Maxwell-Lorentz Without Self Interactions

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In this model, the universe consists of a finite number of point charges. Each charge’s 4–acceleration is determined by the sum of the electromagnetic fields of all the other particles. The approach is covariant so it applies to any spacetime. There is no “external” electromagnetic field. The model is described by an action, or equivalently a Lagrangian. As well as deriving the dynamic equations for the particles and fields, the Lagrangian can also be used to derive the stress-energy tensor. There are two ways of deriving the stress-energy tensor from a Lagrangian. By varying the Lagrangian with respect to the metric one obtains the “Hilbert” stress-energy tensor. However one can use the Noether fields and the Belenfante- Rosen modification to derive the “Belenfante” stress-energy tensor. Since all the fields in this action are dynamic, i.e. there are no external fields, other than gravity, there are three important consequences [21]: The Hilbert and Belenfante-Rosen definitions of the stress- energy tensor are equal, the stress-energy tensor is symmetric and it is covariantly conserved. Thus all Killing symmetries of the spacetime will give rise to conserved currents. This establishes the claim that for Minkowski spacetime, energy, momentum and angular momentum are conserved.

The letter is organised as follows: We start by giving the Lagrangian of the ML–SI model, the corresponding dynamic equations and the stress-energy tensors. The derivation of these results is placed in the appendix. We then give some predictions of this model. In the next section we list additional consequences of the model. We then conclude.

2 The self-interaction free model

Let M be spacetime with background metric gµν, µ, ν = 0, 1, 2, 3 with signature (−, +, +, +). Consider a collection of N particles {PA, A =1,...,N}, with masses mA and charges qA. The particles travel µ 1 − + 2 along worldlines CA(τ) with proper time parameterisation τ with τA <τ<τA so that the 4-velocity ˙ µ d µ ˙ µ ˙ A A CA(τ) = dτ CA satisfies CA Cµ = −1. One needs to consider N electromagnetic fields Fµν which are A A A A generated by N potentials Aµ so that Fµν = ∂µAν − ∂ν Aµ . µ A The action depends on the N worldlines CA(τ), the N fields Aµ and the metric gµν is given formally by µ A S[CA, Aµ ,gµν] N + m τA A ˙ µ ˙ A = C (τ) Cµ (τ) dτ 2 Z − A XA=1 τA + N τA ˙ µ B (1) − qA C (τ) Aµ |CA(τ) dτ Z − A XA=1 B,BX=6 A τA 1 N − F A F µν ωd4x , 4 Z µν B XA=1 B,BX=6 A M where ω = (− det g) and refers to the double sum over all B excluding A, i.e. = p B,BX=6 A B,BX=6 A A−1 N + . We state that (1) is the formal action as any actual integral would likely diverge. To get BX=1 BX=A+1

1 − + In cases where the particles do not accelerate to lightlike infinity, τA = −∞ and τA = ∞. However we do not want to exclude the case when a particle accelerates to infinity in a finite proper time. 2We use the summation convention on the spacetime index µ =0, 1, 2, 3, no summation on the particle index A. The particle index A is placed either high or low for convenience.

2 a convergent integral it is necessary to integrate over a compact region U ⊂ M so that µ A S[CA, Aµ ,gµν, U] N m = A C˙ µ(τ) C˙ A(τ) dτ 2 Z A µ XA=1 IA(U) N ˙ µ B (2) − qA C (τ) A |CA(τ) dτ Z A µ XA=1 B,BX=6 A IA(U) 1 N − F A F µν ωd4x , 4 Z µν B XA=1 B,BX=6 A U where IA(U) = {τ|CA(τ) ∈ U}. An alternative method is to use a test function, which we do below in (14). We can write the action S in terms of a distributional Lagrangian L

S = Lωd4x , (3) ZU where N + m τA A −1 ˙ µ ˙ A L = ω C Cµ δ x − CA(τ) dτ 2 Z − A XA=1 τA + N τA −1 ˙ µ B − qA ω C Aµ δ x − CA(τ) dτ (4) Z − A XA=1 B,BX=6 A τA 1 N − F A F µν . 4 µν B XA=1 B,BX=6 A µ Varying S with respect to CA gives the Lorentz force equation for the particle A, DC˙ µ m A = q C˙ A F νµ , (5) A dτ A ν B B,BX=6 A D ˙ µ ¨µ µ ˙ ν ˙ ρ where dτ CA = CA +ΓνρCACA. A Varying S with respect to Aµ gives

+ τB νµ −1 µ ∇νF = qB ω C˙ δ(x − CB(τ)) dτ . (6) B Z − B B,BX=6 A B,BX=6 A τB B This is equivalent to the electromagnetic ﬁeld Fµν being generated by the particle PB.

