Energy shifts in general relativity

  • #1
428
23

Main Question or Discussion Point

Hi,

In classical and quantum physics and even in special relativity, shifting the energy of a system by a constant (i.e. resetting the zero point) changes nothing in the dynamics and is not observable. In quantum field theory, we even have to shift by an infinite (but "constant") energy to get reasonable results (normal order).

However, if you do something like that to the stress-energy-tensor, the solution of the Einstein field equations and hence the dynamics of the universe will be different. In fact, Einstein's cosmological constant is such a shift.

So how do we know where to place the zero-level of the energy we need to use? For example, where is the zero-level of the energy of the vibrational modes of a hydrogen molecule?
 

Answers and Replies

  • #2
bcrowell
Staff Emeritus
Science Advisor
Insights Author
Gold Member
6,723
423
In classical and quantum physics and even in special relativity, shifting the energy of a system by a constant (i.e. resetting the zero point) changes nothing in the dynamics and is not observable.
I don't think this is true in SR. Mass-energy equivalence exists in SR, not just GR, so changing the zero point would change the inertia of a system.

For example, where is the zero-level of the energy of the vibrational modes of a hydrogen molecule?
It's what you'd expect, [itex](3/2)\hbar\omega[/itex]. It's not arbitrary, because mass-energy is conserved, so the total mass-energy of the molecule can be determined from the mass-energy balance of the process by which the molecule was formed from the two atoms.
 
  • #3
PeterDonis
Mentor
Insights Author
2019 Award
30,175
9,339
For a problem like analyzing the hydrogen molecule, gravitational effects can be ignored, so you can just set the zero point of energy wherever it's convenient. (This turns out to work for *lots* of problems.)

For a problem like analyzing the gravity of an isolated system, like the Earth or the solar system, the standard thing to do is to set the potential energy at infinity equal to zero. This amounts to assuming a zero cosmological constant, i.e., to assuming that the energy of the vacuum is zero. This is, of course, the "natural" assumption in the presence of gravity, which everybody made before we understood that there was even a question to be asked in this area. Experimentally, it works well because any cosmological constant that is present in our universe (see below) is far too small to have any measurable effect on the dynamics of objects on the scale of the Earth or the solar system (or even our galaxy, as far as we can tell).

The only type of problem where the "natural" assumption turns out not to work is cosmology. Our current best model of cosmology includes two eras where the vacuum's nonzero energy affects the dynamics: the inflationary era, and the current dark energy dominated era. In the inflationary era, the effective cosmological constant was extremely large, much larger than the density of matter or radiation; in our current era, the effective cosmological constant is extremely small, but because it does not decrease as the universe expands, it is now larger than the density of matter and radiation, which *do* decrease as the universe expands.
 
  • #4
Bill_K
Science Advisor
Insights Author
4,155
195
Take a look at the book "Quantum Fields in Curved Space", by Birrell and Davies. A major fraction of the book is devoted to the question of calculating a finite, renormalized <Tμν>.
 
  • #5
428
23
It's what you'd expect, [itex](3/2)\hbar\omega[/itex]. It's not arbitrary, because mass-energy is conserved, so the total mass-energy of the molecule can be determined from the mass-energy balance of the process by which the molecule was formed from the two atoms.
But this zero-point energy depends on my choice of the zero-point of the potential, i.e. [itex]\frac{1}{2} m \omega^2 (r-r_{eq})^2[/itex]. We choose the potential minimum to be zero at the equilibrium distance. However this is completely arbitrary, in classical and quantum mechanics all dynamics and observables would be exactly the same if we chose the potential [itex]\frac{1}{2} m \omega^2 (r-r_{eq})^2+C[/itex] for an arbitrary [itex]C[/itex].
 
  • #6
BruceW
Homework Helper
3,611
119
To begin with, for things like the dirac field for electrons, and the field for photons, e.t.c. we explicitly choose the vacuum state (i.e. eigenstate of zero particles) to have zero energy. This makes sense right, if there are no particles around, then there's nothing which can give up its energy to do work or whatever.

But then (as has been mentioned), to explain inflation, we want to say that there is some (as of yet unidentified) quantum field, which interacts very weakly with the known fields, so that in effect we have a non-zero vacuum energy. Technically, this is not the vacuum energy, because this inflaton field is non-zero, so we don't actually have a vacuum. But since this inflaton field is weakly interacting with the known fields, it acts effectively like a restructuring of the vacuum state.
 
