SR Dynamic Implications

Return to Special Relativity

In many texts mass has been defined in a circular manner. Some such texts have asserted a four-vector momentum in the form of p^m = mU^m as a premise which doesn't work for massless particles and then defined mass as the contraction of that vector or visa-versa. In order to avoid circularity and to include massless particles and also in order to facilitate a smoother transition to relativistic quantum mechanics this text will take a newer though not unique approach. For example, here there will be two different momentum four-vectors distinguished by capitalization p^m, and P^m. The lower case will be the momentum four-vector of the first kind and the upper case will be the momentum four-vector of the second kind. This is done in part because a particle's mass will be defined as the contraction of the momentum four-vector of the first kind which is the momentum four-vector referred to in classical relativistic (non-quantum relativistic) texts. The momentum four vector of the second kind is here defined mainly because its elements are what will correspond to quantum operators in relativistic quantum mechanics. (Some authors choose the capitals the other way around)

From experiments or due to quantum mechanics we know that the magnitude of the three component momentum of a particle can be related to a wavelength (whether or not the particle has mass).

(3.1.1a)

and the relation between the three element momentum and the three element k (whether or not the particle has mass) is

(3.1.1b)

Also, a fourth element corresponding to a time coordinate can be related to a frequency (whether or not the particle has mass) and that element times c we will term relativistic energy E_R.

(3.1.2a)

(3.1.2b)

where w can be related to the wavelength by

(3.1.2c)

and from bc = dw/dk:

w² = c²k² + constant

(3.1.2d)

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

We will start with the premise that this four component definition of momentum constitutes a four-vector that will be called the momentum four-vector of the first kind p^m.

Next consider the introduction of a four-vector potential f^m to which the test particle responds with a charge q. It does not matter at this point if this charge is electric, only that the vector potential to which it responds is a four-vector. We will define the momentum four-vector of the second kind by

P^m = p^m + (q/c)f^m

(3.1.3a)

and we will call its time element P₀ total energy E

As an artifact from physics texts that do not make this distinction, when the potential is not zero one might think of the total energy E as p⁰/c even though it is really P₀/c which is E_R + qf in special relativity. At the same time, one would think of the relativistic momentum as pⁱ. As a result one may think of energy E due to containing a potential as something that can have an arbitrary constant added to it, but would think of momentum as something that can not. This superficially seems to draw a distinction between time and space, as energy corresponds to a time element and momentum corresponds to spatial elements. However here where the distinction between momentum four-vectors is made, one finds that there is no such distinction between time and space. This is because it is in P^m that such an arbitrary constant can be added to the potential, and it is also in that four-vector that such arbitrary constants can be added to the spatial components of the vector potential. One can do this so long as one demands that they transform as the coordinates do, as a four-vector.

The special relativistic definition for the mass of a particle given those relations is

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

(3.1.4a)

or

m = [(E_R/c²)² - (p/c)²]^1/2 = E₀/c²

(3.1.4b)

This could just as well be expressed as

m²c² = |h_mnp^mpⁿ|

(3.1.4c)

In the second we define E₀ as the relativistic energy evaluated at zero velocity, E₀ = E_R|_v=0. Magnitude bars are included above merely so that the choice of sign convention for the metric's signature is arbitrary.

Though all 3.1.4 are equivalent given the above relations, the definition in terms of the momentum four-vector of the second kind equation 3.1.4a is preferable in quantum mechanics discussions because that is what yields relativistic quantum mechanics. For example, when one replaces the elements of the momentum four-vector of the second kind with the energy and momentum operators of quantum mechanics, and operates that on the wave function it yields the Klein-Gordon equation with the inclusion of a nonzero vector potential

h^mn[P_opm - (q/c)f_m][P_opn - (q/c)f_n]Y = m²c²Y

(3.1.5a)

which can also be written

[(H - f)² - (P_opc - qf)²]Y = m²c⁴Y

(3.1.5b)

where and

The above definition of mass 3.1.4b, that a mass is rest energy m = E₀/c² , E₀ = E_R|_v=0. or that its is center of momentum frame relativistic energy m = E_cm/c², E_cm = p⁰_cmc|_vcm=0, in the case of a system of particles, is the definition that we will use throughout the rest of the special relativity site where ever the letter m or the word mass is used unqualified. This is the m that goes into the relativistic version of Newton's second law in the form

F^l = mA^l

(four-force = mass times four-acceleration see - Eqn 3.2.3)

This mass is an invariant. It does not change with speed! Equations 3.1.4 is called the mass-shell condition, because they are of isomorphic form to the equation of a spherical shell. Under the above definition of mass, a photon does not have mass. Due to quantum mechanical issues, virtual particles do not tend to have the expected value of energy for a given momentum. So sometimes it is said a particle lays off shell.

The coordinate velocity of a particle is simply given by

u^m = dx^m/dt

(3.1.6)

We write Four-Vector Velocity or Proper Velocity

U^m = dx^m/dt

(3.1.7)

where t is called proper time, which is just time according to the frame of the particle at its location.

