A The g_ij as potentials for the gravitational field

[dx]

TL;DR Summary

invitation to discuss general covariance

The equation of motion for a particle in a gravitational field is

aⁱ = -Γⁱ_jk v^j v^k

In inertial coordinates the Lorentz force is

maⁱ = qFⁱ_j v^k

So it seems like F corresponds to Γ. Just like F is expressed in terms of the derivatives of A, the christoffel symbols are expressed in terms of derivatives of g_ij:

Γⁱ_jk = 1/2 g^li [ g_il,k + g_kl,i - g_il,l ]

Fⁱ_j = g^li [ A_j,l - A_l,j ]

So apparently g_ij are the gravitational potentials just like A is the electromagnetic potential. What is the precise form of this analogy? What is the analogue of the gauge invariance under A → A + dχ? Is it general covariance? And where does R_ijkl fit into all of this?

 

[PeterDonis]

it seems like F corresponds to Γ.

No, it doesn't. The Christoffel symbols are not tensors. The tensor in GR that corresponds to F in electromagnetism, at least if you're looking at field equations and equations of motion, is the Einstein tensor G. (Note that this tensor is the contracted double dual of the Riemann tensor, which answers your question about how the Riemann tensor comes into all this.) G is the tensor that obeys a field equation analogous to Maxwell's Equations, which connects the field with the source (the stress-energy tensor T, which is the analogue of the Maxwell 4-current J) and imposes automatic conservation of the source (covariant divergence equals zero) because the field tensor obeys the necessary identities.

So apparently g_ij are the gravitational potentials just like A is the electromagnetic potential.

In some ways this is a workable analogy, but in other ways it isn't. For example, as noted above, the first derivatives of the metric g cannot be used to form a tensor that obeys a field equation; you have to go to the second derivatives.

Also, when you start looking at gauge invariance, which you also asked about, it turns out that the connection coefficients act like potentials, and the curvature acts like the field. From this viewpoint, the electromagnetic potential A is a connection on an appropriate fiber bundle (a U(1) bundle), and F = dA is the corresponding curvature. In this analogy, A is analogous to the Christoffel symbols, and F is analogous to the Riemann tensor, which is the curvature derived from the connection defined by the Christoffel symbols. Gauge invariance arises in both cases because there is not one unique connection, but a whole equivalence class of connections, that all give rise to the same curvature.

 

[haushofer]

How GR can be interpreted as a gauge theory like Maxwell's electromagnetism, see

https://www.physicsforums.com/insights/general-relativity-gauge-theory/

Maybe this analogy is also useful:

https://en.m.wikipedia.org/wiki/Gravitoelectromagnetism

 

[pervect]

Summary:: invitation to discuss general covariance

The equation of motion for a particle in a gravitational field is

aⁱ = -Γⁱ_jk v^j v^k

In inertial coordinates the Lorentz force is

maⁱ = qFⁱ_j v^k

So it seems like F corresponds to Γ. Just like F is expressed in terms of the derivatives of A, the christoffel symbols are expressed in terms of derivatives of g_ij:

Γⁱ_jk = 1/2 g^li [ g_il,k + g_kl,i - g_il,l ]

Fⁱ_j = g^li [ A_j,l - A_l,j ]

Christoffel symbols are somewhat similar to forces, as they look a little bit like forces in the equations of motion as you point out, but they aren't exactly identical to forces. As others have mentioned, they transform differently - and not as tensors. There are only 3 components in a 3-force, 4 if you generalize to a 4-force. There are 40 Christoffel symbols by my count (taking into account the symmetry of the last two indices in the assued absence of torsion), a lot more components than forces have.

So apparently g_ij are the gravitational potentials just like A is the electromagnetic potential. What is the precise form of this analogy? What is the analogue of the gauge invariance under A → A + dχ? Is it general covariance? And where does R_ijkl fit into all of this?

I'm not sure of the precise form of the analogy, but certainly for linearized gravity, the metric coefficients play the role of potentials. And for linearized graity, a change of gauge is represented by a coordinate change, a diffeomorphism, that leaves the perturbed metric $h_{uv}$ small. See for instance https://en.wikipedia.org/w/index.php?title=Linearized_gravity&oldid=982540808#Gauge_invariance

I'm not sure of the status of the analogy in the strong field. GR isn't linear in strong fields, so I think that would limit the applicability of any analogy.

 

[dx]

No, it doesn't. The Christoffel symbols are not tensors.

As others have mentioned, they transform differently - and not as tensors.

This is true, but they are still linear operators like tensors. The difference comes from the way they are defined. The tensor F is a geometric coordinate independent object like a vector, which has different components depending on the coordinates used to define the components. The affine connection's components on the other hand describe how the coordinate system is twisting and turning from point to point (imagine the present remarks in the context of a flat space with various nonlinear coordinates for definiteness and clarity.) This is not entirely unexpected, since a change of coordinate system in general relativity is equivalent to a change in the gravitational force field, e.g. an inertial frame free of forces is equivalent to an accelerated frame which can be thought of as an inertial frame filled with a uniform gravitational force field. I am talking about forces rather than using the word 'field' to be as unambiguous as possible, since for the purpose of this discussion we want to leave open the question of which symbol corresponds to the curvature, connection, potential, field etc. I believe the central issue, which Einstein emphasizes again and again is the following: Every observation or measurement ultimately rests on the coincidence of two independent events at the same space-time point. The introduction of a system of reference serves no other purpose than to facilitate the description of the totality of such coincidences. Imagine a measuring stick which measures the height of a man. Now a smooth deformation or diffeomorphism will not change the point where the head of the man meets the measuring stick, since both are deformed by the diffeomorphism.

Also, I would like to say that what I am after in this thread is not the details of the analogy (and it may not be an 'analogy' at all in the sense of a precise duality where objects on one side correspond to an analogous object on the other side. It could be more like a 'deformation') Anyway, what I am after is the significance of the diffemorphism invariance.

here are only 3 components in a 3-force, 4 if you generalize to a 4-force. There are 40 Christoffel symbols by my count (taking into account the symmetry of the last two indices in the assued absence of torsion), a lot more components than forces have.

I believe you are mistaken here. F, which has 16 components, is not the force. However, eF_tx = eE_x is the electric force in the x direction. Like I said above, imagine the present remarks in the context of a flat Minkowski spacetime with various nonlinear coordinates for the purpose of simplicity. We easily see in this context that the numbers -mΓⁱ_jk are indeed forces. They are simply the fictitious forces introduced by the non-inertial coordinates.

