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I Geodesics and affine parameterisation

  1. Nov 1, 2016 #1
    As I understand it, a curve ##x^{\mu}(\lambda)## (parametrised by some parameter ##\lambda##) connecting two spacetime events is a geodesic if it is locally the shortest path between the two events. It can be found by minimising the spacetime distance between these two events: $$s_{AB}=\int_{A}^{B}\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}=\int_{\lambda_{A}}^{\lambda_{B}}\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}}d\lambda$$ Minimising ##s_{AB}## (i.e. requiring that ##\delta s_{AB}=0##) results in the following equation of motion: $$\frac{d^{2}x^{\mu}}{d\lambda^{2}}+\Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda}=a(\lambda)\frac{dx^{\mu}}{d\lambda}\qquad (1)$$ where ##a(\lambda)=\left(-\frac{1}{L}\frac{dL}{d\lambda}\right)## and ##L=\frac{ds}{d\lambda}=\sqrt{g_{\mu\nu}(x(\lambda))\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}}## is the appropriate Lagrangian.
    I've read that if ##\lambda## is an affine parameter then the right hand side of ##(1)## is zero and we have that $$\frac{d^{2}x^{\mu}}{d\lambda^{2}}+\Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda}=0\qquad (2)$$ Why is this the case though?!

    How does one show that if the parameter is affine then the right-hand side of ##(1)## vanishes?

    I can see that if ##\lambda\propto s##, where ##s## is the arc-length of the curve, then
    $$\frac{d^{2}x^{\mu}}{d\lambda^{2}}+\Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda}=\left(-\frac{1}{L}\frac{dL}{d\lambda}\right)\frac{dx^{\mu}}{d\lambda}=0$$ since ##L=\frac{ds}{d\lambda}=\text{constant}## and so ##\frac{dL}{d\lambda}=0##, so I guess that my confusion is over the definition of an affine parameter. Is it any parameter, ##\lambda## that is proportional to the arc-length, ##s## of the curve, such that ##\lambda=as+b##, where ##a## and ##b## are constants?!
     
    Last edited: Nov 1, 2016
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  3. Nov 1, 2016 #2

    andrewkirk

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    That first formula is not the canonical definition of a spacetime geodesic. A spacetime geodesic is defined as a curve that parallel transports its velocity vector, which is specified in the formula labelled as (2) above. Its length-extremisation properties are derived as theorems, not a definition.

    There was an interesting discussion a while ago about the extremising properties of geodesics. The actual situation is quite complicated. The thread is here.

    IIRC, the extremisation property of a geodesic only holds for timelike-separated events. There is no upper or lower bound on the length of a curve separating spacelike-separated events. For timelike-separated events, a purely timelike curve connecting them has maximal absolute value of length iff it is a geodesic.

    Note that those square roots in the OP formula need an absolute value taken inside them in order to avoid having to square root a negative in some cases.

    That definition will do, except in the case of lightlike-separated points, for which one cannot parametrise by curve length because that is uniformly zero. But that exception doesn't bother us because the activity of seeking an extremal lightlike curve is trivial, since all lightlike curves have the same zero length.
     
    Last edited: Nov 1, 2016
  4. Nov 1, 2016 #3

    PeterDonis

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    This isn't quite right. You can have null curves that are not geodesics. You just can't parametrize them by arc length, so to figure out whether or not a null curve is a geodesic, you have to find some other parameter to check whether it is extremal.
     
  5. Nov 2, 2016 #4
    As I understand it, a geodesic is a curve whose tangent vector is parallel transported along itself, i.e. $$\nabla_{\dot{\gamma}}\dot{\gamma}=0$$ where ##\gamma(t)## is a putative geodesic and ##\dot{\gamma}(t)## its tangent vector.

    However, I thought that, at least for time-like and null geodesics, one could simply start from the spacetime interval between to finitely separated events and a variational principle, as I put in my OP to derive the equations of motion for a geodesic?!

