Sect. 7.4: Vectors, etc.A combination of Marion, Jackson & Goldstein!

• Consider 4d Minkowski space: x0 = ct, r = (x1,x2,x3)• Define: β = (v/c), γ = [1 - β2]-½

• We’ve seen the Lorentz Transformation: For v || x this is: x0´ = γ(x0 - βx1), x1´ = γ(x1 - βx0), x2´ = x2, x3´ = x3

– We’ve also seen the general case!

• We’ve also seen how the 3d velocity component transforms. For v || x this is (Also saw general case!):

u1 = [u1´ + v]/[1 + (vu1´)/c2]

ui = c(dxi/dx0) = γ-1(ui´)/[1 + (vu1´)/c2] (i = 2,3) • These expressions give us a start on Relativistic Kinematics.

However, we want to do Relativistic Dynamics! We need to discuss the Relativistic versions of several other quantities: Momentum, Energy, Force, ...

Relativistic MomentumMainly from Marion

• Recall the concept of Proper Time:• S Lab frame, S´ Moving frame.

Time intervals dt measured in the lab frame are different from the time intervals dt´ measured in moving frame. Distinguish them: Time measured in the rest frame of a body (S´if the body moves with v in the lab frame) Proper time τ. Time measured in the lab frame (S) Lab time t. For a body moving with v we showed: dτ γ-1dt where: γ [1 - β2]-½ , β (v/c)

dτ < dt “Time dilation”

• In kinematics & dynamics, we usually can’t go wrong if, instead of taking time derivatives with respect to the lab time t, we take time derivatives with respect to the proper time τ.

• Lets do velocity first. Consider a particle observed moving with velocity u in the lab frame.

• Define: The Relativistic Velocity (a 3d vector!) of the particle:

uR (dr/dτ) = γ(dr/dt) = γu– All observers don’t agree on the value of u = (dr/dt) (we’ve seen

how it transforms!). However, all observers will agree on the value of uR (dr/dτ). dτ = proper time

– Note: Here, γ [1 - β2]-½ , β (u/c). Notice that the particle velocity u in the lab frame enters here. We DISTINGUISH this from the γ entering the Lorentz Transformation from the moving to the lab frame! (where β (v/c).)

• Relativistic Velocity:

uR (dr/dτ) = γ(dr/dt) = γu• The Newtonian definition of momentum is p = mu

A logical generalization of this to Special Relativity is:

• Define: the Relativistic Momentum (a 3d vector!) of the particle: p muR = γmu [= m(dr/dτ)]

Writing γ out explicitly this is

p = (mu)[1 - (u2/c2)]-½ This reduces to the classical result for u2/c2 << 1

• Relativistic Momentum (a 3d vector!) of the particle: p = γmu [= m(dr/dτ)] = (mu)[1 - (u2/c2)]-½ (1)• Consider the particle mass m. In old fashioned Relativity treatments: At

speeds u c the mass was u dependent & increased with u. They talked of u dependent “Relativistic” mass m & “rest mass” m0. Then, (1) was written using the classical definition as p = mu, but m was Relativistic & had the form:

m = m0[1 - (u2/c2)]-½ • It’s conventional now to consider the mass m as the same as it is in

Newtonian mechanics; as an invariant, intrinsic property of a body. So, the mass m in (1) is same as Newtonian mass and it is the MOMENTUM which is Relativistic!

• Using the definition (1) of Relativistic Momentum, it can be shown (Ch. 11, Marion) that momentum is conserved in a collision between 2 objects!

Relativistic Energy• Kinetic Energy: We define Relativistic Kinetic Energy T (single

particle) using work, just as in the Newtonian mechanics of Ch. 1:

• If the particle is acted on by a total external force F, the Work done on the particle in moving it from position 1 to position 2 in space is defined as the line integral (ds = differential path length)

(assume mass m = constant)

W12 ∫Fds (limits: from 1 to 2)• We also have the work-energy principle, which tells us that

W12 ∫Fds T2 - T1 = T (2)– We haven’t specified the form of T yet! It will not have classical form!

• For the force F in (2), insert Newton’s 2nd Law (defer a detailed discussion of Relativistic forces to later). Assume that, in the lab frame, we still have: F = (dp/dt). However, in this use the Relativistic Momentum p = γmu instead of the classical one.

