The Relativistic Point Particle

A relativistic particle gives a small but complete example of the ideas behind the string worldsheet: an arbitrary coordinate labels its history, an auxiliary metric imposes a constraint, and gauge fixing leaves a modulus that must still be integrated. We will derive the scalar propagator from this worldline description. The central distinction is between the particle’s spacetime proper time, which has units of length, and the Schwinger modulus, which has units of length squared.

The global conventions fix the spacetime signature and units. Familiarity with classical mechanics and Gaussian integrals is enough for the classical discussion and the free path integral below.

The worldline and its proper length

A trajectory in flat spacetime is a map $X^\mu(\lambda)$ from an interval into spacetime. Initially take the parameter to be dimensionless, $0\leq\lambda\leq1$, and write $\dot X^\mu=dX^\mu/d\lambda$. For a timelike curve, the positive proper-time element and total proper time are

$$ds=\sqrt{-\eta_{\mu\nu}dX^\mu dX^\nu}, \qquad \ell=\int_0^1d\lambda\,\sqrt{-\dot X^2}.$$

The action of a free particle of mass $m>0$ is

$$S_{\rm pp}=-m\ell =-m\int_0^1d\lambda\,\sqrt{-\dot X^2}.$$

The action measures worldline length, just as the Nambu–Goto action will measure worldsheet area. In a curved spacetime one replaces $\eta_{\mu\nu}$ by $G_{\mu\nu}(X)$; here we keep the target spacetime flat so that the quantum calculation is exactly Gaussian.

Coordinate labels are gauge choices

Consider a smooth, orientation-preserving change of parameter,

$$\lambda'=f(\lambda), \qquad f(0)=0,\quad f(1)=1,\quad f'(\lambda)>0.$$

The embedding is a scalar under this change: $X'^\mu(\lambda')=X^\mu(\lambda)$. Thus

$$\frac{dX'^\mu}{d\lambda'}=\frac{\dot X^\mu}{f'}, \qquad d\lambda'\sqrt{-\left(\frac{dX'}{d\lambda'}\right)^2} =d\lambda\sqrt{-\dot X^2}.$$

The action is unchanged. The freedom is local along the curve, so this is a gauge symmetry: different parameterized maps can represent the same geometric history.

For a future-directed timelike trajectory, $X^0$ increases monotonically. We can instead use the target time $t=X^0$ as the parameter, with its own endpoint interval. Writing $\mathbf v=d\mathbf X/dt$, we obtain

$$L_t=-m\sqrt{1-\mathbf v^2}, \qquad \mathbf p=\frac{m\mathbf v}{\sqrt{1-\mathbf v^2}}.$$

The Hamiltonian for evolution in target time is

$$H_t=\mathbf p\cdot\mathbf v-L_t =\frac{m}{\sqrt{1-\mathbf v^2}} =\sqrt{\mathbf p^2+m^2}.$$

This is the positive-energy branch of relativistic mechanics. The choice $t=X^0$ is useful classically, but treating all spacetime coordinates symmetrically will be more convenient for the covariant propagator.

The einbein and the mass-shell constraint

The square root can be removed by introducing an independent positive function $e(\lambda)$, called the einbein:

$$S[X,e]=\frac12\int_0^1d\lambda \left(e^{-1}\dot X^2-e m^2\right).$$

Reparametrization invariance requires

$$e'(\lambda')=\frac{e(\lambda)}{f'(\lambda)}, \qquad e'(\lambda')\,d\lambda'=e(\lambda)\,d\lambda.$$

The factor $e\,d\lambda$ is an invariant measure along the parameter interval. The kinetic term is invariant as well: its two velocity factors supply $(f')^{-2}$, canceled by the factors from $d\lambda'$ and $e'^{-1}$.

An einbein is the one-dimensional analogue of a vielbein. We may introduce an auxiliary Lorentzian metric $\gamma_{\lambda\lambda}=-e^2$ and rewrite the action as

$$S=-\frac12\int d\lambda\,\sqrt{-\gamma} \left(\gamma^{\lambda\lambda}\dot X^2+m^2\right).$$

This metric is independent of the induced metric $\dot X^2$ on the curve. Its normalization in this action gives $[e]=L^2$ for dimensionless $\lambda$; consequently $e\,d\lambda$ is not the spacetime proper-time element $ds$. This auxiliary-metric formulation and its massless extension are discussed in Tong, §1.1.2.

