRSpriggs (talk) 17:18, 21 August 2012 (UTC)

I asked for a citation because it seemed somewhat dubious to me, although not impossible – it is common enough that the eponym of a concept is not its originator. It was only when my request for a citation was removed with the somewhat unsatisfactory summary Plenty of references at end that I started to look for a source for the statement. Call it an overabundance of caution, but sometimes reliable sources have errors which then may get copied to other sources – in this case I saw several with suspiciously similar wordings – so I thought it safer to raise the question first here to see if someone could shed more light on this.  --Lambiam 22:02, 21 August 2012 (UTC)

Lagrangian density

I think we should have an article discuss about Lagrangian density. While waiting for the consensus, I will add some info to this part.

• General form of Lagrangian density: ${\mathcal {L}}={\mathcal {L}}(\varphi _{i},\varphi _{i,\mu })$ where $\varphi _{i,\mu }\equiv {\frac {\partial \varphi _{i}}{\partial x^{\mu }}}\equiv \partial _{\mu }\varphi _{i}$ (see 4-gradient)
• The relationship between ${\mathcal {L}}$ and $L$ is: $L=\int {{\mathcal {L}}\,dxdydz}\,.$ (see above), same as $m=\int \rho dV$.
• In field theory, the independent variable t was replaced by an event in spacetime (x, y, z, t) or still more generally by a point s on a manifold. (see above)

123.21.142.19 (talk) 05:31, 14 May 2013 (UTC)

Standard Model

A concise version of the Lagrangian density for the Standard Model would be a nice addition to the article. Placed just after the entry for Quantum Chromodynamics, I added

The Lagrangian density for the Standard Model can be expressed in many different ways, most of which consume a page or more. However a simple form of it is:

${\mathcal {L}}_{\mathrm {SM} }=-{1 \over 4}F_{\mu \nu }F^{\mu \nu }+i{\bar {\psi }}{D}\!\!\!\!/\ \psi +\psi _{i}y_{ij}\psi _{j}\phi +h.c.+|D_{\mu }\phi |^{2}-V(\phi )$

Unfortunately, someone decided that this does not belong and removed it. Quite frustrating, given the importance of Standard Model and the lack of a concise form elsewhere in Wikipedia. — Preceding unsigned comment added by 69.19.254.81 (talk) 19:09, 1 July 2013 (UTC)

I suspect that Xxanthippe would be more willing to accept this additional material, if you provide a reliable secondary source for it. Please see WP:RS. It would also be nice if you could give at least a brief explanation of the meaning of each of the terms in the formula. For example, you might say "The first term covers gauge bosons, treating them in the first instance as if they were all photons.". JRSpriggs (talk) 07:26, 2 July 2013 (UTC)

First Example from Classical Mechanics - needs work

I can not follow the first example. Specifically, the third equation in the Cartesian section: Then dL/dx_i = -dV/dx_i {where I use d for partial derivative}

This seems to be pulled out of thin air. I assume it is correct, but HOW / WHY it follows from the system's Lagrangian and the Euler-Lagrange Equations is a mystery to me. I am no Lebnitz, but this should either be obvious, or trivially dervived from the starting point and it is not. Would someone be so kind as to expand this example so that it actually is useful?

In addition, at the end of the next section on Spherical coordinates this text appears: "Here the set of parameters si is just the time t, and the dynamical variables ϕi(s) are the trajectories \scriptstyle\vec x(t) of the particle.

Despite the use of standard variables such as x, the Lagrangian allows the use of any coordinates, which do not need to be orthogonal. These are "generalized coordinates". "

This text should be deleted, it serves no useful purpose here and imho adds irrelevancies. First s_i does NOT appear in the section, at all. Second phi_i also isn't used in the section. The last sentence is a non sequitur, statements about generalized coordinates do not belong here.

ThanksAbitslow (talk) 22:50, 21 December 2013 (UTC)

What is the correct Lagrangian for an isolated test particle in general relativity?

What is the correct Lagrangian for an isolated test particle in general relativity?

