The action of a free relativistic particles can be given by $$S=\frac{1}{2}\int d\tau \left(e^{-1}(\tau)g_{\mu\nu}(X)X^\mu(\tau)X^\nu(\tau)-e(\tau)m^2\right),\tag{1.8}$$ with signature $(-,+,\ldots,+)$. If we then make an infinitesimal transformation of the parametrization parameter $\tau$ this would be $$\tau\to\tau^\prime=\tau-\eta(\tau),\tag{1.10}$$ for an infinitesimal parameter $\eta(\tau)$.

Of course we can describe the system as it pleases us so we know that $$X^{\mu^\prime}(\tau^\prime)=X^\mu(\tau).$$ From this relation we see that $X^{\mu^\prime}(\tau)$ must be $$X^{\mu^\prime}(\tau) \approxeq X^{\mu^\prime}(\tau^\prime+\eta(\tau)) \approxeq X^{\mu^\prime}(\tau^\prime)+\eta(\tau)\frac{d X^{\mu^\prime}(\tau^\prime)}{d\tau^\prime} \approxeq X^{\mu}(\tau)+\eta(\tau)\frac{d X^{\mu}(\tau)}{d\tau}$$ Which all can be summarized as $$\delta X^{\mu}(\tau)=X^{\mu^\prime}(\tau)-X^{\mu}(\tau)=\eta(\tau)\frac{d X^{\mu}(\tau)}{d\tau}.\tag{1.10}$$ Now this is all well I hope. But if one does the same argument for $e(\tau)$ one gets the wrong transformation. It is a scalar function so it has to obey $$e^\prime(\tau^\prime)=e(\tau).$$ Which would give the same transformation.

The right transformation are written in David Tong's lectures on string theory on page 13, eq. (1.10). The transformation is $$\delta e(\tau)=\frac{d}{d\tau}\left(\eta(\tau) e(\tau)\right).\tag{1.10}$$ Could someone show me how this is done and elaborate a little on how one knows how different object transforms?


1 Answer 1


I) Passive picture. The einbein $e$ is not an invariant but transforms as

$$ e~=~e^{\prime} \frac{d\tau^{\prime}}{d\tau}\tag{1} $$

under a reparametrization of the world-line (WL) parameter

$$ \tau\longrightarrow \tau^{\prime}=f(\tau).\tag{2} $$

In other words, $\omega:= e \mathrm{d}\tau\in \Gamma(T^{\ast}I) $ is a one-form on the 1-dimensional WL manifold $I$. The particle position

$$ x^{\mu}~=~x^{\prime \mu}\tag{3} $$

is invariant, while the particle velocity transforms as

$$ \dot{x}^{\mu}~=~\dot{x}^{\prime \mu}\frac{d\tau^{\prime}}{d\tau}.\tag{4}$$

These transformation rules (1)-(4) can be seen in many ways. One way is that the action

$$ S~=~\int \! \mathrm{d}\tau ~L , \qquad L~:=~\frac{\dot{x}^2}{2e}-\frac{e m^2}{2},\tag{5}$$

should be invariant under reparametrizations (2). See also this related Phys.SE post.

Let us call variations in the WL manifold $I$ for horizontal and variations in the target space (TS) for vertical. Then the infinitesimal horisontal variation

$$\delta_h\tau ~:=~\tau^{\prime} - \tau ~=~-\eta, \tag{6}$$

where $\eta$ is an infiniteimal parameter. More generally, the infinitesimal horisontal variation is

$$ \delta_h~\stackrel{(6)}{=}~ -\eta\frac{d}{d\tau}.\tag{7}$$

The total infinitesimal variations are

$$\begin{align} \delta x~:=~&x^{\prime}(\tau^{\prime})-x(\tau)~\stackrel{(3)}{=}~0, \cr \delta e~:=~&e^{\prime}(\tau^{\prime})-e(\tau)~\stackrel{(1)+(6)}{=}~e\frac{d\eta}{d\tau} . \end{align}\tag{8}$$

