5

5.1 Four-Vector Momentum

Refer back to chapter 3 for details not repeated in this section. Consider the displacement between events along a world line .

We know how this transforms between any two frames because of the chain rule.

but as we have seen, that is how a tensor transforms and so this constitutes a displacement four-vector.

We have also noted that any tensor multiplied or divided by an invariant remains a tensor. Therefor is a tensor. This is the motivation for defining the velocity four-vector by Eqn 3.1.7

The relationship between this four-vector and coordinate velocity is found using the chain rule

(5.1.1)

Note u^{0} = c.

The momentum four-vector of the first kind is written

and is a rank one tensor with four elements. The time element p^{0} is still just the relativistic energy of the particle

p^{0} = E_{R}/c

51

52 Chapter 5 General Relativity Dynamic Implications

The other elements p^{i} are the relativistic momentum.

These hold true whether the particle has mass or not. For example photons do not have mass, but they do have momentum and energy. The magnitude of the three component momentum of a particle can still *locally* be related to a wavelength (**whether or not the particle has mass**)

and the three component momentum

The time element can still *locally* be related to a frequency (**whether or not the particle has mass**)

Where the frequency is *locally* related to the wavelength

And given we also locally have:

The integration constant will turn out to be proportional to the square of the mass.

Over the years I've come across several widely used definitions for mass. The more modern and most elegant and useful general relativistic definition I have come across for the mass of a particle given the above relations is the positive root for m in the equation

.

(5.1.3a).

Here it is the basically the length of the momentum four-vector. In a later section we will restore the definition of mass as inertia ( The m in the four-vector equation for massive particles, , four-vector force equals mass times four-vector acceleration, is the same m as the invariant defined above. ) . Generally the definition of mass will be local rest frame relativistic energy which is the same as equation 5.1.3a and is the definition that we will use throughout the rest of this sight where ever the letter m or the word mass is used unqualified. Under this definition, mass is an invariant. It does **not** change with speed, nor with location in a gravity field!

Equation 5.1.3a is called the mass-shell condition.

In the presence of a vector potential the mass can be equivalently related to the momentum four-vector of the second kind or cannonical momentum

.

(5.1.3b).

Since often varies throughout a system of particles it isn't always useful to define a system mass. Instead it is useful to define a local proper frame mass density and a local proper frame total of masses density in the following ways. (Note-As discussed in 3.1 the system mass is not equal to the "total" of masses) The zero subscript on the first represents that it is the density of the mass(as defined above) according to an observer in a *local frame* **moving with** that bit of mass. We note "moving with" because this is a density and density would change with Lorenz contraction. This is the that goes into the energy tensor of mass.

(5.1.4)

5.1 Four-Vector Momentum 53

For more general stress-energy tensors it is common to define as Eqn 3.1.16

If is to be positive then must be greater than 0. For this not to be the case is called a violation of the weak energy condition. More generally speaking a the weak energy condition is

for any timelike vector . Matter may only violate this condition within limits set by the Pfenning inequality [Ford, L. H., M. J. Pfenning, 1998] .

Massless fields such as the electromagnetic field do contribute to the stress energy tensor and thus to gravitation, but not by the above equation. The total of masses density is related to the stress energy tensor in general by

(5.1.5)

But, under this definition of a mass density, photons do not have a *total* mass. The electromagnetic field's stress-energy tensor results in . As this is the definition of massless, this is also zero for any other *massless* field. Perhaps this is strange terminology, but it is important to understand what is meant when a field is called massless, , , Vs what is meant when a system of massless particles is said to have nonzero mass, , . A system of captive particles has a *system *mass defined by its center of momentum frame energy, which can be nonzero even when the *total* of masses of the particles is zero, .

As we have discussed, a tensor multiplied by an invariant is still a tensor.

m is itself an invariant and is a four vector. So know that is also a four vector. This four vector also has units of momentum and its spatial components reduce to the known expressions for momentum in a Newtonian limit. Now lets consider what happens if we contract this tensor.