I claim it is:

$L(t)=-mc^{2}{\frac {d\tau (t)}{dt}}=-mc{\sqrt {-g_{\mu \nu }(x(t)){\frac {dx^{\mu }(t)}{dt}}{\frac {dx^{\nu }(t)}{dt}}}}\,.$

Differential 0celo7 (talk · contribs) claims it is:

$L={\frac {1}{2}}g_{\mu \nu }{\frac {dx^{\mu }}{d\tau }}{\frac {dx^{\nu }}{d\tau }}\,.$

0celo7's version is obviously wrong. Not only does it have the wrong units, but it is equal to a constant ${\frac {-c^{2}}{2}}$ which means that varying it will produce only an identity 0=0 that holds for all world paths. The fact that he gets a different (and accidentally correct) result is due to his application of the Euler-Lagrange equations in an inappropriate way.

To see why my version is correct, we must first recall that L=T−V was devised for classical (Newtonian) physics so it only holds in the non-relativistic limit. So let us transform to a locally inertial frame of reference in which the particle is instantaneously at rest (and parameterize the particle's path with t=x⁰). Let L₀ be the Lagrangian at this event in this reference frame. Then L₀=T₀−V₀=2T₀−H₀ where H₀ is the total energy of the particle. Then since v=0, we get L₀=2T₀−H₀=2·0−mc²=−mc². If we transform coordinates to an arbitrary frame of reference, we get:

$L=-mc^{2}{\frac {d\tau }{dt}}\,.$

At low velocities in an inertial frame, this is:

$L=-mc^{2}{\sqrt {1-{v^{2} \over c^{2}}}}=-mc^{2}+{\frac {mv^{2}}{2}}+O(v^{4})=-V+T+{\text{relativistic corrections}}\,$

in agreement with the non-relativistic formula. JRSpriggs (talk) 06:52, 18 December 2014 (UTC)

JRSpriggs Not sure how this "talk" thing works so I apologize for the formatting. Units are irrelevant because I am using natural units. Both Lagrangians are correct you know. In fact, one usually sees it as

$L={\sqrt {-\langle {\dot {\gamma }},{\dot {\gamma }}\rangle }}$

(dot denoting derivative along the curve), which is equal to 1. There is nothing wrong with this, we vary pretending we are using some arbitrary parameterization and then fix it after we have the geodesic equation. Sources for this action: Straumann, Weinberg, Zee. Then comes the other action. Both Zee and Straumann say it is an acceptable action. Indeed, it produces the same Euler-Lagrange equations. Additionally, look at Eq. 2.5 in Becker, Becker & Schwarz. Its that action, which is well-suited for string theory. (One caveat, you have to throw away the second term because it is multiplied by the auxiliary field.)

Whom did you learn GR from? This is not an insult or anything, I'm just wondering which text does not exploit L=1 or -1/2 to make your life easier when solving geodiesc PDEs.

Differential 0celo7 (talk) 10:54, 18 December 2014 (UTC)

My knowledge goes only as far as that there is no such thing as the correct Lagrangian. To save trouble, my advice is to include references, the more in number and the more reputable, the better. A well-sourced (and reasonably well-written) relevant addition to any article is practically impossible to revert (and get away with it). Differential 0celo7's version can apparently be backed up with solid refs. JRSprigg's version can probably be found in Bartin Zweibach's, A first course in string theory (not entirely sure, don't have it where I'm at now) + another million books. Why not have both in the article? YohanN7 (talk) 13:46, 19 December 2014 (UTC)

I see Differential 0celo7 put a lengthy rebuttal post on the JRSpriggs talk page (http://en.wikipedia.org/wiki/User_talk:JRSpriggs#Lagrangian_of_a_free_particle_in_GR). I suspect the formatting would be lost if I just pasted it in here so I'll let Differential 0celo7 move it. Sounds like a good argument though. I vote for returning to the previous version plus additional Differential 0celo7 notes and sources. Spriggs hasn't convinced me he has a lock on this and I'm always open to reading opposing opinions as long as they are well documented. Spriggs claims to be logician so he could put some additional rigor in his arguments by looking at 7's argument more objectively. Keep in mind here, four years of grad school has shown me even well-known physics Ph.D.s don't all agree in this arena! This is tough stuff. Felix01 (talk) 19:42, 19 December 2014 (UTC)

