Starting GR

Return to Modern Relativity

Lets say that there is a space-lab out in the depths of space sealed up so that there is no way for its crew to see anything outside of the lab. There are two experimentalists, Terrance and Stella, inside the space-lab. In this environment they are weightless and Terrance is still with respect to the ship walls. Stella is also initially still with respect to the lab walls, but she can maneuver around without touching the walls because she wears a rocket pack. They both also carry with them cesium watches that keep time accurate to within a millionth of a second and a computer that can read off such small time differences in their displays. They then do the following experiment. They synchronize their watches to start and they start at the same location within the space-lab. Terrance stays there and Stella travels away and back to him along any number of paths so long as she arrives back when his watch says an hour has gone by to within its millionth of a second accuracy. The watch's times are then compared and a path is sought for which as much time as possible goes by on Stella's watch. Finally they experimentally discover what we knew from special relativity which is that the path that maximized her watches time was simply where she stayed put weightless next to Terrance and didn't go anywhere else. Every other path she took she underwent special relativistic time dilation while in motion with respect to Terrance.

Next we shift perspectives to a third party, Lois, who is for the moment moving in a state of constant velocity through the ship. According to Lois the path that Stella followed that maximized Stella's time between the events of the experiment's start and stop next to Terrance was a path of constant velocity. So we see that in special relativity the paths things tend to take which are paths of constant velocity are also the paths that maximize proper time intervals between events along the path.

Next they do another experiment. Lois releases two balls of different mass. They are both unacted on by forces in the ship so they just keep their same motion of constant velocity right along with Lois without deviating away from each-other.

38 Chapter 4 Starting GR

Next we go to a fourth observer, Clark Kent, who is far out in the depths of space, but can see through the walls of the space-lab into the experiments. He also sees that their space-lab is falling toward a planet which they didn't realize because they were in free fall and couldn't see outside their lab. According to Clark the path of maximal proper time that Stella took between the events of the beginning and the ending of her experiment was not a path of a constant velocity state at all, but was the path of a body accelerating in the presence of a gravitational field. So we note that the path that things tend to follow in gravitational fields are still paths of maximal proper time even though they are not paths for a constant velocity state.

He also notices that the balls of the experiment though they have different masses, accelerate at the same rate.

Through this mind experiment we have discovered the core essence of general relativity.  

The equivalence principle comes in different strengths. 

The weak version of the equivalence principle boils down to the equivalence of gravitational and inertial mass. "Gravitational mass" and "inertial mass" are Newtonian concepts refering to variables that enter into equations for Newtonian physics. In Newtonian the gravitational force f from a point active gravitational mass M₁ acting on a point passive gravitational mass M₂ at a distance r comes from

f_r = -GM₁M₂/r²

(4.1.1)

In Newtonian physics we also write the relation between the f_r acting on an inertial mass M_2i and a_r as

f_r = M_2ia_r

(4.1.2)

Putting these together we have

a_r = (-GM₁/r²)(M₂/M_2i)

(4.1.3)

We noted the balls of different masses fell at the same rate of acceleration according to Clark. In order for this acceleration to be independent of the ball mass as Clark saw that it was, with the correct choice for the value of G it becomes clear that the gravitational mass M₂ must be equivalent to inertial mass M_2i. Then we have

a_r = -GM₁/r²

 (4.1.4)

In general relativity we will have an invariant definition of mass just as we have defined mass as invariant in the special relativity chapters. There will also be a four-vector force equation for general relativity in the form

F^l = mA^l

where m is the mass as invariant for general relativity.

Gravitation acting alone corresponds to F^l = 0. This yields:

mA^l = 0

The Acceleration four-vector for general relativity is a combonation of two parts discussed in more detail later resulting in

mdU^l/dt + mG^l_mnU^mUⁿ = 0

The m in the term on the left corresponds to the "inertial mass" in Newtonian physics. The m in the term on the right corresponds to the passive "gravitational mass" in Newtonian physics. As these are really the same thing that was just multiplied through it is obvious that indeed the inertial and gravitational masses are identically equivalent.

