On the Euclidean version of the photon number integral

The photon number integral¹ encodes the average number of photons radiated by a charged particle moving on a prescribed timelike trajectory in Minkowski space-time. Its Euclidean analog is well defined for any smooth closed curve $C$ without self-intersections. For such curves, we shall study it in terms of a given parametrization,

where the coordinates $xj(t)$ are smooth $2π$-periodic functions on the real line, and the tangent $ẋ(t)$ does not vanish for any $t∊[−π,π)$⁠. The integral is now given by

Here, the “transverse tangent” $ẋT(tj)$ stands for the tangent at the point $x(tj)$ with the component along the direction to the other point removed. (This prescription arises from the transverse polarization of photons.¹) That is, if

is the difference vector connecting the two points, then

where

It readily follows from this that $∣ẋT(tj)∣$ is proportional to $S(t1,t2)$ as $t1−t2→0$⁠. Hence, the integrand remains bounded for $t1−t2→0$⁠. It is also easy to see that the integral does not depend on the particular parametrization chosen. Specifically, if

is a diffeomorphism of the circle $S1$⁠, then the integral takes the same value when $x(t)$ is replaced by $x(ϕ(t))$⁠. To express this invariance, one can also write

but we shall use (1) throughout.

Some more invariance properties are easily verified, namely, scale invariance and invariance under the Euclidean group (translations and rotations). It is not at all obvious, however, whether the integral is invariant under the inversion

assuming the origin is not on the curve. Even so, this is true, as first argued in Ref. 2. One purpose of this paper is to reconsider this property, for which we supply a rigorous proof.

At this point, we would like to mention the two references,^3,4 where in particular the conformal invariance of similar integrals is studied. The integrands in these papers differ from that of $nC$ (and from each other) in that the divergent behavior of $ẋ1⋅ẋ2∕S2$ for $t1−t2→0$ is regularized by making different subtractions.

The main new result of this paper is an explicit evaluation of integral (2) for the ellipse. Indeed, thus far it was only possible to evaluate it explicitly for the circle, where it yields the value $2π2$⁠. Some numerical evaluations for the ellipse were given in Ref. 5, and these agree with the closed formula,

where $a$ and $b$ are the major and minor axes.

We present the proof of this formula in Sec. II. In Sec. III, we reobtain invariance under inversion (hence, invariance under the conformal group). In Sec. IV, we study the behavior of the integral in $RN$⁠, $N>2$⁠, when the distance between two parts of the curve goes to 0. Depending on the angle under which a self-intersection develops, we find that the local contribution to the integral can diverge to plus or minus ∞. It follows, in particular, from this result that for $N>2$⁠, there is no lower bound on the integral $nC$⁠. Thus, $nC$ cannot be used to study knots and links in the same way as in Ref. 3. (The key difference is that the regularization of $ẋ1⋅ẋ2∕S2$ in Ref. 3 yields a positive integrand that diverges to ∞ whenever a self-intersection develops.)

We study the ellipse in the form

An obvious parametrization of form (1) is then in terms of polar coordinates,

To evaluate the ellipse integral, it is convenient to employ an alternative way of writing (2) for a curve in $R3$⁠. Clearly, this can then be used for any plane curve by embedding $R2$ into $R3$ in the obvious way, namely, by setting $z(t)=0$⁠. The point is that we can apply the vector identity

to obtain

Comparing the right-hand side to (4), we see it equals $(ẋ1T⋅ẋ2T)∣Δ∣2$⁠. Thus, (2) can also be written as

Returning to the above ellipse, we have

while for $Δ$⁠, we get

where we used some well-known trigonometric identities. Thus, we obtain

with a factorization in terms of sum and difference angles.

Since all vectors are in the $(x,y)$-plane, the cross products are in the $z$-direction. In particular, from (11), we deduce

In our parametrization, therefore, the “angular momentum” (with respect to the origin—not to the focus of the ellipse) is constant. Likewise, we obtain

Therefore, the left-hand side of (13) becomes

Substituting this and (17) in (14), we finally obtain

It remains to calculate the integral on the right-hand side of (21). Its integrand is smooth and $2π$ periodic in $t1$ and $t2$⁠, so we can transform to sum and difference variables to get

where

The integral $J$ can be readily calculated via integrals we have occasion to invoke in Sec. IV, too. We first note that it can be rewritten as

where

Now $T(β)$ can be calculated by a contour integration, the result being

From (24) and (25), we then infer

and substituting this in (22), we obtain the explicit result,

This is the new result (9) announced above. We proceed to comment on its features. It depends on the dimensionless ratio $ξ$ and also exhibits symmetry under $ξ→ξ−1$⁠, i.e., the exchange of $a$ and $b$⁠. The dependence on the ratio reflects the dimensionless character of $n$ and that it only depends on the shape of a curve and not on its absolute scale. The exchange of $a$ and $b$ is equivalent to the rotation of the ellipse by 90° and so invariance under this exchange was to be expected on grounds of rotational invariance.

