"Is it possible to obtain the metric tensor field from

a potential ?"

In Einstein's gravitational theory the metric tensor IS the potential.

"Are there underlying symmetries such as gauge symmetries

in this field theory ? "

General coordinate transformations are analogous to gauge transformations.

Through geodesic motion, which realizes the equivalence principle. The statement is that the action for matter couples to gravity through the energy-momentum tensor only and, regarding gravity itself, if the gravitational action, as a functional of the metric, is expanded in derivatives, the zeroth order term is the cosmological constant and the second order term is the Ricci scalar. These terms give rise to second order field equations. (It turns out that it is possible to consider more general actions, that, nevertheless, give rise to second order field equations, but these face quite severe constraints to reconcile solar system observations with cosmological data.) To obtain the Newtonian potential one then expands the metric about a flat background and keeps only the time-time component of the metric.

In a weak gravity field φ with g00 as the metric tensor element corresponds to fourth dimension (time) we have a good approximation in general relativity for potential via the equation : g00=1-2φ/c^2

This may be interpreted roughly as "motion" along the time axis when the space-time continuum undergoes a curvature determined by Ricci curvature and metric tensor. For more powerful gravity field the relation will no remain linear.

Let us count number of fields in a phi^4 field theory - it is just one.

In QED there are two fields, the electron and the photon and there

is an interaction term in the Lagrangian density, between the fermion

current and the gauge field. Thinking in that line how to count

number of different fields in Einstein gravity. Stam, in his comment

has said that there is an action in Einstein gravity, therefore I presume

that there is also a Lagrangian density. Now how many different

types of fields enter the Lagrangian density ?

In ordinary field theories we allow terms up to mass dimension

four. What is the status if Lagrangian density in Einstein's gravity.

This is standard material of a course in general relativity and the answer can be found there, e.g. http://arxiv.org/pdf/gr-qc/9712019.pdf

The gravitational action involves the metric tensor, i.e. a symmetric tensor, but there are symmetries, so not all D(D+1)/2 possible components are ``physical''. (It turns out to be a massless, spin-2 field.)

One can, similarly, count the dimensions of the couplings from the action. Since it's only possible to add quantities of the *same* dimension, one deduces that, in four dimensions, for example, working in units where the action is dimensionless (i.e. hbar=1), the cosmological constant has dimensions of (length)^(-4) and the Ricci scalar term has a coefficient that must be dimensionful, namely (length)^(-2).

So, at the classical level, what's important is the ratio of the cosmological term to the properly rescaled Newton's constant and this is a free parameter.

This theory is *not* renormalizable as a quantum field theory of point particles--the coupling constant is dimensionful, similar to what happened in the Fermi theory of weak interactions. What are the degrees of freedom of the *quantum* theory is an open problem.

It is *not* true that in field theory one ``allows'' terms of dimension up to four. It is the symmetries that dictate the terms and their structure, given the degrees of freedom available. *If* the symmetries do allow terms of mass dimension up to four *and* these terms satisfy the constraints that the symmetries impose, beyond the classical equations of motion, *then* quantum corrections, in perturbation theory, can be expressed as a renormalization of the terms of the classical Lagrangian.

Stam

Thank you very much for the nice reference. I am happy to confine

myself to 4D only. So 10 components of a symmetric tensor is

more than enough. I feel more comfortable with Lagrangian density

than the action. Here one can see symmetries of each term in

the Lagrangian density. Does the concept of spin-2 field arise

after quantization ? Suppose I do not quantize gravity and remain

at the level of a classical theory, then will it be possible to attach

an idea of spin to the metric tensor field ?

Once more, cf. Sean Carroll's notes, the chapter on gravitational radiation. In the same way that one finds that electromagnetic radiation is described by a massless spin-1 field, one finds that gravitational radiation is described by a massless spin-2 field, which means, among other things, that the fundamental mode of radiation isn't a dipole, but a quadrupole. The gravitational waves that were detected in the variation of the period of the binary pulsar and for whose discovery Hulse and Taylor won the Nobel Prize in Physics in 1993 are, of course, classical.

The problem of quantum gravity has not been resolved, because all modern attempts have resulted in failures. If one understands EFEs and the geodesic equations, it becomes clear that the Newtonian portion of gravity arises from time dilation. Thus, there is no need for exotic explanations to understand what is occurring. To draw analogies from classical electrodynamics, the stress-energy tensor plays the role of charge density (but with energy density, momentum and pressure). Identical to a field theory that produces waves from a varying multipole (accelerating charged particle -> electromagnetic radiation), the same thing will happen when you formulate GR in a similar manner.

For any theory of general relativity however, it will always be possible to obtain the underlying potential. In EFEs, one would simply use the time components in the geodesic equations to find acceleration. Since F=ma the derivation is relatively simple, although proper time should be used.

The binary pulsars also only provide indirect evidence due to the large amount of unknown variables that have been fitted. It may turn out that the parameters of the system are different from that required by EFEs to match the signal. Furthermore, direct methods at detecting GWs are becoming at odds with predictions, as one should have been detected by now according to theory and expected detection rates.

Michael and Stam:

Let us discuss a issue regarding the point of view.

