Gravitation's Source
6.1 Curvature, Parallel transport, and the Riemann tensor
It is often said that mass curves space-time, but interestingly there is more than one definition of curvature and only one is what is meant by that statement. Curvature in terms of what is usually called Riemanian space-time curvature is one definition. Another definition of curvature is what is usually called Gaussian curvature. I'll demonstrate the differences. First lets consider the definition for curvature given by Gaussian curvature. Here's a classic thought experiment. Imagine a two dimensional creature living on the surface of a sphere. It will start at a point we'll call the north pole and move directly away staying on the sphere's surface for some distance r. Then it will orbit the pole once staying a distance r along the ball away from the pole and measure the circumference C. It expects to find the circumference to be C = 2pr, But to its surprise measures that C is less than 2pr. After careful thought it realized that its two dimensional world must be a positively curved space imbedded in a third dimension. Because of that, the R that it should have used to get C = 2pR was not the distance traveled long the surface, but was actually a distance that cut through the third dimension going through the interior of the ball. Limiting ourselves to the consideration of space curvature the Gaussian curvature K of a two dimensional cross section of space is defined as
(6.1.1)
From here we can make a more general definition of curvature in terms of excess radius. Whenever a space-time is represented in someone's pseudo Cartesian coordinate frame and there remains a non-zero affine connection then the space-time has a frame dependent space and/or time curvature according to the coordinate frame of that observer and geodesic acceleration will be experienced. In other words dxl/dt2 will not be zero for an object in free fall. We make the relation between geodesic acceleration and excess radius because of the form of the covariant derivative. We realize that the extra terms involving the affine connections in the covariant derivative operator can be
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68 Chapter 6 Gravitation's Source
interpreted as excess radius due to the curvilinear nature of our space-time coordinates. We also note from the equation of geodesic motion that any frame with nonzero affine connections observes geodesic acceleration of particles in free fall. So we can make the generalization for a frame dependent space and/or time curvature to say that a frame of space-time has this kind of curvature whenever it contains regions where geodesic acceleration is experienced. Now it is important to note that The affine connections can always be transformed away locally by transforming to any local free fall frame. Therefor the very presence of this kind of curvature is frame dependent and this is what is meant should a physicist say that space-time is locally flat. The affine connections can be transformed away and the metric tensor reduced to that of special relativity.
Here is an example of a space-time with this kind of frame dependent space and time curvature that has zero Riemannian space-time curvature(i.e. Rlmrn = 0)
ds2 = (1 + 2az/c2)dct2 - dx2 - dy2 - dz2/(1 + 2az/c2)
(a = constant for a vacuum field solution with a zero Riemann tensor)
Next lets consider Riemannian space-time curvature. When a physicist refers to spacetime curvature unqualified he should be referring to this kind of curvature. Lets say the two dimensional creature on the sphere carries an arrow pointing parallel to the surface in the direction of its motion as it moves from the north pole to the equator. Once it reaches the equator it moves along the equator one quarter the way around without twisting the arrow so that it stayed pointing away from the pole. The creature then returns to the pole without twisting the arrow. Upon the creatures return the creature realizes the arrow is now pointing 90 degrees away from where it initially pointed. From this the creature again infers that it is living in a universe curved within a higher dimensional space. From here we can generalize to define Riemannian space-time curvature in terms of the parallel transport of four vectors. Lets say we parallel transport a four vector along an infinitesimal path dr1 then along dr2 then along -dr1 and then -dr2 to close a little path then, as we show in the section on parallel transport and the Riemann tensor, the total infinitesimal change in a four vector Tl is
dTl = TmRlmrndr1ndr2r
(6.1.2)
Rlmrn is called the Riemann tensor. Going by this definition of curvature, Riemannian space-time curvature is the same thing as having any non-zero elements of Rlmrn. An example of a space-time with Riemannian space-time curvature would be the Schwarzschild solution
ds2 = (1 - 2GM/rc2)dct2 - dr2/(1 - 2GM/rc2) - r2(dq2 + sin2qdf2)
This space-time also has Gaussian curvature.