I posted this on JRSprigg's talk page:

OK, here I will give a detailed discussion. First, some notes on Lagrangian and actions, as well as the geodesic equation:

• The Lagrangian is unphysical.
• The action is unphysical.
• Lagrangians and actions are equivalent if they generate the same equations of motion.
• δS=0 implies the Euler-Lagrange equations, not the other way around.
• Overall factors in the Lagrangian are irrelevant. This includes signs and constants.
• The geodesic equation does not rest on an Euler-Lagrange style argument. It can be derived by other methods.
• The mass shell condition is a consequence of the geodesic equation.

The first three are basic tenets of the Lagrangian prescription. See any textbook on QFT and you will see Lagrangians and actions mangled and beaten into shape to extract physical information. Now, because of the linearity of integration and functional differentiation, if the Lagrangian is multiple of something else, then that multiple cancels in δS=0, leaving the equations of motion unaffected.

Some comments on your "talk" post:

• How does one derive the geodesic equation from your Lagrangian? That Lagrangian IS correct, in a sense, as I will show later. However, it is not in a form that I can see leading to the properly parametrized geodesic equation. (A multiple of τ is the correct parameterization.)
• The units really don't matter. I could be using units in which m=c=1. This does not affect anything.
• The Newtonian argument breaks down when we consider massless particles. For such particles your action is zero! By your reasoning, this leads to trivial equations of motion.

First I will show that the geodesic equation is not strictly a consequence of δS=0. For this, I will adapt the discussion found in Weinberg's Gravitation and Cosmology. Let ξ be a locally inertial coordinate system (free-falling) and τ be the proper time along the worldline (the time on a clock as measured by the particle in its rest frame). According to the Principle of Equivalence, there is such a coordinate system in which the equation of motion for SR holds:

${\frac {d^{2}\xi ^{\alpha }}{d\tau ^{2}}}=0$

In natural units, the proper time in this frame is

$d\tau ^{2}=-\eta _{\alpha \beta }d\xi ^{\alpha }d\xi ^{\beta }$

Now suppose we use any other coordinate system x, which may be a cartesian system in a laboratory, curvilinear, accelerated, or whatever. The freely falling coordinates ξ are functions of the x, thus the first equation becomes

$0={\frac {d}{d\tau }}\left({\frac {\partial \xi ^{\alpha }}{\partial x^{\mu }}}{\frac {dx^{\mu }}{d\tau }}\right)={\frac {\partial \xi ^{\alpha }}{\partial x^{\mu }}}{\frac {d^{2}x^{\mu }}{d\tau ^{2}}}+{\frac {\partial ^{2}\xi ^{\alpha }}{\partial x^{\mu }\partial x^{\nu }}}{\frac {dx^{\mu }}{d\tau }}{\frac {dx^{\nu }}{d\tau }}$

A line of algebra and the chain rule leads to the geodesic equaton

${\frac {d^{2}x^{\lambda }}{d\tau ^{2}}}+\Gamma _{\;\;\mu \nu }^{\lambda }{\frac {dx^{\mu }}{d\tau }}{\frac {dx^{\nu }}{d\tau }}=0$

where

$\Gamma _{\;\;\mu \nu }^{\lambda }={\frac {\partial x^{\lambda }}{\partial \xi ^{\alpha }}}{\frac {\partial ^{2}\xi ^{\alpha }}{\partial x^{\mu }\partial x^{\nu }}}$

is the affine connection. The proper time may also be written in an arbitrary coordinate system:

$d\tau ^{2}=-\eta _{\alpha \beta }{\frac {\partial \xi ^{\alpha }}{\partial x^{\mu }}}dx^{\mu }{\frac {\partial \xi ^{\beta }}{\partial x^{\nu }}}dx^{\nu }=-g_{\mu \nu }dx^{\mu }dx^{\nu }$

where g is the metric tensor

$g_{\mu \nu }=\eta _{\alpha \beta }{\frac {\partial \xi ^{\alpha }}{\partial x^{\mu }}}{\frac {\partial \xi ^{\beta }}{\partial x^{\nu }}}$

