Translation:On the Dynamics of the Theory of Relativity

On the Dynamics of the Theory of Relativity.

by.

The dynamics of the mass point was already dealt with by in his first foundational work on the relativity principle, as well as shortly afterwards by. The most important result of their investigation were the known formulas for the dependence of longitudinal and transverse mass from velocity (since then experimentally confirmed at different occasions using the electron). The assumption that 's dynamics remains in the limiting case of infinitely small velocities, was used as the starting point. Later, extended the theory to the thermodynamic side, and there he completely derived the mechanical inertia from energy (and pressure). There, he founded his consideration upon the principle of least action, however, he additionally had to introduce an assumption about the transformation of forces.

Nevertheless, there are still unsolved problems in dynamics. For example, asks whether the dynamics of the mass point is still valid for the electron, when one doesn't ascribe to it – as it ordinarily happens – radial symmetry, but perhaps an elliptic shape. affirms this, because in the limiting case of infinitely small velocities, 's mechanics must be valid under all circumstances. However, this assumption is surely not accurate in this generality, as we will see later. Also considers it necessary to ascribe spherical symmetry to the electron,  otherwise – in contradiction with experience – also a transverse component of force is connected to a longitudinal acceleration.

The experiment is in close relation with this. According to the electron theory, a uniformly moving charged capacitor experiences a torque by the electromagnetic forces. and tried to confirm this (as the consequence of Earth's motion) by a bifilarly hung capacitor, but they couldn't find any rotation out of its position at rest. The theory of relativity of course can explain this result very easily by the fact, that Earth (relative to which the capacitor is at rest) is a valid reference system. But how is the theory formed, when one chooses another reference system? The electromagnetic torque is also existent according to the theory of relativity. Yet why is there no torque?

The answer to this is given by. In the co-moving system the electrostatic forces are canceled by molecular cohesion, otherwise the capacitor wouldn't be in equilibrium. The resultant from electric and molecular forces is thus zero at any point. If both kinds of forces are transformed in the same way to other reference frames, then this resultant remains zero in all systems, and there is no reason for a rotation. Although it's unquestionable that this answer hits the truth, it isn't entirely satisfactory in so far as it is connected to molecular theory, with which this problem has nothing to do per se.

Already with respect to 's mechanics it has often been point out, that it is more logical to place the dynamics of continua in front of that of the mass point. It seems to me that there is a hint in the two mentioned problems, that the advantages of the mentioned way before the opposite one (to derive the dynamics of continua from that of the mass point), are even much greater in the theory of relativity as in the old theory. Therefore, in the following we want to investigate the concept of elastic stresses in its connection with momentum and energy.

Introductory remarks.
In the classic theory of elasticity, the stresses form a tensor of symmetric $$\mathbf{p}$$, i.e. the quantities

$\begin{array}{ccccc} \mathbf{p}_{xx}, & & \mathbf{p}_{xy}, & & \mathbf{p}_{xz},\\ \mathbf{p}_{yx}, & & \mathbf{p}_{yy}, &  & \mathbf{p}_{yz},\\ \mathbf{p}_{zx}, & & \mathbf{p}_{zy}, &  & \mathbf{p}_{zz},\end{array}$

between which the three relations $$\mathbf{p}_{jk}=\mathbf{p}_{kj}$$ hold, are calculated (when the axis-cross $$x, y, z$$ is rotated) in the same way, as the squares ($$x^2$$ etc.) and the products ($$xy$$ etc.) are calculated from the coordinates. At first, the property of symmetry vanishes in relativity theory. Thus we arrive at the concept of the unsymmetrical tensor, for which, at a rotation expressed by the scheme

$\begin{array}{c

the following transformation formulas should apply:

By that, one confirms that one can form a vector product $$[\mathfrak{A}\mathbf{t}]$$ from $$\mathbf{t}$$ and an arbitrary vector $$\mathfrak{A}$$, whose definition reads:

$[\mathfrak{A}\mathbf{t}]_{k}=\mathfrak{A}_{x}\mathbf{t}_{kx}+\mathfrak{A}_{y}\mathbf{t}_{ky}+\mathfrak{A}_{z}\mathbf{t}_{kz},\ k=x,y,z.$

It's known that one can interpret the operations

$\frac{\partial}{\partial x},\ \frac{\partial}{\partial y},\ \frac{\partial}{\partial z}.$

as components of a symbolic vector, whose vector product with $$\mathbf{t}$$ is denoted by us as the divergence of $$\mathbf{t}[\mathfrak{div}\mathbf{t}]$$; its components are to be formed according to the scheme