+ τB νµ −1 µ ∇νF = qB ω C˙ δ(x − CB(τ)) dτ . (7) B − B ZτB Differentiating the Lagrangian density Lω with respect to the metric ∂(Lω) ∂L ∂ω T µν =2 ω−1 =2 +2 Lω−1 , (8) ∂gµν ∂gµν ∂gµν gives the “Hilbert” stress-energy tensor

+ N τA µν −1 (4) ˙ µ ˙ ν T = ω mA δ x − CA(τ) CA CA dτ Z − XA=1 τA (9) N µ νρ 1 µν ρσ B + FAρFB − 4 g FA Fρσ . XA=1 B,BX=6 A

3 This is the same as the “Belenfante-Rosen” stress-energy tensor

N ∂L T µ = L δµ − 2 F A ν ν νρ ∂F A XA=1 µρ N ∂L N ∂L − AA + C˙ µ (10) ν ∂AA A ∂C˙ ν XA=1 µ XA=1 A N µ ∂L − δ δ x − CA(τ) , ν ∂ δ(x − C (τ)) XA=1 A where we deﬁne ∂ δ x − CA(τ) L δ(x − CA(τ)) dτ ∂ δ(x − C (τ)) Z A (11) = L δ(x − CA(τ)) dτ . Z This last term of (10) may look contrived. However it arises from diffeomorphism invariance of the action. The details are outlined in the appendix. The stress-energy tensor is a total stress-energy tensor, therefore it has the symmetry of the indices

T µν = T νµ , (12) and the divergenceless condition, which is also known as being covariantly conserved

µν ∇µT =0 . (13) The derivations of (5)-(13) are given in the appendix.

We observe that both the Lagrangian L and the stress-energy tensor T µν are well deﬁned distributions. That is for any test function ϕ and test tensor ϕµν , the integrals

4 µν 4 Lϕd x and T ϕµν d x are finite. (14) ZM ZM This assumes that no two worldlines ever intersect. To see this observe that, away from all the particles, A there are only the electromagnetic fields Fµν , which are all finite. Thus if ϕ and ϕab have support which does not intersect any worldline, the integrals in (14) are finite. Along a particle PA, the line integrals are clearly bounded. Choose an adapted coordinate system 0 3 0 i R (x ,...,x ) such that CA(τ)= τ and CA(τ)=0 for i =1, 2, 3. Let r ∈ be the spatial distance given by 1 2 2 2 3 2 1/2 A −2 r = ((x ) +(x ) +(x ) ) . Then approximately Fµν (r) ≈ qA r . That is it does not go to infinity faster −2 2 A B than r . However in polar coordinates ω = r sin θ so Fµν (r)ω is bounded. Also Fµν (r) for PB =6 PA are all bounded so the integrals in (14) are all bounded. µν This contrasts with the standard Lagrangian for electromagnetism which contains Fµν F for a single µν −4 field. In this case Fµν F ≈ r which is not defined as a distribution.

3 Predictions of the ML–SI model

The principle prediction of this model is that no radiation reaction observed. As stated in the introduction, forthcoming laser-plasma experiments will have electromagnetic ﬁelds of sufﬁcient intensities that radiation reaction, it is exist, should be detectable. Thus if charged particle motion is consistent with the Lorentz force and not consistent with the models which include radiation reaction, then this would provide strong evidence for ML–SI. Another interesting prediction is for the case when the universe has a non-trivial topology. In Minkowski, Friedmann–Lemaître–Robertson–Walker or many other spacetimes it is impossible for a charged particle

4 PA to interact with its own electromagnetic radiation after it has been produced. By contrast if the universe is a 3–torus times time, then PA will see multiple copies of itself and will respond many times as its radiation goes round the torus. This model predicts that whereas PA will respond multiple times to the radiation of the other particles PB =6 PA, it will never respond to its own radiation. The same observation could, in principle, be made in our universe by looking at how charged particles behave in the ergosphere A of a rotating black hole. In this case part of the radiation Fµν will rotate faster than PA and hence have the opportunity to interact again with PA. However, in the ML–SI model it will not do so. This contrasts with the case when a charged particle PA sees itself in a perfect electrical conductor. A perfect electrical conductor is, itself, an approximation of a real metal made of electrons and protons. These will respond to the initial charged particle and in turn construct a field which can be seen by PA. This model also predicts that, in a universe with just a single charged particle, there would be no electromagnetic field produced and the particle will simply undergo geodesic motion. In Minkowski spacetime, µ the value of the single homogeneous field Ahom,A, where A =1, is irrelevant as it does not affect the motion of the particle PA.