    How does one derive the geodesic equation from the parallel transport definition?

    Is it simply: $$\nabla_{\dot{\gamma}}\dot{\gamma}=\nabla_{\dot{\gamma}}\left(\frac{dx^{\mu}}{dt}e_{\mu}\right)=\frac{d}{dt}\left(\frac{dx^{\mu}}{dt}\right)e_{\mu}+\frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt}\nabla_{\nu}e_{\mu}\\ =\frac{d^{2}x^{\mu}}{dt^{2}}e_{\mu}+\frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt}\Gamma^{\alpha}_{\nu\mu}e_{\alpha}=\left(\frac{d^{2}x^{\mu}}{dt^{2}}+\frac{dx^{\alpha}}{dt}\frac{dx^{\beta}}{dt}\Gamma^{\mu}_{\alpha\beta}\right)e_{\mu}=0\\ \Rightarrow\qquad\frac{d^{2}x^{\mu}}{dt^{2}}+\frac{dx^{\alpha}}{dt}\frac{dx^{\beta}}{dt}\Gamma^{\mu}_{\alpha\beta}=0$$ where ##\lbrace x^{\mu}\rbrace## are the coordinates of the curve ##\gamma## and ##\dot{\gamma}=\frac{d}{dt}=\frac{dx^{\mu}}{dt}e_{\mu}##, where ##e_{\mu}=\frac{\partial}{\partial x^{\mu}}## are the corresponding coordinate basis vectors.

    What is the definition of an affine parameter? Can one just define it as a parameter that satisfies $$\frac{d}{d\lambda}\left(\text{ln}\left(\bigg\lvert\sqrt{g_{\mu\nu}(x(\lambda))\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}}\bigg\rvert\right)\right)=0$$
     
    Last edited: Nov 2, 2016
  6. Nov 2, 2016 #5

    stevendaryl

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    Regardless of the parametrization, what does it mean for a null curve to be extremal? What is being extremized?
     
  7. Nov 2, 2016 #6

    PeterDonis

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    As I understand it, null geodesics are actually "saddle points" of the action, not minima or maxima, so "extremized" is not really the right term; "stationary" would be better, but I don't know what the "-ized" form of that is. :wink:

    Also, when looking at the action integral for a null curve, the "nearby" curves will not, in general, be null (unlike the timelike/spacelike case, where "nearby" curves will also be timelike/spacelike). So it's not really correct to think of a null geodesic as a curve that is "stationary-ized" with respect to other "nearby" null curves. What is "stationary-ized" can still be thought of as the arc length--the null geodesic with arc length zero is a stationary point between curves with negative squared arc length and positive squared arc length. You just can't write that down mathematically using arc length as a curve parameter.
     
  8. Nov 2, 2016 #7

    TeethWhitener

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    I'm pretty sure this was a problem in Wald's GR book (chapter 3, problem 5). Given ##T^a \nabla_a T^b=\alpha(\lambda) T^b##, where ##T## is the tangent along a curve parameterized by ##\lambda##, you can always reparametrize the curve by ##\mu(\lambda)## to get ##T^a \nabla_a T^b=0##. In that case, ##\mu## is the affine parameterization. To show this, rewrite the (not-affinely-parameterized) geodesic equation using derivatives with respect to ##\mu(\lambda)## and the chain rule. This equation:
    $$\frac{d^2 x^{\alpha}}{d \lambda^2}+\Gamma^{\alpha}_{\gamma \beta} \frac{dx^{\gamma}}{d \lambda} \frac{dx^{\beta}}{d \lambda}=\alpha(\lambda) \frac{dx^{\alpha}}{d \lambda}$$
    becomes
    $$\frac{d^2 x^{\alpha}}{d \mu^2}\left({\frac{d\mu}{d\lambda}}\right)^2+\frac{dx^{\alpha}}{d\mu}\frac{d^2\mu}{d\lambda^2}+\Gamma^{\alpha}_{\gamma \beta} \frac{dx^{\gamma}}{d \mu} \frac{dx^{\beta}}{d \mu}\left({\frac{d\mu}{d\lambda}}\right)^2=\alpha(\lambda) \frac{dx^{\alpha}}{d \mu}\frac{d\mu}{d\lambda}$$
    Rearranging gives:
    $$\left({\frac{d\mu}{d\lambda}}\right)^2 \left[{\frac{d^2 x^{\alpha}}{d \mu^2}+\Gamma^{\alpha}_{\gamma \beta} \frac{dx^{\gamma}}{d \mu} \frac{dx^{\beta}}{d \mu}}\right] = \frac{dx^{\alpha}}{d \mu} \left[{\alpha(\lambda) \frac{d\mu}{d\lambda} - \frac{d^2\mu}{d\lambda^2}}\right] $$
    So all we have to do is choose a ##\mu(\lambda)## that satisfies
    $$\alpha(\lambda) \frac{d\mu}{d\lambda} = \frac{d^2\mu}{d\lambda^2}$$
    and we recover the geodesic equation, affinely parameterized.
    Caveat: I am an absolute amateur when it comes to general relativity, so this could all be completely incorrect.
     
