Stress-energy tensor and energy/momentum conservation clarification

In summary: Schutz would want to include something like this when it complicates the derivation.In summary, Schutz wants us to show that if the energy and momentum of a body are not conserved, for example because the body interacts with some other system, then we can define a nonzero relativistic force four-vector ##F^{\alpha}## which is defined by $$T^{\alpha \beta}{ }_{, \beta}= F^{\alpha}$$Without the extra term, the equation is ## \frac {\partial T^{0 0}} {\partial t} = -T^{0 i}{ }_{,i} + F^{0} $$ which says that the rate
  • #1
kmm
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I've been working through Bernard Schutz's book on GR and have run into some confusion in chapter 4 problem 20 part b. In this chapter, the stress-energy tensor for a general fluid was introduced and was used to derive the general conservation law for energy/momentum, where we found that $$T^{\alpha \beta}{ }_{, \beta}= 0$$

Schutz wants us to show that if the energy and momentum of a body are not conserved, for example because the body interacts with some other system, then we can define a nonzero relativistic force four-vector ##F^{\alpha}## which is defined by $$T^{\alpha \beta}{ }_{, \beta}= F^{\alpha}$$

To get this equation, it appears that we would have to add an extra term to the equation for conservation of energy, which is $$ \frac {\partial T^{0 0}} {\partial t} = -T^{0 i}{ }_{,i} + F^{0}$$ Without the extra term, the equation is ## \frac {\partial T^{0 0}} {\partial t} = -T^{0 i}{ }_{,i}## which says that the rate of change of energy per unit volume is equal to the net influx of energy per unit time per unit volume of the fluid element in question. In this case, I don't see what adding ## F^{0} ## really adds to this equation since I assumed that any influx of energy would be given by the term ## -T^{0 i}{ }_{,i} ##. If I'm looking at some fluid element within the fluid, and the body of fluid collides with some other body, changing it's total energy and momentum, then any change in energy of the fluid element would be due to local interactions with it's neighboring elements and the influx of energy would be given by ## -T^{0 i}{ }_{,i} ##.

Analogously, with the equation for momentum, we have $$ \frac {\partial T^{i 0}} {\partial t} = -T^{i j}{ }_{,j} + F^{i} $$
Again, without the term ##F^{i}##, this equation says that the time rate of change of the ith direction of momentum per unit volume of the fluid element is equal to the net influx of ith direction of momentum per unit time per unit volume of the fluid element. It appears that ##F^{i}## is just the ith component of the net external force. But it seems that any influx of momentum is given by ##-T^{i j}{ }_{,j}##, so as before, I'm not sure what ##F^{i}## adds. But, I see that combining these equations would give the equation Schutz wanted us to derive.

The only resolution I've come up with is that ## -T^{0 i}{ }_{,i} ## and ##-T^{i j}{ }_{,j}## only describe the net influx of energy and momentum, respectively, independent of any external source, and so we have to tack on the extra terms describing the contributions of the external source. These terms then only describe the interactions of the fluid element with neighboring fluid elements. But as I described, it seems that any external influence on the system would be conveyed by internal fluid elements on the fluid element in question. At the same time, I see that the total energy and momentum added to the system would have to be accounted for, and from that perspective, the added terms, ##F^{\alpha}## make sense to me. I am getting confused when I'm applying this to a fluid element within the system, which was the perspective from which Schutz derived ##T^{\alpha \beta}{ }_{, \beta}= 0##. I'm not sure what I'm misunderstanding. I appreciate any help!
 
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  • #2
kmm said:
it seems that any external influence on the system would be conveyed by internal fluid elements on the fluid element in question
This can't possibly be true for all fluid elements. Think of the ones at the boundary of the fluid.

Also, consider cases like a fluid of charged particles in an external electromagnetic field. The field acts directly on the particles, not indirectly by means of inter-particle interactions.
 
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  • #3
PeterDonis said:
This can't possibly be true for all fluid elements. Think of the ones at the boundary of the fluid.