4 Other Consequences of the ML–SI model

The Cauchy problem: The ML–SI is also a well defined Cauchy problem. We can find Cauchy surfaces on µ ˙ µ A which we can prescribe the initial positions CA and velocities CA as well as the electromagnetic field Fµν . The standard theories about the Cauchy problem for Maxwell’s equations and the Lorentz force equation, then state we can find the subsequent values.

tot A type of regularisation: We may think of ML–SI as a regularisation of the Lagrangian. Let Fµν = N A A=1 Fµν then for points away from any particles we may replace (4) and (9) with P 1 N L = F tot F µν − 1 F A F µν , (15) 4 µν tot 4 µν A XA=1 and µν 1 µ ρν 1 µν ρσ tot T = 2 FtotρFtot + 8 g Ftot Fρσ N (16) 1 µ ρν 1 µν ρσ A − 2 FAρFA + 8 g FA Fρσ . XA=1 However (15) and (16) do not extend to the worldlines as each term on the right hand sides diverge as ≈ r−4 as one approaches the worldline. Thus they are not individually distributions.

Liénard-Wiechart ﬁelds: In Minkowski spacetime we can solve Maxwell’s equations (7) using the Liénard-Wiechart potentials.

q C˙ µ(τ ) Aµ (x)= Aµ + A A R , (17) A hom,A ˙ ν A CA(τR) x − Cν (τR) µ µν where τR is the retarded time and Ahom,A is a solution to the source free Maxwell equation ∇νFhom,A = 0 hom,A hom,A hom,A with Fµν = ∂µAν − ∂νAµ .

Violation of weak energy condition: Finally, we observe that the ML–SI stress-energy tensor does not satisfy the weak energy condition. In Minkowski spacetimes, let the particle pass PA through the origin, µ ˙ µ µ 1 2 3 CA(τ0) = (0, 0, 0, 0) with CA(τ0) = (1, 0, 0, 0). Consider an observer at the point x = (r, x , x , x ) 1 2 2 2 3 2 1/2 ˙ µ ˙ ν 00 E E B B where r = ((x ) +(x ) +(x ) ) is small. In this case Tµν CACA = T = A · A + A · A E B E E where A and A are the electric and magnetic ﬁelds due to particle PA and A = B,B=6 A B and B B 00 A = B,B=6 A B are the ﬁelds due to all the other particles. Now T is dominatedP by the Coulomb P

5 00 −2 00 term, T ≈ −qA r rˆ · EA. Now choosing r = ±EA, depending on the sign of qA, then T < 0. The violation of the weak energy condition does not cause problems for the classical theory, but may cause issues for the corresponding quantum theory. This violation is also true for other classical theories such as Bopp-Podolski [22].

5 Discussion and Conclusion

In this letter we have presented the case why Maxwell-Lorentz without self interaction is the best model for the dynamics of charged particles. It has many advantages. The action (1) is a total action, that is all the fields are dynamic. As a result all the equations of motion are derivable from the action and there are no requirements for constitutive relations or equations of state. Another advantage is that the stress-energy tensor derived from varying the Lagrangian with respect to the metric is equivalent to that derived from the Noether current with the Belenfante-Rosen procedure. This tensor is symmetric in its indices and divergenceless and thus gives rise to conserved quantities whenever a Killing vector is present. The ML–SI makes predictions about the behaviour of electrons in plasmas which may be testable soon. The Lagrangian is also useful in that it only requires a finite quantity of information to specify, namely the masses and charges of the particles. Although, in general this quantity is large. This contrasts with for example, the cold plasma model of charge, which requires a infinite quantity of information. This model is certainly simpler than the different models which include radiation reaction. Some of these can also be derived from an action, but which may involve doubling the phase space [23]. The main disadvantage is the violation of the weak stress-energy tensor, which although not relevant for the classical theory, poses challenges for quantisation. Another criticisms of the ML–SI model is the N–fold increase in electromagnetic fields [19]. Thus to set up the Cauchy problem requires N initial conditions for the N particles as well as N initial conditions µ for the N fields. In Minkowski spacetimes, this can be seen by the N homogeneous fields Ahom,A given in µ (17). However if we could argue that one can set the homogeneous fields Ahom,A =0, then we can replace Maxwell’s equation (6) with the Liénard-Wiechart fields. Thus we are left with only differential difference equations. Thus the ML–SI approach is similar to the Feynmann-Wheeler theory [24, 25], in that particles only communicate with other particles. However since, in the Minkowski case, only the retarded potential is used, the system is manifestly casual. Even the results of future experiments indicate the existence of radiation reaction, the ML–SI model is a useful tool. It may be extended for the interaction of point dipoles and quadrupoles [26–29]. Here the multipoles only respond to the electromagnetic fields of the other particles. This is particularly important because the question of the self-interaction of multipoles is very challenging. Although these multipole models are not derived from a Lagrangian, that they can be described by a function distributional stress- energy tensor is very useful.