W12 ∫Fds T2 - T1 = T (2)F = (dp/dt), p = γmu

F = [d(γmu)/dt] (3)• For simplicity, evaluate (2) using (3) & assuming the particle starts

from rest (T1 = 0) & that u is parallel to F. ds = udt

W ( W12) = T ( T2) = ∫[d(γmu)/dt]udt = m∫ud(γu)Limits u = 0 u γ [1 - β2]-½

• Integrate by parts: T = γmu2 - m∫u(du)[1 - (u2/c2)]-½ = γmu2 + mc2[1 - (u2/c2)]½ - mc2

Finally, do some algebra to get:

The Relativistic Kinetic Energy T = γmc2 - mc2 = (γ - 1) mc2 = {[1 - (u2/c2)]-½ -1}mc2

• Relativistic KE T = γmc2 - mc2 = (γ - 1) mc2 = {[1 - (u2/c2)]-½ -1}mc2 (4)

• Note: Given the classical Newtonian KE of (½)mu2, a “guess” for T might have been (½)mu2 or, perhaps (following the example of relativistic momentum) (½)γmu2. But T in (4) is like neither of these! (or perhaps (½)γ2mu2 ?).

• Student exercise: Prove from (4) that in the limit (u2/c2) << 1, the classical result is obtained: T (½)mu2

• Rest Energy: Define the 2nd term in (4) as the rest energy E0 of the particle: E0 mc2 (5)

• It’s convenient to rewrite (4) as: γmc2 = T + mc2 = T + E0

• Now define the Total Energy of the particle as:

E γmc2 = T + E0 (6)

• Relativistic KE: T = γmc2 - mc2 (4)• Rest energy: E0 mc2 (5)• Total Energy: E γmc2 = T + E0 (6)• Note that when u = 0, γ = 1, T = 0 & E = E0

• (4), (5), (6) are the origins of the famous Einstein result that mass & energy are equivalent. The well-known (& correct) interpretation: mass is another form of energy.

We must combine the laws of energy conservation & mass conservation into a single conservation law: Equation (6)

• All of this is well-verified experimentally, of course, in the nucleus, where mass of the constituent particles is converted to energy that binds them together. If the constituents break apart, a HUGE amount of energy can be released!

• An alternate formulation: Quite useful for doing collision & other kinematic problems. Bear in mind definitions: Rest energy: E0 mc2, Total Energy: E γmc2

• Philosophical remark: Physicists generally believe that momentum is (in some sense) a more fundamental concept than kinetic energy. For example, there is no general law of conservation of kinetic energy & there is for conservation of momentum.

Search for a relation between mass & energy that includes momentum rather than kinetic energy!

• Start with Relativistic Momentum: p = γmu • Square this, multiply by c2 & manipulate: p2c2 = γ2m2u2 c2 = γ2m2c4(u2/c2) = γ2m2c4β2 (a)

Now, γ [1 - β2]-½ , so β2 = 1 - γ-2

(a) becomes: p2c2 = γ2m2c4(1 - γ-2) = γ2m2c4 - m2c4 (b)

• Definitions: Rest energy: E0 mc2,

Total Energy: E γ mc2

(b) becomes: p2c2 = E2 - (E0)2 (c)

• Write (c) as The Energy-Momentum conservation law:

E2 = p2c2 + (E0)2 (6´)• (6´): A very useful kinematic relationship between the particle momentum,

total energy, & rest energy. Used often in relativistic collision problems. • Note: A photon has no mass, so for a photon, (6´) becomes E = pc

4-VectorsSome combination of Jackson & Marion!

• Recall the discussion of 4d Minkowski space & the Lorentz Transformation. The position coordinates in this 4d space:

x0 ct, x1 x, x2 y, x3 z or: (x0,x1,x2,x3)• An invariant is the 4d “distance” (from say, the origin):

s2 (ct)2 - x2 - y2 - z2 = (x0)2 - (x1)2 - (x2)2 - (x3)2 (s´)2 • Note the minus signs in front of the spatial parts! These are necessary

to preserve causality in the resulting equations. However, they make Minkowski space non-Euclidean!

• Reminder: The Lorentz Transformation between inertial frames (for v || x) is: (x´ = Lx). Leaves s2 invariant!

x0´ = γ(x0 - βx1), x1´ = γ(x1 - βx0), x2´ = x2, x3´ = x3 • General orientation of v, we can write some previous expressions as (using r

= (x1,x2,x3)): x0´ = γ(x0 - βr),

x||´ = γ(x|| - βx0), x´ = x , ( || & are orientations relative to v)

• Mathematically, we can define vectors & tensors in 4d Minkowski spacetime in analogy with vectors & tensors in ordinary 3d space. There are actually 2 kinds of vectors & tensors in this space: Covariant & Contravariant, as discussed in detail in Goldstein & Jackson. To understand Goldstein’s discussion, you need the distinction between them. Also need the metric tensor, etc. However, to understand the essence of vectors in Minkowski space, and to get at the PHYSICS of Relativity, we needn’t make the distinction now. Hold off on this until a little later.

