Particle Energy in Schutz's "A First Course in General Relativity

In summary, there is no consistent definition of energy in general relativity due to the lack of conservation of energy in this theory. The energy of a particle is only conserved in the presence of a time-like killing vector in a static field, and can only be properly defined at an asymptotic null-infinity. In the Schwarzschild metric, p0 is a conserved quantity and is often referred to as the "energy" of the particle, but it is important to note that it cannot be determined by simply summing up p0. In general spacetimes, p0 is not necessarily conserved, even in special relativity. Its conservation depends on whether work is done on or by the particle, and in a non-static
  • #1
hellfire
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In the first chapters of Schutz's "A first course in general relativity", the energy of a particle is defined to be E = p0 (e.g. in 2.20). But in the chapter 7.4 discussing "conserved quantities" in curved spacetimes, the energy is defined to be E = - p0 = - g0upu. After reading 7.4 it is not clear for me why p0 is used for the definition of E instead of p0, as done in the first chapters of the book. Any help would be appreciated.
 
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  • #2
Edit Added for rigour: The following is all measured in the rest frame of the coordinate system.
Ua = [1,0,0,0]
Otherwise E = - paUa


It is confusing because in GR there is no consistent definition of energy. This is because in general energy is not conserved.

The energy of a particle is only conserved if there is a time-like killing vector (static field) and the energy of a gravitating mass and its gravitational field can be only properly defined at an asymptotic null-infinity.

Basically you have to reach a region where the GR metric becomes Minkowskian - i.e. a SR metric - before you can proceed.

p0 is a natural candidate for the energy of a particle as it is the time component of the 4-momentum, however it is not conserved even if there is a time-like killing vector.

p0 is not an obvious choice until you realize that it is conserved because U0 is a killing vector in a static field and
p0 = mU0.

But note that if there is another moving mass nearby the field is not static and this definition falls apart. It won't even do for the Earth's field because of the presence of the Moon, let alone the Sun!

Note The theory of Self Creation Cosmology addresses this problem and defines the energy of a particle as [p0p0]1/2, which in SCC is conserved.

I hope this helps.

Garth
 
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  • #3
hellfire said:
In the first chapters of Schutz's "A first course in general relativity", the energy of a particle is defined to be E = p0 (e.g. in 2.20). But in the chapter 7.4 discussing "conserved quantities" in curved spacetimes, the energy is defined to be E = - p0 = - g0upu. After reading 7.4 it is not clear for me why p0 is used for the definition of E instead of p0, as done in the first chapters of the book. Any help would be appreciated.
The first section that you refer to is about SR where, in the coordinates he uses (Minkowski coordinates) p0 = p[suo]0[/sup] so there is no difference and typically people think of the time component as energy (although they shouldn't imho).

As far as why p[suo]0[/sup] is energy please see - http://www.geocities.com/physics_world/gr/conserved_quantities.htm

Pete
 
  • #4
For an object in geodesic motion, pa ka or pakb will be a conserved quantity of the motion if and only if ka (in the first case) or ka (in the second) is a Killing vector.

In the exterior region of the Schwarzschild metric, it turns out that the Killing vector ka is a unit time like vector, i.e. ka = (1,0,0,0) regardless of the coordinates. This is the underlying reason that P0 is a conserved quantity. It's a particular feature of the Schwarzschild metric that ka = (1,0,0,0) is a Killing vector, while ka = (1,0,0,0) is NOT a Killing vector. (Well, actually, there are a fair number of metrics that have the property that ka = (1,0,0,0) is a Killing vector - any static metric where the gij are not functions of time will have this property, at least in the vacuum region of the metric.)

Because it's a conserved quantity in the Schwarzschild metric, P0 is often called the "energy" of the particle, though some authors are more careful and call it the "energy parameter". The important thing to know is that this is a conserved quantity for a particle in geodesic motion in the Schwarzschild metric (or in general in a static metric). It is important to also realize that one can NOT determine the total energy of a system simply by summing up P0.

In general spacetimes, P0 is not necessarily conserved - in fact, the constant density solution for a spherically symmetric star deosn't allow P0 as a conserved quantity in the interior region of the star. This last observation comes from my own calculations, but one can confirm that (1,0,0,0) is not a Killing vector in the interior region with Killing's equation.
 
