Return to Modern Relativity

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

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

dx '^m = (¶x'^m/¶xⁿ)dxⁿ

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 dx^m/dt is a tensor. This is the motivation for defining the velocity four-vector U^m by Eqn 3.1.7

U^m = dx^m/dt

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

U^m = (dx^m/dt)(dt/dt)

U^m = (dt/dt)u^m

(5.1.1)

Note u⁰ = c.

The momentum four-vector of the first kind is written

p^m

and is a rank one tensor with four elements. The time element p⁰ is still just the relativistic energy of the particle

p⁰ = E_R/c

 51

52 Chapter 5 GR Dynamic Implications

The other elements pⁱ 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 bc = dw/dk we also locally have:

w² = c²k² + constant

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

m²c² º |g_mnp^mpⁿ|.

(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, F^l = mA^l, 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

m²c² = g^mn[P_m - (q/c)f_m][P_n - (q/c)f_n].

(5.1.3b).

In general relativity since g_mn 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 r₀ and a local proper frame total of masses density r_Tot 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 r₀ that goes into the energy tensor of mass.

T^mn = r₀U^mUⁿ

(5.1.4)

5.1 Four-Momentum 53

For more general stress-energy tensors it is common to define r₀ as Eqn 3.1.16

r₀ º T^mnU_mU_n /c⁴

If r₀ is to be positive then T^mnU_mU_n 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

T^mnV_mV_n ³ 0

for any timelike vector V_m . Matter may only violate this condition within limits set by the Pfenning inequality. See-

http://xxx.lanl.gov/abs/gr-qc/9805037

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

r_Tot º g_mnT^mn/c²

(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 r_Tot = 0. 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, m_tot = 0, r_Tot = 0, Vs what is meant when a system of massless particles is said to have nonzero mass, m ¹ 0, r₀ ¹ 0. A system of captive particles has a system mass defined by its center of momentum frame energy, m = E_cm/c² which can be nonzero even when the total of masses of the particles is zero, å m_i = 0.

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

m is itself an invariant and U^m is a four vector. So know that mU^m 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.

g_mn(mU^m)(mUⁿ)= m²g_mnU^mUⁿ = m²c²

But refering to our definition of mass 5.1.3a

(m²c² = g_mnp^mpⁿ)

we see that this becomes

g_mn(mU^m)(mUⁿ) = g_mnp^mpⁿ

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

p^m = mU^m

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 r_Tot for g_mn = h_mn where diag[T^mn] = [T⁰⁰, a , b , d] to show that a + b + d = T⁰⁰ is the case for all massless fields in a local or SR limit.

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54 Chapter 5 GR Dynamic Implications

In special relativity the Minkowski or four-vector force on a particle is given by the proper time derivative of the four-vector momentum. F^l_s.r. = dP^l/dt. In general relativity 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.

F^l = Dp^l/dt

(5.2.1a)

F^l = dp^l/dt + G^l_mnU^mpⁿ

(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

F'^m_felt = (¶x'^m/¶xⁿ)Fⁿ

and the inverse transformation is given by

F^m = (¶x^m/¶x'ⁿ)F'ⁿ_felt

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

|F| = (-g_mnF^mFⁿ)^1/2 = m(-g_mnA^mAⁿ)^1/2

(5.2.2)

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

f^l_grav = - G^l_mnu^mPⁿ

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

dp^l/dt = (dt/dt)(¶x^l/¶x'ⁿ)F'ⁿ_felt + f^l_grav

and making a definition for fⁱ_ext

fⁱ_ext º (dt/dt)(¶xⁱ/¶x'ⁿ)F'ⁿ_felt

we can write the equation of motion

dpⁱ/dt = fⁱ_ext + fⁱ_grav

(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

fⁱ_ext = F'ⁱ_felt

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

dpⁱ/dt = F'ⁱ_felt + fⁱ_grav

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

dpⁱ/dt = fⁱ_grav = - Gⁱ_mnu^mpⁿ

Now if we are in free fall with the object then dpⁱ /dt = 0 therefor in a local free fall frame we also have

G^l_mn = 0

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

F^l = dp^l/dt

But this is just the equation for the Minkowski force of special relativity. So we see that dynamics reduces to that of special relativity for a local free fall frame.