• In 3d space, we define a vector as a set of 3 numbers

A = (A1,A2,A3) which has specified transformation properties under rotations R: A´ = RA. In particular, the 3 components of A transform with R the same way as the 3 position coordinates r = (x1,x2,x3) transform. Many books talk about this in detail.

• In analogy to this, in 4d Minkowski spacetime, we DEFINE a 4-VECTOR as a set of 4 numbers

AA = (A0,A1,A2,A3) (A0,A) (A = (A1,A2,A3)) which has specified transformation properties under the Lorentz transformation L: A´ = LA. That is, the 4 components of A transform with L in the same way that the 4 position coordinates (x0,x1,x2,x3) transform. A 4-vector A behaves under L in the same way that we’ve discussed for (x0,x1,x2,x3)

• Explicitly, DEFINE a 4-VECTOR as a set of 4 numbers A = (A0,A1,A2,A3) (A0,A) which transforms under the Lorentz transformation L:

A´ = LA as (for v || x):

A0´ = γ(A0 - βA1), A1´ = γ(A1 - βA0)

A2´ = A2, A3´ = A3 • For a general orientation of v, this becomes:

A0´ = γ(A0 - βA), A||´ = γ(A|| - βA0)

A´ = A , ( || & : A orientation relative to v)

• 4-VECTOR = 4 numbers A = (A0,A1,A2,A3) (A0,A) which transform under Lorentz transform L: A´ = LA in the same way as the position coordinates (x0,x1,x2,x3)

• Now PHYSICS with 4-Vectors!

• Define the 4-Position: (x0, ct)

X (x0,x1,x2,x3) (x0,r)• Define the 4-Velocity as the derivative with respect to the Proper

Time of the 4-Position:– Recall the Proper Time τ. For a body moving with velocity u:

dτ dt(γ-1) where: γ [1 - β2]-½, β (u/c)

V (dX/dτ) = [(dx0/dτ),(dr/dτ)]

Note that: (dx0/dτ) = c(dt/dτ) = γc

(dr/dτ) uR = γu = Relativistic (3 vector) velocity

4-Velocity: γ [1 - β2]-½, β (u/c)

V (dX/dτ) = γ (c,u) = (v0,uR) (1)

where: v0 γc

• Define the 4-Momentum as the mass times the 4-Velocity: (use (1))

P mV = γ(mc,mu) = (p0,γmu) = (p0,p) (2)

where: p = γmu = Relativistic (3-vector) momentum.

p0 γmc. Total Energy: E γmc2, so we have: p0 (E/c)

4-Momentum: P mV = [(E/c),p] (3)with: p = γmu, E = γmc2

In 4d spacetime, momentum & energy are the spacelike (p) & timelike (p0) components of the 4-Momentum!

4-Position: X = (x0,r) = (ct,r)

4-Velocity: V = (v0,uR) = γ(c,u)

4-Momentum: P = (p0,p) = γm(c,u) = [(E/c),p]• Recall that each of these transforms

under a Lorentz transformation identically to X. 2 inertial frames, S Lab frame, S´ moving frame. For simplicity, let the velocity v of S´ be || x.

• Consider a particle of mass m moving in S with velocity u. So v u. We must distinguish between

γv [1 - (βv)2]-½, β v (v/c) & γu [1 - (βu)2]-½, βu (u/c)

4-Position: X = (x0,r) = (ct,r)

4-Velocity: V = (v0,uR) = γu(c,u)

= γu(c,u1,u2,u3) • Transformation of the components of X from S to S´:

x0´ = γv(x0 - βv x1), x1´ = γv(x1 - βvx0), x2´ = x2, x3´ = x3

or: ct´ = γv(ct - βvx1), x1´ = γv(x1 - βvx0), x2´ = x2, x3´ = x3 • Transformation of components of V from S to S´:

v0´= γv(v0 - βvuR1), uR1´= γv(uR1- βvv0), uR2´ = uR2, uR3´ = uR3

or: γu´c = γvγu(c - βvu1), γu´u1´ = γvγu(u1 - βvc)

u2´ = u2, u3´ = u3

These are equivalent (again) to the Einstein velocity addition formula! Notice the product γvγu! Also, first eqtn gives how γu itself transforms! Note: SOMETIMES, u = v γv= γu

4-Momentum: P = (p0,p) =

γum(c,u) = [(E/c),p] = [(E/c),p1,p2,p3]

= [(E/c),γumu1,γumu2,γumu3] • Transformation of components of P from S to S´:

p0´ = γv(p0 - βvp1), p1´ = γv(p1- βvp0)

p2´ = p2, p3´ = p3

or: E´ = γv(E - βvp1), p1´ = γv[p1 - (βv/c)E]

p2´ = p2, p3´ = p3

These relations are particularly useful in discussing Relativistic particle collisions. Note that the definitions pi = γumui bring in the product γvγu!