  • #5
pervect said:
In general spacetimes, P0 is not necessarily conserved
. On that point it is important to note that even in SR P0 is not always conserved. If a charged particle is in an Electrtic field then P0 its not conserved.

Pete
 
  • #6
pmb_phy said:
. On that point it is important to note that even in SR P0 is not always conserved. If a charged particle is in an Electrtic field then P0 its not conserved.

Pete
Pete - Is not pervect referring to P0 not being conserved when no work is done on or by the particle?

This occurs in a general non-static space-time metric, such as the Earth's space-time distorted by the Moon and Sun.

Garth
 
  • #7
Garth said:
Pete - Is not pervect referring to P0 not being conserved when no work is done on or by the particle?

This occurs in a general non-static space-time metric, such as the Earth's space-time distorted by the Moon and Sun.

Garth
When the gravitational force does work on the particle in a static gravitational field then P0 is conserved.

Pete
 
  • #8
Pete - True, but where do we find a static gravitational field?

Garth
 
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  • #9
Garth said:
Pete - True, but where do we find a static gravitational field?

Garth
You're sitting in one.

Pete
 
  • #10
Pete - No I'm not - I'm sitting on the Earth, orbiting the Sun, orbiting the Milky Way Galaxy, and with the Moon orbiting it. I see tides rise and fall twice a day. This space-time gravitational field is only approximately static - which in my book is not static!

Garth
 
  • #11
I hope this discussion has been helpful to hellfire, and that we're not over-Killing the thread :-)
 
  • #12
Garth said:
Pete - No I'm not - I'm sitting on the Earth, orbiting the Sun, orbiting the Milky Way Galaxy, and with the Moon orbiting it. I see tides rise and fall twice a day. This space-time gravitational field is only approximately static - which in my book is not static!

Garth
Do you know what a static g-field is?

We're speaking of concepts and not actual situations. If you want to have it your way and be exact then we can't describe anything exactly.

Pete
 
  • #13
pervect said:
I hope this discussion has been helpful to hellfire, and that we're not over-Killing the thread :-)
It is helpful, thank you. The first three posts are a clear answer to my question, but of course you are free to discuss this - or related aspects - more; I will keep my attention...
 
  • #14
pmb_phy said:
Do you know what a static g-field is?
Yes - and as the Earth's gravitational potentials are perturbed by the presence of the Moon's and Sun's gravitational fields, I'm not sitting in one!
pmb_phy said:
We're speaking of concepts and not actual situations. If you want to have it your way and be exact then we can't describe anything exactly.

Pete
Pete - we have had this discussion before. Energy is only conserved in GR under very special circumstances, which do not exist in the real universe. Here we live in a dynamical universe not a static one, therefore we had better accommodate ourselves to the realisation Einstein and others had about GR, and proved by Emmy Noether, that GR is an example of an improper energy theorem and therefore energy is not in general conserved.

Garth
 
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  • #15
We've had this dicussion before? Really? Oh yeah, I guess we have :-)
(kidding).

Anyway, to make my usual comments, there are at least two different but very closely related standard notions of energy in GR (Bondi energy and ADM energy), both of which are derived from the notion of asymptotically flat space-time. There is a detailed discussion of this in Wald's "General Relativity", and I'm sure a search of the forum will turn up a lot of past discussion on this topic as well.

Since the universe appears to be expanding, one can indeed raise significant questions about whether or not asymptotically flat space-time actually exists in the real universe.

As an alternative of modifying GR so that energy is conserved (Garth's Self-Creation cosmology), it's worthwhile asking the question of whehter the universe really does conserve energy on a cosmological scale. An alternate solution to the problem is to say that energy conservation is only approximate - that in cases where one can ignore cosmological expansion, one can come up with a conserved energy based on approximating the universe as being asymptotically flat, but that in other cases, where cosmological expansion is significant and asymptotic flatness breaks down, energy conservation also breaks down.
 
  • #16
Garth said:
Pete - we have had this discussion before.
So you agree that nothing in the universe can be described precisely and that when we speak of physics here we know that and are not worried about it and are usually speaking about ideal situations. E.g. If, relative to ther surface of the earth, I drive my car 50 mph in a straight line for one hour then I will have traveled 50 miles.

