Four-gradient

Divergence is a vector operator that produces a signed scalar field giving the quantity of a vector field's source at each point. Note that in this metric signature [+,−,−,−] the 4-Gradient has a negative spatial component. It gets canceled when taking the 4D dot product since the Minkowski Metric is Diagonal[+1,−1,−1,−1].

The 4-divergence of the 4-position ${\displaystyle X^{\mu }=\left(ct,{\vec {\mathbf {x} }}\right)}$ gives the dimension of spacetime:

$${\displaystyle {\boldsymbol {\partial }}\cdot \mathbf {X} =\partial ^{\mu }\eta _{\mu \nu }X^{\nu }=\partial _{\nu }X^{\nu }=\left({\frac {\partial _{t}}{c}},-{\vec {\nabla }}\right)\cdot (ct,{\vec {x}})={\frac {\partial _{t}}{c}}(ct)+{\vec {\nabla }}\cdot {\vec {x}}=(\partial _{t}t)+(\partial _{x}x+\partial _{y}y+\partial _{z}z)=(1)+(3)=4}$$

The 4-divergence of the 4-current density

$${\displaystyle J^{\mu }=\left(\rho c,{\vec {\mathbf {j} }}\right)=\rho _{o}U^{\mu }=\rho _{o}\gamma \left(c,{\vec {\mathbf {u} }}\right)=\left(\rho c,\rho {\vec {\mathbf {u} }}\right)}$$

gives a conservation law – the conservation of charge:[1]^{: 103–107}

$${\displaystyle {\boldsymbol {\partial }}\cdot \mathbf {J} =\partial ^{\mu }\eta _{\mu \nu }J^{\nu }=\partial _{\nu }J^{\nu }=\left({\frac {\partial _{t}}{c}},-{\vec {\nabla }}\right)\cdot (\rho c,{\vec {j}})={\frac {\partial _{t}}{c}}(\rho c)+{\vec {\nabla }}\cdot {\vec {j}}=\partial _{t}\rho +{\vec {\nabla }}\cdot {\vec {j}}=0}$$

This means that the time rate of change of the charge density must equal the negative spatial divergence of the current density ${\displaystyle \partial _{t}\rho =-{\vec {\nabla }}\cdot {\vec {j}}}$.

In other words, the charge inside a box cannot just change arbitrarily, it must enter and leave the box via a current. This is a continuity equation.

The 4-divergence of the 4-number flux (4-dust) ${\displaystyle N^{\mu }=\left(nc,{\vec {\mathbf {n} }}\right)=n_{o}U^{\mu }=n_{o}\gamma \left(c,{\vec {\mathbf {u} }}\right)=\left(nc,n{\vec {\mathbf {u} }}\right)}$ is used in particle conservation:[4]^: 90–110

$${\displaystyle {\boldsymbol {\partial }}\cdot \mathbf {N} =\partial ^{\mu }\eta _{\mu \nu }N^{\nu }=\partial _{\nu }N^{\nu }=\left({\frac {\partial _{t}}{c}},-{\vec {\nabla }}\right)\cdot \left(nc,n{\vec {\mathbf {u} }}\right)={\frac {\partial _{t}}{c}}\left(nc\right)+{\vec {\nabla }}\cdot n{\vec {\mathbf {u} }}=\partial _{t}n+{\vec {\nabla }}\cdot n{\vec {\mathbf {u} }}=0}$$

This is a conservation law for the particle number density, typically something like baryon number density.

The 4-divergence of the electromagnetic 4-potential ${\textstyle A^{\mu }=\left({\frac {\phi }{c}},{\vec {\mathbf {a} }}\right)}$ is used in the Lorenz gauge condition:[1]^{: 105–107}

$${\displaystyle {\boldsymbol {\partial }}\cdot \mathbf {A} =\partial ^{\mu }\eta _{\mu \nu }A^{\nu }=\partial _{\nu }A^{\nu }=\left({\frac {\partial _{t}}{c}},-{\vec {\nabla }}\right)\cdot \left({\frac {\phi }{c}},{\vec {a}}\right)={\frac {\partial _{t}}{c}}\left({\frac {\phi }{c}}\right)+{\vec {\nabla }}\cdot {\vec {a}}={\frac {\partial _{t}\phi }{c^{2}}}+{\vec {\nabla }}\cdot {\vec {a}}=0}$$

This is the equivalent of a conservation law for the EM 4-potential.

The 4-divergence of the transverse traceless 4D (2,0)-tensor ${\displaystyle h_{TT}^{\mu \nu }}$ representing gravitational radiation in the weak-field limit (i.e. freely propagating far from the source).

The transverse condition

$${\displaystyle {\boldsymbol {\partial }}\cdot h_{TT}^{\mu \nu }=\partial _{\mu }h_{TT}^{\mu \nu }=0}$$

is the equivalent of a conservation equation for freely propagating gravitational waves.

