Relativistic Momentum

10 Key excerpts on "Relativistic Momentum"

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  An Introduction to Mechanics

  • Daniel Kleppner, Robert Kolenkow(Authors)
  • 2013(Publication Date)
  • Cambridge University Press

    (Publisher)

  RELATIVISTIC DYNAMICS 13 13.1 Introduction 478 13.2 Relativistic Momentum 478 13.3 Relativistic Energy 481 13.4 How Relativistic Energy and Momentum are Related 487 13.5 The Photon: A Massless Particle 488 13.6 How Einstein Derived E = mc 2 498 Problems 499 478 RELATIVISTIC DYNAMICS 13.1 Introduction In Chapter 12 we saw how the postulates of special relativity lead to new kinematical relations for space and time. These relations can naturally be expected to have important implications for dynamics, particularly for the meaning of momentum and energy. In this chapter we examine the modifications to the Newtonian concepts of momentum and energy required by special relativity. The underlying strategy is to ensure that momentum and energy in an isolated system continue to be conserved. This approach is often used in extending the frontiers of physics: by reformulating conservation laws so that they are preserved in new situa-tions, we are led to generalizations of familiar concepts. We can also be led to the discovery of unfamiliar concepts, for instance the concept of massless particles that can nevertheless carry energy and momentum. x y A B 13.2 Relativistic Momentum To investigate the nature of momentum in special relativity, consider a glancing elastic collision between two identical particles A and B in an isolated system. We want the total momentum of the system to be con-served, as it is in non-relativistic physics. We shall view the collision in two frames: A ’s frame, the frame moving along the x axis with A so that A is at rest while B approaches along the x direction with speed V , and then in B ’s frame, which is moving with B in the opposite direction so that B is at rest and A is approaching. (The term “frame” is used synony-mously with “reference system.”) We take the collisions to be completely symmetrical. Each particle has the same y speed u 0 in its own frame be-fore the collision, as shown in the sketches.

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  Introduction to Electrodynamics

  • David J. Griffiths(Author)
  • 2017(Publication Date)
  • Cambridge University Press

    (Publisher)

  13 Relativistic Momentum is the spatial part of a 4-vector, p μ ≡ mη μ , (12.47) 13 Older treatments introduce the so-called relativistic mass, m r ≡ m/  1 − u 2 /c 2 , so p can be writ- ten as m r u, but this unhelpful extra terminology has gone the way of the two-dollar bill. 536 Chapter 12 Electrodynamics and Relativity and it is natural to ask what the temporal component, p 0 = mη 0 = mc  1 − u 2 /c 2 (12.48) represents. Einstein identified p 0 c as relativistic energy: E ≡ mc 2  1 − u 2 /c 2 ; (12.49) p μ is called the energy-momentum 4-vector (or the momentum 4-vector, for short). Notice that the relativistic energy is nonzero even when the object is stationary; we call this rest energy: E rest ≡ mc 2 . (12.50) The remainder, which is attributable to the motion, is kinetic energy E kin ≡ E − mc 2 = mc 2  1  1 − u 2 /c 2 − 1  . (12.51) In the nonrelativistic régime (u  c) the square root can be expanded in powers of u 2 /c 2 , giving E kin = 1 2 mu 2 + 3 8 mu 4 c 2 + · · · ; (12.52) the leading term reproduces the classical formula. So far, this is all just notation. The physics resides in the experimental fact that E and p, as defined by Eqs. 12.46 and 12.49, are conserved: In every closed 14 system, the total relativistic energy and momen- tum are conserved. Mass is not conserved—a fact that has been painfully familiar to everyone since 1945 (though the so-called “conversion of mass into energy” is really a conversion of rest energy into kinetic energy). Note the distinction between an invariant quantity (same value in all inertial systems) and a conserved quantity (same value before and after some process). Mass is invariant but not conserved; energy is conserved but not invariant; electric charge is both conserved and invariant; velocity is neither conserved nor invariant.

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  Introduction to Mass in Physics

  • (Author)
  • 2014(Publication Date)
  • Learning Press

    (Publisher)

  When the relative velocity is zero, γ is simply equal to 1, and the relativistic mass is reduced to the rest mass as one can see in the next two equations below. As the velocity increases toward the speed of light c , the denominator of the right side approaches zero, and consequently γ approaches infinity. In the formula for momentum the mass that occurs is the relativistic mass. In other words, the relativistic mass is the proportionality constant between the velocity and the momentum. Newton's second law remains valid in the form the derived form is not valid because in is generally not a constant (see the section above on transverse and longitudinal mass). Modern view The invariant mass is the ratio of four-momentum to four-velocity: and is also the ratio of four-acceleration to four-force when the rest mass is constant. The four-dimensional form of Newton's second law is: Many contemporary authors such as Taylor and Wheeler avoid using the concept of relativistic mass altogether: The concept of relativistic mass is subject to misunderstanding. That's why we don't use it. First, it applies the name mass - belonging to the magnitude of a 4-vector - to a ________________________ WORLD TECHNOLOGIES ________________________ very different concept, the time component of a 4-vector. Second, it makes increase of energy of an object with velocity or momentum appear to be connected with some change in internal structure of the object. In reality, the increase of energy with velocity originates not in the object but in the geometric properties of spacetime itself. The relativistic energy-momentum equation Dependency between the rest mass and E , given in 4-momentum ( p 0 , p 1 ) coordinates; p 0 c = E The relativistic expressions for E and p obey the relativistic energy-momentum equation : where the m is the rest mass. The equation is also valid for photons, which have m=0:

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  A Student's Guide to Special Relativity

  • Norman Gray(Author)
  • 2022(Publication Date)
  • Cambridge University Press

    (Publisher)

  7 Dynamics In the previous chapter, we have learned how to describe motion; we now want to explain it. In newtonian mechanics, we do this by defining quantities such as momentum, energy, force and so on. To what extent can we do this in the context of relativity, with our new 4-vector tools? Aims: you should: 7.1. understand relativistic energy and momentum, the concept of energy- momentum as the magnitude of the momentum 4-vector, and conser- vation of the momentum 4-vector; and 7.2. understand the distinction between invariant, conserved and constant quantities. 7.1 Energy and Momentum We can start with momentum. We know that in newtonian mechanics, momentum is defined as mass times velocity. We have a velocity, so we can try defining a momentum 4-vector, for a particle of mass  moving at velocity , as  =  = (1, ). (7.1) Since  is a scalar, and  is a 4-vector,  must be a 4-vector also. Remember also that  is a function of the 3-vector : (). In the rest frame of the particle, this becomes  = (1, ): it is a 4-vector whose length ( √  ⋅ ) is , and which points along the particle’s worldline. That is, it points in the direction of the particle’s movement in spacetime. Since this is a vector, its magnitude and its direction are frame-independent 125 126 7 Dynamics P 1 P 2 P 3 P 4 Figure 7.1 The momenta involved in a collision. quantities, so a particle’s 4-momentum vector always points in the direction of the particle’s worldline, and the 4-momentum vector’s length is always . We’ll call this vector the momentum (4-)vector, but it’s also called the 4- momentum or the energy-momentum vector, and Taylor & Wheeler (coining the word in an excellent chapter on it) call it the momenergy vector in order to stress that it is not the same thing as the energy or momentum (or mass) that you are used to. Note that here, and throughout, the symbol  denotes the mass as measured in a particle’s rest frame.

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  Classical Field Theory

  • Joel Franklin(Author)
  • 2017(Publication Date)
  • Cambridge University Press

    (Publisher)

  In between the two regimes, the period depends on both m k and a in some complicated manner (see [7] for details). v (in units of c) v (in units of c) v (in units of c) x x x Fig. 1.13 v(x ) from (1.67) for a = 1, 2, and 3 (left to right, in units of m/k c). Notice that for “large a”(a = 3) the speed v ∼ c even for x close to ±a. 22 Special Relativity a −a t x(t) Fig. 1.14 The cosine is associated with non-relativistic harmonic motion, while the straight line represents the motion of the mass moving in the ultra-relativistic regime. Problem 1.24 Rewrite the relativistic energy of a free particle: E = m c 2 1 − v 2 c 2 (1.69) in terms of Relativistic Momentum (i.e., solve p = m v 1− v 2 c 2 for v in terms of p and insert in E). Problem 1.25 If you take the derivative on the left in Newton’s second law: d dt ⎛ ⎝ m v(t) 1 − v(t) 2 c 2 ⎞ ⎠ = F, (1.70) show that you get m d 2 x(t) dt 2 = 1 − v(t) 2 c 2 3/2 F. (1.71) We can gain some (special) relativistic intuition by thinking of the right-hand-side as an “effective” force (notice that as v → ±c, the effective force goes to zero) for the “usual” form of Newton’s second law. Problem 1.26 Take the time derivative of H = m c 2 1 − v(t) 2 c 2 + U(x(t)) (1.72) and show that you recover Newton’s second law (i.e., dp dt = − dU dx where p is the Relativistic Momentum). Problem 1.27 For a positive charge q pinned at the origin and a negative charge, −q, with mass m that starts from rest a distance a away, use conservation of energy to find v(r) 23 Energy and Momentum (r is the distance to the origin for the negative charge) and sketch |v(r)| to establish that v < c for all locations as the negative charge falls toward the positive one. 1.7.1 Energy–Momentum Transformation The momentum of a particle moving with constant speed v in the ˆ x-direction is p = m v 1 − v 2 c 2 = m v γ , (1.73) and its energy is 10 E = m c 2 γ .