"Experimental research gives us experimental numbers

where as theoretical research explains them."

Recent results from BICEP2 experiment (B modes) is perhaps

pointing out towards a quantum theory of gravity.

If it really does after joint analysis with PLANCK data

is made, Do you think theoretical research can

form a consistent field theory such as quantum gravity ?

Or one has to appeal to string theory, which is

not a field theory.

I do not think quantum gravity or string theory are viable explanations. The standard model isn’t far off, but in general needs new foundations. There is also uncertainty as to the properties the final theory should have such as gravitational waves, i.e. the possibility of false positives from indirect experiments cannot be ruled out. For example, the slope of galaxy counts at low redshift supports a static metric. However, there are many variables involved in galactic evolution and therefore this is taken to be indirect evidence of either a static metric, evolution relative to an expanding metric or massive local hole. In particular, the BICEP2 results could suffer from local contamination, incorrect cosmological assumptions or misinterpretation of data due to gravitational lensing mixing E-modes and B-modes. The direct detection of gravitational waves with laser interferometry for example is therefore crucial to proving their existence.

Alright, let us come back to our discussion on fields theories.

At the classical level observables have continuous values

whereas in the quantum level some of them are discrete.

For example, discrete energy eigenvalues and discrete values

of angular momenta. Now the metric tensor may have a

spin in the classical level, but, when we state that it is a

spin 2 object, we mean that the spin projections

are -2,-1,0,1,2 in the units of hbar. Such values are

seen in quantum theories. Therefore the question naturally

arises that at the classical level, how can one quantify spin

of the metric tensor field.

First of all: spin *is* a quantum observable. But this has consequences that can be seen at the classical level, in this case in the helicities for the gravitational waves. This means, in particular, that the quadrupole moment is the lowest possible moment for such radiation and corresponding selection rules, just as the fact that electromagnetic radiation is described by a spin 1 field implies that the dipole moment is the lowest possible moment and selection rules. While the quantization of spin and angular momentum is a quantum effect, the consequences for the propagation of coherent superpositions of a macroscopic number of quanta can be described classically-and is in courses and books on general relativity, in the chapters on gravitational radiation.

Cf. also the Nobel Prize lectures by Hulse and Taylor.

Incidentally it *isn't* true that *all* quantum observables have discrete spectrum, so focusing on this aspect is a distraction.

Stam:

I agree that quadrupole moment is the lowest possible moment

for gravity waves whereas dipole moment is the lowest possible

one for electromagnetic waves.

This comparison between electromagnetic waves and gravity

waves may give us more intuition. I am adding some more.

1) gravity is a force gradient field whereas electromagnetism

is a force field. Therefore gravity waves are two index objects

one coming from the force vector another coming from the

gradient. Where as electromagnetic field is a force field

and consequently has only one index (vector field)

2) It is difficult to detect gravity waves but they can penetrate

matter efficiently. Electromagnetic fields, on the other

hand are easy to detect but they cannot penetrate matter.

Regarding 1): No, that's true in the Newtonian limit. Gravitational waves, however, are relevant in the relativistic case: to obtain them, linearize the Einstein-Hilbert action about some background. (The two-index nature stems from the fact that it's the metric-and that contains the information that the spin is 2.)

Regarding 2) the difficulty is, simply, due to the fact that GMm, that is proportional to the coupling, is so much smaller than all other ``usual'' scales. Mass and (electric) charge are distinct attributes and there isn't any useful relation, beyond that obtained from solving the ordinary wave equation, that can be formulated in any generality.

I look at general relativity in two perspectives. The first is on a per particle basis, where I attach a manifold to each particle’s field(s); i.e. electromagnetic, gravitational potential, ect. I then use an iterative method to find out the deformation of each particle’s manifold in order follow distance as defined by the space-time metric from the presence of all other particles under consideration. This is the equivalence principle on a per particle basis. The problem is that this does not tell you how a system evolves over time beyond the geodesic equations; i.e. it is still possible to time-step with dv/dt, but any radiation must be incorporated through ones preferred field theories. However, coordinate and curvature singularities cannot form without an infinite amount of energy in this case; assuming each particle is defined by a smooth Riemann manifold.

The second is that of localized fields traveling through other localized fields. Similar to attaching a manifold to each particle, it is possible to conceptualize particles as extended bodies. In this perspective, there is no need for faster than light virtual particles to mediate force. Instead, the use of a space-time metric allows the dynamics of these fields to be reduced to point-like objects. Therefore, the Newtonian portion of GR can be viewed as each particle trying to retain its Euclidean state in the local frame of reference due to time-dilation. This of course would originate from some underlying field theory.

Michael,

Could you please explain the second perspective that you

have. Specifically these two points.

1) "Use of space time metric allows the dynamics of these fields

to be reduced to point like objects"

2) "each particle trying to retain its Euclidean state in the local

frame of reference due to time dilation"

Thanking you in advance.