    Last edited: Nov 2, 2016
  9. Nov 2, 2016 #8

    andrewkirk

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    I find the terminology 'affine parameter' terribly misleading and unnecessarily confusing but unfortunately its use is so embedded in GR that there seems little hope of expunging it. What is 'affine' is either a parameterisation of a path (a 'curve'), or a relationship between two parameterisations of a path. It is meaningless to say that ##\lambda## is affine since ##\lambda## is only a symbol, and a symbol cannot be affine or non-affine.

    It's useful here to use terminology that distinguishes between 'curve', which is a function that maps a real interval ##I## to a manifold ##M##, and 'path', which is a path-connected set of points on a manifold ##M##. There is a many-to-one relationship between curves and paths, corresponding to different ways that the path can be traced out. Any path is the image of each of the many curves that 'trace out that path'. A curve whose image is path P can be called a 'parameterisation of P'.

    We could say that the relationship between two curves ##f:I_1\to M## and ##g:I_2\to M## that are parameterisations of the same path ##P## is affine if there is an affine function ##A:I_1\to I_2## such that ##f=g\circ A##. The affine function ##A## is determined by two constants ##a,b\in\mathbb R## such that ##A(s)=as+b##. Intepreted that way, the use of the word 'affine' tells us nothing about geodesics other than that, if ##f## is a geodesic then so must be ##g## - a theorem that is readily proved by differentiation.

    Or we could say that a curve ##f:I\to M## is 'affine' iff it is a geodesic, ie iff it satisfies the equation that is represented in the OP by (2), but which is less ambiguously written as:
    $$
    (f^{\mu})''(\lambda)+\Gamma^{\mu}_{\alpha\beta}(f(\lambda))(f^{\alpha})'(\lambda)(f^{\beta})'(\lambda)=0\qquad (2)
    $$
    where ##f^\nu:I\to\mathbb R## is the ##\nu##th coordinate function of ##f## and primes ##'## indicate differentiation.

    That is the usual definition of 'geodesic' in GR and also in the books on manifolds that I have. Some texts define 'geodesic' in terms of extremal properties, an approach that can work OK for Riemannian manifolds but is really not good for pseudo-Riemannian ones like spacetime. For such a text, a geodesic is a curve that satisfies an equation like (1), and it is called an 'affine geodesic' if it also satisfies the equation (2).

    In short, if one must use the term 'affine' in this context, I suggest it be used only to refer to any curve that is a geodesic, and to note that if ##f## is a geodesic then any curve that is the composition of ##f## with an affine function from a real interval to a subinterval of the domain of ##f## is also a geodesic. Use of the term 'affine parameter' is best avoided altogether, but when reading a text that uses it, interpret it as a reference to an affine curve, ie geodesic (using the preferred definition of geodesic, ie (2)).
     
