The Energy-Momentum Tensor - Durklnh23/presentations/energy-momentum-slides.pdf · to the...

Outline The Basics The Canonical Energy Momentum Tensor The Belinfante Tensor The Relativist’s Energy-Momentum Tensor Conclusions

The Energy-Momentum Tensor

Alan Reynolds

Department of Mathematical SciencesDurham University

October 31, 2016


Outline

• The Basics• Why a tensor?• What do all the bits mean?

• The Canonical Energy-Momentum Tensor• As a Noether current• Active transformations• As a Noether current - revisited• An example: Electromagnetism

• The Belinfante Tensor• A symmetric tensor?• Creating a symmetric energy-momentum tensor

• The Relativist’s Energy-Momentum Tensor• Passive transformations• Variation with respect to the metric• Electromagnetism revisited• Issues


What’s in a name?

• Energy-momentum tensor• Stress-energy tensor• Stress-energy-momentum tensor• Energy tensor• SEM• ...


Why a tensor?

E

E

Rest frame

Moving frame

• A particle has energy and momentum.• If a particle has energy (mass) E in its rest frame,

then in a frame moving at relative velocity v , theenergy is given by γE , where γ = 1/

√1− v2.

Energy is a component of a four-vector — thefour-momentum.

• Given a field or a fluid, we must consider the energydensity.

• If we switch from the rest frame to a moving frame,the energy in a volume element increases by a factorof γ, while the volume decreases by the same factor.The energy density increases by a factor of γ2.

• This suggests that the energy density is an element

of a(

20

)tensor.


What do all the bits mean?

t

x

Well,• T 00: Energy density,• T 0i : Energy flux across a surface of constant x i ,• T i0: Momentum density in i direction,• T ij : Flux of i momentum across a surface of

constant x j ,

or we can be more brief and simply say• Tαβ : Flux of α momentum across a surface of

constant xβ .

Note that, for this case at least, the energy-momentum tensor is symmetric —the energy flux across a surface of constant x i is the momentum density inthe i direction.1

1See B. Schutz, " A First Course in General Relativity", chapter 4, for a pedagogical introductionto the energy-momentum tensor, using ‘dust’ as an example.


Noether current recap

• A continuous ‘active’ transformation that leaves the equations of motionunchanged leads to a conserved current.

• To leave the equations of motion unchanged we require

∆S = 0 ⇔ L → L′ = L+ ε∂µFµ

for some Fµ. If the field transforms as

φ→ φ′ = φ+ ε∆φ,

then the Noether current is given by

jµ =∂L

∂(∂µφ)∆φ− Fµ.

• The Noether current is conserved, i.e. ∂µjµ = 0.


Energy-momentum as a Noether current — A typical presentationConsider a translation invariant theory and consider the translation

xµ → xµ − εaµ

as an active transformation. Then

φ(x)→ φ′(x) = φ(x + εa) = φ(x) + εaµ∂µφ(x),

so that ∆φ = aµ∂µφ(x). Likewise,

L → L+ εaµ∂µL,

i.e. Fµ = aµL. Hence, the Noether current is

jµ =∂L

∂(∂µφ)∆φ− Fµ = aν

(∂L

∂(∂µφ)∂νφ− δµνL

).

Since aµ is arbitrary (and constant), we get ∂µTµν = 0 where

Tµν =

∂L∂(∂µφ)

∂νφ− δµνL.


Active transformations: Misconceptions

• It is, perhaps, unhelpful to think of active and passive transformations asbeing opposites.

• It isn’t really the ’activeness’ of the transformation that is really importanthere.

• The kind of active transformation we want does not change thecoordinates. Only the field is transformed. We translate (or rotate) thecontents of the universe, but leave everything else — the coordinatesystem and the ‘theory’ — alone.


Active transformations: Scalar field

x

x'

• An active transformation is one “inwhich the field is truly shifted”(Tong).

• But it is only the field that ischanged.

• We want the value of the new fieldat x ′ to be equal to that for the oldfield at x , i.e.