$\mathfrak{div}_{k}\mathbf{t}=\frac{\partial\mathbf{t}_{kx}}{\partial x}+\frac{\partial\mathbf{t}_{ky}}{\partial y}+\frac{\partial\mathbf{t}_{kz}}{\partial z},\ k=x,y,z.$|undefined

An unsymmetrical tensor is, for example, the tensor product $$\mathfrak{AB}$$ of two vectors $$\mathfrak{A}$$ and $$\mathfrak{B}$$. Its components are

$\mathfrak{AB}_{jk}=\mathfrak{A}_{j}\mathfrak{B}_{k},\ j,k=x,y,z,l.$

We call $$K$$ a valid space-time-system according to the relativity principle; several of that kind are distinguished by us by appended indices $$\left(K',\ K^{0}\right)$$. The passage from $$K$$ to $$K'$$ (Lorentz transformation) is represented by us – in accordance with – as an imaginary rotation of the four-dimensional axis-cross $$x, y, z, l$$, where $$l=ict$$; that is, by the scheme:

$\begin{array}{c

There, the coefficients $$\alpha_{m}^{(n)}$$ satisfy the known reality- and orthogonality conditions.

Furthermore it is known, that the concepts of vector and tensor can be transferred into a four-dimensional form. A four-vector $$F$$ – the six-vector is not needed in the following – has the direction sense of a directed distance; i.e. its four components $$F_{x},F_{y},F_{z},F_{l}$$ are transformed with a Lorentz transformation like the corresponding coordinates. By a world tensor, however, we understand the totality of 16 components $$T_{jk}(j,k=x,y,z,l)$$, which are transformed as the squares and products of $$x, y, z, l$$. Between them, there are always six symmetry relations $$T_{jk}=T_{kj}$$. Analogous to the fact that (with respect to tensor $$\mathbf{p}$$ in three dimensions) the divergence $$\mathfrak{div}\mathbf{p}$$ becomes a space vector, here the divergence $$\Delta ivT$$ is a four-vector with components

$\Delta iv_{k}T=\frac{\partial T_{kx}}{\partial x}+\frac{\partial T_{ky}}{\partial y}+\frac{\partial T_{kz}}{\partial z}+\frac{\partial T_{kl}}{\partial l},\ k=x,y,z,l$|undefined

§ 1. Transformation of force; energy- and momentum theorem.
From the investigations of, and  we can see, that the ponderomotive force $$\mathfrak{F}$$ related to unit volume (force density) of electrodynamics, can be supplemented to a four-vector $$F$$, when one adds $$F_{l}=i/c(\mathfrak{qF})$$ as fourth component to the three spatial components $$F_{x}=\mathfrak{F}_{x}$$ etc., where $$\mathfrak{q}$$ is the velocity of the point of contact of $$\mathfrak{F}$$, thus $$(\mathfrak{qF})$$ denotes the work per volume- and time unit. Furthermore it was explained by the same authors, that the four-force $$F$$ defined in this way, has the relation

to the world tensor $$T$$, where the components of $$T$$ have simple physical meanings. Namely the nine components, in which $$l$$ doesn't arise as index, give the three-dimensional tensor $$\mathbf{p}$$ of 's stresses $$\left(T_{jk}=\mathbf{p}_{jk},\ j,k=x,y,z\right)$$; $$-T_{ll}$$ is the density $$W$$ of electromagnetic energy, and the other six components are in relation to the momentum density $$\mathfrak{g}$$ and the energy current ( vector) $$\mathfrak{S}$$ by:

In this interpretation of $$T$$, equation (1) indeed contains the momentum theorem when applied to the spatial components:

though when applied to the temporal components, it contains the energy theorem:

It connects both of them in a form which is invariant with respect to the Lorentz transformation.

Now, and  have already announced, that all ponderomotive forces must be transformed by the Lorentz transformation in the same way, as in electrodynamics. Thus it must be possible in all areas of physics, to combine the force density with its power to a four-force. Now, with respect to any four-force $$F$$ defined as a function of the world-points, there are infinitely many world tensors, which are related to it by relation (1)

$F=-\Delta ivT\,$

We assume, that in any field of physics there is one of those world tensors, whose components have the corresponding meaning, as the components of the mentioned electrodynamic tensor; i.e., that for example also in dynamics, the energy density is defined by $$-T_{ll}$$, the momentum density and the energy current according to (1a) by $$T_{xl}-T_{lx}$$ etc., while $$T_{xx}$$ is related to the elastic stresses. That we assume tensor $$T$$ as being symmetric, can appear to be arbitrary at this place; because also the divergence of an unsymmetrical world tensor would be a four-vector. Yet exactly this part of the assumption can be explained later (see § 4.). The most important physical consequence from the symmetry relations

