Chapter 7: Section 7.4: Vectors. More from Marion

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

• Consider 4d Minkowski space x0 ct, r
  (x1,x2,x3)
• Define ß (v/c), ? 1 - ß2-½
• Weve seen the Lorentz Transformation For v x
  this is
• x0 ?(x0 - ßx1), x1 ?(x1 - ßx0), x2
  x2, x3 x3
• Weve also seen the general case!
• Weve 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) ?-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, ...

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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 (Sif
  the body moves with v in the lab frame) ? Proper
  time ? t. Time measured in the lab frame (S) ?
  Lab time ? t. For a body moving with v we showed
  dt ? ?-1dt where ? ? 1 - ß2-½ , ß ? (v/c)
• ? dt lt dt ? Time dilation

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• In kinematics dynamics, we usually cant 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 t.
• 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/dt) ?(dr/dt) ?u
• All observers dont agree on the value of u
  (dr/dt) (weve seen how it transforms!). However,
  all observers will agree on the value of uR ?
  (dr/dt). dt proper time
• Note Here, ? ? 1 - ß2-½ , ß ? (u/c). Notice
  that the particle velocity u in the lab frame
  enters here. We DISTINGUISH this from the ?
  entering the Lorentz Transformation from the
  moving to the lab frame! (where ß ? (v/c).)

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• Relativistic Velocity
• uR ? (dr/dt) ?(dr/dt) ?u
• 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 ?mu m(dr/dt)
• Writing ? out explicitly this is
• p (mu)1 - (u2/c2)-½
• This reduces to the classical result for u2/c2
  ltlt 1

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• Relativistic Momentum (a 3d vector!) of the
  particle
• p ?mu m(dr/dt) (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 m01 - (u2/c2)-½
• Its 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!

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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 ? ?F?ds (limits from 1 to
  2)
• We also have the work-energy principle, which
  tells us that
• W12 ? ?F?ds ? T2 - T1 ?T
  (2)
• We havent specified the form of T yet! It will
  not have classical form!
• For the force F in (2), insert Newtons 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 ?mu instead of the
  classical one.

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• W12 ? ?F?ds ? T2 - T1 ?T
  (2)
• F (dp/dt), p ?mu
• ? F d(?mu)/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(?mu)/dt?udt
  m?ud(?u)
• Limits u 0 ? u ?
  ? 1 - ß2-½
• Integrate by parts T ?mu2 - m?u(du)1 -
  (u2/c2)-½
• ?mu2 mc21 - (u2/c2)½ - mc2
• Finally, do some algebra to get
• The Relativistic Kinetic Energy
• T ?mc2 - mc2 (? - 1) mc2 1 - (u2/c2)-½
  -1mc2

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• Relativistic KE T ?mc2 - mc2 (? - 1) mc2
• 1 - (u2/c2)-½ -1mc2 (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)?mu2. But T in (4) is like neither
  of these! (or perhaps (½)?2mu2 ?).
• Student exercise Prove from (4) that in the
  limit (u2/c2) ltlt 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)
• Its convenient to rewrite (4) as ?mc2 T
  mc2 T E0
• Now define the Total Energy of the particle as
• E ? ?mc2 T E0
  (6)

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• Relativistic KE T ?mc2 - mc2 (4)
• Rest energy E0 ? mc2
  (5)
• Total Energy E ? ?mc2 T E0
  (6)
• Note that when u 0, ? 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!

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

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• Start with Relativistic Momentum p ?mu
• Square this, multiply by c2 manipulate
• p2c2 ?2m2u2 c2 ?2m2c4(u2/c2)
  ?2m2c4ß2 (a)
• Now, ? ? 1 - ß2-½ , so ß2 1 - ?-2
• ? (a) becomes p2c2 ?2m2c4(1 - ?-2) ?2m2c4 -
  m2c4 (b)
• Definitions Rest energy E0 ?mc2,
• Total Energy E ? ? 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

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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 L?x).
  Leaves s2 invariant!
• x0 ?(x0 - ßx1), x1 ?(x1 - ßx0),
  x2 x2, x3 x3
• General orientation of v, we can write some
  previous expressions as (using r (x1,x2,x3))
  x0 ?(x0 - ß?r),
• x ?(x - ßx0), x? x? , ( ? are
  orientations relative to v)

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• 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
  Goldsteins 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 neednt make the distinction now.
  Hold off on this until a little later.