  10. Nov 3, 2016 #9

    TeethWhitener

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    I think you're saying the same thing (in significantly more elegant language), but this how I interpreted "affine:" I thought it simply referred to the fact that if ##\lambda## parameterizes the curve ##x^{\alpha}(\lambda)## to satisfy the geodesic equation:
    $$\frac{d^2 x^{\alpha}}{d\lambda^2} + \Gamma^{\alpha}_{\gamma \beta}\frac{dx^{\gamma}}{d\lambda}\frac{dx^{\beta}}{d\lambda} = 0$$
    then reparameterizing the curve with ##\mu(\lambda) = a\lambda + b## leaves the geodesic equation invariant. Basically, that the geodesic equation is invariant under affine transformations of the parameter.
     
  11. Nov 3, 2016 #10

    vanhees71

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    Indeed, the geodesic is defined as a curve that parallel transports its tangent vectors, which is described by (2) in #1. To get it as a variational problem with the square-root form you must make sure that the world-line parameter is an affine parameter, i.e., for which ##\mathrm{d} s/\mathrm{\d \lambda}=\text{const}## or ##g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=\text{const}## along the solution of the equations of motion ("trajectories").

    Now there is a very simple trick. Instead of using the square-root form of the action given in #1, you use another action with the Lagrangian
    $$L=\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}.$$
    The nice thing with this is that then automatically your world-line parameter is an affine parameter, because since the Lagrangian doesn't explicitly depend on ##\lambda##, the Hamiltonian is conserved along the trajectory, i.e.,
    $$H=p_{\mu} \dot{x}^{\mu}-L=\text{const}, \quad p_{\mu}=\frac{\partial L}{\partial \dot{x}^{\mu}}=\dot{x}_{\mu}.$$
    Consequently you have
    $$H=L,$$
    i.e., ##g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=\text{const}## along the trajectory.

    Indeed, it is easy to see that the Euler-Lagrange equation of this action principle gives the correct geodesic equation (2) in #1. For massive particles you can choose ##\tau=s/c## as the affine parameter by simply choosing the constant as ##c^2##. For light-like (null) geodesics you can't do this of course, but the general parameter ##\lambda## is affine, and you get a null geodesic from this variational principle.

    It's a bit more cumbersome with more general equations of motion, i.e., when there are other forces than gravity present, where you have additionally and interaction Lagrangian. In this case you must use the square-root form of the "kinetic term" in the Lagrangian. But even then you can impose the constraint to have a geodesic parameter using a Lagrange parameter without changing the dynamics. Then you get the action principle with the quadratic form in ##\dot{x}^{\mu}## as the kinetic term, which is more convenient than the general form.
     
  12. Nov 3, 2016 #11
    So as a "working definition", is an affine parametrisation of a spacetime curve, ##s## defined as one for which ##\frac{ds}{d\lambda}=0##?
     
  13. Nov 3, 2016 #12

    vanhees71

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    No its' ##\mathrm{d}s/\mathrm{d} \lambda=\text{const}##. Only for null geodesics you have ##\mathrm{d} s/\mathrm{d} \lambda=0##.
     
  14. Nov 3, 2016 #13
    Whoops, sorry.

    So an affine parametrisation of a spacetime curve then is simply one for which ##\frac{ds}{d\lambda}=\text{constant}##?!
     
  15. Nov 3, 2016 #14

    vanhees71

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  16. Nov 3, 2016 #15
  17. Nov 3, 2016 #16

    pervect

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    Have you written out, or seen written out in some text, the Euler-Lagrange equations that extremize the action for L that you wrote down? It's a bit long for me to want to write it all out myself, but I believe I've seen something along these lines in MTW's text, "Gravitation". Online, you can look at Sean Carrol's derivation in https://arxiv.org/abs/gr-qc/9712019 in his lecture notes (search for geodesics within the document, you should find the right section).