$T_{xl}=T_{lx},\ T_{yl}=T_{ly},\ T_{zl}=T_{lz}$

is the law of inertia of energy, which we find from (1a) immediately in its most general form

In order to interpret the four-force, it is to be noticed, that in pure electromagnetic processes, no ponderomotive force $$\mathfrak{F}$$ arises, since otherwise processes of different kind must necessarily arise in consequence of this. Thus also the four-force is given by

$F=-\Delta ivT=0\,$

Equation (2) then says, that the electromagnetic stress force $$-\mathfrak{div}\mathbf{p}$$ and the electromagnetic inertial force $$-\dot{\mathfrak{g}}$$ are in equilibrium. Also with respect to purely dynamic processes, we will have to set the dynamic four-force equal to zero. Because also 's dynamics – maybe in the clearest way in the principle of – says that the inertial force resisting the increase of mechanical momentum, exactly compensates the force exerted by the elastic stresses – other forces don't exist at all in pure dynamics. In this way, one has as the fundamental equation of dynamics which is invariant against Lorentz transformation and encloses the momentum and energy theorem, the relation

where the components of $$T$$ have the given physical meaning.

If dynamic, electrodynamic or other processes are interacting, then (as it can be easily seen) the relation

is taking the place of (4). The summation is to be carried out over all four-forces, i.e. the dynamic, electrodynamic ones etc., in other words over all world tensors.

§ 2. Transformation of momentum, energy and stresses.
The transformation formulas for momentum, energy density and the stresses $$\mathbf{p}$$ in the passage from one valid reference system $$K$$ to another system $$K'$$, is definitely determined by deriving them from the components of the world tensor $$T$$. However, we don't want to write them for the most general Lorentz transformation, but we assume that the spatial axes-cross $$x, y, z$$ in $$K$$ and $$x', y', z'$$ in $$K'$$ are mutually parallel, and that velocity $$\mathfrak{v}$$ of $$K'$$ with respect to $$K$$, lies in the direction of the $$x$$-axis. In this case, the Lorentz transformation reads

$\begin{array}{c} x=\frac{x'+i\beta l'}{\sqrt{1-\beta^{2}}},\ y=y',\ z=z',\ l=\frac{l'+i\beta x'}{\sqrt{1-\beta^{2}}},\\ \\\beta=\frac{v}{c}.\end{array}$|undefined

If one transforms the components of $$T$$ as $$x^2$$ etc., then one finds, by introducing $$Q,\mathfrak{g,S}$$ and $$\mathbf{p}$$ according to § 1 (see (1a) and (3a)), instead of them the following formulas

{{MathForm1|(6)|$$\left\{ \begin{array}{rl} \mathfrak{g}_{x}= & \frac{1}{c^{2}}\mathfrak{S}_{x}=\frac{\left(c^{2}+v^{2}\right)\mathfrak{g}'_{x}+v\left(\mathbf{p}'_{xx}+W'\right)}{c^{2}-v^{2}},\\ \\\mathfrak{g}_{y}= & \frac{1}{c^{2}}\mathfrak{S}_{y}=\frac{c^{2}\mathfrak{g}'_{y}+v\mathbf{p}'_{xy}}{c\sqrt{c^{2}-v^{2}}},\\ \\W= & \frac{Q'+\beta^{2}\mathbf{p}'_{xx}+2v\mathfrak{g}'_{x}}{1-\beta^{2}},\\ \\\mathbf{p}{}_{yy}= & \mathbf{p}'_{yy},\ \mathbf{p}{}_{yz}=\mathbf{p}'_{yz},\\ \\\mathbf{p}{}_{xx}= & \frac{\mathbf{p}'{}_{xx}+2v\mathfrak{g}'_{x}+\beta^{2}W'}{1-\beta^{2}},\\ \\\mathbf{p}{}_{xy}= & \frac{\mathbf{p}'_{xy}+v\mathfrak{g}'_{y}}{\sqrt{1-\beta^{2}}}.\end{array}\right.$$}}

The formulas (not written down here) for $$\mathfrak{g}_{x},\mathbf{p}{}_{xx},\mathbf{p}{}_{zz},\mathbf{p}{}_{zx}$$ and $$\mathbf{p}{}_{zy}$$ can be found by replacing index $$y$$ by $$z$$ at the suitable places.

The most important question now is, which energy forms we want to combine for the formation of density $$W$$. The answer is: All of those who show no current (and thus also no momentum) in a valid system (the rest system $$K^{0}$$ ). To those belong: heat, chemical energy, elastic energy, inner energy of atoms, and maybe also new and unknown energy forms. Electromagnetic energy has to be excluded in general, since it can also flow in the rest system $$K^{0}$$. If this doesn't occur in a special case, then one can include it in $$W$$ as well; tensor $$\mathbf{p}$$ then encloses also 's stresses besides the elastic ones, and vector $$\mathfrak{g}$$ also the electromagnetic momentum besides the mechanical momentum.