Through time dilation we can relate the two

U^m = gu^m

(3.1.8)

Consider the following expression

h_mnmU^mmUⁿ

3.1 Mass, Energy and the Universe's Speed Limit 23

h_mnmU^mmUⁿ = [(mU⁰)² - (mU¹)² - (mU²)² - (mU³)²]

h_mnmU^mmUⁿ = m²[(dct/dt)² - (dx/dt)² - (dy/dt)² - (dz/dt)²]

h_mnmU^mmUⁿ = m²(dt/dt)²[c² - (dx/dt)² - (dy/dt)² - (dz/dt)²]

h_mnmU^mmUⁿ = m²g²c²(1 - v²/c²)

h_mnmU^mmUⁿ = m²c²g²g ^-2

h_mnmU^mmUⁿ = m²c²

Next we refer to the definition of mass 3.1.4c to arrive at

h_mnmU^mmUⁿ = h_mnp^mpⁿ

Final examination of this reveals the relation between four-vector velocity and the four-vector momentum of the first kind.

p^m = mU^m

(3.1.9)

We can then from equation 3.1.8 discover the relation between four-momentum and coordinate velocity for massive particles

p^m = gmu^m

(3.1.10a)

or write the relation for particles that may or may not have mass

p^m = (E_R/c)(u^m /c)

(3.1.10b)

The g term is physically associated to the velocity term through time dilation. In the past a few physicists starting with Planck, Lewis, and Tolman, not Einstein, have miss-associated the g term with the mass defining a new kind of mass

M = gm ¬ Bad

(3.1.11)

This M is then inappropriately called "relativistic mass". In the absence of a potential, the zero^th element of the momentum four-vector is defined as the energy divided by c, resulting in

p⁰ = Mu⁰

E/c = Mc

M = E/c² ¬ Bad

(3.1.12)

Though much more complicated in the long run, the math is consistent and leads to consistent predictions concerning observation and so one might argue that the physics is therefor correct. But, in keeping with Occam's razor this definition and method must be done away. The m in this method is then inappropriately qualified and called the "rest mass". It is wrong to do this for the following reason. Calling m the "rest mass" infers to the listener that m is not the mass according

24 Chapter 3 SR Dynamic Implications

to other frames for which it is not at rest. We have already noted that m is an invariant as it is the same value as calculated according to any frame. It is not just the value for the rest frame. The relativistic mass method also leads to many erroneous conclusions. By that method light has zero "rest mass". For one of many examples, it has been argued that since light is not at rest in any frame, that the question of whether it has mass at rest or "rest mass" is unanswerable. No. m = 0 is observed as the contraction of a photon's four-momentum according to any frame, not just the "rest frame".

In short the terms "relativistic mass" and "rest mass" need to be done away and the real mass m which is actually observed is an invariant. It does not change with speed. Also, by this, the physically correct definition a photon, or anything that travels at the Lorentz invariant speed c, has zero mass.

We have

p⁰ = E_R/c.

We have also demonstrated the relation between Four-Momentum and Four-Velocity Eqn 3.1.9 resulting in

p⁰ = mU⁰.

Putting these together we have

E_R/c = mU⁰

E_R/c = m(dct/dt)

E_R = (dt/dt)mc²

E_R = gmc² ¬ Good

(3.1.13)

This is the mass - relativistic energy relationship for a massive particle. Now this energy does not go to zero as v goes to zero so we see that a massive particle still has energy even when it is at rest. This tells us that mass is equivalent to rest energy meaning relativistic energy at zero velocity

E₀ = E_R|_v=0 = mc² ¬ Good

(3.1.14)

3.1 Mass, Energy and the Universe's Speed Limit 25

The kinetic energy of a particle is the amount of energy that is associated with its motion only. Therefor

E_K = E_R - E₀

(3.1.15a)

This results in

E_K = (g - 1)mc²

(3.1.15b)

The stress energy tensor is a tensor that contains information about the density of energy, momentum, stresses, etc.. contained in the space. The energy tensor mass alone is Eqn. 5.1.4

T^mn = r₀U^mUⁿ

The T⁰⁰ component of this is

T⁰⁰ = (dt/dt)²r₀c²

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)r₀

So this becomes

T⁰⁰ = (dt/dt)rc²

But this is just the coordinate frame energy density.

The simplest consistent general relativistic definition of coordinate frame energy density is then just

coordinate frame energy density º T⁰⁰

(6.3.6)

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

r₀ = T^mnU_mU_n /c⁴

(3.1.16)

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

(3.1.17)

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

Other elements have other interpretations. 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. T^ij is the xⁱ component of momentum per area in the x^j direction or describes a shearing from stresses. T⁰ⁱ is the volume density of the i^th component of momentum flow.

Next we will discuss the concept of system mass. We have seen that for a single particle mass is equivalent to rest energy. Eqn 3.1.14

E₀ = mc ².

For a system of particles the best concept for system mass m is defined as center of momentum frame energy E_cm.

E_cm = mc².

(3.1.18)

The system mass does not turn out to be equal to the total or sum of masses m_tot of the constituent parts. Instead it is the total energy summed for all of the constituent parts according to the center of momentum frame.

Consider for a moment a Lorenz invariant for the system consistent with the mass shell condition.

First define

p_sys' = [E_cm/c , 0, 0, 0]

(3.1.19)

as a four-element vector for the inertial center of momentum frame. Then define P_sys as the Lorentz transform of this for any frame of interest.

p_sys = Lp_sys'

(3.1.20)

Due to relative simultaneity p_sys as defined here is not always equal to the "simultaneous" sum of the four-momentum of the constituent parts when there are external forces acting at various locations on the system. The system mass is defined as the following invariant.

m²c² = h_mnp_sys^m p_sysⁿ

(3.1.21)

or for the scenario described above,

m = [(E_Rsys/c²)² - (p_sys/c)²]^1/2 = E_cm/c²

(3.1.22a)

Considering the time element of equation 3.1.21 restores the relation

E_Rsys = g_cmmc²

(3.1.22b)

Proof that this definition of system four momentum is the same as the sum of the four-momenta of the systems components goes as follows. Start with the sum of four-momentum for an arbitrary frame.

p_sys = S_i p_i

Lorentz boost to another frame

Lp_sys = LS_i p_i

Interchange sum and transformation symbols

Lp_sys = S_i Lp_i

The Lorentz transform of each four-momentum is the four momentum according to the new frame

