I am trying to derive the geodesic equation using variational principle. My Lagrangian is $$ L = \sqrt{g_{jk}(x(t)) \frac{dx^j}{dt} \frac{dx^k}{dt}}$$ Using the Euler-Lagrange equation, I have got this. $$ \frac{d^2 x^u}{dt^2} + \Gamma^u_{mk} \frac{dx^m}{dt} \frac{dx^k}{dt} = \frac{\frac{dx^u}{dt}}{g_{pq}\frac{dx^p}{dt}\frac{dx^q}{dt}} \frac{d}{dt}\left[g_{rs}\frac{dx^r}{dt}\frac{dx^s}{dt}\right]$$  How do I prove the right hand side to be zero to get the geodesic equation?  Basically, I can rewrite the above mentioned equation as  $$ \frac{d^2 x^\mu}{dt^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{dt} \frac{dx^\lambda}{dt} = \frac{dx^\mu}{dt} \frac{d\ \log(L)}{dt}$$  I need to know why is the following true? $$\frac{d\ \log(L)}{dt} =0$$  I know that the derivation might be simpler using a different Lagrangian but I want to do it using this one.
65341
I would like to recover the (timelike) geodesics equations via the variational principle of the following action:  $$ \mathcal{S}[x] = -m \int d\tau = -m \int \sqrt{-g_{\mu\nu}\,dx^{\mu}\,dx^{\nu}} $$  Using an arbitrary auxiliary parameter $\lambda$, then one is able to rewrite the action and obtains: $$ d\tau = \sqrt{-g_{\mu\nu}\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}} d\lambda $$ $$ -m \int d\tau = -m \int (d\tau / d\lambda) d\lambda =  -m\int\sqrt{-g_{\mu\nu}\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}} d\lambda $$  Now we can vary the paths $x^{\mu}(\tau) \rightarrow x^{\mu}(\tau) + \delta x^{\mu}(\tau)$.  I need the following equation to be true, so i get the geodesics equations, but I don't know how one can say the following: $$ \int d\tau \,\delta g_{\mu\nu}\frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau} = \int d\tau\, g_{\mu\nu, \rho} \frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau} \delta x^{\rho} $$ I know there is a simpler action, which leads to the same equation of motion, but there one has to do the same manipulation, which I sadly don't understand.  Thanks for helping me!
99993
I (desperately) need help with the following:  What is the null geodesic for the space time $$ds^2=-x^2 dt^2 +dx^2?$$  I don't know how to transform a metric into a geodesic...! There is no need to start from the Lagrangian. I know that $$0=g_{ij}V^iV^j$$ where $V^i={dx^i\over d\lambda}$  where $\lambda$ is some parameter. But I don't know what that parameter is nor how to find the geodesic.  Many thanks!! Please help!
1034
(Just some recalls)  We have an action on which we want to apply Least action principle.  $$ S=\int_{t_i}^{t_f} L(q,\dot{q},t)dt$$  We assume that $t \mapsto q(t)$ is the function that will extremise the action, thus if we replace :   $$q(t) \rightarrow q(t)+\delta q(t)$$  with $ \delta q(t_i)=\delta q(t_f)=0$  We will have at first order in $\delta q$ the variation of the action that will be zero.  $$ \delta S = \int_{t_i}^{t_f} \frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot{q}} \delta \dot{q} = 0  $$  And we say that because : $ \delta \frac{dx}{dt} = \frac{d \delta x}{dt} $, we can integer by part (the boundary term vanish because of (*)): $$ \delta S = \int_{t_i}^{t_f} (\frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} )\delta q  = 0  $$    My question :  It makes sense to me to say that, as :  $$q(t) \rightarrow q(t)+\delta q(t)$$  then by derivation I will have :  $$\dot{q}(t) \rightarrow \dot{q}(t)+\frac{d}{dt} \delta q(t)$$  We immediately find that $ \delta \frac{dx}{dt} = \frac{d \delta x}{dt} $, so it seems to be very general for any transformation.  But my teacher said it could'nt be the case for any transformation. I don't see why ??  What's more, just to understand :  In a Lagrangian, we consider $q(t)$ and $\dot{q}(t)$ as independant variables, because at a given point, if I have a value of a function, its derivative can take any values.  But the functions $q$ and $\dot{q}$ are not independant functions (they are linked by derivative operation).  So, we always have $\delta q(t)$ and $\delta \dot{q}(t)$ that are independant, and to proove that  $ \delta \frac{dx}{dt} = \frac{d \delta x}{dt} $, we use the "trick" that, the functions associated are not independant.  In a sense, we "go" in the function space where functions are not independent and then we go back in "variable" spaces where they are independent.  Am I right ? And why did my teacher said that $ \delta \frac{dx}{dt} = \frac{d \delta x}{dt} $ is not always true, was him wrong?