4.1 The Conceptual Premises For GR 39

The semi-strong level of the equivalence principle comes from the realization that the crew never knew that they were actually falling in a gravitational field. The experiments of a local free fall frame have results indistinguishable from the same experiments done in inertial frames. This is an equivalence of inertial and local free fall frames. We could also extend this to the realization that if the lab had rocket engines burning, keeping them at a constant proper acceleration, they wouldn't have known the difference between being accelerated by the rocket engines or sitting on the surface of a planet in the presence of a gravitational field.

The strong level of the equivalence principle comes from the realization that any local free fall frames are equivalent for doing the physics. The laws of physics were the same for Lois as they were for Terrance. When the equivalence principle is mention unqualified it is usually this level of equivalence that is being referred to.

Above this strength we find the level of equivalence that is really required to result in the form of general relativity that we have today. This is sometimes called the general principle of relativity and sometimes the general principle of covariance. That is simply the statement that the general laws of physics are frame covariant. In other words the equation form that the laws of physics take are the same, invariant, according to every frame whether accelerated or not, whether in the presence of a gravitational field or not, whether rotating or not. To ensure this we must model the general laws of physics with tensor equations. The equations for the general laws of physics are then unchanged by transformations.

Problem 4.1.1

If a laser is mounted on the bottom of an elevator in free fall, would a passenger notice any red shift?

Problem 4.1.2

If a laser were mounted on the side of an elevator in free fall, would a passenger notice any bend in the beam? What about an observer standing on the ground outside?

Problem 4.1.3

If a spaceship orients a laser in the direction of its acceleration, do the passengers observe a red shift? What does the result mean for the rate clocks high in the ship run compared to clocks low in the ship? What would this mean for clocks hovering at different heights over a planet?

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40 Chapter 4 Starting GR

What defines a vector in any physics is its vector transformation properties. Not everything that merely has a magnitude and a direction is a vector, even in non-relativistic physics. For instance angular displacement is not really a vector because it doesn't always obey the vector property

A + B = B + A.

The vectors of relativity obey tensor transformation properties. In general, a four-vector is a rank one tensor. In element notation is has only one index, so it is a tensor with only four elements.

Some of the things we like to think of as individual properties of nature are incomplete as physical properties being only a component of a tensor. For instance, the electric field by itself does not obey tensor transformation properties. The magnetic field by itself also does not obey tensor transformation properties. In the context of this text a pseudovector will be anything that has multiple elements like a vector, but lacks any of the tensor transformation properties. These two pseudo-vectors can be combined into a unified field called the electromagnetic field tensor. Thus we see that the electric and magnetic fields are actually incomplete parts of the actual unified field called the electromagnetic field. This is a rank two tensor.

In the same sense, momentum by itself is not a complete physical quantity as it does not obey tensor transformation properties and so it is not really a vector in the relativistic sense. But, when we combine it with a fourth element, energy, we get a tensor called the momentum four-vector.

Likewise there are displacement four-vectors, velocity four-vectors, acceleration four-vectors, force four-vectors, etc...

According to a general principle of relativity the laws of physics are frame covariant. Therefor when modeling the general laws of physics with equations we must use expressions that are also frame covariant. For instance, if we use one coordinate system to write an equation like

F(ct,x,y,z) - G(ct,x,y,z) = 0,

Then in any other coordinate system it should also be

F ' (ct', x',y',z' ) - G ' (ct', x',y',z' ) = 0

It should not change its basic form. For example, it should not become

F ' (ct', x',y',z' ) - G ' (ct', x',y',z' ) = H ' (ct', x',y',z' )

If such an equation does transforms like this then it is not one of the fundamental equations of physics.

4.2 Tensors in GR 41

Here we will define a tensor in terms of its transformation properties. A contravariant tensor will be any quantity that transforms between frames according to

T ' ^m = (¶x' ^m /¶xⁿ )T ⁿ

(4.2.1)

A covariant tensor will be any quantity that transforms between frames according to

T ' _m = (¶x ⁿ /¶x' ^m )T _n

(4.2.2)

There are also mixed tensors. For example

T ' ^m _n = (¶x' ^m /¶x^s )(¶x ^r /¶x' ⁿ )T ^s _r

(4.2.3)

From these transformation properties we can deduce that for an individual particle,

1.) A sum or difference of tensors is still a tensor.