The 4-divergence of the stress–energy tensor ${\displaystyle T^{\mu \nu }}$ as the conserved Noether current associated with spacetime translations, gives four conservation laws in SR:[4]^{: 101–106}

The conservation of energy (temporal direction) and the conservation of linear momentum (3 separate spatial directions).

$${\displaystyle {\boldsymbol {\partial }}\cdot T^{\mu \nu }=\partial _{\nu }T^{\mu \nu }=T^{\mu \nu }{}_{,\nu }=0^{\mu }=(0,0,0,0)}$$

It is often written as:

$${\displaystyle \partial _{\nu }T^{\mu \nu }=T^{\mu \nu }{}_{,\nu }=0}$$

where it is understood that the single zero is actually a 4-vector zero ${\displaystyle 0^{\mu }=(0,0,0,0)}$.

When the conservation of the stress–energy tensor (${\displaystyle \partial _{\nu }T^{\mu \nu }=0^{\mu }}$) for a perfect fluid is combined with the conservation of particle number density (${\displaystyle {\boldsymbol {\partial }}\cdot \mathbf {N} =0}$), both utilizing the 4-gradient, one can derive the relativistic Euler equations, which in fluid mechanics and astrophysics are a generalization of the Euler equations that account for the effects of special relativity. These equations reduce to the classical Euler equations if the fluid 3-space velocity is much less than the speed of light, the pressure is much less than the energy density, and the latter is dominated by the rest mass density.

In flat spacetime and using Cartesian coordinates, if one combines this with the symmetry of the stress–energy tensor, one can show that angular momentum (relativistic angular momentum) is also conserved:

$${\displaystyle \partial _{\nu }\left(x^{\alpha }T^{\mu \nu }-x^{\mu }T^{\alpha \nu }\right)=\left(x^{\alpha }T^{\mu \nu }-x^{\mu }T^{\alpha \nu }\right)_{,\nu }=0^{\alpha \mu }}$$

where this zero is actually a (2,0)-tensor zero.

The Jacobian matrix is the matrix of all first-order partial derivatives of a vector-valued function.

The 4-gradient ${\displaystyle \partial ^{\mu }}$ acting on the 4-position ${\displaystyle X^{\nu }}$ gives the SR Minkowski space metric ${\displaystyle \eta ^{\mu \nu }}$:[3]^: 16

$${\displaystyle {\begin{aligned}{\boldsymbol {\partial }}[\mathbf {X} ]=\partial ^{\mu }[X^{\nu }]=X^{\nu _{,}\mu }&=\left({\frac {\partial _{t}}{c}},-{\vec {\nabla }}\right)\left[\left(ct,{\vec {x}}\right)\right]=\left({\frac {\partial _{t}}{c}},-\partial _{x},-\partial _{y},-\partial _{z}\right)[(ct,x,y,z)],\\[3pt]&={\begin{bmatrix}{\frac {\partial _{t}}{c}}ct&{\frac {\partial _{t}}{c}}x&{\frac {\partial _{t}}{c}}y&{\frac {\partial _{t}}{c}}z\\-\partial _{x}ct&-\partial _{x}x&-\partial _{x}y&-\partial _{x}z\\-\partial _{y}ct&-\partial _{y}x&-\partial _{y}y&-\partial _{y}z\\-\partial _{z}ct&-\partial _{z}x&-\partial _{z}y&-\partial _{z}z\end{bmatrix}}={\begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}}\\[3pt]&=\operatorname {diag} [1,-1,-1,-1]=\eta ^{\mu \nu }.\end{aligned}}}$$

For the Minkowski metric, the components ${\displaystyle \left[\eta ^{\mu \mu }\right]=1/\left[\eta _{\mu \mu }\right]}$ (${\displaystyle \mu }$ not summed), with non-diagonal components all zero.

For the Cartesian Minkowski Metric, this gives ${\displaystyle \eta ^{\mu \nu }=\eta _{\mu \nu }=\operatorname {diag} [1,-1,-1,-1]}$.

Generally, ${\displaystyle \eta _{\mu }^{\nu }=\delta _{\mu }^{\nu }=\operatorname {diag} [1,1,1,1]}$, where ${\displaystyle \delta _{\mu }^{\nu }}$ is the 4D Kronecker delta.

The Lorentz transformation is written in tensor form as[4]^: 69

$${\displaystyle X^{\mu '}=\Lambda _{\nu }^{~\mu '}X^{\nu }}$$

and since ${\displaystyle \Lambda _{\nu }^{\mu '}}$ are just constants, then

$${\displaystyle {\dfrac {\partial X^{\mu '}}{\partial X^{\nu }}}=\Lambda _{\nu }^{\mu '}}$$

Thus, by definition of the 4-gradient

$${\displaystyle \partial _{\nu }\left[X^{\mu '}\right]=\left({\dfrac {\partial }{\partial X^{\nu }}}\right)\left[X^{\mu '}\right]={\dfrac {\partial X^{\mu '}}{\partial X^{\nu }}}=\Lambda _{\nu }^{\mu '}}$$

This identity is fundamental. Components of the 4-gradient transform according to the inverse of the components of 4-vectors. So the 4-gradient is the "archetypal" one-form.

