Special conformal transformations in electrodynamics: mathematical foundations

Abstract

We review the theory of special conformal transformations, deriving the transformation formulas for some important tensors and tensor densities, to the point where we are able to understand the conformal invariance of Maxwell’s equations. The invariant relativistic action reveals that the rest mass of a particle is a scalar density, while its electric charge is a scalar.

Keywords:
Weyl space; special conformal transformations; electrodynamics

1. Introduction

The Special Relativity Theory has made it clear that Maxwell’s equations are invariant under rotations (homogeneous Lorentz transformations [¹]) and translations of the reference frame. The Lorentz group, as well as the larger Poincaré group, conserve line elements $d{s}^2=d{x}^2+d{y}^2+d{z}^2+{(icdt)}^2$. Cunningham [²] and Bateman [³, ⁴] have discovered that in flat Minkowski space Maxwell’s equations are also invariant under the transformations of the larger conformal group. The conformal group is the largest group that conserves null line elements $d{s}^2=d{x}^2+d{y}^2+d{z}^2+{(icdt)}^2=0$, and consists of spacetime translations (1), proper and improper homogeneous Lorentz transformations (2), dilatation (or scale) transformations (3), and special conformal (or acceleration) transformations (4) [⁵,⁶,⁷,⁸]

(1) $${x^{\prime }}^{\mu }=x^{\mu }+t^{\mu },$$

(2) $${x^{\prime }}^{\mu }=M_{\nu }^{\mu }{x}^{\nu },$$

(3) $${x^{\prime }}^{\mu }=s{x}^{\mu },$$

(4) $${x^{\prime }}^{\mu }=\frac{x^{\mu }+b^{\mu }{x}^2}{1+2b\cdot x+b^2{x}^2},$$

where $b\cdot x=g_{\alpha \beta }{b}^{\alpha }{x}^{\beta }$, $b^2=g_{\alpha \beta }{b}^{\alpha }{b}^{\beta }$, and $x^2=g_{\alpha \beta }{x}^{\alpha }{x}^{\beta }$. We are concerned with conformal transformations restricted to the flat Minkowski space $(x,y,z,ict)$, where the metric tensor is given by the Kronecker delta

(5) $$(g_{\alpha \beta })=(g^{\alpha \beta })=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right).$$

Equivalently, one could drop the imaginary unit from the time coordinate and use metric tensors of signature $(+,+,+,-)$ or $(-,-,-,+)$.

The special conformal transformation (4) can be decomposed into an inversion in a hypersphere of radius L and center at the origin (6), followed by a translation by a constant vector $L^2{b}^{\mu }$ (7), and then a second conformal inversion in a hypersphere of the same radius and center at the new origin (8)

(6) $${x^{″}}^{\mu }=L^2\frac{x^{\mu }}{x^2},$$

(7) $${x^{‴}}^{\mu }={x^{″}}^{\mu }+L^2{b}^{\mu },$$

(8) $${x^{\prime }}^{\mu }=L^2\frac{{x^{‴}}^{\mu }}{{x^{‴}}^2},$$

where L has units of length and $b^{\mu }$ has reciprocal units of length.

It is also possible to decompose an inversion into a translation, a reflection (an improper Lorentz transformation), a special conformal transformation, and the opposite translation [⁸].

2. The Transformation of the Line Element

Consider the special conformal transformation (4) with the translation vector $L^2{b}^{\mu }$. The inverse special conformal transformation has the translation vector in the opposite direction, $b_{inv}^{\mu }=-b^{\mu }$, which gives

(9) $$x^{\mu }=\frac{{x^{\prime }}^{\mu }-b^{\mu }{x^{\prime }}^2}{1-2b\cdot x^{\prime }+b^2{x^{\prime }}^2}.$$

In order to prove this, the scalar products $b\cdot x^{\prime }$ and ${x^{\prime }}^2$ are calculated as

(10) $$b\cdot x^{\prime }=g_{\alpha \beta }{b}^{\alpha }{x^{\prime }}^{\beta }=\frac{b\cdot x+b^2{x}^2}{1+2b\cdot x+b^2{x}^2},$$

(11) $$x^{\prime 2}=g_{\alpha \beta }{x}^{\prime \alpha }{x}^{\prime \beta }=\frac{x^2}{1+2b\cdot x+b^2{x}^2},$$

and then are substituted into the right side of (9), together with the expression of ${x^{\prime }}^{\mu }$ from equation (4).