Also, consider cases like a fluid of charged particles in an external electromagnetic field. The field acts directly on the particles, not indirectly by means of inter-particle interactions.
I had considered the fact that the external forces can be like those of the electric field on charged particles, however, why wouldn't those external forces simply be a part of the flux? Why would only the inter-particle reactions contribute to the flux of momentum and energy on a fluid element and not the external forces? I had the same thought regarding the fluid elements on the boundary. I also couldn't help but notice that with external forces included, this equation looks a lot like the differential form of Gauss's law in electrostatics, where you would get a divergence of the electric field only at points where there is a nonzero charge density. In this case, the external forces play the role of the "charge density". I just don't understand why the external forces don't contribute to the flux on the fluid element.

When we derived the conservation law, we looked at differentials of the flux across a fluid element, so it seems to me that if we had some other external force on a fluid element, not due to other fluid elements, then those forces would then contribute to those differentials.
 
  • #4
kmm said:
why wouldn't those external forces simply be a part of the flux?
Because they are not part of the stress-energy tensor you wrote down. The fluxes in your stress-energy tensor only include fluxes of the fluid particles themselves, not of external fields or field-particle interactions. That's why you have to include an external force term.

A different approach, which you will find in some other textbooks, is to add the stress-energy tensor of the external field (say an EM field) to the stress-energy tensor of the fluid, so you have a total stress-energy tensor for the entire system (now assumed to be a closed system). Then, in flat spacetime (or in a local inertial frame in curved spacetime in Minkowski coordinates), the equation ##T^{\alpha \beta}{}_{,\beta} = 0## will apply to that total stress-energy tensor--and if you expand out the individual terms in that equation, you will find that they contain the effects of "forces" being exchanged between the field and the fluid.

Schutz's approach simply models all of the effects on the fluid due to the EM field in the "external force" ##F^\alpha## instead of going to the additional trouble of writing down a stress-energy tensor for the field and the field-fluid interactions. His approach works as long as you can treat the external force as given and don't have to actually solve for its behavior.
 
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  • #5
PeterDonis said:
Because they are not part of the stress-energy tensor you wrote down. The fluxes in your stress-energy tensor only include fluxes of the fluid particles themselves, not of external fields or field-particle interactions. That's why you have to include an external force term.
OK, so this is probably the source of my confusion. The stress-energy tensor is only defining the fluid itself, which includes the internal interactions and random motions of particles giving rise to pressure within the fluid and so to the fluxes of the fluid particles themselves. I think the problem was that I made a leap in my thinking about the stress-energy tensor as describing any flux of energy or momentum including those from external sources, but it isn't. When we are looking at changes in energy or momentum of a fluid element, we must include all sources of that change, which in this case we now take into account by adding ##F^{\alpha}##. I see now, as you had said earlier in post #2, I would not have included ##F^{\alpha}## in calculating the change of energy and momentum for fluid elements at the center of the body if the external force only acted at the boundary of the body, since the stress-energy tensor already describes the internal interactions that would occur between the internal elements. I would only include ##F^{\alpha}## at the boundary elements. If we had a fluid of charged particles then, I would have to include ##F^{\alpha}## for all the elements. Do I seem to be making any mistakes here?
 
  • #6
kmm said:
I think the problem was that I made a leap in my thinking about the stress-energy tensor as describing any flux of energy or momentum including those from external sources, but it isn't.
Not the way Schutz is doing it in that section, no.

kmm said:
I would not have included ##F^{\alpha}## in calculating the change of energy and momentum for fluid elements at the center of the body if the external force only acted at the boundary of the body, since the stress-energy tensor already describes the internal interactions that would occur between the internal elements. I would only include ##F^{\alpha}## at the boundary elements. If we had a fluid of charged particles then, I would have to include ##F^{\alpha}## for all the elements. Do I seem to be making any mistakes here?
All of this looks correct to me.

There is also, as I said, the alternative of writing down a stress-energy tensor for the full fluid-field system that includes all of the stress-energy present, including energy stored in the field (and the field, in this formulation, would also have pressure and other nonzero SET components). Then what Schutz is calling the "external force" would show up as certain terms in the conservation law for the full stress-energy tensor.
 