The scalar product of 4-velocity ${\displaystyle U^{\mu }}$ with the 4-gradient gives the total derivative with respect to proper time ${\displaystyle {\frac {d}{d\tau }}}$:[1]^: 58–59

$${\displaystyle {\begin{aligned}\mathbf {U} \cdot {\boldsymbol {\partial }}&=U^{\mu }\eta _{\mu \nu }\partial ^{\nu }=\gamma \left(c,{\vec {u}}\right)\cdot \left({\frac {\partial _{t}}{c}},-{\vec {\nabla }}\right)=\gamma \left(c{\frac {\partial _{t}}{c}}+{\vec {u}}\cdot {\vec {\nabla }}\right)=\gamma \left(\partial _{t}+{\frac {dx}{dt}}\partial _{x}+{\frac {dy}{dt}}\partial _{y}+{\frac {dz}{dt}}\partial _{z}\right)=\gamma {\frac {d}{dt}}={\frac {d}{d\tau }}\\{\frac {d}{d\tau }}&={\frac {dX^{\mu }}{dX^{\mu }}}{\frac {d}{d\tau }}={\frac {dX^{\mu }}{d\tau }}{\frac {d}{dX^{\mu }}}=U^{\mu }\partial _{\mu }=\mathbf {U} \cdot {\boldsymbol {\partial }}\end{aligned}}}$$

The fact that ${\displaystyle \mathbf {U} \cdot {\boldsymbol {\partial }}}$ is a Lorentz scalar invariant shows that the total derivative with respect to proper time ${\displaystyle {\frac {d}{d\tau }}}$ is likewise a Lorentz scalar invariant.

So, for example, the 4-velocity ${\displaystyle U^{\mu }}$ is the derivative of the 4-position ${\displaystyle X^{\mu }}$ with respect to proper time:

$${\displaystyle {\frac {d}{d\tau }}\mathbf {X} =(\mathbf {U} \cdot {\boldsymbol {\partial }})\mathbf {X} =\mathbf {U} \cdot {\boldsymbol {\partial }}[\mathbf {X} ]=U^{\alpha }\cdot \eta ^{\mu \nu }=U^{\alpha }\eta _{\alpha \nu }\eta ^{\mu \nu }=U^{\alpha }\delta _{\alpha }^{\mu }=U^{\mu }=\mathbf {U} }$$

or

$${\displaystyle {\frac {d}{d\tau }}\mathbf {X} =\gamma {\frac {d}{dt}}\mathbf {X} =\gamma {\frac {d}{dt}}\left(ct,{\vec {x}}\right)=\gamma \left({\frac {d}{dt}}ct,{\frac {d}{dt}}{\vec {x}}\right)=\gamma \left(c,{\vec {u}}\right)=\mathbf {U} }$$

Another example, the 4-acceleration ${\displaystyle A^{\mu }}$ is the proper-time derivative of the 4-velocity ${\displaystyle U^{\mu }}$:

$${\displaystyle {\begin{aligned}{\frac {d}{d\tau }}\mathbf {U} &=(\mathbf {U} \cdot {\boldsymbol {\partial }})\mathbf {U} =\mathbf {U} \cdot {\boldsymbol {\partial }}[\mathbf {U} ]=U^{\alpha }\eta _{\alpha \mu }\partial ^{\mu }\left[U^{\nu }\right]\\&=U^{\alpha }\eta _{\alpha \mu }{\begin{bmatrix}{\frac {\partial _{t}}{c}}\gamma c&{\frac {\partial _{t}}{c}}\gamma {\vec {u}}\\-{\vec {\nabla }}\gamma c&-{\vec {\nabla }}\gamma {\vec {u}}\end{bmatrix}}=U^{\alpha }{\begin{bmatrix}\ {\frac {\partial _{t}}{c}}\gamma c&0\\0&{\vec {\nabla }}\gamma {\vec {u}}\end{bmatrix}}\\[3pt]&=\gamma \left(c{\frac {\partial _{t}}{c}}\gamma c,{\vec {u}}\cdot \nabla \gamma {\vec {u}}\right)=\gamma \left(c\partial _{t}\gamma ,{\frac {d}{dt}}\left[\gamma {\vec {u}}\right]\right)=\gamma \left(c{\dot {\gamma }},{\dot {\gamma }}{\vec {u}}+\gamma {\dot {\vec {u}}}\right)=\mathbf {A} \end{aligned}}}$$

or

$${\displaystyle {\frac {d}{d\tau }}\mathbf {U} =\gamma {\frac {d}{dt}}(\gamma c,\gamma {\vec {u}})=\gamma \left({\frac {d}{dt}}[\gamma c],{\frac {d}{dt}}[\gamma {\vec {u}}]\right)=\gamma (c{\dot {\gamma }},{\dot {\gamma }}{\vec {u}}+\gamma {\dot {\vec {u}}})=\mathbf {A} }$$

The Faraday electromagnetic tensor ${\displaystyle F^{\mu \nu }}$ is a mathematical object that describes the electromagnetic field in spacetime of a physical system.[1]^{: 101–128}[5]^: 314[3]^: 17–18[6]^: 29–30[7]^: 4