φ′(x ′) = φ(x).

• If x ′ = Rx then we need φ′(Rx) =φ(x) and hence

φ′(x) = φ(R−1x).


Active transformations: Vector field

x

x'

• A rotation of a vector field results ina rotation of each individual vector,as well as the change of location.

• We get

φ′(x ′) = Rφ(x).

and hence

φ′(x) = Rφ(R−1x).

• Of course, other fields mightundergo different transformations,i.e.

φ′(x) = LRφ(R−1x),

where the LR are elements of arepresentation of the group ofrotations.


Energy-momentum as a Noether current, revisitedConsider a transformation of the field, φ→ φ′, such that

φ′(x ′) = φ(x),

where x ′µ = xµ − εaµ corresponds to a translation of the field. Then

φ′(x) = φ(x + εa) = φ(x) + εaµ∂µφ(x)

and ∆φ = aµ∂µφ(x).

We do not change the form of the Lagrangian, L(φ, ∂µφ), but its value atposition x will be different for the new field:

L(φ′(x), ∂µφ′(x)) = L(φ(x + εa), ∂µφ(x + εa)

= L(φ(x), ∂µφ(x)) + εaµ∂µL(φ(x), ∂µφ(x)).

Hence Fµ = aµL. As a result, the Noether current is

jµ =∂L


(∂L


).


Tµν =

∂L∂(∂µφ)

∂νφ− δµνL.



φ′(x ′) = φ(x),

where(((((((x ′µ = xµ − εaµ corresponds to a translation of the field. Then

φ′(x) = φ(x + εa) = φ(x) + εaµ∂µφ(x)






jµ =∂L


(∂L


).


Tµν =

∂L∂(∂µφ)

∂νφ− δµνL.



φ′(x ′) = φ(x),

where corresponds to a translation of the field. Then

φ′(x) = φ(x + εa) = φ(x) + εaµ∂µφ(x)






jµ =∂L


(∂L


).


Tµν =

∂L∂(∂µφ)

∂νφ− δµνL.



φ′(x ′) = φ(x),

where x ′µ = xµ − εaµ corresponds to a translation of the field. Then

φ′(x) = φ(x + εa) = φ(x) + εaµ∂µφ(x)






jµ =∂L


(∂L


).


Tµν =

∂L∂(∂µφ)

∂νφ− δµνL.


A non-translation invariant Lagrangian

Consider the Lagrangian

L(φ(x), ∂µφ(x), xµ) = xµxµφ∗(x)φ(x).

If we perform the same active transformation, i.e. translate the field only, thenthe value of the Lagrangian at point x becomes

L(φ′(x), ∂µφ′(x), xµ) = xµxµφ

∗(x + εa)φ(x + εa)

6= L(φ(x), ∂µφ(x)) + εaµ∂µL(φ(x), ∂µφ(x)).

So, we now know where the derivation of the energy-momentum tensor failsfor non-translation invariant theories.


An example: ElectromagnetismFor electromagnetism2, without sources, the action is

S =

∫d4x

(−1

4FµνFµν

),

where Fµν = ∂µAν − ∂νAµ. Now

∂Fµν

∂(∂ρAσ)= δρµδ

σν − δρνδσµ ,

∂Fµν

∂(∂ρAσ)= ηρµησν − ηρνησµ,

so

∂L∂(∂ρAσ)

= −14(δρµδ

σν − δρνδσµ

)Fµν − 1

4Fµν (ηρµησν − ηρνησµ)

= −14

Fρσ +14

Fσρ − 14

Fρσ +14

Fσρ = Fσρ.

The energy-momentum tensor is therefore

Tµν =

∂L∂(∂µAρ)

∂νAρ − δµνL = Fρµ∂νAρ +14

FρσFρσδµν .

2Peskin and Schroeder, exercise 2.1


A symmetric tensor?

• For fluids or dust, the energy-momentum tensor defined in terms ofmometum flux is symmetric.