From equation (9) it follows that

(12) $$x^2=g_{\alpha \beta }{x}^{\alpha }{x}^{\beta }=\frac{{x^{\prime }}^2}{1-2b\cdot x^{\prime }+b^2{x^{\prime }}^2},$$

and then, from equations (11) and (12), one can see right away that

(13) $$(1+2b\cdot x+b^2{x}^2)(1-2b\cdot x^{\prime }+b^2{x^{\prime }}^2)=1.$$

Under the special conformal transformation (4) the coordinates of a second point y transform according to

(14) $${y^{\prime }}^{\mu }=\frac{y^{\mu }+b^{\mu }{y}^2}{1+2b\cdot y+b^2{y}^2},$$

and we also have

(15) $${y^{\prime }}^2=g_{\alpha \beta }{y^{\prime }}^{\alpha }{y^{\prime }}^{\beta }=\frac{y^2}{1+2b\cdot y+b^2{y}^2}.$$

The scalar product $x^{\prime }\cdot y^{\prime }$ is calculated as

(16) $$\begin{array}{c}{x}^{\prime }\cdot y^{\prime }=g_{\alpha \beta }{x^{\prime }}^{\alpha }{y^{\prime }}^{\beta }\\ =\frac{x\cdot y+b\cdot x{y}^2+b\cdot y{x}^2+b^2{x}^2{y}^2}{(1+2b\cdot x+b^2{x}^2)(1+2b\cdot y+b^2{y}^2)},\end{array}$$

and the distance between the two points

(17) $${(x-y)}^2=g_{\alpha \beta }(x^{\alpha }-y^{\alpha })(x^{\beta }-y^{\beta })=x^2-2x\cdot y+y^2$$

transforms according to

(18) $$\begin{array}{c}{(x^{\prime }-y^{\prime })}^2=g_{\alpha \beta }({x^{\prime }}^{\alpha }-{y^{\prime }}^{\alpha })({x^{\prime }}^{\beta }-{y^{\prime }}^{\beta })\\ \text{=}{x^{\prime }}^2-2x^{\prime }\cdot y^{\prime }+{y^{\prime }}^2\\ \text{=}\frac{x^2}{1+2b\cdot x+b^2{x}^2}\\ -2\frac{x\cdot y+b\cdot x{y}^2+b\cdot y{x}^2+b^2{x}^2{y}^2}{(1+2b\cdot x+b^2{x}^2)(1+2b\cdot y+b^2{y}^2)}\\ +\frac{y^2}{1+2b\cdot y+b^2{y}^2}\\ \text{=}\frac{{(x-y)}^2}{(1+2b\cdot x+b^2{x}^2)(1+2b\cdot y+b^2{y}^2)}.\end{array}$$

Equation (18) shows that, under special conformal transformations, a light cone transforms into a light cone. Any two space-time points connected by a light signal, for which ${(x-y)}^2=0$, will remain connected by a light signal, because ${(x^{\prime }-y^{\prime })}^2=0$ as well after the transformation.

Finally, in the limiting case when x and y are separated by an infinitesimal d istance, ${(x-y)}^2={(ds)}^2$, ${(x^{\prime }-y^{\prime })}^2={(d{s}^{\prime })}^2$, and we obtain [⁹]

(19) $$(ds\text{'})=\frac{{(ds)}^2}{{(1+2b\cdot x+b^2{x}^2)}^2},$$

from which, assuming that $1+2b\cdot x+b^2{x}^2>0$, we get the formula

(20) $$d{s}^{\prime }=\frac{ds}{1+2b\cdot x+b^2{x}^2},$$

which clearly shows that only the length of infinitesimal null line elements is conserved by special conformal transformations. Although the length of infinitesimal segments is not conserved in general, ratios of such segments at the same spacetime point stay the same, which means that (locally) angles are conserved. This is the origin of the “conformal” name given to these transformations.

3. The Need for a Metric Tensor Density

When calculating the scalar products in equations (10), (11), (15), (16), and (18) we have used the same metric tensor (5), essentially making the hidden assumption that ${g^{\prime }}_{\alpha \beta }=g_{\alpha \beta }$. In other words, as Rohrlich noticed, “in using conformal transformations one forces a Minkowski space onto a noninertial observer” [¹⁰].

However, if we look at equation (4) as a coordinate transformation within the framework of General Relativity, we will get a different m etric t ensor. I n General Relativity the metric tensor transforms like

(21) $${g^{\prime }}_{\mu \nu }^{(GR)}(x^{\prime })=\frac{\partial x^{\alpha }}{\partial {x^{\prime }}^{\mu }}\frac{\partial x^{\beta }}{\partial {x^{\prime }}^{\nu }}{g}_{\alpha \beta }(x).$$

The infinitesimal displacements always transform like contravariant vectors

(22) $$d{x^{\prime }}^{\mu }=\frac{\partial {x^{\prime }}^{\mu }}{\partial x^{\sigma }}d{x}^{\sigma },$$

(23) $$d{x^{\prime }}^{\nu }=\frac{\partial {x^{\prime }}^{\nu }}{\partial x^{\rho }}d{x}^{\rho }.$$