    Neither of this is probably exactly what you want, but it should get you close. For instance Caroll starts out with a general parameterization, then makes the additional assumption that the time parameterization is affine, using this assumption to show algebraically that this results in the version of the geodesic equations where the RHS is zero.

    I suspect this may be close to what you are looking for, though perhaps you might additionally want to review where the Euler-Lagrange equations came from in the first place (?). If this is the case, I think Goldstein's "Classical Mechanics" (to give but one example) would have a discussion on this. It should be easy to find a disucssion on this in most sufficiently advanced physics textbooks.

    I rarely use the non-affine parameterization, but I'd suggest considering a simple case to see the effect of non-affine parameterizations. Consider the flat space-time case, then he geodesic equations with the RHS equal to zero lead to a straight line motion with the points evenly spaced in time (a constant coordinate velocity in the same direction ). Consider what happens if you change the time parameterization in this simple case - the proper acceleration of the object is still zero, but the coordinate acceleration of the object is non-zero (it's velocity doesn't stay constant with the new time parmeterization, the direction however does stay contant). However, the resultant coordinate acceleration will always be in the direction of motion of the object - because in both cases the object doesn't change the direciton, the time reparameteriztation doesn't affect that.
     
  18. Nov 4, 2016 #17

    vanhees71

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    Well, one must admit that this issue is sometimes a bit confused in the
    literature. Usually you start with the "square-root form" of the action
    principle in manifest covariant form, because it has the advantage to be
    completely independent of the choice of the parametrization of the world
    line. If you then choose also the "interaction part" (due to external
    fields, e.g., the em. field, which is the most simple example) as a
    homogeneous function of the generalized "velocities" ##\dot{x}^{\mu}##
    (where the dot means the derivative wrt. the world-line parameter
    ##\lambda##). So lets concentrate on this example. The pertinent
    Lagrangian is (natural units ##c=1##)
    \begin{equation}
    \label{1}
    L=-m \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}-q A_{\mu}
    \dot{x}^{\mu}=L_0+L_{\text{int}}.
    \end{equation}
    Now it is immediately clear that here you can not (!!!) choose the
    proper time or any other affine parameter of the particle as a
    world-line parameter, because it's not a free variable. In this
    formalism you have to keep ##\lambda## arbitrary.

    The advantage of this form of the action principle is that the choice of
    the parameter is completely unconstraint but nevertheless you get
    consistent equations of motion, i.e., the on-shell condition is
    automatically fulfilled. To see this we define the "mechanical momentum"
    by
    \begin{equation}
    \tilde{p}_{\mu} =-\frac{\partial L_0}{\partial
    \dot{x}^{\mu}}=-\frac{m}{\sqrt{g_{\nu \nu} \dot{x}^{\mu}
    \dot{x}^{\nu}}} g_{\mu \nu} \dot{x}^{\nu}=-m g_{\mu \nu}
    \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}.
    \end{equation}
    This indeed implies that
    \begin{equation}
    \label{3}
    g^{\mu \nu} \tilde{p}_{\mu} \tilde{p}_{\nu}=m^2=\text{const}.
    \end{equation}
    The equations of motion are then given by
    \begin{equation}
    -\frac{\mathrm{d}}{\mathrm{d} \lambda} \frac{\partial L}{\partial
    \dot{x}^{\mu}} + \frac{\partial L}{\partial x^{\mu}}=0.
    \end{equation}
    This gives
    \begin{equation}
    \label{5}
    \dot{\tilde{p}}_{\mu} + \frac{\partial L_0}{\partial x^{\mu}}=K_{\mu},
    \end{equation}
    where
    \begin{equation}
    K_{\mu}=-\frac{\partial L_{\text{int}}}{\partial
    x^{\mu}}+\frac{\mathrm{d}}{\mathrm{d} \lambda} \frac{\partial
    L_{\text{int}}}{\partial \dot{x}^{\mu}}
    \end{equation}
    is the Minkowski force. We note that (\ref{5}) is consistent since due
    to
    \begin{equation}
    \label{7}
    \dot{x}^{\mu} \frac{\partial L_0}{\partial \dot{x}^{\mu}}=L_0, \quad
    \dot{x}^{\mu} \frac{\partial L_{\text{int}}}{\partial \dot{x}^{\mu}}=L_{\text{int}}
    \end{equation}
    both
    \begin{equation}
    \label{8}
    \dot{x}^{\mu} \frac{\mathrm{d}}{\mathrm{d} \lambda} \frac{\partial
    L_0}{\partial \dot{x}^{\mu}}= \dot{x}^{\mu} \frac{\partial
    L_0}{\partial x^{\mu}} \quad \text{and} \quad \dot{x}^{\mu} \frac{\mathrm{d}}{\mathrm{d} \lambda} \frac{\partial
    L_{\text{int}}}{\partial \dot{x}^{\mu}}= \dot{x}^{\mu} \frac{\partial
    L_{\text{int}}}{\partial x^{\mu}}
    \end{equation}
    are identically (i.e., not only for the trajectories!) fulfilled.