But refering to our definition of mass 5.1.3a

we see that this becomes

So we find that for particles with mass the relationship between four momentum and four velocity is

Exercises

**Problem **5.1.1

The frequency and direction of travel of a neutrino, a nearly massless particle traveling near the speed of light, is somehow measured by an experimenter that is near the particle during the measurement while hovering near a black hole. How would a remote observer use this information and a coordinate transformation to calculate the remote frame energy and momentum components of the momentum four-vector? How does the local experimenter calculate the covariant momentum four-vector? How does the remote observer calculate the covariant momentum four-vector from the previous contravariant momentum four-vector calculation?

**Problem **5.1.2

Calculate for where to show that is the case for all massless fields in a local or special relativistic physics theory limit.

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54 Chapter 5 General Relativity Dynamic Implications

5.2 Geodesic and Nongeodesic Motion

In Einstein's special relativistic physics theory the Minkowski or four-vector force on a particle is given by the proper time derivative of the four-vector momentum. . In Einstein's general relativistic physics theory we also define a force four-vector. We define four-vector force as the *covariant* derivative of the four-vector momentum of the first kind with respect to proper time.

(5.2.1a)

(5.2.1b)

We can make a relation between the force four-vector and the force "felt" by the object being pushed by doing a transformation to the accelerated frame of the object in question

and the inverse transformation is given by

Note - The magnitude of the force felt can also be written

(5.2.2)

[Fokker, A. D., 1965]

As shown in section 6.2, the pseudo-force of gravitation can be expressed

So from the equation for the four-force we can write the equation of motion in the form

and making a definition for

we can write the equation of motion

(5.2.3)

5.2 Geodesic and Nongeodesic Motion 55

As long as the observer of the coordinate frame is local to the object in question and the object is moving only along the direction of the external force, we can write

and so as long as the observer of the coordinate frame is local to the object in question and the object is moving only along the direction of the external force, then we can write the equation of motion

Note that in this local frame, if there are no four-forces acting on the object then we have

Now if we are in free fall with the object then therefor in local free fall coordinates we also have

So for a local free fall coordinate system the equation of motion for a nearby particle is Eqn 3.2.2

But this is just the equation for the Minkowski force of special relativity. So we see that dynamics reduces to that of Einstein's special relativistic physics theory for local free fall.

In fact we will define a *local free fall frame* for Einstein's general relativity theory as one with its observer based local to the events in question and in which physics reduces to the dynamics of Einstein's special relativity theory.

Using this definition be aware that for the particle dynamics to be that of Einstein's special relativity theory the particle displacements considered must remain local not only in distance but also in time. For example

Imagine drilling a hole through the center of the earth and drop two noninteracting particles in initially separated just a small amount in height. Neglecting air resistance they will undergo simple harmonic motion with the same period. Thus to someone in local free fall in space to the particles they will be seen to oscillate about a central point which is not special relativistic dynamics at all. Thus the time interval considered was too long to consider this frame local to the particles for the full period of oscillation. If the time interval we were interested in was much smaller than the period of oscillation we still could have considered the frame to be local free fall. Otherwise the particle's displacements though small become great enough so that the free fall frame can no longer be considered to be local.

56 Chapter 5 General Relativity Dynamic Implications

In local free fall we have both

and

In a local frame that is not in free fall we have

but

So whether or not it is in free fall we define a *local frame* as one which has its observer based close enough to the events in question so that the metric is reduced to that of special relativity there.

From our equivalence section we have demonstrated that objects not acted on by forces aside from gravitation should follow paths of maximal proper time. As an equation we can write the statement

(5.2.4)

Since we can write this equation in terms of a path known as a geodesic

This equation gives gravitation its modern interpretation. Space-time is given an intrinsic geometry dependent on the matter within the space according to Einstein's field equations, and particles tend to follow geodesic paths in the spacetime. This concept of geodesic motion explains inertia and replaces the more primitive concept that objects in motion tend to remain in a state of constant velocity. This is how Einstein's general relativity theory produces a *macroscopic* explanation of gravitation. We can now parameterize the equation as follows.

or

5.2 Geodesic and Nongeodesic Motion 57

Next note that the same path that maximizes is could just as well obtained by the equation

(5.2.5)

We then use the definition of four velocity Eqn 3.1.7

and use Eqn. 4.3.1

to obtain

(5.2.6)

and refer to the Euler-Lagrange equation to obtain the result

(5.2.7)

After differentiation and use of identities and simplifications are carried out we find the result to be

(5.2.8)

This is known as the geodesic equation [Einstein A., L. Infield, and B. Hoffmann, 1938] .