Applying the 4-gradient to make an antisymmetric tensor, one gets:

$${\displaystyle F^{\mu \nu }=\partial ^{\mu }A^{\nu }-\partial ^{\nu }A^{\mu }={\begin{bmatrix}0&-E_{x}/c&-E_{y}/c&-E_{z}/c\\E_{x}/c&0&-B_{z}&B_{y}\\E_{y}/c&B_{z}&0&-B_{x}\\E_{z}/c&-B_{y}&B_{x}&0\end{bmatrix}}}$$

where:

• Electromagnetic 4-potential ${\displaystyle A^{\mu }=\mathbf {A} =\left({\frac {\phi }{c}},{\vec {\mathbf {a} }}\right)}$, not to be confused with the 4-acceleration ${\displaystyle \mathbf {A} =\gamma \left(c{\dot {\gamma }},{\dot {\gamma }}{\vec {u}}+\gamma {\dot {\vec {u}}}\right)}$
• The electric scalar potential is ${\displaystyle \phi }$
• The magnetic 3-space vector potential is ${\displaystyle {\vec {\mathbf {a} }}}$

By applying the 4-gradient again, and defining the 4-current density as ${\displaystyle J^{\beta }=\mathbf {J} =\left(c\rho ,{\vec {\mathbf {j} }}\right)}$ one can derive the tensor form of the Maxwell equations:

$${\displaystyle \partial _{\alpha }F^{\alpha \beta }=\mu _{o}J^{\beta }}$$

$${\displaystyle \partial _{\gamma }F_{\alpha \beta }+\partial _{\alpha }F_{\beta \gamma }+\partial _{\beta }F_{\gamma \alpha }=0_{\alpha \beta \gamma }}$$

where the second line is a version of the Bianchi identity (Jacobi identity).

A wavevector is a vector which helps describe a wave. Like any vector, it has a magnitude and direction, both of which are important: Its magnitude is either the wavenumber or angular wavenumber of the wave (inversely proportional to the wavelength), and its direction is ordinarily the direction of wave propagation

The 4-wavevector ${\displaystyle K^{\mu }}$ is the 4-gradient of the negative phase ${\displaystyle \Phi }$ (or the negative 4-gradient of the phase) of a wave in Minkowski Space:[6]^: 387

$${\displaystyle K^{\mu }=\mathbf {K} =\left({\frac {\omega }{c}},{\vec {\mathbf {k} }}\right)={\boldsymbol {\partial }}[-\Phi ]=-{\boldsymbol {\partial }}[\Phi ]}$$

This is mathematically equivalent to the definition of the phase of a wave (or more specifically a plane wave):

$${\displaystyle \mathbf {K} \cdot \mathbf {X} =\omega t-{\vec {\mathbf {k} }}\cdot {\vec {\mathbf {x} }}=-\Phi }$$

where 4-position ${\displaystyle \mathbf {X} =\left(ct,{\vec {\mathbf {x} }}\right)}$, ${\displaystyle \omega }$ is the temporal angular frequency, ${\displaystyle {\vec {\mathbf {k} }}}$ is the spatial 3-space wavevector, and ${\displaystyle \Phi }$ is the Lorentz scalar invariant phase.

$${\displaystyle \partial [\mathbf {K} \cdot \mathbf {X} ]=\partial \left[\omega t-{\vec {\mathbf {k} }}\cdot {\vec {\mathbf {x} }}\right]=\left({\frac {\partial _{t}}{c}},-\nabla \right)\left[\omega t-{\vec {\mathbf {k} }}\cdot {\vec {\mathbf {x} }}\right]=\left({\frac {\partial _{t}}{c}}\left[\omega t-{\vec {\mathbf {k} }}\cdot {\vec {\mathbf {x} }}\right],-\nabla \left[\omega t-{\vec {\mathbf {k} }}\cdot {\vec {\mathbf {x} }}\right]\right)=\left({\frac {\partial _{t}}{c}}[\omega t],-\nabla \left[-{\vec {\mathbf {k} }}\cdot {\vec {\mathbf {x} }}\right]\right)=\left({\frac {\omega }{c}},{\vec {\mathbf {k} }}\right)=\mathbf {K} }$$

with the assumption that the plane wave ${\displaystyle \omega }$ and ${\displaystyle {\vec {\mathbf {k} }}}$ are not explicit functions of ${\displaystyle t}$ or ${\displaystyle {\vec {\mathbf {x} }}}$.

The explicit form of an SR plane wave ${\displaystyle \Psi _{n}(\mathbf {X} )}$ can be written as:[7]^: 9

$${\displaystyle \Psi _{n}(\mathbf {X} )=A_{n}e^{-i(\mathbf {K_{n}} \cdot \mathbf {X} )}=A_{n}e^{i(\Phi _{n})}}$$

where ${\displaystyle A_{n}}$ is a (possibly complex) amplitude.