In fact we will define a local free fall frame as a frame with its observer based local to the events in question and in which physics reduces to the dynamics of special relativity.

Using this definition be aware that for the particle dynamics to be that of special relativity 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 a free fall frame local 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 a free fall frame local to the particles. Otherwise the particle's displacements though small become great enough so that the free fall frame can no longer be considered to be local to them.

56 Chapter 5 GR Dynamic Implications

In a local free fall frame we have both

G^l_mn = 0

and

g_mn = h_mn

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

g_mn = h_mn

but

G^l_mn ¹ 0

So whether or not a frame is in free fall we define a local frame as a frame 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 demonstraded 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

dòdt = 0

(5.2.4)

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

dòds= 0

This equation gives gravitation its modern interperetation. 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 consept 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 general relativity produces a macroscopic explanation of gravitation. We can now parameterise the equation as follows.

dò(ds/dt)dt = 0

or

dò[(ds/dt)²]^1/2dt = 0

5.2 Geodesic and Nongeodesic Motion 57

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

dò[(ds/dt)²]dt = 0

(5.2.5)

We then use the definition of four velocity Eqn 3.1.7

U^m = dx^m/dt

and use Eqn. 4.3.1

ds² = g_mndx^mdxⁿ

to obtain

dòg_mnU^mUⁿdt = 0

 (5.2.6)

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

(d/dt)[(¶/¶U^l)(g_mnU^mUⁿ)] = (¶/¶x^l)(g_mnU^mUⁿ)]

(5.2.7)

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

d²x^l/dt² + G^l_mn(dx^m/dt)(dxⁿ/dt) = 0

(5.2.8)

This is known as the geodesic equation.

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.

F^l = dp^l/dt + G^l_mnU^mpⁿ

In the case of gravitation acting alone F^l = 0 and we have

dp^l/dt + G^l_mnU^mpⁿ = 0

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

p^m = mU^m

58 Chapter 5 GR Dynamic Implications

so we have

dU^l/dt + G^l_mnU^mUⁿ = 0

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

U^m = dx^m/dt

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

d²x^l/dt² + G^l_mn(dx^m/dt)(dxⁿ/dt) = 0

Problem 5.2.1

An observer is held still in a spacetime given by

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

Where a 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 d²z/dt² using the geodesic equation Eqn 5.2.8. Hint - Start with

1 = (1 + 2az/c²)(dctdct)² - (dz/dct)²/(1 + 2az/c²)

And integrate the geodesic equation d²ct/dct² term just to get an expression for dct/dt . Then differentiate the equation above with this inserted.

Problem 5.2.3

Work out a time travel equation for dt/dt by integrating d²x⁰/dct² from the geodesic equation Eqn.5.2.8 for the Schwarzschild spacetime

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

For simplicity consider only radial motion. Notice the change in sign of dt/dt for r < 2GM/c². What does this mean?

Problem 5.2.4

A particle follows geodesic motion in the spacetime

ds² = dct² - dr² - r²(dq² + sin²qdf²)

Find d²r/dt² for equatorial motion. Is it what you'd expect?

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5.3 Types of 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.

F^l = dp^l/dt + G^l_mnU^mpⁿ

We can express this as

F^l = m(dU^l/dt + G^l_mnU^mUⁿ)

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

A^l = dU^l/dt + G^l_mnU^mUⁿ

(5.3.1)

So we can write the nongeodesic equaiton as

F^l = mA^l

(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, |A| = (-g_mnA^mAⁿ)^1/2. 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 a 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 a of 5.3.3 is also equal to the proper acceleration. a may sound like a strange thing to define at all at first, but it is useful in equivalence problems and in special relativity. In special relativity 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 special relativity, and as observed from local free fall frames, the ordinary force on a particle is the mass times a as defined in these writings.