Next we check to see if this is consistent with our equation for four-vector force (or the equation of nongeodesic motion for a test mass) Eqn 5.2.1b.

In the case of gravitation acting alone for Einstein's general relativistic physics theory and we have

We then recall the relation between four-vector momentum and four velocity Eqn. 3.1.9

58 Chapter 5 General Relativity Dynamic Implications

so we have

Now we again recall the definition of four-vector velocity Eqn. 3.1.7

and sure enough it results in the geodesic equation Eqn 5.2.8.

Exercises

**Problem **5.2.1

An observer is held still in a spacetime given by

Where is a constant. Calculate the force the observer feels using Eqn 5.2.2.

**Problem **5.2.2

If a particle moves parallel to the z axis for the spacetime of problem 5.2.1, calculate using the geodesic equation Eqn 5.2.8. Hint - Start with

And integrate the geodesic equation term just to get an expression for . Then differentiate the equation above with this inserted.

**Problem **5.2.3

Work out a time travel equation for by integrating from the geodesic equation Eqn.5.2.8 for the Schwarzschild spacetime

For simplicity consider only radial motion. Notice the change in sign of for . What does this mean?

**Problem **5.2.4

A particle follows geodesic motion in the spacetime

Find for equatorial motion. Is it what you'd expect?

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5.3 Acceleration 59

We have defined mass in such a way that it is an invariant. However, in the Newtonian physics from which the concept of mass came, mass was defined as the inertia of an object or in other words it was the resistance to an objects change in motion. As an equation that means that mass is the proportionality constant between force and acceleration.

f = ma

We have seen that the equation of nongeodesic motion is Eqn 5.2.1.

We can express this as

We now note that the expression in parenthesis is the *covariant* derivative of four-vector velocity. This is a tensor and we will call it four-vector acceleration

(5.3.1)

So we can write the nongeodesic equation for Einstein's general relativity theory as

(5.3.2)

Now we see that mass is the resistance of a particle to deviation from geodesic motion.

So if we think of relativistic inertia as the resistance to deviation from geodesic motion we restore the concept of mass as inertia for particles with mass.

In these writings, proper acceleration A**'** means the acceleration according to a local inertial frame according to which the test mass is instantaneously at rest. Its magnitude is equal to the *invariant acceleration*, . This is the amount of acceleration "felt" by the accelerated observer and is also the magnitude of the coordinate acceleration according to an inertial frame in which the accelerated or "proper frame" observer is instantaneously at rest. We also define an acceleration because 5.3.3 is useful. In the case that the force is in the direction of the motion and the affine connections vanish, the acceleration of 5.3.3 is also equal to the proper acceleration. may sound like a strange thing to define at all at first, but it is useful in equivalence problems and in Einstein's special relativity theory. In Einstein's special relativistic physics theory the "ordinary force" on a particle is the derivative of the momentum with respect to coordinate time, not proper time. As we've seen from the section on the four-vector momentum four-vector velocity relationship, the momentum is actually the mass times the four-vector velocity. This means that in Einstein's special relativistic physics theory, and as observed from local free fall, the ordinary force on a particle is the mass times as defined in these writings.

60 Chapter 5 General Relativity Dynamic Implications

In any coordinate frame *in which the Affine connections vanish* and for which the object moves along the direction of the force applied, the of another observer's motion is the Weight "felt" by that observer divided by m. For example, you may hear something like, "the g-force of a rocket accelerating in space is 4 g's". The kind of acceleration that is being referred to as felt is actually proper acceleration. In other words if a passenger were to be pressed up against a weight scale and the passenger's mass was known to be m, the weight scale would read a weight W of

And according to an observer of local free fall instantaneously moving along the same direction this would be equal to

where for this 4 g case,

Note - The magnitude of the force felt can more generally be written by Eqn 5.2.2

Let be given by

(5.3.3)

So, in terms of ordinary force we have

(5.3.4)

In terms of four-force it goes as follows. We have seen that the equation of non-Geodesic motion is Eqn 5.2.1

Or

In terms of Four Velocity this becomes

(5.3.5)

5.3 Acceleration 61

Making use of the chain rule this is

Or

(5.3.6)

This is what we define as so.