It is a plausible conjecture that for any plane curve the minimum value of $n$ obtains for the circle, and it is indeed true that the minimum of (29) holds for $ξ=1$⁠. The indefinite increase in $n$ as the ellipse becomes more and more eccentric manifests the increased “radiation” from the acute ends. (We stress, however, that since here we are in Euclidean space the notion of “photon” should not be taken too literally.)

Finally, in Ref. 5 $n$ was evaluated numerically for ellipses of various eccentricities $ϵ$⁠. Taking into account that $ξ=(1−ϵ2)1∕2$⁠, we find agreement between the numerical calculations and (29).

Consider a circle $C$ in $R2$ with center $c=(x0,y0)$ and radius $R$⁠, which does not pass through the origin. It is given by the equation

with the difference

being nonzero. Substituting

in (30), we obtain the equation

In the degenerate case $d=0$⁠, this yields a line. For $d≠0$⁠, one easily verifies that (33) describes a circle $C̃$ with center $c̃$ and radius $R̃$ given by

The upshot of this calculation is that the inversion of the circle $C$ yields a new circle $C̃$⁠, assuming $C$ does not pass through the origin. In this case, therefore, one obtains the same value $2π2$ for the curve and its image under inversion.

Next, consider the ellipse $E$ given by (10) with $a>b$⁠. Its image $Ẽ$ under inversion (32) is given by the equation

Thus, the new curve $Ẽ$ is not an ellipse. Rather, (35) yields a so-called limaçon. At face value, it would seem untractable to calculate $n$ for this curve, but in fact this can be done, the answer being the same as for the ellipse $E!$

The reason why this is true is that this amounts to a special case of the invariance under inversion for any curve $C$ in the class defined in the Introduction, it being assumed in addition that $C$ does not pass through the origin. This striking property was first observed in Ref. 2. We proceed to reobtain this feature rigorously, the analysis in Ref. 2 being a bit formal. The key algebraic identities can be found in Ref. 2, however. In our given parametrization (1), the first one reads

where $S$ is given by (5). Its validity can be established by a straightforward calculation using in particular,

The point of identity (36) is that it expresses the integrand of $n$ as a function of $lnS$ only. Thus, one needs to only note the behavior of $lnS$ under the inversion $I$ [cf. (8)],

and calculate $tj$ partials to get

From identity (36), one then concludes

where

The upshot is that the invariance of $n$ under inversion is equivalent to vanishing of the $I$-integral.² 

Now the first term in $I$ is of the form $−f′(t1)f′(t2)∕4$⁠, where

Since $∣x(t)∣$ does not vanish by assumption, $f(t)$ is a smooth $2π$-periodic function. Hence, it is immediate that the integral of the first term vanishes.

For the second and third terms one would also be inclined to integrate the partials of $lnS$ directly. However, the analytical difficulty is that they both have a singularity for $t1−t2→0$⁠, so that one cannot integrate these terms without further ado. On the other hand, their sum is a continuous function on $[−π,π]2$⁠, as we shall show shortly. Therefore the integral of the sum can be obtained as the limit of an integral where an $ϵ$ neighborhood of the “diagonal” is omitted. In the latter integral, we are entitled to integrate directly, and then take $ϵ$ to 0. From suitable bounds, it then follows that the limit vanishes.

We proceed to fill in the details of this reasoning. To this end, we need only consider the remaining integrand,

where $K$ is given by

and $Kj$ stands for $∂jK$⁠. To begin with, we have expansions

Using (37), this entails

and likewise

Thus, we obtain

Hence, $R$ is continuous on all of the integration region $[−π,π]2$⁠. [Recall that we are dealing with smooth $2π$-periodic functions $x1(t),…,xN(t)$⁠, so that the above expansions also apply to $ϵ$-neighborhoods of $t1=−π,t2=π$ and $t1=π,t2=−π$⁠, where $Δ$ vanishes too.]