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  • #7
PeterDonis said:
There is also, as I said, the alternative of writing down a stress-energy tensor for the full fluid-field system that includes all of the stress-energy present, including energy stored in the field (and the field, in this formulation, would also have pressure and other nonzero SET components). Then what Schutz is calling the "external force" would show up as certain terms in the conservation law for the full stress-energy tensor.
Great. This has definitely cleared things up for me. Thanks for the help!
 
  • #9
PeterDonis said:
Not the way Schutz is doing it in that section, no.All of this looks correct to me.

There is also, as I said, the alternative of writing down a stress-energy tensor for the full fluid-field system that includes all of the stress-energy present, including energy stored in the field (and the field, in this formulation, would also have pressure and other nonzero SET components). Then what Schutz is calling the "external force" would show up as certain terms in the conservation law for the full stress-energy tensor.
In general you can say that the equations of motion for a (special) relativistic fluid is given by the conservation of total energy and momentum. E.g., you can derive the expression for the electromagnetic energy-momentum tensor by applying this to the motion of the fluid.

In the most simple case you can assume "dust matter", i.e., you consider an ultra-dilute plasma, for which the pressure can be entirely neglected ("collisionless plasma"). Then the energy-momentum tensor reads
$$T_{\text{mech}}^{\mu \nu}=m n u^{\mu} u^{\nu},$$
where ##m## is the (invariant) mass of the fluid particles, ##n## the proper number density (i.e., the number of particles per unit volume as measured in the local restframe of the fluid), and ##u^{\mu}## the flow-four-velocity field. We also assume that there are no particle annihilation-creation processes, i.e., the particle number is conserved, which implies that
$$\partial_{\mu} (n u^{\mu})=0.$$
The "material proper-time derivative" for an arbitrary field quantity ##A(x)## is obviously given by ##\mathrm{D}_{\tau} A(x)=u^{\mu} \partial_{\mu} A(x)##.
The equation of motion for our dilute plasma then reads
$$m u^{\nu} \partial_{\nu} u^{\mu}=q n F^{\mu \nu} u_{\nu}.$$
On the righ-hand side is the Lorentz four-force acting on the charged particles of charge ##q##.

Now we have from the conservation of energy and momentum for the closed system consisting of the plasma and the em. field
$$\partial_{\mu} T_{\text{mech}}^{\mu \nu} = -\partial_{\mu} T_{\text{em}}^{\mu \nu}.$$
Using the continuity equation for particle number, we get for the left-hand side
$$\partial_{\mu} T_{\text{mech}}^{\mu \nu} =m n u^{\mu} \partial_{\mu} u^{\nu}=n q F^{\nu \mu} u_{\mu},$$
where we've used the equation of motion above. Now we use the inhomogeneous Maxwell equations to eliminate the current on the righ-hand side:
$$n q u^{\mu} = \partial_{\nu} F^{\nu \mu},$$
which gives
$$\partial_{\mu} T_{\text{mech}}^{\mu \nu} =F^{\nu \mu} \partial_{\rho} {F^{\rho}}_{\mu} = \partial_{\rho} (F^{\nu \mu} {F^{\rho}}_{\mu})-{F^{\rho}}_{\mu} \partial_{\rho} F^{\nu \mu}.$$
The homogeneous Maxwell equations read
$$\partial^{\rho} F^{\nu \mu} +\partial^{\nu} F^{\mu \rho} + \partial^{\mu} F^{\rho \nu}=0$$
The 2nd part of the previous equation thus is
$${F^{\rho}}_{\mu} \partial_{\rho} F^{\nu \mu} =F_{\rho \mu} \partial^{\rho} F^{\nu \mu} = \frac{1}{2} F_{\rho \mu} (\partial^{\rho} F^{\nu \mu} - \partial^{\mu} F^{\nu \rho}) = \frac{1}{2} F_{\rho \mu} (\partial^{\rho} F^{\nu \mu} + \partial^{\mu} F^{\rho \nu})=-\frac{1}{2} F_{\rho \mu} \partial^{\nu} F^{\mu \rho} = \frac{1}{4} \partial^{\nu} (F_{\rho \sigma} F^{\rho \sigma}) = \frac{1}{4} \partial_{\rho} (F_{\mu \sigma} F^{\mu \sigma} \eta^{\rho \nu}).$$
Thus we have
$$\partial_{\mu} T_{\text{mech}}^{\mu \nu} = -\partial_{\rho} \left (\frac{1}{4} F_{\mu \sigma} F^{\mu \sigma} \eta^{\rho \nu}-F^{\nu \mu} {F^{\rho}}_{\mu} \right).$$
From this we get
$$T_{\text{em}}^{\mu \nu} = \frac{1}{4} F_{\rho \sigma} F^{\rho \sigma} \eta^{\mu \nu} + {F^{\mu}}_{\rho} F^{\rho \nu}.$$
 