6.1 Curvature, Parallel transport, and the Riemann tensor 69
We have introduced the covariant derivative operator Eqn 4.4.1 with Eqn 4.4.2
DTl = dTl + GlmnTmdxn
The first term dTl is the variation of the vector. The second term GlmnTmdxn is a variation in the form the vector takes that is due to the curvilinear nature of the spacetime coordinates. If the vector is parallel transported and the vector is varied so that DTl = 0. Then we have
dTl = - GlmnTmdxn
Imagine parallel transporting a vector along an infinitesimal closed path that you might imagine as a rectangle within the space-time moving along a segment we'll call dr1, then along dr2, then along -dr1 and finally along -dr2 starting from the point x0. From the second term we find that the total variation in the vector is given by
dTl = - Glmr(x0)Tr(x0)dr1m - Glmr(x0 + dr1)Tr(x0 + dr1)dr2m + Glmr(x0 + dr2)Tr(x0 + dr2)dr1m + Glmr(x0)Tr(x0)dr2m
This can be rewritten
dTl = [Glmr(x0 + dr2)Tr (x0 + dr2) - Glmr(x0)Tr(x0)]dr1m - [Glmr(x0 + dr1)Tr(x0 + dr1) - Glmr(x0)Tr(x0)]dr2m
Realizing that these terms constitute difference quotients we have
dTl = (GlmrTr),sdr2sdr1m - (GlmrTr),ndr1ndr2m
Using the product rule and again using dTl = - GlmnTmdxn and renaming indices we come to
dTl = Tm(Glmn,r - Glmr,n + GlsrGsmn - GlsnGsmr)dr1ndr2r
At this point we define the Riemann tensor by
Rlmrn = Glmn,r - Glmr,n + GlsrGsmn - GlsnGsmr
(6.1.3)
So that we have Eqn 6.1.2
dTl = TmRlmrndr1ndr2r
Whenever matter is present, the Riemann tensor is not zero. Though the affine connections are analogous to the Newtonian concept of the gravitational field, they can always at least locally be transformed away. The Riemann tensor is a tensor and the affine connections or Cristoffel symbols are not. One property of tensors is that if it is not zero according to one frame, it is not zero according to any frame. Spacetime curvature, or a nonzero Riemann tensor, is always present where there is any kind of matter, whether or not the affine connection has been transformed away. Therefor what matter intrinsically produces is a field of curvature, not a field of acceleration. In the sense of spacetime curvature, gravitation can never be transformed away, even though the Newtonian acceleration field can always be transformed away. This is why from a general relativistic perspective, instead of thinking of mass producing an acceleration g field it is better to think of matter curving spacetime and objects seem to accelerate because they follow geodesics in that spacetime. So, in GR the field should not be thought of as an acceleration field, the affine connections, as in a Newtonian analogy. In GR the field should be thought of as a curvature field, the Riemann tensor. This is what makes sense of the word field, in "Einstein's field equations". Whenever there is matter present, Enstein's equations imply that the Riemann tensor is not zero at its location. The Riemann tensor being the field of spacetime curvature that is referred to by the name.
Note the antisymmetry in the last two indices
Rlmrn = -Rlmnr
(6.1.4)
70 Chapter 6 Gravitation's Source
We will now move on to work out the Bianchi Identities
Rlmrn;s + Rlmns;r + Rlmsr;n = 0
(6.1.5)
We start with the expression for the Riemann tensor Eqn 6.1.3
Rlmrn = Glmn,r - Glmr,n + GlsrGsmn - GlsnGsmr
Then transform to a local free fall frame so that the affine connections vanish
Rlmrn = Glmn,r - Glmr,n
Now we take a derivative
Rlmrn,s = Glmn,rs - Glmr,ns
Now we permutate indices and add.
Rlmrn,s + Rlmns,r + Rlmsr,n = Glmn,rs - Glmr,ns + Glms,nr - Glmn,sr + Glmr,sn - Glms,rn
Since we can switch order of partial differentiation we find
Rlmrn,s + Rlmns,r + Rlmsr,n = Glmn,rs - Glmr,ns + Glms,nr - Glmn,rs + Glmr,ns - Glms,nr.
So
Rlmrn,s + Rlmns,r + Rlmsr,n = 0
Now we replace ordinary differentiation with covariant differentiation and we have the Bianchi identities Eqn. 6.1.5.
Rlmrn;s + Rlmns;r + Rlmsr;n = 0
Exercises
Problem 6.1.1
Calculate K for the 2d creature in the above problem.
Problem 6.1.2
Calculate, or have a computer calculate the Riemann curvature tensors for both spacetimes mentioned in this section. (GRTensorII for Maple6 is a free downloadable tensor calculus package for Maple or Mathematica that does this well.)
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6.2 Mach, Gravitational Forces, and Tidal Gradients
71There are different interpretations of Mach's principle. See-
http://www.bun.kyoto-u.ac.jp/~suchii/mach.pr.html
Here we are only concerned with the one that is physically correct.
Even before Einstein philosophers such as Newton Vs Leibniz debated whether space/time were absolute Vs relative. Newton thought that there was an absolute space, independent of the matter within it, which alone determines which motions are inertial Vs accelerated. Leibniz thought that space was merely descriptive of the distribution of bits matter relative to each other and thus depended on the matter itself. Mach was a mentor of Einstein's who thought that it must be the distribution and behavior of other matter which is ultimately to be responsible for what frames we find should here be treat as inertial Vs accelerated. This is Mach's principle. If this is the case, then Newton's absolute space is incorrect, whereas Leibniz relative space would be a consistent description of the universe. As we will show, Mach's principle is not only consistent with the universe containing matter and obeying general relativity, but is also implied by those two conditions.