To prove that $R$ has vanishing integral, it therefore suffices to show

where

Thus, we are not only excising $ϵ$-neighborhoods of the diagonal and the corners $(−π,π)$ and $(π,−π)$ of the $(t1,t2)$-square but also two horizontal strips of height $ϵ$ (see Fig. 1). These strips allow us to avoid dealing with the corners and to choose the order of the integrations depending on which of the two terms in $R(t1,t2)$ is in question.

Specifically, we rewrite integral (52) as

The first line refers to the upper left area of the central square in Fig. 1 and the second line to the lower right area. We can now do the inner integrations to get

Rearranging, we obtain

The first integral results from combining the first and third lines of the previous equation. In the second integral, we have arranged for symmetric limits of integration, while the last two integrals are only over $ϵ$-regions.

It remains to estimate the various terms occurring in (56). The integration region of the terms on the third and fourth lines has a measure of $2ϵ$⁠, whereas the integrand is $O(ln(1∕ϵ))$ on the region [cf. (45), (16), and (7)]. Thus the $ϵ→0$ limit of these terms vanishes. Also, from (47) and a similar expansion, we deduce

and from

we get

Recalling $f′(t)$ is smooth, it now follows that the limits of the terms on the first and second lines also vanish. Thus, the $I$-integral vanishes. We have therefore completed our proof of the invariance of $nC$ under inversion, assuming the curve $C$ does not pass through the origin (the center of inversion).

In this section, we take $N>2$ and study what happens when we let the distance $d$ between two parts of the non-self-intersecting curve $C$ go to zero. Since the tangents to the two parts are already transverse at the points of minimal distance, their inner product does not generally vanish as $d$ goes to zero. However, the distance $S=d$ does go to zero, so the integrand of $n$ diverges for the two points of closest approach.

Of course, this is a priori compatible with the contribution of the two parts to the $n$-integral remaining finite. Here, we analyze what happens under the simplifying assumption that the angle $ϕ$ between the tangents at the points $P1$ and $P2$ of closest approach remain constant as the distance $d$ between $P1$ and $P2$ vanishes. Since we are dealing with smooth curves and we are studying a local behavior, we may also assume that the two parts are straight. Finally, using a change of parameters and an eventual rotation and translation, we may and will study the case of two line pieces in $R3$ given by

so that $∣t1∣$ and $∣t2∣$ are the distances from

Since we now work in $R3$⁠, we can use the vector product form of the integrand featuring in (14). As we have

this readily yields

Now we have two distances $d$ and $t0$ in our problem, and the $n$-integral is dimensionless (scale invariant). Therefore, the latter can only depend on the ratio

Indeed, from the change of variables,

we get the integrand

We are concerned with the divergence or convergence behavior of the integral of $Mμ,ϕ$ over $[−1,1]2$ as $μ$ vanishes, and since the only possible divergence of the integrand arises from the origin, we may as well study the integral over the unit disk in the $(t,t′)$-plane. The crux of this is that we can then pass to polar coordinates and use integrals occurring in Sec. II to do the angular integration explicitly. This yields an integral over the radius whose behavior for $μ→0$ can be easily determined. The details now follow.

To start with, the integral over the unit disk is given by

where $A$ is the angular integral,

Next, we rewrite (69) as

where we have introduced

Recalling (26), we now see that we have

It is easy to check that evaluation (27) of $T(β)$ is also valid for $α∊(−β,0]$⁠. Using it, we get

If we now substitute (71) and simplify the result, we obtain

where

The upshot is that the unit disk integral (68) is given by

Thus, for $μ→0$⁠, we get the asymptotic behavior,

As a consequence, we obtain divergence to ∞ for $∣ϕ∣∊[0,π∕2)$ and divergence to $−∞$ for $∣ϕ∣∊(π∕2,π]$⁠. Note that it is already clear from (69) that when the line pieces are orthogonal, the local contribution to $n$ vanishes identically in $μ$⁠.

The results reported in this paper were obtained during a stay of S.R. at the Max-Planck-Institute for Physics in Munich (Heisenberg Institute). He would like to thank the Institute for its hospitality and financial support, and E. Seiler for his invitation and for useful discussions.