6 Acknowledgements

The author acknowledges support provided by STFC (Cockcroft Institute, ST/G008248/1 and ST/P002056/1), and EPSRC (Lab in a Bubble, EP/N028694/1). He would also like to thank Prof. Robin Tucker, Dr. Spyri- don Talaganis and Alex Warwick for their helpful advice.

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8 A Proofs of the formulae

We use the standard results

∂ω 1 −1 ∂(det g) 1 −1 µν 1 µν = − 2 ω = − 2 (det g)ω g = 2 ωg (18) ∂gµν ∂gµν so

∂ ω = ∂ (− det g)= − 1 ω−1∂ (det g)= − 1 ω−1(det g) gνρ ∂ g = 1 ωgνρ ∂ g µ µ 2 µ 2 µ νρ 2 µ νρ (19) 1 pσν ν 1 σρ ν ρ ν = 2 ωg (∂µgνσ + ∂ν gµσ − ∂σgµν )= δρ 2 ωg (∂µgνσ + ∂νgµσ − ∂σgµν)= ω δρ Γνµ = ω Γνµ and

ρσ ∂g ρα ∂gαβ βσ ρα µ ν βσ ρµ νσ = −g g = −g (δαδβ)g = −g g . (20) ∂gµν ∂gµν

A B A A A For the following, let Aµ = B,B=6 A Aµ , F µν = ∂µAµ − ∂ν Aµ then (1) becomes P N + N τA m A ˙ µ ˙ A ˙ µ A 1 A µν 4 S = CA Cµ − qA CA Aµ |CA(τ) dτ − Fµν F A ωd x . (21) Z − 2 4 Z XA=1 τA XA=1 M

1 ˙ µ ˙ A ˙ µ A Proof that varying (1) with respect to CA gives (5). Let LA = 2 mA CA Cµ − qA CA Aµ |CA(τ). From the standard Euler-Lagrange formula, from (21) we have ∂L d ∂L d 0= A − A = 1 m C˙ ν C˙ ρ ∂ g − q C˙ ν ∂ AA − m C˙ ν g − q AA ∂Cµ dτ ˙ µ 2 A A A µ νρ A A µ ν dτ A A µν A µ A ∂CA 1 ˙ ν ˙ ρ ˙ ν A ¨ν ˙ ν ˙ ρ ˙ ρ A = 2 mA CA CA ∂µgνρ − qA CA ∂µAν − mA CA gµν − mA CAC ∂ρgµν + qA C ∂ρAµ ˙ ν A ¨ν 1 ˙ ν ˙ ρ = −qA CA F µν − mA CA gµν + 2 mA CA CA ∂ρgµν + ∂νgµρ − ∂µgνρ ˙ ν A ¨ν ˙ ν ˙ ρ σ = −qA CA F µν − mA CA gµν + CA CA Γ νρgµσ ˙ ν A ¨ν ˙ σ ˙ ρ ν = qA CA F νµ − mA CA gµν + CA CA Γ σρgµν ˙ ν A ˙ ρ ˙ ν = qA CA F νµ − mA gµν CA∇ρCA hence (5) follows.