From (21)-(23), it follows that the length of the line element is invariant

(24) $$\begin{array}{c}{(d{s^{\prime }}^{(GR)})}^2={g^{\prime }}_{\mu \nu }^{(GR)}(x^{\prime })d{x^{\prime }}^{\mu }d{x^{\prime }}^{\nu }\\ =\frac{\partial x^{\alpha }}{\partial {x^{\prime }}^{\mu }}\frac{\partial x^{\beta }}{\partial {x^{\prime }}^{\nu }}{g}_{\alpha \beta }(x)\frac{\partial {x^{\prime }}^{\mu }}{\partial x^{\sigma }}d{x}^{\sigma }\frac{\partial {x^{\prime }}^{\nu }}{\partial x^{\rho }}d{x}^{\rho }\\ =\frac{\partial x^{\alpha }}{\partial x^{\sigma }}\frac{\partial x^{\beta }}{\partial x^{\rho }}{g}_{\alpha \beta }(x)d{x}^{\sigma }d{x}^{\rho }\\ ={\delta }_{\sigma }^{\alpha }{\delta }_{\rho }^{\beta }{g}_{\alpha \beta }(x)d{x}^{\sigma }d{x}^{\rho }\\ =g_{\sigma \rho }(x)d{x}^{\sigma }d{x}^{\rho }={(ds)}^2,\end{array}$$

a well known result that is very different from what we have in (19).

In order to understand this apparent paradox, starting from equation (9) we calculate

(25) $$\begin{array}{}\begin{array}{c}\frac{\partial x^{\alpha }}{\partial {x^{\prime }}^{\mu }}=\frac{{\delta }_{\mu }^{\alpha }-2b^{\alpha }{x^{\prime }}_{\mu }}{1-2b\cdot x^{\prime }+b^2{x^{\prime }}^2}\\ -\frac{({x^{\prime }}^{\alpha }-b^{\alpha }{x^{\prime }}^2)(-2b_{\mu }+2{x^{\prime }}_{\mu }{b}^2)}{{(1-2b\cdot x^{\prime }+b^2{x^{\prime }}^2)}^2},\end{array}\end{array}$$

(26) $$\begin{array}{c}\frac{\partial x^{\beta }}{\partial x^{\prime \nu }}=\frac{{\delta }^{\beta }{}_{\nu }-2b^{\beta }{x}_{\nu }^{\prime }}{1-2b\cdot x^{\prime }+b^2{x}^{\prime 2}}\\ -\frac{\left(x^{\prime \beta }-b^{\beta }{x}^{\prime 2}\right)\left(-2b_{\nu }+2x_{\nu }^{\prime }{b}^2\right)}{{\left(1-2b\cdot x^{\prime }+b^2{x}^{\prime 2}\right)}^2}.\end{array}$$

Direct substitution of (25), (26), and (5) into (21) produces, after some lengthy algebraic calculations, a very simple result

(27) $$\begin{array}{c}{g^{\prime }}_{\mu \nu }^{(GR)}(x^{\prime })=\frac{g_{\mu \nu }}{{(1-2b\cdot x^{\prime }+b^2{x^{\prime }}^2)}^2}\\ ={(1+2b\cdot x+b^2{x}^2)}^2{g}_{\mu \nu },\end{array}$$

which clearly shows that the ${g^{\prime }}_{\alpha \beta }=g_{\alpha \beta }$ assumption does not hold true.

In order to save the ${g^{\prime }}_{\alpha \beta }=g_{\alpha \beta }$ assumption, we have to replace the transformation law (21) with [⁷, ⁸]

(28) $${g^{\prime }}_{\mu \nu }(x^{\prime })=\frac{1}{{(1+2b\cdot x+b^2{x}^2)}^2}\frac{\partial x^{\alpha }}{\partial {x^{\prime }}^{\mu }}\frac{\partial x^{\beta }}{\partial {x^{\prime }}^{\nu }}{g}_{\alpha \beta }(x),$$

which means that we have to replace the pseudo-Riemannian space of General Relativity with the Weyl space of Conformal Relativity. In this conformally flat space a tensor density of weight N transforms according to

(29) $${T^{\prime }}_{\nu }^{\mu }=W^N\frac{\partial {x^{\prime }}^{\mu }}{\partial x^{\alpha }}\frac{\partial x^{\beta }}{\partial {x^{\prime }}^{\nu }}{T}_{\beta }^{\alpha },$$

where W is the Jacobian determinant of the coordinate transformation from x to $x^{\prime }$. Scalars, vectors, and higher rank tensors have zero weight.

The Jacobian determinant of the special conformal transformation (4) is

(30) $$\begin{array}{c}W=\det \left(\frac{\partial x}{\partial x^{\prime }}\right)=\frac{1}{{(1-2b\cdot x^{\prime }+b^2{x^{\prime }}^2)}^4}\\ ={(1+2b\cdot x+b^2{x}^2)}^4.\end{array}$$

As discussed by Nicholas Wheeler [⁸], the easiest way to calculate the Jacobian W is to multiply the Jacobian W₁ of the inversion (6), the Jacobian W₂ of the translation (7), and the Jacobian W₃ of the inversion (8).