• We would like a symmetric tensor for many reasons:• General relativity requires a symmetric tensor,• Symmetric tensors are simpler to manipulate,• etc.

• ButTµν =

∂L∂(∂µφ)

∂νφ− ηµνL

is not obviously symmetric.• The energy-momentum tensor for electromagnetism,

Tµν = Fρµ∂νAρ +14

FρσFρσηµν

certainly doesn’t look symmetric.


Creating a symmetric energy-momentum tensor — the Belinfante tensor

• If Kλµν is antisymmetric in the first two indices, then adding ∂λKλµν tothe energy-momentum tensor does not change its conservationproperties, since if

T̃µν = Tµν + ∂λKλµν ,

then∂µT̃µν =��∂µTµν +

XXXXX∂µ∂λKλµν = 0.

• For electromagnetism3, choose Kλµν = FµλAν . After a few calculations,we get

T̃µν = Tµν + ∂λKλµν =14

FρσFρσηµν + FλµF νλ + Aν∂λFµλ.

• But we can use the equations of motion, ∂λFµλ, to eliminate the lastterm to leave

14

FρσFρσηµν + FλµF νλ ,

which is manifestly symmetric.

3See Wikipedia (!) for the general case.


Symmetric ‘on shell’

• We used the equations of motion to show that the newenergy-momentum tensor is symmetric.

• In other words, we have shown that we can convert the canonicalenergy-momentum tensor to an alternative form that is equal, classicallyto an identically symmetric energy-momentum tensor.

• Classically, this distinction is of no importance.• Quantum mechanically, the two energy-momentum tensors are different

and result in different Ward identities.• Apparently, though, the change in the Ward identities is of no physical

significance.4

4See Di Francesco et al., "Conformal Field Theory", section 2.5.


Passive transformations

• A passive transformation is one “in which the mapping x → x ′ is viewedsimply as a coordinate transformation”5.

• Wikipedia (!) suggests that, given a rotation matrix R, a passivetransformation consists of rotating the basis vectors of our coordinatesystem using R. Hence with initial basis vectors eµ we obtain new basisvectors e′µ = Reµ. In the usual way, we find coordinates transform as

x ′µ = (R−1)µνxν ,

or simply x → x ′ = R−1x . Hence, in a sense, we obtain the oppositetransformation to that in the active case. But...

5Di Francesco et al.


Opposites?

• As physicists, we tend to consider coordinate transformations withoutthinking a great deal about basis vectors. Our starting point is usuallysomething like x → x ′ = Rx . If the same coordinate transformation isalso used to generate the active transformation, then the field transformsthe same way in both cases.

• An active transformation transforms the fields only. Any explicitdependence the Lagrangian has on x is unchanged. However, if weperform a mere change of coordinates, then the explicit dependence ofthe Lagrangian on x ′ will be different.


Variation with respect to the metric

Just when you think you might be getting the hang of the energy-momentumtensor, someone will say that it is defined as something like

Tab = −αM

8π1√−g

δSM

δgab

orTµν = − 2√

−gδLmatter

δgµν

orTαβ = − 4π√

−g∂S∂gαβ

.

They might also describe this as the ‘gravitational’ or ‘relativists’ definition ofenergy-momentum. But where on earth did these come from?


An intermediate, new definition6

• “Promote the constant parameter ε that appears in the symmetry to afunction of the spacetime coordinates".

• The change in the action should then be of the form

δS =

∫d4x Jα∂αε,

or in the case of a translation

δS =

∫d4x Jαβ∂αεβ .

• Terms in ε (undifferentiated) should cancel — these are the terms thatoccur when ε is a constant and we have assumed that the theory issymmetric under such transformations.

• But, on-shell, equations of motion are found by assuming that δS is zerofor any infinitesimal variation of the fields.

• Integrate by parts to get ∂αJα = 0, or ∂αJαβ = 0.

6Here we attempt to follow David Tong’s Part III lecture notes for String Theory.