 

[vanhees71]

Mathematically I'd say in electromagnetism the gauge field $A_{\mu}$ is defining the gauge-covariant derivative, i.e., it's a connection (on a fiber bundle) and thus it's analogous to the Christoffel symbols in GR, defining the connection on the corresponding pseudo-Riemannian manifold.

From this point of view the Faraday tensor $F_{\mu \nu}$ is connected with the commutator of gauge-covariant derivatives and thus a curvature tensor on the fiber bundle. That makes it analogous to the pseudo-Riemannian curvature tensor in GR.

 

[pervect]

I believe you are mistaken here. F, which has 16 components, is not the force. However, eF_tx = eE_x is the electric force in the x direction. Like I said above, imagine the present remarks in the context of a flat Minkowski spacetime with various nonlinear coordinates for the purpose of simplicity. We easily see in this context that the numbers -mΓⁱ_jk are indeed forces. They are simply the fictitious forces introduced by the non-inertial coordinates.

I think you need a few more qualifiers here - but if you have an orthonormal basis $\hat{t}, \hat{x}, \hat{y}, \hat{z}$ and a stationary test charge, one of the components of F does turn out to be equal to the force.

In general, though, you need to write $e F^{ab} u_b$, where $u_b$ is the 4-velocity of the test charge. When the test charge is stationary, and $g_{00}$ is unity, $u_b$ has components (1,0,0,0). If $g_{00}$ is not unity or the charge is moving, $u_b$ won't have compnents (1,0,0,0).

And of course if the test charge is moving, you have contributions from the components of the Faraday tensor F which represent the magnetic fields to the forces.

Specific components of the Christoffel symbols in an orthonormal basis, namely $\Gamma^{\hat{x}}{}_{\hat{t}\hat{t}}, \Gamma^{\hat{y}}{}_{\hat{t}\hat{t}}, \Gamma^{\hat{z}}{}_{\hat{t}\hat{t}}$ representing some a specific frame are formally similar to forces in that frame.

However, calling this particular subset of the Christoffel symbols forces is a bit risky , because the transformation laws for the Christoffel symbols are generally different. If you stick to one set of coordinates and/or basis vectors, it can be useful to pick some particular frame and it's corresponding basis vectors , compute the above symbols, and get some physical insight. But if one wishes to use multiple or general frames, one has to be more careful.

To give a specific example, one of the predictions of GR is the deflection of light being doubled from a Newtonian-like prediction. Interpreting this extra deflection as due to a doubling of the "force" is problematical.

 

[vanhees71]

It's not a doubling of the force but the fact that it's the curvature of spacetime not only space that's relevant for the geodesics. Einstein's result of 1912 was indeed wrong, because he had not yet the full equations of GR at hand at this time. In this sense it's somewhat lucky that the light-bending measurement couldn't be done by Fröhlich in 1914, because then he'd found out that Einstein was wrong by a factor 2 in his prediction of the bending! So it was a great success in 1919 when Einstein's then of course correct prediction of the amount of light bending was confirmed by Eddington et al.

 

[dx]

I think you need a few more qualifiers here - but if you have an orthonormal basis $\hat{t}, \hat{x}, \hat{y}, \hat{z}$ and a stationary test charge, one of the components of F does turn out to be equal to the force.

In general, though, you need to write $e F^{ab} u_b$, where $u_b$ is the 4-velocity of the test charge. When the test charge is stationary, and $g_{00}$ is unity, $u_b$ has components (1,0,0,0). If $g_{00}$ is not unity or the charge is moving, $u_b$ won't have compnents (1,0,0,0).

The fact that u_b doesn't have components (1, 0, 0, 0) does not seem to be relevant to me. Maybe I am misunderstanding your point, but the schematic form of F

$$
F =
\begin{pmatrix}
\mathrm{0} & \mathrm{-E} \\
\mathrm{E} & \mathrm{B}
\end{pmatrix}
$$

$$
*F =
\begin{pmatrix}
\mathrm{0} & \mathrm{*B} \\
\mathrm{-*B} & \mathrm{*E}
\end{pmatrix}

$$

retains its meaning independently of coordinates. Here E is a 1-form and B is a 2-form. In terms of these forms, for example ∇⋅E = 1/ε₀ ρ becomes

$$-*\mathrm{d*E} = \frac{1}{\epsilon_0} \rho$$

To give a specific example, one of the predictions of GR is the deflection of light being doubled from a Newtonian-like prediction. Interpreting this extra deflection as due to a doubling of the "force" is problematical.

Could you explain this in a little more detail? A gravitational general relativistic effect can always be interpreted as the result of fictitious forces, because fictitious forces cannot be distinguished from gravitational fields.

 

[pervect]

re: the doubling of the deflection of light

It's not a doubling of the force but the fact that it's the curvature of spacetime not only space that's relevant for the geodesics.

I'd agree. A detailed calculation of the deflection of light can in my opinion shed some light on the limitations on the "force" paradigm, and why we often say that gravity is not a force. I actually suspect it's possible for a sufficiently determined person who is very attached to the idea of forces to "patch up" the force paradigm to handle light, but it's pretty clear that the the analogy to the coulomb force fails to explain the extra deflection, and the analogy to the magnetic force in this case also yields no extra deflection.

The conventional explanation of the extra deflection of light doesn't involve any extra forces, but as you say, the curvature of space, with a particular conventional choice of the space-time split.

So it's not a "force" that deflects light - but spatial curvature.

Actually, no part of the usual calculations of the deflection of light use the idea of forces at all, as far as I know. So the usual calculations (for instance using PPN, though one can use more exact schemes as well) also illustrates that one doesn't actually need forces to calculate "stuff".

To even attempt to apply forces at all, one would probably want to consider the deflection of rapidly moving particles, and calculate the deflection of the particles as a function of velocity. Then one could take the limit as the velocity of the particles approached "c". I believe one would still conclude that spatial curvature, rather than force, was the best explanation of the extra deflection of rapidly moving particles.

 

[pervect]

The fact that u_b doesn't have components (1, 0, 0, 0) does not seem to be relevant to me. Maybe I am misunderstanding your point, but the schematic form of F

$$
F =
\begin{pmatrix}
\mathrm{0} & \mathrm{-E} \\
\mathrm{E} & \mathrm{B}
\end{pmatrix}
$$

$$
*F =
\begin{pmatrix}
\mathrm{0} & \mathrm{*B} \\
\mathrm{-*B} & \mathrm{*E}
\end{pmatrix}

$$

retains its meaning independently of coordinates.

Well, rather than attempt a lengthly exposition, I'll suggest considering what F looks like in cylindrical coordiantes, $t, r, \theta, z$. What you should eventually conclude is that the components of F do not follow your simple schema.