  • #7
PeterDonis
Mentor
Insights Author
2019 Award
30,175
9,339
this inflaton field is non-zero, so we don't actually have a vacuum.
I think a better way of stating the definition of "vacuum" in quantum field theory is that all fields are in their ground states, i.e., their states of lowest energy. However, in curved spacetime, there is not necessarily a unique state which has "lowest energy" everywhere and at all times. (This is one way of stating the issue raised by the OP.)

As I understand it (but my understanding is very heuristic, so I may be significantly oversimplifying in what follows), the inflationary era was an era in which the inflaton field was in a state called the "false vacuum", which was a vacuum state in that era, but which was no longer a vacuum state when the universe underwent a phase transition. After that transition, the inflaton field changed state to the state it is in now, called the "true vacuum", which (as far as we can tell) is its lowest energy state now and for the rest of the universe's history; but this state has much *less* energy than the "false vacuum" state, so the phase transition resulted in a lot of energy being transferred into the other quantum fields (which were in their vacuum states before the transition), the ones we now observe as the matter and radiation in the universe. (Of course the energy density in these fields now is much smaller than it was right after the phase transition, due to the expansion of the universe since then.)
 
  • #8
bcrowell
Staff Emeritus
Science Advisor
Insights Author
Gold Member
6,723
423
But this zero-point energy depends on my choice of the zero-point of the potential, i.e. [itex]\frac{1}{2} m \omega^2 (r-r_{eq})^2[/itex]. We choose the potential minimum to be zero at the equilibrium distance. However this is completely arbitrary, in classical and quantum mechanics all dynamics and observables would be exactly the same if we chose the potential [itex]\frac{1}{2} m \omega^2 (r-r_{eq})^2+C[/itex] for an arbitrary [itex]C[/itex].
Mass in SR isn't additive in general, but mass is additive for systems so far apart that they can't interact. This means that you don't want to pick the equilibrium as the zero, you want the zero of PE to be at infinity.
 
  • #9
martinbn
Science Advisor
1,845
558
What do you mean when you say "if you do something like that to the stress-energy-tensor"? What is a constant tensor? The cosmological constant is constant, but the term you add to Einstein's equation is a multiple of the metric.
 
  • #10
BruceW
Homework Helper
3,611
119
I think a better way of stating the definition of "vacuum" in quantum field theory is that all fields are in their ground states, i.e., their states of lowest energy. However, in curved spacetime, there is not necessarily a unique state which has "lowest energy" everywhere and at all times. (This is one way of stating the issue raised by the OP.)
whoops, yeah, sorry. I was using bad terminology. OK, using this correct terminology, our usual fields (like the electron and photon fields) have zero vacuum expectation value, and the inflaton field has non-zero vacuum expectation value. So in loose terminology, for the usual fields, the vacuum state has zero field, while for the inflaton field, the vacuum state has non-zero field.

I am less certain about the answer than I was initially, but anyway, after some more thinking, I think I understand what's going on. I think it is related to how the fields can represent particles. For the usual fields, around their vacuum state, the field just represents these particles (photons or electrons or whatever). Therefore, when there is just one electron, we want the field to have energy of just one electron, and when the field is zero, we want the field to have zero energy, since there are no electrons.

But this inflaton field around its vacuum state does not directly correspond to particles. It corresponds to some field (call it ##\nu##), plus some fields that correspond to particles. So, at the vacuum state, we want the energy due to the particle fields to equal zero (since there are no particles at the vacuum state). But, this means the ##\nu## field is still left over. And this field will have some non-vanishing energy.

So, in essence, I think the inflaton field has nonzero energy at its vacuum state because the energy due to its particles is zero, but it still has some energy due to the part of the field that does not directly correspond to particles like the usual fields do.

Also, I'm talking about the inflaton field (which is the cause of inflation), but also I think it is standard to say that the potential energy of the inflaton field is equal to the cosmological constant... I still think this is a bit suspect, but oh well.
 
  • #11
PeterDonis
Mentor
Insights Author
2019 Award
30,175
9,339
our usual fields (like the electron and photon fields) have zero vacuum expectation value, and the inflaton field has non-zero vacuum expectation value.
Yes. However, note that this still depends on picking a "zero point" of energy so that the vacuum expectation values come out this way--i.e., so that the VEV of the "usual" fields is in fact zero. Doing that, in general, still requires subtracting the "zero-point energy" term, which is formally infinite, from the VEV.

For the usual fields, around their vacuum state, the field just represents these particles (photons or electrons or whatever). Therefore, when there is just one electron, we want the field to have energy of just one electron, and when the field is zero, we want the field to have zero energy, since there are no electrons.

