Restoring units, the decaying Yukawa potential is $V(r)=C{e}^{-r/{\lambda }_m}/r$ with range ${\lambda }_m=\hslash /(mc)$. For $r_{\ast }=10^{-18}\mathrm{m}$,The geometric $1/r$ factor belongs in the potential ratio; the exponential alone is the additional suppression relative to massless exchange. Using $\hslash c=197.32698\mathrm{MeV}\mathrm{fm}$ givesThe following are ratios of potential magnitudes to the magnitude at $r_{\ast }$, with the same coupling held fixed:

For example, the pion's nuclear exponential factor is $e^{-0.68346}=0.505$, whereas the Higgs nuclear exponent is about $-632.83$. The huge atomic suppressions are most reliably reported through logarithms, not through floating-point exponentiation that silently underflows to zero. These are potential energy ratios, not force ratios or absolute comparisons of different interaction strengths.

Use $\hslash =c=1$ and the mostly-plus Minkowski metric. Then the source's Klein-Gordon equation is $({\partial }_t^2-{\nabla }^2+m^2)\varphi =0$. For a plane wave $\varphi =C{e}^{-i\omega t+i\mathbf{k}\cdot \mathbf{x}}$, substitution givesThus a real propagating wave vector requires $|\omega |\ge m$, or $|E|\ge m{c}^2$ on restoring units. The positive-energy branch has a rest-energy threshold $m{c}^2$. At $|\omega |=m$, $\mathbf{k}=0$; frequencies below the mass threshold give evanescent spatial dependence rather than real-wave-vector propagation. Both signs of $\omega$ occur in the classical equation; a real scalar field combines complex-conjugate modes. Its group velocity is $\mathbf{k}/\omega$ and is subluminal, though the positive-branch phase speed exceeds one when $m>0$.

For a static spherically symmetric scalar field, the equation outside the source isSetting $\psi =r\varphi$ gives ${\psi }^{\prime \prime }-m^2\psi =0$, soDecay at infinity selects $B=0$, the Yukawa potential. It is singular at the origin and describes a point source distributionally; a source-free solution regular at the origin and decaying at infinity must be zero. For $m=0$, $\varphi =A/r+B$, and decay sets the constant $B$ to zero.

The scalar contraction of the electromagnetic field tensor isWith orientation ${ϵ}^{0123}=+1$, define the dual electromagnetic field tensor by ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. Direct substitution givesThe first is a Lorentz scalar; the second is a pseudoscalar, invariant under proper Lorentz transformations such as boosts and rotations. Under spatial inversion $\mathbf{E}$ changes sign while $\mathbf{B}$ does not, so $\mathbf{E}\cdot \mathbf{B}$ reverses sign. For example, $\mathbf{E}=\mathbf{B}=\hat{\mathbf{z}}$ has product $+1$ before inversion and $-1$ afterwards. Thus the unrestricted statement including improper transformations needs this qualification.

For the spacetime integral, assume a globally regular electromagnetic four-potential on ordinary Minkowski space and sufficient boundary decay. Using $\nabla \cdot \mathbf{B}=0$ and ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$,ConsequentlyThe magnetic helicity boundary term controls the integral. If its surface flux vanishes and the magnetic helicity $H_B(t)=\int \mathbf{A}\cdot \mathbf{B}{d}^{3}x$ vanishes at both temporal ends, thenCompactly supported smooth potentials are one sufficient version of the intended falloff assumption. The local identity is exact, but vanishing fields at spatial infinity alone is insufficient. For a concrete counterexample, take $\varphi =0$ and, in dimensionless coordinates,where $a$ changes smoothly from zero to one and $\dot{a}$ has compact temporal support. Both electric fields and magnetic fields vanish rapidly at spatial infinity, butThe counterexample leaves a final magnetic field and therefore fails the extra temporal boundary condition. It shows exactly why the printed integral cannot follow from a merely spatial reading of “at infinity”; vanishing of the full boundary term is the needed hypothesis.

Use $c=1$ and the mostly-plus Minkowski metric fixed in part a. The electromagnetic four-potential and its covariant form areDefine the electromagnetic field tensor by $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$. Since $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$,Raising both indices changes the signs of its time-space entries. The lower-index signs agree with ^[ref]; the signature and index placement must be kept consistent.

For a passive boost to a frame moving with $v\hat{\mathbf{x}}$, transform the tensor asFor example, $F^{\prime 02}=\gamma F^{02}-\gamma v{F}^{12}=\gamma (E_y-v{B}_z)$ and $F^{\prime 12}=\gamma (B_z-v{E}_y)$. Reading off all entries givesFor an arbitrary boost velocity $\mathbf{v}$, rotational covariance gives the Lorentz transformation of electromagnetic fieldsHere all velocities are in $c=1$ units. Equivalently the parallel components are unchanged and the perpendicular components have the cross-product expressions multiplied by $\gamma$.