The basic point I'm making is that tensors are expressed in terms of basis vectors. Basis vectors are not coordinates, but there is a set of basis vectors, called a coordinate basis, that is associated with a specific choice of coordinates.

Additionally, an important sub-point is that coordinate basis vectors don't necessarily have unit length. For instance, for the line element $dr^2 + r^2 d\theta^2$, the coordinate basis vector with components (0,1), usually dentoed as $\partial / \partial \theta$, has a length of "r", not a length of 1.

So the point I'm making is to consider F in cylindrical coordinates, once with the coordinate basis vectors, usually denoted as $\partial / \partial r, \partial / \partial \theta$, with the length of $\partial / \partial \theta$ being "r" and not unity. Additionally, you can if you chose use cylindrical coordianates, but not use coordinate basis vectors, but the unit basis vectors $\hat{r}, \hat{\theta}$, with $\hat{\theta} = (1/r) \partial / \partial \theta$. You'll probably not find an easy textbook explanation of how to calculate the Christoffel symbols in a non-coordinate basis, though. If you do, please share :).

Could you explain this in a little more detail? A gravitational general relativistic effect can always be interpreted as the result of fictitious forces, because fictitious forces cannot be distinguished from gravitational fields.

You can't readily explain gravitational time dilation in terms of forces, nor can you readily explain spatial curvature in terms of forces. But tensors and the associated Christoffel symbols can explain both.

To give another specific example, look at the behavior of "forces" in Einstein's elevator, ideally using tensors and Christoffel symbols. It's a flat space-time problem, no curvavture is needed. So it won't address the curavture issue, but it will address the time dilation issue. You'll see that you have some Christoffel symbols that repreent your force, but you'll have some additional Christoffel symbols related to time dilation and not to forces.

The metric for Einstein's elevator is the well-known Rindler metric, which has many forms. One of the most concise is

$$ds^2 = -a^2 z^2 dt^2 + dx^2 + dy^2 + dz^2$$

however, the conciseness comes at the expense of the origin of the coordinate system being displaced.

 

[PeterDonis]

they are still linear operators like tensors

No, they aren't. Tensors are linear operators on the tangent space at a particular event. Connection coefficients are maps from vectors in the tangent space at one event to vectors in the tangent space at other nearby events, along a particular curve. These are very different things.

F, which has 16 components

No, it doesn't. The tensor F is an antisymmetric 2nd-rank tensor, and so has only 6 independent components. These are easily interpreted, in a particular frame, as the components of the electric and magnetic force field vectors in that frame.

 

[PeterDonis]

They are simply the fictitious forces introduced by the non-inertial coordinates.

You can adopt a similar interpretation locally in curved spacetime. For example, if you set up Fermi normal coordinates centered on a particular object's worldline, with spatial axes fixed to the object, then there will be acceleration and rotation terms in the metric that tell you about the non-inertial motion of the object, and these terms, in the equation of motion of a test object in these coordinates, will appear as "fictitious forces" like "acceleration due to gravity", "centrifugal acceleration", and "Coriolis acceleration".

However, this interpretation will not work globally in a curved spacetime; in general you can't interpret terms in the metric in some global coordinate chart, or the terms in the equation of motion of a test object that they give rise to, this way.

A gravitational general relativistic effect can always be interpreted as the result of fictitious forces, because fictitious forces cannot be distinguished from gravitational fields.

As above, such an interpretation will work locally in GR, but not globally. Locally, the effects of tidal gravity--spacetime curvature--can be ignored in a small enough region of spacetime, and then the equivalence principle, which is what you are stating here, applies. But globally, the effects of tidal gravity/spacetime curvature cannot be ignored, and the equivalence principle does not apply. For example, if you are allowed to make measurements over a large enough region of spacetime, you can tell the difference between accelerating at 1 g in a rocket in free space and sitting at rest on the surface of the Earth with 1 g acceleration, by measuring the presence of tidal gravity in the latter case and its absence in the former.

Since light bending by an object like the Sun is a global effect, not a local effect, it cannot be interpreted solely in terms of "fictitious forces"; the effects of spacetime curvature cannot be ignored. In the most convenient coordinates for doing the analysis, those effects appear as "space curvature" doubling the "Newtonian" deflection (which can be derived by a local analysis using the equivalence principle), as others have said.

 

[dx]

The conventional explanation of the extra deflection of light doesn't involve any extra forces, but as you say, the curvature of space, with a particular conventional choice of the space-time split.

So it's not a "force" that deflects light - but spatial curvature.

Since light bending by an object like the Sun is a global effect, not a local effect, it cannot be interpreted solely in terms of "fictitious forces"; the effects of spacetime curvature cannot be ignored. In the most convenient coordinates for doing the analysis, those effects appear as "space curvature" doubling the "Newtonian" deflection (which can be derived by a local analysis using the equivalence principle), as others have said.

The equation

aⁱ = -Γⁱ_jk v^j v^k

is perfectly capable of explaining the deflection of light by the sun. The tensor

$$ R^i_{jkl} = \Gamma^i_{jk,l} - \Gamma^i_{jl,k} + \Gamma^m_{jk}\Gamma^i_{km} - \Gamma^m_{jl}\Gamma^m_{km}$$

and its geometric interpretation is of secondary importance to me. Obviously, Einstein needed the full field equations in order to calculate the appropriate Γⁱ_jk with the appropriate curvature, but the geodesic equation aⁱ = -Γⁱ_jk v^j v^k is exactly valid as a description of light. It applies to massless particles also.

I want to say that what I am proposing as a topic of discussion in this thread is not a crackpot or unorthodox idea. I want to discuss the gauge invariance of the field equations and what it might mean. To quote Feynman,

Consider an inertial reference frame on a flat Minkowski space. It makes no sense for us to claim that the choice of coordinates that someone else makes is crazy and cockeyed, just because it does not look like an inertial frame to us. If he insists on treating it as inertial, and ascribes curvature to the fields, he is perfectly justified in doing so, and it is our reference frame that looks cockeyed to him[...] The fact that a spin-two field has a geometrical interpretation is not really necessary or essential to physics. It might be that the whole coincidence might be understood as some kind of gauge invariance

 

[martinbn]

The tensor in GR that corresponds to F in electromagnetism, at least if you're looking at field equations and equations of motion, is the Einstein tensor G. (Note that this tensor is the contracted double dual of the Riemann tensor, which answers your question about how the Riemann tensor comes into all this.) G is the tensor that obeys a field equation analogous to Maxwell's Equations, which connects the field with the source (the stress-energy tensor T, which is the analogue of the Maxwell 4-current J) and imposes automatic conservation of the source (covariant divergence equals zero) because the field tensor obeys the necessary identities.