    With that it is easy to see that instead of the Lagrangian (\ref{1}) we
    can as well use
    \begin{equation}
    \label{9}
    L_2=\frac{m}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu} -
    L_{\text{int}}=L_{20} - L_{\text{int}}.
    \end{equation}
    with the same interaction Lagrangian, which still is a homogeneous
    function of rank 1 in ##\dot{x}^{\mu}## (!) as in (\ref{1}), provided we
    can choose ##\lambda## such that
    \begin{equation}
    \label{10}
    g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=1=\dot{\tau}^2,
    \end{equation}
    i.e., that ##\lambda## is an affine parameter.

    This constraint is indeed automatically consistent with the equations of
    motion derived from the action principle with the Lagrangian (\ref{9}),
    on the one hand, it is not explicitly dependent on ##\lambda## and thus
    the ``Hamiltonian''
    \begin{equation}
    \label{11}
    H_2=\dot{x}^{\mu} \frac{\partial L_2}{\partial \dot{x}^{\mu}}-L_2=\text{const}.
    \end{equation}
    along the solutions of the equations of motion, and on the other hand,
    since ##L_{\text{int}}## is a homogeneous function of rank 1 with respect
    to ##\dot{x}^{\mu}## and thus
    \begin{equation}
    \label{12}
    H_2=\dot{x}^{\mu} \frac{\partial L_{20}}{\partial \dot{x}^{\mu}}-L_{20}=L_{20}.
    \end{equation}
    so indeed we can fulfill (\ref{10}) as a constraint, and thus
    ##\tilde{p}_{\mu}=\partial L_{20}/\partial \dot{x}^{\mu}##.

    Now the equation of motion according to ##L_2## reads
    \begin{equation}
    \label{13}
    \dot{\tilde{p}}_{\mu}-\frac{\partial L_{20}}{\partial
    x^{\mu}}=\frac{\mathrm{d}}{\mathrm{d} \lambda} \frac{\partial
    L_{\text{int}}}{\partial \dot{x}^{\mu}}-\frac{\partial
    L_{\text{int}}}{\partial x^{\mu}} = K_{\mu}.
    \end{equation}
    But now since
    \begin{equation}
    \label{14}
    L_{20}=\frac{L_0^2}{2m} \; \Rightarrow \; \frac{\partial
    L_{20}}{\partial x^{\mu}}=\frac{L_0}{m} \frac{\partial L_0}{\partial x^{\mu}}.
    \end{equation}
    Now since the constraint (\ref{10}) is fulfilled, we have ##L_0=-m##, and
    thus (\ref{13}) is indeed the same equation as (\ref{5}).

    Thus we can use ##L_2## as the Lagrangian, which implies that ##\lambda## is
    an affine parameter and can be chosen as ##\lambda=\tau##, and with this
    choice we can use the interaction Lagrangian from (\ref{1}).
     
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