  • #10
I like to explain things with the aid of simpler analogies, so here is one. Instead of a fluid, consider a single particle. To further simplify, let the particle be nonrelativistic, falling in a Newtonian gravitational field. Clearly, the momentum and the kinetic energy of the particle are not conserved. The energy conservation can be saved by defining the total energy of the particle as the sum of its kinetic and potential energy. However, we still have a non-conservation of the momentum, because there is no such thing as "potential momentum" (*) that could compensate the change of the ordinary (kinetic) momentum. Thus, there is no way to accommodate all the conservation laws into a description of the particle alone; the external gravitational force does not conserve the momentum of the particle. To accommodate all conservation laws, one must consider the momentum (and the energy) of the full system, which involves not only the particle, but also the source of the gravitational field, such as the planet Earth.

(*) Why is there potential energy, but no potential momentum? Where does this difference come from? That's an interesting question by itself. Perhaps this is because, in non-relativistic mechanics, we treat time (related to energy) and space (related to momentum) differently. Indeed, in relativistic mechanics where time and space are treated on an equal footing, there is neither potential momentum nor potential energy. It would be interesting to put this idea into a more precise form.
 
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  • #11
Of course, in the relativistic context you also have to take into account that there cannot be any "actions at a distance" as in Newtonian mechanics, where you can have closed systems of a finite number of interacting point particles, where all 10 conservation laws of the spacetime symmetry of Newtonian mechanics hold.

In relativistic physics, the only successful way to describe closed dynamical systems is in terms of local field theories. The paradigmatic example is charged matter and the electromagnetic field. The field-theoretical description of matter is continuum mechanics, and indeed here you get a consistent system of equations of motion, where all 10 conservation laws of Poincare symmetry hold.

Point-particle mechanics cannot be formulated consistently in the relativistic realm. If you try that for electrodynamics you run into all the trouble with "radiation reaction", and in fact there are no-go theorems like the one by Leutwyler, according to which the only consistent relativistic Hamiltonian system of point particles are non-interacting particles.
 
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  • #12
Demystifier said:
in relativistic mechanics where time and space are treated on an equal footing, there is neither potential momentum nor potential energy.
This is not correct; in stationary spacetimes in relativity there is a well-defined potential energy, but not potential momentum. The reason for the difference is that the timelike Killing vector field of the spacetime defines a constant of geodesic motion that is used to define potential energy, but there is no corresponding constant of the motion for momentum.
 
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  • #13
PeterDonis said:
This is not correct; in stationary spacetimes in relativity there is a well-defined potential energy, but not potential momentum. The reason for the difference is that the timelike Killing vector field of the spacetime defines a constant of geodesic motion that is used to define potential energy, but there is no corresponding constant of the motion for momentum.
What if there is also a spacelike Killing vector field, in addition to the timelike one? The momentum in the direction of the spacelike Killing vector is conserved, but I wouldn't say that this defines a potential momentum. So if you are right that there is potential energy, it looks as if space directions and time directions behave differently even in relativistic mechanics. Do I miss something?
 