A general wave ${\displaystyle \Psi (\mathbf {X} )}$ would be the superposition of multiple plane waves:

$${\displaystyle \Psi (\mathbf {X} )=\sum _{n}[\Psi _{n}(\mathbf {X} )]=\sum _{n}\left[A_{n}e^{-i(\mathbf {K_{n}} \cdot \mathbf {X} )}\right]=\sum _{n}\left[A_{n}e^{i(\Phi _{n})}\right]}$$

Again using the 4-gradient,

$${\displaystyle \partial [\Psi (\mathbf {X} )]=\partial \left[Ae^{-i(\mathbf {K} \cdot \mathbf {X} )}\right]=-i\mathbf {K} \left[Ae^{-i(\mathbf {K} \cdot \mathbf {X} )}\right]=-i\mathbf {K} [\Psi (\mathbf {X} )]}$$

or

$${\displaystyle {\boldsymbol {\partial }}=-i\mathbf {K} }$$

which is the 4-gradient version of complex-valued plane waves

In special relativity, electromagnetism and wave theory, the d'Alembert operator, also called the d'Alembertian or the wave operator, is the Laplace operator of Minkowski space. The operator is named after French mathematician and physicist Jean le Rond d'Alembert.

The square of ${\displaystyle {\boldsymbol {\partial }}}$ is the 4-Laplacian, which is called the d'Alembert operator:[5]^: 300[3]^: 17‒18[6]^: 41[7]^: 4

$${\displaystyle {\boldsymbol {\partial }}\cdot {\boldsymbol {\partial }}=\partial ^{\mu }\cdot \partial ^{\nu }=\partial ^{\mu }\eta _{\mu \nu }\partial ^{\nu }=\partial _{\nu }\partial ^{\nu }={\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}-{\vec {\nabla }}^{2}=\left({\frac {\partial _{t}}{c}}\right)^{2}-{\vec {\nabla }}^{2}.}$$

As it is the dot product of two 4-vectors, the d'Alembertian is a Lorentz invariant scalar.

Occasionally, in analogy with the 3-dimensional notation, the symbols ${\displaystyle \Box }$ and ${\displaystyle \Box ^{2}}$ are used for the 4-gradient and d'Alembertian respectively. More commonly however, the symbol ${\displaystyle \Box }$ is reserved for the d'Alembertian.

Some examples of the 4-gradient as used in the d'Alembertian follow:

In the Klein–Gordon relativistic quantum wave equation for spin-0 particles (ex. Higgs boson):

$${\displaystyle \left[({\boldsymbol {\partial }}\cdot {\boldsymbol {\partial }})+\left({\frac {m_{0}c}{\hbar }}\right)^{2}\right]\psi =\left[\left({\frac {\partial _{t}^{2}}{c^{2}}}-{\vec {\nabla }}^{2}\right)+\left({\frac {m_{0}c}{\hbar }}\right)^{2}\right]\psi =0}$$

In the wave equation for the electromagnetic field (using Lorenz gauge ${\displaystyle ({\boldsymbol {\partial }}\cdot \mathbf {A} )=\left(\partial _{\mu }A^{\mu }\right)=0}$):

• In vacuum:
  $${\displaystyle ({\boldsymbol {\partial }}\cdot {\boldsymbol {\partial }})\mathbf {A} =({\boldsymbol {\partial }}\cdot {\boldsymbol {\partial }})A^{\alpha }=\mathbf {0} =0^{\alpha }}$$

  
• With a 4-current source, not including the effects of spin:
  $${\displaystyle ({\boldsymbol {\partial }}\cdot {\boldsymbol {\partial }})\mathbf {A} =({\boldsymbol {\partial }}\cdot {\boldsymbol {\partial }})A^{\alpha }=\mu _{0}\mathbf {J} =\mu _{0}J^{\alpha }}$$

  
• With quantum electrodynamics source, including effects of spin:
  $${\displaystyle ({\boldsymbol {\partial }}\cdot {\boldsymbol {\partial }})\mathbf {A} =({\boldsymbol {\partial }}\cdot {\boldsymbol {\partial }})A^{\alpha }=e{\bar {\psi }}\gamma ^{\alpha }\psi }$$

  

where:

• Electromagnetic 4-potential ${\displaystyle \mathbf {A} =A^{\alpha }=\left({\frac {\phi }{c}},\mathbf {\vec {a}} \right)}$ is an electromagnetic vector potential
• 4-current density ${\displaystyle \mathbf {J} =J^{\alpha }=\left(\rho c,\mathbf {\vec {j}} \right)}$ is an electromagnetic current density
• Dirac Gamma matrices ${\displaystyle \gamma ^{\alpha }=\left(\gamma ^{0},\gamma ^{1},\gamma ^{2},\gamma ^{3}\right)}$ provide the effects of spin

In the wave equation of a gravitational wave (using a similar Lorenz gauge ${\displaystyle \left(\partial _{\mu }h_{TT}^{\mu \nu }\right)=0}$)[6]^{: 274–322}

$${\displaystyle ({\boldsymbol {\partial }}\cdot {\boldsymbol {\partial }})h_{TT}^{\mu \nu }=0}$$

where ${\displaystyle h_{TT}^{\mu \nu }}$ is the transverse traceless 2-tensor representing gravitational radiation in the weak-field limit (i.e. freely propagating far from the source).