Now we can write (28) as

(31) $${g^{\prime }}_{\mu \nu }(x^{\prime })=W^{-\frac{1}{2}}\frac{\partial x^{\alpha }}{\partial {x^{\prime }}^{\mu }}\frac{\partial x^{\beta }}{\partial {x^{\prime }}^{\nu }}{g}_{\alpha \beta }(x),$$

which shows that in Conformal Relativity the covariant metric tensor density g_αβ has a weight of $-1/2$.

Suppose that we start with a covariant tensor component $S_{\mu }$. If we raise an index, $S^{\alpha }=S_{\mu }{g}^{\mu \alpha }$, and then we lower the same index, $S_{\nu }=S^{\alpha }{g}_{\alpha \nu }$, we get back the same tensor component. As a result

(32) $$g^{\mu \alpha }{g}_{\alpha \nu }={\delta }_{\nu }^{\mu }.$$

The Kronecker delta ${\delta }_{\nu }^{\mu }$ is a universal tensor of weight 0,

(33) $${{\delta }^{\prime }}_{\nu }^{\mu }=\frac{\partial {x^{\prime }}^{\mu }}{\partial x^{\alpha }}\frac{\partial x^{\beta }}{\partial {x^{\prime }}^{\nu }}{\delta }_{\beta }^{\alpha }=\frac{\partial {x^{\prime }}^{\mu }}{\partial x^{\alpha }}\frac{\partial x^{\alpha }}{\partial {x^{\prime }}^{\nu }}=\frac{\partial {x^{\prime }}^{\mu }}{\partial {x^{\prime }}^{\nu }}={\delta }_{\nu }^{\mu },$$

and as a result the contravariant metric tensor density $g^{\alpha \beta }$ seen in equation (32) must have a weight of 1/2

(34) $${g^{\prime }}^{\mu \nu }(x^{\prime })=W^{\frac{1}{2}}\frac{\partial {x^{\prime }}^{\mu }}{\partial x^{\alpha }}\frac{\partial {x^{\prime }}^{\nu }}{\partial x^{\beta }}{g}^{\alpha \beta }(x),$$

a result consistent with ${g^{\prime }}^{\mu \nu }=g^{\mu \nu }$, as can be checked by direct substitution of (5), (30), and

(35) $$\begin{array}{c}\frac{\partial {x^{\prime }}^{\mu }}{\partial x^{\alpha }}=\frac{{\delta }_{\alpha }^{\mu }+2b^{\mu }{x}_{\alpha }}{1+2b\cdot x+b^2{x}^2}\\ -\frac{(x^{\mu }+b^{\mu }{x}^2)(2b_{\alpha }+2x_{\alpha }{b}^2)}{{(1+2b\cdot x+b^2{x}^2)}^2},\end{array}$$

(36) $$\begin{array}{c}\frac{\partial {x^{\prime }}^{\nu }}{\partial x^{\beta }}=\frac{{\delta }_{\beta }^{\nu }+2b^{\nu }{x}_{\beta }}{1+2b\cdot x+b^2{x}^2}\\ -\frac{(x^{\nu }+b^{\nu }{x}^2)(2b_{\beta }+2x_{\beta }{b}^2)}{{(1+2b\cdot x+b^2{x}^2)}^2},\end{array}$$

into (34). An expression equivalent to (35) was reported by Ryder [⁹].

Due to the non-zero weights of the contravariant and covariant metric tensor densities, the contravariant and covariant components of a tensor or tensor density do not transform with the same weights [⁹, ¹¹]. This is because in all tensorial equations the sum of the weights must be the same on both sides. For example, the equations

(37) $$A_{\mu }=g_{\mu \alpha }{A}^{\alpha },$$

(38) $$B^{\mu }=g^{\mu \alpha }{B}_{\alpha },$$

show that, if $A^{\alpha }$ is a contravariant component of a tensor, then $A_{\mu }$ transforms like a covariant component of a tensor density of weight $-1/2$, and if $B_{\alpha }$ is a covariant component of a tensor, then $B^{\mu }$ transforms like a contravariant component of a tensor density of weight $1/2$.