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  • #14
PeterDonis said:
This is not correct; in stationary spacetimes in relativity there is a well-defined potential energy, but not potential momentum. The reason for the difference is that the timelike Killing vector field of the spacetime defines a constant of geodesic motion that is used to define potential energy, but there is no corresponding constant of the motion for momentum.
In other words, for a momentum conservation law you'd also need invariance under spatial translations, i.e., corresponding coordinates such that the metric components don't depend on at least one of the spatial coordinates. The same you can, however, say about invariance under rotations and angular momentum. It depends on the Lie algebra generated of the Killing fields, which corresponds to the Lie algebra of the corresponding symmetry group of the given spacetime. A good introduction to this subject is in

H. Stephani, Relativity, 3rd edition, Cambridge University Press (2004) (chpt. 33)
 
  • #15
Demystifier said:
The momentum in the direction of the spacelike Killing vector is conserved, but I wouldn't say that this defines a potential momentum.
Nor does it define a "kinetic momentum" as normally understood. Mathematically, I suppose you could use the integral curves of the spacelike KVF to split up the conserved momentum into "kinetic" and "potential" parts, similar to the way integral curves of the timelike KVF are used to split up the conserved energy into "kinetic" and "potential" parts. But physically, you are right, they are not treated the same.

Demystifier said:
it looks as if space directions and time directions behave differently even in relativistic mechanics
In some ways, yes, they do. Timelike and spacelike vectors are not the same (and null vectors are different again). Relativity does not say that all vectors must be alike in all respects.
 
  • #16
PeterDonis said:
integral curves of the timelike KVF are used to split up the conserved energy into "kinetic" and "potential" parts.
Can you explain how this is done by equations? Or point to a reference where this is done?
 
  • #17
PeterDonis said:
Nor does it define a "kinetic momentum" as normally understood. Mathematically, I suppose you could use the integral curves of the spacelike KVF to split up the conserved momentum into "kinetic" and "potential" parts, similar to the way integral curves of the timelike KVF are used to split up the conserved energy into "kinetic" and "potential" parts. But physically, you are right, they are not treated the same.In some ways, yes, they do. Timelike and spacelike vectors are not the same (and null vectors are different again). Relativity does not say that all vectors must be alike in all respects.
A time-like Killing vector is the generator of temporal translations and thus the corresponding conserved quantity is rather an energy than a momentum. I have no clue, what you mean by "potential" or "momentum potential" in this context.

Take the example of the electromagnetic field in flat space. There is no potential for the Lorentz force acting on charged matter although there's conservation of energy and momentum for the closed system consisting of this charged matter an the em. field.
 
  • #18
Demystifier said:
Can you explain how this is done by equations?
I'll give it a shot from memory. Consider a spacetime with a timelike Killing vector field ##\xi^a##. An object with 4-momentum ##p_a## has energy at infinity ##\mathscr{E} = p_a \xi^a##.

Now consider an integral curve of the KVF; an observer whose worldline is such an integral curve has 4-velocity ##u^a = \xi^a / \sqrt{\xi^a \xi_a}##. (The factor in the denominator is often called the "redshift factor".) Call this observer a "stationary" observer. Then the energy ##E## of the above object as measured by the stationary observer is ##E = p_a u^a = p_a \xi^a / \sqrt{\xi^a \xi_a}##. Note that ##E## includes the rest energy ##m## of the object, so the kinetic energy of the object, as measured by the stationary observer, will be ##K = E - m##. (As usual I am using units in which ##G = c = 1##.)

We can thus split up the energy at infinity as follows:

$$
\mathscr{E} = m + K + U
$$

where ##U = p_a u^a \left( \sqrt{\xi^a \xi_a} - 1 \right)##.

If we consider the special case of Schwarzschild spacetime in the weak field limit, in standard Schwarzschild coordinates, we have ##\sqrt{\xi^a \xi_a} = \sqrt{1 - 2M / r} \approx 1 - M / r##, so ##U = p_a u^a \left( - M / r \right)##, in accordance with our Newtonian expectation.
 