In any coordinate frame in which the Affine connections vanish and for which the object moves along the direction of the force applied, the a 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 four g's". The kind of

60 Chapter 5 GR Dynamic Implications

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

W = ma

where

a = 4g = 4(9.80m/s²)

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

|F| = m(-g_mnA^mAⁿ)^1/2

Let a^l be given by

a^l º dU^l/dt

(5.3.3)

So, in terms of ordinary force we have

a^l = dU^l/dt = (dp^l/dt)/m

a^l = f ^l/m

(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

F^l = dp^l/dt + G^l_mnp^mUⁿ

Or

dp^l/dt = F^l - G^l_mnp^mUⁿ

In terms of Four Velocity this becomes

dU^l/dt = F^l/m - G^l_mnU^mUⁿ

(5.3.5)

5.3 Types of Acceleration 61

Making use of the chain rule this is

(dU^l/dt)(dt/dt) = F^l/m - G^l_mnU^mUⁿ

Or

dU^l/dt = (dt/dt)(F^l/m - G^l_mnU^mUⁿ)

(5.3.6)

This is what we define as a so.

a^l = dU^l/dt = (dt/dt)(F^l/m - G^l_mnU^mUⁿ)

Continuing on with the simplification

a^l = (dt/dt)F^l/m - (dt/dt)G^l_mnU^mUⁿ

a^l = (dt/dt)F^l/m - G^l_mnu^mUⁿ

(5.3.7)

Next lets derive the equation for coordinate acceleration.

a^l = (dt/dt)²[F^l - (u^l/c)F⁰]/m - G^l_mnu^muⁿ + (u^l/c)G⁰_mnu^muⁿ

(5.3.8)

Refer back to Eqn (5.3.6)

dU^l/dt = (dt/dt)(F^l/m - G^l_mnU^mUⁿ)

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

(d/dt)(dx^l/dt) = (dt/dt)(F^l/m - G^l_mnU^mUⁿ)

Again using the chain rule:

(d/dt)[(dx^l/dt)(dt/dt)] = (dt/dt)(F^l/m - G^l_mnU^mUⁿ)

62 Chapter 5 GR Dynamic Implications

Now using the product rule we have

(d²x^l/dt²)(dt/dt) + (dx^l/dt)(d/dt)(dt/dt) = (dt/dt)(F^l/m - G^l_mnU^mUⁿ)

(d²x^l/dt²)(dt/dt) + (dx^l/dt)(dt/dt)(d²t/dt²) = (dt/dt)(F^l/m - G^l_mnU^mUⁿ)

or

d²x^l/dt²= (dt/dt)²(F^l/m - G^l_mnU^mUⁿ) - (dt/dt)²(dx^l/dt)²(d²t/dt²)

We now write coordinate acceleration a^l

a^l = d²x^l/dt²= (dt/dt)²(F^l/m - G^l_mnU^mUⁿ) - (dt/dt)²(dx^l/dt)(d²t/dt²)

Continuing on with the simplification...

a^l = (dt/dt)²(F^l/m) - G^l_mn(dt/dt)U^m(dt/dt)Uⁿ - (dt/dt)²u^l(d²t/dt²)

a^l = (dt/dt)²(F^l/m) - G^l_mnu^muⁿ - (dt/dt)²u^l(d²t/dt²)

from Eqn 5.3.5 we have

d²t/dt² = (1/c)(F⁰/m - G⁰_mnU^mUⁿ)

Inserting this we have

a^l = (dt/dt)²(F^l/m) - G^l_mnu^muⁿ - (dt/dt)²u^l(1/c)(F⁰/m - G⁰_mnU^mUⁿ)

a^l = (dt/dt)²(F^l/m) - G^l_mnu^muⁿ - (dt/dt)²(u^l/c)(F⁰/m) + (u^l/c)G⁰_mn(dt/dt)U^m(dt/dt)Uⁿ

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

a^l = (dt/dt)²[F^l - (u^l/c)F⁰]/m - G^l_mnu^muⁿ + (u^l/c)G⁰_mnu^muⁿ

Problem 5.3.1

Consider a particle in an SR 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 d²z/dt ², yet another kind of acceleration.