Lp_sys = S_i p_i'

But the right side is the net four momentum according to the new frame

Lp_sys = p_sys'

This proves that the system net four-momentum is indeed a four-vector itself and yields

E_Rsys = g_cmmc²

where m is the system's mass and is its center of momentum frame energy as well as

p = g_cmmu_cm

(3.1.22c)

and

m²c² = h_mnp_sys^mp_sysⁿ = E_Rsys²/c² - p_sys² = E_cm²/c²

The reason that the mass of 3.1.21/3.1.22a is not the same as the "total" of constituent masses, m_tot , is that the sum of masses of the constituent parts does not always equal the center of momentum frame energy. For example, a system of massless particles have a zero mass shell condition when they all move the same direction while the system has a nonzero mass shell condition when they move in different directions. One advantage the definition of center of momentum frame energy for mass has over "total mass" is that by this definition, not only is mass an invariant, but this mass of a system is also conserved. Note that the concept of captive mass is equivalent to center of momentum frame energy m and not the total of masses m_tot. In order to increase the system mass m one must increase the total center of momentum frame energy E_cm equivalently. This demonstrates that mass defined by m is conserved in the same way that energy is. Transferring energy from some external matter to change the center of momentum frame energy of an object will increase its individual system mass, but when you extend the system to include the matter from which the energy was transferred, it will always be found that center of momentum frame energy or system mass m is ultimately conserved.

A sum of invariants is also an invariant and so one could just as well write the total of masses m_tot as a sum of the constituent parts. For a system of n particles this could be written

m_tot = (h_mnp₁^m p₁ⁿ)^1/2 + (h_mnp₂^m p₂ⁿ)^1/2 + ... + (h_mnp_n^m p_nⁿ)^1/2

(3.1.23)

or

m_tot = m₁ + m₂ + ... + m_n

(3.1.24)

The subscript indicates the particle number. Again, the major problem with thinking of a system mass as this is that this total of masses is not conserved. However, part of the reason this is brought up is that people do tend to think that mass is that kind of sum. This leads to another missunderstanding as to what is meant by "mass to kinetic energy conversion". Consider for example a massive particle that decays into two massless photons. Because system energy is conserved and the system energy for the center of momentum frame is the system mass, the system mass did not change in the decay. What did change was that the energy initially was associated with rest, the rest energy of the particle, but finally was associated with motion, the kinetic energies of the photons. In that light one should really not say there is mass-energy conversion. The energy and system mass for the system is conserved. One should instead say that energy associated with the resting particles of the initial state becomes associated with the motion of the particles in the final state. Since this is cumbersome the term mass-energy conversion is used, but be wary that what it refers to is that a change in the sum of masses can be the cause of the change in kinetic energies of the remaining masses. Just remember that the system mass of a closed system doesn't change or "convert" into anything.

Sometimes it is more useful to define an invariant mass density instead. Just as there are two ways to describe the masses of a system above, m and m_tot, there are two important ways of describing an invariant mass density. The first is the r₀ in the following relation equation 5.1.4

T^mn = r₀U^mUⁿ

This definition most closely corresponds to the mass m description of system mass above. It relates the stress energy tensor for matter composed of non-interacting constituents to the four-velocity of the unpressurized "fluid" at any given location.

The total energy for the system is conserved and could instead be defined by the following volume integral

E_sys = ò ò ò T⁰⁰dxdydz

(3.1.25a)

For the example stress energy tensor 5.1.4 this would become

E_sys = ò ò ò g_fluid²r₀c² dxdydz

(3.1.25b)

One can also instead define the system momentum from the next integral

p_sys = ò ò ò (T ⁰ⁱ e_i/c) dxdydz

(3.1.26a)

For the example stress energy tensor 5.1.4 this would become

p_sys = ò ò ò (g_fluid²r₀uⁱe_i)dxdydz

(3.1.26b)

e_i is a unit vector in the direction of the i^th momentum component.

One then still can define the system's mass as the center of momentum frame energy. It is the energy for the frame according to which

p_sys = 0. So we still have

E_cm = mc².

The other invariant mass density concept corresponding to the total of masses m_tot would be

r_Tot = h_mnT^mn/c²

(3.1.27)

This kind of mass density is an invariant, but its volume integral is not conserved. This kind of mass is what is meant when one refers to a field such as the electromagnetic field as a massless field, or when one refers to any system as massless. This is zero for any system of massless particles.

Of the two descriptions of system mass, the mass m concept is far more useful.

Refer to Eqn 3.1.11

E_R = gmc²

where g was given by

g = (1 - v²/c²)^-1/2.

Notice that the energy becomes divergent at v = c for nonzero mass. Thus no matter how hard or how long you push on a mass, you can never impart enough energy to it so that it reaches the speed c. The only way such a thing can travel at the speed c and still have finite energy is if it had zero mass. In that case, instead of a mass energy relation, there is a energy momentum relation from Eqn 3.1.5 resulting in,

E = E_R = pc,

(3.1.28)

where E and p are related to frequency and wavelength.

One might consider the case of a particle that instead of being pushed beyond the speed c, moves faster than c upon its creation. Such a hypothetical particle is called a tachyon. Notice that if v is greater than c then g is imaginary. Since imaginary energy makes no physical sense, we would expect that the mass would also have to be imaginary so that the energy(and momentum) would be real.

The primary problem with the existence of such a particle is that it could be use to violate the principle of causality. The principle of causality is simply the statement that effect never precedes cause. Imagine setting up a tachyon emitter and a tachyon receiver at different points along an S frame x axis. Lets say that the signal travels arbitrarily fast so that the event of transmission and the event of reception are virtually simultaneous. Next recall that events simultaneous in one frame are not all simultaneous in other frames. We could then easily pick a frame to look at the situation in which the event of reception precedes the event the event of transmission. This is a violation of causality.