No problem,

1. If we consider the equivalence principle on a per particle basis, then the field(s) of each particle should follow that of the space-time metric. Thus it is possible to define a contravariant space-time metric (g_uv) in these regards. It is also possible to attribute another manifold to each particle (g^uv) that is a covariant metric. The orthogonality of these two metrics results in the Dirac delta (http://mathworld.wolfram.com/images/equations/MetricTensor/NumberedEquation15.gif). What this is essentially doing is making sure that the particle (g^uv) relative to the space-time metric is in it's Euclidean state at all points in space. This greatly simplifies the problem of tracking the motion of a localized wave or field to that of tracking the classical position of it. For example, if you had a wave traveling on a flat Euclidean space, you wouldn't need to determine the trajectory at each point of the wave. You would simple treat it as an envelope moving as a rigid body. This is the benefits of working with general relativity in the perspective of a global space-time metric, i.e. everything can be treated as point-like sources.

2. Let's say that particle's are actually localized waves with some type of underlying time-dependence. In the case that a single particle is being influenced by the field of some other particle, the time-dilation it experiences will vary with distance. It appears that the Newtonian part of gravity is to compensate for the time dilation, where the same trick of reducing the dynamics to point-like objects applies here. I see it as the underlying field trying to stay in phase with itself (kind of like a spring-mass system at the Planck scale), although I do not have the unified field theory to give any examples beyond the space-time metric and geodesic equations.

The metric isn't a particle (other than the graviton). The equivalence principle states that a test, massive, spinless, non-relativistic particle moves along the geodesics of the manifold, whose metric is defined by a given metric tensor. That's all. The mathematical translation of this sentence is the geodesic equation.

In fact the geodesic equation does not depend on the mass of the test particle -the non-relativistic approximation appears as the fact that the geodesic equation singles out the proper time as parameter for the world line. Beyond the non-relativistic approximation, one needs to parametrize the worldline more carefully and all this is covered in general relativity courses. Once more the metric *isn't* the test particle-that's the error here. The metric is *fixed* (that's what ``test particle'' means-it doesn't change the metric.

It is possible to describe test particles with spin moving on a given manifold-but the calculation is considerably more involved.

(To take into account how the manifold is affected, it's necessary to solve the Einstein equations with source the energy-momentum tensor of the particle in question, along with the geodesic equation of motion for the particle.)

I think you are misunderstanding what I'm saying. There is a space-time metric in the modern sense and an additional metric that can be attached to each particle. These can be treated as g_uv for the space-time metric in each particle's perspective and then p^uv for the attached manifold to each particle so that diracdelta = g_uv p^uv. The idea is that the classical fields (electromagnetic, gravitational potential, ect. -> p^uv) follow space as defined by the g_uv metric, allowing per particle solutions through the equivalence principle.

In fact, if you attach a spinning manifold (v=c, although not rigid) to a spinning electron, you arrive at the correct Bohr magneton value at the zitterbewegung radius (r_e) along the spin plane. The Dirac equation can be viewed as a two point dynamical system where one is the classical position of the electron and the other is a point along the spin plane at r_e.

Indeed, but all this has been worked out in the 70s, for instance in the work of P. Howe and collaborators: ``Local supersymmetry for spinning particles''

http://inspirehep.net/record/109550

and ``A Lagrangian Formulation of the Classical and Quantum Dynamics of Spinning Particles''

http://inspirehep.net/record/4176

Thanks for the links. Most of my work on quantum mechanics is from the space-time algebra approach combined with ensemble interpretation. I don't really understand why most sources state that quantum spin cannot be interpreted as a physical spin. There were some attempts at trying to get the Bohr magneton value from an electron following a circular path, but I believe all of those had failed.

I suppose another way to look at what I'm saying is that in EFEs a single particle would have a space-time metric attached to it (Schwarzschild metric). So if we have two particles (or objects), the summed curvature would require for the first objects manifold to follow the space-time metric defined by the other. Also in regards to the equivalence principle, I was referring to Einstein's definition: "The outcome of any local non-gravitational experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime". The dirac delta relation allows this to be implemented by attributing a manifold to each particle relative to the space-time metric induced by all other particles under consideration.

If there is one particle in a gravitational field, the field defines a manifold and the action is the integral of the proper time and this gives an unambiguous definition. The action, which is the integral over the proper time, can be expressed in terms of the energy-momentum tensor of the particle and the metric of the manifold. The ``backreaction'' can be computed by solving Einstein's equations. If there are many particles things can get complicated-the case of an infinite number of particles is, of course, a string. So one way to see what's happening is to study the Nambu-Goto action, or, perhaps, better, the Polyakov action. Discretizing the spatial coordinate will give rise to an action of a finite number of particles, interacting with a gravitational field.

The reason spin is a quantum attribute is that, first of all, if one writes the coupling of the worldline to an electromagnetic field, classically, the spin variables don't appear.

In non-relativistic quantum mechanics on needs to add them by hand and the g-factor remains arbitrary. In relativistic quantum mechanics they appear naturally, since spin, along with mass, are the Casimirs of the Poincaré group and the g-factor is fixed, within the one-particle approximation. When fluctuations must be taken into account, then the g factor receives corrections, the so-called ``g-2'' terms (more precisely (g-2)/2 terms).