If we now put the rest system instead of system $$K'$$, then formulas (6) become simplified due to $$\mathfrak{g}^{0}=0$$ in the following way:

{{MathForm1|(7)|$$\left\{ \begin{array}{ccc} \mathfrak{g}_{x}=\frac{q}{c^{2}-q^{2}}\left(\mathbf{p}_{xx}^{0}+W^{0}\right), & & \mathfrak{g}_{y}=\frac{q}{q\sqrt{c^{2}-q^{2}}}\mathbf{p}_{xy}^{0},\\ \\W=\frac{c^{2}W^{0}+q^{2}\mathbf{p}_{xx}^{0}}{c^{2}-q^{2}},\ &  & \mathbf{p}_{yy}=\mathbf{p}_{yy}^{0},\ \mathbf{p}_{yx}=\mathbf{p}_{yx}^{0},\\ \\\mathbf{p}_{xx}=\frac{c^{2}\mathbf{p}_{xx}^{0}+q^{2}W^{0}}{c^{2}-q^{2}},\ &  & \mathbf{p}_{xy}=\frac{c}{\sqrt{c^{2}-q^{2}}}\mathbf{p}_{xy}^{0}.\end{array}\right.$$}}

Here, $$v$$ is replaced by $$q$$, since $$\mathfrak{q}$$ denotes the velocity of the body in system $$K$$. We can also write the equations for $$W$$ and $$\mathfrak{g}$$ in a vectorial way as follows:

{{MathForm1|(8)|$$\left\{ \begin{array}{l} W=\frac{c^2}{c^{2}-q^{2}}\left(W^{0}+\frac{1}{c^{2}}\left(\mathfrak{q}\left[\mathfrak{q}\mathbf{p}^{0}\right]\right)\right),\\ \\\mathfrak{g}=\frac{\mathfrak{q}}{c^{2}-q^{2}}\left\{ W^{0}+\frac{1}{q^{2}}\left(\mathfrak{q}\left[\mathfrak{q}\mathbf{p}^{0}\right]\right)\right\} +\frac{1}{c\sqrt{c^{2}-q^{2}}}\left\{ \left[\mathfrak{q}\mathbf{p}^{0}\right]-\frac{\mathfrak{q}}{q^{2}}\left(\mathfrak{q}\left[\mathfrak{q}\mathbf{p}^{0}\right]\right)\right\} .\end{array}\right.$$}}

This can be easily proven, by assuming the $$x$$-axis as being parallel to $$\mathfrak{q}$$; the vector product $$\left[\mathfrak{q}\mathbf{p}^{0}\right]$$ then obtains the components

$\left[\mathfrak{q}\mathbf{p}^{0}\right]_{x}=q\mathbf{p}_{xx}^{0},\ \left[\mathfrak{q}\mathbf{p}^{0}\right]_{y}=q\mathbf{p}_{xy}^{0},$

so that one obtains the formulas under (7) again. Equations (8) contain in the most general way, the derivation of energy and momentum from velocity and the inner state of the body specified by $$W^{0}$$ and $$\mathbf{p}^{0}$$. Especially it must be emphasized, that the second summand in the equation for $$\mathfrak{g}$$ represents a vector perpendicular to $$\mathfrak{q}$$; namely its scalar product with $$\mathfrak{q}$$ is zero. Thus the momentum density is composed of a component parallel to $$\mathfrak{q}$$ and proportional to $$q/c^{2}-q^{2}$$, and a component perpendicular to $$\mathfrak{q}$$ and proportional to $$q/c\sqrt{c^{2}-q^{2}}$$. The latter vanishes only then, when velocity $$\mathfrak{q}$$ lies in one of the axis-directions of the ellipsoid of the rest-stresses $$\mathbf{p}^{0}$$; then we choose one of them as $$x$$-axis, and if $$\mathfrak{q}$$ is parallel to $$x$$, then $$\left[\mathfrak{q}\mathbf{p}^{0}\right]=\mathfrak{q}_{x}\mathbf{p}_{xx}^{0}$$, thus parallel to $$\mathfrak{q}$$.

