In the Annus Mirabilis, 1905, one of Einstein's accomplishments was to establish the theory of special relativity. What was special was that all observers must travel at a constant speed, neither accelerating or decelerating. For such an observer, the speed of light is a constant. Different observers will see different wavelengths and frequencies, but the product of wavelength with the frequency is identical. The wavelength and frequency are said to be Lorentz covariant, meaning we know how they change for different observers. The speed of light is Lorentz invariant. It is one of my pet peeves that invariants should always be paired with their corresponding covariant quantities or else an incomplete story is being told.

Newton's law of gravity does a remarkable job in describing the motion of the planets. It is all that is needed by today's rocket ships unless those devices also carry atomic clocks or other tools of exceptional accuracy. Here is Newton's law in potential form:

From the perspective of special relativity, the equation suffers a fatal flaw: if there is a change in the mass density rho, then that must propagate everywhere instantaneously. Oops.

Einstein set out to fix this flaw. The struggle took him ten years ("Subtle is the Lord..." by Abraham Pais http://www.amazon.com/Subtle-Is-Lord-Einstein-Paperbacks/dp/0192851381 is the was to get the real details on the subject). The math was hard then and remains hard today. At a far away level, it sounds easy - describe all physics the same way whether one is accelerating or not. It is the details of Riemann geometry that are daunting. Einstein got a private tutor and collaborator for the subject, his school buddy Marcel Grossmann. He also traded letters on his math struggles with the leading math minds of his day, including David Hilbert. Einstein came to the field equations not from an action, but from thinking all about the physics. Hilbert figured out the action that generates the Einstein field equations. That is where the derivation begins:

1. Start with the Hilbert action:

Note the square root of the determinant of the metric as part of the volume element. That is required so the volume element can be in curved spacetime. It plays a critical role in the derivation, so I wish I had a better handle on why that factor in that form is required so that the differential volume element transforms like a tensor.

2. Vary with respect to the metric tensor :

3. Pull back the factor of the square root of the metric and use the product rule on the term with the Ricci scalar R:

4. Focus on the first term, using the definition of a Ricci scalar as a contraction of the Ricci tensor:

A total derivative does not make a contribution to the variation of the functional, so can be ignored in our quest to find an extremum. This is Stokes theorem in action.

<SIDEBAR>

Show that the variation in the Ricci tensor is a total derivative.

Since I don't understand this all in detail, I will try to get you in the neighborhood of getting it.

SB1. Start with the Riemann curvature tensor:

Lots of stuff there, but here is a simplifying viewpoint. One is comparing two paths, that is why there is a subtraction here. The two paths are found by switching the order of the mu and the nu. This is a really complicated structure, but that should be obvious :-)

SB2: Vary the Riemann curvature tensor with respect to the metric tensor:

Lots of terms, but remember the mu <-> nu exchange is responsible for half of them.

One cannot take a covariant derivative of a connection since it does not transform like a tensor. Apparently the difference of two connections does transform like a tensor. I say "apparently" because this is an example where I have to rely on authority, I don't appreciate the details.

SB3: Calculate the covariant derivative of the variation of the connection:

Notice that the third terms of these two expressions are identical because the mu and nu are neighbors in the connection.

Again, this is a step whose details I don't understand enough to clarify should others have questions.

SB4: Rewrite the variation of the Riemann curvature tensor as the difference of two covariant derivatives of the variation of the connection written in step SB3.

SB5: Contract the result of SB4

SB6: Contract the result of SB5:

This now looks to my eye like a total derivative, so will not contribute to the action.

<END SIDEBAR>

Since that was such a long sidebar, what has been done is the first of three terms in the variation is the Ricci tensor.

5. Focus on evaluating the variation of the second term in the action. Transform the coordinate system to one where the metric is diagonal and use the product rule:

Notice there was a flip of the metric in the variation which required one more sign change. That is the kind of detail I always trip on.

6. Define the stress energy tensor as the third term:

That factor of a minus a half? I don't get it. Bet it comes out of some classical limit. Hopefully I can research that later in the week.

7. The variation of the Hilbert action will be at an extremum when the integrand is equal to zero:

or

Fini.

But not fini. This was a math exercise. Note how little physics was involved. There are a huge number of physics issues one could go into. As an example, these equations bind to particles with integral spin which is good for bosons, but there are quite a few fermions that also participate in gravity. To include those, one can consider the metric and the connection to be independent of each other. That is the Palatini approach.

Doug

Next Monday/Tuesday: Dot and cross products, differences and overlaps with quaternions

## Comments

Note the square root of the determinant of the metric as part of the volume element. That is required so the volume element can be in curved spacetime. ... I wish I had a better handle on why that factor in that form is required so that the differential volume element transforms like a tensor.I've seen you bring this up before. This is not an issue of curved spacetime. I'm hesitant to just give you the answer, since you seem to have become dependent on people doing that. So let me translate it to something that may look familiar or at least simpler.

In some inertial cartesian coordinates you have the spacetime volume element:

*dx dy dz dt*Now let's do a Lorentz transformation (say a "boost" along the x direction), to get some new coordinate system. How do you know that the volume element is invariant? How do you know that it is still just

*dx' dy' dz' dt'*and without any other constant factors like gamma or something? This is worth working out later, but let's simplify it further to help bridge the gap to physical intuition.