39829
I feel like this question has been asked before but I can't find it. would the Euler Lagrange equation for, say, the standard model Lagrangian be $$\frac{\partial L}{\partial \phi}=\partial_\mu \frac{\partial L}{\partial (\partial_\mu \phi)}$$ Where $\phi$ is whatever field is an question and $\mu$ is (I believe) being summed from 0 to 3. Or, ist the correct equation $$\frac{\partial L}{\partial \phi}=D_\mu \frac{\partial L}{\partial (D_\mu \phi)}$$ Where $D$ is the covariant derivative of the theory. My intuition tells me its the second eqn but i just wanted to be sure, and I think I once saw someone say that the two were equivalent.
10376
I have seen numerous 'derivations' of the Maxwell Lagrangian,  $$\mathcal{L} ~=~ -\frac{1}{4}F_{\mu \nu}F^{\mu \nu},$$  but every one has sneakily inserted a factor of $-1/4$ without explaining why. The Euler-Lagrange equations are the same no matter what constant we put in front of the contraction of the field strength tensors, so why the factor of $-1/4$?
110404
In reading Weinstock's Calculus of Variations, on  he explains how Schrodinger apparently first derived the Schrodinger equation from variational principles.  Unfortunately I don't think page 262 is showing so I'll explain the gist of it:  "In his initial paper" he considers the reduced Hamilton-Jacobi equation   $$\frac{1}{2m}\left[\left(\frac{\partial S}{\partial x}\right)^2 \ + \ \left(\frac{\partial S}{\partial y}\right)^2 \ + \ \left(\frac{\partial S}{\partial z}\right)^2\right] \ + \ V(x,y,z) \ - \ E \ = \ 0$$   for a single particle of mass $m$ in an arbitrary force field described by a potential $V = V(x,y,z)$.  With a change of variables $S \ = \ K\log(\Psi)$, (where $K$ will turn out to be $\hbar=h/2\pi$) it reduces to  $$\frac{K^2}{2m}\left[\left(\frac{\partial \Psi}{\partial x}\right)^2 \ + \ \left(\frac{\partial \Psi}{\partial y}\right)^2 \ + \ \left(\frac{\partial \Psi}{\partial z}\right)^2\right] \ + \ (V \ - \ E)\Psi^2 \ = 0.$$  Now instead of solving this he, randomly from my point of view, chose to integrate over space  $$I  = \iiint_\mathcal{V}\left(\frac{K^2}{2m}\left[\left(\frac{\partial \Psi}{\partial x}\right)^2 \ + \ \left(\frac{\partial \Psi}{\partial y}\right)^2 \ + \ \left(\frac{\partial \Psi}{\partial z}\right)^2\right] + (V - E)\Psi^2\right)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$   then Weinstock extremizes this integral which gives us the Schrodinger equation.  Apparently as the book then claims on page 264 it is only after this derivation that he sought to connect his idea's to deBroglie's wave-particle duality.  Thus I have three questions,  1) What is the justification for Feynman's famous quote:     Where did we get that [Schrödinger's equation] from? It's not possible to derive it from anything you know. It came out of the mind of Schrödinger.   The Feynman Lectures on Physics   in light of the above derivation. I note that all the derivations I've seen of the Schrodinger equation doing something like using operators such as $i\hbar\partial/\partial t = E$ to derive it always mention it's merely heuristic, yet what Schrodinger apparently originally did seems like a roundabout way of solving the Hamilton-Jacobi equation with no heuristic-ness in sight. What subtleties am I missing here? Why would I be a fool to arrogantly 'correct' someone who says Schrodinger is not derivable from anything you know?  