But this inflaton field around its vacuum state does not directly correspond to particles. It corresponds to some field (call it ##\nu##), plus some fields that correspond to particles. So, at the vacuum state, we want the energy due to the particle fields to equal zero (since there are no particles at the vacuum state). But, this means the ##\nu## field is still left over. And this field will have some non-vanishing energy.
Heuristically, this looks fine to me. The only caveat I would make is that the concept of "particle" is itself heuristic; I'm not sure that all field states even for the "usual" fields correspond to what we would call "particle" states. One obvious point is that states that actually occur in some situations are not eigenstates of the particle number operator, so they do not have a definite number of particles. But also, the whole notion of "particles" comes from perturbation theory, and there are situations where non-perturbative effects come into play that can't be described using perturbation theory.
 
  • #12
PeterDonis
Mentor
Insights Author
2019 Award
30,175
9,339
I think it is standard to say that the potential energy of the inflaton field is equal to the cosmological constant
I think a better way to put this is that the vacuum expectation value of the inflaton field is equal to the cosmological constant.
 
  • #13
bcrowell
Staff Emeritus
Science Advisor
Insights Author
Gold Member
6,723
423
Mass in SR isn't additive in general, but mass is additive for systems so far apart that they can't interact. This means that you don't want to pick the equilibrium as the zero, you want the zero of PE to be at infinity.
Oops, this wasn't quite right. The systems have to be noninteracting and also at rest relative to one another.
 
  • #14
BruceW
Homework Helper
3,611
119
I think a better way to put this is that the vacuum expectation value of the inflaton field is equal to the cosmological constant.
right, the field is an operator, so we need to talk about expectation values of the operator, when acting on some vacuum state. I keep using bad terminology and forgetting this. Also, it will be the expectation value of the potential energy of the inflaton field I think. (maybe with a minus sign, since we will need to move this term from one side of the Einstein field equations to the other side, so it can be interpreted as the cosmological constant).
 
  • #15
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,369
974
It is usually assumed that (dark energy/cosmological constant) and inflation are unrelated, but there are some models that do relate the two concepts. When they related, dark energy comes from dynamic quintessence, not a cosmological constant. For example, see the 3-parargraph section 23.5.2 "Dark energy from a rolling scalar field" in the book "The Primordial Density Perturbation: Cosmology, Inflation, and the Origin of Structure" by Lyth and Liddle (2009).

I am working on a much longer more technical post.
 
  • #16
BruceW
Homework Helper
3,611
119
yeah, I've seen in several places that authors seem to say that it is completely natural to say that the cosmological constant is caused by a constant homogeneous scalar field. This is already a bit suspicious, I feel. And then worse than that, authors go on to say that this field is not constant with time, but changes as the temperature of the universe changes. And that this field is responsible for both inflation and the cosmological constant. But the cosmological constant is constant. The field might be very slowly changing now that the universe is quite big. But still, the cosmological constant should be exactly constant, not approximately constant.
 
  • #17
BruceW
Homework Helper
3,611
119
But also, the whole notion of "particles" comes from perturbation theory, and there are situations where non-perturbative effects come into play that can't be described using perturbation theory.
without perturbation, I wouldn't even know how to calculate the vacuum expectation value. I definitely need to learn more, hehe.
 
  • #18
PeterDonis
Mentor
Insights Author
2019 Award
30,175
9,339
without perturbation, I wouldn't even know how to calculate the vacuum expectation value.
Actually, this is kind of backwards; you need to have some idea of what the VEV is before you can even *do* perturbation theory.

For example, in the case of the inflaton field, you need to first have some model for the potential energy of the field such that its ground state (i.e., the state with lowest energy, the "vacuum") has a nonzero expectation value for the field. Only then can you do perturbation theory *around* the ground state and see that the low energy spectrum consists of scalar particles (at least for the simplest type of inflaton field).

The type of situation I was referring to when I mentioned non-perturbative effects is different, though. A simple example is a scalar field ##\phi## in 1+1 dimensional spacetime that has a potential energy function which allows two possible VEV's, call them ##v## and ##-v##, with an infinite potential energy barrier between them. Then it is possible to have a state called a "kink" in which ##\phi \rightarrow -v## as ##x \rightarrow - \infty## and ##\phi \rightarrow v## as ##x \rightarrow \infty##.

This state is not an ordinary "vacuum" state: it's easy to show that it has higher energy than a state with ##\phi = v## everywhere or ##\phi = -v## everywhere; the latter two states are the "vacuum" states we expect. However, there is no way for this state to decay to a state of constant ##\phi = +/- v##, because of the infinite potential energy barrier between the two VEVs. So there is no way to discover the "kink" state by doing perturbation theory around either of the ordinary vacuum states (in other words, there is no way to produce the "kink" state by applying particle creation operators to either of the ordinary vacuum states). The presence of the "kink" state is an intrinsically non-perturbative effect.
 