For the same reason, if $A^{\alpha }$ is a contravariant component of a tensor, then the contraction $A^{\alpha }{A}_{\alpha }$ transforms like a scalar density of weight $-1/2$, and if $B_{\alpha }$ is a covariant component of a tensor, then the contraction $B^{\alpha }{B}_{\alpha }$ transforms like a scalar density of weight $1/2$. We can easily prove this, using the formulas $\frac{\partial x^{\alpha }}{\partial {x^{\prime }}^{\mu }}\frac{\partial {x^{\prime }}^{\mu }}{\partial x^{\beta }}=\frac{\partial x^{\alpha }}{\partial x^{\beta }}={\delta }_{\beta }^{\alpha }$ and $\frac{\partial {x^{\prime }}^{\alpha }}{\partial x^{\nu }}\frac{\partial x^{\nu }}{\partial {x^{\prime }}^{\beta }}=\frac{\partial {x^{\prime }}^{\alpha }}{\partial {x^{\prime }}^{\beta }}={\delta }_{\beta }^{\alpha }$.

(39) $$\begin{array}{c}{A^{\prime }}^{\mu }{A^{\prime }}_{\mu }={A^{\prime }}^{\mu }{A^{\prime }}^{\nu }{g^{\prime }}_{\mu \nu }\\ =\frac{\partial {x^{\prime }}^{\mu }}{\partial x^{\alpha }}{A}^{\alpha }\frac{\partial {x^{\prime }}^{\nu }}{\partial x^{\beta }}{A}^{\beta }{W}^{-\frac{1}{2}}\frac{\partial x^{\rho }}{\partial {x^{\prime }}^{\mu }}\frac{\partial x^{\sigma }}{\partial {x^{\prime }}^{\nu }}{g}_{\rho \sigma }\\ ={\delta }_{\alpha }^{\rho }{A}^{\alpha }{\delta }_{\beta }^{\sigma }{A}^{\beta }{W}^{-\frac{1}{2}}{g}_{\rho \sigma }\\ =W^{-\frac{1}{2}}{A}^{\alpha }{A}^{\beta }{g}_{\alpha \beta }=W^{-\frac{1}{2}}{A}^{\alpha }{A}_{\alpha }\end{array}$$

(40) $$\begin{array}{c}{B^{\prime }}_{\mu }{B^{\prime }}^{\mu }={B^{\prime }}_{\mu }{B^{\prime }}_{\nu }{g^{\prime }}^{\mu \nu }\\ =\frac{\partial x^{\alpha }}{\partial {x^{\prime }}^{\mu }}{B}_{\alpha }\frac{\partial x^{\beta }}{\partial {x^{\prime }}^{\nu }}{B}_{\beta }{W}^{\frac{1}{2}}\frac{\partial {x^{\prime }}^{\mu }}{\partial x^{\rho }}\frac{\partial {x^{\prime }}^{\nu }}{\partial x^{\sigma }}{g}^{\rho \sigma }\\ ={\delta }_{\rho }^{\alpha }{B}_{\alpha }{\delta }_{\sigma }^{\beta }{B}_{\beta }{W}^{\frac{1}{2}}{g}^{\rho \sigma }\\ =W^{\frac{1}{2}}{B}_{\alpha }{B}_{\beta }{g}^{\alpha \beta }=W^{\frac{1}{2}}{B}_{\alpha }{B}^{\alpha }\end{array}$$

Since the Jacobian is $W={(1+2b\cdot x+b^2{x}^2)}^4$, we recognize the transformation of the square of the line element (19) as a special case of equation (39).

If T_αβ is a covariant component of a tensor, then $T^{\mu \nu }=g^{\mu \alpha }{g}^{\nu \beta }{T}_{\alpha \beta }$ is a contravariant component of a tensor density of weight 1, and the contraction $T_{\alpha \beta }{T}^{\alpha \beta }$ transforms like a scalar density of weight 1.

4. The Invariance of Maxwell’s Equations

When going from Special Relativity to Conformal Relativity, we need to establish which scalars, vectors, and higher rank tensors of Special Relativity remain (weight-less) tensors in Conformal Relativity, and which tensors turn into tensor densities, and with what weights.

Nicholas Wheeler [⁸] has proved that, under unrestricted coordinate transformations, we have the following results:

1. If $\varphi$ is a scalar, then the gradient ${\partial }_m\varphi$ is a covariant vector.

2. If X_m is a covariant vector, then ${\partial }_n{X}_m-{\partial }_m{X}_n$ is an antisymmetric covariant tensor.

3. If X_jk is an antisymmetric covariant tensor, then ${\partial }_i{X}_{jk}+{\partial }_j{X}_{ki}+{\partial }_k{X}_{ij}$ is a covariant tensor.

4. If $X^m$ is a contravariant vector density of weight 1, then the divergence ${\partial }_m{X}^m$ is a scalar density of weight 1.

5. If $X^{mn}$ is an antisymmetric contravariant tensor density of weight 1, then ${\partial }_m{X}^{mn}$ is a contravariant vector density of weight 1.

6. As a consequence of the previous two statements, if $X^{mn}$ is an antisymmetric contravariant tensor density of weight 1, then ${\partial }_m{\partial }_n{X}^{mn}$ is a contravariant scalar density of weight 1. Because ${\partial }_m{\partial }_n$ is symmetric and $X^{mn}$ is antisymmetric, we have ${\partial }_m{\partial }_n{X}^{mn}=0$.