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• 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
  R?A. 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
• A (A0,A1,A2,A3) ? (A0,A) (A (A1,A2,A3))
  which has specified transformation properties
  under the Lorentz transformation L A L?A.
  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 weve discussed for
  (x0,x1,x2,x3)

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• 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 L?A as (for v x)
• A0 ?(A0 - ßA1), A1 ?(A1 - ßA0)
• A2 A2, A3 A3
• For a general orientation of v, this becomes
• A0 ?(A0 - ß?A), A ?(A - ßA0)
• A? A? , ( ? A orientation
  relative to v)

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• 4-VECTOR 4 numbers A (A0,A1,A2,A3) ? (A0,A)
  which transform under Lorentz transform L A
  L?A 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 ? t. For a body moving
  with velocity u
• dt ? dt(?-1) where ? ? 1 - ß2-½, ß ? (u/c)
• V ? (dX/dt) (dx0/dt),(dr/dt)
• Note that (dx0/dt) c(dt/dt) ?c
• (dr/dt) ? uR ?u Relativistic (3 vector)
  velocity

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• ? 4-Velocity ? ? 1 - ß2-½, ß ? (u/c)
• V ? (dX/dt) ? (c,u) (v0,uR)
  (1)
• where v0 ? ?c
• Define the 4-Momentum as the mass times the
  4-Velocity (use (1))
• P ? mV ?(mc,mu) (p0,?mu)
  (p0,p) (2)
• where p ?mu Relativistic (3-vector)
  momentum.
• p0 ? ?mc. Total Energy E ? ?mc2, so we
  have p0 ? (E/c)
• ? 4-Momentum P ? mV (E/c),p
  (3)
• with p ?mu, E ?mc2
• ? In 4d spacetime, momentum energy are the
  spacelike (p) timelike (p0) components of the
  4-Momentum!

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• 4-Position X (x0,r) (ct,r)
• 4-Velocity V (v0,uR) ?(c,u)
• 4-Momentum P (p0,p)
• ?m(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
• ?v ? 1 - (ßv)2-½, ß v ? (v/c) ?u ?
  1 - (ßu)2-½, ßu ? (u/c)

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• 4-Position X (x0,r) (ct,r)
• 4-Velocity V (v0,uR) ?u(c,u)
• ?u(c,u1,u2,u3)
• Transformation of the components of X from S to
  S
• x0 ?v(x0 - ßv x1), x1 ?v(x1 -
  ßvx0), x2 x2, x3 x3
• or ct ?v(ct - ßvx1), x1 ?v(x1 -
  ßvx0), x2 x2, x3 x3
• Transformation of components of V from S to S
• v0 ?v(v0 - ßvuR1), uR1 ?v(uR1- ßvv0),
  uR2 uR2, uR3 uR3
• or ?uc ?v?u(c - ßvu1), ?uu1
  ?v?u(u1 - ßvc)
• u2 u2, u3 u3
• These are equivalent (again) to the Einstein
  velocity addition formula! Notice the product
  ?v?u! Also, first eqtn gives how ?u itself
  transforms! Note SOMETIMES, u v ? ?v ?u

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• 4-Momentum P (p0,p)
• ?um(c,u) (E/c),p (E/c),p1,p2,p3
• (E/c),?umu1,?umu2,?umu3
• Transformation of components of P from S to S
• p0 ?v(p0 - ßvp1), p1 ?v(p1- ßvp0)
• p2 p2, p3 p3
• or E ?v(E - ßvp1), p1 ?vp1 - (ßv/c)E
• p2 p2, p3 p3
• These relations are particularly useful in
  discussing Relativistic particle collisions. Note
  that the definitions pi ?umui bring in the
  product ?v?u!

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• 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)
  ?u(c,u) (its 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),?umu (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

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• 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 EM course (EM II).

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• Thinking of Newtons 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)
  ?u(c,u)
• Define the 4-Acceleration as the derivative with
  respect to the Proper Time of the 4-Velocity
• A ? (dV/dt) (dv0/dt),(duR/dt) (A0,AR)
• Here, A0 (dv0/dt) cd(?u)/dt
  c?ud(?u)/dt
• AR (duR/dt) d(?uu)/dt ?ud(?uu)/dt
• ?uud(?u)/dt (?u)2 a
• Where weve defined
• (du/dt) ? a The ordinary Newtonian
  acceleration
• Note ?u ? 1 - (u/c)2-½ ? d(?u)/dt
  (?u)3 (ua/c2)
• Note also that ßu (u/c)

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• So, the 4-Acceleration has the (messy!) form
• A ? (A0,AR)
• With, A0 (?u)4ßua, AR (?u)4(ßu)2a(u/u)
  (?u)2a
• It turns out that the 4-Acceleration has limited
  use.
• To talk about forces, lets start with an analogy
  to Newtons 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/dt) (dp0/dt),(dp/dt) ? (K0,K)
• Here K0 (dp0/dt) (?u/c)(dE/dt) (p
  ?umu)
• Note that (dE/dt) Energy loss or gain rate
  (power).
• Also K (dp/dt) ?u(dp/dt).

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• 4-Force (or Minkowski Force)
• K ? (dP/dt) (dp0/dt),(dp/dt) ? (K0,K)
• K0 (dp0/dt) (?u/c)(dE/dt), K (dp/dt)
  ?u(dp/dt)
• We want to write ?u(dp/dt) ? ?uF, with F
  (dp/dt) Newtonian force. Also remember, p
  ?umu the (3d) relativistic momentum. So K
  ?u(dp/dt) can be written in terms of the
  acceleration just discussed.
• Goldstein, however, claims that the exact form of
  the force
• K ?u (dp/dt) depends on WHICH OF THE 4
  FUNDAMENTAL FORCES of nature we are dealing with
  how they transform under a Lorentz
  Transformation. For EM 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).