The infinitesimal vertical variations



$$\begin{align} \delta_v x~:=~&x^{\prime}(\tau)-x(\tau) ~\stackrel{(7)+(8)+(9)}{=}~\eta\frac{d x^{\mu}}{d\tau}, \cr \delta_v e~:=~&e^{\prime}(\tau)-e(\tau) ~\stackrel{(7)+(8)+(9)}{=}~\frac{d}{d\tau}(\eta e) . \end{align}\tag{10}$$

II) Active picture. From the perspective of the 1-dimensional WL manifold $I$, the infinitesimal transformation can e.g. be encoded via Lie derivatives ${\cal L}_Y$ wrt. a vector field

$$ Y ~=~\eta \frac{d}{d\tau}~\in~ \Gamma(TI) \tag{11} $$

on the 1-dimensional WL manifold $I$. The Lie derivatives are

$$ {\cal L}_Y x^{\mu}~=~Y[x^{\mu}]~=~\eta \frac{dx^{\mu}}{d\tau},\tag{12} $$

$$\begin{align} ({\cal L}_Ye)\mathrm{d}\tau~:=~&{\cal L}_Y\omega ~=~\{\mathrm{d}, i_Y\}\omega~=~\mathrm{d}i_Y\omega \cr ~=~&\mathrm{d}(i_Y\omega)~=~\mathrm{d}(\eta e) ~=~\mathrm{d}\tau\frac{d}{d\tau}(\eta e),\end{align} \tag{13} $$

and hence

$$ {\cal L}_Ye ~\stackrel{(8)}{=}~\frac{d}{d\tau}(\eta e).\tag{14} $$

The above should be compared with eq. (1.10) in Ref. 1 $$ \tau\to \tau^{\prime}~=~\tau-\eta, \qquad \delta_v x^{\mu}~=~\eta\frac{d x^{\mu}}{d\tau}, \qquad \delta_v e ~=~\frac{d}{d\tau}(\eta e), \tag{1.10}$$ respectively.

III) Classical BV formulation. Let us mention for completeness that the gauge transformation $\delta$ can be encoded as a BRST transformation, cf. e.g. Ref. 2 and this Phys.SE post. Roughly speaking, the Grassmann-even gauge parameter $\eta$ is then replaced by a Grassmann-odd Faddeev-Popov (FP) ghost $C$. (Actually, the gauge parameter $\eta$ will more precisely be replaced with the combination $e^{1-r}C$, where $r\in\mathbb{R}$ is a power, to be more general, cf. eq. (26) below.) To minimize the appearances of time derivatives, instead of using the Lagrangian (5), it becomes a bit simpler to start from the Hamiltonian Lagrangian

$$L_H~:=~ p_{\mu} \dot{x}^{\mu} - H, \qquad H~:=~ eT, \qquad T~:=~\frac{1}{2}(p^2+m^2), \qquad p^2~:=~ g^{\mu\nu}(x)~p_{\mu} p_{\nu}, \tag{21} $$

cf. e.g. this Phys.SE post. Here we will use the Batalin-Vilkovisky (BV) formalism, cf. Ref. 3. The fields

$$ \phi^{\alpha} ~=~ \{ x^{\mu};~p_{\mu};~ e;~C;~ \bar{C};~B\} \tag{22}$$

are positions $x^{\mu}$; momenta $p_{\mu}$; einbein $e$; FP ghost $C$; FP antighost $\bar{C}$; and Lautrup-Nakanishi (LN) Lagrange multiplier $B$, respectively. They are WL tensors of contravariant orders $0$; $0$; $-1$; $r$; $0$; and $0$, respectively. Each field $\phi^{\alpha}$ has a corresponding antifield $\phi^{\ast}_{\alpha}$ of opposite Grassmann parity. The corresponding BV action$^1$

$$\begin{align} S_{BV}~=~&\int \! \mathrm{d}\tau ~L_{BV} , \cr L_{BV}~=~&L_H +\left(x^{\ast}_{\mu} \dot{x}^{\mu}+p_{\ast}^{\mu} \dot{p}_{\mu} +r C^{\ast}\dot{C} \right)e^{r-1}C +\underbrace{e^r C\dot{e}^{\ast}}_{\sim~e^{\ast}\frac{d}{d\tau}( e^r C)} + B \bar{C}^{\ast},\end{align} \tag{23} $$

satisfies the classical master equation

$$ (S_{BV},S_{BV})~=~0, \tag{24}$$

with antibracket $(\cdot,\cdot)$ on Darboux-form, i.e. the non-zero fundamental antibrackets read

$$ (\phi^{\alpha}(\tau),\phi^{\ast}_{\beta}(\tau^{\prime})) ~=~\delta^{\alpha}_{\beta}~\delta(\tau\!-\!\tau^{\prime}). \tag{25}$$

