The action of a free relativistic particles can be given by $$S=\frac{1}{2}\int d\tau \left(e^{-1}(\tau)g_{\mu\nu}(X)X^\mu(\tau)X^\nu(\tau)-e(\tau)m^2\right).$$ If we then make an infinitesimal transformation of the parametrization parameter $\tau$ this would be $$\tau\to\tau^\prime=\tau-\eta(\tau),$$ 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 $$\delta X^{\mu}(\tau)=X^{\mu^\prime}(\tau)-X^{\mu}(\tau)=\eta(\tau)\frac{d X^{\mu}(\tau)}{d\tau}.$$ 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 $$\delta e(\tau)=\frac{d}{d\tau}\left(\eta(\tau) e(\tau)\right).$$ Could someone show me how this is done and elaborate a little on how one knows how different object transforms?


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.

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

$$ Y ~=~\eta \frac{d}{d\tau}~\in~ \Gamma(TI)\tag{6} $$

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

$$ {\cal L}_Y x^{\mu}~=~Y[x^{\mu}]~=~\eta \frac{dx^{\mu}}{d\tau},\tag{7} $$

$$ ({\cal L}_Ye)\mathrm{d}\tau~:=~{\cal L}_Y\omega ~=~\{\mathrm{d}, i_Y\}\omega~=~\mathrm{d}i_Y\omega $$ $$ ~=~\mathrm{d}(i_Y\omega)~=~\mathrm{d}(\eta e) ~=~\mathrm{d}\tau\frac{d}{d\tau}(\eta e),\tag{8} $$

and hence

$$ {\cal L}_Ye ~\stackrel{(8)}{=}~\frac{d}{d\tau}(\eta e).\tag{9} $$

Formula (6), (7) and (9) correspond to eq. (1.10) in Ref. 1 $$ \tag{1.10} \tau\to \tilde{\tau}=\tau-\eta, \qquad \delta x^{\mu}~=~\eta\frac{d x^{\mu}}{d\tau}, \qquad \delta e ~=~\frac{d}{d\tau}(\eta e), $$ 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. (16) 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

$$L_H~:=~ p_{\mu} \dot{x}^{\mu} - H, \qquad H~:=~ eT, \qquad T~:=~\frac{1}{2}(p^2+m^2), \qquad p^2~:=~ g^{\mu\nu}(x)~p_{\mu} p_{\nu}, \tag{11} $$

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

$$ \phi^{\alpha} ~=~ \{ x^{\mu};~p_{\mu};~ e;~C;~ \bar{C};~B\} \tag{12}$$

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 $0$; $0$; $-1$; $r$; $1$; and $1$, respectively. Each field $\phi^{\alpha}$ has a corresponding antifield $\phi^{\ast}_{\alpha}$ of opposite Grassmann parity. The corresponding BV action$^1$

$$ S_{BV}~=~\int \! \mathrm{d}\tau ~L_{BV} , $$ $$ L_{BV}~=~L_H +\left(x^{\ast}_{\mu} \dot{x}^{\mu}+p_{\ast}^{\mu} \dot{p}_{\mu} +r C^{\ast}\dot{C} \right)e^{r-1}C +\underbrace{e^r C\dot{e}^{\ast}}_{\sim~e^{\ast}\frac{d}{d\tau}( e^r C)} + B \bar{C}^{\ast},\tag{13} $$

satisfies the classical master equation

$$ (S_{BV},S_{BV})~=~0, \tag{14}$$

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

$$ (\phi^{\alpha}(\tau),\phi^{\ast}_{\beta}(\tau^{\prime})) ~=~\delta^{\alpha}_{\beta}~\delta(\tau\!-\!\tau^{\prime}). \tag{15}$$

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

$${\bf s}x^{\mu}~=~e^{r-1} C \dot{x}^{\mu}, \qquad {\bf s}p_{\mu}~=~e^{r-1} C \dot{p}_{\mu}, \qquad {\bf s}e~=~ \frac{d}{d\tau}( e^r C), $$ $$ {\bf s}C~=~ re^{r-1} C\dot{C},\qquad {\bf s}\bar{C}~=~ - B,\qquad {\bf s}B ~=~0, \tag{16} $$

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

$$ \psi ~:=~\int \! \mathrm{d}\tau~\bar{C}\left(\frac{\xi}{2}B +\chi(e) +\epsilon \dot{e}\right), \tag{17} $$ where $\xi,\epsilon\in\mathbb{R}$ are gauge-fixing parameters. 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

$$ L_{\rm gf}~=~ \left. L_{BV} \right|_{\phi^{\ast}~=~\frac{\delta \psi}{\delta \phi}}~=~ L_H + \overbrace{\underbrace{\left(\chi^{\prime}\bar{C}-\epsilon\dot{\bar{C}}\right) \frac{d}{d\tau}(e^r C)}_{ \sim~ \bar{C} \left(\frac{\chi^{\prime}}{2}+\epsilon\frac{d}{d\tau}\right)\frac{d}{d\tau}(e^r C) + e^r C\left(\frac{\chi^{\prime}}{2}-\epsilon\frac{d}{d\tau}\right)\frac{d}{d\tau}\bar{C} }}^{\text{Faddeev-Popov term}} + \overbrace{B \left(\frac{\xi}{2}B +\chi(e) +\epsilon \dot{e}\right)}^{\text{gauge-fixing term}} , \tag{18} $$