We understand that this is not literally possible because of miniscule variations in various variables. But we can say these things and talk about it and not have to worry that rather than 50 miles I might actually have traveled 50 miles and 1 nanometer.

We know you care about minor variations like that. However the situations we discuss here we almost never care about them.

Pete
 
  • #17
Pete - The perturbations of the Moon on the Earth's spherically symmetric and 'approximately static' gravitational potentials cause significant tides under which the Earth rotates so that a lot of energy can be extracted. Hardly just a "nanometer" or two.

But also the Earth is in an elliptical orbit around the Sun. As the Sun's dimensionless gravitational potential at the Earth is about 10-8, the six-monthly variation in its potential is about +/-10-10. Measured from the Earth the Sun's total energy varies, and is not conserved, by about 1 part in 10-10, which considering the Sun is quite a lot!

Garth
 
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  • #18
To focus my earlier response a bit, I'll add a few more comments to the specific problem raised - that of the Earth, sun, & moon. If we isolate the solar system from the rest of the galaxy, by assuming that the solar system is embedded in an asymptotically flat space time, we can use the definitions of energy I mentioned earlier to come up with a theoretical conserved energy for the solar system (or the simpler 3-body problem of just the Earth, Moon & sun if one desires).

One does have to take into account the gravitational radiation that the Earth, moon, and other planets are emitting to come up with a truly conserved energy with this approach.

Actually carrying out the calculations would be extremely difficult, as I don't believe there are any analytical solutions to even the two-body problem in GR, much less the three-body problem. Approximate numerical solutions do exist, using the Newtonian solutions for the orbits would be an excellent starting approximation for further numerical refinement.

One will undoubtedly find many physical effects which are hard to model and a lot more significant than the effects predicted by GR in the issue of accurately modelling even the three-body system of the Earth, moon, & sun as they actually exist. If we replace the Earth moon & sun by black holes of the same mass, the GR analysis would be a lot simpler - this may or may not be regarded as cheating. An actual analysis that would represnet the actual Earth, moon & sun rather than their idealizations would have to include such diverse factors as:

the solar wind and light pressure from the Sun's radiation, the magnetic field of the Earth and it's interaction with the solar wind

the Sun's gradual loss of mass through solar radiation

the evolution of the sun as it burns it's nuclear fuel (which effects the other two factors above)

tidal interactions due to the extended nature of the planets & sun

so it would be a very difficult problem.
 
  • #19
It would indeed be a difficult problem to accurately model these effects, however that is not what I see the issue here to be.

The Sun losing mass through solar wind or losing energy through electromagnetic radiation, or the system losing energy through gravitational radiation can all be accommodated by the conservation of energy. Energy lost on one side of the equation is energy gained somewhere else.

My point is the principle of relativity and the equivalence of different inertial frames of measurement.

For example as the Earth approaches the Sun moving from aphelion to perihelion the red shift of the Sun's radiation decreases by about 1 part in 10-10. Therefore, other things remaining equal, as measured in the Earth's frame of reference the Sun would appear to heat up by that amount. The amount of heat energy stored by the Sun would appear to have increased as measured from the inertial frame of reference of the Earth and not the Sun. And this is only a tiny part of the problem.

To simplify; consider a massive gravitating body, A with mass M, alone in space, static, with no radiation, solar wind, gravitational wave or other loses of mass or energy. Consider a point mass, a test particle, falling from rest at infinity onto A. Because A has a static gravitational field there is a time-like Killing vector, A's 4-velocity U0, and the total energy of the test particle with 4-velocity one-form V0, here defined by the 'energy parameter' P0 = mV0, is conserved. Conserving the energy parameter and comparing it at infinity where v = 0 with its value at altitude r one obtains
v = [2GM/r]1/2 :- as in Newton where KE + PE = constant.

However from the test particles point of view, in its freely falling inertial frame of reference, the metric components of A's gravitational field are not time independent, they change with time as the particle descends. Therefore there is no time-like Killing vector and A's total energy is not conserved in that freely falling frame of reference.