The Grassmann-odd nilpotent BRST transformation ${\bf s}~=~(S_{BV},\cdot)$ reads

$$\begin{align} {\bf s}x^{\mu}~=~&e^{r-1} C \dot{x}^{\mu}, \qquad {\bf s}p_{\mu}~=~e^{r-1} C \dot{p}_{\mu}, \qquad {\bf s}e~=~ \frac{d}{d\tau}( e^r C), \cr {\bf s}C~=~& re^{r-1} C\dot{C},\qquad {\bf s}\bar{C}~=~ - B,\qquad {\bf s}B ~=~0, \end{align} \tag{26} $$

which should be compared with eq. (1.10). The BV gauge-fixing fermion $\psi$ can be chosen on the form

$$ \psi ~:=~\int \! \mathrm{d}\tau~\bar{C}\left(\frac{\xi\rho}{2}B +\chi(e) +\epsilon \frac{d(e/\rho)}{d\tau}\right), \tag{27} $$

where $\xi,\epsilon\in\mathbb{R}$ are gauge-fixing parameters, and $\rho$ is a fiducial einbein. Moreover, $\chi(e)=(e\!-\!e_0)\chi^{\prime}$ is a gauge-fixing condition (which we will assume is affine in $e$, so that the derivative $\chi^{\prime}$ is constant). The gauge-fixed Lagrangian becomes

$$\begin{align} L_{\rm gf}~=~ &\left. L_{BV} \right|_{\phi^{\ast} ~=~\frac{\delta \psi}{\delta \phi}}\cr ~=~& L_H + \overbrace{\underbrace{\left(\chi^{\prime}\bar{C}-\frac{\epsilon}{\rho}\dot{\bar{C}}\right) \frac{d}{d\tau}(e^r C)}_{ \sim~ \bar{C} \left(\chi^{\prime}+\frac{d}{d\tau}\frac{\epsilon}{\rho}\right)\frac{d}{d\tau}(e^r C) ~\sim~ e^r C\left(\chi^{\prime}-\frac{d}{d\tau}\frac{\epsilon}{\rho}\right)\frac{d}{d\tau}\bar{C} }}^{\text{Faddeev-Popov term}} + \overbrace{B \left(\frac{\xi\rho}{2}B +\chi(e) +\epsilon \frac{d(e/\rho)}{d\tau}\right)}^{\text{gauge-fixing term}}, \end{align} \tag{28} $$

where the $\sim$ symbol means equality up to total time derivative terms. The physical quantities do not depend on the choice of the gauge-fixing fermion $\psi$, as long as certain rank conditions are met.

IV) Quantum master equation. The odd Laplacian

$$ \Delta~=~(-1)^{|\alpha|}\int\! \mathrm{d}\tau~ \frac{\delta_L}{\delta\phi^{\alpha}(\tau)} \frac{\delta_L}{\delta\phi^{\ast}_{\alpha}(\tau)} ~=~(-1)^{|\alpha|}\iint\! \mathrm{d}\tau~\mathrm{d}\tau^{\prime}~ \delta(\tau\!-\!\tau^{\prime})\frac{\delta_L}{\delta\phi^{\alpha}(\tau)} \frac{\delta_L}{\delta\phi^{\ast}_{\alpha}(\tau^{\prime})} \tag{29} $$

is a singular object, which strictly speaking needs to be regularized. We calculate formally

$$ \Delta S_{BV}~\stackrel{(23)+(29)}{=}~ 2(n\!-\!r)\iint\! \mathrm{d}\tau~\mathrm{d}\tau^{\prime}~ e(\tau)^{r-1}C(\tau)~ \delta(\tau\!-\!\tau^{\prime}) \frac{d}{d\tau}\delta(\tau\!-\!\tau^{\prime}) $$ $$+r \iint\! \mathrm{d}\tau~\mathrm{d}\tau^{\prime}~ e(\tau)^{r-1}\dot{C}(\tau)~\delta(\tau\!-\!\tau^{\prime})^2~\neq~0, \tag{30} $$ where $n$ is the target space (TS) dimension. This shows that the BV action (23) does not satisfy the quantum master equation; only the classical master equation. We will discuss appropriate modifications of the BV action (23) in Section VII.