Note I've read a ton of the threads on this topic on this forum &amp; none even go near the calculus of variations, in fact the above contradicts  that even refers to Schrodinger's Nobel prize lecture, so hopefully it's not a duplicate.  2) Is the mathematical trick Schrodinger has used something you can use to solve problems?   3) Why can't you use this exact derivation in the relativistic case?  Edit: On the topic of complex numbers:  On page 276, 14 pages after he explains what I posted, &amp; after he goes through how Schrodinger linked his work with DeBroglie's work, it is only then that Weinstock says:     In a more complete study of quantum mechanics than the present one the   admissibility of complex eigenfunctions $\Psi$ is generally shown to be   necessary. If $\Psi$ is complex, the quantity $|\Psi|^2$ is employed as the   position probability-density function inasmuch as $\Psi^2$ is not restricted   to real nonnegative values.   Apparently Schrodinger was able to do what I have posted using real-valued functions &amp; have K as I have defined it, without i. If you're following what Weinstock is saying he shows how the hydrogen atom energy levels are explainable without complex numbers, i.e. he is able to derive a physical interpretation of the eigenvalues (discrete energy levels) of the Schrodinger equation that were in accord with experiment (see Section 11.3 Page 279 on). As far as I understand it it is in trying to find a physical interpretation of the eigenfunctions that one is forced into complex numbers, though apparently, according to the book, it can be shown to be necessary. That is something to take into account!
69548
How can I prove the product rule of derivatives using the first principle?  $$\frac{d (f(x) g(x))}{d x}    = \left( \frac{d f(x)}{d x} g(x) + \frac{d g(x)}{d x} f(x) \right)$$  Sorry if i used the wrong symbol for differential (I used \delta), as I was unable to find the straight "d" on the web.
53337
I was trying to derive equation (65) in the following review:    This slightly unusual then usual classical mechanics because it includes a variation of time also, $\delta t$.  Usually one would define, $$\delta S~=~ \int \left[ L \left(\tilde{q}(t),\dot{\tilde{q}}(t),t\right) -  L \left(q(t),\dot{q}(t),t\right)\right] dt , $$ where,  $$\tilde{q}~=~q+\delta q. $$  We have then (before applying int by parts),  $$ \delta S ~=~ \int \left(\frac{\partial L}{\partial q}\delta q +\frac{\partial L}{\partial \dot q}\delta \dot q\right)dt.$$  How does one proceed if both $q$ and $t$ vary and further that $q$ depends on a varying $t$?  Is the definition now  $$\delta S~=~ \int \left[ L \left(\tilde{q}(t+\delta t),\dot{\tilde{q}}( t+\delta t), t+\delta t\right) -  L \left(q(t),\dot{q}(t),t\right)\right] dt~?  $$  If so how does one proceed?
13611
I am reading about Lagrangian mechanics.  At some point the difference between the temporal derivative of a variation and variation of the temporal derivative is discussed.   The fact that the two are the same is presented in the book I am reading as a rule, commutativity, and possible non-commutative rules are mentioned too.  I do not get it: given a path $q(t)$ and its variation $\delta q(t)$, the equivalence between the variation of the derivative $\delta \dot{q}$ and derivative of the variation $\dot{\delta q}$ seems to me a fact descending straight from calculus, not an arbitrary choice.  References:   B.D. Vujanovic and T.M. Atanackovic,, (2004); p.12. C. Lanczos, The Variational principles of Mechanics.