With these results in mind, we look at Maxwell’s equations (in Gaussian units) [¹²]

(41) $${\partial }_{\mu }{F}^{\mu \nu }=-\frac{4\pi }{c}{j}^{\nu },$$

(42) $${\partial }_{\mu }{F}_{\nu \lambda }+{\partial }_{\nu }{F}_{\lambda \mu }+{\partial }_{\lambda }{F}_{\mu \nu }=0,$$

where the field strength $F^{\mu \nu }$ is antisymmetric and $F_{\mu \nu }=g_{\mu \alpha }{g}_{\nu \beta }{F}^{\alpha \beta }$.

As noticed by Nicholas Wheeler [⁸], the unrestricted form invariance under coordinate transformations of Maxwell’s equations is assured when the contravariant electromagnetic field strength $F^{\mu \nu }$ and the contravariant electric current density $j^{\nu }$ are tensor densities of weight 1, and when the covariant field strength F_μν is a tensor. This behavior is in perfect agreement with the $-1/2$ weight of the covariant metric tensor density g_αβ. Since the contravariant current density $j^{\mu }$ transforms like a vector density of weight 1, it follows that the covariant current density $j_{\mu }=g_{\mu \alpha }{j}^{\alpha }$ transforms like a vector density of weight 1/2. In summary, we have

(43) $${F^{\prime }}_{\mu \nu }(x^{\prime })=\frac{\partial x^{\alpha }}{\partial {x^{\prime }}^{\mu }}\frac{\partial x^{\beta }}{\partial {x^{\prime }}^{\nu }}{F}_{\alpha \beta }(x),$$

(44) $${F^{\prime }}^{\mu \nu }(x^{\prime })=W\frac{\partial {x^{\prime }}^{\mu }}{\partial x^{\alpha }}\frac{\partial {x^{\prime }}^{\nu }}{\partial x^{\beta }}{F}^{\alpha \beta }(x),$$

(45) $${j^{\prime }}_{\mu }(x^{\prime })=W^{\frac{1}{2}}\frac{\partial x^{\alpha }}{\partial {x^{\prime }}^{\mu }}{j}_{\alpha }(x),$$

(46) $${j^{\prime }}^{\mu }(x^{\prime })=W\frac{\partial {x^{\prime }}^{\mu }}{\partial x^{\alpha }}{j}^{\alpha }(x).$$

Substitution of (35) and (36) into (44) produces, after some lengthy calculations, a formula that shows how the electromagnetic field changes under special conformal transformations. A very large number of terms disappear because the antisymmetric $F^{\alpha \beta }$ is contracted with the symmetric expressions $x_{\alpha }{x}_{\beta }$, $b_{\alpha }{b}_{\beta }$, and $x_{\alpha }{b}_{\beta }+x_{\beta }{b}_{\alpha }$. The final formula is

(47) $$\begin{array}{c}{F^{\prime }}^{\mu \nu }={\sigma }^2{F}^{\mu \nu }\\ +2\sigma (F^{\mu \alpha }{b}^{\nu }-F^{\nu \alpha }{b}^{\mu })(x_{\alpha }\sigma -b_{\alpha }{x}^2-x_{\alpha }{x}^2{b}^2)\\ +2\sigma (F^{\mu \alpha }{x}^{\nu }-F^{\nu \alpha }{x}^{\mu })(-b_{\alpha }-x_{\alpha }{b}^2)\\ +4\sigma F^{\alpha \beta }{b}_{\alpha }{x}_{\beta }(b^{\mu }{x}^{\nu }-b^{\nu }{x}^{\mu }),\end{array}$$

where $\sigma \equiv 1+2b\cdot x+b^2{x}^2$. This result was first reported by Barut and Haugen [⁷], who have derived equation (47) in a very original way, by using a six-dimensional linear representation of the conformal group.

With the help of equations (43) and (44) we can derive the electromagnetic field produced by an electric charge in hyperbolic motion, starting from the electric field of a source particle at rest, and vice versa [¹³].

From (22), (23), and (43) it follows that $F_{\alpha \beta }d{x}^{\alpha }d{x}^{\beta }$ is a scalar, a conformal invariant.

(48) $$\begin{array}{c}{F^{\prime }}_{\mu \nu }d{x^{\prime }}^{\mu }d{x^{\prime }}^{\nu }=\frac{\partial x^{\alpha }}{\partial {x^{\prime }}^{\mu }}\frac{\partial x^{\beta }}{\partial {x^{\prime }}^{\nu }}{F}_{\alpha \beta }\frac{\partial {x^{\prime }}^{\mu }}{\partial x^{\rho }}d{x}^{\rho }\frac{\partial {x^{\prime }}^{\nu }}{\partial x^{\sigma }}d{x}^{\sigma }\\ ={\delta }_{\rho }^{\alpha }{\delta }_{\sigma }^{\beta }{F}_{\alpha \beta }d{x}^{\rho }d{x}^{\sigma }=F_{\alpha \beta }d{x}^{\alpha }d{x}^{\beta }\end{array}$$

Codirla and Osborn [¹⁴] mention that equation (48) is equivalent to equation (43) that describes the transformation of the electromagnetic field.