Denote by $\partial _{\mu }$ the operator $\partial /\partial x^{\mu }$. You can have the pleasure of verifying that, given the previous equations

$\partial _{\lambda }g_{\mu \nu }=\Gamma _{\;\;\lambda \mu }^{\rho }g_{\rho \nu }+\Gamma _{\;\;\lambda \nu }^{\rho }g_{\rho \mu }$

Add to this equation the same equation with μ and λ interchanged and subtract the same equation with ν and λ interchanged. Remembering that the affine connection as defined above is guaranteed to be symmetric, this leads to the usual Christoffel symbols

$\Gamma _{\;\;\mu \nu }^{\lambda }={\frac {1}{2}}g^{\lambda \rho }(\partial _{\mu }g_{\nu \rho }+\partial _{\nu }g_{\mu \rho }-\partial _{\rho }g_{\mu \nu })$

So it is easy to see that the geodesic equation does not rest upon any particular variational principle of any particular Lagrangian, because you don't need the action principle to derive it.

From here on I will use some coordinate free notation mixed in with the usual Ricci notation. Let γ(τ) be the worline of the particle, parameterized by the proper time τ. Let (M, g) be a 4-dimensional pseudo-Riemannian manifold with a Lorentz metric g and inner product $\langle .,.\rangle =g(.,.)$. Let $\nabla$ be the Levi-Civita connection on M. Let

${\dot {\gamma }}={\frac {d}{d\tau }}$

be the tangent vector to the curve. In this notation, the geodesic equation is

$\nabla _{\dot {\gamma }}{\dot {\gamma }}=0$

i.e. the tangent vector is autoparallel along the geodesic. Next we use the Ricci identity. See any text on differential geometry for a proof. For any three vectors X, Y and Z

$X\langle Y,Z\rangle =\langle \nabla _{X}Y,Z\rangle +\langle Y,\nabla _{X}Z\rangle$

So let $X=Y=Z={\dot {\gamma }}$. Then

${\frac {d}{d\tau }}\langle {\dot {\gamma }},{\dot {\gamma }}\rangle ={\dot {\gamma }}\langle {\dot {\gamma }},{\dot {\gamma }}\rangle =2\langle \nabla _{\dot {\gamma }}{\dot {\gamma }},{\dot {\gamma }}\rangle =0$

where the last equality follows from the geodesic equation. We integrate the very first term along the worldline P. Thus

$\langle {\dot {\gamma }},{\dot {\gamma }}\rangle ={\text{const}}$

The convention is to take this constant to be -1. This comes about as follows. Suppose a particle in SR travels along a worldline P. Then the change in proper time is given by the line integral

$\Delta \tau =\int _{P}d\tau =\int _{P}{\sqrt {-\eta _{\alpha \beta }d\xi ^{\alpha }d\xi ^{\beta }}}$

Now suppose we turn on the gravitational field. Then by the Equivalence Principle (I will drop the P from now on)

$\Delta \tau =\int {\sqrt {-g_{\mu \nu }dx^{\mu }dx^{\nu }}}$

You object that the integral is not in the form ∫f(x)dx. So we choose to parameterize using the proper time and write

$\Delta \tau =\int {\sqrt {-\langle {\dot {\gamma }},{\dot {\gamma }}\rangle }}\,d\tau$

Comparing with above, we conclude that the choice

$\langle {\dot {\gamma }},{\dot {\gamma }}\rangle =-1$

is consistent. But what if we choose some other parameter λ? As of this moment we have no restrictions on what can be a parameter. Thus

$\Delta \tau =\int {\sqrt {-\langle {\bar {\gamma }},{\bar {\gamma }}\rangle }}\,d\lambda$

where

${\bar {\gamma }}={\frac {d}{d\lambda }}$

Now I want to bring the action principle into the fold. I think that we can both agree that in SR, if we care about overall signs and factors (c=1 still remains though), the free action for a point particle of mass m is