Okay, now for some calculus. In simple Euclidean 3-space, the volume element is:

*dx dy dz*If we instead use spherical coordinates, it is

*r*^{2 }sinθ dr dθ dφYou've probably seen this many times, but how do you actually calculate that? When I was first taught this, we just built up the volume element using simple arguments of arclengths, etc. This is probably how you saw it before as well, but is essentially cheating for this discussion because you are using your knowledge of the "physical" lengths to derive something about the lengths. Given only the original elements, and the coordinate transformation, how do you get the new volume element? Maybe try something even simpler such as the transformation

**x' = 2x**

After playing with it awhile (I encourage you to do so), and possibly rereading some calculus textbook discussion, you will eventually come up with the Jacobian. I may have ruined it just by uttering the word.

Okay, now return to the spacetime. For the action to be invariant it is not sufficient to have a Lagrangian to be a scalar density. We'd need the volume element to be invariant as well. Sticking with cartesian coordinates hides some factors. So let's consider a general coordinate transformation (this has nothing to do with curved spacetime, we can still be doing SR), we see

except for the special case

So before with

**we clearly are not working with an invariant. It just looked like it because it didn't change between two inertial coordinate systems. That fact is not enough to show invariance (sound familiar to any other discussions lately? ::smile:: ). Anyway, we need to add a scaling term that we are leaving out. If you want, you can use spherical coordinates or simple scaled coordinates to help you work this one out.**

*dx dy dz dt*So what is this scaling factor "a"?

Hint, if you look at the transformation law for the metric tensor, the answer becomes pretty clear. This isn't worth working out in great detail. If you understand the steps above, the answer should should be reachable without wading in the swamp of detailed calculations (plugging it into Mathematica doesn't teach you anything either; take the time to learn what the math means).

As a historical side note:

Einstein originally limited his field equations to coordinate systems which set this "scaling factor" to 1. When Schwarzschild worked out his solution to GR, he purposely used a "scaled" coordinate system (not the one denoted to his name now) to preserve this feature. (He also made a mistake with the event horizon that Hilbert (I think?) corrected later in a brief footnote when he used it in some paper.) With the Hilbert action it became clear one could easily lift this restriction.

I'll try to get to your other questions later.

EDIT: Wow, that was a record number of typos. Probably still haven't caught them all. I hope no one was reading that in the mean time.

One cannot take a covariant derivative of a connection since it does not transform like a tensor. ...You already wrote out the Riemann tensor. The Christoffel symbols can be written in terms of the metric. So you can write the whole thing in terms of the metric. It is messier, but you could proceed that way without making any 'leaps' of faith.

In fact, this leap of faith to the covariant derivative needs to be undone later, because when you say:

A total derivative does not make a contribution to the variation of the functional, so can be ignored in our quest to find an extremum. This is Stokes theorem in action.Note that Stokes theorem doesn't work for covariant derivatives. It is just for the usual derivatives. This is part of the issue of defining mass in a local volume in GR, as I think David or someone else described earlier.

I forgot the rest of the steps, but wikipedia comments

unfortunately that is not obvious to me at this time of night. Can anyone explain why this should be clear?

After some sleep I'll work it out to convince myself, but I don't think I'll be able to give a simple answer.

According to this:

http://arxiv.org/pdf/gr-qc/0406088v4.pdf

The vacuum lagrangian for GR can also be written as (eq. 32):

L = -1/(k Λ) Sqrt[ det(R) ]

where Λ is the cosmological constant.

What in the world!?

And with Λ right next to k, how can we separate the gravitational constant from Λ ?? This also seems to disallow the possibility of Λ=0.

David, Henry, Doug, anyone? They call this the "aﬃne formulation in General Relativity". It is really equivalent? How? I really don't understand this paper.

So what happenned to all the unfinished discussion about coordinate transformations of quaternions from the last thread? I hope that isn't just being dropped, and everyone is deciding to leave that in the past. Next week's topic on cross products seems unrelated. I think everyone was getting close to the real issues at hand in for Doug's quest in this blog.

Doug, using David's suggested notation, can you explain what you mean by a coordinate transformation of quaternions?

but I have to think about it in terms of managing the changes between two different basis vectors. I am certain in all the years I held a paying job, a Jacobian was never used.If you are talking about transforming from a 4-vector basis to a quaternion basis, are you sure the question is well defined? The bases satisfy different algebras. Biquaternions represent the Lorentz Group, but can you show me a transformation matrix from biquaternions to 4-vectors?

You can see that rotating the boosted V is the same as boosting by a rotated v, since quaternionic rotation both respects products and quaternionic conjugation. This is a direct proof that the boosts and rotations in the stand-up representation properly commute, so that all 6 generators properly work together.I think this is a nifty way to write down the Lorentz group, and I am surprised that it is new. It is a spinorial trick specific to the four-dimensional Lorentz transformations, but it is a nice one, and it might be useful for something.

Most frigging impressive, the work done by Ron Marion and Qmechanic at http://physics.stackexchange.com/! This feels like the first time when others more skilled than I figured out an issue without much of a bumbling comment by yours truly. They even reached out to the gamma matrices, the correct ones as it were. While I am aware of a connection between the two, I could not discuss the issue in a correct technical way due to my informal training.

**Anyone who has found this discussion of the way to do boosts with quaternions suspect should read that thread.**

On but not fini, indeed I don't think modern science has proved much with fermion spin under gravity, (most of our particle accelerators are horiozontal) are have only seen spin 1/2, half integral spins, are there any other kind? Are there real spin 3/2 gravitinos, and could we ever see them.