Further conditions on ${\displaystyle h_{TT}^{\mu \nu }}$ are:

• Purely spatial:
  $${\displaystyle \mathbf {U} \cdot h_{TT}^{\mu \nu }=h_{TT}^{0\nu }=0}$$

  
• Traceless:
  $${\displaystyle \eta _{\mu \nu }h_{TT}^{\mu \nu }=h_{TT\nu }^{\nu }=0}$$

  
• Transverse:
  $${\displaystyle {\boldsymbol {\partial }}\cdot h_{TT}^{\mu \nu }=\partial _{\mu }h_{TT}^{\mu \nu }=0}$$

  

In the 4-dimensional version of Green's function:

$${\displaystyle ({\boldsymbol {\partial }}\cdot {\boldsymbol {\partial }})G\left[\mathbf {X} -\mathbf {X'} \right]=\delta ^{(4)}\left[\mathbf {X} -\mathbf {X'} \right]}$$

where the 4D Delta function is:

$${\displaystyle \delta ^{(4)}[\mathbf {X} ]={\frac {1}{(2\pi )^{4}}}\int d^{4}\mathbf {K} e^{-i(\mathbf {K} \cdot \mathbf {X} )}}$$

In vector calculus, the divergence theorem, also known as Gauss's theorem or Ostrogradsky's theorem, is a result that relates the flow (that is, flux) of a vector field through a surface to the behavior of the vector field inside the surface. More precisely, the divergence theorem states that the outward flux of a vector field through a closed surface is equal to the volume integral of the divergence over the region inside the surface. Intuitively, it states that the sum of all sources minus the sum of all sinks gives the net flow out of a region. In vector calculus, and more generally differential geometry, Stokes' theorem (also called the generalized Stokes' theorem) is a statement about the integration of differential forms on manifolds, which both simplifies and generalizes several theorems from vector calculus.

$${\displaystyle \int _{\Omega }d^{4}X\left(\partial _{\mu }V^{\mu }\right)=\oint _{\partial \Omega }dS\left(V^{\mu }N_{\mu }\right)}$$

or

$${\displaystyle \int _{\Omega }d^{4}X\left({\boldsymbol {\partial }}\cdot \mathbf {V} \right)=\oint _{\partial \Omega }dS\left(\mathbf {V} \cdot \mathbf {N} \right)}$$

where

• ${\displaystyle \mathbf {V} =V^{\mu }}$ is a 4-vector field defined in ${\displaystyle \Omega }$
• ${\displaystyle {\boldsymbol {\partial }}\cdot \mathbf {V} =\partial _{\mu }V^{\mu }}$ is the 4-divergence of ${\displaystyle V}$
• ${\displaystyle \mathbf {V} \cdot \mathbf {N} =V^{\mu }N_{\mu }}$ is the component of ${\displaystyle V}$ along direction ${\displaystyle N}$
• ${\displaystyle \Omega }$ is a 4D simply connected region of Minkowski spacetime
• ${\displaystyle \partial \Omega =S}$ is its 3D boundary with its own 3D volume element ${\displaystyle dS}$
• ${\displaystyle \mathbf {N} =N^{\mu }}$ is the outward-pointing normal
• ${\displaystyle d^{4}X=(c\,dt)\left(d^{3}x\right)=(c\,dt)(dx\,dy\,dz)}$ is the 4D differential volume element

The Hamilton–Jacobi equation (HJE) is a formulation of classical mechanics, equivalent to other formulations such as Newton's laws of motion, Lagrangian mechanics and Hamiltonian mechanics. The Hamilton–Jacobi equation is particularly useful in identifying conserved quantities for mechanical systems, which may be possible even when the mechanical problem itself cannot be solved completely. The HJE is also the only formulation of mechanics in which the motion of a particle can be represented as a wave. In this sense, the HJE fulfilled a long-held goal of theoretical physics (dating at least to Johann Bernoulli in the 18th century) of finding an analogy between the propagation of light and the motion of a particle

The generalized relativistic momentum ${\displaystyle \mathbf {P_{T}} }$ of a particle can be written as[1]^: 93–96

$${\displaystyle \mathbf {P_{T}} =\mathbf {P} +q\mathbf {A} }$$

where ${\displaystyle \mathbf {P} =\left({\frac {E}{c}},{\vec {\mathbf {p} }}\right)}$ and ${\displaystyle \mathbf {A} =\left({\frac {\phi }{c}},{\vec {\mathbf {a} }}\right)}$

This is essentially the 4-total momentum ${\displaystyle \mathbf {P_{T}} =\left({\frac {E_{T}}{c}},{\vec {\mathbf {p_{T}} }}\right)}$ of the system; a test particle in a field using the minimal coupling rule. There is the inherent momentum of the particle ${\displaystyle \mathbf {P} }$, plus momentum due to interaction with the EM 4-vector potential ${\displaystyle \mathbf {A} }$ via the particle charge ${\displaystyle q}$.