We note that the desire to conserve energy may lead us to choose A's centre of mass as a preferred frame of reference but this would not be consistent with the principles of GR, instead it would be a Machian approach to gravitational theory which is the route Self Creation Cosmology has taken.
 
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  • #20
Garth said:
Measured from the Earth the Sun's total energy varies, and is not conserved, by about 1 part in 10-10, which considering the Sun is quite a lot!

Garth
Do you understand my point?

Pete
 
  • #21
Pete - we are only dealing with a weak field and a nearly circular orbit, but nevertheless 1 part in 10-10 of the Sun's total energy is (OOM)

10-10 x Mc2 = 10-10x2x1033x1021 ergs ~= 1044 ergs = Solar output for about 3,000yrs, so do you see my point?

When we deal with strong fields and widely changing radii from the central mass then things get considerably worse!

Garth
 
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  • #22
Garth - You didn't answer my question. Did you understand my point?

Pete
 
  • #23
Pete - If your point was that the conservation of the total energy of the Sun is broken by only one part in 1010, then yes. But my point is that this is still an incredible amount of energy to go unaccounted for, and it could be an order of magnitude worse, even in the Sun's weak field, if we were on asteroid Icarus, say, rather than the Earth!
 
  • #24
The fact remains that one can define the energy of an isolated system in GR with a flat background space-time, by at least two means. The fact that there ARE two different ways of defining energy may be somewhat disturbing, but they differ primarily in how they account for the energy in gravitational radiation, and it can be shown that except for this accounting issue, they otherwise agree.

One of the simplest is the Bondi energy, which utilizes the existence of asymptotic time translation symmetry (asymptotic Killing vectors). It's a LOT easier to compute the Bondi energy in the center of mass frame, but it's not strictly necessary - the Bondi energy has meaning even for an observer who is not in the center of mass system. The energy and momentum of the system will transform as a 4-vector, so of course the energy will depend on what frame the observer is in (but this is nothing new).

All of these definitions have the unusual property that the energy in a general metric cannot be strictly localized. Different observers will agree on the total energy of the system, but not about its distribution. The exception to this "non-localization" rule is for systems that have a static metric - in this case, everyone agrees not only about the value of the system energy, but on its location.

The universe as a whole doesn't necesarily have a flat background, nor does a large enough piece of it. But there aren't any inusrmountable problems with energy conservation in a tiny piece of the universe, like the solar system, even with standard GR.
 
  • #25
hellfire - You asked about the energy of a particle. There is much to be learned about asking about the energy of things which are not particles. Here is an interesting idea I came up with

Let there be a square box of length/width/height L which is at rest in the inertial frame S where one face of the box is parallel to the yz-plane. At t = 0 let two photons of equal energy be emitted from the box. One is emitted from the face which faces the +x direction and thereupon travels in the +x direction. Let the other photon be emitted from the opposite face and thereafter move in the -x direction. The total momentum of the photons will be zero and therefore the box will remain at rest. Let E0i be the initial proper energy of the box and E0f the final proper energy of the box. Therefore for t < 0 -> E = E0i, for t > 0 E = E0f. Take note of the fact that there are only two values of energy of the box as observed in S.

Now change your frame of reference to S' which is in standard configuration with S and thus is moving in the +x direction with respect to the box. Since the two events in S (two photons being emitted) were simultaneous in S they will not be simultaneous in S'. First the box will emit one photon, then later it will emit another photon. But there is a finite time between these two events in S'. Therefore there will be three values of energy as observed in S'. No single 4-momentum can be written for this box. :biggrin:

Pete
 
  • #26
And the moral of the story is (or at least should be), that when the box emits a photon, the atom in the box emitting the photon (or whatever else is emitting the photon) recoils in the other direction. Thus the total momentum of the box+photons is always constant.
 
  • #27
pervect said:
And the moral of the story is (or at least should be), that when the box emits a photon, the atom in the box emitting the photon (or whatever else is emitting the photon) recoils in the other direction. Thus the total momentum of the box+photons is always constant.
The moral of this story is that in frame S' the observer measures three distinct values for E whereas observers in S measure only two distinct values of E.