Do you know of any references that consider multiple particle/object solutions with Einstein’s field equations? I’ve researched this topic for a while and could not find anything beyond very simple approximations.

If we have some object defined by a Schwarzschild metric (for simplicity), placing a particle upon it will deform the particle’s field(s) with respect to the Schwarzschild reference frame (n_uv). This is not stated explicitly in EFEs because the field(s) (electromagnetic, gravitational potential, ect.) are following the space-time metric as defined by the larger object. So it is possible to take the coordinates that one classically uses to define fields (i.e. 1/r, 1/r^2, ect) and keep track of the n_uv deformation upon a separate manifold. I’m unsure how you would find the summed curvature with this method in EFEs, but it works very well when using a locally isotropic metric with numerical methods (so perhaps a change in coordinates).

My perspective of GR and the standard model however arises from a more fundamental foundation. The thing is that particles, virtual particles and scalar fields have no physical meaning beyond measurements, i.e. they are simply symbols on paper. So one can continue to add mathematical constructs to a theory to match measurements, but this says nothing about what is actually taking placing. Realistically there is only one physical material in existence, i.e. space itself. This is why I can assure that gravitational waves will never be directly detected, because the only way to derive a physical theory that is consistent with observations is by making space at the Planck scale analogous to a 3d spring-mass system. Thus gravity is due to the underlying vacuum energy density from these fluctuations of space and all other forces arise from evolving the underlying spring-mass system forwards in time.

@Michael.

I am trying to look at the perspective about which you have

explained in detail in your post. But I have to understand the

starting point of your post.

"If we consider the equivalence principle on a per particle

basis, then the field(s) of each particle should follow that

of the space-time metric."

Here fields correspond to gravitational field produced

by a massive object at a certain point. Space time metric

exists irrespective of whether the particle is located at this

point or not. Here what is the meaning of the

word "follow". Does "follow" mean that particles are allowed

to move along certain directions ?

Robert,

Thanks for the link, I’ve read over a few pages but need to go over it again (currently finishing an article due this weekend).

“The case of two black holes that satisfies this isometry condition (often called inversion symmetry) is usually referred to as ‘Misner initial data’. It has an analytic representation in terms of an infinite series expansion.”

http://relativity.livingreviews.org/open?pubNo=lrr-2000-5&page=articlesu7.html

This seems similar to what I do on a per particle basis assuming the isometry condition is what I believe it to be. They are basically finding the initial configuration of the summed curvature and then numerically evolving it. It should therefore be possible to do per particle solutions with EFEs, which would be interesting to see if coordinate singularities arise (although I’m unsure if anyone has ever developed single particle solutions to EFEs beyond assuming the Schwarzschild metric).

Biswajoy,

If we consider a large static potential such as the Schwarzschild metric, then placing a particle upon it will have several consequences. It’s trajectory will follow geodesics as expected. However, the Schwarzschild metric will also have a line element (ds) that defines distance relative to local stationary observers. For example, an observer falling into a black hole will follow distance as defined by the line element. Yet particles are not simply point-like objects, this is known from classical electrodynamics to wave-particle duality. So along with the trajectory following geodesics as defined by the space-time metric, it is also possible to determine how the underlying field(s) will deform with respect to (n_uv). This can be the electric field, gravitational potential or even a transformation that acts upon these fields similar to the Lorentz. The simplest way to view this is by making each particle’s manifold (p^uv) the inverse of the space-time metric (g_uv) induced by the presence of all other particles under consideration, i.e. p^uv g_uv = dirac delta. Thus relative to the space-time metric every particle is in it’s Euclidean state, which means the outcome of any experiment will be independent of relative motion or gravitational curvature.

Michael,

You have three tensors. n_uv --> Schwarzchild metric

g_uv --> Space time metric

p^uv --> Particles manifold

you have stated that

p^uv g_uv=dirac delta

what about the tensor product

p^uv n_uv

Is it also well defined ?

Biswajoy,

Realistically there should be three indices so that g_ik p^ij = delta^j_k, but since the main concept is g_uv and p^uv being inverses of each other I skipped over this detail (it should also be the kronecker delta rather than dirac, but these essentially have identical concepts). If the Schwarzschild reference frame is a flat Lorentzian manifold (n_uv) the diagonal terms will be [-c^2, 1, 1, 1]. I usually work in three dimensions when visualizing how the underlying fields deform, so the –c^2 can be ignored in this case. Therefore, p^uv n_uv would need to be p^ik n_ij for an equivalent comparison. For j=1 and k=1 (working with indices 1, 2 and 3) there would be (p^11 * n_11) + (p^21 * n_21) + (p^31 * n_31) = p^11. For non-diagonal terms (for example, j=1 and k=2) you would get (p^12 * n_11) + (p^22 * n_21) + (p^32 * n_31) = p^12. In other words, p^ik n_ij = p^jk unless I’ve made a mistake.