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

$$ \Delta~=~(-1)^{|\alpha|}\int\! \mathrm{d}\tau~ \frac{\delta_L}{\delta\phi^{\alpha}(\tau)} \frac{\delta_L}{\delta\phi^{\ast}_{\alpha}(\tau)} ~=~(-1)^{|\alpha|}\iint\! \mathrm{d}\tau~\mathrm{d}\tau^{\prime}~ \delta(\tau\!-\!\tau^{\prime})\frac{\delta_L}{\delta\phi^{\alpha}(\tau)} \frac{\delta_L}{\delta\phi^{\ast}_{\alpha}(\tau^{\prime})} \tag{19} $$

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

$$ \Delta S_{BV}~\stackrel{(13)+(19)}{=}~ 2(n\!-\!r)\iint\! \mathrm{d}\tau~\mathrm{d}\tau^{\prime}~ e(\tau)^{r-1}C(\tau)~ \delta(\tau\!-\!\tau^{\prime}) \frac{d}{d\tau}\delta(\tau\!-\!\tau^{\prime}) $$ $$+r \iint\! \mathrm{d}\tau~\mathrm{d}\tau^{\prime}~ e(\tau)^{r-1}\dot{C}(\tau)~\delta(\tau\!-\!\tau^{\prime})^2~\neq~0, \tag{20} $$ where $n$ is the target space (TS) dimension. This shows that the BV action (13) does not satisfy the quantum master equation; only the classical master equation. We will discuss appropriate modifications of the BV action (13) in Section VII.

V) Classical BFV formulation. We identify $p_e\approx\epsilon B$ with the canonical momentum of the einbein $e$, 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

$$ \{x^{\mu}(\tau), p_{\nu}(\tau^{\prime})\}_{PB} ~=~\delta^{\mu}_{\nu}~\delta(\tau\!-\!\tau^{\prime}), \qquad \{e(\tau)^rC(\tau), \bar{P}(\tau^{\prime})\}_{PB} ~=~\delta(\tau\!-\!\tau^{\prime}), $$ $$ \{e(\tau), B (\tau^{\prime})\}_{PB} ~=~\frac{1}{\epsilon}\delta(\tau\!-\!\tau^{\prime}), \qquad \{\bar{C}(\tau), P(\tau^{\prime})\}_{PB} ~=~\frac{1}{\epsilon}\delta(\tau\!-\!\tau^{\prime}).\tag{21} $$

Note the non-Darboux form

$$ \{C(\tau), \bar{P}(\tau^{\prime})\}_{PB} ~=~e(\tau)^{-r}\delta(\tau\!-\!\tau^{\prime}), \qquad \{ B (\tau), C(\tau^{\prime})\}_{PB} ~=~\frac{r}{\epsilon} \frac{C(\tau)}{e(\tau)} \delta(\tau\!-\!\tau^{\prime}), \tag{22} $$

to ensure that

$$ \{e(\tau)^r C(\tau), B(\tau^{\prime})\}_{PB}~=~0. \tag{23} $$

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

$${\bf s}x^{\mu}~=~e^r C g^{\mu\nu}(x)p_{\nu} ~\approx~ e^{r-1} C \dot{x}^{\mu}, \qquad {\bf s}p_{\mu} ~=~ -\frac{1}{2}e^r C \partial_{\mu}g^{\nu\lambda}(x)~p_{\nu}p_{\lambda} ~\approx~ e^{r-1} C \dot{p}_{\mu}, $$ $${\bf s}e~=~P~\approx~ \frac{d}{d\tau}( e^r C) , \qquad {\bf s}C~=~r\frac{C}{e}P ~\approx~ re^{r-1} C\dot{C},\qquad {\bf s}\bar{C}~=~ - B,\qquad {\bf s} B ~=~0, \tag{24} $$

which should be compared with eq. (16). Here the $\approx$ symbol means equality modulo eqs. of motion. The BRST transformation (24) is generated by

$$ \mathbb{Q}~:=~ \int \! \mathrm{d}\tau ~Q, \qquad \{\mathbb{Q}, \mathbb{Q}\}_{PB}~=~0,\tag{25}$$


$$ -Q~:=~T e^r C + \epsilon B P ~\approx~ T e^r C + \epsilon B \frac{d}{d\tau}( e^r C)\tag{26}$$

is the BRST charge. The BFV action becomes

$$ S_{BFV} ~=~ \int \! \mathrm{d}\tau~\left(\dot{x}^{\mu}p_{\mu}+e^r C\dot{\bar{P}} \right) -\left\{ \psi, \mathbb{Q} \right\}_{PB} ~=~ \int \! \mathrm{d}\tau ~L_{BFV} , \tag{27} $$