Pete
 
  • #28
Momentum (and energy) are conserved quantities, that don't have "three different values" - as long as one is careful to satisfy the continuity conditions. Your example is on a par with the question of "what happens to the gravity of the sun if it suddenly disappears", or "what happens to the electric field of a charge if it suddenly disappears". The answer is that mass or charge cannot suddenly dissappear, or (in your example) appear. If a photon appears, it appears because something has emitted it, and that something must lose the appropriate amount of energy, and gain the appropriate amount of momentum in order for the photon to be emitted.

So a correct explanation of what the stationary observer sees is

the left side of the box emits a photon to the right, causing the left side of the box to recoil in the opposite direction of the photon's movement (to the left).

the right side of the box emits a photon to the left at the same time, causing the right side of the box to recoil in the opposite direction (to the right).

The box, since it cannot be perfectly rigid, stretches a bit. After a long enough time period has elapsed, it will settle down and resume its original shape.

The momentum and the energy of the box + photon system stays constant through this entire process.
 
  • #29
pervect said:
So a correct explanation of what the stationary observer sees is

the left side of the box emits a photon to the right, causing the left side of the box to recoil in the opposite direction of the photon's movement (to the left).

the right side of the box emits a photon to the left at the same time, causing the right side of the box to recoil in the opposite direction (to the right).

The box, since it cannot be perfectly rigid, stretches a bit. After a long enough time period has elapsed, it will settle down and resume its original shape.

The momentum and the energy of the box + photon system stays constant through this entire process.
I'm sorry to inform you that total momentum and total energy was assumed to be conserved in the first place. But you gave an answer to a question nobody asked. Nobody said the box had to recoil. For all you know there was an electron positron pair inside the box which just annihilatied and shot through tiny holes in sides of the box.

The quesition was - What is the energy of the box as a function of time.

Pete
 
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  • #30
If a photon just happens to fly through a hole in the box, it's difficult for me to see how one can attribute the energy to the box. One can attribute the energy to the volume inside the box, in which case one has shown that the energy in a unit volume is a function of the frame of reference. That dependence of the energy and momentum in a unit volume on the frame of reference of the observer is why the stress-energy tensor is a second rank tensor - one can specify a volume by the time vector that's orthogonal to it, and the stress energy tensor maps this vector-valued volume into the energy-momentum in that volume.
 
  • #31
pervect said:
If a photon just happens to fly through a hole in the box, it's difficult for me to see how one can attribute the energy to the box.
In that case if someone asked you what the mass of a gas was you'd be unable to answer them?

The idea is pretty simple: In the rest frame there is one event, box radiates energy. After the box radiates energy its mass has decreased. In the moving frame there are two events separated in time and the box has three different values of enegy.

A simpler value is if you have a long hot, very thin, rod lying on the x-axis in S in free space. Assume the rod cools uniformly. By symettry the rod will not move. Then at any point in time in S the rod will have one and only one value of energy. In S' it will also me true but the value will be be E = gamma*E_o


[qupte]One can attribute the energy to the volume inside the box, in which case one has shown that the energy in a unit volume is a function of the frame of reference. That dependence of the energy and momentum in a unit volume on the frame of reference of the observer is why the stress-energy tensor is a second rank tensor [/qupte]That tensor is second rank because mass-energy has two gamma factors.

Think more about simultaineity.

Pete
 
  • #32
pmb_phy said:
In that case if someone asked you what the mass of a gas was you'd be unable to answer them?

If the question is stated unambiguously, there is no problem computing the mass of a gas. When the volume of the gas becomes large enough, though, one will start to run into other more interesting problems in the computation of the mass, where a simple approach of integrating the energy density over the volume will not give the correct answer for the system mass.

The idea is pretty simple: In the rest frame there is one event, box radiates energy. After the box radiates energy its mass has decreased. In the moving frame there are two events separated in time and the box has three different values of enegy.
[/quote

Last I heard the box was gaining energy, now its losing energy? Oy vey! Anyway, so far we've just been arguing about what constitutes the box and what does not constitute the box, and clarifying the issue that if the box gains energy, it gains energy because of some transaction with an entity that is not the box that is losing energy. The rest is accounting details which revolve around what constitutes the box "now", and exactly when we make the bookeeping transaction that deducts the energy from the "non-box" account, and adds it to the "box" account, or vica versa.

Think more about simultaineity.

