# How to get the fourth component of EOM in a relativistic formulation of a charged particle in an electromagnetic field?

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.
• Note that the Euler-Lagrange equation is nonsense for $i=0$. What would $\frac{\partial}{\partial\dot t}$ even mean?
– Chris
Commented Dec 31, 2017 at 9:07
• @Chris: Dot could in principle be differentiation wrt. a world-line parameter, which may be different from coordinate time $x^0=ct$. Commented Jan 6, 2018 at 17:07

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.

1. Susskind's statement is the following.

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

Proof: By linearity it is enough to consider the case where $$W^{\mu}~=~\begin{bmatrix} 0 \cr 0 \cr 0 \cr 0 \end{bmatrix}.$$ 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 by assumption, 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$$^2$$ \begin{align} S~:=~&\int\!d\tau~L , \cr L~:=~&L_0 - U +L_{\rm gf}, \end{align}\tag{C} with \begin{align} L_0~:=~& \frac{\dot{x}^2}{2e}-\frac{e m^2}{2}, \cr \dot {x}^2~:=~& g_{\mu\nu} \dot {x}^{\mu} \dot {x}^{\nu}, \cr \dot {x}^{\mu} ~:=~&\frac{dx^{\mu}}{d\tau},\end{align}\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$$^3$$ \begin{align} 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}, \cr F_{\mu\nu}~:=~&\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}. \end{align}\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 \begin{align} 0~\approx~&\frac{\delta S}{\delta x^{\mu}} ~=~ -\dot{p}^{\rm kin}_{\mu} +\frac{\partial_{\mu}g_{\nu\lambda}}{2e}\dot {x}^{\nu} \dot {x}^{\lambda} +F_{\mu} + \color{red}{B}\frac{\partial \chi}{\partial x^{\mu}} \cr ~=~& -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}^{\lambda} +F_{\mu} + \color{red}{B}\frac{\partial \chi}{\partial x^{\mu}}, \end{align} \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. $$\tau$$: \begin{align} \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)\cr ~\stackrel{(I)}{\approx}~&\frac{\dot{x}^{\mu}}{e}\color{red}{B}\frac{\partial \chi}{\partial x^{\mu}}. \end{align}\tag{M}

8. Static gauge $$\tau=x^0$$. This is OP's gauge. In the 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.

--

$$^1$$ 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.

• Notes for later: What happens if we have quadratic gauge-fixing instead of a delta function? Commented Oct 30, 2018 at 9:20
• In the first point, you say that for $V^\mu=[V^0, 0, 0, 0]^T$, an arbitrary Lorentz transformation (which I know as any matrix $\Lambda$ satisfying $\Lambda^T\eta\Lambda=\eta$) does not change the three 0's. However I doubt that. Can you please prove?
– Atom
Commented Jul 13, 2020 at 14:51
• It is not true in general. That's merely Susskind's assumption. Commented Jul 13, 2020 at 15:04
• So Susskind is wrong? And the only way to get to the 4th equation is to use the gauge invariance argument?
– Atom
Commented Jul 13, 2020 at 15:19
• Susskind's statement in the video is correct. Commented Jul 13, 2020 at 15:34