Worse yet, it then leads to grandfather paradox's. The grandfather paradox is the idea that a time traveler goes back in time and kills his grandfather before his father was conceived. To set up a grandfather paradox with tachyons we simply set the receiver in motion away from the

26 Chapter 3 SR Dynamic Implications

transmitter and give it a relay transmitter. We call the frame in which it is stationary the S' frame. We also connect a receiver to the S frame transmitter. We then program the S frame transmitter so that in say 1hr it will send a signal unless it's receiver receives a signal. To begin lets say that it receives no signal and so it sends one. The signal arrives at the relay, which is moving away and sets off the relay transmitter. Now the relay transmitter sends a return signal, but the return signal travels back to the S frame receiver/transmitter setup virtually instantaneously according to the relay's S' frame. Due to the Lack of simultaneity between the frames the return signal will be received back at the S frame transmitter at a time prior to the original transmission. But because we programmed it not to send a signal if it receives one it will now not send a signal. But then there is no signal to receive and so it sends one......

It is sometimes said that special relativity says that nothing can travel faster than the speed of light. As discussed, what it really implies is that long as we restrict our physics to special relativity, and we wish to preserve causality, information can not travel faster than c. Likewise, as long as we restrict our physics to special relativity, nothing with mass can travel at the speed c.

There have been experiments done in which the physicists involved say that they have indeed been able to get electromagnetic waves to propagate information faster than c through a dispersive medium.

In particular the controversy is over gain assisted faster than c group velocity transmission demonstrated in anomalous dispersive media.

http://www.iitk.ac.in/infocell/Archive/dirjuly3/science_light.html

If their claims that it was the "information" that has indeed been transferred at faster than c speeds are correct, then SR implies that we can find a frame according to which the reception of a signal at one end of the apparatus precedes the transmission of the signal at the other end. This would indeed be a causality violation and brings to question the validity of the principle of causality and causes us to reevaluate the (im)possibility of a physical grandfather paradox. However, it is conceivable that the universe may be structured in such a way that such a causality violation is attainable, but that grandfather paradoxes will still not be allowed. For example, if one of their dispersive media faster than c experiments were devised in attempt to simulate the relay-transmitter paradox discussed above, one could hypothetically set such a receiver in motion such a medium, but that medium is what determines the speed of the electromagnetic waves according to its rest frame. A relay transmitter sending the signal through the same medium would not end with the signal arriving at a time prior to transmission.

The following is an explanation why the experiments are not completely convincing of faster than c "information" transfer.

As an example, in a 6.0cm medium a laser pulse has been transmitted that transversed the distance at a speed of 310c. This is a group wave speed, not a phase wave speed. The below figures are a recreation representative of a receiver's data for two pulses. The blue dotted curve represents the intensity Vs time curve for the reception of the 310c speed pulse and the red curve represents the intensity Vs time curve for the arrival of a c speed pulse sent at the same time.

Figures 3.1.1a,b

One can see on the close up second graph from the horizontal shift that the 310c curve arrived 62ns earlier than the c speed pulse. The question then arises whether this experiment is an example of a causality violation. In order for causality to be violated one must have faster than c "information" transfer. Under typical "long" transmissions one can consider the information transfer speed to be the group wave speed or the speed of the energy carried by the pulse. It has long been known that phase wave speeds often exceed c which is why it is often pointed out that the energy transmission in ordinary wave-guides occurs at the group speed which is less than c. As such, the information transfer speed is less than c in ordinary wave guides. The reason that the group speed exceeding c for this experiment is not convincing of faster than c information transfer and the reason why the information transfer for this experiment can not be taken to be the group speed is because of the following. Notice that the 62ns time shift between the two pulses is much less than the time it takes to receive the entirety of a pulse itself. Thus the time it took from the time the pulse began to enter the medium until the time the receiver read the entirety of the information was the sum of group transfer and read times. The full width at half-max FWHM of a pulse here is approximately 4.0ms. Take that to be read time. The group transfer time was

6.0cm/310c = 0.65ps. Taking the information transfer time to be the sum of these, approximately just 4.0ms, one finds that the information speed was

6.0cm/4.0ms = 5.0x10^-5c, a mere measly fraction of the vacuum speed of light. There are two ways one might modify this experiment so that if successful it would clearly demonstrate faster than c information transfer. First, one might made the medium of transfer much longer so that in the information transfer time it is the read time that is insignificant instead. The reason that this may be an impossible task is that due to the dispersive nature of the media itself, even with the gain assistance, there will be a trade off between the signal degradation and length. There may be a limiting trade off so that in a long enough transmission line so that the information transfer time yields a faster than c speed the signal would have been lost. Second, one might try to significantly narrow the pulse so that the information transfer time is approximately just the group transfer time. The reason that this may be an impossible task two fold. There is a narrow frequency range at which the light must be sent through the medium in order for it to transfer with a faster than c group speed. This in itself puts a limit on how narrow the pulse may be. Also, the narrower the pulse is made the more rapidly it will tend to widen itself as it travels across the medium. By the time it gets to the other end the read time will always be longer than the send time so no matter how short the send time is made one will have to contend with a longer read time.

In conclusion, though such experiments do successfully demonstrate faster than c group transfer, they do not conclusively demonstrate the faster than c information transfer, which they would have to in order to show a causality violation.

Problem 3.1.1

Consider monochromatic plane waves of light with a wavelength of 500nm. What is the frequency of the light?