Pete

In the limit as the size of the box approaces zero, the difference in simultaneity approaches zero, and it becomes a non-issue. This is one of the reasons the approach of describing a system in terms of a stress-energy tensor is very useful. All the trivial accounting details disappear, and one can concentrate on the physics.
 
  • #33
pervect said:
In the limit as the size of the box approaces zero, the difference in simultaneity approaches zero, and it becomes a non-issue.

Relativity applies to everthing, not just objects of no spatial extent.

Pete
 
  • #34
I'm not quite sure what your point is supposed to be?

The point I'm making is that energy does not appear from nowhere. Thus, when energy disappears from one set of books, it is moved to another set of books, instantaneously. There is no conceptual problem in making this transaction instantaneous, because the "transaction" of moving the energy from one set of books to another set of books, (possibly involving a physical change in form of the energy) occurs at one specific spatial location, not an extended region.

The specific time and the specific location at which the bookeeping transaction occurs depends on the coordinate system of the bookeeper, which is arbitrary.

If you wish to move the energy of a large object from one set of books to another set of books, you may/will have to subdivide the large transaction into a number of smaller transactions, which occur over an extended time period (in some frames of reference).

Because relativity is a classical theory, one is assumed to be able to divide up space-time into as small a region(s) as are needed to maintain the desired accuracy.

Thus in relativity we have the stress energy tensor, which talks about the density of energy and momentum at a point in space, which is assumed to be a continuous function.

The tricky part in GR of accouting for energy is not in the issue of dividing up energy into small enough chunks, it's another issue entirely.
 
  • #35
I wasn't trying to make any point whatsoever regarding the conservation of energy. The point I am making is a pretty simple point. If the object is a particle the the term "proper mass" has a well defined meaning. However if the object is not a particle, has spatial extent, and whose mass is changing then the term "proper mass" is not well defined. On can't simply choose any frame they wish to measure the proper mass. In particular one can't always write a meaninful 4-vector for an object nor write E = gamma*E_o

Its all spelled out much clearer here
http://www.geocities.com/physics_world/sr/invariant_mass.htm

To make this much easier to understand consider tube of very tiny cross-section which is laying on the x-axis in S. There is a deviceon each side which will emit flashes of light at regular time intervals in directions perpendicular to the x-axis and in equal amounts. Therefore a series of flashes will be emitted from near each out to the sides with no change in motion of the rod as a whole. If you let "m" = proper mass then the proper mass always has a well-defined quantity in this frame. However if you move to S' which is moving in the +x direction then the proper mass will not awlays have a value found by the rest observer.

Pete

ps - Do you have Ohanian's GR text? Tolman?

Recall why I brought this up. I told hellfire this; hellfire - You asked about the energy of a particle. There is much to be learned about asking about the energy of things which are not particles. Here is an interesting idea I came up with.

I find several of my relativist friends perplexed by this problem for a bit and hence its fun to think about. Plus it instills further that simultaneity is always there waiting to play games with our heads. :smile:
 
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FAQ: Particle Energy in Schutz's "A First Course in General Relativity

What is particle energy in Schutz's "A First Course in General Relativity"?

Particle energy refers to the energy of a single particle in a gravitational field, as described by Einstein's theory of general relativity. It is a measure of the particle's ability to do work and is affected by the curvature of spacetime.

How is particle energy related to mass in general relativity?

In general relativity, mass and energy are equivalent and can be converted into each other through the famous equation E=mc^2. This means that a particle's energy is directly related to its mass and vice versa.

What is the role of particle energy in gravitational fields?

Particle energy plays a crucial role in gravitational fields as it determines the trajectory of the particle and how it interacts with the curvature of spacetime. The more energy a particle has, the more it will be affected by the gravitational field.

How is particle energy affected by the curvature of spacetime?

The curvature of spacetime, caused by the presence of massive objects, affects the energy of a particle by changing its trajectory and altering its speed. This is because the curvature of spacetime is directly related to the strength of the gravitational field.

Can particle energy be negative in general relativity?

Yes, in general relativity, particle energy can be negative. This is due to the fact that energy and mass are equivalent, and mass can be negative in certain situations. However, negative energy particles are not commonly observed in nature and are mainly theoretical constructs.

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