2.) A product of tensors is still a tensor.

3.) A tensor multiplied or divided by an invariant is still a tensor.

[note - these rules apply only when the tensors involved describe that which is observed, not the state of the observer himself. So for example let F_mn be a tensor describing something observed like say the electromagnetic field and Uⁿ is the four-vector velocity of the observer (c,0,0,0). It turns out that the electric field given by

E_m = F_m0 = F_mnUⁿ/c

is NOT a tensor. As Uⁿ is the four-vector velocity of whoever is the observer everyone uses (c,0,0,0) as a result and the expression does not transform as a four-vector. E'_m = F'_m0 ¹ (¶x^l/¶x'^m)F_l0. If Uⁿ were the four-vector velocity of one "particular" observer then the expression would transform as a tensor, but then it wouldn't represent the electric field to anyone except that observer and it would then only when F_mn is the electromagnetic field already expressed according to his own frame. Likewise the magnetic field

B_m = - (1/2)e_m0^lrF_lr/c = - (1/2)e_mn^lrF_lrUⁿ/c²

where Uⁿ is the four-velocity of the observer (c,0,0,0) is also not a tensor.]

In relativity we must write the fundamental equations of physics as tensor equations such as

T ^{ms ...}_{nr ...} = 0

(4.2.4)

because this remains frame covariant. For instance, using the above transformation properties, it is easy to show that in any other frame this equation remains in the same form

T ' ^{ms ...}_{nr ...} = 0

Problem 4.2.1

Use the general relativistic definition of a tensor to show that for an individual particle,

1.) A sum or difference of tensors is still a tensor.

2.) A product of tensors is still a tensor.

3.) A tensor multiplied or divided by an invariant is still a tensor.

[note - these rules apply only when the tensors involved describe that which is observed, not the state of the observer himself. So for example let F_mn be a tensor describing something observed like say the electromagnetic field and Uⁿ is the four-vector velocity of the observer (c,0,0,0). It turns out that the electric field given by

E_m = F_m0 = F_mnUⁿ/c

is NOT a tensor. As Uⁿ is the four-vector velocity of whoever is the observer everyone uses (c,0,0,0) as a result and the expression does not transform as a four-vector. E'_m = F'_m0 ¹ (¶x^l/¶x'^m)F_l0. If Uⁿ were the four-vector velocity of one "particular" observer then the expression would transform as a tensor, but then it wouldn't represent the electric field to anyone except that observer and it would then only when F_mn is the electromagnetic field already expressed according to his own frame. Likewise the magnetic field

B_m = - (1/2)e_m0^lrF_lr/c = - (1/2)e_mn^lrF_lrUⁿ/c²

where Uⁿ is the four-velocity of the observer (c,0,0,0) is also not a tensor.]

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42 Chapter 4 Starting GR

Recall that for special relativity the invariant interval can be expressed in the form Eqn 2.2.3

ds² = dct² - dx² - dy² - dz²

Or in a more compact notation it can be written Eqn 2.2.5

ds² = h_mndx^mdxⁿ

If we were to express this in a curvilinear coordinate system it will take on a form different from the top equation. For example do the following transformation to cylindrical coordinates

x = rcosq

y = rsinq

The invariant interval will then take the form

ds² = dct² - dr² - r²dq² - dz²

Notice that in curvilinear coordinate systems functions of the coordinates may appear as coefficients of the differential quantities within the interval such as the

-r² appears front of the dq² term above. Another possibility is the appearance of cross terms such as a dctdz term. To write this as a more compact and general form it is expressed

ds² = g_mndx^mdxⁿ

(4.3.1)

When there is matter or fields of any type in the space it effects the form that g_mn can take globally. So the popular interpretation for gravitation is simply that matter gives space-time an intrinsic curvature. In a situation where matter curves the space-time one can not globally transform g_mn to h_mn. However one can always do the transformation locally.