$S=-m\int {\sqrt {1-{\vec {v}}^{2}}}\,dt=-m\int d\tau$

The GR generalization is

$S=-m\int d\tau =-m\int {\sqrt {-\langle {\bar {\gamma }},{\bar {\gamma }}\rangle }}\,d\lambda$

The principle of least action is of course δS=0. Now I hope you see that the factor -m does not do anything at all! Thus the variational principle is simply

$\delta \int {\sqrt {-\langle {\bar {\gamma }},{\bar {\gamma }}\rangle }}\,d\lambda =0$

I also hope you see that $\langle {\bar {\gamma }},{\bar {\gamma }}\rangle$ is in no way restricted or fixed and can thus S can be varied. In this article this variation is performed. I will not do it here. The result is

${\frac {d^{2}x^{\lambda }}{d\lambda ^{2}}}+\Gamma _{\;\;\mu \nu }^{\lambda }{\frac {dx^{\mu }}{d\lambda }}{\frac {dx^{\nu }}{d\lambda }}=0\quad {\text{or}}\quad \nabla _{\bar {\gamma }}{\bar {\gamma }}=0$

Once again we can easily show that

$\langle {\bar {\gamma }},{\bar {\gamma }}\rangle ={\text{const}}$

from which it follows that $\lambda \propto \tau$. This follows only from the equations of motion and is not obvious before. As a constraint derived from the equations of motion, it cannot be applied to the Lagrangian! This would be like doing QFT on the mass shell only. (In fact, as I will show, the mass shell condition is a consequence of the geodesic equation.)

I said that the Lagrangian you presented was not incorrect. This is because you have defined the action as

$S=\int L\,dt$

whereas I have defined it as

$S=\int L\,d\tau$

This is a simple change of parametrization. However, your Lagrangian is not invariant under diffeomorphisms. The coordinate t has a nontrivial transformation rule as opposed to λ, which allows for invariant parameterization. Some food for thought: try deriving the geodesic equation from your Lagrangian. Maybe you can surprise me and get it.

The current discussion is adapted from Becker, Becker & Schwarz String Theory and M-Theory. The action -m ∫dτ has some problems with it:

• In the path integral sense, it is a disaster. The square root makes this very hard to quantize.
• It is completely unsuited to handling massless particles for obvious reasons.

These problems can be circumvented by introducing an auxiliary (bosonic) field α(τ) in the modified action

$S={\frac {1}{2}}\int d\lambda \,\left({\frac {\langle {\bar {\gamma }},{\bar {\gamma }}\rangle }{\alpha }}-m^{2}\alpha \right)$

First we determine the equations of motion for α(τ):

${\frac {\delta S}{\delta \alpha }}=-{\frac {1}{2}}\left({\frac {\langle {\bar {\gamma }},{\bar {\gamma }}\rangle }{\alpha ^{2}}}+m^{2}\right)=0\longrightarrow \langle {\bar {\gamma }},{\bar {\gamma }}\rangle =-m^{2}\alpha ^{2}$

Plugging this back into S, we see that we recover the original action. Now, it may be shown that this modified action is parameterization invariant if α has certain transformation properties. (BBS exercise 2.3) This leads to a sort of gauge invariance. The gauge α=1 is nice. In this gauge

$\langle {\bar {\gamma }},{\bar {\gamma }}\rangle =-m^{2}$

is the familiar mass shell condition. This means ${\bar {\gamma }}$ is the momentum vector. The additive constant in the Lagrangian can be just thrown away because it contributes nothing to the equations for γ. Thus the action is effectively

$S={\frac {1}{2}}\int \langle {\bar {\gamma }},{\bar {\gamma }}\rangle \,d\lambda$

Now this makes perfect sense for massless particles. For massive particles, we have τ=mλ. So, up to an overall factor (which we just throw away, the 1/2 is kept for historical reasons)

$S={\frac {1}{2}}\int \langle {\dot {\gamma }},{\dot {\gamma }}\rangle \,d\tau$

As I showed in the post you deleted, this leads to the geodesic equation

$\nabla _{\dot {\gamma }}{\dot {\gamma }}=0$

The really nice thing about this action is that it is computationally simple to work with.