99047
I am currently working on some differential geometry and I came across the following question.     Let $H = \{(x,y)\in \mathbb{R}^2|y&gt;0\}$, this is the hyperbolic plane (the upper half-plane model). The first fundamental form is   $$ \frac{dx^2+dy^2}{y^2}. $$   Write down the geodesic equation of the hyperbolic plane and show that geodesics in $H$ are either straight lines with constant $x$ or a half circle with center on the $x$-axis.   I have found that the geodesic equations are $$ \frac{d}{dt}\bigg[ \frac{\overset{.}{x}}{y^2} \bigg] =0 $$ $$ \frac{d}{dt}\bigg[ \frac{\overset{.}{y}}{y^2} \bigg] = \frac{-1}{y^3}(\overset{.}{x}^2+\overset{.}{y}^2). $$  From this its easy to see that $\overset{.}{x} = Cy^2$ for some constant $C$, and so if $C=0$ we find the straight-line solutions. However I am struggling to show that there are also circular solutions. Using the second geodesic equation I am also able to show that  $$ \overset{..}{y} = \frac{\overset{.}{y}^2}{y}-\frac{\overset{.}{x}^2}{y}. $$   However I have not been able to make any more progress after this.
14757
of a particle moving in the presence of a scalar field $\varphi (x)$ is given by $$ S = -m\int e^{\varphi (x)}\sqrt{\eta_{\alpha \beta}\frac{dx^{\alpha}}{d \lambda}\frac{dx^{\beta}}{d \lambda}}d\lambda , $$ where $\lambda$ is the parametrization of the worldline of the particle, ignoring the free field part $\int \eta_{\alpha \beta}\partial^{\alpha}\varphi \partial^{\beta} \varphi d^{4}x$.   How does one derive the equations of motion in terms of the parameter  $$d\tau = \sqrt{\eta_{\alpha \beta}\frac{dx^{\alpha}}{d \lambda}\frac{dx^{\beta}}{d \lambda}} d\lambda.   $$ $u^{\alpha} = \frac{dx^{\alpha}}{d\tau} \Rightarrow u_{\alpha}u_{\beta}\eta^{\alpha \beta} = 1$?  My attempt:  $$ \delta S = 0 \Rightarrow \int \left( \frac{\partial (e^{\varphi } \sqrt{...})}{\partial x^{\alpha}}\delta x^{\alpha} +\frac{\partial (e^{\varphi } \sqrt{...})}{\partial \left( \frac{d x^{\alpha}}{d \lambda } \right)}\frac{d}{d\lambda} \delta x^{\alpha} \right)d\lambda = |d\tau = \sqrt{...}d\lambda | = $$ $$ = \int \left(\sqrt{...}e^{\varphi}\partial_{\alpha}\varphi - \frac{d}{d \lambda} \left(\frac{d x_{\alpha}}{d\tau}e^{\varphi}\right)  \right)\delta x^{\alpha}d \lambda  =  $$ $$ =\int \left( \partial_{\alpha}\varphi - \frac{d^{2}x_{\alpha}}{d \tau^{2}} - \frac{dx_{\alpha}}{d\tau} \frac{d \varphi }{d\tau }\right) \delta x^{\alpha} e^{\varphi} \sqrt{...}d\lambda =  $$ $$ =\int \left( \partial_{\alpha}\varphi - \frac{d u_{\alpha}}{d \tau} - u_{\alpha} u_{\beta} \partial^{\beta} \varphi \right)\delta x^{\alpha} e^{\varphi} \sqrt{...}d\lambda \Rightarrow $$ $$ \partial_{\alpha}\varphi - \frac{d u_{\alpha}}{d \tau} - u_{\alpha} u_{\beta} \partial^{\beta} \varphi = 0 \Rightarrow \partial_{\alpha} \varphi = e^{-\varphi}\frac{d }{d \tau}\left( e^{\varphi } u_{\alpha}\right). $$ Unfortunately, this equation doesn't look like the equation , $$ \frac{d (\varphi u_{\alpha})}{d \tau} = -\partial_{\alpha } \varphi.  $$ I can explain the part of differences by renaming the function, $e^{\varphi } \to \varphi $, in the expression for action (then my equation reduces to the form $ \partial_{\alpha} \varphi = \frac{d }{d \tau}\left( \varphi  u_{\alpha}\right)$), but I can't explain why my equation has the wrong sign.