An example of Mach's principle goes as follows. We might look at the earth from an external perspective which doesn't rotate among the average motions of the distant stars in which case we describe the physics of the earth with simple Newtonian mechanics and Newtonian gravitation. But if we were to rotate with the earth and we rotate our coordinates with the earth, then we say that according to our coordinates, we are not actually rotating at all, but it is the stars instead that revolve around the earth. We also experience inertial forces. These forces are even today sometimes called fictitious forces, but according to Mach's principle they must be gravitational forces. In this sense the centrifugal force, for example, is not any more or less fictitious than gravity. It is just a case of a frame dependent gravitational force, but since it is the same thing as a frame dependent gravitational force, and it is indeed frame dependent, it would be better to think of gravitational force itself as an inertial or fictitious force than to think of an inertial force as real. In the end, according to an earth-based frame, the stars rotating around the earth induce a gravitational acceleration field, which is what causes the bulging of the oceans at the equator.
Initially Mach's principle was just a philosophical guess on the part of Mach. However, it is a correct description within the context of general relativity. The actual universe we have is a universe with matter. When you have matter in a universe and you're doing general relativity you can't arbitrarily say one guy is in an inertial frame and another guy is not. It is in special relativity where you must arbitrarily decide, before the physics is to be done, whose frames are to be taken as inertial Vs accelerated. In general relativity the space-time metric itself depends on matter within it as described through Einstein's field equations. You have different solutions for space-times with different matter distributions with Einstein's field equations. For example, when you have a universe with a uniform matter distribution then it has the Robertson-Walker space-time as the solution. In this space-time we can easily distinguish which frames are comoving frames by the form that the metric tensor takes. Locally, physics within a comoving frame reduces to that of an inertial frame as the affine connections vanish. Also local constant velocity transformations from comoving frames result in the physics of inertial frames. However, when you transform to a local frame that is accelerated with respect to a comoving frame we find that the affine connections do not vanish locally. This is what distinguishes an inertial frame from an accelerated one, the local vanishing of the affine connections according to pseudo Cartesian coordinate systems.
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This is a demonstration of how the matter in the universe determined which frames were inertial and which were not and that is the full realization of Mach's Principle. The inertial forces arising in accelerated frames, or gravitational forces, I'll refer to as pseudo forces because they are not four-vectors and can always be locally transformed away going to a local inertial frame.
In special relativity the ordinary force on a particle is the coordinate time derivative of the four momentum. In the absence of four-forces, what ordinary forces remain as observed by one attempting special relativity will be known as Mach forces or gravitational pseudo forces. We earlier arrived at the Eqn 5.3.7
al
= (dt/dt)Fl/m - GlmnumUnTherefor the Mach's forces of gravitation, or the gravitational pseudo forces, are given by the case that Fm = 0 and are
f l = mal = - mGlmnumUn
or more simply
f l grav º - Glmnumpn
(6.2.1)
Though Mach's principle really was intended to deal with the inertial forces associated with a accelerating or rotating frame, we here have classified all gravitational forces or forces of affine connection as Mach forces or pseudo-forces because they are all the same thing as inertial forces. Whether we have a vacuum field solution or not, whether we have Riemannian space-time curvature or not, all the inertial forces or all the gravitational forces are directly due to nonzero affine connections.
We have seen that the equation of nongeodesic motion is Eqn 5.2.1
Fl = dpl/dt + GlmnpmUn
Fl is the net four-force acting on the particle. For gravitation alone the net four force is zero. In other words the true force of gravitation is
Fl = 0.
The tensor tidal gradient of this is then
Fl ;r = 0.
So the true forces and tidal gradients of gravitation are zero, and what we are left with is actually geodesic motion and geodesic deviation or the other gravitational pseudo forces as defined below. In special relativity the Riemann tensor is zero. In gravitation its not always zero, and
6.2 Mach, Gravitational Forces, and Tidal Gradient 73
when it is not, the metric can not be globally transformed to hmn and one is always left with a gravitational pseudo force somewhere in the universe. Therefor is has become fashionable in modern terminology to refer to gravitational pseudo forces as "real" when they are associated with nonzero elements of the Riemann tensor. For instance geodesic deviation as discussed in a later section is a gravitational pseudo-force tidal gradient, which is related to nonzero elements of the Riemann tensor. Often modern relativists will refer to this as a "real" tidal force gradient, even though as shown above the real tensor force gradient for gravitation is always zero.
However as we have seen from Eqn. 6.2.1, there are gravitational pseudo-forces given by
f l = -Glmnumpn
and the tidal gradient of these are then given by
f l,r = -(Glmnumpn),r
. We have seen that the expression for the pseudoforces of gravitation is
f l = -Glmnumpn
The exact expression for the tidal gradient of the Mach force is then
f l , r = -(Glmnumpn),r
(6.2.2a)
So far no approximations have been made and this expression is correct and does not involve the Riemann tensor. However, when certain approximations can be made and a particular frame is chosen, the tidal gradient can be related to certain elements of the Riemann tensor.