Vary before fixing the gauge

With the endpoints of $X$ fixed, the two independent variations give

$$\frac{d}{d\lambda}\left(e^{-1}\dot X^\mu\right)=0, \qquad -\frac{\dot X^2}{e^2}-m^2=0.$$

The first equation states that the momentum is constant. The second is a constraint, because $e$ has no derivative term. For $m>0$ on the positive-einbein branch,

$$e=\frac{\sqrt{-\dot X^2}}{m}.$$

Substituting this expression into the action gives

$$\frac12\left(\frac{\dot X^2}{e}-em^2\right) =-m\sqrt{-\dot X^2}.$$

The two actions therefore describe the same classical massive particle. The same equation also gives the precise relation between their invariant measures:

$$ds=m e\,d\lambda, \qquad \ell=m\int_0^1e\,d\lambda.$$

These relations use the einbein equation of motion. They do not identify the auxiliary measure with the length of every off-shell path in a quantum path integral.

Canonical form and the meaning of zero Hamiltonian

The covariant canonical momentum is $P_\mu=e^{-1}\dot X_\mu$. The Legendre transform gives

$$H_{\rm can}=\frac e2(P^2+m^2), \qquad S=\int_0^1d\lambda \left[P_\mu\dot X^\mu-\frac e2(P^2+m^2)\right].$$

In this first-order form $e$ is explicitly a Lagrange multiplier. Its variation imposes

$$P^2+m^2=0.$$

Thus $H_{\rm can}$ vanishes on constrained trajectories. This does not mean that the particle has zero energy: $H_{\rm can}$ generates changes of the arbitrary label $\lambda$, whereas $H_t=\sqrt{\mathbf p^2+m^2}$ generates evolution in the physical target time after temporal gauge has been chosen.

Replacing $P_\mu$ by $-i\partial_\mu$ gives the wave equation

$$(-\Box+m^2)\psi=0, \qquad \Box=\eta^{\mu\nu}\partial_\mu\partial_\nu.$$

This is the Klein–Gordon equation. The constraint alone does not choose a Green function or its boundary condition; the propagator will require a prescription for the modulus integral.

The massless limit

Setting $m=0$ directly in $-m\int ds$ makes that action identically zero; it cannot supply the massless particle’s equations. By contrast,

$$S_{m=0}=\frac12\int d\lambda\,e^{-1}\dot X^2$$

still gives both $\dot X^2=0$ and $d(e^{-1}\dot X^\mu)/d\lambda=0$. The nontrivial classical trajectories are null straight lines in flat spacetime. One cannot solve for $e$ by dividing by $m$, but the constraint and gauge symmetry remain meaningful. In particular, a null trajectory has zero spacetime proper time without forcing its positive einbein to vanish.

Gauge fixing leaves one modulus

Return to the fixed parameter interval $[0,1]$. For any smooth $e(\lambda)>0$, define

$$\mathcal T=\int_0^1d\lambda\,e(\lambda), \qquad f(\lambda)=\frac{1}{\mathcal T}\int_0^\lambda du\,e(u).$$

Then $f(0)=0$, $f(1)=1$, and $f'=e/\mathcal T>0$. The transformed einbein is constant:

$$e'(\lambda')=\frac{e}{f'}=\mathcal T.$$

This explicit map shows exactly what gauge fixing removes. It removes the variation of $e$ along the interval, but preserves its invariant integral $\mathcal T$. No endpoint-preserving reparametrization can change that integral. Equivalently, the coordinate $u=\int_0^\lambda e(v)\,dv$ sets the einbein to one but changes the coordinate interval to $0\leq u\leq\mathcal T$; the same modulus survives in its endpoint.

An endpoint-preserving change of parameter replaces $e(\lambda)>0$ by $e'=\mathcal T$, with $\mathcal T=\int_0^1e\,d\lambda$. The auxiliary measure and the geometric worldline are unchanged. The drawing is schematic; label spacing is a coordinate choice, not a physical velocity measurement.