A dynamical background metric

• Now consider the same theory, but coupled to a dynamical backgroundmetric, and view x ′ = x + ε as a “diffeomorphism”.

• The idea is that any sensible theory should not depend on the choice ofcoordinates.

• We will need to use methods from general relativity, since the metric (orits dependence on the coordinates) will change.

• We must introduce, temporarily at lease, the usual√−g into the

measure, and replace partial derivatives with covariant ones.• The assumption is that the various terms of δSdiff should come from two

different sources — the change in the field and the change in the metric.


A dynamical background metric• So

0 = δSdiff =

∫d4x Jαβ∂αεβ +

∫d4x

δSδgαβ

δgαβ .

• Evaluating δgµν :

g′µν =∂xα

∂x ′µ∂xβ

∂x ′νgαβ

= (δαµ − ∂µεα)(δβν − ∂νεβ)gαβ

= gµν − ∂µεν − ∂νεµ,

i.e. δgµν = −(∂µεν + ∂νεµ).• Hence we get ∫

d4x Jαβ∂αεβ = 2∫

d4xδSδgαβ

∂αεβ .

This means that Jαβ and 2 δSδgαβ

must have equal divergences.

• Since we know Jαβ is conserved, we can define

Tαβ =δSδgαβ

and know that this is also conserved.


Electromagnetism revisited

• Returning to the electromagnetism example, we adapt the action tocurved spacetime.

S =

∫d4x

√−g(−1

4gµρgνσFµνFρσ

),

where Fµν = ∇µAν −∇νAµ.• A bunch of calculations then gives us

δAδgαβ

= −18ηαβFµνFµν −

12

FασF βσ .

• Di Francesco et al. suggest a normalization factor of -2, giving us

Tαβ =14ηαβFµνFµν + FασF β

σ ,

which is precisely what we obtained for the Belinfante tensor.


Issues

• Remember, we assumed that δSdiff consisted of two parts — one fromvarying the fields and the other from varying the metric. If these are theonly contributions, then the above analysis is correct. But are they?

• Consider electromagnetism again. Calculating δSdiff, we get

δSdiff =

∫d4x δ

√−g(−1

4FµνFµν

)+

∫d4x

√−g(−1

4δgµρgνσFµνFρσ

)+

∫d4x

√−g(−1

4gµρδgνσFµνFρσ

)+

∫d4x

√−g(−1

4gµρgνσδFµνFρσ

)+

∫d4x

√−g(−1

4gµρgνσFµνδFρσ

).


Issues• At first sight, this looks fine — the first three lines come from varying the

metric, while the last two appear to come from varying the field.• However, the field we vary to find the equation of motion is Aµ, not Fµν .• Moreover, as Fµν is defined in terms of derivatives of Aµ, we must

consider the possibility that we get contributions from the change, underthe coordinate transformation, of the derivative.

• Under the active transformation, the derivative operator does notchange. Looking at just ∇µAν , we get

∇µA′ν = ∂µAν − ∂µερ∂ρAν − ερ∂µ∂ρAν − ∂µAρ∂νερ − Aρ∂µ∂νε

ρ.

• Under the passive transformation, the derivative operator (or itsdependence on the coordinates) changes. We get

∇′µA′ν = ∂µAν − ∂µερ∂ρAν − ερ∂µ∂ρAν − ∂µAρ∂νερ.

• It looks like we might get a third contribution to δSdiff. But, inelectromagnetism we get a stroke of luck. The derivatives of Aµ arecombined antisymmetrically to obtain Fµν , which results in this thirdcontribution cancelling. (Or is there some deeper reason for this ‘luck’?)


Conclusions

• Active and passive transformations are often presented in a less thanclear manner, yet an understanding of precisely what sort oftransformation is taking place is often necessary.

• Rather than being one object, there are multiple definitions of theenergy-momentum tensor.

• These different definitions lead to different tensors — for instance, thecanonical energy-momentum tensor need not by symmetric.

• A ‘relocalization procedure’ can be used to convert between forms(though I haven’t described the details.)

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