  • #19
PeterDonis
Mentor
Insights Author
2019 Award
30,175
9,339
yeah, I've seen in several places that authors seem to say that it is completely natural to say that the cosmological constant is caused by a constant homogeneous scalar field. This is already a bit suspicious, I feel.
At the risk of somewhat anticipating George Jones' forthcoming technical post, I think this is really a matter of terminology. A scalar field is similar to a cosmological constant in that it induces a term in the Lagrangian which is just a scalar multiple of the metric. The only difference is that a scalar field can vary from event to event in spacetime, where a cosmological constant is, well, constant (as you note).

But physically, the measurements that we currently interpret as showing a very small nonzero cosmological constant could also be interpreted (at least AFAIK) as showing a very small nonzero scalar field whose variation in spacetime is very, very small. Our data at this point is simply not accurate enough to distinguish these two possibilities. The authors in question should be more precise in their terminology, but I believe this is basically what they're talking about.
 
  • #20
PeterDonis
Mentor
Insights Author
2019 Award
30,175
9,339
Also, it will be the expectation value of the potential energy of the inflaton field I think.
No; the potential energy is not an operator. It's a term in the Lagrangian that describes the system. There are many possible operators whose expectation values *could* be taken (on the vacuum state or any other state), but normally when the term "vacuum expectation value" is used, the operator in question is something like the particle number operator that arises in the perturbation theory around the vacuum state.
 
  • #21
BruceW
Homework Helper
3,611
119
No; the potential energy is not an operator. It's a term in the Lagrangian that describes the system. There are many possible operators whose expectation values *could* be taken (on the vacuum state or any other state), but normally when the term "vacuum expectation value" is used, the operator in question is something like the particle number operator that arises in the perturbation theory around the vacuum state.
why isn't the potential energy an operator? Also, I think the VEV is usually assumed to be using the field operator itself. But in this case, we were talking about the cosmological constant, and so (I think) the VEV of the potential energy is used. Even if the potential energy isn't an operator, the part of the field which does not correspond to particles is classical anyway (as far as I can tell), so we can talk about the potential energy associated with it, and not worry about making averages.
 
  • #22
PeterDonis
Mentor
Insights Author
2019 Award
30,175
9,339
why isn't the potential energy an operator?
Um, because a term in the Lagrangian is not the same as an operator? See further comments below.

Also, I think the VEV is usually assumed to be using the field operator itself.
Usually that's what's meant, yes, although it is somewhat unclear in the literature I've read exactly *which* "field operator" is being referred to, since there are several possible candidates. I assumed in an earlier post that the operator in question was the particle number operator, but another possibility is the Hamiltonian--see below.

But in this case, we were talking about the cosmological constant, and so (I think) the VEV of the potential energy is used.
I should have been clearer about distinguishing the potential energy I was talking about from the potential energy I think you are talking about. The potential energy I was talking about is a term in the Lagrangian. The potential energy I think you are talking about is a part of the Hamiltonian operator, which is not the same thing, but is related, yes.

Suppose the field is in its vacuum state, and we evaluate the Hamiltonian. We expect the "kinetic energy" part of the Hamiltonian to be zero (note that we might have to subtract off an infinite correction term to obtain this result) since the field is in its vacuum state. But the Hamiltonian can have a "potential energy" part that is nonzero in this state, *if* the Lagrangian contains an appropriate potential energy term.

In classical physics, the distinction between the Lagrangian and the Hamiltonian, while not exactly trivial, is not at all the same as the distinction in quantum field theory, since in classical physics the Lagrangian and the Hamiltonian are the same sort of thing. In QFT, they're not. The Lagrangian is not an operator; it's a fundamental equation describing the theory. The Hamiltonian is an operator, a fundamentally different sort of thing.

Even if the potential energy isn't an operator, the part of the field which does not correspond to particles is classical anyway (as far as I can tell), so we can talk about the potential energy associated with it, and not worry about making averages.
No, you can't. In QFT everything observable has to be an operator; you can't have a part that is "classical". The "part of the field that does not correspond to particles" is just a way of referring to the fact that operators like the Hamiltonian can have a nonzero expectation value even when the field is in its vacuum state.
 
  • #23
BruceW
Homework Helper
3,611
119
In classical physics, the distinction between the Lagrangian and the Hamiltonian, while not exactly trivial, is not at all the same as the distinction in quantum field theory, since in classical physics the Lagrangian and the Hamiltonian are the same sort of thing. In QFT, they're not. The Lagrangian is not an operator; it's a fundamental equation describing the theory. The Hamiltonian is an operator, a fundamentally different sort of thing.
I think the Lagrangian is an operator too.