The constant-einbein action is

$$S[X;\mathcal T]=\frac{1}{2\mathcal T}\int_0^1d\lambda\,\dot X^2 -\frac{m^2\mathcal T}{2}.$$

The parameter $\mathcal T$ is often called proper time in the Schwinger representation. Its dimension is $L^2$, and it equals $\ell/m$ only when the massive timelike einbein constraint holds. We will call it the Schwinger modulus to keep it distinct from the spacetime proper time $\ell$.

Classically one must retain the constraint derived before gauge fixing. Fixing $e$ and varying only $X$ would give a straight-line equation but lose the mass-shell condition. Quantum mechanically the positive modulus remains an integration variable after the local gauge redundancy has been removed.

A fixed modulus gives the heat kernel

To calculate the free Gaussian integral, continue to Euclidean target coordinates and a positive Euclidean einbein. The oscillatory Lorentzian integral is defined by analytic continuation with its causal prescription; rotating a target coordinate alone is not a claim that a Lorentzian Gaussian becomes convergent for every real einbein. The resulting Euclidean action in constant gauge is

$$S_E[X;\mathcal T] =\frac{1}{2\mathcal T}\int_0^1d\lambda\,\dot X_E^2 +\frac{m^2\mathcal T}{2}, \qquad \mathcal T>0.$$

For endpoints $x_i,x_f\in\mathbb R^D$, the fixed-modulus kernel is

$$K(x_f,x_i;\mathcal T) =\int_{X(0)=x_i}^{X(1)=x_f}\mathcal D X\, e^{-S_E[X;\mathcal T]}.$$

Here the measure is normalized to give the identity kernel as $\mathcal T\to0^+$. The particle is free, so this normalization and the Gaussian calculation determine $K$ completely.

Separate the classical path from fluctuations

Write $\Delta x=x_f-x_i$ and

$$X(\lambda)=x_i+\lambda\Delta x+\eta(\lambda), \qquad \eta(0)=\eta(1)=0.$$

The cross term vanishes because $\int_0^1\dot\eta\,d\lambda=0$. Hence

$$S_E=\frac{|\Delta x|^2}{2\mathcal T} +\frac{m^2\mathcal T}{2} +\frac{1}{2\mathcal T}\int_0^1d\lambda\,\dot\eta^2.$$

This already determines the endpoint dependence. To fix the fluctuation prefactor without a formal infinite product, use the momentum eigenstates of the Euclidean Hamiltonian $\widehat H=(-\nabla^2+m^2)/2$:

$$K(x_f,x_i;\mathcal T) =\langle x_f|e^{-\mathcal T\widehat H}|x_i\rangle =\int\frac{d^Dp}{(2\pi)^D} e^{ip\cdot\Delta x-\mathcal T(p^2+m^2)/2}.$$

Each momentum integral is elementary:

$$\int_{-\infty}^{\infty}\frac{dp}{2\pi} e^{-\mathcal T p^2/2+ip a} =\frac{1}{\sqrt{2\pi\mathcal T}} e^{-a^2/(2\mathcal T)}.$$

Multiplying the $D$ factors gives

$$K(x_f,x_i;\mathcal T) =\frac{1}{(2\pi\mathcal T)^{D/2}} \exp\left[-\frac{|\Delta x|^2}{2\mathcal T} -\frac{m^2\mathcal T}{2}\right].$$

The result has dimension $L^{-D}$, as a position-space kernel should. It is the massless diffusion kernel multiplied by $e^{-m^2\mathcal T/2}$.

The differential equation fixes the interpretation

Differentiating the Fourier representation gives

$$\left(\partial_{\mathcal T}-\frac12\nabla_f^2 +\frac{m^2}{2}\right)K=0, \qquad K(x_f,x_i;0^+)=\delta^{(D)}(x_f-x_i).$$

The second equality is a distributional limit: integrating the narrow Gaussian against a smooth test function returns that function at $x_i$. Its total integral is

$$\int d^Dx_f\,K(x_f,x_i;\mathcal T) =e^{-m^2\mathcal T/2}.$$

Thus a massive kernel does not conserve the integral of the diffusion density. The kernels nevertheless obey the composition law

$$\int d^Dy\,K(x_f,y;\mathcal T_2) K(y,x_i;\mathcal T_1) =K(x_f,x_i;\mathcal T_1+\mathcal T_2),$$

which follows either from multiplication of $e^{-\mathcal T\widehat H}$ or from a Gaussian convolution.