Thus, the Lagrangian of general relativity can be taken to be

$L={\frac {1}{2}}g_{\mu \nu }{\dot {x}}^{\mu }{\dot {x}}^{\nu }$

Because this leads to the correct equations of motion, it is an acceptable Lagrangian.

I hope this clears up any confusion.

Now, as to my rebuttal on his comment that followed this:

I will extract the salient points from JRSprigg's comment:

• You cannot ignore constants because if we wish to add other terms the proportionality factors will be off.
• Constants are needed for a correct energy momentum tensor.
• My Lagrangian leads to the geodesic equation.
• I have not seen a fully satisfactory derivation of the massless particle equations of motion.
• Using the proper-time τ as a parameter for the particle's path is inappropriate because the value of τ is path-dependent.

1) I kept your first point in mind when typing my previous response. This IS correct, but we are talking about a free particle action here, not an interaction. For that case, yes, constants do matter.

2) Your second point is correct, but we are not talking about a matter Lagrangian here, but rather a particle Lagrangian.

3) You did prove me wrong there. You did find a version of the geodesic equation.

4) You have, actually! In my first derivation of the geodesic equation, simply use the affine parameter $\sigma =\xi ^{0}$ instead of proper time. Mass is not mentioned anywhere, making this a perfectly valid derivation.

5) This is quite the bold claim! If you make a claim that contradicts every authoritative text on the subject, I hope you understand that you must provide extraordinary proof that supports your claim. If you do not think your claim is contradictory, then I posit you have not read the authoritative texts. Indeed, see Lemma 1.4.5 of Riemannian Geometry and Geometric Analysis by Jurgen Jost:

"Each geodesic is parametrized proportionally to arc length."

And of course τ is the natural worldline arc length.

As someone said above, sources are key. So to refute your last claim, I will cite every text I know that uses proper time parameterization (in no particular order):

Carroll, Spacetime and Geometry (p. 107 uses both the Lagrangian you claim is false and proper time.)

Cahill, Physical Mathematics (p. 446)

Weinberg, Gravitation and Cosmology (p. 71)

Weinberg, Cosmology (p. 4)

Zee, Einstein Gravity in a Nutshell (p.289, p. 130 for my "false Lagrangian")

Zee, Quantum Field Theory in a Nutshell (p. 84)

Straumann, General Relativity (p. 22, p. 25 for my "false Lagrangian")

Becker, Becker & Schwarz, String Theory and M-Theory (p. 19)

Hartle, Gravity (p. 173)

Wald, General Relativity (p. 44)

If you can refute these sources, I will happily delete the section myself.

Differential 0celo7 (talk) 20:41, 19 December 2014 (UTC)

But how boring wouldn't it be if we all agreed? YohanN7 (talk) 04:04, 20 December 2014 (UTC)

I will copy-over here my response on my talk page since I do not think 0celo7's summary of it does it justice:

BEGIN COPY

It would have been better to hold this discussion at Talk:Lagrangian#What is the correct Lagrangian for an isolated test particle in general relativity? so that other interested people could find it and participate.

Much of what you say is true and already well known to me. However, I disagree on some important points.

I disagree about ignoring constant factors because we may wish to add Lagrangians for various parts of a system together to get the Lagrangian for the total system and this will fail unless their proportions are correct. Also you need the constant factors to get the correct stress-energy tensor from the Lagrangian.

Your Lagrangian was either pulled out of the air and justified by its results (which I think should be avoided when possible) or it is an attempt to generalize ${\frac {mv^{2}}{2}}$ in an inappropriate way.

I do not disagree with the geodesic equation, but only with your method of deriving it.

I have shown how to derive a variation of the geodesic equation (using momentum) from my Lagrangian at User:JRSpriggs/Force in general relativity#Derivation from Lagrangian.

I have not yet seen a fully satisfactory derivation of the equation of motion for massless particles except by taking a limit of the equation for massive particles as the mass approaches zero. However, the only particles which might be massless are the photon and the graviton (if it exists).