If we integrate equations (8) over the volume

$V=V^{0}\frac{\sqrt{c^{2}-q^{2}}}{c}$|undefined

of a body, in which the velocity is spatially constant, then we find for its energy

$E=\int WdV=\frac{\sqrt{c^{2}-q^{2}}}{c}\int WdV^{0}$|undefined

and its momentum

$\mathfrak{G}=\int\mathfrak{g}dV=\frac{\sqrt{c^{2}-q^{2}}}{c}\int\mathfrak{g}dV^{0}$|undefined

the relations

{{MathForm1|(8a)|$$\left\{ \begin{array}{l} E=\frac{c}{\sqrt{c^{2}-q^{2}}}\left\{ E^{0}+\frac{1}{c^{2}}\left(\mathfrak{q}\left[\mathfrak{q},\ \int\mathbf{p}^{0}dV^{0}\right]\right)\right\} ,\\ \\\mathfrak{G}=\frac{\mathfrak{q}}{c\sqrt{c^{2}-q^{2}}}\left\{ E^{0}+\frac{1}{q^{2}}\left(\mathfrak{q}\left[\mathfrak{q},\ \int\mathbf{p}^{0}dV^{0}\right]\right)\right\} +\frac{1}{c^{2}}\left\{ \left[\mathfrak{q},\ \int\mathbf{p}^{0}dV^{0}\right]-\frac{\mathfrak{q}}{q^{2}}\left(\mathfrak{q}\left[\mathfrak{q},\ \int\mathbf{p}^{0}dV^{0}\right]\right)\right\} .\end{array}\right.$$}}

There, $$E^{0}$$ is the rest energy, $$\int\mathbf{p}^{0}dV^{0}$$ like $$\mathbf{p}^{0}$$ is a symmetrical tensor.

When a body is accelerated in a adiabatic-isopiestic way (i.e. with constant $$E^0$$ and $$\mathbf{p}^{0}$$), namely in longitudinal direction, then by (8a) the momentum increase is generally not at all in the common direction of velocity and acceleration. Its transverse component is rather equal to

$\frac{1}{c^{2}}\left\{ \left[\mathfrak{\dot{q}},\ \int\mathbf{p}^{0}dV^{0}\right]-\frac{\mathfrak{\dot{q}}}{q^{2}}\left(\mathfrak{q}\left[\mathfrak{q},\ \int\mathbf{p}^{0}dV^{0}\right]\right)\right\} $|undefined

and not at all vanishes in the limiting case $$q=0$$. Already at this example one recognizes, that 's dynamics is not at all generally valid in this limiting case.

This (at first maybe strange) behavior becomes easily understandable in the sense of the inertia of energy (3a). The first summand in equation (8) for $$\mathfrak{g}$$, $$\tfrac{q}{c^{2}-q^{2}}W^{0}$$, represents the convection current of energy, the other ones the energy current also known in the classical theory of elasticity in the motion of stressed bodies. That the latter by no means must have the direction of velocity, but can also be perpendicular to it, shows in a very illustrative way the example of a rotating and torqued drive shaft, in which the energy transfer happens in a direction parallel to the rotation axis.

The equations become essentially simplified, when the stresses $$\mathbf{p}^{0}$$ form a pressure $$p^{0}$$ which is equal in all directions. Then according to (8)

If besides $$q$$, also $$p^{0}$$ is spatially constant, then 's equations follow from (8a):

§ 3. Absolute and relative (elastic) stresses.
Fundamental equation (4) says, when applied to spatial components, that

$$\mathfrak{\dot{g}}$$ is the derivative of $$\mathfrak{g}$$ and $$t$$ and is formed for a point fixed in space, i.e. at constant $$x, y, z$$. Therefore, $$\mathbf{p}$$ is not the tensor of elastic stresses, because they must (as in the previous theory) be in connection with the change of momentum of a certain body element $$\delta V$$. If we denote this change for the time interval $$dt$$ by $$\mathfrak{\underline{\dot{g}}}\delta Vdt$$, then it's known that the relations hold:

$\mathfrak{\dot{g}}_{x}=\mathfrak{\underline{\dot{g}}}+\frac{\partial}{\partial x}\left(\mathfrak{g}_{x}\mathfrak{q}_{x}\right)+\frac{\partial}{\partial y}\left(\mathfrak{g}_{x}\mathfrak{q}_{y}\right)+\frac{\partial}{\partial z}\left(\mathfrak{g}_{x}\mathfrak{q}_{z}\right)$|undefined

etc., or written in a vectorial way (see the "Introductory remarks")

If we now introduce the unsymmetrical tensor

then one finds from (11) and (11a)

By this equation, tensor $$\mathbf{t}$$ proves to be the tensor of elastic stresses. Because if one integrates (13) over a finite body, it follows for its momentum change

($$d\sigma$$ is the surface element, $$n$$ its normal), where the components of vector $$\mathfrak{t}_{n}$$ have to be formed according to the scheme

The momentum increase of the body is on the left-hand side of (13a), thus the surface integral on the right-hand side is the force exerted upon it by the stresses, i.e. $$\mathfrak{t}_{n}d\sigma$$ is the force acting upon $$d\sigma$$.