From (43)–(44) it follows that $F^{\alpha \beta }{F}_{\alpha \beta }$ is a scalar density of weight 1.

(49) $$\begin{array}{c}{F^{\prime }}^{\mu \nu }{F^{\prime }}_{\mu \nu }=W\frac{\partial {x^{\prime }}^{\mu }}{\partial x^{\alpha }}\frac{\partial {x^{\prime }}^{\nu }}{\partial x^{\beta }}{F}^{\alpha \beta }\frac{\partial x^{\rho }}{\partial {x^{\prime }}^{\mu }}\frac{\partial x^{\sigma }}{\partial {x^{\prime }}^{\nu }}{F}_{\rho \sigma }\\ =W{\delta }_{\alpha }^{\rho }{\delta }_{\beta }^{\sigma }{F}^{\alpha \beta }{F}_{\rho \sigma }=W{F}^{\alpha \beta }{F}_{\alpha \beta }\end{array}$$

This result was also derived by Barut and Haugen [⁷] using a six-dimensional linear representation of the conformal group.

5. The Invariance of the Relativistic Action

The conclusions reached by Nicholas Wheeler can also be reached by looking at the expression of the conformally invariant relativistic action. This action (in Gaussian units) is [¹², ¹⁵]

(50) $$I=-c{\displaystyle \int m_o}ds-\frac{1}{16\pi c}{\displaystyle \int F_{\mu \nu }}{F}^{\mu \nu }{d}^{4}x+\frac{1}{c^2}{\displaystyle \int j_{\mu }}{A}^{\mu }{d}^{4}x.$$

The relativistic action (50) has three terms, each of them being a product of some tensors or tensor densities. In each of these three terms, when we add the weights of all the tensor densities from that product, we get a zero sum. This guarantees the conformal invariance of the relativistic action.

Looking at the first term in (50), since the line element ds transforms like a scalar density of weight $-1/4$, we conclude that the rest mass m_o transforms like a scalar density of weight 1/4. This fact was discovered by Schouten and Haantjes [¹⁶] in 1936.

Since under a general coordinate transformation $d^{4}x=W{d}^4{x}^{\prime }$, it follows that the four-dimensional volume element $d^{4}x$ transforms like a scalar density of weight $-1$.

(51) $$d^4{x}^{\prime }=W^{-1}{d}^{4}x$$

Looking at the second term in (50), since the volume element $d^{4}x$ transforms like a scalar density of weight $-1$, and since the weight of $F^{\mu \nu }$ is one unit higher than that of F_μν, we conclude that the contravariant electromagnetic field strength $F^{\mu \nu }$ transforms like a tensor density of weight 1, and that the covariant F_μν transforms like a tensor.

Since the gradient ${\partial }_{\mu }$ transforms like a covariant vector, and since by definition $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$, we conclude that the covariant four-potential $A_{\mu }$ transforms like a vector, and that the contravariant $A_{\mu }$ transforms like a vector density of weight $1/2$.

Looking at the third term in (50), since the volume element $d^{4}x$ transforms like a scalar density of weight $-1$, and since the contravariant electromagnetic four-potential $A^{\mu }$ transforms like a vector density of weight 1/2, we conclude that the covariant current density $j_{\mu }$ transforms like a vector density of weight 1/2, and that the contravariant $j_{\mu }$ transforms like a vector density of weight 1.

When we have a point particle with electric charge q, the relativistic action (50) becomes [¹⁵]

(52) $$I=-c{\displaystyle \int m_o}ds-\frac{1}{16\pi c}{\displaystyle \int F_{\mu \nu }}{F}^{\mu \nu }{d}^{4}x+\frac{1}{c}{\displaystyle \int q}{A}^{\mu }d{x}_{\mu }.$$

We know that $d{x}^{\mu }$ transforms like a contravariant vector. Since $d{x}_{\mu }$ transforms like a covariant vector density of weight $-1/2$, and since $A^{\mu }$ transforms like a contravariant vector density of weight 1/2, we conclude that the electric charge q transforms like a scalar.

Since $ds=icd\tau$, where $d\tau$ is the infinitesimal proper time, the four-velocity $v_{\mu }=d{x}_{\mu }/d\tau$ transforms like a covariant vector density of weight $-1/4$, and the four-momentum $p_{\mu }=m_0{v}_{\mu }$ transforms like a covariant vector density of weight 0. As pointed out by Ryder [⁹], the four-momentum transforms like the gradient, and also like the four-potential. This is consistent with the expression of the momentum operator in quantum mechanics, $\widehat{p}=-i\hslash \nabla$, and also with the expression of the canonical momentum in analytical mechanics, which (in Gaussian units) is $\overrightarrow{P}=\overrightarrow{p}+(q/c)\overrightarrow{A}$.