First we will make the approximation that speeds are non-relativistic. In this approximation the above expression reduces to
f l , r » -(Gl00ER),r
(6.2.2b)
We will also make an approximation that gravitational potentials are weak. Combining this with the low velocity approximation we have
ER » E0 = mc2
So we now have
f l ,r = -mc2Gl00,r
(6.2.2c)
We will next take a look at the equation for the Riemann tensor Eqn. 6.1.3.
Rlmrn = Glmn,r - Glmr,n + GlsrGsmn - GlsnGsmr
74 Chapter 6 Gravitation's Source
Now we will make the particular choice of frame referred. Transform to a local free fall frame so that the affine connections(but not their derivatives) vanish. In this choice of coordinate system the Riemann tensor becomes
Rlmrn = Glmn,r - Glmr,n
Now consider the following elements of the Riemann tensor
Rl0r0 = Gl00,r - Gl0r,0
As long as we can also make the approximation that the affine connections are also time independent which they are only right at the local free fall frame observer's location for most space-times, then this becomes
Rl0r0 = Gl00,r
Now we compare this to the expression above for the tidal gradient and we can finally make the relation
f l ,r = -mc2Rl0r0
(6.2.3)
With all the approximations made and in forcing a particular frame in which to use and in restricting the relation to immediately about free fall frames which all must be done to make the relation, we see that this is a very noncovariant way of looking at things, especially when the two expressions at the top of the page are so much more exact and simpler. Still, making this relationship between the Riemann Tensor and this particular frame's pseudo tidal gradients has become a very popular thing to do.
Exercises
.Problem 6.2.1
Calculate the tidal gradient around a test mass held still around the origin Eqn. 6.2.2 for the spacetime
ds2 = (1 + az/c2)2dct2 - dx2 - dy2 - dz2
The Riemann tensor is zero for this spacetime. Considering Eqn. 6.2.3, How can this be so? What does this say about a general relation between the Riemann tensor and tidal gradients?
Problem 6.2.2
If you know the remote frame expression for the Riemann tensor of a black hole and a probe is in free fall, what would you have to do to the tensor to get the tidal gradient experienced using Eqn. 6.2.3?
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6.3 Stress Energy of Matter and Einstein's Field Equations
75We have expressed the Riemann tensor in the form of Eqn. 6.1.3.
Rlmrn = Glmn,r - Glmr,n + GlsrGsmn - GlsnGsmr
We will define the Ricci tensor as a contraction of the Riemann tensor over the first and third indices as follows
Rmn = Rlmln
(6.3.1)
Since the Riemann tensor is antisymetric in the last two indices, the Ricci tensor could just as well be defined as a contraction over the first and fourth indices. This will only change the sign in front of the Ricci tensor and scalar in expressions in which they appear.
The Ricci scalar is the full contraction of a tensor and so it is an invariant. We define it as the contraction of the Ricci tensor as follows.
R = gmnRmn
(6.3.2)
Recall the Bianchi identity Eqn. 6.1.5
Rlmrn;s + Rlmns;r + Rlmsr;n = 0
Contract this expression over m and n and then again contract it over l and r and recall the antisymetry of the Riemann tensor in the last two indices to arrive at
R; s - Rrs;r - Rns;n = 0
Then rename indices
R;s - Rms;m - Rms;m = 0
Simplify
Rms;m - (1/2)R;s = 0
Insert a Delta Kronecker
Rms;m - (1/2)dmsR;m = 0
Multiply through by gsn and we have
(Rmn - (1/2)gmnR);m = 0
(6.3.3)
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We could also express this as
(Rmn - (1/2)gmnR);n = 0
(Summation still implied) (6.3.4)
The expression in the parentheses is called the Einstein tensor
Gmn º Rmn - (1/2)gmnR
(6.3.5)
The stress energy tensor is a tensor that contains information on the density of energy, momentum, stresses, etc.. contained in the space. The energy tensor mass alone is Eqn. 5.1.4
Tmn = r0UmUn
The T00 component of this is
T00 = (dt/dt)2r0c2
Consider for a moment the case where there is negligable gravitational time dilation and length contraction. r0 is the mass density according to a frame moving with that bit of mass, but because of special relativistic Lorenz length contraction on the local mass the coordinate frame mass density is then
r
= (dt/dt)r0So this becomes
T00 = (dt/dt)rc2
But this is just the coordinate frame energy density.
The simpelest consistent general relativistic definition of coordinate frame energy density is then just
coordinate frame energy density º T00
(6.3.6)
Other elements have other interperetations. For instance Tii is a flow of momentum per area in the xi direction or the pressure on a plane whose normal is in the xi direction. Tij is the xi component of momentum per area in the xj direction or describes a shearing from stresses. T0i is the volume density of the ith component of momentum flow.