To prove the vanishing component theorem for a Lorentz four-vector, first suppose $Y^0=0$ in every inertial frame. An infinitesimal boost along spatial direction $i$ gives $Y^{\prime 0}=Y^0-\beta Y^i+O({\beta }^2)$. It too vanishes for every $\beta$, so each $Y^i=0$. If instead a specified spatial component vanishes in every frame, spatial rotations first imply that all three spatial components vanish in a chosen frame, and a boost in the specified direction then forces $Y^0=0$. ThereforeA physical consequence is that conservation of energy in every inertial frame implies full four-momentum conservation: apply the result to the difference between outgoing and incoming total four-momenta.

The statement fails for general tensors. A nonzero antisymmetric rank-two tensor has $F^{00}=0$ in every frame, because antisymmetry is preserved. For an off-diagonal component, the nonzero Minkowski metric is another counterexample: ${\eta }^{01}=0$ in all orthonormal Lorentz frames, while $\eta$ is certainly not zero. The transformation law can preserve structural zero components without forcing the entire tensor to vanish.

An invariant has the same value when the same physical event or state is described in different inertial frames. A conserved quantity has the same value as the physical system evolves, subject to its equations and isolation conditions. These are different comparisons. The invariant mass of a heated object can change with time, while an isolated system's conserved energy has different numerical values in boosted frames.

Under a Lorentz transformation $x^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{x}^{\nu }$, a Lorentz four-vector transforms asA covector transforms by the inverse matrix, $W_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{W}_{\nu }$. A general tensor has one corresponding factor for each index; for example,Contraction with the Minkowski metric produces Lorentz scalars; it does not by itself make those scalars conserved.

Let $t$ label the reception event in the source reference frame. Its distance from the source and the relevant retarded time areThe received phase is $2\pi f_0{t}_e$. Because the observer's proper time obeys $d\tau =dt/\gamma$, its measured frequency isEquivalently, write $\beta =v/c$ and $s(t)=vt/r(t)$. The incoming Doppler factor is $D_{\mathrm{in}}=\gamma (1-\beta s)$, obtained also by contracting the photon four-momentum with the observer's four-velocity. The frequency decreases monotonically, withIn particular, the instant of closest approach is not the zero-shift point: a moving receiver looking perpendicular to its motion sees a factor $\gamma$, not $1/\gamma$. The latter factor applies to a different moving-emitter geometry. The graph uses source-frame reception time, with observer proper time $\tau =t/\gamma$.

Use the inverse standard Lorentz boost, $t={\gamma }_v(t^{\prime }+v{x}^{\prime }/c^2)$ and $x={\gamma }_v(x^{\prime }+v{t}^{\prime })$. Label two rod endpoints by a separation $\Delta x^{\prime }=L$; their world lines satisfy $y^{\prime }=u{t}^{\prime }$. A rod's orientation in $S$ must compare endpoint events simultaneous in $S$, not in $S^{\prime }$. ConsequentlyThenThe signed counterclockwise angle therefore satisfiesFor positive $u,v$, the rod slopes downward to the right, so the acute clockwise angle is $\arctan ({\gamma }_{v}uv/c^2)$. This is a relativity of simultaneity effect in addition to the longitudinal length contraction; the rod is not being mechanically bent by the coordinate change.

For causal reachability with periodic time, let $P_t=2\pi T$. The lower-right panel of part c identifies its top and bottom edges. The vertical curve at fixed $x$ is a closed timelike curve: after a positive proper duration $P_t$, it returns to the same spacetime event. The quotient is a mathematically well-defined flat Lorentzian manifold, but it violates chronology and admits no usual global causal ordering or ordinary global initial-value evolution.

These are physical causality problems, rather than an algebraic contradiction in the metric. Fields and particle histories must satisfy periodic consistency conditions; an arbitrary initial state followed by irreversible evolution will generally fail to return to itself after a temporal lap. The fact that the causal past and causal future overlap everywhere, established below, explains why the ordinary three-way causal partition degenerates.

For causal reachability on a spatial circle, put $P=2\pi R$ and use a representative $\bar{x}\in [-P/2,P/2)$ of the spatial circle. The lower-left panel of part c is a fundamental strip with its vertical edges identified. On the covering space the charged world line is still the solution of part b; on the quotient it isA jump between opposite strip edges is a coordinate seam, not a discontinuity of the physical world line. At its $n$th complete lap, $x=nP$, soFor $a=q{E}_x/m>0$, the speed increases on each return while remaining below $c$. Its kinetic energy increases byThere is no single-valued electrostatic potential energy $-q{E}_x\bar{x}$ on a circle: the nonzero electromotive force is $\oint \mathbf{E}\cdot d\mathbf{l}=E_{x}P$. A globally defined magnetic vector potential one-form $A_x=-E_{x}t$ instead makes the driving explicit. In an embedded circuit, Faraday's law realizes this electromotive force through changing magnetic flux. More generally an external apparatus maintaining the field supplies the increasing particle energy. In an isolated dynamical field-plus-particle system, backreaction would change the field. Thus this is not a cyclic machine extracting work from a conservative potential, and it does not violate conservation of energy.