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

$$ \psi ~:=~\int \! \mathrm{d}\tau \left(\bar{C} \left(\frac{\xi}{2}B +\chi(e) +\epsilon \dot{e}\right) -\bar{P}e\right),\tag{28} $$

and where the BFV Lagrangian reads$^2$

$$ L_{BFV}~=~\left(p_{\mu}\dot{x}^{\mu}+ e^r C\dot{\bar{P}} \right) + \epsilon\left( B \dot{e} + \bar{C} \dot{P}\right) + \left(-eT +\bar{C}\chi^{\prime} P +B \left(\frac{\xi}{2}B+\chi(e)\right) -\bar{P}P \right)$$ $$ ~\sim~ L_H+ \underbrace{\epsilon\left( B \dot{e} + \bar{C} \dot{P}\right)}_{\text{kinetic term}}+ \bar{P}\left( \frac{d}{d\tau}( e^r C)-P\right) + \underbrace{\bar{C} \chi^{\prime} P}_{\text{FP term}} + \underbrace{B \left(\frac{\xi}{2}B+\chi(e)\right)}_{\text{gauge-fixing term}} \tag{29} .$$

VI) Dirac bracket. Let us integrate out the two FP momenta $P$ and $\bar{P}$. Then the BFV Lagrangian (29) becomes the gauge-fixed Lagrangian (18) from Section III. The corresponding two 2nd class constraints

$$ \Theta~:=~ P - \frac{d}{d\tau}( e^r C)~\approx~0, \qquad \bar{\Theta}~:=~ \bar{P} - \chi^{\prime} \bar{C}+\epsilon\dot{\bar{C}}~\approx~0,\tag{30} $$

has non-zero Poisson bracket

$$ \Delta(\tau,\tau^{\prime} ) ~:=~ \{\Theta(\tau), \bar{\Theta}(\tau^{\prime}) \}_{PB} ~=~ -\left(\frac{\chi^{\prime}}{\epsilon}+2\frac{d}{d\tau} \right) \delta(\tau\!-\!\tau^{\prime}),\tag{31} $$

with inverse

$$ \Delta^{-1}(\tau,\tau^{\prime} ) ~=~ - \frac{1}{4} \exp\left[\frac{(\tau^{\prime}-\tau)\chi^{\prime}}{2\epsilon}\right] {\rm sgn}(\tau\!-\!\tau^{\prime}).\tag{32} $$

Therefore the Dirac bracket becomes

$$ \{e(\tau)^rC(\tau), \bar{C}(\tau^{\prime})\}_{DB} ~=~ \frac{1}{4\epsilon} \exp\left[\frac{(\tau^{\prime}-\tau)\chi^{\prime}}{2\epsilon}\right] {\rm sgn}(\tau\!-\!\tau^{\prime}).\tag{33} $$

Alternatively, the Poisson structure (33) could be deduced from the FP term in the gauge-fixed Lagrangian (18).

Note the non-Darboux form

$$ \{C(\tau), \bar{C}(\tau^{\prime})\}_{DB} ~=~\frac{e(\tau)^{-r}}{4\epsilon} \exp\left[\frac{(\tau^{\prime}-\tau)\chi^{\prime}}{2\epsilon}\right] {\rm sgn}(\tau\!-\!\tau^{\prime}) , $$ $$ \{ B (\tau), C(\tau^{\prime})\}_{DB} ~=~\frac{r}{\epsilon} \frac{C(\tau)}{e(\tau)} \delta(\tau\!-\!\tau^{\prime}), \tag{34}$$

to ensure that

$$ \{e(\tau)^r C(\tau), B(\tau^{\prime})\}_{DB}~=~0.\tag{35} $$

VII) Quantum BV formulation. Eqs. (20), (22) & (34) suggest that we should put $r=0$, so let us do this from now on. Inspired by the BFV-BRST transformations (24), we modify the BV Lagrangian (13) into

$$ \tilde{L}_{BV}~=~L_H +x^{\ast}_{\mu} g^{\mu\nu}(x)p_{\nu}C -\frac{1}{2}p_{\ast}^{\mu} \partial_{\mu}g^{\nu\lambda}(x)~p_{\nu}p_{\lambda} C +e^{\ast}\dot{C} + B \bar{C}^{\ast}. \tag{36} $$

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

$$ (\tilde{S}_{BV}, \tilde{S}_{BV})~=~0~=~\Delta\tilde{S}_{BV}. \tag{37} $$

The modification (36) does not alter the gauge-fixed Lagrangian (18) 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.


$^1$ 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 (27) comes only from the gauge-fixing fermion (28). The $\epsilon$-dependence can be removed via redefinition

$$ \epsilon B~\longrightarrow~ B, \qquad \epsilon\bar{C}~\longrightarrow~ \bar{C}, \qquad \frac{\chi}{\epsilon} ~\longrightarrow~ \chi, \qquad \frac{\xi}{\epsilon^2} ~\longrightarrow~ \xi .\tag{38} $$

In the limit $\epsilon\to 0$, the infinities on the rhs. of the Poisson brackets (21) 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 '17 at 20:09
  • $\begingroup$ References: 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. $\endgroup$ – Qmechanic Dec 29 '18 at 16:40
  • $\begingroup$ 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 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 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 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.