V) Classical BFV formulation. We identify the einbein $e\equiv \rho\lambda $, where $\lambda$ is a Lagrange multiplier. We identify $p_{\lambda}\approx\epsilon B$ with the canonical momentum of the Lagrange multiplier $\lambda$, and we identify the antifield $e^{\ast}\equiv \bar{P}$ with the FP ghost momentum. Introduce an ultra-local Poisson bracket $\{\cdot,\cdot\}_{PB}$ with the following canonical pairs

$$\begin{align} \{x^{\mu}(\tau), p_{\nu}(\tau^{\prime})\}_{PB} ~=~&\delta^{\mu}_{\nu}\frac{1}{\rho(\tau)}\delta(\tau\!-\!\tau^{\prime}), \qquad \{e(\tau)^rC(\tau), \bar{P}(\tau^{\prime})\}_{PB} ~=~\frac{1}{\rho(\tau)}\delta(\tau\!-\!\tau^{\prime}), \cr \{\lambda(\tau), B (\tau^{\prime})\}_{PB} ~=~&\frac{1}{\epsilon\rho(\tau)}\delta(\tau\!-\!\tau^{\prime}), \qquad \{\bar{C}(\tau), P(\tau^{\prime})\}_{PB} ~=~\frac{1}{\epsilon\rho(\tau)}\delta(\tau\!-\!\tau^{\prime}).\end{align}\tag{31} $$

Note the non-Darboux form

$$ \{C(\tau), \bar{P}(\tau^{\prime})\}_{PB} ~=~\frac{e(\tau)^{-r}}{\rho(\tau)}\delta(\tau\!-\!\tau^{\prime}), \qquad \{ B (\tau), C(\tau^{\prime})\}_{PB} ~=~\frac{r}{\epsilon} \frac{C(\tau)}{e(\tau)} \delta(\tau\!-\!\tau^{\prime}), \tag{32} $$

to ensure that

$$ \{e(\tau)^r C(\tau), B(\tau^{\prime})\}_{PB}~=~0. \tag{33} $$

The BRST transformation ${\bf s}~=~\{\mathbb{Q},\cdot\}_{PB}$ (which is independent of the $\epsilon$-parameter) reads

$$\begin{align}{\bf s}x^{\mu}~=~&e^r C g^{\mu\nu}(x)p_{\nu} ~\approx~ e^{r-1} C \dot{x}^{\mu}, \cr {\bf s}p_{\mu} ~=~& -\frac{1}{2}e^r C \partial_{\mu}g^{\nu\lambda}(x)~p_{\nu}p_{\lambda} ~\approx~ e^{r-1} C \dot{p}_{\mu}, \cr {\bf s}e~=~&\rho P~\approx~ \frac{d}{d\tau}( e^r C) , \qquad {\bf s}C~=~r\frac{C}{e}\rho P ~\approx~ re^{r-1} C\dot{C},\cr {\bf s}\bar{C}~=~& - B,\qquad {\bf s} B ~=~0, \end{align} \tag{34} $$

which should be compared with eq. (26). Here the $\approx$ symbol means equality modulo eqs. of motion. The BRST transformation (34) is generated by

$$ \mathbb{Q}~:=~ \int \! \rho \mathrm{d}\tau ~Q, \qquad \{\mathbb{Q}, \mathbb{Q}\}_{PB}~=~0,\tag{35}$$


$$ -Q~:=~T e^r C + \epsilon B P ~\approx~ T e^r C + \epsilon B \frac{d}{d\tau}( e^r C)\tag{36}$$