A Proof that varying (1) with respect to Aµ gives (6). Observe from (19)

νµ νµ νµ νµ νµ ρ νµ ρµ ν νρ µ ∂ν(FA ω)=(∂νFA ) ω + FA ∂νω =(∂νFA ) ω + FA Γρν ω = ω ∂νFA + FA Γνρ + FA Γνρ νµ = ω ∇νFA . From (4) we have

N + N + m τA τA A ˙ µ ˙ A ˙ µ A Lω = C Cµ δ x − CA dτ − qB C Aµ δ x − CB dτ 2 Z − A Z − B XA=1 τA XA=1 B,BX=6 A τB 1 N − ∂ AA F µν ω , 2 µ ν B XA=1 B,BX=6 A hence

+ ∂(Lω) ∂(Lω) τA ∂ q C˙ µ δ x C dτ ∂ F νµω 0= A − ν A = − B B ( − B) + ν B ∂A ∂(∂ν A ) Z − µ µ B,BX=6 A τA B,BX=6 A

9 + τA µ νµ = − qB C˙ δ(x − CB) dτ + ∂ν F ω Z − B B B,BX=6 A τA + τA µ νµ = − qB C˙ δ(x − CB) dτ + ω ∇ν F . Z − B B B,BX=6 A τA

Proof that (6) implies (7). Sum (6) over all A (and dividing by the number of particles) gives

+ N τB 1 νµ −1 µ 0= ∇νF − ω qB C˙ δ(x − CA(τ)) dτ N B Z − A XA=1 B,BX=6 A τB + N τB νµ −1 µ = ∇νF − ω qA C˙ δ(x − CA(τ)) dτ . A Z − A XA=1 τA Subtracting (6) gives (7).

Proof that (8) gives (9). Writing (21) with gµν explicit we have

+ + N τA N τA mA µ ν µ A ω L = C˙ C˙ gµν δ x − CA(τ) dτ − qA C˙ A δ x − CA(τ) dτ − A A − A µ 2 ZτA ZτA XA=1 XA=1 (22) 1 N − F A F A gµρ gνσ ω . 4 µν ρσ XA=1

When differentiating with respect to gµν we ignore the constraint gµν = gνµ. I.e. we assume that gµν is independent of gνµ, when µ =6 ν. Thus ∂ ˙ ρ ˙ σ ˙ ρ ˙ σ µ ν ˙ µ ˙ ν (CA CA gρσ)= CA CA δρ δσ = CA CA . ∂gµν Hence differentiating the ﬁrst term of (22) gives the ﬁrst term of (9). The second term of (22) in independent µν of gµν so does not contribute to T . For the last term

∂ A A ρα σβ A A ρµ να σβ ρα σµ νβ (Fρσ F αβ g g )= −Fρσ F αβ (g g g + g g g ) ∂gµν A νσ ρµ A ρν σµ µ νσ µ ρν µ νρ = −(Fρσ F A g + Fρσ F A g )= −(FAσ F A + FAρ F A )= −2FAρ F . Hence the last term of (22) gives

N N 1 ∂ A ρσ µ ρν 1 A ρσ − 2 Fρσ F A ω = ω FAρ F − 2 Fρσ F A . ∂gµν XA=1 XA=1 Hence (9).

Proof that (10) implies (9). Taking each term in (10) in turn. Second term:

N ∂L N −2 F A = F A F µρ . νρ ∂F A νρ A XA=1 µρ XA=1 Third term:

N N + ∂L τA A A −1 ˙ µ − Aν = Aν qA ω C δ x − CA(τ) dτ ∂AA Z − B XA=1 µ XA=1 B,BX=6 A τA

10 + N τA −1 ˙ µ A = qA ω C Aν δ x − CA(τ) dτ . Z − A XA=1 τA Fourth term:

N N + N + ∂L τA τA ˙ µ −1 ˙ µ ˙ A −1 ˙ µ A C = mA ω C Cν δ x − CA(τ) dτ − qA ω C Aν δ x − CA(τ) dτ . A ∂C˙ ν Z − A Z − A XA=1 A XA=1 τA XA=1 τA Fifth term:

N µ ∂L − δν δ x − CA(τ) XA=1 ∂ δ x − CA(τ) N + N + m ω−1 τA τA µ A ˙ µ ˙ A −1 ˙ µ A = −δν C Cµ δ x − CA(τ) dτ + qA ω C Aµ δ x − CA(τ) dτ . 2 Z − A Z − A XA=1 τA XA=1 τA Adding these together gives (9).