Problem 3.1.2

a. How much kinetic energy does a 75 kg man have if he is traveling to another star at (4/5)c ?

b. Though "system mass" i.e. center of momentum frame energy is conserved in an isolated reaction, the sum of masses is not. The sum of kinetic energies of a system's parts after a reaction is the sum of kinetic energies of the system's parts before the reaction plus the difference in the sum of masses that the system decreased by through the reaction. Changes in binding energies of interacting particles in the system are typically accounted for in the sum of masses. For example a hydrogen atom in a system is least massive in its ground state. So, such a reaction is sometimes called a "mass to kinetic" energy conversion. How much of the sum of masses of a system would have to be annihilated in order to produce the same amount of kinetic energy as the man had in part a? (An atomic bomb "converts" about the mass of a coin to kinetic energy in its explosion)

Problem 3.1.3

Show that if the speed of an object is v = c(1 - e) where e is an extremely small number for a high speed, that e can be approximated by

Problem 3.1.4

(Modified from Misner, Thorne, and Wheeler's Gravitation problem 5.4)

a. Use the Lorentz transformation property of a tensor to show that

T ⁰ⁱ = S_k i ^{i k}r₀ u^kc

where i ^{i k} is defined by

i ^{i k} r₀c² = T ' ⁰⁰ d ^{i k} + T ' ^{i k}

Note- r ₀ is defined in equation 3.1.16 for general stress-energy tensors.

b. Derive the equations of part a. from Newtonian considerations plus the conservation of relativistic energy. (Hint: The total energy carried past the observer by a volume of the medium V includes both the rest energy T ' ⁰⁰ V and the work done by forces acting across the volume's face as they push the volume of the medium through a distance.)

c. Refer to equation 3.1.26 a. Considering that force is a change in momentum with respect to time, this results in a force per unit volume f ⁱ of

f ⁱ = dT ⁰ⁱ/dct

Which for constant i ^{i k} results in

f ⁱ = S_k i ^{i k} r₀du^k/dt

This equation suggests that one call r₀S_k i ^{i k} an inertia density matrix. What does this become for the special case of a perfect fluid?

d. Consider an isolated, stressed body at rest and in equilibrium (T ^mn,₀ = 0) according to the laboratory frame. Show that its total inertia matrix I ^{i k} (not moment of intertia) defined by

I ^{i k} = ò ò ò i ^{i k} r₀ dxdydz

is isotropic and results from the mass or center of momentum frame energy of the body

:

I ' ^{i k} c² = d ^{i k}ò ò ò T ' ⁰⁰ dx'dy'dz'

Problem 3.1.5

Show that if v << c

E_K » (1/2)mv²

Problem 3.1.6

What is the speed, frequency, and wavelength of a 2Mev proton?

Problem 3.1.7

According to problem 3.1.4 combining the results of parts c and d results in an isolated stressed body in equilibrium obeying Newton's second law at low speeds

f ⁱ = ò ò ò f ⁱ dxdydz = ò ò ò (T ' ⁰⁰/c²) dx'dy'dz'duⁱ/dt

which can be expressed by the four-vector equation F = mA. This indicates that isolated stressed bodies in equilibrium will obey all the same dynamics as relativistic individual particle dynamics and must therefor obey

E = g_cmE_cm. Consider a box containing a photon gass. The interior wall lengths will be L,W, and H. And the wall thicknesses will all be da.

The proper frame stress-energy tensor of the photon gass is

The proper frame stress energy tensor for two of the walls whose surface normal is (anti)parallel the direction of motion has stresses in the wall both from being pushed on by the gass pressure normally and from being stretched from the edge attachments to the other walls that are also being pushed on by the gass. The average x-x component of the stress energy tensor for these two walls would be proportional to the gass pressure on them. i.e.

T ' ^xx = e T ' ^xx/3

Assuming a material for which e~1 the proper frame stress-energy tensor consistent with the entire system being an isolated stressed body in equilibrium for those two walls will be

The proper frame stress energy tensor for two of the four of the walls whose surface normal is perpendicular the direction of motion is

The proper frame stress energy tensor for the other two of the four of the walls whose surface normal is perpendicular the direction of motion is

Lorentz transform the stress energy tensors and integrate T ⁰⁰ throughout the volume in order to show

E = g_cmE_cm

Don't forget length contraction.

Answer

Problem 3.1.8

For a particle in a four vector potential f^m, the mass will be defined as the following invariant m in:

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

where q is the charge reacting to that field.

In special relativity this results in

(E - qf)² = (Pc - qf)² + m²c⁴

If one were to replace the energy and momentum with their operators and operate this on a wave function the result would be second order differential equations. In order to get a first order differential equation one could take the square root of the equation if the right hand side could be expressed as a perfect square. As it stands one can not do this, but consider the expression multiplied by the 2x2 identity matrix I of 2x2 identity matrices s₀.

(E - qf)²I = (Pc - qf)²I + m²c⁴I

Now one can hope to find a perfect square for the right hand side in the form

(E - qf)I = F×(Pc - qf) + Gmc²

where the three elements of F are Fⁱ, and where Fⁱ and G are 2x2 martices of 2x2 matrices.

Try inserting

and

where s_i is any one of the three conventional Pauli matrices and verify that this yields

(E - qf)² = (Pc - qf)² + m²c⁴

This results in a hamiltonian of

H_Dirac = F×(P_opc - qf) + Gmc² + qfI

(often multiplied identity matrices are suppressed and identity by itself written as just 1)

and the Dirac equation for a nonzero potential:

_______________________________________________________________________________________ 

3.2 The SR Dynamical Equations 27

In special relativity we define a four-vector force as

F^l = dp^l/dt

(3.2.1)

For a particle with mass we have

p^l = mU^l

The Acceleration Four-Vector for special relativity is given by

A^l = dU^l/dt

(3.2.2)

and so we can write the relativistic version of Newton's second law as

F^l = mA^l

(3.2.3)

28 Chapter 3 SR Dynamic Implications

Considering an inertial frame according to which the test mass is instantaneously at rest it is easy to show that

h_mnA^mUⁿ = 0

(3.2.4)

which yields

h_mnF^mUⁿ = 0

(3.2.5)

This serves as the work energy theorem for modern special relativity.