is the BRST charge. The BFV action becomes

$$ S_{BFV} ~=~ \int \! \mathrm{d}\tau~\left(\dot{x}^{\mu}p_{\mu}+e^r C\dot{\bar{P}} \right) -\left\{ \psi, \mathbb{Q} \right\}_{PB} ~=~ \int \! \mathrm{d}\tau ~L_{BFV} , \tag{37} $$

where the BFV gauge-fixing fermion $\psi$ is

$$ \psi ~:=~\int \! \mathrm{d}\tau \left(\bar{C} \left(\frac{\xi\rho}{2}B +\chi(e) +\epsilon \dot{\lambda}\right) -\bar{P}e\right),\tag{38} $$

and where the BFV Lagrangian reads$^2$

$$\begin{align} L_{BFV}~=~&\left(p_{\mu}\dot{x}^{\mu}+ e^r C\dot{\bar{P}} \right) + \epsilon\left( B \dot{\lambda} + \bar{C} \dot{P}\right) + \left(-eT +\bar{C}\rho\chi^{\prime} P +B \left(\frac{\xi\rho}{2}B+\chi(e)\right) -\rho\bar{P}P \right)\cr ~\sim~& L_H+ \underbrace{\epsilon\left( B \dot{\lambda} + \bar{C} \dot{P}\right)}_{\text{kinetic term}}+ \bar{P}\left( \frac{d}{d\tau}( e^r C)-\rho P\right) + \underbrace{\bar{C} \rho\chi^{\prime} P}_{\text{FP term}} + \underbrace{B \left(\frac{\xi\rho}{2}B+\chi(e)\right)}_{\text{gauge-fixing term}} \end{align} \tag{39} .$$

VI) Dirac bracket. Let us integrate out the two FP momenta $P$ and $\bar{P}$. Then the BFV Lagrangian (39) becomes the gauge-fixed Lagrangian (28) from Section III. The corresponding two 2nd class constraints

$$ \Theta~:=~ \rho P - \frac{d}{d\tau}( e^r C)~\approx~0, \qquad \bar{\Theta}~:=~ \rho\bar{P} - \rho\chi^{\prime} \bar{C}+\epsilon\dot{\bar{C}}~\approx~0,\tag{40} $$

has non-zero Poisson bracket

$$ \Delta(\tau,\tau^{\prime} ) ~:=~ \{\Theta(\tau), \bar{\Theta}(\tau^{\prime}) \}_{PB} ~=~ -\left(\frac{\rho(\tau) \chi^{\prime}}{\epsilon}+2\frac{d}{d\tau} \right) \delta(\tau\!-\!\tau^{\prime}),\tag{41} $$

with inverse

$$ \Delta^{-1}(\tau,\tau^{\prime} ) ~=~ - \frac{1}{4} \exp\left[\frac{R(\tau^{\prime},\tau)\chi^{\prime}}{2\epsilon}\right] {\rm sgn}(\tau\!-\!\tau^{\prime}), \qquad R(\tau^{\prime},\tau)~:= \int_{\tau}^{\tau^{\prime}}\!\mathrm{d}\tau^{\prime\prime}\rho(\tau^{\prime\prime}). \tag{42} $$

Therefore the Dirac bracket becomes

$$ \{e(\tau)^rC(\tau), \bar{C}(\tau^{\prime})\}_{DB} ~=~ \frac{1}{4\epsilon} \exp\left[\frac{R(\tau^{\prime},\tau)\chi^{\prime}}{2\epsilon}\right] {\rm sgn}(\tau\!-\!\tau^{\prime}).\tag{43} $$

Due to double time derivatives in the FP term in the gauge-fixed Lagrangian (28), it is not related to the Poisson structure (43) in a simple manner.