Something a bit off topic. This made me think about the analogy, and I do realize that all analogies are limited, but is G the analog of F. From the field equations point of view, it seems to be the right choice. But in vacuum G is zero and F need not be. Wouldn't the Weyl tensor, at least for vacuum solutions, be the analogous to the Maxwell tensor?

 

[PeterDonis]

The equation

aⁱ = -Γⁱ_jk v^j v^k

is perfectly capable of explaining the deflection of light by the sun.

No, it isn't. Integrating this equation as it stands over the global path of the light ray gives an answer for the amount of bending only half as large as the correct answer. (Note, btw, that this is not limited to light; if you try to compute the trajectory of a massive particle flying by the Sun at a speed close to the speed of light, you will have the same problem, the only difference will be that the factor by which your computed answer falls short of the correct answer will not quite be 2. The general form of the extra factor, in barycentric coordinates, is $1 + v / c$, where $v$ is the coordinate velocity of the object.)

 

[PeterDonis]

is G the analog of F. From the field equations point of view, it seems to be the right choice. But in vacuum G is zero and F need not be.

That's because the field equations aren't identical in the two cases. The Maxwell equation that relates F to the source J is $\nabla_\mu F^{\mu \nu} = 4 \pi J^{\nu}$. But the Einstein field equation that relates G to the source T is $G _{\mu \nu} = 8 \pi T_{\mu \nu}$. The latter is a simple equality, but the former is not, it's a differential equation, so it leaves room for F to not vanish even when J vanishes.

Wouldn't the Weyl tensor, at least for vacuum solutions, be the analogous to the Maxwell tensor?

I suppose one could view it that way. However, I don't know if there's any way of splitting F up mathematically into a "Weyl" piece that can be nonzero in the absence of a source, and a "Ricci" or "Einstein" piece that must vanish in the absence of a source, the way one can with the Riemann tensor.

 

[dx]

No, it isn't. Integrating this equation as it stands over the global path of the light ray gives an answer for the amount of bending only half as large as the correct answer.

Einstein's calculation was wrong because he was using Newtonian potentials instead of appropriate relativistic fields.

This is from Rindler's book.

 

[PeterDonis]

Einstein's calculation was wrong

What "Einstein's calculation" are you referring to?

 

[dx]

What "Einstein's calculation" are you referring to?

The one that was wrong by a factor of two.

 

[PeterDonis]

The one that was wrong by a factor of two.

Einstein made that calculation based on a previous version of his field equation that turned out not to be the right one, as @vanhees71 noted in post #8. When he found the final, correct version of his field equation, in 1915, he re-did the calculation and got the correct answer.

None of this changes the fact that your claim in post #14 is wrong, for the reason I gave in post #16.

 

[dx]

Einstein made that calculation based on a previous version of his field equation that turned out not to be the right one, as @vanhees71 noted in post #8. When he found the final, correct version of his field equation, in 1915, he re-did the calculation and got the correct answer.

None of this changes the fact that your claim in post #14 is wrong, for the reason I gave in post #16.

In your post 16, you seem to be claiming that not only light, but also particles (with v ~ c) don't follow the geodesic equation. This is obviously wrong. The fact that light is described by the geodesic equations is standard textbook material as far as I know. If you want I can post the details of how that works, or give a reference. Anyway, there is no way Einstein could have calculated the deflection of light from the equivalence principle before he had the field equations. When people say that the factor of two came out wrong due to the fact that there is no global inertial frame, they are referring to the fact that application of the equivalence principle was only possible in uniform gravitational fields before the Einstein equation was discovered. This is why Einstein tried to use the Newtonian potential -GM/r, and got the factor of two wrong. Your comments about v ~ c particles are also probably connected with some approximation like this.

 

[PeterDonis]

In your post 16, you seem to be claiming that not only light, but also particles (with v ~ c) don't follow the geodesic equation.

I was not making that claim. I was clarifying what "follow the geodesic equation" actually means. Integrating that equation globally involves aspects that do not appear if you only look at the equation in a single local inertial frame. Those aspects include the effects of the Riemann curvature tensor, which you appeared to be saying you could ignore. You can't just use the geodesic equation and ignore the curvature tensor if you are trying to compute a global answer. Possibly I misunderstood your remarks about the curvature tensor being "of secondary importance".

there is no way Einstein could have calculated the deflection of light from the equivalence principle before he had the field equations

Agreed; calculating it from the equivalence principle gives an answer that is too small by a factor of 2.

Einstein tried to use the Newtonian potential -GM/r, and got the factor of two wrong.

I have seen no evidence that Einstein ever did this. His 1912 calculation, referred to previously, that got an answer too small by a factor of 2, was based on an earlier, incorrect version of the field equation, which turned out to give the same answer as you would get from the equivalence principle without using any field equation. It was not based on "using the Newtonian potential".

 

[dx]

You can't just use the geodesic equation and ignore the curvature tensor if you are trying to compute a global answer. Possibly I misunderstood your remarks about the curvature tensor being "of secondary importance".

Okay. I agree you can't ignore the curvature tensor. By "secondary importance" I simply meant that I wanted to leave its geometrical interpretation as an open question for the purpose of this discussion of general covariance. I say we start with Einstein's point-coincidence argument. As I mentioned in one of my earlier posts, an active diffeomorphism of everything in spacetime does not change where a ruler meets the head of a man, so nothing measurable is affected by active diffeomorphisms. Einstein's equations are not only invariant under general coordinate transformations (passive diffeomorphisms) but also under active diffeomorphisms of the fields, so this is indeed some kind of gauge transformation of the fields, rather than just a coordinate change.

I have seen no evidence that Einstein ever did this. His 1912 calculation, referred to previously, that got an answer too small by a factor of 2, was based on an earlier, incorrect version of the field equation, which turned out to give the same answer as you would get from the equivalence principle without using any field equation. It was not based on "using the Newtonian potential".

I haven't actually read the historical papers, but when I say "using the Newtonian potential", I was referring to the fact that time dilation in the gravitational field and such things can be calculated in terms of the gravitational potential. So a half-relativistic attempt at calculating the deflection can be made by taking these dilation effects into account. Using the Newtonian potential to calculate the time dilation and then trying to calculate the deflection does indeed result in an answer off by a factor of two.