93935
In cartesian coordinates, the Laplacian is  $$\nabla^2 = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\qquad(1)$$    If it's converted to spherical coordinates, we get $$\nabla^2=\frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2 \frac{\partial}{\partial r}\right)+\frac{1}{r^2 sin\theta}\frac{\partial}{\partial \theta}\left(sin \theta \frac{\partial}{\partial \theta}\right)+\frac{1}{r^2 sin^2 \theta}\frac{\partial^2}{\partial \phi^2}\qquad(2)$$  I am following the derivation (i.e. the method of conversion from cartesian to spherical) in "Quantum physics of atoms, molecules, solids, nuclei and particles" by Eisberg and Resnick (it's in Appendix M). Their method is to first consider a function of only $r$, $\psi(r)$, then calculate $\frac{\partial^2 \psi}{\partial x^2}$, $\frac{\partial^2 \psi}{\partial y^2}$, $\frac{\partial^2 \psi}{\partial z^2}$. After plugging into (1) and simplifying we end up with $\nabla^2=\frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2 \frac{\partial}{\partial r}\right)$ which is, indeed the first term in (2).  For the second (and third) term they don't show the derivation, they just say "the second and third terms can be obtained by taking $\psi = \psi(\theta)$ and $\psi = \psi(\phi)$.  I am trying to complete the derivation for the other terms, but I don't know how to proceed. Here is what I have so far: $$\frac{\partial \psi}{\partial x} = \frac{\partial \psi}{\partial \theta} \frac{\partial \theta}{\partial x}\qquad (3)$$  $$\frac{\partial\theta}{\partial x} = \frac{xz}{(x^2 + y^2 +z^2)^{3/2}\sqrt{1-\frac{z^2}{x^2+y^2+z^2}}}\qquad (4)$$  putting (4) into (3) and substituting for $r$:  $$\frac{\partial\psi}{\partial x} =\frac{\partial \psi}{\partial\theta} \frac{xz}{r^{3}\sqrt{1-\frac{z^2}{r^2}}}$$  Now, making the conversions $x=rsin\theta cos\phi$ and $z=r cos\theta$ and simplifying I get  $$\frac{\partial\psi}{\partial x}= \frac{\partial\psi}{\partial\theta} \frac{cos \theta cos \phi}{r} $$  My question is how do I proceed?
36921
The other day, I was listening to  on the Lagrangian for a charged particle in an electromagnetic field, and at one point in the video, the lecturer mentions that we can add any total time derivative of a function $f(q, t)$ to the Lagrangian without altering its equations of motion.   This is nothing new to me, and I understand it fully, but shortly afterwards (approximately two minutes after the linked starting point), he goes on to say that you can, in fact, add a total time derivative of a function $f(q, \dot{q}, t)$, given certain conditions. This definitely surprised me, and I would love to know more about it, but the lecturer quickly moves on, so my question is as follows: under what conditions can one add the total time derivative of a function which depends on the particle's generalized velocities in addition to its generalized coordinates and time without affecting the particle's equations of motion?
17652
I've just finished a Classical Mechanics course, and looking back on it some things are not quite clear. In the first half we covered the Lagrangian formalism, which I thought was pretty cool. I specially appreciated the freedom you have when choosing coordinates, and the fact that you can basically ignore constraint forces. Of course, most simple situations you can solve using good old $F=ma$, but for more complicated stuff the whole formalism comes in pretty handy.  Then in the second half we switched to Hamiltonian mechanics, and that's where I began to lose sight of why we were doing things the way we were. I don't have any problem understanding the Hamiltonian, or Hamilton's equations, or the Hamilton-Jacobi equation, or what have you. My issue is that I don't understand why would someone bother developing all this to do the same things you did before but in a different way. In fact, in most cases you need to start with a Lagrangian and get the momenta from $p = \frac{\partial L}{\partial \dot{q}}$, and the Hamiltonian from $H = \sum \dot{q_i}p_i - L$. But if you already have the Lagrangian, why not just solve the Euler-Lagrange equations?  I guess maybe there are interesting uses of the Hamiltion formalism and we just didn't do a whole lot of examples (it was the harmonic oscillator the whole way, pretty much). I've also heard that it allows a somewhat smooth transition into quantum mechanics. We did work out a way to get Schrödinger's equation doing stuff with the action. But still something's not clicking.  My questions are the following: Why do people use the Hamiltonian formalism? Is it better for theoretical work? Are there problems that are more easily solved using Hamilton's mechanics instead of Lagrange's? What are some examples of that?