6.3 Stress Energy of Matter and Einstein's Field Equations 77
Any matter contained in the space provides contributions to the elements of the tensor. The stress-energy tensor of pressure is
Tmn = (p/c2)UmUn - gmnp
(6.3.7)
where p is the local pressure.
Combining the stress-energy tensor of pressure with the energy tensor of mass results in the stress-energy tensor for an ideal fluid
Tmn = (r0 + p/c2)UmUn - gmnp
(6.3.8)
The contraction of this results in
r
Totc2 = r0c2 - 3p(6.3.9)
which verifies that the density r0, the total mass density rTot, and the pressure p are indeed all invariants as they are all contractions of tensors.
Electromagnetic fields don't have mass as previously defined but they do carry energy, momentum and stresses that contribute to the stress-energy tensor. The symmetric stress energy tensor of an electromagnetic field is given by
Tmn = e[FmlFln + (1/4)gmnFlsFls]
(6.3.10a)
which in terms of E&B in Cartesian inertial frame coordinates for flat spacetime is
T 00 = (1/2)e(E2 + c2B2)
(6.3.10b)
T 0i = e(E´cB)i
(6.3.10c)
T ij = - e[EiEj + c2BiBj - (1/2)dij(E2 + c2B2)]
(6.3.10d)
Where Fmn is the electromagnetic field tensor. In local Cartesian coordinates it is given by
(6.3.11)
In special relativity conservation of energy/momentum can be expressed by the statement
Tmn,n = 0
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In general relativity this must be expressed as a tensor equation. As a result there isn't a true conservation of momentum/energy law. Instead the general relativistic conservation of energy/momentum law takes the form
Tmn;n = 0
(6.3.12)
Or we could write it(with summation still implied)
Tmn;n = 0
(6.3.13)
Continuing on we can present a construction of the Einstein field equations.
Rmn - (1/2)gmnR - lgmn = (8pG/c4)Tmn
(6.3.14)
which can also be expressed
Rmn + gmnl = (8pG/c4)[Tmn - (1/2)gmnT]
(6.3.15)
As we have seen from Eqn. 6.3.4 the covariant divergence of the Einstein tensor is zero.
[Rmn - (1/2)gmnR];n = 0
(Summation still implied)
We have also noted that the conservation of momentum/energy law for general relativity takes the form Eqn. 6.3.13
Tmn;n = 0
(Summation still implied)
And finally we have seen that the covariant divergence of the metric tensor is zero. Eqn. 4.3.2
gmn;n = 0
(Summation still implied)
Prior to relativity fields were modeled according to the Poisson equation
Ñ
2f = kr(6.3.16)
6.3 Stress Energy of Matter and Einstein's Field Equations 79
f
is the potential from which the field was obtained and r was the field's source density. The expressionÑ
2finvolves taking second order partial derivatives of f . Einstein knew that the gravitational field should be explained through the tendency for particles to follow geodesics in a curved space-time because only this behavior would be consistent with the weak equivalence principle. He also knew that the equations for the general laws of physics should be tensor equations in order to satisfy the strong equivalence principle. The curvilinear nature of the space-time comes from the metric tensor and the sources for the gravitational field were known to be contained in the stress energy tensor. Keeping all these things in mind Einstein was able to model gravitation in equations analogous to the Poisson equation. The Einstein tensor is a set of second order differential equations for the metric tensor like
Ñ
2fis a second order differential expression for the potential. The stress energy tensor describes the density of the gravitational sources just as r is the field's source density. So the obvious model to start with is
Rmn - (1/2)gmnR = kTmn
(6.3.17)
Fortunately this equation is consistent with the relations above. If the covariant divergence of one side is zero so is the covariant divergence of the other. Experimentally this form of Einstein's field equations check out as a good model and we find that
k = 8pG/c4.
(6.3.18)
Where G is the gravitational constant of the universe.
Since the covariant divergence of the metric tensor is also zero we could write a more general and still consistent model of gravitation as Eqn 6.3.14.
Rmn - (1/2)gmnR - lgmn = (8pG/c4)Tmn
These are called Einstein's field equations and the constant l (to be experimentally determined) is called the cosmological constant. Einstein originally included it in hopes of finding a steady state universe solution. When it was found that it would lead to a gravitationally repulsive effect (if positive) and was as near to zero as could then be determined anyway, he decided that cluttering his equations with its presence was his greatest blunder. Though small, there is some experimental evidence that it is not zero after all and so it was a good thing that he included it.