From (12) and the transformation formulas for $$\mathbf{p}$$ and $$\mathfrak{g}$$ given in (7), one easily finds the following for $$\mathbf{t}$$:

{{MathForm1|(15)|$$\left\{ \begin{array}{c} \mathbf{t}_{xx}=\mathbf{p}_{xx}^{0},\ \mathbf{t}_{yy}=\mathbf{p}_{yy}^{0},\ \mathbf{t}_{yz}=\mathbf{p}_{yz}^{0},\\ \\\mathbf{t}_{xy}=\frac{c}{\sqrt{c^{2}-q^{2}}}\mathbf{p}_{xy}^{0},\ \mathbf{t}_{yx}=\frac{\sqrt{c^{2}-q^{2}}}{c}\mathbf{p}_{yx}^{0}.\end{array}\right.$$}}

Contrary to (8), however, $$\mathfrak{q}$$ is set parallel to $$x$$. If the inner stress state is a pressure $$p^{0}$$ equal to all sides, then

$\mathbf{p}_{xx}^{0}=\mathbf{p}_{yy}^{0}=\mathbf{p}_{zz}^{0}=p^{0},\ \mathbf{p}_{xy}^{0}=\mathbf{p}_{yz}^{0}=\mathbf{p}_{zx}^{0}=0$

According to (15), exactly the same relations hold for $$\mathbf{t}$$. The relative pressure (equal at all sides)

$p=p^{0}\,$

is thus an invariant of the Lorentz transformation.

That the relative stresses $$\mathbf{t}$$ must be denoted as elastic stresses, not the absolute stresses $$\mathbf{p}$$, is also demonstrated in the transformation formulas (15) and (7). Namely, $$\mathbf{t}_{jk}$$ are only connected (except with $$\mathfrak{q}$$) with the rest stresses $$\mathbf{p}_{jk}^{0}$$, while $$\mathbf{p}_{jk}$$ are also connected with $$W^{0}$$. Thus the latter change their value and meaning, when one for example would separate the heat from $$W^{0}$$, as it would become necessary under consideration of thermal conduction, while this is not the case with respect to the first ones.

§ 4. The surface theorem.
In the previous theory of elasticity, the symmetry of the stress tensor is in closest connection with the so-called surface theorem, which is stating the conservation of angular momentum. If we want here to search for the deeper meaning of the non-symmetry of tensor $$\mathbf{t}$$, we consequently have to transfer this theorem into relativity theory at first.

As in the previous theory we define as angular momentum, contained in a certain spatial area, the integral

extended over this area; $$\mathfrak{r}$$ is the radius vector, directing from an arbitrary fixed point to $$dS$$. If we ask after the change of $$\mathfrak{L}$$ with time, then the surface of this area is to be considered as invariable during the differentiation. Consequently it is according to (11)

where $$\mathfrak{p}_{n}$$ is the vector with components

However, now we are asking after the angular momentum of a material volume element $$dV$$; it is

however, if we calculate here the derivative with respect to time, then it is to be considered, that not only $$\mathfrak{g}$$, but also $$V$$ changes in quantity and location, thus vector $$\mathfrak{r}$$ as well. If we assume momentum $$\mathfrak{g}dV$$ as being constant, then this derivative is evidently equal to

$[\mathfrak{\dot{r},\ g}dV]=[\mathfrak{q,\ g}dV]$

since $$\mathfrak{\dot{r}=q}$$. In general, however,

$[\mathfrak{r,}\frac{d}{dt}(\mathfrak{g}dV)]=[\mathfrak{r\underline{\dot{g}}}]dV$|undefined

enters as a summand. With respect to a moving bodies, it is consequently according to (13)

The first part of the space integral on the right-hand side can only be transformed by partial integration, and one finds in this way:

$-\int[\mathfrak{r,\ \mathfrak{div}}\mathbf{t}]_{x}dV=\int\left[\mathfrak{rt}_{n}\right]_{x}d\sigma+\int\left(\mathbf{t}_{xy}-\mathbf{t}_{yz}\right)dV,$|undefined

thus under consideration of (12), according to which

$\mathbf{t}_{xy}-\mathbf{t}_{yz}=\mathfrak{g}_{y}\mathfrak{q}_{z}-\mathfrak{g}_{z}\mathfrak{q}_{y}=-[\mathfrak{qg}]$

it is

$-\int[\mathfrak{r\ \mathfrak{div}}\mathbf{t}]dV=\int\left[\mathfrak{rt}_{n}\right]d\sigma-\int[\mathfrak{qg}]dV$|undefined