We again express the invariant interval in the form

ds² = g_mndx^mdxⁿ

Given that the interval is invariant we know that

g_mndx^mdxⁿ = g'_lrdx'^ldx'^r

We also know that dx^m transforms according to the calculus chain rule

dx'^m = (¶x'^m/¶xⁿ)dxⁿ

4.3 The Metric and Invariants of GR 43

This results in

g_mndx^mdxⁿ = (¶x'^l/¶x^m)(¶x'^r/¶xⁿ)g'_lrdx^mdxⁿ

And therefor

g_mn = (¶x'^l/¶x^m)(¶x'^r/¶xⁿ)g'_lr

Now this is how a rank 2 covariant tensor transforms. Therefor if ds² is to be invariant then g_mn is a rank 2 covariant tensor. This has been given the name "the metric tensor"

As we shall cover in the sections on gravitational pseudo forces the metric tensor is analogous to the gravitational potential for non-relativistic physics. In non-relativistic physics the gravitational force or other fields are often describable as the gradient of a potential. In later sections the gravitational pseudo forces will be related to affine connections which contain the metric tensor and its first order derivatives.

For special relativity we have

h_mn,_n= 0

We can always transform to a local frame according to which the metric is h_mn so we know so far that for a local frame also

h_mn,_n= 0

Now consider the transformation to be to a local free fall frame so that the affine connections vanish. In that case we also have

h_mn;_n= 0

Now transform this result to an arbitrary frame and we also find

g_mn;_n = 0

(4.3.2)

(Summation still implied on all four above)

Next consider the quantity

g_mrg^rn

as arrived at for any point in spacetime by a transformation to an arbitrary set of Coordinates from a local Cartesian coordinate frame:

g_mrg^rn = (¶x^a/¶x'^m)(¶x^b/¶x'^r)h_ab(¶x'^r/¶x^l)(¶x'ⁿ/¶x^s)h^ls

Rearrange terms

g_mrg^rn = (¶x^a/¶x'^m)(¶x^b/¶x'^r)(¶x'^r/¶x^l)(¶x'ⁿ/¶x^s)h_abh^ls

Yielding

g_mrg^rn = (¶x^a/¶x'^m)d^b_l(¶x'ⁿ/¶x^s)h_abh^ls

Simplify

g_mrg^rn = (¶x^a/¶x'^m)(¶x'ⁿ/¶x^s)h_abh^bs

From the matrix equation for h_mn it is easy to verify the next step

g_mrg^rn = (¶x^a/¶x'^m)(¶x'ⁿ/¶x^s)d_a^s

Simplify

g_mrg^rn = (¶x^a/¶x'^m)(¶x'ⁿ/¶x^a)

44 Chapter 4 Starting GR

This yields

g_mrg^rn = d_mⁿ

(4.3.3)

Contract this and we have

g_mrg^rm = d_m^m

Which results in

g_mng^mn = 4

(4.3.4)

The covariant metric tensor also acts as a lowering index operator and the contravariant metric tensor acts as a raising index operator. For example,

T_m = g_mnTⁿ

and (4.3.5)

T^m = g^mn T_n

It is easy to verify this property based how contravariant and covariant tensors are defined by how they transform. For example consider the following expression,

(¶x ^l /¶x' ^m )(g_lnTⁿ)

based on how tensors transform this becomes

(¶x ^l /¶x' ^m )(¶x' ^a /¶x^l )(¶x' ^b /¶xⁿ )g'_ab(¶x ⁿ /¶x' ^r )T ' ^r = (¶x ^l /¶x' ^m )(g_lnTⁿ)

Rearranging:

(¶x ^l /¶x' ^m )(¶x' ^a /¶x^l )(¶x' ^b /¶xⁿ )(¶x ⁿ /¶x' ^r )g'_abT ' r = (¶x^l /¶x' ^m )(g_lnTⁿ)

Recognizing these result in delta Kroneckers and collecting the priming it becomes,

d_m^ad^b_r(g_abT^r)' = (¶x ^l /¶x' ^m )(g_lnTⁿ)

This simplifies to

(g_mrT^r)' = (¶x ^l /¶x' ^m )(g_lnTⁿ)

But then we recognize that this is how a covariant tensor transforms and so we name T_m by calling it,

T_m = g_mnTⁿ

4.3 The Metric and Invariants of GR 45

Thus we've verified the lowering index property of the covariant metric tensor. Verifying the raising index property of the contravariant metric tensor is easier at this point. Start with the expression,

g^mnT_n

We've named our previous expression T_n and so we insert it.

g^mng_nrT^r = g^mnT_n

But we've already verified that g^mng_nr = d^m_r so we have

d^m_rT^r = g^mnT_n

Which results in

T^m = g^mnT_n

This verifies the raising of index property of the contravariant metric tensor.