Problem 5.3.3

Refer to Eqn 4.4.8 and Eqn 5.3.1 and show that A^m = Uⁿ(U^m ;_n)

_______________________________________________________________________________________ 

5.4 More About Motion 63

We have arrived at Eqn. 5.3.8

a^l = (dt/dt)²[F^l - (u^l/c)F⁰]/m - G^l_mnu^muⁿ + (u^l/c)G⁰_mnu^muⁿ

In the case of zero four-vector force this becomes

a^l = - G^l_mnu^muⁿ + (u^l/c)G⁰_mnu^muⁿ

Notice that this equation is now mass independant. 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

d²x^l/dt² + G^l_mn(dx^m/dt)(dxⁿ/dt) = 0

The problem that arises is that we'ld 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

q = t/m

(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

dp^l/dq + G^l_mnp^mpⁿ = 0

(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 GR Dynamic Implications

ct = (c²/a + x')sinh(act'/c²)
x = (c²/a + x')cosh(act'/c²) - c²/a
y = y'
z = z'

(a is equal to the proper acceleration for this case of motion. In this case, it is also related to the coordinate acceleration "a" by a = g³a)
The invariant interval according to the accelerated frame

ds² = (1 + ax'/c²)²dct'² - dx'² - dy'² - dz'²

(5.4.3)

transforms form into the inertial frame to Eqn 2.2.3

ds² = dct² - dx² - dy² - dz²

Unfortunately the coordinate transformation above does not work for a time dependant a. 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.

ct = gct' + gbx'
x = gx' + gbct'
y = y'
z = z'

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

ct = ò^ct'gdct' + gbx'
x = gx' + ò^ct'gbdct'
y = y'
z = z'

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

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

Consider two inertial parallel Cartesian coordinate frames 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 frame 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 gbct' and gct' terms with ò^ct'gbdct' and ò^ct'gdct' and expressing the g and b as functions of proper time ct'.  

Consider an accelerated frame S' coordinate system moving with respect to an inertial frame with a proper time dependent velocity of v = bc. The appropriate transformation between the two is

ct = ò^ct'gdct' + gb×r'

(5.4.4a)

r = ò^ct'gbdct' + r' + (g - 1)(b×r'/b²)b

(5.4.4b)

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 frame transformation, but general relativity allows us to use any global transformation we want. As such, this is not a unique choice for an accelerated observers extended coordinate frame, but is a simple and fairly general choice.

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

Consider the S ' frame observer to start at (x,y) = (0,-r₀) and circles the x,y origin without "spin" with the velocity of

b = (v/c)[cos(wgt')e_x + sin(wgt')e_y]

This S ' frame 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:

r = xe_x + ye_y + ze_z

(5.4.5a)

r' = x'e_x + y'e_y + z'e_z

(5.4.5b)

Using the above, equation 5.4.4a yields

t = gt' + g(v/c²)[x'cos(wgt') + y'sin(wgt')]

equation 5.4.4b yields

xe_x + ye_y + ze_z = ò^ct'g(v/c)[cos(wgt')e_x + sin(wgt')e_y]dct' + x'e_x + y'e_y + z'e_z + (g - 1)((v/c)[x'cos(wgt') + y'sin(wgt')]/b²)(v/c)[cos(wgt')e_x + sin(wgt')e_y]

simplified

t = gt' + g(v/c²)[x'cos(wgt') + y'sin(wgt')]

x = x'[gcos²(wgt') + sin²(wgt')]+ r₀sin(wgt') + (g - 1)y'sin(wgt')cos(wgt')

y = y'[cos²(wgt') + gsin²(wgt')] - r₀cos(wgt') + (g - 1)x'sin(wgt')cos(wgt')

z = z'

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

dct = gdct' + d(gbx')
dx = d(gx') + gbdct'

dct = gdct' + bx'dg + gx'db + gbdx'
dx = x'dg + gdx' + gbdct'

dct = gdct' + bx'(dg /db)(db /dct')dct' + gx'(db /dct')dct' + gbdx'
dx = x'(dg /db)(db/dct')dct' + gdx' + gbdct'

dct = gdct' + b²x'g ³(db/dct')dct' + gx'(db/dct')dct' + gbdx'
dx = bx'g ³(db/dct')dct' + gdx' + gbdct'