Integrating the modulus gives the propagator

A fixed-modulus kernel is not yet a relativistic Green function. For $m^2>0$, the Euclidean inverse of $-\nabla^2+m^2$ is

$$G_E(x_f,x_i)=\frac12\int_0^\infty d\mathcal T\, K(x_f,x_i;\mathcal T).$$

In momentum space the statement is simply

$$\frac12\int_0^\infty d\mathcal T\, e^{-\mathcal T(p^2+m^2)/2} =\frac{1}{p^2+m^2}.$$

The factor $1/2$ is required by the $1/2$ in the Hamiltonian; omitting it doubles the Green function. If we instead use $t_{\rm S}=\mathcal T/2$, the formula takes the familiar form $A^{-1}=\int_0^\infty dt_{\rm S}\,e^{-t_{\rm S}A}$ with $A=-\nabla^2+m^2$. Both $t_{\rm S}$ and $\mathcal T$ have units of length squared. This operator identity fixes the normalization of the gauge-fixed interval path integral. Worldline representations of propagators and effective actions are developed in Schubert, §§1 and 3.

The source equation is an especially useful check. Since $(-\nabla_f^2+m^2)K=-2\partial_{\mathcal T}K$,

$$\begin{aligned} (-\nabla_f^2+m^2)G_E &=-\int_0^\infty d\mathcal T\, \partial_{\mathcal T}K\\ &=K(0^+)-K(\infty) =\delta^{(D)}(x_f-x_i). \end{aligned}$$

The short-modulus endpoint produces the delta-function source. The large-modulus boundary term vanishes for positive $m^2$; massless limits require an infrared check. At separated points the massless integral converges for $D>2$, whereas in $D\leq2$ it needs an infrared prescription. Coincident-point ultraviolet singularities are understood distributionally rather than as finite values of $G_E$.

For the free scalar in Euclidean space, $K(\mathcal T)$ is the kernel of $e^{-\mathcal T A/2}$, where $A=-\nabla^2+m^2$. The integral $\frac12\int_0^\infty d\mathcal T$ gives $A^{-1}$, and the boundary value $K(0^+)=\delta$ supplies the Green-function source. Positive $m^2$ ensures decay at large modulus; the diagram represents operator relations, not a spacetime trajectory.

The Lorentzian causal prescription

Returning to Lorentzian momentum, the Feynman two-point function has the mostly-plus-signature form

$$\widetilde G_F(p)=\frac{-i}{p^2+m^2-i0} =\frac12\lim_{\epsilon\to0^+} \int_0^\infty d\mathcal T\, e^{-i\mathcal T(p^2+m^2-i\epsilon)/2}.$$

The exponent includes $-\epsilon\mathcal T/2$, so the regulator damps large $\mathcal T$ and fixes the pole prescription. Since $p^2=-(p^0)^2+\mathbf p^2$, this is the usual $i/[(p^0)^2-\mathbf p^2-m^2+i0]$.

A propagator is an inverse differential operator with specified boundary conditions, not simply a projection onto on-shell momenta. Integrating an oscillatory exponential over the entire real modulus axis would instead produce a delta function of $p^2+m^2$. The positive half-line and its convergence prescription are therefore part of the physical definition of the propagator.

From a worldline to a worldsheet

The particle construction isolates the steps that will reappear for strings:

Particle	String
Embedding $X^\mu(\lambda)$ of a line	Embedding $X^\mu(\tau,\sigma)$ of a surface
Action proportional to proper length	Nambu–Goto action proportional to area
Independent einbein $e$	Independent worldsheet metric $h_{ab}$
Einbein variation imposes $P^2+m^2=0$	Metric variation imposes worldsheet stress-tensor constraints
Interval gauge fixing leaves $\mathcal T$	Worldsheet gauge fixing can leave moduli depending on topology and insertions

The string has an additional feature: in two dimensions its auxiliary-metric action has Weyl symmetry. The consequences of that symmetry and its quantum anomaly will occupy much of the course. The immediate next step is the Nambu–Goto string and its symmetries.