In the perspective of EFEs, the space-time metric is taken to be something that is physically real. For example, it is the medium that gravitational waves move upon similar to the electromagnetic field. However, I strongly believe that this is a flawed idea and there is no reason to assume that gravity should be similarly defined by some type of field equation with point-like sources producing gravitational radiation. The concept of each particle's field(s) following the space-time metric is built into general relativity, so it is really a two sided thing. If one were to apply the foundations of general relativity with classical assumptions such as the equivalence principle and 1/r per particle potentials, then there are no gravitational waves or singularities. Any decrease in a particle's momentum without collision is instead released via braking radiation, i.e. bremsstrahlung.

Michael, I have something to support your belief. I am quoting

a sentence from your post

"However, I strongly believe that this is a flawed idea and there is no

reason to assume that gravity should be similarly defined by some

type of field equation with point-like sources producing

gravitational radiation."

I was recently studying, qualitatively, why gravitational radiation

is emitted. Consider a mass distribution. In general it will have

monopole term, dipole term, quadrupole term and higher

terms. The monopole term gives "total mass-energy", dipole

term gives "center of mass-energy". These two moments are

connected with conserved quantities. Total mass-energy is

conserved hence, monopole term does not give radiation

(radiation will carry out mass-energy) whereas

the "center of mass-energy" is fixed in CM frame. So dipole

term cannot radiate in CM frame. Because existence of radiation

is independent of frames, dipole term will not radiate in other

frames also. Quadrupole term is not related to a conserved

quantity, therefore it will cause gravitational radiation which

will carry out some energy.

Therefore, source of gravitational waves are not point-like sources

as you have said very correctly. The source is a mass distribution

with non-vanishing quadrupole moment. Furthermore this

quadrupole moment should vary with time; then gravitational waves

will be formed.

reference:

http://www.astro.umd.edu/~miller/teaching/astr498/lecture24.pdf

Biswajoy,

It is interesting when considering the physical significance of various multipoles. The source terms in EFEs are similar in concept to a charge density in electrodynamics. For example, a charge density is defined by the amount of electric charge per unit space. Although the end-result is a smooth field, the underlying foundations are that of point-like sources.

I think the central aspect to focus on would be how gravitational waves are produced on a per particle basis. It is possible to work in terms of densities, but these are just representations for systems of point like particles. The conclusion I have reached is that both singularities and gravitational waves are a consequence of applying point-like sources to a field equation (in the form of a density), similar to that of electrodynamics. Although there is indirect evidence for gravitational waves, it is also important to realize that the theoretical rate for direct detections has decreased by several orders of magnitude over the last few decades. This decrease in theoretical detection rates was a consequence of null detections rather than advancement in theory, i.e. the most recent experiments were constructed on the basis that they would detect something. Since all indirect evidence has a substantial amount of tunable parameters and bias, odds are now favoring that they do not exist (the last calculation I made was with the updated detection rates two years ago, providing a 58.8% chance that gravitational waves do not exist). The probability is now higher since experiments have continued to run since. With advanced LIGO, it won't be long until they are conclusively ruled out (of course assuming that the theoretical rates are not modified once again by several orders of magnitude).

I do not think Einstein's theory of gravity can be interpreted as a field theory. Unless relativistic mass is introduced in Newton's law of gravitation, a Schrodinger like wave equation cannot be developed from it. This is not possible in Einstein's geometrical theory. I have attached a theory which does this and in which quantum spin is interpreted as a physical spin. I agree with Michael when he said "Realistically there is only one physical material in existence, i.e. space itself." I will add that, this physical material of space is perfectly motionless and does not interact with the matter of our universe. Hence it can never be detected. Because it is motionless, it cannot be studied with the Nambu-Goto action, or, the Polyakov action. This is the underlying field of the universe. In the attached theory some 10^64 pairs of photon like particles called Savitons arise simultaneously. Each particle has the energy of order 10^19 GeV. So all these particles constitute pure kinetic energy from which entire table of standard model particles is obtained using a single formula. And this represents only the tip of the iceberg. The same formula can generate hundreds and thousands of other particles.

Article Periodic quantum gravity and cosmology

In GR (General relativity), the metric tensor itself plays the role of the potential. The analogy goes as follows. The motion of particle in gravitational field is governed by the geodesic equation. In coordinates, this equation has the form

acceleration = connection,

where connection is represented by the Christoffell symbols (denoted by capital Gamma usually). These coefficients are determined from the condition that the metric tensor be covariantly constant through the spacetime. This requirement then leads to the expression for the Christoffell symbols in the form

Gamma = some derivatives of the metric tensor.

This is an analogy to classical Newtonian expression, where

acceleration = force = derivatives of the potential.

Hence, comparing relativistic and classical expression for the acceleration, we conclude that the Christoffell symbols are analogic to the force. SInce they are given by the derivatives of the metric tensor (similarly: classical force is derivative of potential), the metric tensor plays the role of the potential. Instead of single scalar field, however, now we have ten components of metric tensor, i.e. ten "potentials".

From the connection the Riemann curvature tensor can be derived. Again, it is given by derivatives of the Christoffell symbols (plus some non-linear terms), so that the Riemann tensor is analogous to tidal forces in classical physics.

There is a huge gauge freedom associated with the metric tensor, for any coordinate transformation is merely a gauge freedom: it does not change physics, only the observer. In other language, any diffeomorphism of the spacetime onto itself is a gauge transformation, because the two spacetimes are equivalent from the physical point of view.