Continuing on with the simplification

(5.3.7)

Next lets derive the equation for coordinate acceleration.

(5.3.8)

Refer back to Eqn (5.3.6)

From here we go on to work out the coordinate acceleration of the mass m.

Again using the chain rule:

62 Chapter 5 General Relativity Dynamic Implications

Now using the product rule we have

or

We now write coordinate acceleration

Continuing on with the simplification...

from Eqn 5.3.5 we have

Inserting this we have

We now arrive at the final general expression for coordinate acceleration for general relativity in terms of Four-force. Eqn 5.3.8

Exercises

**Problem **5.3.1

Consider a particle in an special relativistic physics spacetime in a constant ordinary force constant mass problem. Find the coordinate, proper, and four-vector accelerations. Which is a constant?

**Problem **5.3.2

What are the coordinate, proper and four-vector accelerations for the object in problem 5.2.2. Notice that none of these are equivalent , yet another kind of acceleration.

**Problem **5.3.3

Refer to Eqn 4.4.8 and Eqn 5.3.1 and show that

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We have arrived at Eqn. 5.3.8

In the case of zero four-vector force this becomes

Notice that this equation is now mass [Roll, P. G., R. Krotkov, and R. H. Dicke, 1964] independent [Braginsky, V. B., and V. I. Panov, 1971] . Therefor, even for a massless particle such, as a photon, traveling in vacuum this equation remains valid for describing the motion. One merely needs to determine the initial conditions(including the initial coordinate velocity) to predict its motion. However, this is not a tensor equation and some desire to write the equation of motion in terms of a tensor equation. This can be done in the following manner.

We have seen the equation for geodesic motion Eqn.5.2.8

[Infeld, L., and J. Plebanski, 1960]

The problem that arises is that we'd like our general equation of geodesic motion to be something that will work for massless particles. The terms in this equation diverge and so it isn't directly applicable to massless particles. We can change the form into a geodesic equation directly applicable to massless particles in the following way. First we introduce **the path parameter** q. For massive particles it is related to the proper time by

(5.4.1)

Using this and the definition for four-velocity and the four-velocity four-momentum relationship it is easy to show that the geodesic equation becomes

(5.4.2)

Now the entire left hand side of the equation taken together is still a tensor and the right hand side is zero and so we still have a tensor equation for geodesic motion. The advantage is that the terms in this tensor geodesic equation do not diverge for a massless particle.

The time dependant transformation equations we present are not actually Lorentz. We are calling them that because they look and work so much like the Lorentz transformation equations. Special relativity can deal with accelerated frames to an extent. Meisner Thorne &Wheeler *Gravitation *Pg173 show how a coordinate transformation relates an accelerated frame's coordinates to an inertial frame's coordinates for the case that the proper acceleration is kept constant. The transformation (our version of their equation 6.17) can be written by Eqn. 3.3.9

64 Chapter 5 General Relativity Dynamic Implications

( is equal to the proper acceleration for this case of motion. In this case, it is also related to the coordinate acceleration by )

The invariant interval according to the accelerated frame

(5.4.3)

transforms form into the inertial frame to Eqn 2.2.3

Unfortunately the coordinate transformation above does not work* for a time dependant* . However, there is a simple coordinate transformation for which this invariant interval is the correct result. It looks a lot like the Lorentz transformation equations. The Lorentz transformation equations can be written from Eqn. 1.1.1

The transformation from an arbitrarily time dependant accelerated frame to one inertial frame can be written by Eqn.3.3.8.

and are to be expressed as functions of proper time ct**'**

*These are Lorentz transformation equations modified to handle an accelerated frame*.

This coordinate transformation transforms Eqn.5.4.3 into the form of Eqn 2.2.3.