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  • #19
vanhees71 said:
A time-like Killing vector is the generator of temporal translations and thus the corresponding conserved quantity is rather an energy than a momentum. I have no clue, what you mean by "potential" or "momentum potential" in this context.

Take the example of the electromagnetic field in flat space. There is no potential for the Lorentz force acting on charged matter although there's conservation of energy and momentum for the closed system consisting of this charged matter an the em. field.
In order to generalize the mechanics of a free particle to a particle coupled to the electromagnetic field, we make the substitution:$$\left(E,\boldsymbol{p}\right)\rightarrow\left(E+q\Phi,\boldsymbol{p}+\frac{q}{c}\boldsymbol{A}\right)$$If we refer to ##E+q\Phi## as "kinetic + potential energy", isn't it just as consistent to call ##\boldsymbol{p}+\frac{q}{c}\boldsymbol{A}## the "kinetic + potential momentum"?
 
  • #20
renormalize said:
In order to generalize the mechanics of a free particle to a particle coupled to the electromagnetic field, we make the substitution:$$\left(E,\boldsymbol{p}\right)\rightarrow\left(E+q\Phi,\boldsymbol{p}+\frac{q}{c}\boldsymbol{A}\right)$$If we refer to ##E+q\Phi## as "kinetic + potential energy", isn't it just as consistent to call ##\boldsymbol{p}+\frac{q}{c}\boldsymbol{A}## the "kinetic + potential momentum"?
Well, the Hamilton equations of motion give
$$m\dot{\bf x}=\boldsymbol{p}+\frac{q}{c}\boldsymbol{A}$$
which is why the expression on the right-hand side is called kinematic momentum (not kinetic + potential momentum as you suggested). In this context, ##\boldsymbol{p}## is called canonical momentum (not kinetic momentum as you suggested).
 
  • #21
PeterDonis said:
Now consider an integral curve of the KVF; an observer whose worldline is such an integral curve has 4-velocity ##u^a = \xi^a / \sqrt{\xi^a \xi_a}##.
And this explains why the idea does not work for spacelike KVF, because then the 4-velocity would be spacelike, which is unphysical. This seems to explain why even in relativistic mechanics we can have potential energy, but no potential 3-momentum.
 
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  • #22
renormalize said:
In order to generalize the mechanics of a free particle to a particle coupled to the electromagnetic field, we make the substitution:$$\left(E,\boldsymbol{p}\right)\rightarrow\left(E+q\Phi,\boldsymbol{p}+\frac{q}{c}\boldsymbol{A}\right)$$If we refer to ##E+q\Phi## as "kinetic + potential energy", isn't it just as consistent to call ##\boldsymbol{p}+\frac{q}{c}\boldsymbol{A}## the "kinetic + potential momentum"?
First of all it should be ##\hat{H}_0 \rightarrow \hat{H}_0+q \Phi(t,\hat{\vec{x}})## and ##\hat{\vec{p}} \rightarrow \hat{\vec{p}}-q \vec{A}(t,\hat{\vec{x}})## ("minimal substitution").

Further ##\hat{\vec{p}}## in quantum mechanics is always the canonical momentum, while
$$\hat{\vec{\pi}}=\hat{\vec{p}}-q \vec{A}(t,\hat{\vec{x}})=m \hat{\vec{v}}$$
is the kinetic momentum. Note that in the position representation it's always the canonical (!) momentum which has to be defined as ##\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}##.

This becomes clear, when you evaluate
$$\hat{\vec{v}}=\mathring{\hat{\vec{x}}}=\frac{1}{\mathrm{i} \hbar} [\hat{\vec{x}},\hat{H}],$$
with
$$\hat{H}=\frac{1}{2m} [\hat{\vec{p}}-q \vec{A}(t,\hat{\vec{x}})]^2 + q \Phi(t,\hat{\vec{x}}).$$
Calculating also
$$\mathring{\hat{\vec{p}}}= \frac{1}{\mathrm{i} \hbar} [\hat{\vec{p}},\hat{H}]$$
shows that this is the correct Hamiltonian for a particle in the electromagnetic field.
 

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