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We consider in Lorentz spacetime, $(x^0,x^1,x^2,x^3)=(t,x,y,z)$, choose the unit of time such that $c=1$.

Given a four vector $A_\mu$, and let the Lagrangian $$L(x^i,\dot x^i,t)=-m\sqrt{1-\dot x_i\dot x^i}+qA_0+qA_i\dot x^i,\tag{1}$$ where we use Einstein's convention for summation. (See video at 35:15 with electron charge $q=-e$.)

By the Euler-Lagrange equation $$\frac{d}{dt}\frac{\partial L}{\partial\dot x^i}-\frac{\partial L}{\partial x^i}=0\tag{2}$$ we can get three equations, and if we use the fact that $$\frac{d}{d\tau}=\frac{1}{\sqrt{1-\dot x_i\dot x^i}}\frac{d}{dt}\tag{3}$$ then these equations are, for $i=1,2,3$ $$m\frac{d^2x^i}{d\tau^2}=q(\frac{\partial A_\mu}{\partial x^i}-\frac{\partial A_i}{\partial x^\mu})\frac{dx^\mu}{d\tau}\tag{4}$$

My question follows, can we obtain a fourth equation which is just letting $i=0$?

The video at 59:30 says that this fourth equation is from the property that the action is invariant under Lorentz transformation. But I think this is not very convincing. I tried to view the first three equations as two four vectors having three equation space components, but I do not think this fact can lead to the equality for the last time component.

References:

  1. L. Susskind, Special Relativity, video lecture 7, May 21, 2012.
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  • $\begingroup$ Note that the Euler-Lagrange equation is nonsense for $i=0$. What would $\frac{\partial}{\partial\dot t}$ even mean? $\endgroup$ – Chris Dec 31 '17 at 9:07
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    $\begingroup$ @Chris: Dot could in principle be differentiation wrt. a world-line parameter, which may be different from coordinate time $x^0=ct$. $\endgroup$ – Qmechanic Jan 6 '18 at 17:07
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The action you start with is $$ S = \int d\tau L $$ and $$ L = - m \sqrt{ - \eta_{\mu\nu} {\dot x}^\mu {\dot x}^\nu } +q {\dot x}^\mu A_\mu \, . $$ This action has a gauge symmetry, which is reparameterization invariance, $\tau \to \tau'(\tau)$. In order to write down your Lagrangian you choose the gauge $\tau = x^0 = t$. Once in this gauge, you can derive equations of motion for $x^i$ but not $x^0$ because you have gauge-fixed it. These are the ones you have written down.

However, since you are fixing a gauge, there's going to be a gauge constraint which is essentially the equation of motion derived w.r.t. $x^0$ in the first action. This is the equation you are missing.

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  1. Susskind's statement is the following.

    If three components of a Lorentz covariant 4-vector equation $V^{\mu}=W^{\mu}$ are equal, then the fourth components must be equal as well.

    Proof: By linearity it is enough to consider the case where $W^{\mu}=0$. Now we perform an arbitrary Lorentz transformation on a 4-vector $$ V^{\mu}~=~\begin{bmatrix} V^{0} \cr 0 \cr 0 \cr 0 \end{bmatrix}. $$ Since this does not change the three 0s, it must be because $V^0=0$ is zero to begin with. $\Box$

  2. Another idea to get 4 Euler-Lagrange (EL) eqs. is to have 4 independent dynamical variables $x^{\mu}$, cf. Prahar's answer. The main point is that OP's original action $S[x^1,x^2,x^3]$ can be viewed as a gauge theory $S[x^0,x^1,x^2,x^3]$ with a certain gauge-fixing condition $$\chi~\approx~0. \tag{A}$$ Now strictly speaking, we should introduce a Lagrange multiplier $\color{red}{B}$ to implement the gauge-fixing condition (A), i.e. include a gauge-fixing term $$ L_{\rm gf}~:=~\color{red}{B} ~\chi \tag{B} $$ in the Lagrangian. But this raises a question.

    How do we know that the presence of the Lagrange multiplier $\color{red}{B}$ does not affect the EL eqs.?

    Formal Answer: Because of the (second) Noether identity, cf. e.g. Ref. 1.

    Here we want to explicitly verify this fact for OP's model (1).