Consider two inertial parallel Cartesian coordinate systems in motion relative to each other. An appropriate transformation between the two for an arbitrary direction of motion is simply Lorentz transformation. Next consider the case that one is accelerates while the other is inertial. A transformation from the accelerated system to the inertial frame consistent with the above transformation can then be constructed simply by doing a Lorentz

5.4 More About Motion 65

transformation, and replaceing the resulting and terms with and and expressing the and as functions of proper time ct**'**.

Consider an accelerated S**'** coordinate system moving with respect to an inertial frame S with a proper time dependent velocity of . An appropriate choice of accelerated frame coordinates can be described by the transformation given by

(5.4.4a)

(5.4.4b)

Using the mathematics of general relativity one can do physics just fine for any choice of coordinates. The very premise is that physics does not depend on frame. As such this choice of coordinates to assign to an accelerated observer is not unique, but it is a natural choice in the sense that they reduce to Lorentz transformation for arbitrary direction in the case of constant velocity and in the sense that they reduce to Rindler's accelerated observer coordinates [Rindler, W., 1969] in the case of constant proper acceleration.

In using this transformation one must arbitrarily choose an inertial frame from which to do it. One might get the impression that this choice prefers one particular inertial frame. It really does not. We could have chosen any inertial frame and the transformation would be perfectly useful. What this choice does instead is it fixes the choice of coordinate that you use to describe the accelerated frame extending to remote locations. Locally inertial frames exist in which one can always use Lorentz transformation, and this is an example of a Lorentz like accelerated transformation, but Einstein's general relativistic physics theory allows us to use any global transformation we want. As such, this is not a unique choice for an accelerated observer's extended coordinates, but is a simple and fairly general and the most natural choice.

Example for the use of 5.4.4,

*S*^{ }**'*** accelerated circular motion with respect to S which will be the inertial center of momentum frame.*

Consider the S^{ }**'** observer to start at (x,y) = (0,-r_{0}) and circles the x,y origin without "spin" with the velocity of

This S^{ }**'** observer will also choose coordinates such that he is always at (x**'**,y**'**) = (0,0)

The radial 3-vectors for 5.4.4 are defined as:

(5.4.5a)

(5.4.5b)

Using the above, equation 5.4.4a yields

equation 5.4.4b yields

simplified

It may be of interest to note that the same length contraction equation falls out of the arbitrary acceleration case as comes from constant velocity motion.

Proof:

Differentiate eqn 3.3.8

substitute

Consider a firecracker pops at each end of the accelerating ship. It is obvious that the ship frame length between them is

. If they go off simultaneous according to the inertial frame then will be the inertial frame distance L between them. Therefor consider the case ,

which then implies

Exercises

**Problem **5.4.1

Show that locally for massless particles.

**Problem **5.4.2.

Show that Eqn. 3.3.8 is consistent with Eqn. 1.3.1 for the scenario described leading to 1.3.1.

**Problem** 5.4.3

Show that

is a vacuum field solution with a zero Riemann tensor if where are functions of proper time, and and by using software to calculate the zero Riemann tensor.

This interval therefor represents the spacetime according to a choice of frame for an observer who undergoes arbitrarily time dependant accelerations in arbitrary directions in an otherwise special relativistic physics theory spacetime.

**Problem** 5.4.4

Consider equations 3.3.8 relating inertial unprimed coordinates to accelerated primed coordinates

**a.** Show that these coordinate transformations result in

where is equal to the proper acceleration, and is the x component of coordinate velocity for an object according to the inertial frame.

**b.** Show that for u_{x} describing an accelerated frame clock at the accelerated frame origin (x**'** = 0) this reduces to

resulting in ordinary special relativistic time dilation.

**c.** Show that for u_{x} describing an inertial frame clock (u_{x} = 0) this reduces to

(Notice from this one that mutual time dilation only occurs when the clock is instantaneously at x**'** = 0, but that inertial frame clocks far in the direction of acceleration undergo rapid advancement. This is the break in symmetry between frames that results in both frames agreeing that in a round trip the accelerated frame observer ages less and by an agreed upon amount.)