If one substitutes this value, then it follows from (18)

The surface integral on the right-hand side is the torque exerted by the surrounding upon the body, because $$\mathfrak{t}_{n}d\sigma$$ is the force acting upon $$d\sigma$$. If we conversely calculate the torque exerted by the body upon its surrounding, then we also find $$\int\left[\mathfrak{rt}_{n}\right]d\sigma$$, but now the normal has the opposite direction than before, consequently $$\mathfrak{t}_{n}$$ as well as $$\int\left[\mathfrak{rt}_{n}\right]d\sigma_{n}$$ is oppositely equal to the previous values. If no other torque is acting upon the surrounding as the calculated one, then for the change of its angular momentum $$\mathfrak{L}^{a}$$ the relation holds

$\frac{d\mathfrak{L}^{a}}{dt}=-\frac{d\mathfrak{L}}{dt}$|undefined

thus

The surface theorem thus also holds in the theory of relativity. This can of course also be concluded from (17).

If we apply equation (19) upon an infinitesimal material parallelepiped, where we locate the coordinate axes parallel to its edges, then the previous theory concludes as follows: $$\mathfrak{g}$$ is parallel to $$\mathfrak{q}$$, thus $$[\mathfrak{qg}]=0$$ and according to (18)

$\frac{d\mathfrak{L}}{dt}=[\mathfrak{q,\underline{\dot{g}}}]dV$|undefined

Since we are allowed to displace the origin of $$\mathfrak{r}$$ into the volume of $$dV$$, then $$[\mathfrak{rg}]dV$$ becomes as small as $$dV$$ in the limiting passage of higher order. Torque $$\int\left[\mathfrak{rt}_{n}\right]d\sigma$$, however, becomes in its $$x$$-component equal to $$\left(\mathbf{t}_{yz}-\mathbf{t}_{zy}\right)dV$$. Consequently, it must be $$\mathbf{t}_{zy}=\mathbf{t}_{yz}$$ etc.. Although in relativity theory, $$[\mathfrak{r\mathfrak{\underline{\dot{g}}}}]dV$$ is also to be neglected in the limit, yet it follows according to (18)

$\frac{\partial\mathfrak{L}}{\partial t}=\left[\mathfrak{qg}\right]dV$,|undefined

thus

$\mathbf{t}_{yz}-\mathbf{t}_{zy}=\left[\mathfrak{qg}\right]_{x}$

in agreement with (12). The stress tensor $$\mathbf{t}$$ is unsymmetrical because of the reason, that a stressed body element requires a torque to maintain its velocity.

In the rest system $$K^{0}$$, tensor $$\mathbf{t}^{0}$$ must be symmetric due to the surface theorem; at the same time, the momentum density $$\mathfrak{g}^{0}$$ and the energy current $$\mathfrak{S}^{0}$$ is zero in $$K^{0}$$. Consequently, in $$K^{0}$$ (see. (1a)) the symmetry equations hold

$T_{jk}=T_{kj}\ \left(k=x^{0},y^{0},z^{0},l^{0}\right)$

By that, however, the symmetry of world-tensor $$T$$ is proven per se; in every system the corresponding symmetry equations must hold.

With respect to processes not entirely dynamic, in which equation (5) replaces equation (4)

$\Sigma F=-\Delta iv(\Sigma T)=0$

we conclude in analogy to (11), that

$\Sigma\mathit{\mathfrak{\dot{g}}}=-\mathfrak{div}\Sigma\mathbf{p}$|undefined

which is the equation containing the momentum theorem. If we then define the total angular momentum

$\Sigma\mathfrak{L}=\int[\mathfrak{r},\Sigma\mathfrak{g}]dS$

then in analogy to (17)

$\frac{\partial}{\partial t}\Sigma\mathfrak{L}=\int[\mathfrak{r},\Sigma\mathfrak{p}_{n}]d\sigma$

From that, as above, the theorem of conservation of angular momentum $$\Sigma\left(\mathfrak{L}+\mathfrak{L}^{a}\right)=const.$$ follows. The summations are to be extended over all forms of momentum (mechanical, electromagnetic, etc.) and over all angular momentums. The surface theorem thus obtains a meaning, which surpasses that of dynamics, and which is valid for the whole of physics.