Note the non-Darboux form

$$\begin{align} \{C(\tau), \bar{C}(\tau^{\prime})\}_{DB} ~=~&\frac{e(\tau)^{-r}}{4\epsilon} \exp\left[\frac{R(\tau^{\prime},\tau)\chi^{\prime}}{2\epsilon}\right] {\rm sgn}(\tau\!-\!\tau^{\prime}) , \cr \{ B (\tau), C(\tau^{\prime})\}_{DB} ~=~&\frac{r}{\epsilon} \frac{C(\tau)}{e(\tau)} \delta(\tau\!-\!\tau^{\prime}), \end{align} \tag{44}$$

to ensure that

$$ \{e(\tau)^r C(\tau), B(\tau^{\prime})\}_{DB}~=~0.\tag{45} $$

VII) Quantum BV formulation. Eqs. (30), (32) & (44) suggest that we should put $r=0$, so let us do this from now on. Inspired by the BFV-BRST transformations (34), we modify the BV Lagrangian (23) into

$$ \tilde{L}_{BV}~=~L_H +x^{\ast}_{\mu} g^{\mu\nu}(x)p_{\nu}C -\frac{1}{2}p_{\ast}^{\mu} \partial_{\mu}g^{\nu\lambda}(x)~p_{\nu}p_{\lambda} C +e^{\ast}\dot{C} + B \bar{C}^{\ast}. \tag{46} $$

One may show that the quantum master equation is now satisfied$^1$

$$ (\tilde{S}_{BV}, \tilde{S}_{BV})~=~0~=~\Delta\tilde{S}_{BV}. \tag{47} $$

The modification (46) does not alter the gauge-fixed Lagrangian (28) apart from putting $r=0$.


  1. David Tong, Lectures on String Theory, arXiv:0908.0333.

  2. J. Polchinski, String Theory, Vol. 1, 1998; Section 4.2.

  3. M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994; Chapter 17.

  4. F. Bastianelli, Constrained hamiltonian systems and relativistic particles, 2017 lecture notes.

  5. A. Cohen, G. Moore, P. Nelson & J. Polchinski, Nucl. Phys. B267 (1986) 143; Chapter 2.

  6. C.M. Hull & J.-L. Vazquez-Bello, arXiv:hep-th/9308022; Chapter 2.

  7. P. van Nieuwenhuizen, Lecture notes.


$^1$ We ignore boundary terms. Effectively this means that we impose pertinent boundary conditions, and limit gauge symmetry to the bulk.

$^2$ The $\epsilon$-dependence in the BFV action (37) comes only from the gauge-fixing fermion (38). The $\epsilon$-dependence can be removed via redefinition

$$ \epsilon B~\longrightarrow~ B, \qquad \epsilon\bar{C}~\longrightarrow~ \bar{C}, \qquad \frac{\chi}{\epsilon} ~\longrightarrow~ \chi, \qquad \frac{\xi}{\epsilon^2} ~\longrightarrow~ \xi .\tag{48} $$

In the limit $\epsilon\to 0$, the infinities on the rhs. of the Poisson brackets (31) should be interpreted as zero, i.e. the corresponding canonical variables become decoupled.

  • $\begingroup$ Comment to the answer (v12): The expression (20) vanishes in WL dimensional regularization, cf. Ref. 3. $\endgroup$
    – Qmechanic
    May 14, 2017 at 20:09
  • $\begingroup$ Minor correction. We assume $r=0$ for simplicity. There is a gauge-for-gauge symmetry $\delta C=C_0$ ghost-zeromode. Define BRST transformation ${\bf s}C=\gamma_0$ ghost-for-ghost zeromode. $\endgroup$
    – Qmechanic
    Jan 13, 2019 at 19:11
  • $\begingroup$ Comments for later: It seems the BRST transformation (16) for $x^{\mu}$ should transform into $p$ rather than $\dot{x}$. (Of course, that's the same on-shell.) This is precisely what is done in eq. (36). $\endgroup$
    – Qmechanic
    Jan 15, 2019 at 12:56
  • $\begingroup$ Comments for later: The abelianization $r=0$ (changing the gauge algebra) may alter the partition function, even for Hamiltonian BFV. $\endgroup$
    – Qmechanic
    Jan 21, 2019 at 14:08
  • 1
    $\begingroup$ Hi @Jens Wagemaker. Thanks for the feedback. The difference of order ${\cal O}(\eta^2)$ is ignored, since $\eta$ is infinitesimal. $\endgroup$
    – Qmechanic
    2 days ago

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