PeterDonis said:
No, you can't. In QFT everything observable has to be an operator; you can't have a part that is "classical". The "part of the field that does not correspond to particles" is just a way of referring to the fact that operators like the Hamiltonian can have a nonzero expectation value even when the field is in its vacuum state.
The field operator is ##\phi## and ##\nu## is just some number that we take away from ##\phi## so that ##\phi - \nu## is the usual type of field, which corresponds to the particles. So ##\nu## is a classical field, since it's just a number. That is how I had understood it. (this is just for complex scalar fields, I think there are more complicated cases).
 
  • #24
BruceW
Homework Helper
3,611
119
Actually, this is kind of backwards; you need to have some idea of what the VEV is before you can even *do* perturbation theory.
uhhh... I dunno. any vacuum state is a minimum energy state, but is also a state with zero particles ##|0\rangle##. So then, if you write out the field operator as creation and annihilation operators, then when this acts on a state of zero particles ##|\phi |0\rangle## there will be a bunch of terms give zero, and if you act on the left with the same state (to give an expectation) ##\langle 0|\phi |0\rangle## there will be a bunch more terms that give zero. And what is left of ##\phi## is the bit that does not correspond to particles, because the part that did correspond to particles just got made zero when taking expectation with the zero particle state. So this is what I meant about needing to think about particles, to be able to work out the VEV.

PeterDonis said:
The type of situation I was referring to when I mentioned non-perturbative effects is different, though. A simple example is a scalar field ##\phi## in 1+1 dimensional spacetime that has a potential energy function which allows two possible VEV's, call them ##v## and ##-v##, with an infinite potential energy barrier between them. Then it is possible to have a state called a "kink" in which ##\phi \rightarrow -v## as ##x \rightarrow - \infty## and ##\phi \rightarrow v## as ##x \rightarrow \infty##.
ok. I don't know about this kind of stuff. I only read about quantum scalar fields which have a Lagrangian that does not vary spatially. I think my brain might explode if I tried to add that complexity :)
 
  • #25
PeterDonis
Mentor
Insights Author
2019 Award
30,175
9,339
any vacuum state is a minimum energy state, but is also a state with zero particles ##|0\rangle##. So then, if you write out the field operator as creation and annihilation operators...
Just to note, *every* operator can be expressed in terms of creation and annihilation operators, including the Hamiltonian.

Also, my earlier phrasing was ambiguous (more on ambiguity of phrasing below); I should have said you have to have picked out a vacuum *state* before you can do perturbation theory. Different operators may have different expectation values in that state, and you don't necessarily have to know what all those values are to pick the vacuum state.

...when this acts on a state of zero particles ##|\phi |0\rangle## there will be a bunch of terms give zero, and if you act on the left with the same state (to give an expectation) ##\langle 0|\phi |0\rangle## there will be a bunch more terms that give zero. And what is left of ##\phi##
It's not what's left of ##\phi##, the field, it's what's left of the expectation value once you've eliminated all the terms that give zero when acting on the vacuum state. Your notation is sloppy in this respect; you should be writing ##\langle 0| \hat{\phi} |0\rangle##, where ##\hat{\phi}## is whatever operator you are calling the "field operator", not the field itself. You can't act on a part of the field ##\phi## with an operator and leave another part alone; you can only act on ##\phi## (more precisely, you can only act on a particular state in the spectrum of the field ##\phi##, such as ##|0\rangle##). All you are really saying is that ##\langle 0| \hat{\phi} |0\rangle = v## instead of ##0##, where ##v## is the VEV of the operator ##\hat{\phi}##.

(Btw, being sloppy in this respect is common; I've seen a number of texts blur the distinction between the field ##\phi## and operators that act on states in its spectrum, and say things like "the VEV of ##\phi## instead of the more precise phrasing I used just now. In fact I probably did it myself in earlier posts in this thread. :redface:)

quantum scalar fields which have a Lagrangian that does not vary spatially
I think you mean fields that have a *vacuum expectation value* that does not vary spatially? I'm not aware of any field with a Lagrangian that doesn't have a derivative term, and the presence of a derivative term means the field varies in spacetime, and variation in spacetime means spatial variation in at least some frame.
 

Related Threads on Energy shifts in general relativity

Replies
8
Views
1K
Replies
4
Views
4K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
981
Replies
1
Views
1K
Replies
6
Views
2K
Replies
14
Views
532
  • Last Post
2
Replies
35
Views
10K
Replies
12
Views
2K
Top