§ 5. Completely static system.
As it followed from equation (8) and (8a), the dynamics of relativity theory is generally quite complicated. Yet the relations become simple again with respect to a complete static system. By that, we understand such one, which is in static equilibrium in a valid reference system $$K^{0}$$, without interacting with other bodies; thus for example an electrostatic field including all carriers of charge. In this field, the momentum density (related to its rest system) is everywhere zero and its energy is rigidly connected at its place. In every other reference system $$K$$, the total energy (including the electromagnetic energy when it is present) shares the motion. Consequently we can understand in formulas (8a) under $$E^{0}$$ the total energy, under $$\mathfrak{G}$$ the total momentum, and under $$\mathbf{p}^{0}$$ the sum of elastic stresses and stresses.

First, we consider the state in $$K^{0}$$. Since $$\mathfrak{g}^{0}=0$$, it follows from (11) for an arbitrary limited space

Now we choose the boundary, so that it consists of an arbitrary cross-section of the system, and of an area completely outside of the system. Since it was presupposed that the system is not in interaction with other bodies, it thus can be viewed as being in vacuum, then for the second part $$\mathfrak{p}_{n}^{0}=0$$, thus also for the cross-section alone

Now we choose a plane $$x^{0}=const.$$ as cross-section. Then it follows, by applying the vector equation (20b) upon the coordinate directions, and by expressing the components of $$\mathfrak{p}_{n}^{0}$$ by $$\mathbf{p}_{x^{0}x^{0}},\ \mathbf{p}_{x^{0}y^{0}}$$ etc. according to (17a):

$\begin{array}{c} \int\mathfrak{p}_{nx^{0}}^{0}d\sigma^{0}=\int\mathbf{p}_{x^{0}x^{0}}^{0}dy^{0}dz^{0}=0,\\ \\\int\mathfrak{p}_{ny^{0}}^{0}d\sigma^{0}=\int\mathbf{p}_{x^{0}y^{0}}^{0}dy^{0}dz^{0}=0,\\ \\\int\mathfrak{p}_{nz^{0}}^{0}d\sigma^{0}=\int\mathbf{p}_{x^{0}z^{0}}^{0}dy^{0}dz^{0}=0.\end{array}$|undefined

If we multiply $$dx^{0}$$ here, and integrate over the total volume $$V^{0}$$ of the system, then we find

$\int\mathbf{p}_{x^{0}x^{0}}^{0}dV^{0}=\int\mathbf{p}_{x^{0}y^{0}}^{0}dV^{0}=\mathbf{p}_{x^{0}z^{0}}^{0}dV^{0}=0$|undefined

Six additional equation are obtained by us, when we choose $$y^{0}=const$$ or $$z^{0}=const$$ as cross-section. All of them can be summarized in the tensor equation

For a completely static system, equation (8a) thus pass into:

i.e., a completely static system behaves (when in uniform motion) as a mass point of rest mass

$m^{0}=\frac{E^{0}}{c^{2}}.$|undefined

The same, however, also holds for quasi-stationary (adiabatic isopiestic) acceleration; because they are denoted as quasi-stationary, when the inner state $$\left(E^{0},\mathbf{p}^{0}\right)$$ is not noticeably changed. However this system may be formed, its longitudinal mass is always

$m_{l}=\frac{\partial G}{\partial q}=\frac{c^{3}m^{0}}{\left(\sqrt{c^{2}-q^{2}}\right)^{3}}$|undefined

its transverse mass

$m_{t}=\frac{G}{q}=\frac{cm^{0}}{\sqrt{c^{2}-q^{2}}}$|undefined

in the limiting case $$q=0$$ it satisfies 's mechanics.

One such system is for example formed by the electron with its field; however it may be formed, in quasi-stationary motion it must satisfy the dynamics of the mass point, so that from attempts of this kind one cannot draw conclusions about its form, its charge distribution, and also not as to whether it has another momentum besides its electromagnetic momentum. In this way, actually answered 's question correctly. That felt compelled to assume spherical symmetry, lies in the fact that he didn't consider the mechanical momentum of the electron, which can only be assumed as zero at certain assumptions concerning the form and charge distribution, for example when one assumes spherical symmetry; but not generally. Because in the electron, there are stresses of other kinds (which can provisionally be denoted as elastic ones) besides the electromagnetic ones, and these will in general supply a transverse component to the mechanical momentum according to (8a), which cannot be made to zero (unlike the longitudinal component) by a suitable choice of $$E^{0}$$. In the case of spherical symmetry, the transverse component of course vanishes.

A complete static system is also the capacitor of the experiment with its field. The total system requires (at uniform velocity) a torque as little as a mass point. The torque, exerted by the electromagnetic forces upon the capacitor itself, it exactly that one, which is required by this elastically stressed body according to § 4. Neither the electromagnetic momentum of the field, nor the mechanical momentum of the body, have in this case the direction of velocity, but the total momentum consisting of both of them has surely that direction, as it follows from (22).

(Received April 30, 1911)