**d.** If X is the inertial frame distance between the clock fixed to the origin of S and the clock fixed to the origin of S^{ }**'** , show that part c results in

**Problem** 5.4.5

Refer to the example for use of equation 5.4.4 in terms of a rotating observer. Consider a second double primed accelerated observer who is rotating about the same central point in the same plane at the same speed, but in the opposite direction. By changing the appropriate signs, the transformation from this new rotating observer's system to the inertial frame can be found based on the other rotating transformation results. Use the transformations to relate the two accelerated coordinates to each other and consider the double prime observer's watch as observed by the single primed observer in order to show that the time it displays according to the single prime observer is given by

. Note when the watches re-synch and show that at the meetings where they pass each other one observer observes the other to be time dilated according to special relativistic time dilation.

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5.5 Conserved Parameters of Geodesic Motion

Energy and momentum in classical Newtonian physics and in special relativity are conserved quantities. It is often debated whether they are conserved in general relativity [Press, W. H., and J. M. Bardeen, 1971] in part because what one usually thinks of as energy and momentum - components of a momentum four vector - are not what are always conserved in Einstein's general relativistic physics theory. However the Noether theorem tells us how to find relativistic energy and momentum parameters that actually are conserved for Einstein's general relativistic physics [Nordtvedt, K., Jr., and C. M. Will, 1972] for metrics with isometries, and this section is a discussion of how to find these conserved parameters of geodesic motion [Newman, E. T., and R. Penrose, 1965] . Corresponding to the derivation of the equation for geodesic motion from the principle of maximal proper time we *can* choose to consistently define the **Lagrangian for gravitation** as in

.

(5.5.1)

though in some texts it is defined without the 2 exponent. From this one can show with some differential calculus that

(5.5.2)

for a coordinate .

If for an example, the metric and therefor are independent of the coordinate , then

and

and are constants.

They are then conserved parameters of geodesic motion.

A unit Killing four-vector will here be defined as a unit four-vector that is along an isommetry direction of the metric. For example, if the metric tensor is independent of the coordinate then the unit four-vector in the direction is a unit Killing four-vector and will result in

(5.5.3)

Given the above statement that would be constant we also have

(5.5.4a)

We may also define a **rank 2** **Killing tensor** in conjunction with this equation, which also corresponds to an isometry transformation yielding

(5.5.5)

**In general** a four-vector will be a **Killing four-vector** if and only if it obeys

.

(5.5.6)

{some authors will also call the vector satisfying where f is a function of the coordinates a conformal Killing vector or if f is a nonzero constant a homothetic Killing vector. Those that do will call what I am just calling a Killing vector as a proper Killing vector}

and for a Killing four-vector **in general** we will have

(5.5.4b)

**Problem** 5.5.1

Use Killing vectors to show that special relativistic energy and momentum are conserved parameters of motion for

ds^{2} = dct^{2} - dx^{2} - dy^{2} - dz^{2}

**Problem** 5.5.2

Find isometries for various spacetimes of your choice and use them to find conserved parameters of geodesic motion.

Citations References

Braginsky, V. B., and V. I. Panov, 1971, "Verification of the Equivalence of Inertial and Gravitational Mass, " *Sov. Phys.-JETP 34, *464-466

Einstein A., L. Infield, and B. Hoffmann, 1938, "The Gravitational Equations and the Problem of Motion," *Ann. Math. 39, *65-100

Fokker, A. D., 1965, *Time Space, Weight and Inertia, *Pergamon Press. London

Ford, L. H., M. J. Pfenning, 1998, "Quantum Inequality Restrictions on Negative Energy Densities in Curved Spacetimes," *Gen. Rel and Quant. Cos. Dissertation*

Infeld, L., and J. Plebanski, 1960, *Motion and relativity, *Pergamon, New York

Newman, E. T., and R. Penrose, 1965, "10 Exact Gravitationally-Conserved Quantities," *Phys. Rev. Lett. 15, *23

Press, W. H., and J. M. Bardeen, 1971, "Nonconservation of the Newman-Penrose Conserved Quantities," *Phys. Rev. Lett. 27, *1303-13061-233

Rindler, W., 1969, *Essential Relativity: Special, General, and Cosmological, *Van Nostrand, New York

Roll, P. G., R. Krotkov, and R. H. Dicke, 1964, "The equivalence of Inertial and Passive Gravitational Mass," *Ann. Phys. 26, *442-517

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