Does it answer your question?

Martin,

On a mathematical framework without considering the complications of gravitational waves and singularities this is correct. When calculating the Christoffel symbols from the space-time metric, only the time components play a role in the Newtonian part (in terms of potential). The spatial components instead vary the path similar to a wave moving through a variable medium. One can therefore say that gravity is due to time-dilation without resorting to any exotic explanations, although the spatial components are a different story.

However, in regards to the initial question this doesn’t really explain how to interpret Einstein’s field equations as a field theory. This is really two-folds; i.e. there is one field (stress-energy tensor) that acts as the source, which further relates to the underlying point-like sources via densities (energy density, momentum density, ect.). Then there is a second field, the space-time metric, that behaves in a manner similar to the electromagnetic field within classical electrodynamics (beyond the obvious differences of dipole radiation and quadrupole radiation). So the best way to interpret Einstein's perspective on gravity is through the classical electrodynamics and charge density analogy.

Thank you Martin

for writing from an intuitive point of view. If we take Newtonian

limit, Christoffell symbols should reduce to force.

In Newtonian limit what will happen to remaining nine

components of Metric tensor ? One (combination) of them

will be gravitational potential I guess ?

Please also explain to us the idea of diffeomorphism.

@Deleted,

thanks for the comment. Yes, you are right that I did not address the question of how to interpret GR as a field theory and your remarks are correct and elucidating. I have just two other remarks.

1. The interpretation of energy-momentum tensor as true energy and momentum densities is quite delicate question, since there is no universal definition of what the energy is in general curved spacetime. Thus, it is correct to say that energy-momentum tensor is the source for gravity, but that's all which can be said in general.

2. Important difference to electrodynamics is that energy-momentum tensor itself contains metric tensor, whereas the 4-current in Maxwell's equations does not contain electromagnetic fields. This makes the theory more complicated.

But I agree that the analogy between GR and Maxwell's theory is very close. This analogy is even more manifest in the 2-spinor formalism (Penrose). Maxwell's equations then have the form

spinor divergence of 2-valent spinor = source,

while gravitational field equations (Bianchi identities in this case) have the form

spinor divergence of 4-valent spinor = source.

@Biswajoy

Yes, you are right, in the Newtonian limit, the metric tensor is equal to the sum of flat (Minkowski) metric and small perturbation, which is a diagonal tensor with Newtonian potential on the diagonal.

Diffeomorphism is 1-1 mapping of the manifold onto itself. In a sense it is a generalization of translations from Euclidean space. Diffeomorphism simply shifts all points of the manifold to another points, but in a bijective way.

Any such mapping can be described in terms of the coordinates. Thus, a diffeomorphism can be regarded as the transformation of coordinates. More precisely, diffeomorphisms are active transformations, while the change of coordinates is passive transformation, but they are equivalent.

In GR we require that the theory be invariant under coordinate transformations. That means that if we describe geometrical objects living on the manifold and then change the coordinates, the objects themselves cannot be affected. The change of coordinates corresponds to the change of the observer, it is not related to the change of the object we study.

Since diffeomorphism is equivalent to a coordinate change, the original manifold and the manifold transformed by the diffeomorphism must be physically equivalent. The two manifolds just correspond to the two spacetimes as seen by two different observers, not to two different spacetimes.

One thing was a bit confusing for me time ago. Sometimes the spacetime possesses isometries. Isometries (generated by Killing vector field) are mappings which preserve the metric tensor. For example, in the static spacetime (e.g. Schwarzschild), if we perform a translation in time direction, the metric tensor does not change because it does not depend on time. We say that vector in the direction of t-axis is the Killing vector. The existence of the Killing vectors is connected with symmetries of the spacetime. Minkowski spacetime has 10 Killing vectors: 4 translations, 3 rotations and 3 boosts.

Isometries are therefore special diffeomorphisms which preserve the metric tensor. General diffeomorphism does not have this property and in general spacetime, such isometry does not exist at all.

But since each diffeomorphism is essentially the coordinate transformation, ANY diffeomorphism is just a gauge transformation. It can change the metric tensor, so it is not an isometry, but physically it corresponds to the same spacetime.

So, in GR, when we talk about the gauge invariance of the theory, we mean

a) arbitrary coordinate transformation

b) invariance of the theory with respect to arbitrary diffeomorphism.

This makes GR similar to other gauge theories, electromagnetism and Yang-Mills theory. However, here the gauge group is infinite-dimensional which makes the thing very complicated. For example, you cannot define the angular momentum (unless there are Killing vectors).

I am not sure whether I explained it clearly :)

Martin,

Yes, I can understand your passage. A few questions are there

1) By "invariance of the theory" do you mean that an action

is invariant with respect to arbitrary space-time coordinate

transformation ?

2) In a matter free theory of GR space-time is not changing.

Then will isometry transformations exist which will preserve

the metric tensor ? For example, in field theories, we can write

down a pure QED theory without any interaction terms.

Dear professor,

1) yes, by invariance I mean exactly that: the action is invariant under the coordinate transformation or, equivalently, it is invariant under arbitrary diffeomorphism.