Consider where h_mnF^mUⁿ = 0 leads:

h_mnF^mUⁿ = 0

h_mn(dp^m/dt)Uⁿ = 0

g²h_mn(dp^m/dt)uⁿ = 0

h_mn(dp^m/dt)uⁿ = 0

(dp⁰/dt)u⁰ - (dp/dt)× u = 0

u⁰ = c and p⁰c = E_R so

(dE_R/dt) - (dp/dt)× u = 0

dE_R = (dp/dt)× dx

dE_R = dE_k so

òdE_K = ò(dp/dt)× dx

W = ò(dp/dt)× dx

If we define another kind of force that is not a four-vector which we will call ordinary force as

f = dp/dt

(3.2.6)

then we arrive at

W = òf×dx

(3.2.7)

and there we see the work energy theorem of pre-modern relativistic physics.

So Newton's second law for special relativity in terms of ordinary force is 3.2.6

f ⁱ = dp ⁱ/dt

Using this force definition has its purposes, but in a lot of ways thinking of relativistic physics in terms of nontensor quantities very much complicates things. For example let us work out the relation between ordinary force and coordinate acceleration.

we can write 3.2.6 as

f ⁱ = m(d/dt)(dx ⁱ/dt)

We then define a^l by

a^l = (d/dt)(dx^l/dt)

(3.2.8a)

Which for acceleration in the direction of motion results in

a = g³a

(3.2.8b)

and for that case of motion a will be equal to the proper acceleration A' which is the acceleration as observed from a frame according to which the particle is instantaneously at rest. The magnitude of this can be calculated from any frame as it is an invariant, |A'| = (-h_mnA^mAⁿ)^1/2. This is the amount of acceleration "felt" by the accelerated observer and according to the inertial frame in which the accelerated or "proper frame" observer is instantaneously at rest a = A'. We define a according to 3.2.8 as well because it is useful. a as calculated from any inertial frame according to which the force is in the direction of the motion turns out to be equal to the proper acceleration.)

This also restores a Newtonian form

f ⁱ = ma ⁱ

 (3.2.9)

( Note also - When the force is in the same direction as the motion, then the force felt by the object being pushed is equal to the ordinary force. In that case we have F' ^l _felt = f ^l = ma^l )

3.2 The SR Dynamical Equations 29

we can eliminate dt, in terms of dt from time dilation

f ⁱ = m(d/dt)(gdx ⁱ/dt)

Use of the chain rule and simplification results in

f = gm[a + g²u(u×a)/c²]

(3.2.10)

where a^l is the coordinate acceleration.

The four-vector force for special relativity is sometimes called the Minkowski force and is related to the electromagnetic field tensor F^mn

(3.2.11)

by

F^l = qh_mn(U^m/c)F^nl

(3.2.12)

From this we can work out the relation between the components of the electromagnetic field, the coordinate velocity and the ordinary force, which yields

f = q(E + u´B)

(3.2.13)

In the case that the force is in the direction of the motion 3.2.10 yields,

f ⁱ = g³maⁱ

(3.2.11)

Note that no matter what finite value the ordinary force is, as u approaches c, g diverges and so the acceleration must vanish.

We expect this as nothing with mass can be pushed all the way up to the speed of light.

We have seen how c is a speed limit for the universe. Because of this, we must answer a question concerning velocity addition. Lets say an S' frame observer observes an object at speed u'. An S

30 Chapter 3 SR Dynamic Implications

frame observer observes the S frame observer to be moving at speed v. u' can be any value less than c and v can be any value less than c. People tend to come to the wrong conclusion that the S frame observer observes that the object moves at a speed u = u' + v and that this speed should therefor be any speed less than 2c. They are using the wrong velocity addition formula. Consider the following Lorentz coordinate transformation equations in differential form.

dx = g(dx' + bdct')

and

dct= g(dct' + bdx')

To obtain the correct velocity addition equation divide equations and simplify.

dx/dct = [g(dx' + bdct')]/[g(dct' + bdx')]

simplified

dx/dt = [dx'/dt' + v]/[1 + (dx'/dt')v/c²]

Now making the replacements u = dx/dt and u' = dx'dt' we arrive at

u = (u' + v)/(1 + u'v/c²)

(3.2.12)

This is the correct equation to use for that velocity addition. u' and v can be any values less than c but the result will always be that the speed of the object according to the S frame, u , will always be less than c. One can also use the same method to find the velocity addition equations for the case that the object moves in the y or z directions.

Problem 3.2.1

Find h_mnU^mUⁿ, h_mnU^mAⁿ, and h_mnU^mFⁿ, h_mnp^mpⁿ using the fact that the contraction of any tensor is an invariant. How is the last related to the definition of mass? How is the result for h_mnU^mFⁿ related to the time derivative of the work energy relation, (dp/dt)×v = dKE/dt?

Problem 3.2.2

Show that the momentum for a charged particle in cyclotron motion or orbital radius R in a magnetic field is given by p = qBR.

Problem 3.2.3

Use Eqn 3.2.10 to show that the motion of a charged particle in an electromagnetic field is given by

a = (q/m)g ^-1[E + u´B - (u/c)(u×E/c)]

Problem 3.2.4

Refer to Eqn 3.2.2 and show that A^m = Uⁿ(U^m,_n)

Problem 3.2.5

Consider an S' frame moving in the x direction with speed v with respect to an S frame.