 

[PeterDonis]

when I say "using the Newtonian potential", I was referring to the fact that time dilation in the gravitational field and such things can be calculated in terms of the gravitational potential

So can the "space curvature" that provides the factor of 2 correction. The general weak field approximation to the metric that is applicable to this case is (in units where $G = c = 1$):

$$
ds^2 = - \left( 1 - \frac{2M}{r} \right) dt^2 + \left( 1 + \frac{2M}{r} \right) \left( dr^2 + r^2 d\Omega^2 \right)
$$

As you can see, the "Newtonian potential" appears in both the time and space parts of the metric. The error in the earlier version of the field equation that Einstein used in 1912 was basically that that version did not give the $1 + 2M / r$ factor in front of the spatial part of the metric in this approximation; that was what caused the answer he got then to be a factor of 2 too small.

 

[haushofer]

Something a bit off topic. This made me think about the analogy, and I do realize that all analogies are limited, but is G the analog of F. From the field equations point of view, it seems to be the right choice. But in vacuum G is zero and F need not be. Wouldn't the Weyl tensor, at least for vacuum solutions, be the analogous to the Maxwell tensor?

The analogue of F would be the field strength of the spin connection omega. It can be constructed from the [M,M] commutator of the Lorentz algebra.

 

[PeterDonis]

By "secondary importance" I simply meant that I wanted to leave its geometrical interpretation as an open question for the purpose of this discussion of general covariance.

Ok, that helps to clarify things.

Einstein's equations are not only invariant under general coordinate transformations (passive diffeomorphisms) but also under active diffeomorphisms of the fields, so this is indeed some kind of gauge transformation of the fields, rather than just a coordinate change.

There is a huge amount of literature on this, some of which has been discussed in previous PF threads, though it's been a while since we've had one.

I think the general takeaway from all those discussions is that, as you say, none of these transformations (either passive or active diffeomorphisms) change any actual physical observables, and that means there is room for a number of different viewpoints on what the transformations "mean", none of which can be tested against each other by experiment because all of the viewpoints agree on the results of all experimental tests. So it comes down to a matter of personal preference on how you want to describe what the transformations "mean".

 

[PeterDonis]

a half-relativistic attempt at calculating the deflection can be made by taking these dilation effects into account

But from a relativity standpoint, any such attempt should be immediately suspect, because "time" and "space" are frame-dependent, so if we are looking for a generally covariant expression, "time dilation" can't just be there on its own. There has to be some corresponding effect on the space part of the metric.

One way of describing the error in the 1912 version of the field equation that Einstein used for his earlier wrong calculation of Mercury's perihelion precession is that it failed to recognize the above point. Recognizing it forced Einstein to go back and rework the field equation until he found a fully generally covariant form for it, which in turn led to a weak field approximation that, as expected, included effects on both time and space, both proportional to the Newtonian potential, and which gave the correct result for the precession.

 

[pervect]

The equation

aⁱ = -Γⁱ_jk v^j v^k

is perfectly capable of explaining the deflection of light by the sun.

Agreed. But calculating the deflection requires, as others have said, is a more involved process than the shortcuts you are apparently taking.

One can calculate the deflection in Schwarzschild coordiantes, but it doesn't give as much insight into the cause as using PPN coordinates.

A bit of background. The PPN paramter $\gamma$ is described as:

\gamma How much space curvature $g_{ij}$ is produced by unit rest mass

There is another nonzero PPN paramter $\beta$ in GR, but it doesn't affect the deflection of light.

The details of the calculation can be found in MTW's (Misner, Thonre, Wheeler) "Gravitation", pg 1102, though you'll need to go to the earlier pages to get the PPN metric. Which is a bit involved, as PPN as described in MTW is used to accommodate any metric theory of gravity to allow a comparsion framework for multiple theories. However, GR has only two nonzero PPN paramters, $\beta$ and $\gamma$, and I'll give the metric coefficients for those at the end of the post.

Using the geodesic equation, one can calculate the trajectory of the light path in the $r, \phi$ plane that light takes:

$$\frac{b}{r} = \sin \, \phi + \frac{\left( 1+\gamma \right) M}{b} \left( 1 - \cos \phi \right)$$

That's just from the textbook, if you don't trust the textbook, you'll probably have to wade through it on your own, but it should be possible by solving the geodesic equations.

b can be regarded as the "impact parameter". It is approximately the distance of closest approach, which is

$$r_{min} = b\left[ 1 - \frac{ (1 + \gamma) M}{b}\right]$$

This can be expanded in a series form around around $\phi=0$, which will be regard as the ingoing ray, and around $\phi=\pi$, which will be regard as the outgoing ray:

$$\frac{b}{r} = \phi + O(\theta^2) \quad \frac{b}{r} = \frac{2 \left(1+ \gamma \right) M}{b} + \left( \pi - \phi \right) + O((\pi - \phi)^2)$$

The end result is that the spatial curvature term $\gamma=1$ doubles the deflection of the light for small angles $\phi$ compared to a hypothetical deflection for a metric theory with no spatial curvature. An example of such a theory would be the Newtonian approximation to GR which has no spatial curvature. This approximation is also discussed in MTW.

The metric elements for the Newtonian approximation in geometric units where G=c=1 are:

$$g_{00} = -1 + 2U \quad g_{ij} = \delta_{ij}$$

The metric elements for the PPN approximation in geometric units are

$$g_{00} = -1 + 2U - 2\beta U^2 \quad g_{ij} = \delta_{ij}(1+2\gamma U)$$

where U = M/r, and which can be regarded as the Newtonian potential energy. Also $\delta_{ij} =$ 1 if i=j and 0 otherwise

This is of course a lot of work. You'll most likely need to consult a textbook (such as MTW) along with a lot of time to go through it all.

But it is a textbook description of why we say that "spatial curvature" causes the doubling of the deflection of light. Implicit in talking about "spatial curvature" is a specific time slicing. The time slicing used in PPN and in the Schwarzschild metric are the same, they're the natural choice for the split which make the metric coordinate expressions time independent.

 

[dx]

...means there is room for a number of different viewpoints on what the [gauge] transformations "mean"

Agreed. But calculating the deflection requires, as others have said, is a more involved process than the shortcuts you are apparently taking.

I'm not sure why you guys feel that the deflection of light is terribly relevant for this discussion ("cannot be ascribed to a force", connection of time dilation to forces etc.), but I have some remarks about this which may be relevant. The geodesic equation can also be written in the form

(d/ds) p_i = 1/2 ∂g^jk/∂xⁱ p_j p_k

where vⁱ = g^ijp_j. Comparing the two equations

maⁱ = -eFⁱ_jk vⁱ

d/ds ( g_ij v^j) = 1/2 g_kl,i v^k v^l

we see that there is the interesting fact that that the second equation not only has gradients of the potentials, but also the potentials themselves. This would be connected to the time dilation. Also the geodesic equation in this form is more obviously reparametrization invariant, and can be applied to light more easily. Also as I said, when I use the idea of forces, I am not saying curvature is irrelevant or wrong. This is partly just to get a handle on the confusion about which symbol is the gravitational field. The gravitational potentials are of course g_ij, which is clear from the fact that gravity is a spin-2 field. haushofer says F corresponds to something to do with the spin connection. There was a comment about the Weyl tensor as opposed to the Einstein tensor etc. But independent of all this, the 'force' concept seems to be the most unambiguous thing to define which won't be affected by the confusion of which symbol is what. Guage transformations have to do with transformation of the potentials, like A. and the closest thing to the potentials are the forces, which are the gradients of the potentials. So I feel we can talk about the potentials and forces without reference to the ill-understood 'analogy'

 

[PeterDonis]

I'm not sure why you guys feel that the deflection of light is terribly relevant for this discussion

Because you keep talking about it.

the confusion about which symbol is the gravitational field

There is no single quantity that corresponds to "the gravitational field" in GR. That's why, for clarity and precision, one should use math, not vague ordinary language, to describe what one is talking about.

the 'force' concept seems to be the most unambiguous thing to define

The way "force" is defined in GR, gravity is not a force; a force is something that is actually felt as a force. An object moving along a geodesic worldline, which is what you have been discussing, feels no force at all and is in free fall. So nothing you are discussing has anything to do with "force" in the GR sense.

You appear to be using "force" in a different sense, but that sense doesn't work for what you are trying to do. See below.

and the closest thing to the potentials are the forces, which are the gradients of the potentials

If the potentials in GR are the metric coefficients, their gradients can all be made to vanish at a particular event, or along a particular worldline, by an appropriate choice of coordinates. So the intuitive Newtonian idea of "the force is the gradient of the potential" doesn't work in GR.

Note that it doesn't work in general in electromagnetism either. The potential in EM is $A_\mu$, but the force is not the gradient of $A_\mu$, except in the special case of a static Coulomb potential.

 

[dx]

The way "force" is defined in GR, gravity is not a force; a force is something that is actually felt as a force. An object moving along a geodesic worldline, which is what you have been discussing, feels no force at all and is in free fall. So nothing you are discussing has anything to do with "force" in the GR sense.

But in any reference system which is not the 'inertial' free fall frame, there is an equivalent description in terms of fictitious forces. As this idea is embedded in the equivalence principle, there should be some way to retain this idea at least in some sense. See below.

You appear to be using "force" in a different sense, but that sense doesn't work for what you are trying to do. See below.If the potentials in GR are the metric coefficients, their gradients can all be made to vanish at a particular event, or along a particular worldline, by an appropriate choice of coordinates. So the intuitive Newtonian idea of "the force is the gradient of the potential" doesn't work in GR. Note that it doesn't work in general in electromagnetism either. The potential in EM is $A_\mu$, but the force is not the gradient of $A_\mu$, except in the special case of a static Coulomb potential.

I was using the word 'gradient' in a slightly wider sense than what you are talking about here. By 'gradient' I mean some combination of the space and time derivatives of the potentials. Both E = ∇φ and B = ∇×A are such things. (or in more pretty fancy language, F = dA)

In the corresponding equations for gravity

d/ds ( g_ij v^j) = 1/2 g_kl,i v^k v^l

d/ds vⁱ = -1/2[ g_il,k + g_kl,i - g_il,l ] v^j v^k

There are two candidates for the 'gradient of the potential', i.e.

g_il,k + g_kl,i - g_il,l

g_kl,i

to be the analog of the 'gradient' A_j,l - A_l,j which gives the electric and magnetic fields. The fact that this equation can be written in two ways, and there are two candidates for the 'gradient', is obviously connected to that fact that the idea of 'force' is more subtle in general relativity, but as I said before, provided we take the subtleties into account, we may be able to retain some force like interpretation for these things. In fact, these subtleties and what they mean is one thing I want to discuss.

 

[PeterDonis]

in any reference system which is not the 'inertial' free fall frame, there is an equivalent description in terms of fictitious forces

Locally, you could construct a local non-inertial frame centered on the worldline of an object with nonzero proper acceleration (such as a rocket whose engines are firing), and in that frame, the equation of motion for a geodesic would include a "fictitious force" term corresponding to the acceleration of the rocket. (Note that this means "gravity" itself, in GR, is a "fictitious force".)

However, in a curved spacetime (GR), this does not work globally. Even in flat spacetime, a non-inertial frame in which such a "fictitious force" interpretation works (such as Rindler coordinates centered on an object with constant proper acceleration) cannot cover the entire spacetime, only a region of it.

Note also that these non-inertial frames are carefully constructed ones; I don't think you could plausibly defend a "fictitious force" interpretation for any non-inertial frame, since many non-inertial frames will look very, very different from the ones you are intuitively imagining.

we may be able to retain some force like interpretation for these things

No, you can't, because none of "these things" are tensors. You have asked about general covariance, and a key message of general covariance is that anything that depends on your choice of coordinates has no physical meaning, since you can always make it disappear by some choice of coordinates, and general covariance means any choice of coordinates is just as valid, physically, as any other. The physical meaning of the theory is contained in invariants, not coordinate-dependent quantities. The only invariants that correspond to "forces" in GR are quantities associated with non gravitational forces and proper accelerations--things that are actually felt as forces.

 

[pervect]

Locally, you could construct a local non-inertial frame centered on the worldline of an object with nonzero proper acceleration (such as a rocket whose engines are firing), and in that frame, the equation of motion for a geodesic would include a "fictitious force" term corresponding to the acceleration of the rocket. (Note that this means "gravity" itself, in GR, is a "fictitious force".)

Yes, I agree. The other noteworthy effect is that such an accelerated frame, constructed in the standard manner, will have clocks ticking at different rates depending on their "heights". This is a new effect in special relativity that is not present in Newtonian physics. It's an effect that does not fall into the "force" paradigm.

The idea that the only difference between an accelerating frame and a non-accelerating one is the addition of fictitious forces works for Newtonian physics, but it doesn't work for special relativity.

 

[haushofer]

If the potentials in GR are the metric coefficients, their gradients can all be made to vanish at a particular event, or along a particular worldline, by an appropriate choice of coordinates. So the intuitive Newtonian idea of "the force is the gradient of the potential" doesn't work in GR.

Why is that an argument? The same argument applies to the Newtonian potential, i.e. the equivalence principle also holds there, albeit restricted to spatial accelerations.

 

[PeterDonis]

The same argument applies to the Newtonian potential

No, it doesn't. You can't make the gradient of the Newtonian potential vanish by changing coordinates in Newtonian physics.

the equivalence principle also holds there

The equivalence principle holds in Newtonian physics in the sense that gravitational mass is assumed to be equal to inertial mass. But in Newtonian physics, that is an added assumption that is not linked to anything else. In GR it is not an assumption at all; it is an unavoidable consequence of spacetime geometry.

 

[dx]

Einstein's viewpoint, which is not generally the one adopted in textbooks, is that even in a flat spacetime, if you choose curvilinear coordinates, there is no way to say that these coordinates are the truly non-inertial ones. The idea of an 'inertial frame' has absolutely no objective significance. Since the physical equations are generally covariant, any reference frame has equal right to claim that it is the 'nicest' one. The laws have the exact same form in any system of coordinates, so just by making measurements, you cannot say 'this system of coordinates is truly curvilinear.' Rather, we say, "This system is also inertial, except it is filled with a uniform gravitational field." The simple minded idea that 'gravity is simply the curvature of spacetime' I think, though convenient and good enough for most purposes, does not do justice to general relativity. Such an approach must be supplemented by caveats about gauge invariance, in order to make any sense. The reason I say this is the following: Take two systems of reference A and B. In both A and B, the equation of motion of particles is

maⁱ = -mΓⁱ_jk vⁱ v^j

If, in A, the connection coefficients all happen to be zero, we want to say that it is the 'more nice' system of coordinates. We want to say that the field Γⁱ_jk = 0 is somehow special, because it seems to be distinguished from the non-zero fields by being connected to the physical 'flatness' of this particular spacetime solution. This, according to Einstein, is foreign to the spirit of general relativity. The spirit of general relativity is, 'all coordinate systems are equally good.' Even though the system A seems more special, Einstein would consider both systems as having equal right to claiming that it is the 'most nice.' He came to this conclusion from something called the 'hole paradox' and the associated 'point coincidence argument.' The upshot of these arguments is that the coordinate system is simply a book keeping device in order to label these coincidences, and no way of doing this can claim to be 'more natural.'

No, it doesn't. You can't make the gradient of the Newtonian potential vanish by changing coordinates in Newtonian physics.

Both vanishing Γⁱ_jk and the non-zero Γⁱ_jk have equal right and equal standing as representing the physical situation, and neither can claim to be nicer than the other. I realize that this is not stressed in textbooks (which invariably never discuss gauge invariance at all), but let me share the following quote to support my claims about Einstein's views.

How sound, however, Mach's critique is in essence can be seen particularly dearly from the following analogy. Let us imagine people construct a mechanics, who know only a very small part of the Earth's surface and who also can not see any stars. They will be inclined to ascribe special physical attributes to the vertical dimension of space (direction of the acceleration of falling bodies) and, on the ground of such a conceptual basis, will offer reasons that the Earth is in most places horizontal. They might not permit themselves to be influenced by the argument that as concerns the geometrical properties space is isotropic and that it is therefore supposed to be unsatisfactory to postulate basic physical laws, according to which there is supposed to be a preferential direction; they will probably be inclined (analogously to Newton) to assert the absoluteness of the vertical, as preyed by experience as something with which one simply would have to come to terms. The preference given to the vertical over all other spatial directions is precisely analogous to the preference given to inertial systems over other rigid co-ordination

 

[PeterDonis]

The idea of an 'inertial frame' has absolutely no objective significance.

This is obviously false. An inertial frame is a frame in which a body with zero proper acceleration is either at rest or moving with constant coordinate velocity (zero coordinate acceleration). This is a perfectly objective criterion, and I have seen nothing to indicate that Einstein was unaware of it or did not appreciate its significance. The only difference between flat and curved spacetime in this respect is that inertial frames can be global only in flat spacetime; in curved spacetime they can only be local.

we say, "This system is also inertial, except it is filled with a uniform gravitational field."

I don't think anyone in the GR literature has ever said this, including Einstein. Einstein did describe frames "filled with a uniform gravitational field", but I am not aware that he ever called them "inertial".

Note that in Newtonian physics, the definition of an "inertial frame" is different from its definition in relativity, because in Newtonian physics, gravity counts as a force. So it makes sense in Newtonian physics to say that, for example, a frame in which you, sitting on the Earth's surface, are at rest is "inertial", because you can account for the coordinate acceleration of a dropped rock by saying that the force of gravity is pulling on it. But in GR, you can't do this; a frame in which you, sitting on the Earth's surface, are at rest is not inertial in GR, which includes not being "inertial with a uniform gravitational field".

let me share the following quote

From where? Please give a reference.

to support my claims about Einstein's views

It doesn't. Einstein is not saying that non-inertial frames are indistinguishable from inertial frames, or that there is any such thing as "an inertial frame with a uniform gravitational field", instead of simply calling such a frame non-inertial. He is only saying that the laws of physics should not be any different in inertial frames and non-inertial frames.

 

[dx]

This is obviously false. An inertial frame is a frame in which a body with zero proper acceleration is either at rest or moving with constant coordinate velocity (zero coordinate acceleration). This is a perfectly objective criterion, and I have seen nothing to indicate that Einstein was unaware of it or did not appreciate its significance.

The following quote is from "The Meaning of Relativity".

"The weakness of the principle of inertia lies in this, that it involves an argument in a circle: a mass moves without acceleration if there are no forces; we know that there are no forces only by that fact that it moves without acceleration" - Einstein

From where? Please give a reference.

"Autobigraphical Notes"

 

[PeterDonis]

The following quote is from "The Meaning of Relativity".

Einstein is talking about the Newtonian definition of "inertia" in terms of "a body that is not subject to any forces". Also, here Einstein is using "acceleration" to mean "coordinate acceleration". In short, Einstein is pointing out inherent problems with the Newtonian definition of "inertia".

In relativity, however, "inertial frame" is not defined this way. It is defined as I said: a body with zero proper acceleration is at rest or moving at a constant coordinate velocity. "Zero proper acceleration" is a direct observable: you can measure it with an accelerometer. It does not require any definition of "force" or any choice of coordinates, and so there is no circularity in defining an inertial frame this way: we can pick out which objects are "moving inertially" (have zero proper acceleration) without having to already know what an "inertial frame" or a "force" is and without having to measure any coordinate acceleration.

 

[pervect]

I'm not sure why you guys feel that the deflection of light is terribly relevant for this discussion ("cannot be ascribed to a force", connection of time dilation to forces etc.),

It's mainly a few specific example of where the force paradigm you seem to be attached to breaks down. Pointing out abstractly that it breaks down is one thing, but I thought a few more specific examples would help. So far, I can't say that I feel like you appreciate the point in quesiton.

The other example where the force paradigm breaks down is "gravitational time dilation". That phenomenon is one that occurs in accelerated frames in SR and some situations in GR, but does not naturally (or unnaturally, as far as I know) arise from any amount of musing about forces. It's an example of why there is more to Christoffel symbols than "forces".

but I have some remarks about this which may be relevant. The geodesic equation can also be written in the form

(d/ds) p_i = 1/2 ∂g^jk/∂xⁱ p_j p_k

Why $p_i$ and not $u^i$? $u^i$ being the 4-velocity. I believe it's also necessary that one impose $|u^i u_i| = 1$, but I'd have to look up the details.

 

[haushofer]

No, it doesn't. You can't make the gradient of the Newtonian potential vanish by changing coordinates in Newtonian physics.

You mean globally? Under an acceleration

<br /> x^{&#039;i} = x^i + \xi^i (t)<br />

the gradient transforms as

<br /> \partial^{&#039;i} \phi&#039;= \partial^{i} \phi - \ddot{\xi}^i (t)<br />

such that locally (where we can approximate $\partial^{i} \phi$ to be constant) we can put

<br /> \partial^{&#039;i} \phi&#039; = 0 \ \ \ \ (locally \ in \ space!)<br />

by solving, for a given $\partial^i \phi (x)$,

<br /> \partial^{i} \phi = \ddot{\xi}^i (t)<br />

for the acceleration $\xi^i (t)$. But maybe I'm misunderstanding your statement.

 

[PeterDonis]

You mean globally?

No, I mean locally. Consider just the radial direction for simplicity. If $x$ is the spatial coordinate in a frame in which the Earth is at rest, then $x' = x + v t$ is a Galilean transformation, and $v$ is the relative velocity, not an acceleration. An object that is free-falling downward due to the Earth's gravity will only be at rest in the transformed frame for a single instant. The gradient transforms like this:

$$
\frac{\partial}{\partial x'} = \frac{d x}{dx'} \frac{\partial}{\partial x} + \frac{d t}{dx'} \frac{\partial}{\partial t}
$$

Since $dx / dx' = 1$ and $dt / dx' = 0$, the gradient is unchanged by this transformation.

 

[PeterDonis]

Under an acceleration

<br /> x^{&#039;i} = x^i + \xi^i (t)<br />

the gradient transforms as

<br /> \partial^{&#039;i} \phi&#039;= \partial^{i} \phi - \ddot{\xi}^i (t)<br />

What kind of transformation is this? Certainly not a Galilean transformation. But the Newtonian laws of physics are only invariant under Galilean transformations.

 

[dx]

Why $p_i$ and not $u^i$? $u^i$ being the 4-velocity. I believe it's also necessary that one impose $|u^i u_i| = 1$, but I'd have to look up the details.

It is essentially to distinguish between group velocity and phase velocity. The group velocity v = g(_,p) where p is the phase velocity. There are several ways to work with these problems, but for the point I am trying to make, it is convenient to treat the motion of particles in the WKB approximation, or the 'geometrical optics limit' of waves described by the 'dispersion relation'

g(p, p) = 1

So the group velocity

d/ds = -g^ijp_j (∂/∂xⁱ)

represents the 4-velocity of the particle, i.e. the tangent vector to the curve s → xⁱ(s), i.e. vⁱ = -g^ijp_j. Differentiating the dispersion relation, we get

∂_k (g^ij p_i p_j) = g^ij_,k p_i p_j + 2g^ijp_i ∂_kp^j

Because p is a phase velocity, we have ∂_kp_j = ∂_jp_k, so we get the equation

∂_jp_k (g^ij p_i) = -1/2 g^ij_k p_ip_j

The left hand side is the derivative of p_k in the direction of the tangent vector d/ds = -g^ijp_j (∂/∂xⁱ), so we are left with

d/ds ( g_ij p^j) = 1/2 g_kl,i p^k p^l

The only reason I was talking about phase velocity and group velocity is to distinguish between 1-forms p_j and vectors vⁱ = -g^ijp_j. This is what allows us to write ∂_kp_j = ∂_jp_k. If you keep this distinction in mind, you can forget about the whole 'geometrical optics limit' interpretation.

 

[PeterDonis]

It is essentially to distinguish between group velocity and phase velocity.

We are not talking about waves, we are talking about particles. There is no such thing as "group velocity" vs. "phase velocity" for particles.

The only reason I was talking about phase velocity and group velocity is to distinguish between 1-forms p_j and vectors vⁱ = -g^ijp_j.

That distinction has nothing to do with "phase velocity" vs. "group velocity" for particles. See above.

Also, there is no minus sign in the transformation from 1-form to vector.

 

[dx]

The WKB approximation treats particles as wave-packets, so there is indeed a distinction. phase velocity is a 1-form. But as I said, this is not the essential thing in my post. An object with a superscript is a tangent vector, and an object with a subscript is a 1-form. This is all that is needed to understand my post.

 

[PeterDonis]

The WKB approximation treats particles as wave-packets

This is the relativity forum, not the quantum physics forum. Discussion of this kind of thing belongs in the quantum physics forum.

this is not the essential thing in my post

Yes, and it is also off topic for this forum, as above.

An object with a superscript is a tangent vector, and an object with a subscript is a 1-form.

Yes, anyone participating in an "A" level thread in the relativity forum already knows this.

 

[dx]

Yes, and it is also off topic for this forum, as above.

The algebra is easier to do in the WKB approximation picture, talking about 'dispersion relation' etc. (It is useful conceptually also, but I won't use this if you feel it is off topic) The 'dispersion relation' is of course nothing but the mass-shell condition, so you can do it that way also. The equation I wrote can be derived by a different procedure, but I wanted to avoid that. It has nothing to do with the WKB thing, it is equivalent to the usual form of the geodesic equation, and general relativity books derive it in a slightly different way.

 

[PeterDonis]

The algebra is easier to do in the WKB approximation picture.

It's not the "WKB approximation picture". It's the "1-form picture". This is the relativity forum.

The equation I wrote can be derived by a more complicated procedure, but I wanted to avoid that.

This is an "A" level thread in the relativity forum. That means all participants are assumed to know what the geodesic equation is and how to raise and lower indexes and what vectors and 1-forms are. There is no need to belabor any of these points.

What @pervect was asking was why you prefer the 1-form picture to the vector picture for the geodesic equation. The only substantive answer you have provided so far that is even applicable to a discussion in this forum is "the algebra is easier to do", which is just your subjective opinion and doesn't make much sense to me--raising and lowering indexes is a simple operation and doesn't change the algebraic complexity of any equations.