2). Unfortunately, I don't really understand the question. What do you mean by "the spacetime is not changing"? Once you solve the Einstein equations, the spacetime is given. Do you mean the evolution of the geometry in time? Usually (in globally hyperbolic spacetimes) you can foliate the spacetime by 3d-hypersurfaces an each hypersurface represents the space (rather than spacetime) at given instant of time. There are solutions which are static or stationary, for which each slice (hypersurface) has the same geometry. Then there exists an isometry which preserves the metric tensor and this isometry can be identified as a translation in the time-like direction.

I cannot see, however, the relation to the matter content of the space. GR is a non-linear theory and hence you can have wavelike solution in the matter free spacetime.

I don't understand the analogy with QED you mention. By interaction you mean, e.g. interaction of EM field with the scalar field or the Dirac field? Of course, you can have free field in the flat spacetime. But GR itself is non-linear, so that the gravitational field is coupled to itself. There is nothing like "free" gravitational field in the sense that governing equations are linear. It can be an approximation, when you NEGLECT nonlinearities, but these are always present.

As I understand it, in quantum field theory you can have either free field (Maxwell equations without sources) or field interacting with another field (Dirac field). There is, however, no analogue of free field in GR.

Or did I misunderstood your question?

I wish you a nice day,

with the best regards

Martin

Martin, Thank you very much for your post.

In QED theory without electrons on can write a Lagrangian

density using only kinetic terms. That means no interaction

terms are present.

In gravity space-time is deformed near a massive body.

If I consider a hypothetical model where massive bodies

are absent, then will isometry transformations be present ?

I am asking this because isometry will preserve metric tensor.

Thank you Robert.

Could you please tell us about exact and broken

symmetries of space times.

We already know that FRW universe is isotropic and

homogeneous. But due to the presence of structures

isotropy and homogeneity is broken.

A broken symmetry in this context would be small deviations

from isotropy and homogeneity. This type of small deviations

can be treated by perturbation theory, see reference [4]. I am quoting

a few lines from wikipedia which can be useful to our readers in

this forum.

"The gauge-invariant perturbation theory is based on developments

by Bardeen (1980),[1] Kodama and Sasaki (1984)[2] building on the

work of Lifshitz (1946).[3] This is the standard approach to perturbation

theory of general relativity for cosmology.[4] This approach is widely

used for the computation of anisotropies in the cosmic microwave

background radiation[5] as part of the physical cosmology program

and focuses on predictions arising from linearisations that preserve

gauge invariance with respect to Friedmann-Lemaître-Robertson

-Walker (FLRW) models"

1. Bardeen J (1980) Phys. Rev. D, 22, 1882

2. Kodama H, Sasaki M, (1984) Prog. Theor. Phys. Supplement, 78, 1-166

3. Lifshitz E M (1946) J. Phys. (USSR), 10, 116

4. Mukhanov, V. F., and Feldman, H. A., and Brandenberger, R. H.: "Theory of Cosmological Perturbations", Physics Reports (1992)

5. Hu W, Sugiyama N (1995). Phys. Rev. D 51 (6): 2599.

Dear all

Respect to the title of this discussion:

"How to interpret Einstein's theory of gravity as a field theory ? "

Although the precious subjects has been already discussed , it seems that the role of the "Tetrad"( vierbein) formalism has been missed in the answers.

these local fields(tetrad) are connecting the inertial( Lorentz) basis to space-time basis( or local inertial to general manifold coordinate equivalently) and hence allows embedding of gauge field theories in curved space time to approach a gauge field of gravity.This method has been attempted in a vast majorities of related texts.(e.g. S.Weinberg "The Quantum Theory of Fields: Supersymmetry) so redirecting the discussion into this field may be worthy.

Regards

"Is it possible to obtain the metric tensor field from

a potential ?"

In Einstein's gravitational theory the metric tensor IS the potential.

"Are there underlying symmetries such as gauge symmetries

in this field theory ? "

General coordinate transformations are analogous to gauge transformations.

@Eric; them we can say that this is a "tensor potential"

Scalar potential in electrostatics gives the electric field

by e= -grad phi, Vector potential in electrodynamics gives

the magnetic field B= curl A.

Then by which relation we may get gravitational field from

the tensor potential. Is there a straightforward relation?

Thanks for your answer.

Dear Biswajoy, what do you mean by "gravitational field"? Which quantity is it for you?

Metric tensor represents the geometry in given coordinate system and as I explained above, it plays the role of the potential. But the metric tensor itself IS in some sense gravitational field, because it tells you how the time flows, what are the distances, angles, etc.