The relativistic Hamilton–Jacobi equation is obtained by setting the total momentum equal to the negative 4-gradient of the action ${\displaystyle S}$.

$${\displaystyle \mathbf {P_{T}} =-{\boldsymbol {\partial }}[S]=\left({\frac {E_{T}}{c}},{\vec {\mathbf {p_{T}} }}\right)=\left({\frac {H}{c}},{\vec {\mathbf {p_{T}} }}\right)=-{\boldsymbol {\partial }}[S]=-\left({\frac {\partial _{t}}{c}},-{\vec {\boldsymbol {\nabla }}}\right)[S]}$$

The temporal component gives: ${\displaystyle E_{T}=H=-\partial _{t}[S]}$

The spatial components give: ${\displaystyle {\vec {\mathbf {p_{T}} }}={\vec {\boldsymbol {\nabla }}}[S]}$

where ${\displaystyle H}$ is the Hamiltonian.

This is actually related to the 4-wavevector being equal the negative 4-gradient of the phase from above. ${\displaystyle K^{\mu }=\mathbf {K} =\left({\frac {\omega }{c}},{\vec {\mathbf {k} }}\right)=-{\boldsymbol {\partial }}[\Phi ]}$

To get the HJE, one first uses the Lorentz scalar invariant rule on the 4-momentum:

$${\displaystyle \mathbf {P} \cdot \mathbf {P} =(m_{0}c)^{2}}$$

But from the minimal coupling rule:

$${\displaystyle \mathbf {P} =\mathbf {P_{T}} -q\mathbf {A} }$$

So:

$${\displaystyle {\begin{aligned}\left(\mathbf {P_{T}} -q\mathbf {A} \right)\cdot \left(\mathbf {P_{T}} -q\mathbf {A} \right)=\left(\mathbf {P_{T}} -q\mathbf {A} \right)^{2}&=\left(m_{0}c\right)^{2}\\\Rightarrow \left(-{\boldsymbol {\partial }}[S]-q\mathbf {A} \right)^{2}&=\left(m_{0}c\right)^{2}\end{aligned}}}$$

Breaking into the temporal and spatial components:

$${\displaystyle {\begin{aligned}&&\left(-{\frac {\partial _{t}[S]}{c}}-{\frac {q\phi }{c}}\right)^{2}-({\boldsymbol {\nabla }}[S]-q\mathbf {a} )^{2}&=(m_{0}c)^{2}\\&\Rightarrow &({\boldsymbol {\nabla }}[S]-q\mathbf {a} )^{2}-{\frac {1}{c^{2}}}(-\partial _{t}[S]-q\phi )^{2}+(m_{0}c)^{2}&=0\\&\Rightarrow &({\boldsymbol {\nabla }}[S]-q\mathbf {a} )^{2}-{\frac {1}{c^{2}}}(\partial _{t}[S]+q\phi )^{2}+(m_{0}c)^{2}&=0\end{aligned}}}$$

where the final is the relativistic Hamilton–Jacobi equation.

The 4-gradient is connected with quantum mechanics.

The relation between the 4-momentum ${\displaystyle \mathbf {P} }$ and the 4-gradient ${\displaystyle {\boldsymbol {\partial }}}$ gives the Schrödinger QM relations.[7]^: 3–5

$${\displaystyle \mathbf {P} =\left({\frac {E}{c}},{\vec {p}}\right)=i\hbar {\boldsymbol {\partial }}=i\hbar \left({\frac {\partial _{t}}{c}},-{\vec {\nabla }}\right)}$$

The temporal component gives: ${\displaystyle E=i\hbar \partial _{t}}$

The spatial components give: ${\displaystyle {\vec {p}}=-i\hbar {\vec {\nabla }}}$

This can actually be composed of two separate steps.

First:[1]^: 82–84

$${\displaystyle \mathbf {P} =\left({\frac {E}{c}},{\vec {p}}\right)=\hbar \mathbf {K} =\hbar \left({\frac {\omega }{c}},{\vec {k}}\right)}$$

which is the full 4-vector version of:

The (temporal component) Planck–Einstein relation ${\displaystyle E=\hbar \omega }$

The (spatial components) de Broglie matter wave relation ${\displaystyle {\vec {p}}=\hbar {\vec {k}}}$

Second:[5]^: 300

$${\displaystyle \mathbf {K} =\left({\frac {\omega }{c}},{\vec {k}}\right)=i{\boldsymbol {\partial }}=i\left({\frac {\partial _{t}}{c}},-{\vec {\nabla }}\right)}$$

which is just the 4-gradient version of the wave equation for complex-valued plane waves

The temporal component gives: ${\displaystyle \omega =i\partial _{t}}$

The spatial components give: ${\displaystyle {\vec {k}}=-i{\vec {\nabla }}}$

In quantum mechanics (physics), the canonical commutation relation is the fundamental relation between canonical conjugate quantities (quantities which are related by definition such that one is the Fourier transform of another).