µν Proof of (13). Given test functions φν then for any tensor T then

µν 4 µν 4 ρν µ 4 µρ ν 4 (∇µT ) φν ωd x = (∂µT ) φν ωd x + T Γµρ φν ωd x + T Γµρ φν ωd x ZM ZM ZM ZM µν 4 ρν 4 µρ ν 4 = − T ∂µ(φν ω) d x + T (∂ρω)φν d x + T Γµρ φν ωd x ZM ZM ZM µν 4 µρ ν 4 µν 4 = − T (∂µφν) ωd x + T Γµρ φν ωd x = − T (∇µφν) ωd x . ZM ZM ZM For the ﬁrst term of (9) we have

+ N τA −1 (4) ˙ µ ˙ ν 4 ∇µ ω mA δ x − CA(τ) C CA dτ φν ωd x Z Z − A M XA=1 τA + N τA (4) ˙ µ ˙ ν 4 = − mA δ x − CA(τ) C CA dτ (∇µφν) d x Z Z − A M XA=1 τA N + N + τA τA Dφ ˙ µ ˙ ν ˙ ν ν = − mA C CA (∇µφν) dτ = − mA CA dτ Z − A Z − dτ XA=1 τA XA=1 τA N + τA dφ ˙ ν ν ρ ˙ µ = − mA CA − Γµν C φν dτ Z − dτ A XA=1 τA + N τA ¨ν ρ ˙ ν ˙ µ = mA CA +Γµν CA CA φν dτ Z − XA=1 τA N + τA ˙ ν DCA = mA φν dτ . Z − dτ XA=1 τA Hence using the Lorentz force equation (5)

N + N + τA τA DC˙ µ −1 (4) ˙ µ ˙ ν −1 A ∇µ ω mB δ x − CA(τ) C CA dτ = ω mA δ x − CA(τ) dτ Z − A Z − dτ XA=1 τA XA=1 τA + N τA −1 ˙ A νµ = ω qA Cν F δ x − CA(τ) dτ . (23) Z − A XA=1 τA

11 For the second term of (9) we have

µ νρ 1 µν ρσ A µ νρ µ νρ 1 µν ρσ A 1 µν ρσ A ∇ν F AρFA − 4 g FA F ρσ = ∇νF Aρ FA + F Aρ ∇νFA − 4 g ∇ν FA F ρσ − 4 g FA ∇νF ρσ µ νρ A µσ νρ 1 µν ρσ A = F Aρ ∇νFA + ∇νF σρ g FA − 2 g FA ∇ν F ρσ µ νρ 1 µσ A νρ ρν A = F Aρ ∇νFA + 2 g 2∇νF σρ FA − FA ∇σF ρν µ νρ 1 µσ A νρ A ρν νρ A = F Aρ ∇νFA + 2 g ∇νF σρ FA + ∇ρF σν FA + FA ∇σF ρν µ νρ 1 µσ νρ A A A = F Aρ ∇νFA + 2 g FA ∇νF σρ + ∇ρF νσ + ∇σF ρν µ νρ = F Aρ ∇νFA . Hence from Maxwell (7)

+ N N τB µ νρ 1 µν ρσ A µ νρ −1 ˙ ρ µ ∇ν F ρF − g F F ρσ = F ρ ∇νF = qB ω C F ρ δ(x − CB(τ)) dτ A A 4 A A A Z − B A XA=1 XA=1 τB + N τB −1 ˙ B ρµ = − qB ω Cρ F δ(x − CB(τ)) dτ . Z − A XA=1 τB Combining this with (23) gives (13).

A.1 Diffeomorphism invariance and the Noether formulation of stress-energy tensor The stress-energy can be derived in two ways from a Lagrangian. Either from variation with respect to the metric (8) or by using the Noether theorem, with the Belenfante-Rosen procedure (10). That these two are the same is due to: one, the action is diffeomorphism invariant and two, there are no background fields, other than the metric. These two conditions imply that the action is a total action, i.e. all the fields it depends on are dynamic. Total actions also lead to the dynamical equations, the index symmetry of the stress-energy tensor (12) and that it is divergenceless (13). As stated in the conclusion, however, such total actions are actually quite rare. The effect of background fields is detailed in [21]. The details of (10) require discussion about diffeomorphisms which is beyond the scope of this article. However (10) has been demonstrated directly above. To see where the last term of (10) originates from it is sufficient to observe that for a partial Lagrangian

τ + Spart = L(C˙ µ,Cµ) dτ Zτ − and corresponding distributional Lagrangian

τ + part µ µ L = L(C˙ ,C ) δ(x − CB(τ)) dτ , Zτ − then the corresponding stress-energy tensor is given by the fourth term of (10), i.e.

τ + part µ µ ∂L (T ) ν = C˙ δ(x − CB(τ)) dτ . ν Zτ − ∂C˙ part µ However, the ﬁrst term of (10), L δν , does not contribute to the stress-energy tensor. This is removed due to (11), which gives the following ∂Lpart Lpart = δ x − C(τ) . ∂ δ(x − C(τ)) Hence (10).