With the exception of the locations of physical singularities, the space-time for the universe in which we live is an everywhere locally Lorentzian spacetime. A locally Lorentzian spacetime is a spacetime for which we can locally transform g_mn to h_mn where h_mn is given by Eqn 2.2.4

A locally Euclidean Space-time is a spacetime for which we can locally transform g_mn to w_mn where w_mn is given by

(4.3.6)

In other words all the dimensions of a Euclidean "spacetime" are spacelike.

Either type of spacetime can have Riemannian Curvature as these are only locally Euclidean, or Lorentzian.

46 Chapter 4 Starting GR

Note- Sometimes it is said that our Universe is everywhere locally Euclidean. This basically means that we can do local transformations to arrive at

(4.3.7)

This is correct, but to prevent confusion it is really more appropriate to say that our universe is everywhere locally Lorentzian.

Our universe is also described as being a globally Riemannian spacetime. This means that it globally takes the quadradic form of Eqn. 4.3.1

ds² = g_mndx^mdxⁿ

and is the same thing as saying it is everywhere locally Lorentzian.

An invariant as defined for this text is a quantity whose value does not depend on speed, location with respect to gravitational sources etc... nor upon whose frame it was calculated from. Invariants are said to be invariant to frame transformations, or frame invariant. This does not imply that the value of an invariant must be the same everywhere (for example invariant "densities") nor that it must be conserved. In this context an invariant can be thought of as short for invariant scalar though there are tensor expressions such as the delta kronecker tensor whose elements are all frame invariant. Some people also think of tensors in general as invariants as they represent physical entities and physical entities will not depend in any intrinsic way on our choice of frame. From this perspective the "elements of a tensor" are thought of as "projections of the tensor" onto a coordinate dependent template. The paradigm for this text will instead be that the tensor is the template onto which the projections have been made. It is not invariant, but transforms according to the transformation properties of an infinitesimal displacement vector. Some relativity authors use the word scalar to be short for invariant scalars or what are just called invariants in this text. This is popular, but extremely inappropriate. The reason that it is inappropriate is that if people continue to redefine things without good reason so that they have a different meaning for whatever theory comes along then when they are used in general, eventually a student will practically have to learn a different dialect of the spoken language for every theory encountered. This is complication beyond reason. Here are a few examples of invariants

1. c The local vacuum speed of light
2. m Mass
3. p The pressure scalar [p = (1/3)(T^mnU_mU_nc ^-2 - g_mnT^mn), for example the pressure of a gass]
4. t The proper time between events along a world line.
5. q Charge

An example of how one of these invariants might not be conserved would be to consider the pressure of the gas after a balloon is popped in space. As it expands the pressure decreases and so it is not conserved.

An example from special relativity of a quantity that is conserved, but not invariant would be the total energy of a particle E.

An example of a quality that is both invariant and conserved would be total charge q.

4.3 The Metric and Invariants of GR 47

Consider the transformation of the full contraction of a tensor T^m.

g'_mnT '^mT 'ⁿ = [(¶x^l/¶x'^m)(¶x^r/¶x'ⁿ)g_lr][(¶x'^m/¶x^a)T^a][(¶x'ⁿ/¶x^b)T^b]

g'_mnT '^mT 'ⁿ = (¶x^l/¶x'^m)(¶x'^m/¶x^a)(¶x^r/¶x'ⁿ)(¶x'ⁿ/¶x^b)g_lrT^aT^b

g'_mnT '^mT 'ⁿ = d^l_ad^r_bg_lrT^aT^b

g'_mnT '^mT 'ⁿ= g_lrT^lT^r

So we note that the full contraction of a tensor is an invariant.

Problem 4.3.1

Find h_lrP^lP^r. If the contraction of a tensor is an invariant and this was a local result for the contraction of P^m, what does this tell you about mass in general relativity?