43500
The Lagrangian density for a single real scalar field theory is \begin{equation}\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}-V(\phi)\end{equation} I have often seen this written \begin{equation}\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-V(\phi)\end{equation} which appears to differ in the kinetic term.  Specifically \begin{equation}(\partial_{\mu}\phi)^{2}=(\partial_{\mu}\phi)(\partial^{\mu}\phi)\neq\partial_{\mu}\phi\partial^{\mu}\phi=(\partial_{\mu}\phi)(\partial^{\mu}\phi)+\phi\partial_{\mu}\partial^{\mu}\phi\end{equation} where the product rule has been used in the last equality. Is this just sloppy notation (which is acceptable because only first derivatives in the field are considered anyway), or is my maths very bad?
38589
I'll be generous and say it might be reasonable to assume that nature would tend to minimize, or maybe even maximize, the integral over time of $T-V$.  Okay, fine.  You write down the action functional, require that it be a minimum (or maximum), and arrive at the Euler-Lagrange equations.  Great.  But now you want these Euler-Lagrange equations to not just be derivable from the Principle of Least Action, but you want it to be equivalent to the Principle of Least Action.  After thinking about it for awhile, you realize that this implies that the Principle of Least Action isn't really the Principle of Least Action at all:  it's the "Principle of Stationary Action".  Maybe this is just me, but as generous as I may be, I will not grant you that it is "natural" to assume that nature tends to choose the path that is stationary point of the action functional.  Not to mention, it isn't even obvious that there is such a path, or if there is one, that it is unique.  But the problems don't stop there.  Even if you grant the "Principle of Stationary Action" as fundamentally and universally true, you realize that not all the equations of motions that you would like to have are derivable from this if you restrict yourself to a Lagrangian of the form $T-V$.  As far as I can tell, from here it's a matter of playing around until you get a Lagrangian that produces the equations of motion you want.  From my (perhaps naive point of view), there is nothing at all particularly natural (although I will admit, it is quite useful) about the formulation of classical mechanics this way.  Of course, this wouldn't be such a big deal if these classical ideas stayed with the classical physics, but these ideas are absolutely fundamental to how we think about things as modern as quantum field theory.  Could someone please convince me that there is something natural about the choice of the Lagrangian formulation of classical mechanics (I don't mean in comparison with the Hamiltonian formulation; I mean period), and in fact, that it is so natural that we would not even dare abandon these ideas?
47885
For a point charge that moves with the trajectory $ \vec r(t)$, we know that it has the singular charge and current densities  $$ \rho (\vec x, t) = q \delta^3(\vec x - \vec r(t)) $$ $$ \vec J(\vec x, t) = q \frac{d\vec r}{dt} \delta ^3 (\vec x - \vec r (t))$$  How do you prove that these densities satisfy the continuity equation?  $$ \frac{\partial \rho}{\partial t} + \nabla \cdot \vec J = 0 $$  I considered using the property of the dirac delta  $$ \delta (f(x)) = \sum_i^N  \frac{\delta (x - x_i)}{|\frac{df}{dx}|_{x = x_i}}$$  where $x_i$s are the roots of $f(x)$. In the problem at hand, we see that the dummy variable x is t, and $f(x) = \vec x - \vec r(t)$ so that the only root is $t_0$ which makes $\vec r(t_0) = \vec x$.  However I just don't know how to apply the derivative to the dirac delta. I know that   $$ \delta '[f] = f'(0)$$  Any help please? Thank you very much!