Show that the ordinary force on a particle moving with velocity u with respect to S transforms between the two frames according to

Problem 3.2.6

Consider the ordinary force as given by the column vector

and the coordinate acceleration given by the column vector

.

Show that according to equation 3.2.10, these can be related by

if we define *matrix mass* by

________________________________________________________________________________________

3.3 Rotations, Rockets, and Frequency Shifts 31

We have shown that velocities do not add linearly in special relativity. For motion along one direction velocities were adding nonlinearly according to Eqn 3.2.12

u = (u' + v)/(1 + u'v/c²)

Rapidity q as a function of v is given by

tanhq = v/c

(3.3.1)

This definition is useful as it simplifies much of dynamics equations. It does this because, unlike velocity, rapidity does add linearly.

q_u = q_u' + q_v

Recall the Lorentz transformation matrix

Rapidity also has the following relations to g and gb

g = coshq

(3.3.2a)

gb = sinhq

From these, the Lorentz transformation matrix becomes

(3.3.2b)

Comparing this to an ordinary rotation matrix makes it clear why Lorentz transformations can be thought of as a rotation in space-time. At this point the relation between Lorentz transformation and rotation may still seem to be a superficial one, but once one becomes familiar with spinor calculus a much more intimate relation is revealed.

32 Chapter 3 SR Dynamic Implications

Here we will derive and discuss the implications of single stage relativistic rocket equations. The non-relativistic rocket equation is

Dv = v_exln(m_i/m)

(3.3.3)

This gives the change in velocity Dv a rocket undergoes accelerating in one direction given a measure of exhaust speed v_ex which is a constant and the initial mass of the rocket m_i and the final mass of the rocket m after some of the ships mass in fuel has been burnt off.

The relativistic version of this equation in terms of rapidity q is similar

Dq = (v_ex/c)ln(m_i/m)

 (3.3.4)

The speed of the rocket is then calculated from the rapidity Eqn 3.3.1.

v = ctanhq

Notice that since tanhq < 1 for any q , v is always less than c no matter how much of the ships mass is burnt off as fuel and no matter how fast the exhaust speed is. We can even consider tachyon exhaust where v_ex > c and yet the rocket still never reaches the speed of light.

 To derive 3.3.4 Start with conservation of momentum and energy relating the initial and final states of the rocket and exhaust for a small element m_fex burned off.

gmv = (m + dm)[gv + d(gv)] + m_fexg_fexv_fex

gmc² = (m + dm)(g + dg)c² + m_fexg_fexc²

Simplified

0 = gvdm + md(gv) + m_fexg_fexv_fex

0 = gdm + mdg + m_fexg_fex

Eliminate m_fexg_fex

0 = gvdm + md(gv) - (gdm + mdg)v_fex

Insert relativistic velocity addition

0 = gvdm + md(gv) - (gdm + mdg)[(v - v_ex)/(1 - vv_ex/c²)]

Simplify

0 = [gvdm + md(gv)] (1 - vv_ex/c²) - (gdm + mdg)(v - v_ex)

Switch variables to rapidity

0 = [sinhqdm + md(sinhq)] [1 - tanhq(v_ex/c)]c - [coshq(dm) + md(coshq)](tanhq - v_ex/c)c

Simplify

0 = [sinhqdm + mcoshqdq] [1 - tanhq(v_ex/c)] - [coshqdm + msinhqdq](tanhq - v_ex/c)

0 = sinhqdm + mcoshqdq - (sinhqdm + mcoshqdq)tanhq(v_ex/c) - (coshqdm + msinhqdq)tanhq + (coshqdm + msinhqdq)(v_ex/c)

0 = sinhqdm + mcoshqdq - (v_ex/c) sinhqtanhqdm - (v_ex/c)msinhqdq - sinhqdm - msinhqtanhqdq + (v_ex/c)coshqdm + (v_ex/c)msinhqdq

0 = mcoshqdq - (v_ex/c) sinhqtanhqdm - msinhqtanhqdq + (v_ex/c)coshqdm

0 = (mcoshq - msinhqtanhq)dq + (v_ex/c)(coshq - sinhqtanhq)dm

0 = m(cosh²q - sinh²q)dq + (v_ex/c)(cosh²q - sinh²q)dm

0 = mdq + (v_ex/c)dm

dq = -(v_ex/c)dm/m

After integration equation 3.3.4 is obtained

Dq = (v_ex/c)ln(m_i/m)

Now consider the ships proper acceleration for motion in one direction refer to equation 3.2.8b

a = g³a = cosh³qdv/dt = cosh³q(dv/dq)(dq/dm)(dm/dt')(dt'/dt)

a = cosh³q(csech²q)(-(v_ex/mc))(dm/dt')sechq

a = (v_ex/m)(dm/dt')

If the proper acceleration is kept constant then integration results in

aDt'/c = (v_ex/c)ln(m_i/m) = Dq

Consider initial conditions of v = 0 at t = t' = 0.

at'/c = q

If the rocket starts at rest and is run at a constant proper acceleration a , then the equation can be written

v = ctanh(at'/c)

(3.3.5)

These initial conditions also result in

v = ctanh[(v_ex/c)ln(m_i/m)]

(3.3.6a)

equivalently

(3.3.6b)

Inverting these results in

m_i/m = exp[(c/v_ex)tanh^-1(v/c)]

(3.3.6c)

equivalently

m_i/m = [g(1 + b)]^c/vex

(3.3.6d)

Running it at a constant proper acceleration also results in

b = tanh(at'/c)

(3.3.7a)

g = cosh(act'/c²)

(3.3.7b)

gb = sinh(act'/c²)

(3.3.7c)