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For l = 0 we have
Rmn - (1/2)gmnR = (8pG/c4)Tmn
(6.3.19)
Contracting the Einstein field equations we have
gmnRmn - (1/2)gmngmn R - lgmngmn = (8pG/c4) gmnTmn
R - (1/2)4R - 4l = (8pG/c4)T
R + 4l = -(8pG/c4)T
(6.3.20)
Replacing this back into the field equations Eqn. 6.3.14 we have
Rmn - (1/2)gmn(-(8pG/c4)T - 4l) - lgmn = (8pG/c4)Tmn
Simplified its Eqn 6.3.15
Rmn + gmnl = (8pG/c4)[Tmn - (1/2)gmnT]
Or if l is zero then its
Rmn = (8pG/c4)[Tmn - (1/2)gmnT]
(6.3.21)
Note that for l = 0 Eqn. 6.3.20 can be expressed
R = -(8pG/c2)rTot
(6.3.22)
Since one can not have a zero Riemann tensor with a nonzero Ricciscalar, it is often said mass curves spacetime (in the sense of Riemannian curvature). Consider also a massless field where rTot = 0. Eqn. 6.3.21 results in
Rmn = (8pG/c4)Tmn
Since one can not have a zero Riemann tensor with a nonzero Ricci tensor we see that any other matter whether massive or not also curves spacetime. Thus the requirement of "gravitational mass" to produce gravitation is a purely Newtonian concept and does not extend to all cases of Einsteinian Gravitation.
Here below we derive the fictitious forces from geodesic motion in a rotating frame. In particular arriving at centrifugal, Coriolis, and transverse forces.
The exact solution for ds2 for an observer who spins about his z axis is
ds2 = [1 - (w/c)2(x2 + y2)]dct2 + 2y(w/c)dctdx - 2x(w/c)dctdy - dx2 - dy2 - dz2
(6.3.23)
And the contravariant metric tensor is
(6.3.24)
The equation of motion for a test particle is
Fl = mAl
and we will consider no real forces acting on the particle,
Fl = 0.
Resulting in
Al = 0
d2xl/dt2 + Glmn(dxm/dt)(dxn/dt) = 0
Glmn
= (1/2)glr(gmr,n + gnr,m - gmn,r)
d2xl/dt2 + (1/2)glr(gmr,n + gnr,m - gmn,r)(dxm/dt)(dxn/dt) = 0
(6.3.25)
First apply this geodesic equation 6.3.25 to the time component.
d2ct/dt2 + (1/2)gctr(gmr,n + gnr,m - gmn,r)(dxm/dt)(dxn/dt) = 0
d2ct/dt2 + (1/2)(gmct,n + gnct,m - gmn,ct)(dxm/dt)(dxn/dt) + (1/2)y(w/c)(gmx,n + gnx,m - gmn,x)(dxm/dt)(dxn/dt) - (1/2)x(w/c)(gmy,n + gny,m - gmn,y)(dxm/dt)(dxn/dt) = 0
d2ct/dt2 + (1/2)(gctct,n + gnct,ct - gctn,ct)(dct/dt)(dxn/dt) + (1/2)(gxct,n + gnct,x - gxn,ct)(dx/dt)(dxn/dt) + (1/2)(gyct,n + gnct,y - gyn,ct)(dy/dt)(dxn/dt) + (1/2)y(w/c)(gctx,n + gnx,ct - gctn,x)(dct/dt)(dxn/dt) + (1/2)y(w/c)(gnx,y - gyn,x)(dy/dt)(dxn/dt) - (1/2)x(w/c)(gcty,n + gny,ct - gctn,y)(dct/dt)(dxn/dt) - (1/2)x(w/c)(gny,x - gxn,y)(dx/dt)(dxn/dt) = 0
d2ct/dt2 + (1/2)(gctct,ct)(dct/dt)2 + (1/2)(gctct,x)(dct/dt)(dx/dt) + (1/2)(gctct,y)(dct/dt)(dy/dt) + (1/2)(gctct,x)(dx/dt)(dct/dt) + (1/2)(gxct,y + gyct,x)(dx/dt)(dy/dt) + (1/2)(gctct,y)(dy/dt)(dct/dt) + (1/2)(gyct,x + gxct,y)(dy/dt)(dx/dt) + (1/2)y(w/c)(2gctx,ct - gctct,x)(dct/dt)2 + (1/2)y(w/c)(gctx,y - gcty,x)(dct/dt)(dy/dt) + (1/2)y(w/c)(gctx,y - gyct,x)(dy/dt)(dct/dt) - (1/2)x(w/c)(2gcty,ct - gctct,y)(dct/dt)2 - (1/2)x(w/c)(gcty,x - gctx,y)(dct/dt)(dx/dt) - (1/2)x(w/c)(gcty,x - gxct,y)(dx/dt)(dct/dt) = 0
d2ct/dt2 + (1/2)(gctct,ct)(dct/dt)2 + (gctct,x)(dct/dt)(dx/dt) + (gctct,y)(dct/dt)(dy/dt) + (gxct,y + gyct,x)(dx/dt)(dy/dt) + (1/2)y(w/c)(2gctx,ct - gctct,x)(dct/dt)2 + y(w/c)(gctx,y - gcty,x)(dct/dt)(dy/dt) - (1/2)x(w/c)(2gcty,ct - gctct,y)(dct/dt)2 - x(w/c)(gcty,x - gctx,y)(dct/dt)(dx/dt) = 0
d2ct/dt2 - (w/c2)(dw/dct)x2(dct/dt)2 - (w/c2)(dw/dct)y2(dct/dt)2 - 2x(w/c)2(dct/dt)(dx/dt) - 2y(w/c)2(dct/dt)(dy/dt) + y2(dw/dct)(w/c2)(dct/dt)2 + xy(w/c)3(dct/dt)2 + 2y(w/c)2(dct/dt)(dy/dt) + x2(dw/dct)(w/c2)(dct/dt)2 - xy(w/c)3(dct/dt)2 + 2x(w/c)2(dct/dt)(dx/dt) = 0
d2ct/dt2 = 0
(6.3.26)
Therefor
dt/dt = constant
Call that constant g'. It is the value of gamma for the test mass according to an inertial frame in which the spin frame observer has no translational motion.