Using the proper-time τ as a parameter for the particle's path is inappropriate because the value of τ is path-dependent. In particular, the limits of integration might have to be changed as you do the variation which is highly undesirable. JRSpriggs (talk) 12:27, 19 December 2014 (UTC)

END COPY

Now I continue with an answer to his reply to the above.

If you know that the constant factor may be needed in some cases, then why did you not include it when you reverted and put your section back into the article?

Let me be clear here, the particle's world-path can use any smooth parameterization. So for massive particles, one can choose a parameterization after the derivation is finished which is equal or proportional to the proper-time τ. But it is wrong to jump the gun and call it τ from the beginning before the path is determined. That would imply that the path is constrained to never change the value of τ which would render the variation ineffective and perhaps impossible (depending on the location of the end-events and the choice of τ at those events). JRSpriggs (talk) 05:42, 20 December 2014 (UTC)

You make a good point there. That's why I called it λ in my careful derivation above. Then I showed it must be proportional to the proper time. This is a mathematical fact. If you really want to argue this you HAVE to provide proof. This fact is stated in numerous (graduate) texts.

I didn't put in the constants because it does not matter in this case at all! I want to emphasize that the energy functional is used for computational ease, both in mathematical and physical literature. I'll edit my post to reflect this.

You have still not said whom you learned general relativity from. I'm curious which text disagrees with everything out there, especially with mathematical literature.

Differential 0celo7 (talk) 11:46, 20 December 2014 (UTC)

OK JR, the time for talking is over. You've run into a subject matter expert here who's outpaced your generalized GR knowledge. Either cite some credible literature published since Feynman passed and show conclusively that it contradicts the Differential 0celo7 conclusion or knock this off. This exchange is becoming boring in a lopsided way. You might have survived a challenge to your User Page but you won't win this one if it gets bumped up for a moderated review. Felix01 (talk) 15:30, 20 December 2014 (UTC)

To Felix01: I do not think that you know me well enough to reach such a conclusion. Let us see how well 0celo7 can answer the following question.

To 0celo7: Regarding massless particles (such as the photon), you said "In my first derivation of the geodesic equation, simply use the affine parameter $\sigma =\xi ^{0}$ instead of proper time.". That would be fine in flat spacetime. What do you do when spacetime is curved (as in GR) so you cannot have $g_{\alpha \beta }=\eta _{\alpha \beta }$ everywhere for any coordinate system? Remember that you specified that $\xi ^{\alpha }$ were coordinates for flat spacetime. JRSpriggs (talk) 17:18, 20 December 2014 (UTC)

To 0celo7: To answer your question, I did not learn relativity from any one person. Relativity has been my main academic interest since about 1960. I have read so many books and papers on it that I could not begin to list them. I have attended many lectures and seminars on it. I have done my own independent calculations and derivations related to it, for example, see Talk:Friedmann equations#Derivation of simplified Friedmann equations. Probably the book which impressed me the most was by Sir Arthur Stanley Eddington. When I have to look something up, I usually look first at MTW.

However, this is just an argument from authority which is why I did not mention it before. It is not really relevant to our disagreement. JRSpriggs (talk) 05:33, 23 December 2014 (UTC)

How come none of the text that is being (re)inserted into the article has citations? This is Wikipedia, not a collection of private lecture notes. Zueignung (talk) 06:26, 23 December 2014 (UTC)

To 0celo7: If you want to see the calculation from the square-root to the geodesic equation in a book, look at MTW pages 316 to 318. Unfortunately, they are not treating it as a Lagrangian per se so they leave out the -mc^2 factor.

To Zueignung: It is very hard to find someone who has written a formula in exactly the way we want it. The number of variations is amazing. That is why the Mathematics project has a relaxed rule concerning sources, see Wikipedia:Scientific citation guidelines#Examples, derivations and restatements. The user must use some judgement and not just blindly copy what he finds in a book. JRSpriggs (talk) 10:28, 23 December 2014 (UTC)

Of course the article should be supported with references. YohanN7 (talk) 17:53, 24 December 2014 (UTC)

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