Problem 4.3.2

Write out P_m for g_mn = h_mn .

Problem 4.3.3

Recall problem 2.2.3. What does dt/dt' turn out to be for the spacetime

ds² = (1 + az/c²)²dct² - dx² - dy² - dz²

Problem 4.3.4

Use Eqn 4.3.3 to find g^mn for the spacetime in Problem 4.3.3. Hint - this is simply a matrix inversion.

Problem 4.3.5

Write out g^mn for

ds² = (A² + g_zzb²(1 - f)²)dct² + 2g_zzb(1 - f)dctdz + g_zzdz² - H²(dx² + dy²)

Hint - g_mn is symmetric and g^mn is a matrix inversion of g_mn.

Problem 4.3.6

Consider the spacetime

ds² = (1 - r₀/r)dct² - dr²/(1 - r₀/r) - r²(dq² + sin²qdf²)

Write out T_m for an arbitrary vector T^m and find T_mT^m.

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48 Chapter 4 Starting GR

We want to make equations for the general laws of physics out of tensor equations. So in developing a differentiation operator for general relativity we must assure that when it is operated on a tensor it results in something that is still a tensor. We find that many of the special relativistic laws of physics are described by equations involving ordinary differentiation and so this operator must also reduce to the ordinary differentiation operator in local free fall frames. Consider the chain rule for the ordinary differentiation of a tensor.

dT^l = (¶T^l/¶x^r)dx^r

Using the transformation property of a contravariant tensor we find

dT^l = {(¶/¶x^r)[(¶x^l/¶x ' ^s)T ' ^s]}dx^r

Using the product rule we come to

dT^l = (¶²x^l/¶x^r¶x ' ^s)T'^sdx^r + (¶x^l/¶x ' ^s)(¶T ' ^s/¶x^r)dx^r

And again from the chain rule we finally have

dT^l = (¶²x^l/¶x^r¶x ' ^s)T ' ^sdx^r + (¶x^l/¶x ' ^s)dT ' ^s

Now if on the right hand side we only had the second term then the differentiation of a tensor would still transform as a tensor, but we have the extra first term so we know it does not. Thus to find a differentiation operator which maps tensors to tensors we introduce a second term in the operation. The new differential operator is called the covariant derivative opperator.

DT^l = dT^l + dT^l

(4.4.1)

For a contravariant vector the second term necessary to keep DT^l a tensor is

dT^l = G^l_mnT^mdxⁿ

(4.4.2)

where the affine connection(sometimes called the Christophel symbol of the second kind) G^l_mn is given by

G^l_mn = (1/2)g^lr(g_mr,_n + g_nr,_m - g_mn,_r)

(4.4.3)

4.4 The Affine Connections and The Covariant Derivative 49

For covariant four vectors we can write it in the same form

DT_l = dT_l + dT_l

(4.4.4)

But here we have

dT_l = - G^m_lnT_mdxⁿ

(4.4.5)

In the case of the differentiation of a multiple mixed rank tensor we find

DT^l..._{k ...} = dT^l..._{k ...} + G^l_mnT^m..._{k ...}dxⁿ +... - G^r_knT^l..._{r ...}dxⁿ -...

(4.4.6)

Also it is important to make note that though the affine connection is a part of a covariant derivative operator, it is not a tensor itself.

So, for example, the covariant derivative of a tensor T^l with respect to some invariant parameter such as dt is

DT^l/dt = dT^l/dt + G^l_mnT^m(dxⁿ/dt)

(4.4.7)

As mentioned, a comma will represent a partial derivative and a semicolon will represent a partial covariant derivative. So for example

T^l;_r = T^l,_r + G^l_mnT^m(¶xⁿ/¶x^r)

This simplifies to (4.4.8)

T^l;_r = T^l,_r + G^l_mrT^m

Problem 4.4.1

Work out the affine connections and verify Eqn 4.3.2 for the spacetime in problem 4.3.3.

Problem 4.4.2

Given that the metric tensor is symmetric, verify that the Affine connections are symmetric in the lower indices. That is to say verify G^l_mr = G^l_rm.