When you want to investigate the motion of a particle in the gravitational field, you need the so called Christoffel symbols. I will not repeat the argumentation again, but these symbols play the role of gravitational force. The relation between the metric tensor and the Christoffel symbols is

\Gamma^{\alpha}_{\mu\nu} = \frac{1}{2}g^{\alpha\beta}(\partial_\mu g_{\beta\nu}+\partial_\nu g_{\beta\mu}-\partial_\beta g_{\mu\nu})

I am using LaTeX notation here, if you are not familiar with it, just insert it to online LaTeX editor on the page

http://www.codecogs.com/latex/eqneditor.php

In the equation above, g_{\mu\nu} is the metric tensor, \partial_\mu is partial derivative with respect to coordinate x^\mu, and g^{\mu\nu} is the inverse metric tensor defined by

g^{\mu\alpha} g_{\alpha\nu} = \delta^\mu_\nu

FInally, capital Gammas are desired Christoffel symbols. As I said, they determine the motion of the particle in gravitational field through the geodesic equation

\ddot{x}^\mu + \Gamma^\mu_{\alpha\beta} \dot{x}^\alpha \dot{x}^\beta = 0

where the dot means derivative with respect to proper time of the particle. In Newton's theory, the acceleration is the second derivative of the position vector, and it is related to the potential by

m a = force = - derivative of the potential

The geodesic equation has also the form

acceleration = - Christoffel symbols,

and since these symbols are composed of derivatives of the metric tensor, we say that metric tensor is the potential.

Notice however, that Christoffel symbols are not tensors and they cannot be identified with the gravitational field. For if you are in the frame freely falling in gravitational field, your metric tensor will look flat (Minkowski) and its derivatives as well (locally). Hence, in such freely falling frame you will find out

Christoffel symbols = 0

and corresponding observer doesn't feel gravitational field at all (locally), Hence, whether Gammas are zero or not depends on the frame and thus are genuinely observer-dependent quantities.

What is again invariant, is the RIemann curvature tensor which measures non-local effects of gravitational field, namely the deviation of gedoesics, but I don't want to go into details here.

So, after this boring and long explanation, I can recapitulate the answer to your question.

You asked for a relation between the potential and gravitational field. Now, what you mean by gravitational field? Metric tensor, Christoffell symbols, or the Riemann tensor? When we say that metric tensor plays the role of the potential, we mean that

Christoffell symbols = derivatives of the metric tensor

and I gave you exact relation above. We can say that metric tensor represents the gravitational field, because everything else derives from it. Christoffell symbols are analogous to gravitational force, but they don't have invariant meaning.

Riemann curvature tensor is derived from Christoffell symbols but in such a way that the result is again invariant geometrical object. Riemann tensor is given by second derivatives of the metric tensor and it is non-linear in its first derivatives. If you wish I can post all desired relations here :)

@Martin,

From the point of view of an experimentalist who can measure

"forces", a gravitational field is just the force on unit mass.

Because we cannot directly measure potentials in electrostatics

or Electromagnetism, experimental verification is done by the

electric field and magnetic fields, by measuring its effects on test

charges. And we know that the last word comes from experiments.

Theory may be very elegant, but it needs experimental verification.

@ Robert,

I understand your point that time-time component of the

metric can be treated as a gravitational potential.

Because I am not very familiar with the term "weak field limit"

could you please explain how to take this limit. Unfortunately

I do not have formal training in general relativity.

General field theories are obtained through the principle of least action, meaning a lagrangian. The equations of General Relativity can be computed this way (the action is the Hilbert action which involves the curvature). This is more illuminating in the "modern" representation of GR in fiber bundlesn, then the formalism is similar for any gauge theories, and GR can be seen as a gauge theory.

The metric tensor gmu nu has 10 components. Of these 10, it is g00 that determines how fast a local clock goes (when its velocity is zero, and is seen from a distance). So g00 takes the place of the gravitational potential. In general relativity, gravity is not a single vector field (3 vector components) but a more complicated object with 36 independent components. The single gravitational potential field is replaced by the metric tensor with 10 components (of which 4 can bet set to zero by choosing your coordinates right). If you think general relativity is not elegant, that's a question of taste. Realising how efficiently it takes care of equivalence of gravitational and inertial mass, while respecting space-time relativity, without any ambiguity, I find the theory extremely elegant, in big contrast with any alternative that has been proposed.

@Biswajoy, you also asked: "In this case will that be a vector potential or a scalar potential? Are there underlying symmetries such as gauge symmetries in this field theory?", so yes, the metric tensor can be regarded as the analogue of the vector potential in electromagnetism, and it has 10 components where the em vector potential has four. And indeed there is a gauge symmetry: the "gauge transformations" in gravity are the transformations that replace the 4 space-time coordinates by 4 other coordinates, which may form a curved grid, so this transformation is a space-time dependent one, just as in em, except that there are now 4 numbers to fix rather than one. The analogy with electromagnetism goes one step further: in electrodynamics, the vector potential has 4 components, of which 1 can be set to zero using gauge transformations, and 1 plays the role of Lagrange multiplier that arranges the Coulomb attraction between charges. This leaves 2 dynamical degrees of freedom that propagate in space-time: the two possible polarisations of electromagnetic radiation). Now in gravity, the metric tensor is a 4x4 matrix, but because it is symmetric ( g mu nu=g nu mu) it has only 10 independent components. Of these, 4 can be set to zero by a "gauge constraint" and 4 others fix the Newtonian attraction between masses, as well as a force between momenta. This also leaves 2 dynamical degrees of freedom for gravitational radiation, which can therefore also be polarised in two ways. 

Prof. t Hooft

I really would like to thank you for giving me these two answers.