• Emphasize: By definition, all 4-Vectors transform under a Lorentz transformation in the same way as the 4-Position X. By its design, this transformation preserves, or leaves invariant, the square of the magnitude of X or the 4-distance from the origin (note again the minus signs!):

X2 = (x0)2 - (x1)2 - (x2)2 - (x3)2 (x0´)2 - (x1´)2 - (x2´)2 - (x3´)2 = (X´)2

It is also true that a Lorentz transformation preserves, or leaves invariant, the magnitude of ALL other 4-Vectors.

• Example: 4-Velocity V = (v0,uR) = (v0,v1,v2,v3) = γu(c,u) (it’s a student exercise to show):

V2 = (v0)2 - (v1)2 - (v2)2 - (v3)2 (v0´)2 - (v1´)2 - (v2´)2 - (v3´)2 = (V´)2

• Example: 4-Momentum P = (p0,p) = [(E/c),p1,p2,p3]= [(E/c),γumu] (student exercise). Gives another useful expression!

P2 = (E/c)2 - (p1)2 - (p2)2 - (p3)2

(E´/c)2 - (p1´)2 - (p2´)2 - (p3´)2 = (P ´)2

• With these expressions we can do Relativistic kinematics. Particle collisions, etc. We will briefly do some of this. What about Relativistic Dynamics?

We need to discuss the Relativistic version of Force! • Unfortunately, this is not so easy! A first point is that, Special Relativity

is for inertial frames only. This excludes gravitational fields explicitly. We need General Relativity for that. Second, the exact form of the 4-Force in relativity depends explicitly on WHICH OF THE 4 FUNDAMENTAL FORCES of nature we are dealing with & how they transform under a Lorentz Transformation. In fact, the exact form is only given in the texts (Goldstein, Jackson) for the electromagnetic force & only after a detailed discussion of how the E & B fields transform under a Lorentz transformation. This is more properly left for an E&M course (E&M II).

• Thinking of Newton’s 2nd Law, F = ma, it might be natural to first ask about form of the 4-Vector acceleration the or 4-Acceleration.

• Start with the 4-Velocity: V = (v0,uR) = γu(c,u)

• Define the 4-Acceleration as the derivative with respect to the Proper Time of the 4-Velocity:

A (dV/dτ) = [(dv0/dτ),(duR/dτ)] = (A0,AR)

Here, A0 = (dv0/dτ) = c[d(γu)/dτ] = cγu[d(γu)/dt]

AR = (duR/dτ) = [d(γuu)/dτ] = γu[d(γuu)/dt]

= γuu[d(γu)/dt] + (γu)2 aWhere we’ve defined:

(du/dt) a = The ordinary Newtonian acceleration

Note: γu [1 - (u/c)2]-½ [d(γu)/dt] = (γu)3 (|u||a|/c2)

Note also that βu = (u/c)

• So, the 4-Acceleration has the (messy!) form:

A (A0,AR)

With, A0 = (γu)4βu|a|, AR = (γu)4(βu)2|a|(u/u) + (γu)2a• It turns out that the 4-Acceleration has limited use.• To talk about forces, lets start with an analogy to Newton’s 2nd Law in

the form F = (dp/dt)

• Define the 4-Force (or the Minkowski Force) as the derivative with respect to the Proper Time of the 4-Momentum:

K (dP/dτ) = [(dp0/dτ),(dp/dτ)] (K0,K)Here: K0 = (dp0/dτ) = (γu/c)(dE/dt) (p = γumu)Note that: (dE/dt) = Energy loss or gain rate (power).

Also: K = (dp/dτ) = γu(dp/dt).

• 4-Force (or Minkowski Force):

K (dP/dτ) = [(dp0/dτ),(dp/dτ)] (K0,K)

K0 = (dp0/dτ) = (γu/c)(dE/dt), K = (dp/dτ) = γu(dp/dt)

We want to write γu(dp/dt) γuF, with F = (dp/dt) = Newtonian force. Also remember, p = γumu = the (3d) relativistic momentum. So K = γu(dp/dt) can be written in terms of the acceleration just discussed.

• Goldstein, however, claims that the exact form of the force

K = γu (dp/dt) depends on WHICH OF THE 4 FUNDAMENTAL FORCES of nature we are dealing with & how they transform under a Lorentz Transformation. For E&M forces, after a detailed discussion of the field transformations, K is shown to be the Relativistic form of the usual Lorentz force. Personal opinion: I think he is talking Philosophically here, rather than Physically. I see nothing wrong with defining the Relativistic (3d) force as F = (dp/dt).