• According to:[7]^: 4
  $${\displaystyle \left[P^{\mu },X^{\nu }\right]=i\hbar \left[\partial ^{\mu },X^{\nu }\right]=i\hbar \partial ^{\mu }\left[X^{\nu }\right]=i\hbar \eta ^{\mu \nu }}$$

  
• Taking the spatial components,
  $${\displaystyle \left[p^{j},x^{k}\right]=i\hbar \eta ^{jk}}$$

  
• Since ${\displaystyle \eta ^{\mu \nu }=\operatorname {diag} [1,-1,-1,-1]}$,
  $${\displaystyle \left[p^{j},x^{k}\right]=-i\hbar \delta ^{jk}}$$

  
• Since ${\displaystyle [a,b]=-[b,a]}$,
  $${\displaystyle \left[x^{k},p^{j}\right]=i\hbar \delta ^{kj}}$$

  
• And, relabeling indices gives the usual quantum commutation rules:
  $${\displaystyle \left[x^{j},p^{k}\right]=i\hbar \delta ^{jk}}$$

  

The 4-gradient is a component in several of the relativistic wave equations:[5]^{: 300–309}[3]^{: 25, 30–31, 55–69}

In the Klein–Gordon relativistic quantum wave equation for spin-0 particles (ex. Higgs boson):[7]^: 5

$${\displaystyle \left[\left(\partial ^{\mu }\partial _{\mu }\right)+\left({\frac {m_{0}c}{\hbar }}\right)^{2}\right]\psi =0}$$

In the Dirac relativistic quantum wave equation for spin-1/2 particles (ex. electrons):[7]^: 130

$${\displaystyle \left[i\gamma ^{\mu }\partial _{\mu }-{\frac {m_{0}c}{\hbar }}\right]\psi =0}$$

where ${\displaystyle \gamma ^{\mu }}$ are the Dirac gamma matrices and ${\displaystyle \psi }$ is a relativistic wave function.

${\displaystyle \psi }$ is Lorentz scalar for the Klein–Gordon equation, and a spinor for the Dirac equation.

It is nice that the gamma matrices themselves refer back to the fundamental aspect of SR, the Minkowski metric:[7]^: 130

$${\displaystyle \left\{\gamma ^{\mu },\gamma ^{\nu }\right\}=\gamma ^{\mu }\gamma ^{\nu }+\gamma ^{\nu }\gamma ^{\mu }=2\eta ^{\mu \nu }I_{4}}$$

Conservation of 4-probability current density follows from the continuity equation:[7]^: 6

$${\displaystyle {\boldsymbol {\partial }}\cdot \mathbf {J} =\partial _{t}\rho +{\vec {\boldsymbol {\nabla }}}\cdot {\vec {\mathbf {j} }}=0}$$

The 4-probability current density has the relativistically covariant expression:[7]^: 6

$${\displaystyle J_{\text{prob}}^{\mu }={\frac {i\hbar }{2m_{0}}}\left(\psi ^{*}\partial ^{\mu }\psi -\psi \partial ^{\mu }\psi ^{*}\right)}$$

The 4-charge current density is just the charge (q) times the 4-probability current density:[7]^: 8

$${\displaystyle J_{\text{charge}}^{\mu }={\frac {i\hbar q}{2m_{0}}}\left(\psi ^{*}\partial ^{\mu }\psi -\psi \partial ^{\mu }\psi ^{*}\right)}$$

Relativistic wave equations use 4-vectors in order to be covariant.[3][7]

Start with the standard SR 4-vectors:[1]

• 4-position ${\displaystyle \mathbf {X} =\left(ct,{\vec {\mathbf {x} }}\right)}$
• 4-velocity ${\displaystyle \mathbf {U} =\gamma \left(c,{\vec {\mathbf {u} }}\right)}$
• 4-momentum ${\displaystyle \mathbf {P} =\left({\frac {E}{c}},{\vec {\mathbf {p} }}\right)}$
• 4-wavevector ${\displaystyle \mathbf {K} =\left({\frac {\omega }{c}},{\vec {\mathbf {k} }}\right)}$
• 4-gradient ${\displaystyle {\boldsymbol {\partial }}=\left({\frac {\partial _{t}}{c}},-{\vec {\boldsymbol {\nabla }}}\right)}$

Note the following simple relations from the previous sections, where each 4-vector is related to another by a Lorentz scalar:

• 4-velocity ${\displaystyle \mathbf {U} ={\frac {d}{d\tau }}\mathbf {X} }$, where ${\displaystyle \tau }$ is the proper time
• 4-momentum ${\displaystyle \mathbf {P} =m_{0}\mathbf {U} }$, where ${\displaystyle m_{0}}$ is the rest mass
• 4-wavevector ${\displaystyle \mathbf {K} ={\frac {1}{\hbar }}\mathbf {P} }$, which is the 4-vector version of the Planck–Einstein relation & the de Broglie matter wave relation
• 4-gradient ${\displaystyle {\boldsymbol {\partial }}=-i\mathbf {K} }$, which is the 4-gradient version of complex-valued plane waves

Now, just apply the standard Lorentz scalar product rule to each one: