RETRACTED: It's a bug of general relativity
By Doug Sweetser | May 1st 2012 12:03 AM | 37 comments | Print | E-mail | Track Comments

Trying to be a semi-pro amateur physicist (yes I accept special relativity is right!). I _had_ my own effort to unify gravity with other forces in...

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[Retracted: In this blog I hope to get to a well known issue in general relativity, namely that gravitational energy needs to be defined for a spacetime volume instead of at a point.]  The [new] goal is to sketch the Riemann curvature tensor...

There are limitations to what one can do in a blog.  The flowchart is from Sean Carroll's lecture notes that I highly recommend.  In this blog, I will presume I am working with a Riemannian manifold, the final step of the geometry construction.  A Riemannian manifold allows one to do differential calculus even in curved spacetime.

There still are choices to be made.  One of the key choices is the sort of connection to use with the manifold, in step 3 of the flow.  Should the connection involve a metric?  Should the connection account for torsion?  Since the blog concerns general relativity, the choices made in that theory are for a connection that is metric compatible and torsion-free.  In practice that means one works with the Christoffel symbol of the second kind (that almost sound like a movie title).  The Christoffel employs three derivatives of a metric:

$\Gamma_{\gamma \alpha \beta} = \frac{1}{2} \left( \frac{\partial g_{\gamma \alpha}}{\partial x^{\beta}} + \frac{\partial g_{\gamma \beta}}{\partial x^{\alpha}} - \frac{\partial g_{\alpha \beta}}{\partial x^{\gamma}} \right)$

The Christoffel symbol is not a tensor even though those look like tensor indices.  It is also "flat" in a technical sense (discussed by Carroll at the bottom of page 73, partly because moving a vector with a given Christoffel keeps it pointed in the same direction).

Where does one use Christoffel symbols?  Start with a 4-potential.  Then take its 4-derivative.  Unfortunately the result is not a second rank tensor.  The reason is that spacetime may be curved.  When that happens, the changes in the potential will remain silent to the changes in the curved spacetime.  A covariant derivative can account for both changes in the potential and in spacetime:

$\nabla_{\mu} A_{\nu} = \partial_{\mu} A_{\nu} - \Gamma^{\sigma}_{\;\mu \nu}A_{\sigma}$

The covariant derivative does transform like a tensor.  One term keeps track of changes in the potential, the other will add up all changes due to curvature.

In EM, the field strength tensor uses this covariant derivative and its antisymmetric twin:

$F_{\mu \nu} = \nabla_{\mu} A_{\nu} - \nabla_{\nu} A_{\mu}$

The Christoffel symbol is symmetric, so it will drop out of this calculation.  In EM, one needs to supply a metric as part of the background mathematical structure.  Whenever I read about the background structure, I like to re-read a post from John Baez on the subject

General relativity on the other hand provides a means for determining what metric applies based on the physical conditions.  The tool for this task is known as the Riemann curvature tensor.  Consider a 4-vector in spacetime.  Move it around a bit, but eventually bring it back to its starting place.  Will the 4-vector remain the same?  It will if spacetime is flat.  If spacetime is curved, then the 4-vector will point in a different direction.

The Riemann curvature tensor quantifies the difference between the starting and ending vectors caused by the curvature of spacetime.  This is done by looking at the difference between two paths:

$R^{\rho}_{\;\sigma \mu \nu}= \partial_{\mu} \Gamma^{\rho}_{\;\nu \sigma} - \partial_{\nu} \Gamma^{\rho}_{\;\mu \sigma} + \Gamma^{\rho}_{\;\mu \lambda} \Gamma^{\lambda}_{\;\nu \sigma} - \Gamma^{\rho}_{\;\nu \lambda} \Gamma^{\lambda}_{\;\mu \sigma}$

This complex creature does transform like a tensor even though the Christoffel symbols do not.  Recall that the Christoffel has first order derivatives of the metric, so the curvature tensor has second order derivatives of the metric.  Solutions to second order differential equations of the metric can be used to define a dynamic metric (not completely however).

A reductionist points to two critical properties: as the name implies, the Riemann curvature tensor transforms like a tensor and this tensor contains within it second order derivatives of the metric.  This is the central tool for general relativity.  The vacuum field equations are the difference between the Ricci tensor, a contraction of the Riemann curvature tensor, and the Ricci scalar, a contraction of the Ricci tensor:

$R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = 0$

The reason one needs both the Ricci tensor and scalar is to conserve energy via the Bianchi identities.  Since these tensors are contractions of the Riemann curvature tensor, they both are tensors that contain second order derivatives of the metric.

Riemann Normal Coordinates

One picks coordinates to make things simple.  The Riemann normal coordinates are nice in the way they ensure the partial derivatives of a metric vanish at a particular point p.  At that point p, all the Christoffel symbols are zero since the Christoffel is a contraction of three partial derivatives of a metric.  It is vital to note the disappearing act only happens at the point p, not its closest neighbor, call it p'.  Curved spacetime remains curved in Riemann normal coordinates.  The Riemann curvature tensor cannot be made zero by the choice of Riemann normal coordinates since the curvature tensor accounts for changes between points p and p'. [Retracted: This underscores the need to look over a volume of spacetime when thinking about general relativity].

In an electric field, every point on the manifold will have a field strength, and from that, a corresponding energy (ignoring issues that arise from singularities).  [Retracted: In contrast, it is more difficult to know what one means for a point in spacetime if curvature necessarily involves a comparison between two points.  Certainly there are limit processes, but that becomes a concern in light of the Riemann normal coordinates since one cannot eliminate spacetime curvature by a choice of coordinates.]

[Retracted: Some consider this issue a feature of general relativity, the need to work with volumes of spacetime to define the energy of a gravitational field.  This is a fundamental way the fundamental force of gravity is different from the other three forces of Nature that can be defined point-wise. ]

The only way to get a paying job in gravity theory research is to presume the Riemann curvature tensor is part of the way Nature works in practice.  Even folks working out in eleven dimensions justify their funding by pointing to the Einstein field equations sitting in their high dimensional stew.  Loop quantum gravity work presumes the Einstein field equations are a point of departure for constructing spin foams at the Planck scale.

Before I present my own evolving position, let me remind the reader of a few things.  This is just a blog.  My views are my views.  I am not asking you to follow.  I write to clarify my own perspective.  Should you disagree, please remain polite so I don't have to deal with violations of the golden rule.

[Moderation aside:  I used to participate in the moderated newsgroup sci.physics.research.  In one points, I described the problem that has been around since 1930, getting general relativity to play nicely with quantum mechanics.  The current efforts try to bend one toward the other with fancy math, taking care to break neither.  I suggested a simpler idea: what if general relativity was wrong?  There is a long tradition to directly confronting old, widely accepted theories in physics.  GR beat Newtonian gravity by experimental tests.  Alas, the post was deemed overly speculative.]

[Retracted: If fields must be well-defined for every point in a spacetime manifold, then the Riemann curvature tensor represents a bug.]  There [may] be another expression that transforms like a tensor yet contains within its bounds a second order derivative of a metric tensor.  It should be obvious, but apparently isn't, that any second order covariant derivative contains second order derivatives of a metric tensor.  The first covariant derivative has a metric compatible connection which contains first order derivatives of a metric.  Take a derivative of that, and one necessarily has a second order derivative of a metric that at least stands a chance of characterizing a dynamic metric.

[Discussing below, so of questionable value: The Lagrangian of general relativity now strikes me as a rigged game.  We need second order derivatives of a metric, so start with the Ricci scalar which contains a second order derivative in its heart.  Vary that with respect to a metric, and voila, the field equations still contain second order derivatives of the metric.  In EM, the fields are first order derivatives of the potential.  The field equations are second order derivatives of the potential.  That is how the game should be played.]

It is vital to look at general relativity with the eyes wide open.  [Clarification: I am biased.  As someone trying to make an alternative proposal for gravity, I am hoping to find a subtle issue that indicates a flaw.  Yet I have said for a few years that the more one learns about GR, the more impressive are its accomplishments]. How good is the data? An honest answer is needed.  Here is the Schwarzschild solution in isotropic coordinates, the sort of coordinates that will be used in experimental tests:

$d \tau^2 = \left( \frac{1 - \frac{G M}{2 c^2 R}}{1 + \frac{G M}{2 c^2 R}}\right)^2 d t^2 - \left( 1 + \frac{G M}{2c^2 R}\right)^4 \left( d R^2 + R^2 d \theta^2 + R^2 \sin^2 \theta ~d \phi^2\right)/c^2$

I always have to look this form up.  The value of GM/c2R is small, even for the Sun at its surface (2.12 x 10-6).

$g_{tt} =1 - 2 \frac{G M}{c^2 R} + 2 \left( \frac{G M}{c^2 R} \right)^2 - \frac{3}{2} \left( \frac{G M}{c^2 R} \right)^3 + \left( \frac{G M}{c^2 R} \right)^4 + O(5)$

wolframalpha

$g_{RR}=1+ 2 \frac{G M}{c^2 R} + \frac{3}{2} \left( \frac{G M}{c^2 R} \right)^2 + \frac{1}{2} \left( \frac{G M}{c^2 R} \right)^3 + \frac{1}{16} \left( \frac{G M}{c^2 R} \right)^4 + O(5)$

wolframalpha

In physics as in beauty contests, the prettiest equations always win.  The exponential metric is prettier:

$d \tau^2 = e^{-2 \frac{G M}{c^2 R}} d t^2 - e^{2 \frac{G M}{c^2 R}} d R^2/c^2 - R^2/c^2 (d \theta^2 + \sin^2 \theta d \phi^2)$

I never have to look this one up.  The Taylor series is better looking too, particularly the way the terms for changes in time are identical to those for changes in space forever.

$g_{tt}=1-2 \frac{G M}{c^2 R}+2\left( \frac{G M}{c^2 R} \right)^2- \frac{4}{3}\left( \frac{G M}{c^2 R} \right)^3 +\frac{2}{3} \left( \frac{G M}{c^2 R} \right)^4+O(5)$

wolframalpha

$g_{RR}=1+2 \frac{G M}{c^2 R}+2\left( \frac{G M}{c^2 R} \right)^2+\frac{4}{3}\left( \frac{G M}{c^2 R} \right)^3 +\frac{2}{3} \left( \frac{G M}{c^2 R} \right)^4+O(5)$

wolframalpha

It should be simple to distinguish between these two: go do the measurement, and the real one will win.  Surely this has been done, right?  The answer is no, it has not been done.  Recall how small GM/c2R is for the Sun.  The classical tests of general relativity have measured these terms of the Taylor series and no others:

$\\ g_{tt} = 1-2 \frac{G M}{c^2 R}+2 \left( \frac{G M}{c^2 R} \right)^2 + O(3)\\ \\ g_{RR} = 1+2 \frac{G M}{c^2 R} + O(2)$

Both metrics are the same.  Why no more?  This was hard enough to do in 1919.  There are some technical questions over whether the data did support the conclusion reached at the time.  Radio wave astronomy made it possible to repeat the light bending tests at any time of year.  The current resolution is at 100 microarcseconds, 10,000 times better than in Einstein's time, but a factor of 100 less than needed to tell the difference between these two metrics.

Formulating a new proposal is not an all or nothing game.  It is an all or -500 game.  One person did not wish me to write this post, so he offered to pay for mental health services for me instead.  Given the high costs of such professionals, I did not take him seriously and deleted his request (he did not log in so i was unable to haggle over the costs).  Let me talk about my hypercomplex gravity failure.  What I wanted were two fields that were like those in EM, but different.  I placed a few conjugate operators in there so the fields were different.  There were multiple flaws leading to the retraction.

The tension in my research on gravity is how to make it different from EM, but not too different.  The hypercomplex gravity proposal was too different.  Now I am feeling a tug the other direction, to someday (not today) come up with an idea that is more like EM, yet all about geometry.  In EM, I see three levels:

The potential:
$A_{\mu}$

The fields:
$F_{\mu \nu}=\nabla_{\mu} A_{\nu} - \nabla_{\nu} A_{\mu}$

The field equations:
$\\ \nabla_{\nu} F_{\mu \nu} = J_{mu}\\ \\ \nabla_{\gamma} F_{\mu \nu} + \nabla_{\nu} F_{\gamma \mu} + \nabla_{\mu} F_{\nu \gamma} = 0$

One generates the source field equations by varying the action with respect to the potential.

In general relativity, everything is different.  There is no potential.  In EM, the action arises from a contraction of the field strength tensor which is a derivative of a potential.  In GR, it is like they know they need to work with the Riemann curvature somehow, and the Ricci scalar does the trick (it certainly does).  Because it is as pure geometry as anything can be, it cannot be linked to the other three forces of Nature.  At least not after generations of work by theoretical physicists.

Because the field strength tensor of EM is antisymmetric, it cannot speak about metric geometry.  Perhaps a proposal that starts with the fields of EM but goes smaller in some technical fashion, the field equations that result can only speak of geometry.  The fields are the same, but the field equations are different.  It is something to dream about.

Doug

Snarky puzzle: Calculate the Christoffel symbol for the exponential metric.

Next Monday/Tuesday: GR vs QFT - DiscussNote: I will set up the question, but not provide my own answer in the blog, saving that for the comments which you can skip ;-)

"In an electric field, every point on the manifold will have a field strength, and from that, a corresponding energy (ignoring issues that arise from singularities). In contrast, it is more difficult to know what one means for a point in spacetime if curvature necessarily involves a comparison between two points. Certainly there are limit processes, but that becomes a concern in light of the Riemann normal coordinates since one cannot eliminate spacetime curvature by a choice of coordinates."

It sounds like your "bug" of general relativity, is that you feel a derivative can't be defined at a single point!? Sure, as you point out, you can choose your coordinate system so the Christoffel symbol is zero at some point. And then you complain this means you can't define the "field" at a point...

"This is a fundamental way the fundamental force of gravity is different from the other three forces of Nature that can be defined point-wise."

If you consider a derivative as non "point-wise" (essentially saying a derivative in non-local), then NONE of the forces or Nature are defined "point-wise" and are all non-local theories. Remember when I asked you to think critically for awhile and work to figure out what you even mean by the "field" of gravity? Can you please take time to think about that question now?

"If fields must be well-defined for every point in a spacetime manifold, then the Riemann curvature tensor represents a bug."

Do you think F_uv of electromagnetics can be well defined at every point? That involves a derivative of the potential. Why then can't the Riemann curvature tensor be well defined at every point? Why is that derivative a "bug"?

"The exponential metric is prettier:"

It is ruled out by any experiment that can measure the Lense–Thirring effect. Said another more constructive way: it only looks pretty to you because you restrict yourself to a highly symmetric static case. Any attempt to generalize this will immediately look much uglier than GR.

You have a knack of avoiding useful questions, and becoming mislead by questions which are not useful in your quest. Try following up on David's breadcrumbs, or some of the questions I gave, until you can train yourself to ask useful questions at which point you'll be able to critically evaluate your own ideas. Currently you are relying very heavily on your readers to do the thinking for you.

"Because it is as pure geometry as anything can be, it cannot be linked to the other three forces of Nature. At least not after generations of work by theoretical physicists."

A mere 5 years after GR, there was a famous proposal for how to combine GR and EM in a geometric theory.
http://en.wikipedia.org/wiki/Kaluza%E2%80%93Klein_theory

The Riemann curvature tensor is an incredibly complicated rank four tensor.  The four derivative of a four potential that gives rise to the fields of EM is by comparison quite a trivial operation.  It strikes me as an oversimplification to say what is true for the simple structure should just as well apply to the far more complicated thing.  As an example, I have written out all the components of the EM field strength tensor, its dual, then done the contraction.  It took an hour or so, but was doable.  I once started off trying to handle writing the Riemann curvature tensor in terms of the metrics (45=1024 metrics in all), but gave up on the task.

The blog does not claim a massive flaw, it is far more subtle.  In the EM case, you can pick a frame of reference where the B field goes to zero for a point, but not a frame of reference where both B and E go to zero.  In GR, you can pick a frame where for any arbitrary point the field is zero, but for any volume, the gravitational field is not zero.

"The exponential metric is prettier:"
It is ruled out by any experiment that can measure the Lense–Thirring effect. Said another more constructive way: it only looks pretty to you because you restrict yourself to a highly symmetric static case. Any attempt to generalize this will immediately look much uglier than GR.

I don't see the link.  One uses the Kerr metric for frame dragging experiments which applies to a spinning mass with angular momentum J.  What I was doing was comparing an apple to an apple.  Yes, I am comparing two metrics that apply only to a highly symmetric, static case.  And when I do so, I am not at peace with the terms claimed by general relativity.  General relativity say the fourth term in the Taylor series expansion for gRR should be a factor of 16 smaller than the similar one for gtt.  I cannot justify that difference for a situation of such high symmetry.  For the exponential metric, they are the same.

A mere 5 years after GR, there was a famous proposal for how to combine GR and EM in a geometric theory.
http://en.wikipedia.org/wiki/Kaluza%E2%80%93Klein_theory

That is part of the history of physics.  I don't think there is a large school of people that think the Kaluza-Klein proposal such as it is does constitute a viable replacement for general relativity and the Maxwell equations.  Folks who work with strings often pitch their program as the modern extension of Kaluza-Klein.

"In GR, you can pick a frame where for any arbitrary point the field is zero, but for any volume, the gravitational field is not zero."

I keep asking you to define the gravitational field. The only way to have that statement make any sense in the context of your article is if you are defining "gravitational field" = "Christoffel symbol".

Is that your definition of "gravitational field"?

It's still not clear why you think the Riemann curvature tensor isn't well defined at a point. I asked: "Do you think F_uv of electromagnetics can be well defined at every point? That involves a derivative of the potential. Why then can't the Riemann curvature tensor be well defined at every point? Why is that derivative a "bug"? "

Your answer is not logical. Your justification is that you think we can take a derivative of F_uv because it is simple, but not the Riemann curvature tensor because it is complicated? How does that have anything to do with the existence of a derivative?

Why in the world would you _want_ to write out all the coordinate components of the Riemann curvature tensor. There is nothing illustrative to be gained in such an exercise. Do you agree a computer could write out all the components? If it can be written in principle, do you agree that in principle it is well defined at every point? Does someone need to write it out before you will believe this? Why?

"In the EM case, you can pick a frame of reference where the B field goes to zero for a point, but not a frame of reference where both B and E go to zero. In GR, you can pick a frame where for any arbitrary point the field is zero, but for any volume, the gravitational field is not zero."

To make a more fair comparison, in EM you can set A_u = 0 at an arbitrary point. Does that mean F_uv is not well defined at that point?

It seems like your confusion comes down to whether it is well defined to take the derivative of a function at a point where the function equals zero.

"What I was doing was comparing an apple to an apple."

No you were not comparing apples to apples. You restricted yourself to a highly symmetric situation so you could write the metric in a form that you consider "simpler", then declared that the Rosen metric was prettier.

Again, the point is that to modify the exponential metric idea to handle anything but this symmetric situations and still match the experimental data, it would necessarily become much much MUCH uglier.

You state in your article that "It is vital to look at general relativity with the eyes wide open." But you seem to be demanding the opposite. You demand we deny a derivative is well defined so that you can say the Riemann curvature tensor is not well defined at a point, and then require everyone squint with eyes barely open to ignore all possible situations except for a highly symmetric one in order to declare "In physics as in beauty contests, the prettiest equations always win. The exponential metric is prettier" You are not approaching this with eyes wide open.

At least take the time to critically think about whether it is possible for a function to be zero at a point, but still have a well defined derivative at that point. Think about it for awhile and see if you can come up with an example function that shows this. I think you'll learn something by trying to think this out yourself. Then, without requiring yourself to write out all the components of the Riemann tensor, can you apply this lesson to the existence of the Riemann tensor?

I keep asking you to define the gravitational field. The only way to have that statement make any sense in the context of your article is if you are defining "gravitational field" = "Christoffel symbol".

???  How many times must I say a Christoffel symbol does not transform like a tensor?

There is a metric field guv.  Go to different places in space-time, and the metric will be measurably different.  What one does in GR is to take the Hilbert action in a vacuum which is just the Ricci scalar R, vary it with respect to the metric, and that generates the Einstein field equations in the vacuum.  I agreed to that line of logic in a previous reply in a different blog.

Why in the world would you _want_ to write out all the coordinate components of the Riemann curvature tensor. There is nothing illustrative to be gained in such an exercise. Do you agree a computer could write out all the components? If it can be written in principle, do you agree that in principle it is well defined at every point? Does someone need to write it out before you will believe this? Why?

Yup, a computer could do, but a human probably would never bother.  To me, that is a reason to be skeptical of people who say GR is simple.  I know I have said the same thing myself based on the form of the Hilbert action.  It is a reason for me to call into doubt the long term viability of GR.

To make a more fair comparison, in EM you can set A_u = 0 at an arbitrary point. Does that mean F_uv is not well defined at that point?

I wound not consider this more fair.  Where Au=0, then there is no electric field nor a magnetic field, so the field strength tensor is zero there.  Likewise, the energy density of the field will also be zero.  The problem which I claim is more subtle than that, but I don't think we can discuss subtle issues if you consider my position to be so trivial.

No you were not comparing apples to apples.

This kind of dialog shows we are just bickering.  I will define apple #1 with greater care, and apple #2.  Apple #1 is for outside a spherically symmetric, non-rotating, electrically uncharged source.  The solution to the Einstein field equations in this highly symmetric case is known as the Schwarzschild metric.  If one writes that in isotrophic coordinates and takes the power series in terms of the geometric mass (GM/c2R), then the fourth term has a gRR term that is a factor of 16 smaller than gtt for no apparent reason.  Apple #2 is for outside a specially symmetric, non-rotating, electrically uncharged source.  The exponential metric is proposed ad hoc.  All the terms of the same power are of the same magnitude, but the signs flip for the gtt terms.

To give a possibly positive spin on what is certain a drag on us both, here is a question.  Feynman, Weinberg, and Hawking will forever have so many more accomplishments that either you or I in the areas of quantum mechanics and gravity.  All three certainly considered for longer periods of time how to get them to work together.  Do you consider their efforts a failure?  If so why?

Doug,
You said before to ask you to write out the math if you start increasingly becoming illogical. So I am asking you to please translate two things into math to think them out better.

You previously said:
"In GR, you can pick a frame where for any arbitrary point the field is zero, but for any volume, the gravitational field is not zero."

What you claimed in your blog could be set to zero with careful coordinate choice was the Christoffel symbol. You did not show or claim this for R, R_uv, R_uvwz or any other tensor. So I took this to mean you were calling the "gravitational field" the Christoffel symbols. Instead of answering my question, you instead give the non-sequitor "??? How many times must I say a Christoffel symbol does not transform like a tensor?", and then ignore the gravitational field question again.

#1] Convert your statement "In GR, you can pick a frame where for any arbitrary point the field is zero, but for any volume, the gravitational field is not zero." into math. In particular clearly define mathematically what you mean by "gravitational field", and then show that you can always choose coordinates to have the field be zero at an arbitrary point.

The next question, you did answer. (Thank you.) Unfortunately, as it becoming the pattern, instead of taking the time to think about a simple counter-example to help you learn your mistake, you have decided to just double down on your misunderstanding and make more mistakes.
I asked "in EM you can set A_u = 0 at an arbitrary point. Does that mean F_uv is not well defined at that point?"

You answered "Where Au=0, then there is no electric field nor a magnetic field, so the field strength tensor is zero there. Likewise, the energy density of the field will also be zero."

This is very sad. Just because Au=0 at a point does NOT mean Fuv is zero at that point. You ignored some very simple math questions, so you keep making this error. So math request two is a rephrasing of a previous question:

#2] If at the point x, the function f(x) = 0, can we say anything about the derivative of f at the point x from this information? More specifically, can one conclude at the point x that f ' (x) = 0, if f(x)=0?

Please take the time to slow down and think.

Agreed.  I had 3 people to respond to.  If I failed to respond to every point, I would face the critique that I ignored someone's reply.  Gotta go, programming to be done.

#1] In Riemann normal coordinates at a point p have all partial derivatives of the metric are zero (but NOT an area, that would be illegal).

$\Gamma_{\gamma \alpha \beta}_{at\,p} = \frac{1}{2} \left( \frac{\partial g_{\gamma \alpha}}{\partial x^{\beta}} + \frac{\partial g_{\gamma \beta}}{\partial x^{\alpha}} - \frac{\partial g_{\alpha \beta}}{\partial x^{\gamma}} \right) = 0$

At the point p, that simplifies the calculation of the Riemann curvature tensor:

\begin{align*} R^{\rho}_{\;\sigma \mu \nu\;at\;p}&= \partial_{\mu} \Gamma^{\rho}_{\;\nu \sigma} - \partial_{\nu} \Gamma^{\rho}_{\;\mu \sigma} + \Gamma^{\rho}_{\;\mu \lambda} \Gamma^{\lambda}_{\;\nu \sigma} - \Gamma^{\rho}_{\;\nu \lambda} \Gamma^{\lambda}_{\;\mu \sigma} \\ &= \partial_{\mu} \Gamma^{\rho}_{\;\nu \sigma} - \partial_{\nu} \Gamma^{\rho}_{\;\mu \sigma} \end{align*}

There is zero reason for this to be zero.

#2] If I am counting correctly, there are about infinite number plus or minus a few of situations where at a particular point x, if f(x)=0, then the derivative of f(x) will not be zero.

\begin{align*} x&=0\\ f(x)& = sin(x) = 0\\ f'(x) &= cos(x) = 1 \end{align*}

Obvious stuff.  I was buried in the bunker of details of the Rienmann curvature tensors while in Riemann normal coordinates.

"Certainly there are limit processes, but that becomes a concern in light of the Riemann normal coordinates since one cannot eliminate spacetime curvature by a choice of coordinates."

The "bug" of GR is that curvature can be defined in a coordinate independent way so that a choice of coordinates can't eliminate it? It would be a bug if the curvature -could- be removed merely with a change of coordinates.

"The covariant derivative does transform like a tensor. One term keeps track of changes in the potential, the other will add up all changes due to curvature."

I don't think it is correct to separate them like that. As you point out yourself, it is possible to choose coordinates so that the second term is zero. So by your prescription, in that coordinate system the covariant derivative of the potential measures just changes of the potential. So it probably isn't the best idea to try to give separate physical meaning to the two pieces.

"If fields must be well-defined for every point in a spacetime manifold, then the Riemann curvature tensor represents a bug."

You feel the Riemann curvature tensor cannot be well defined at every point?

This is like the 'logical disconnect' LFB was complaining about. When you basically are complaining about a math definition, why you do think it is something to motivate your theory search? After all this time, if you come across a worry about a theoretical construct in a well studied and respected physics theory, why don't you immediately first consider that you may need to learn more about the theory?

Last time you blamed the disconnect on us for now writing math equations. That is unfair to blame us for this. As this appears to be a problem before you even interact with us. It is a disconnect in critical thinking. I'm not sure what to suggest to fix this situation. But not blaming it on your readers may be a good start.

"The covariant derivative does transform like a tensor. One term keeps track of changes in the potential, the other will add up all changes due to curvature."
I don't think it is correct to separate them like that. As you point out yourself, it is possible to choose coordinates so that the second term is zero. So by your prescription, in that coordinate system the covariant derivative of the potential measures just changes of the potential. So it probably isn't the best idea to try to give separate physical meaning to the two pieces.

$\nabla_{\mu} A_{\nu} = \partial_{\mu} A_{\nu} - \Gamma^{\sigma}_{\;\mu \nu}A_{\sigma}$

There are two pieces, and they are not the same.  I am curious what words you would use to describe the two pieces.

You feel the Riemann curvature tensor cannot be well defined at every point?

No, the issue is more subtle than that.  It is that curvature can be set to zero for an arbitrary point, but not be set zero for a volume.

"There are two pieces, and they are not the same. I am curious what words you would use to describe the two pieces."

And why do you insist on demanding the two coordinate dependent pieces have a physical description?

"No, the issue is more subtle than that. It is that curvature can be set to zero for an arbitrary point, but not be set zero for a volume."

NO! No, no, no, NO!
You cannot set the curvature to zero for an arbitrary point. You understood this in your article, but instead of fixing mistakes, you are creating new ones.

Just because a function is zero at a point does not mean its derivative is. Translate that to a math equation if you really need to see it as such.

I don't get it, are you claiming the Riemann curvature tensor isn't really a tensor? What exactly are you complaining is a bug (but also a useful feature) of GR? It's hard to follow your thought process.

The title of the blog may be doing a disservice to the blog proper.  The Riemann curvature tensor really is a tensor.  It is not only a tensor, but its two contractions, the Ricci tensor and Ricci scalar, are also tensors.  More importantly, those tensors do lead to the general relativity theory which so far has passed every experimental tests we have been able to develop.  Any replacement would need to pass the weak field tests, the gravity probe b tests, and, at least in my opinion, be consistent with the strong equivalence principle.  Since I don't have a well formed counter proposal,  people can and should defend GR.

If one wants to quantize a field theory, what does one do?  I was surprised to read in a quantum field theory book that a first step is to invert the field equations to generate the propagator.  That step often requires picking a gauge.  That process works for EM.  People who are really good at quantum field theory can probably discuss for hours why to this day there remain problems with doing that for GR.  I believe one of the technical problems is the ability to set the connection to zero at any point.

Right at the end of Raum, Zeit, Materie Hermann Weyl found that the gravitational field involves momentum flux across a surface of some kind. But he was later most surprised to find the metric as he defined it morphing into the Schroedinger wave (see Preface to 4th ed. In Dover edition).

Now in Loop QG, the method entails only surfaces bounded by loops, but the formalism throws up at the same time a quantization of volumes!

And there is now a trend in GR to orbifolds which give "normal" vectors at any point. The more adventurous look to p-adics or non-archimedean points. So there's really no need to get strung-out on first-order tension!

I fully agree about second-order derivatives, but see the problem in relation to torsion. How do you fit spin into the energy-momentum tensor? In what geometry?

I fully agree about second-order derivatives, but see the problem in relation to torsion. How do you fit spin into the energy-momentum tensor? In what geometry?
I am so "no there" yet.  There have been quite the collection of people who worked on torsion, but I have not been able to follow any of that work.
Maybe if we clap twice and say the secret words David will magically appear and Doug will suddenly realize its not a good idea to just randomly claim a derivative isn't well defined.

Doug will you ever be humble enough to consider that when you come across some confusion when studying physics, that maybe, just maybe, you should first consider that you don't understand something, instead of assuming you've discovered something to motivate your search for a new pet theory?
What is the cause of these logical disconnects?
Until you acknowledge the problem, you have no hope in fixing it. Take that first step to recovery.

Doug:

You already have people here that are trying to help you see your own errors.  Listen to them.  Work out the issues they are trying to get you to be able to see for yourself.

I was going to show a correspondence between your "three levels" of Electromagnetism (EM)*, and General Relativity (GR).**  However, that would interfere with you answering the question of what you think is the gravitational "field".

So, I'll focus on this paragraph from your article:

The Lagrangian of general relativity now strikes me as a rigged game.  We need second order derivatives of a metric, so start with the Ricci scalar which contains a second order derivative in its heart.  Vary that with respect to a metric, and voila, the field equations still contain second order derivatives of the metric.  In EM, the fields are first order derivatives of the potential.  The field equations are second order derivatives of the potential.  That is how the game should be played.

Now, in general, if the Lagrangian Density (or the Action, which is the spacetime integral of that density) includes nth order derivatives, what will be the order of the derivatives in the equations obtained by the variational principle?  Or, far more simply, if the Action (functional) involves second order derivatives, what order derivatives will be involved in the equations obtained by the variational principle?  Or still simpler, if the variational principle applied to an Action involving only first order derivatives yields equations with second order derivatives, will the variational principle applied to an action involving second order derivatives yield equations with derivatives of order greater than 2?

Hint:  If you get even just the last question right, then you will see that having an Action involving the Ricci scalar is not such a "rigged game."  Those pesky second derivatives in the Action could be a real problem!

David

*  Actually, the "three levels" for Yang-Mills theory, which includes EM as a special case for a commuting field, can be even more instructional since there is an additional term at the "second level" that disappears for a commuting field.

**  Additionally, one can even find a "level" that is before your "first level" (a "zeroth level", so to speak) for both Yang-Mills and GR.

Here is the Hilbert action in a vacuum:

$S = \int \sqrt{-g} d^4 x R$

where

$g = det(g_{\mu \nu})$

Small technical aside… What's up with the square root?  It is not needed in flat space-time.  The 4-volume element is called a tensor density.  In curved space-time, volumes are bound to change.  The weighting will make this transform like a tensor instead of a density.

In studies of gravity, the metric guv is a field: go to a different location, and the metric will have a different value.  In GR, one varies the Hilbert action with respect to the metric field:

Oh holy wiki crap!  Looks like the "right" way to derive the field equations really is on wikipedia.  I think instead of typing that into this comment, I will make it into a blog. , n steps to deriving the Einstein field equations.  [note, I have done that calculation before as part of the Sean Carroll lecture notes.]

"What's up with the square root? It is not needed in flat space-time."

If you want to use arbitrary coordinate systems, it is there. In flat space-time we can of course just write the action restricting ourselves to a metric of the form (-1,1,1,1) on the diagonal.

"Oh holy wiki crap!"

Focus! You got side tracked and didn't answer David's questions.

David was not asking you to go through all the details of deriving the Einstein field equations. Don't get lost in the weeds. Look at the big picture and you'll learn more. (Actually, if you just blindly work out the specific case of the Hilbert Action and look at the result, you'll completely miss the point since the result will be the Einstein field equation which is indeed a second order differential equation for the metric. David wasn't claiming that it wasn't. Please go back and read what he wrote. Try focussing on what he actually asked.)

Also, diving into the GR derivation on wikipedia will be meaningless if you still are having trouble understanding derivatives.
Consider the function f(x) = x^3 - x. What is the value of f(x) at x=0? (answer: f(x=0)=0) Does this mean the derivative ( f ' (x) ) at x=0 is also zero? (answer: No, as you can check explicitly f ' (x=0) = -1)

I have finally picked up on my error, but it is easy to get lost in the world of Riemann curvature tensors and Riemann normal coordinates.  Before I go and fix the blog, I am questioning a vague technical issue.  I do recall other people taking about energy density in GR, and saying it could not be thought of in a point-wise fashion, it required thinking about volumes of spacetime.  No doubt what I blogged about here has not a thing to do with that, but is that (even roughly) an issue with GR proper?
There are a couple things you may have heard pieces from and are combining. You can undoubtable find better info in your books, but here is the run down:

- First, the stress energy tensor is well defined at every point. The T_00 component is usually referred to as the energy density.

- Momentum-energy is locally conserved in the sense that T^ab ; _a = 0. This is garaunteed before even specifying the details of the matter Lagrangian !! due to the choice of the geometric description of gravity. Einstein used this to guide him to his equations.

- Unfortunately, there is some competing semantics on this. Every physicist will at least know what you mean when you say energy is locally conserved in GR (T^ab;_a = 0), but since this involves the covariant derivative some people push the notion that energy is not conserved even locally. I strongly dislike that point of view as it can lead to confusion in conversations, but be aware of that point of view.
http://blogs.discovermagazine.com/cosmicvariance/2010/02/22/energy-is-no...
http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

- Now in the "volume" vs "point" issue, a little aside first. In flat spacetime in nice inertial coordinates like we normally use when learning EM, we can quickly go from the 'derivative form' to the 'integral form' of Maxwell's equations. The math that allows us to do this doesn't work right when we have covariant derivatives instead of just partial derivatives. This means even if the stress-energy tensor is well defined at every point, we can't just jump to gauss's law like integral instead. The Baez link above has a decent intro to this if you want a better casual intro than I can give here.

- OK for this next point, again we are going to run into some terminology issues here, but in this case I am not sure I will be able to help more than confuse. David can assuredly word this more carefully than I, so please ask him to correct this and give insight. Hilbert had a hunch and Noether did some work to show that GR has enough freedom that we cannot discuss global energy conservation in GR in the general case (even though we can always say locally T^ab;_a = 0). However, if there is enough symmetry in the physical scenario, then this can be used to define some global conservation of energy (this is related to the existence of timelike Killing vectors in some spacetimes).

- And finally, and this really gets into some semantic nonsense (enough that Feynmann was very disgusted with the academic field at the time because of it), there is the issue of energy in the gravitational field. The shortest, albeit conflicting sounding, summary is this: "gravity" doesn't contribute to the stress energy tensor, but "gravity waves" do carry energy. There was a long debate about how to interpret the second part. Because of the first part, Einstein and others argued for awhile that gravity waves may just be mathematical artifacts with no physical consequence. Einstein eventually used a pseudo-tensor to describe a local energy-momentum, in which the 'energy' in a gravitational fields could be seen better. As with anything in math problems, if it is computational useful, it will be used even if physical interpretation is a bit tough. The problem with energy pseudo-tensors is that they are not true geometric entities, and there is not a unique definition for it. http://en.wikipedia.org/wiki/Stress%E2%80%93energy%E2%80%93momentum_pseu... Until you have time to really study GR, I guess the summarized take-away is just that gravity waves are "real" in that they can have physical effects (I've even seen solutions in GR where colliding gravity waves create a black hole.)

- Another common way to phrase some of the math difficulties in GR is the 'mass problem'. How do you even define mass in GR? (For example, have you ever wondered where that M in the Schwarzschild metric comes from? After all it is a vacuum solution. The association of the integration constant to some factor of M is usually fixed by comparing to Newton in the weak field limit.) Its worth thinking about this a bit and reading up on various defintions of mass in GR. http://en.wikipedia.org/wiki/Mass_in_general_relativity

I'm very very worried that I just caused a lot more harm than help. GR is well defined. I stress again, GR is well defined. So one could just do the math and completely avoid all terminology issues if they so chose. But if history shows anything here, it is that you need to be very careful to avoid laziness and sloppiness in translating ideas and expectations from flatspace into the dynamic geometry of GR. If there is ever confusion, clearly and precisely define what each person means (I've seen arguments between students in real life about GR go away immediately once they took the time to define what each meant), and ultimately, refer to the math.

---
David, I'd appreciate it if you can reinforce and/or correct the points I made above.

Bravo.

I used to follow John Baez closely when sci.physics.research was his main outlet.  I am not going to quote extensively, but the stuff about energy being a pseudo-tensor got partially written in my brain.
I think you've done just fine.  Of course, the question of whether you "just caused a lot more harm than help" will only be answered by where Doug goes from here.  :/

David

Doug:

There is an "issue" with trying to pin down the energy-momentum-stress of the "gravitational field" (of course, this goes back to the question of what one means by that term) in full General Relativity (GR).

This "tensor" is rather explicit in linear (spin 2) "gravity", but it also plays some parts in the internal inconsistency of that "theory".

Of course, since we are talking about a tensor quantity (if it could be defined), I also don't see how this can be addressed over "volumes of spacetime", in GR:  Integrals of vectors and other tensors are not well defined for curved spaces (spacetimes).

Besides, for a theory such as GR, which is about the "bending" of spacetime itself, what does one actually mean by the "energy", or "energy-momentum-stress" of such a "thing"—of spacetime itself?

Additionally, when one goes through the work of making linear (spin 2) "gravity" internally consistent, one finds that the "energy-momentum-stress" "tensor" of that theory is incorporated into Gμν = Rμν - (1/2)gμνR.  However, it is incorporated in a rather inseparable fashion, and can be seen as being at the root of the non-linearity of GR.

David

Doug you previously wrote:
"Physics word problems stump me. Please feel free to remind me of this particular weakness I have should we dive into another issue. I don't mind translating (not verified) people's posts into LaTeX."

There is a severe disconnect in logic here.

Here's a copy of some of the math that would help you if you translated it to equations and thought it out. (Reread the previous posts to see why it is relevant.)

---------------
#1] Convert your statement "In GR, you can pick a frame where for any arbitrary point the field is zero, but for any volume, the gravitational field is not zero." into math. In particular clearly define mathematically what you mean by "gravitational field", and then show that you can always choose coordinates to have the field be zero at an arbitrary point.
---------------
#2] If at the point x, the function f(x) = 0, can we say anything about the derivative of f at the point x from this information? More specifically, can one conclude at the point x that f ' (x) = 0, if f(x)=0?
---------------
Consider the function f(x) = x^3 - x. What is the value of f(x) at x=0? (answer: f(x=0)=0) Does this mean the derivative ( f ' (x) ) at x=0 is also zero? (answer: No, as you can check explicitly f ' (x=0) = -1)
---------------
Now, in general, if the Lagrangian Density (or the Action, which is the spacetime integral of that density) includes nth order derivatives, what will be the order of the derivatives in the equations obtained by the variational principle?
---------------

Will do that tonight.  I see an error in my future.
Doug:

May I recommend a highly simplified version of my question(s)?

For the sake of this question, I don't think we really need to look at general Lagrangian densities.  I think (hope) that you will be able to extrapolate from the more simple one dimensional Lagrangian case.

You have already shown that you recognize that for a Lagrangian that depends only on a function and its first derivative, L(f(t), df/dt = f '(t)), that the Euler-Lagrange equation (obtained by varying the action, equal to the time integral of the Lagrangian, with respect to the function) is

So, the first question is, what is the highest order derivative of the function that may be found in this Euler-Lagrange equation, in general?

Now, the second question is, what is the equivalent of the Euler-Lagrange equation (obtained by varying the action, equal to the time integral of the Lagrangian, with respect to the function) when the Lagrangian depends on a function, and its first and second derivatives, L(f(t), f '(t), f ''(t))?

The third and final question (the one that is fully analogous to the simple form of my questions) is, what is the highest order derivative of the function that may be found in this new Euler-Lagrange like equation, in general?

David

To make this even more concrete (the kind of shoes I like to wear on the physics floor, not the swing dance floor), define a Lagrange density like so:
$\mathcal{L}=f^2 + f'^2$

At this time, I prever the derivatives with respect of f on one side of the equals, while the derivatives of the f' are on the other.  Granted it makes no technical difference.

\begin{align*}\frac{d(f^2 + f'^2)}{d f}& = \frac{d}{d t}\,\frac{d(f^2 + f'^2)}{d f'}\\ 2 f &= 2 \frac{d}{d t} f' \\&= 2\frac{d^2 f}{d t^2} \end{align*}

Applying the Euler-Lagrange equation to a functional with first order derivatives found an extremum function that is second order in its derivatives.

Continue the process with a new Lagrangian, one that depends on second order derivatives.  If one thinks of the functional as a function of functions, then I am feeding it a different diet of functions, but it should do the similar thing (+1).  See if the details work out like that:

\begin{align*}\frac{d(f^2 + f'^2 + f")}{d f}& = \frac{d}{d t}\left(\frac{d(f^2 + f'^2 + f")}{d f'}+\frac{d(f^2 + f'^2 + f")}{d f"}\right)\\ 2 f &= 2 \frac{d}{d t} f'+2 \frac{d}{d t} f" \\ &= 2\frac{d^2 f}{d t^2}+2\frac{d^3 f}{d t^3} \end{align*}

Let me do a close-up of one of those derivatives, with the f" in the numerator and f' in the denominator:

\begin{align*}\frac{d(f")}{d f'} = \left(\frac{df'}{d f'} \right)'=0 \end{align*}

Why?  With functionals, those different derivatives of functions are really different variables and are independent of each other.  This does depend on being able to move around the partial derivatives.

Now that I have two point, should I rush to draw the line?  Probably not.  The derivatives with respect to the function or its derivatives change the functional in the process of finding an extremum, but the derivatives that remain are of the same order.  It is the time derivative that does the +1 to the highest order derivative in the functional or Lagrangian.  In general, whatever the highest derivative in the Lagrangian happens to be, the extremum function that results may have at most one higher order derivative.
Doug:

I'm not so sure you did yourself any favors by choosing some particular form for your Lagrangian.  (Technically, you don't have Lagrangian densities, by the way.)

1. You did get the correct answer for the first question.
2. You didn't actually answer the second question.  However, judging from the first line of your big, second from last equation, you didn't get this question right.
3. You correctly found that you get higher order derivatives, but you are actually off by one order.  (Even just dimensional analysis would yield the correct answer, because, other than the signs, it would give the correct forms for the three parts of the answer to the second question.)

David

I was pre-occupied with the retraction, so failed to do this the right way.  The goal is to derive the Euler-Lagrange equation, but with one more term in the Taylor series expansion.  Start with the action:
$S[f]=\int_a^bdt\mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]$

Do the variation on the function f:

\begin{align*} \delta S[f]&=S[f+\delta f]-S[f]\\ &=\int_a^bdt\left(\mathcal{L}\left[f + \delta f, \frac{d (f+ \delta f)}{d t},\frac{d^2 (f+ \delta f)}{d t^2}\right]-\mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right] \right) \end{align*}

Take the Taylor series of the first Lagrangian assuming small variations to second order:

\begin{align*} \mathcal{L}&\left[f + \delta f, \frac{d (f+ \delta f)}{d t},\frac{d^2 (f+ \delta f)}{d t^2}\right]\\&=\mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]+\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial f} \delta f+\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial \frac{d f}{dt}} \frac{d(\delta f)}{d t}\\&\quad+\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial \frac{d^2 f}{d t^2}} \frac{d^2(\delta f)}{d t^2} + O(3) \end{align*}

Plug this back into the action, noticing the first term drops:

$\delta S[f]&=\int_a^bdt\left( \frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial f} \delta f+\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial \frac{d f}{dt}} \frac{d(\delta f)}{d t} +\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial \frac{d^2 f}{d t^2}} \frac{d^2(\delta f)}{d t^2}\right)$

The middle term appears in the first order case.  The trick is to use the product rule of calculus like so:

$\frac{d a}{d t} * b = \frac{d (a * b)}{d t} - a * \frac{d b}{d t}$

The middle term can thus be written as:

$\frac{d}{dt} \left(\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial f} \frac{d(\delta f)}{dt}\right) - \frac{d}{dt} \left(\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial f} \right)\delta f$

S
The first term is a total derivative.  As such, by the fundamental rule of calculus in 1D, or Stoke's theorem, the value will depend only on the boundary.  As we are looking to vary the function inside that boundary, this term will not make a contribution to the variation.

Look at the third term.  Again rewrite that term using the product rule:

$\frac{d}{dt}\left(\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial \frac{d^2 f}{d t^2}} \frac{d(\delta f)}{d t} \right) - \frac{d}{dt}\left(\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial \frac{d^2 f}{d t^2}}\right) \frac{d(\delta f)}{d t}$

The first term depends only on the boundary and thus will not contribute to the variation.  The second term can be rewritten using...the product rule.

$- \frac{d}{dt}\left(\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial \frac{d^2 f}{d t^2}} \frac{d(\delta f)}{d t} \right) +\frac{d^2}{dt^2}\left(\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial \frac{d^2 f}{d t^2}}\right) \delta f$

Lots of applications of the product rule.  Use the product rule an even number of times, then the sign is positive.  Use it an odd number of times, the sign is negative.  For each term, only one lives "inside" the boundary contributing to the variation.

$\delta S[f]&=\int_a^bdt\left( \frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial f} - \frac{d}{dt} \left(\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial f} \right) +\frac{d^2}{dt^2}\left(\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial \frac{d^2 f}{d t^2}}\right) \right)\delta f$

This integral will be at an extremum if the integrand is zero, or:

$\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial f} - \frac{d}{dt} \left(\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial f} \right) +\frac{d^2}{dt^2}\left(\frac{\partial \mathcal{L}\left[f, \frac{d f}{d t},\frac{d^2 f}{d t^2}\right]}{\partial \frac{d^2 f}{d t^2}}\right) \right)=0$

The result can be generalized for more terms of the Taylor series expansion:

$\frac{\partial \mathcal{L}}{\partial f} + \sum_{n=1}(-1)^n\frac{d^n}{dt^n}\left(\frac{\partial \mathcal{L}}{\partial \frac{d^n f}{d t^n}}\right) \right)=0$

Now to repeat with more than the time derivative...maybe not.

The Lagrangian has to depend on n derivatives, and then one takes n time derivatives, so the general result is 2n.
Very good, Doug.  I knew you could do it.  :)
Replied in a new starting spot to reset the conversation at a new place...
It was a good question that I was able to work on while the young lady slept in the car awaiting a hair cut.  It is straight forward to see what is going on for both EM and gravitomagnetism since their structures are so similar.

I am going to have to go back and look at GR and see if I can "get" the Ricci scalar.  The variation is with respect to the metric tensor, guv.  According to the 2n solution here, it all depends on what gets packed into the Ricci scalar, the contraction of the contraction of the Riemann curvature tensor.  I claimed in the main blog that this should come with two derivatives of a metric already.  While the Euler-Lagrange could end up with as many as 4th order derivatives, that is not what is inside the Ricci tensor.  One step forward, two steps back, but at least I hope this are not trivial steps back.
Doug:

You are quite correct that the Ricci scalar contains second order derivatives of the metric tensor (gμν).  You are also quite correct that this seems to conflict with the 2n result you found, since we know that the result of variation of the Hilbert action is a field equation with only second order derivatives of the metric tensor.

So what gives?

This was actually a homework problem in my General Relativity class.  While one can simply go through the variational procedure with the Hilbert action, and find that all the higher order terms cancel, this provides little if any insight.  The homework problem, instead, required us to simply take a closer look at the Hilbert action, and integrate out any and all integrals of total derivatives—in other words, apply Stokes' theorem to the Hilbert action, without any consideration of variations, etc.

I don't know whether you want to take on that "homework" assignment, but it could be instructive.

David

I don't want to make too much out of your comment, but do need to calibrate the significance. To derive the Maxwell equations or the Einstein field equations requires Stokes theorem.  One also need to vary the functional on the road to the Maxwell equations, picking out an extremum.  Now you are saying Stokes is enough in GR, the commonly referred to variation by a metric field is not needed, there will be no extremum of possible functions.  I don't recall ever hearing a statement like that before about GR (and I hope my paraphrase was accurate).

Although it may be a suicide mission, let me take a few steps.  I can write the Hilbert action, from memory no less:

$S = \int \sqrt{-g} dx^4 R$

That gives me one point out of a possible ten.  Looks like one integral to me, but the question talks about multiple integrals.  Let me try to rewrite R (and I have little confidence about the next line, but forward I go):

$S = \int \sqrt{-g} dx^4 g^{\sigma \nu} R^{\mu}_{\;\sigma \mu \nu}$

Now I get to apply my definition of the Riemann curvature tensor:

$S = \int \sqrt{-g} dx^4 g^{\sigma \nu} \left( \partial_{\mu} \Gamma^{\mu}_{\;\nu \sigma} - \partial_{\nu} \Gamma^{\mu}_{\;\mu \sigma} + \Gamma^{\mu}_{\;\mu \lambda} \Gamma^{\lambda}_{\;\nu \sigma} - \Gamma^{\mu}_{\;\nu \lambda} \Gamma^{\lambda}_{\;\mu \sigma} \right)$

This is progress in there are 4 integrals now where there used to be one.  Is this the beast I am suppose to look at with my fancy Stoke theorem glasses?

Doug:

Unfortunately, your paraphrase is incorrect, on the General Relativity side.

When I said you would not be doing any variational work, I was strictly referring to the "homework" assignment.

The "homework" assignment is not about deriving the General Relativistic field equations.  To do that, one would have to vary the action functional by varying the metric tensor, and picking out an extremum.  All quite normal stuff.

No.  The "homework" assignment is, essentially, to use Stokes' Theorem to "simplify" the action by taking out all the integrals of total derivatives, and seeing what's left.

As to your equations...  You have them correct, so far.  So, you're good to go, if you wish.  ;)

Are you certain you can recognize what applies to Stokes' Theorem?

Be careful, and good luck.

David

Glad to hear my interpretation was wrong :-)

I think I will put this one on a back burner for now.