Since $d{p}_{\mu }$ transforms just like $p_{\mu }$, the four-force $F_{\mu }=d{p}_{\mu }/d\tau$ transforms like a covariant vector density of weight $1/4$. This can also be seen from the formula of the electrodynamic four-force, which (in Gaussian units) is $F_{\mu }=(q/c)F_{\mu \nu }{v}^{\nu }$.

6. Concluding Remarks

Starting from the expression (4) of special conformal transformations, we have derived the transformation rules for the line element, the metric tensor density, the electromagnetic field s trength, t he c urrent d ensity, the four-potential, the rest mass, the electric charge, and other related quantities.

The conformal transformation rules from Table 1 guarantee the form invariance of the relativistic action, and thus of the related Maxwell’s equations. As a direct application of these rules one can calculate the electromagnetic field o f a n e lectric c harge i n hyperbolic motion [¹³].

Table 1
The weights of the tensors and tensor densities.

gμν	$g^{\mu \nu }$	$d{x}_{\mu }$	$d{x}^{\mu }$	${\partial }_{\mu }$	${\partial }^{\mu }$	ds	$d^{4}x$	$v_{\mu }$	$v^{\mu }$	$F_{\mu }$
$-\frac{1}{2}$	$\frac{1}{2}$	$-\frac{1}{2}$	0	0	$\frac{1}{2}$	$-\frac{1}{4}$	$-1$	$-\frac{1}{4}$	$\frac{1}{4}$	$\frac{1}{4}$
Fμν	$F^{\mu \nu }$	$j_{\mu }$	$j^{\mu }$	$A_{\mu }$	$A^{\mu }$	m 0	q	$p_{\mu }$	$p^{\mu }$	$F^{\mu }$
0	1	$\frac{1}{2}$	1	0	$\frac{1}{2}$	$\frac{1}{4}$	0	0	$\frac{1}{2}$	$\frac{3}{4}$

When restricting our discussion to Maxwell’s equations, it appears that general relativity is the more encompassing theory, and that special conformal transformations are only some particular transformations among all the possible coordinate transformations accommodated by the General Relativity Theory. It is in this context that Rohrlich writes: “Since general relativity admits all point transformations, the conformal transformations are also included in them, although in somewhat disguised form” [¹²].

However, when the equations of motion of electrically charged particles are also included in our discussion, we are forced to choose between two mutually exclusive options.

We could hold onto the invariance of the rest mass of the particles, in which case we lose the conformal invariance of the equations of motion. It is in this context that Rohrlich writes: “Conformal invariance does not hold for particles of nonvanishing rest mass in Minkowski space” [¹²].

There is also another option, and Rohrlich adds in a footnote: “In a more general space, (Weyl space), one can have a conformal theory also for particles with $m\ne 0$. The rest mass is then no longer an invariant” [¹²].

The assumption of the rest mass invariance is unchallenged in general relativity, where the focus is on the gravitational interaction, and where a test particle subject to a gravitational field moves along the same geodesic regardless of its rest mass. It is for this reason that Fulton, Rohrlich, and Witten write: “The discussion of mass in general relativity is not obvious and straightforward and a simple comparison of results is difficult” [¹⁷].

Since the Weyl space is more general than the pseudo-Riemannian space, it appears that general relativity should be embedded into conformal relativity. In a different context, a similar conclusion was reached by Hanno Essén: “General relativity is in fact a conformally invariant scalar field theory” [¹⁸].

We hope that the readers will find this review useful, and that they will continue to explore the relationship between classical electrodynamics and Conformal Relativity.

References

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  » http://www.reed.edu/physics/faculty/wheeler/documents/
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• ^[11] T. Fulton, F. Rohrlich and L. Witten, Rev. Mod. Phys. 34, 442 (1962).
• ^[12] F. Rohrlich, Classical Charged Particles (Addison-Wesley, Reading, 1965).
• ^[13] C. Galeriu, European Journal of Physics 40, 065203 (2019).
• ^[14] C. Codirla and H. Osborn, Annals of Physics 260, 91 (1997).
• ^[15] L.D. Landau and E.M. Lifshitz, The Classical Theory of Fields (Butterworth-Heinemann, Oxford, 1980).
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• ^[18] H. Essén, International Journal of Theoretical Physics 29, 183 (1990).

Publication Dates

• Publication in this collection
  14 Feb 2025
• Date of issue
  2025

History

• Received
  04 Nov 2024
• Reviewed
  15 Dec 2024
• Accepted
  15 Jan 2025