dct = gdct' + g(b²ax'/c²)dct' + g ^-1 (ax'/c²)dct' + gbdx'
dx = g(bax'/c²)dct' + gdx' + gbdct'

gdct'[1 + (b²ax'/c²) + g ^-2 (ax'/c²)] + gbdx' = dct
dx = gdx' + gb(1 + ax'/c²)dct'

gdct'[1 + (b²ax'/c²) + (1 - b ²)(ax'/c²)] + gbdx' = dct
dx = gdx' + gb(1 + ax'/c²)dct'

gdct'(1 + ax'/c²) + gbdx' = dct
dx = gdx' + gb(1 + ax'/c²)dct'

gb(1 + ax'/c²)dct' = bdct - gb²dx'
dx = gdx' + gb(1 + ax'/c²)dct'

substitute

dx = gdx' + bdct - gb²dx'

dx = g(1 - b²)dx' + bdct

dx = g ^-1 dx' + bdct

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

L₀ = Dx'. If they go off simultaneous according to the inertial frame (Dct = 0) then Dx will be the inertial frame distance L between them. Therefor consider the case dct = 0,

dx = g ^-1 dx'

which then implies

L = g ^-1 L₀

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

ds² = (1 + a×r/c²)²dct² - ds ²

is a vacuum field solution with a zero Riemann tensor if a = a_xe_x + a_ye_y + a_ze_z where a_x, a_y, a_z are functions of proper time, and r = xe_x + ye_y + ze_z and ds ² = dx² + dy² + dz². 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 spacetime.

Problem 5.4.4

Consider equations 3.3.8 relating inertial unprimed frame coordinates to accelerated primed frame coordinates

ct = ò^ct'gdct' + gbx'
x = gx' + ò^ct'gbdct'

a. Show that these coordinate transformations result in

dt/dt' = g ^-1 (1 + ax'/c²)/(1 - bu_x/c)

where a = g²dv/dt' is the proper acceleration, and u_x = dx/dt 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 (u_x = bc) at the accelerated frame origin(x' = 0) this reduces to

dt = gdt'

resulting in ordinary special relativistic time dilation.

c. Show that for u_x describing an inertial frame clock (u_x = 0) this reduces to

dt/dt' = g ^-1 (1 + ax'/c²)

(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 at the origin of the inertial frame and the origin of the accelerated frame (X = ò^ct'gbdct'), and the inertial frame clock under consideration is the one at the origin (x = 0), show that part c results in

dt/dt' = g ^-1 - g ^-2 aX/c²

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 observer's frame to the inertial frame can be found based on the other resuslts. Use the transformations to relate the two accelerated frame coordinates and consider the double prime frame observer's watch as observed by the single primed frame in order to show that the time it displays according to the single prime frame observer is given by

gt' - r₀(v/c²)sin[wg(t" + t')] - gt" = 0

. 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. ANSWER

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Energy and momentum in classical Newtonian physics are conserved quantities. It is often debated whether they are conserved in general relativity 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 general relativity. However there are an energy parameter and momentum parameters that are conserved for general relativity for metrics with isometries and this section is a discussion of how to find these conserved parameters of geodesic motion. 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 L in

L² = g_mnU^mUⁿ.

(5.5.1)

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

(d/dt)(¶L/¶U^z) = (d/dt)(g_znUⁿ)

(5.5.2)

for a coordinate x^z.

If for an example, the metric and therefor L are independent of the coordinate x^z, then

(d/dt)(¶L/¶U^z) = 0

and

¶L/¶U^z and g_znUⁿ 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 x^z then the unit four-vector in the x^z direction K = e_z is a unit Killing four-vector and will result in

g_mnK^m = g_zn

(5.5.3)

Given the above statement that g_znUⁿ would be constant we also have

g_znUⁿ = g_mnK^mUⁿ = constant

(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

g_mkg_rnK^mrU^kUⁿ = constant

(5.5.5)

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

K^m ;_n + Kⁿ ;_m = 0.

(5.5.6) 

and for any Killing four-vector in general we will have

g_mnK^mUⁿ = constant

(5.5.4b)

Problem 5.5.1

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

ds² = dct² - dx² - dy² - dz²

Problem 5.5.2

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