Exercises

1. Recover the nonrelativistic expansion

Expand the temporal-gauge Lagrangian through fourth order in $|\mathbf v|$. Compute its exact momentum and Hamiltonian, and explain why the constant term in the Lagrangian can be discarded for the classical trajectory while the rest energy still appears in $H_t$.

Solution

Using $\sqrt{1-u}=1-u/2-u^2/8+O(u^3)$,

$$L_t=-m+\frac m2\mathbf v^2+\frac m8(\mathbf v^2)^2 +O(|\mathbf v|^6).$$

Direct differentiation of the unexpanded expression gives

$$\mathbf p=\frac{m\mathbf v}{\sqrt{1-\mathbf v^2}}, \qquad H_t=\mathbf p\cdot\mathbf v-L_t =\frac{m}{\sqrt{1-\mathbf v^2}} =\sqrt{\mathbf p^2+m^2}.$$

Adding $m$ to $L_t$ changes the action by $m(t_f-t_i)$ for fixed endpoints and does not change the Euler–Lagrange equation. It shifts the Hamiltonian by $-m$, choosing kinetic energy rather than total relativistic energy as the zero of energy. The covariant mass-shell relation still involves the total energy $E$ and reads $E^2=\mathbf p^2+m^2$.

2. Make a nonconstant einbein constant

Let $e(\lambda)=e_0(1+a\lambda)$ with $e_0>0$ and $a>-1$ on $[0,1]$. Find the modulus and the endpoint-preserving gauge map, verify the transformed einbein explicitly, and explain why one cannot impose $e'=1$ on the same dimensionless interval for every $e_0$.

Solution

Integration gives

$$\mathcal T=e_0\left(1+\frac a2\right), \qquad f(\lambda)=\frac{\lambda+a\lambda^2/2}{1+a/2}.$$

The endpoint values are $0,1$, and

$$f'(\lambda)=\frac{1+a\lambda}{1+a/2}>0, \qquad \frac{e(\lambda)}{f'(\lambda)} =e_0\left(1+\frac a2\right)=\mathcal T.$$

The integral $\int e\,d\lambda$ is invariant. After a length unit has been chosen, $e'=1$ on $[0,1]$ would fix that integral to one and discard all other moduli; it is not an allowed gauge choice for arbitrary $e_0$. Taking $u=\int_0^\lambda e(v)\,dv$ does produce unit einbein, but the new coordinate has units $L^2$ and interval $[0,\mathcal T]$.

3. A null particle with nonzero modulus

For $m=0$ and constant $e=\mathcal T>0$, solve the classical equations. Show that a nonzero target energy is compatible with a vanishing canonical Hamiltonian and vanishing spacetime proper time.

Solution

The momentum is constant, so

$$X^\mu(\lambda)=x_i^\mu+\mathcal T P^\mu\lambda, \qquad P^2=0.$$

For example, $P^\mu=(E,E,0,\ldots,0)$ with $E>0$ is a nonzero null momentum. Then $\dot X^2=\mathcal T^2P^2=0$ and $\ell=0$, while

$$H_{\rm can}=\frac{\mathcal T}{2}P^2=0, \qquad E=P^0>0.$$

The zero Hamiltonian generates gauge evolution in $\lambda$; the nonzero $E$ generates target-time translations. Nullness places no requirement that $\mathcal T$ vanish. For prescribed endpoints a classical solution of this kind exists only when their separation is null (or zero).

4. Check the composition law

Prove the heat-kernel composition law directly in momentum space, including all Fourier factors and the mass dependence.