3.3 Rotations, Rockets, and Frequency Shifts 33

Using the Lorentz like transformation equations

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

(3.3.8)

Results in a good global coordinate transformation from the accelerated ship frame to an inertial frame. Take the inertial frame to be the one in which it starts instantaneously at rest at t' = 0 and these become

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

(3.3.9)  

 There is a difference between what frequency one observes as being emitted from a source and what frequency an observer actually sees as coming from the source. This is true even in nonrelativistic physics. For instance, as a car drives past you will hear a shift in the tone of the engine as it goes from coming toward you to going away from you. This is the frequency you hear. You may use the ordinary Doppler shift formula with the speed it was traveling to then extrapolate what frequency it really emits according to your coordinate frame. This is the frequency you observe. The relativistic Doppler shift formula is really the same thing as the ordinary Doppler shift formula except that it is usually written in terms of the source frame's emitted frequency instead of the observed emitted frequency. Its just that in the nonrelativistic case these are the same. In relativistic Doppler shift, you accounts for the fact that due to time dilation the frequency you observe to be emitted is different then the frequency according to the frame of the object.

If you are at rest with respect to the medium of propagation for a wave, then the ordinary Doppler shift formula is

f = f₀/[1 + (v/c)cosq ]

(3.3.10)

In terms of sound we make the following relations.

f is the frequency you hear (for instance if this was sound).

34 Chapter 3 SR Dynamic Implications

f₀ is the frequency you observe have been emitted according to your frame at the time of the emission. It is the transverse or

q = p /2 frequency for f.

v is the speed the emitter travels with respect to the medium of the waves at the time the wave was actually emitted.

c is the speed of the waves of the medium with respect to the medium.

q is the angle it was traveling off of strait away from you at the time the heard frequency was actually emitted.

f is the frequency you see.

f₀ is the frequency you observe to have been emitted according to your frame at the time of the emission. It is the transverse or

q = p /2 frequency for f.

v is the speed the emitter travels with respect your frame at the time the light was actually emitted.

c is the Lorentz invariant vacuum speed of light.

q is the angle it was traveling off of strait away from you at the time the heard frequency was actually emitted according to your frame. <--- The angle is different according to the other frame and so use of the other frames angle changes the form of the equation.

Now to write it in terms of the frequency emitted according to the frame of the object we start by writing the periods according to the two frames in terms of time dilation.

T₀ = T₀'(1 - v²/c²)^-1/2

We then relate frequency to period.

f₀ = 1/T

f₀' = 1/T₀'

 3.3 Rotations, Rockets, and Frequency Shifts 35

Putting these together results in

f₀ = f₀' (1 - v²/c²)^1/2

Inserting this into the Doppler shift formula results in

f = f₀' (1 - v²/c²)^1/2/[1 + (v/c)cosq ]

(3.3.11)

The wavelength of the light will be l = c/f , or...

l = l ₀' [1 + (v/c)cosq ]/(1 - v²/c²)^1/2

(3.3.12)

Next consider the case that the object travels strait toward the observer. q = p . Then after algebraic simplification these becomes

f = f₀ ' [(c + v)/(c - v)]^1/2

(3.3.13)

l = l ₀' [(c - v)/(c + v)]^1/2

If the object traveled strait away from the observer q = 0, it would have become

f = f₀ ' [(c - v)/(c + v)]^1/2

(3.3.14)

l = l ₀' [(c + v)/(c - v)]^1/2

Problem 3.3.1

Consider an ideal matter/antimatter rocket with v_ex = c. What is the mass ratio m_i/m in order to reach (4/5)c? Even if it were technologically possible to make that much, think about how much anti-matter becomes a serious safety concern.

Problem 3.3.2

Consider an ideal matter/antimatter rocket with v_ex = c. What is the mass ratio in order to accomplish the round trip of problem 1.2.4 ?

Problem 3.3.3

What wavelength do you see from a 557nm proper frame source approaching at (5/13)c?

Problem 3.3.4

a. Show that 3.3.6a yields 3.3.6b

b. Show that inverting 3.3.6b yields 3.3.6d

Problem 3.3.5

Show that the relativistic Doppler formula for a source traveling toward an observer equation 3.3.13 can be written as

v = ctanh[ln(f/f₀')]

or

f/f₀' = exp[tanh^-1(v/c)]

and the case of the source traveling away from the observer 3.3.14 can be written

v = ctanh[ln(f₀'/f)]

or

f₀'/f = exp[tanh^-1(v/c)]

With these forms one can punch numbers easier on a scientific calculator.

Problem 3.3.6

Show that the position of the ship undergoing constant proper acceleration considered in equations 3.3.9 is given as a function of proper time by

x = (c²/a)[cosh(at/c) - 1]

a. Use 3.3.9 to show that the wrist watch time or proper time for the ship frame observer t describing events at the location of the ship is related to coordinate time t by

at/c = sinh(at/c)

b. Use the results of "a" and 3.3.5 to show that

v = at/[(1 + a²t²/c²)^1/2]

and look at the limiting behavior of v for both small and large times for both the velocity equation here and 3.3.5.

c. Show that the position of the ship undergoing constant proper acceleration considered in equations 3.3.9 is given as a function of proper time by

x = (c²/a)[cosh(at/c) - 1]

and use the results of "a" to show that

(1 + ax/c²)² - (at/c)² = 1.

Note that this is the equation of a hyperbola.

Problem 3.3.7

A space ship of constant mass is somehow propelled by an external force. The ordinary force on the ship will be constant and will be in the direction of motion so one may use 3.2.11. Use this and 3.2.9 to show that

v = at/[(1 + a²t²/c²)^1/2]

and use this result to show that

(1 + ax/c²)² - (at/c)² = 1.

Then compare the results to problem 3.3.6.