  3. We prefer to work with a non-square root$^1$ formulation$^2$ $$ S~:=~\int\!d\tau~L , \qquad L~:=~L_0 - U +L_{\rm gf}, \tag{C}$$ with $$ L_0~:=~ \frac{\dot{x}^2}{2e}-\frac{e m^2}{2}, \qquad \dot {x}^2~:=~ g_{\mu\nu} \dot {x}^{\mu} \dot {x}^{\nu}, \qquad \dot {x}^{\mu} ~:=~\frac{dx^{\mu}}{d\tau},\tag{D}$$ and with an einbein $e>0$. Here $\tau$ is the world-line (WL) parameter (which is not necessarily proper time). The velocity-dependent Lorentz potential is $$ U~:=~ - q{\dot x}^{\mu} A_{\mu}, \tag{E} $$ with corresponding generalized Lorentz 4-force$^3$ $$ F_{\mu}~:=~\frac{d}{d\tau} \frac{\partial U}{\partial \dot{x}^{\mu}} - \frac{\partial U}{\partial x^{\mu}}~=~qF_{\mu\nu}\dot {x}^{\nu}, \qquad F_{\mu\nu}~:=~\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}. \tag{F}$$

  4. The canonical/conjugate 4-momentum$^3$ reads $$ p_{\mu}~:=~\frac{\partial L}{\partial \dot{x}^{\mu}}~=~ p^{\rm kin}_{\mu}+ q A_{\mu}, \tag{G}$$ where the kinetic 4-momentum$^3$ reads $$ p^{\rm kin}_{\mu} ~:=~ \frac{g_{\mu\nu}\dot{x}^{\nu}}{e}. \tag{H}$$

  5. Now the EL eqs. for the action (B) read $$ 0~\approx~\frac{\delta S}{\delta x^{\mu}} ~=~ -\dot{p}^{\rm kin}_{\mu} +\frac{\partial_{\mu}g_{\nu\lambda}}{e}\dot {x}^{\nu} \dot {x}^{\lambda} +F_{\mu} + \color{red}{B}\frac{\partial \chi}{\partial x^{\mu}} $$ $$~=~ -g_{\mu\nu}\frac{d}{d\tau}\left(\frac{\dot{x}^{\nu}}{e}\right) -\frac{\Gamma_{\mu,\nu\lambda}}{e}\dot {x}^{\nu} \dot {x}^{\nu} +F_{\mu} + \color{red}{B}\frac{\partial \chi}{\partial x^{\mu}}, \tag{I}$$ $$ 0~\approx~\frac{\delta S}{\delta e} ~=~-\frac{1}{2}\left(\frac{\dot{x}^2}{e^2}+m^2 \right) + \color{red}{B}\frac{\partial \chi}{\partial e}, \tag{J} $$ $$ 0~\approx~\frac{\delta S}{\delta B}~=~\chi. \tag{K}$$

  6. The question we would like to address is the following.

    How do we conclude that the Lagrange multiplier $\color{red}{B}~\approx~0$ vanishes in EL eqs. (I) & (J)?

  7. In anticipation of what to come, let us first differentiate eq. (J) wrt. $\tau$: $$ \frac{d}{d\tau}\left( \color{red}{B}\frac{\partial \chi}{\partial e}\right) ~\stackrel{(J)}{\approx}~\frac{\dot{x}^{\mu}}{e} \left(g_{\mu\nu}\frac{d}{d\tau}\left(\frac{\dot{x}^{\nu}}{e}\right) +\frac{\Gamma_{\mu,\nu\lambda}}{e}\dot {x}^{\nu} \dot {x}^{\nu} \right) ~\stackrel{(I)}{\approx}~\frac{\dot{x}^{\mu}}{e}\color{red}{B}\frac{\partial \chi}{\partial x^{\mu}}. \tag{M}$$

  8. Static gauge $\tau=x^0$. This is OP's gauge. In static gauge $$ \chi~:=~x^0-\tau . \tag{N}$$ Then eq. (M) becomes $$ 0~\stackrel{(M)+(N)}{\approx}~ \frac{\color{red}{B}}{e}, \tag{O}$$ i.e. the Lagrange multiplier $\color{red}{B}$ vanishes! $\Box$

  9. Affine parametrization gauge $e=1/m$.
    $$ \chi~:=~e-1/m. \tag{P}$$ Then eq. (M) becomes $$ \dot{\color{red}{B}}~\stackrel{(M)+(P)}{\approx}~0.\tag{Q} $$ We can only conclude that the Lagrange multiplier $\color{red}{B}$ is a constant. This is related to the fact that this is only a partial gauge fixing. In this gauge the world-line (WL) parameter $\tau$ is an affine function of proper time.

References:

  1. H. Motohashi, T. Suyama & K. Takahashi, Phys. Rev. D 94 (2016) 124021, arXiv:1608.00071.

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$^1$ To achieve OP's square root formulation (1), simply integrate out the $e$ field, cf. e.g. this Phys.SE post. The gauge symmetry is e.g. discussed in this Phys.SE post and links therein.

$^2$ Notation & conventions. We use signature convention $(-,+,+,+)$ and $c=1$. The $\approx$ symbol means equality modulo eoms.

$^3$ The usual notions of 4-momentum & 4-force correspond to the gauge where the world-line (WL) parameter $\tau$ is proper time.

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