Solution

Insert one Fourier representation for each kernel. The integral over the intermediate point is

$$\int d^Dy\,e^{-ip\cdot y+iq\cdot y} =(2\pi)^D\delta^{(D)}(q-p).$$

It removes one momentum integral and one factor $(2\pi)^{-D}$. The remaining expression is

$$\int\frac{d^Dp}{(2\pi)^D} e^{ip\cdot(x_f-x_i)} e^{-(\mathcal T_1+\mathcal T_2)(p^2+m^2)/2},$$

which is precisely $K(x_f,x_i;\mathcal T_1+\mathcal T_2)$. The mass factors multiply to $e^{-m^2(\mathcal T_1+\mathcal T_2)/2}$, as required. The law expresses the sum over every possible intermediate position; no integration over an additional modulus is involved in this fixed-modulus composition.

5. Derive two position-space Green functions

At separation $r=|x_f-x_i|>0$, evaluate $G_E$ for (a) a massless scalar in $D=4$ and (b) a scalar with $m>0$ in $D=3$. For part (b) you may use

$$\int_0^\infty dT\,T^{-3/2}e^{-a/T-bT} =\sqrt{\frac\pi a}\,e^{-2\sqrt{ab}}, \qquad a,b>0.$$

Check the massless limit of the three-dimensional result.

Solution

For $D=4,m=0$,

$$G_E(r)=\frac{1}{8\pi^2} \int_0^\infty\frac{d\mathcal T}{\mathcal T^2} e^{-r^2/(2\mathcal T)}.$$

Set $u=r^2/(2\mathcal T)$. Then $d\mathcal T/\mathcal T^2=-2\,du/r^2$, with reversed endpoints, and

$$G_E(r)=\frac{1}{4\pi^2r^2}\int_0^\infty e^{-u}\,du =\frac{1}{4\pi^2r^2}.$$

For $D=3$, apply the supplied identity with $a=r^2/2$ and $b=m^2/2$:

$$\begin{aligned} G_E(r) &=\frac{1}{2(2\pi)^{3/2}} \sqrt{\frac{2\pi}{r^2}}\,e^{-mr}\\ &=\frac{e^{-mr}}{4\pi r}. \end{aligned}$$

Its massless limit is $1/(4\pi r)$. The dimensions are $L^{-2}$ in four dimensions and $L^{-1}$ in three, agreeing with $[G_E]=L^{2-D}$. Both expressions are singular at $r=0$, where their Green equations hold as distributional identities.

6. Relate the Euclidean saddle to geometric length

For a straight Euclidean path of length $r>0$ and $m>0$, extremize the classical exponent

$$S_{E,\mathrm{cl}}(\mathcal T) =\frac{r^2}{2\mathcal T}+\frac{m^2\mathcal T}{2}.$$

Find the stationary modulus and action. Explain what happens for $m=0$, and why this exercise does not assert that the full modulus integrand is stationary at the same point.

Solution

The derivative vanishes when

$$-\frac{r^2}{2\mathcal T^2}+\frac{m^2}{2}=0, \qquad \mathcal T_* =\frac r m.$$

The second derivative $r^2/\mathcal T^3$ is positive, and substitution gives $S_{E,\mathrm{cl}}(\mathcal T_*)=mr$. Thus $r=m\mathcal T_*$, the Euclidean counterpart of the massive length relation. The minimum is also immediate from the arithmetic–geometric mean inequality applied to the two positive terms.

For $m=0$ and $r>0$, the exponent decreases toward zero as $\mathcal T\to\infty$ and has no finite stationary point. The full kernel contains the fluctuation factor $\mathcal T^{-D/2}$, so the negative logarithm of its modulus integrand also contains $(D/2)\log(\mathcal T/\mathcal T_0)$, where a fixed reference modulus $\mathcal T_0$ only changes an irrelevant additive constant. Its stationary point differs from that of the classical exponent. The geometric saddle controls a semiclassical approximation when $mr$ is large, not an exact localization of the quantum integral.

References

• Schubert, Christian. “Perturbative Quantum Field Theory in the String-Inspired Formalism.” Physics Reports 355 (2001), 73–234. arXiv:hep-th/0101036; doi:10.1016/S0370-1573(01)00013-8. Sections 1 and 3 explain worldline path integrals and Schwinger representations; the loop applications go beyond this page’s open interval.
• Tong, David. Lectures on String Theory. University of Cambridge Part III lecture notes, 2009. arXiv:0908.0333, §1.1, especially §§1.1.1–1.1.2 on quantization and the einbein.