dt/dt = g'
(6.3.27)
Second apply the geodesic equation 6.3.25 to the z component.
d2z/dt2 + (1/2)gzr(gmr,n + gnr,m - gmn,r)(dxm/dt)(dxn/dt) = 0
d2z/dt2 + (1/2)gzz(gmz,n + gnz,m)(dxm/dt)(dxn/dt) = 0
d2z/dt2 + (1/2)gzz(gnz,m)(dz/dt)(dxn/dt) = 0
d2z/dt2 = 0
g
' 2 d2z/dt2 = 0d2z/dt2 = 0
(6.3.28)
This is in exact agreement with Newtonian physics.
Next we apply the geodesic equation 6.3.25 to the x component.
d2x/dt2 + (1/2)gxr(gmr,n + gnr,m - gmn,r)(dxm/dt)(dxn/dt) = 0
d2x/dt2 + (1/2)gxct(gmct,n + gnct,m - gmn,ct)(dxm/dt)(dxn/dt) + (1/2)gxx(gmx,n + gnx,m - gmn,x)(dxm/dt)(dxn/dt) + (1/2)gxy(gmy,n + gny,m - gmn,y)(dxm/dt)(dxn/dt) = 0
d2x/dt2 + (1/2)gxct(gctct,n)(dct/dt)(dxn/dt) + (1/2)gxct(gxct,n + gnct,x - gxn,ct)(dx/dt)(dxn/dt) + (1/2)gxct(gyct,n + gnct,y - gyn,ct)(dy/dt)(dxn/dt) + (1/2)gxx(gctx,n + gnx,ct - gctn,x)(dct/dt)(dxn/dt) + (1/2)gxx(gnx,y - gyn,x)(dy/dt)(dxn/dt) + (1/2)gxy(gcty,n + gny,ct - gctn,y)(dct/dt)(dxn/dt) + (1/2)gxy(gny,x - gxn,y)(dx/dt)(dxn/dt) = 0
d2x/dt2 + (1/2)gxct(gctct,ct)(dct/dt)2 + (1/2)gxct(gctct,x)(dct/dt)(dx/dt) + (1/2)gxct(gctct,y)(dct/dt)(dy/dt) + (1/2)gxct(gctct,x)(dx/dt)(dct/dt) + (1/2)gxct(gxct,y + gyct,x)(dx/dt)(dy/dt) + (1/2)gxct(gctct,y)(dy/dt)(dct/dt) + (1/2)gxct(gyct,x + gxct,y)(dy/dt)(dx/dt) + (1/2)gxx(2gctx,ct - gctct,x)(dct/dt)2 + (1/2)gxx(gctx,y - gcty,x)(dct/dt)(dy/dt) + (1/2)gxx(gctx,y - gyct,x)(dy/dt)(dct/dt) + (1/2)gxy(2gcty,ct - gctct,y)(dct/dt)2 + (1/2)gxy(gcty,x - gctx,y)(dct/dt)(dx/dt) + (1/2)gxy(gcty,x - gxct,y)(dx/dt)(dct/dt) = 0
d2x/dt2 - gxct(w/c)((dw/dct)/c)(x2 + y2)(dct/dt)2 - 2gxctx(w/c)2(dct/dt)(dx/dt) - 2gxcty(w/c)2(dct/dt)(dy/dt) + gxx(y((dw/dct)/c) + x(w/c)2)(dct/dt)2 + 2gxx(w/c)(dct/dt)(dy/dt) + gxy(-x((dw/dct)/c) + y(w/c)2)(dct/dt)2 - 2gxy(w/c)(dct/dt)(dx/dt) = 0
d2x/dt2 - y(w/c)2((dw/dct)/c)(x2 + y2)(dct/dt)2 - 2xy(w/c)3(dct/dt)(dx/dt) - 2y2(w/c)3(dct/dt)(dy/dt) + (-1+y2(w/c)2)(y((dw/dct)/c) +x(w/c)2)(dct/dt)2 + 2(-1+y2(w/c)2)(w/c)(dct/dt)(dy/dt) - xy(w/c)2(-x((dw/dct)/c) + y(w/c)2)(dct/dt)2 + 2xy(w/c)2(w/c)(dct/dt)(dx/dt) = 0
d2x/dt2 - x2y(w/c)2((dw/dct)/c)(dct/dt)2 - y3(w/c)2((dw/dct)/c)(dct/dt)2 - 2xy(w/c)3(dct/dt)(dx/dt) - 2y2(w/c)3(dct/dt)(dy/dt) + y3(w/c)2((dw/dct)/c)(dct/dt)2 + xy2(w/c)4(dct/dt)2 - y((dw/dct)/c)(dct/dt)2 - x(w/c)2(dct/dt)2 - 2(w/c)(dct/dt)(dy/dt) + 2y2(w/c)3(dct/dt)(dy/dt) + x2y((dw/dct)/c)(w/c)2(dct/dt)2 - xy2(w/c)4(dct/dt)2 + 2xy(w/c)3(dct/dt)(dx/dt) = 0
d2x/dt2 - y((dw/dct)/c)(dct/dt)2 - x(w/c)2(dct/dt)2 - 2(w/c)(dct/dt)(dy/dt) = 0
g
' 2 d2x/dt2 - g' 2 y(dw/dt) - g' 2 xw2 - 2g' 2 w(dy/dt) = 0d2x/dt2 - y(dw/dt) - xw2 - 2w(dy/dt) = 0
d2x/dt2 = y(dw/dt) + xw2 + 2w(dy/dt)
(6.3.29)
This is in exact agreement with Newtonian physics.
To get the y component one may go through the same process or just arrive at the answer easily by symmetry. It is:
d2y/dt2 = - x(dw/dt) + yw2 - 2w(dx/dt)
(6.3.30)
This is in exact agreement with Newtonian physics.
The coordinate acceleration for this frame on the test mass is then
a = [y(dw/dt) + xw2 + 2w(dy/dt)]ex + [- x(dw/dt) + yw2 - 2w(dx/dt)]ey
a = y(dw/dt)ex + xw2ex + 2w(dy/dt)ex - x(dw/dt)ey + yw2ey - 2w(dx/dt)ey
a = - (dw/dt)(xey - yex) - w2(- xex - yey) - 2w[(dx/dt)ey - (dy/dt)ex]
a = - (dw/dt)ez´(xex + yey + zez) - wez´w(xey - yex) - 2wez´[(dx/dt)ex + (dy/dt)ey + (dz/dt)ez]
a = - (dw/dt)´r - w´w(xey - yex) - 2w´u
a = - (dw/dt)´r - w´[wez´(xex + yey + zez)] - 2w´u
a = - (dw/dt)´r - w´(w´r) - 2w´u
(6.3.31)
Multiplying by the mass:
ma = - m(dw/dt)´r - mw´(w´r) - 2mw´u
(6.3.32)
So the spinning observer will reckon that there are forces that are really only forces of affine connection which as with all gravitational forces are fictitious because they can be transformed away and because they correspond to a zero force four-vector. In this case these inertial forces are given names:
Coriolis force = - 2mw´u
(6.3.33a)
Centrifugal force = - mw´(w´r)
(6.3.33b)
Transverse force = - m(dw/dt)´r
(6.3.33c)
The latter should not be confused with the real force acting transverse to a test mass's motion which is also sometimes called transverse force.
In a real universe of matter the spacetime's differential geometry is determined by the matter distribution and behavior. Which frames will then yield ds2 consistent with the spin frame observer over a sufficiently local region and thus yields these inertial forces is therefor determined by the matter distribution and behavior. This proves that general relativity implies the version of Mach's principle in which inertial forces are determined by the matter in the universe and in which inertial("comoving" technically) verses accelerated frames are distinguished by the matter's distribution and behavior.
6.3 Stress Energy of Matter and Einstein's Field Equations 81
Exercises
Problem 6.3.1
Calculate the stress-energy tensor with gmn = hmn for Eqn. 6.3.11.
Problem 6.3.2
Show that rTot = 0 for the result of problem 6.3.1.
Problem 6.3.3
Calculate, or have a computer calculate the Einstein tensor and the Ricci scalar for the spacetime.
ds2 = (1 - 2GM/rc2)dct2 - dr2/(1 - 2GM/rc2) - r2(dq2 + sin2qdf2)
(GRTensorII for Maple6 is a free downloadable tensor calculus package for Maple or Mathematica that does this well.)
Is the result what you would expect for a point mass source at the origin?
Problem 6.3.4
We might model a gas of particles, possibly of little or no mass which is then sometimes called a photon gas, by
Tmn = Smn + (p/c2)Ymn - gmnp
where in some particular local (gmn = hmn) frame we have
and
.
For a massless gass, what would the relation between the pressure p and this coordinate frame's value for the energy density
T ' 00 be?