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  Modern Physics

  • Kenneth S. Krane(Author)
  • 2020(Publication Date)
  • Wiley

    (Publisher)

  The momentum is obtained from this result by dividing by the symbol c (not its numerical value), which gives p = 1580 MeV∕c The units of MeV∕c for momentum are often used in rel- ativistic calculations because, as we show later, the quan- tity pc often appears in these calculations. You should be able to convert MeV∕c to kg ⋅ m∕s and show that the two results obtained for p are equivalent. Relativistic Kinetic Energy Like the classical definition of momentum, the classical definition of kinetic energy also causes difficulties when we try to compare the interpretations of 2.7 Relativistic Dynamics 51 different observers. According to O ′ , the initial and final kinetic energies in the collision shown in Figure 2.21a are K ′ i = 1 2 m 1 v ′2 1i + 1 2 m 2 v ′2 2i = (0.5)(2m)(0) 2 + (0.5)(m)(−0.750c) 2 = 0.281mc 2 K ′ f = 1 2 m 1 v ′2 1f + 1 2 m 2 v ′2 2f = (0.5)(2m)(−0.500c) 2 + (0.5)(m)(0.250c) 2 = 0.281mc 2 and so energy is conserved according to O ′ . The initial and final kinetic ener- gies observed from the reference frame of O (as in Figure 2.21b) are K i = 1 2 m 1 v 2 1i + 1 2 m 2 v 2 2i = (0.5)(2m)(0.550c) 2 + (0.5)(m)(−0.340c) 2 = 0.360mc 2 K f = 1 2 m 1 v 2 1f + 1 2 m 2 v 2 2f = (0.5)(2m)(0.069c) 2 + (0.5)(m)(0.703c) 2 = 0.252mc 2 Thus energy is not conserved in the reference frame of O if we use the classical formula for kinetic energy. This leads to a serious inconsistency—an elastic collision for one observer would not be elastic for another observer. As in the case of momentum, if we want to preserve the law of conservation of energy for all observers, we must replace the classical formula for kinetic energy with an expression that is valid in the relativistic case (but that reduces to the classical formula for low speeds). In classical physics, we obtained the expression for kinetic energy by using the work–energy theorem, which states that the change in kinetic energy of a particle is equal to the work W done on the particle by a force F.

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  Gravity from the Ground Up

  An Introductory Guide to Gravity and General Relativity

  • Bernard Schutz(Author)
  • 2003(Publication Date)
  • Cambridge University Press

    (Publisher)

  This is not really a well-defined notion, since photons cannot be brought to rest in order to measure their rest-mass. It is really only a convenient way of speak- ing about photons to explain why they do not fit into the rest of mechanics. But it is also a new perspective that allows us to speculate that perhaps there are other particles that have zero rest-mass as well. They, too, would travel at the speed of light. From this point of view, the “speed of light” is more fun- damental than light itself: it is the speed of all zero-rest-mass particles. The neutrino (see Chapter 11) was at first thought to be massless and to travel at speed c, although observations today suggest that it has a very small mass. When gravity is turned into a quantum theory, some physicists expect that there may be a particle associated with gravitational waves, called the gravi- ton, which will also be massless. But, as we shall discuss in Chapter 27, it may be very different from the photon. We have seen that the momentum carried by a photon is important in astron- omy. The momentum of any particle is its mass times its speed. In relativity, the mass is its total mass m. This is equivalent to its total energy E divided by c 2 . For a photon, whose speed is always c, the momentum is therefore photon momentum = (E/c 2 ) × c = E/c. 9. The Doppler redshift formula changes slightly. The changed notion of time in Because of the time dilation effect, even velocities across the line-of-sight to a body will slow time down and therefore change the apparent frequency of light it emits. So the Doppler formula must be modified to take account of time dilation. special relativity leads to a simple modification of the formula for the redshift of a photon. Remember how we derived the formula, by a visual method using Figure 2.3 on page 15. We counted the number of wave crests that passed by a moving wave-crest counter, and compared that with the number that passed one at rest.

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  Advanced Classical Electromagnetism

  • Robert Wald(Author)
  • 2022(Publication Date)
  • Princeton University Press

    (Publisher)

  Thus, the princi-ples of the previous paragraph lead, in an essentially unique way, to the modification eq. (8.36) of Newton’s second law that is special relativistically covariant. Note that u μ ∂ μ = d / d τ (i.e., u μ ∂ μ is the derivative along the worldline of the particle using the proper time parametrization). Thus, we may rewrite eq. (8.36) in the form d d τ m dx μ d τ = f μ . (8.37) It is also worth noting that if we assume that the 4-force f μ is orthogonal to the 4-velocity u μ , then 170 8 Special Relativity 0 = μ , ν η μν u μ f ν = μ , ν η μν u μ dp ν d τ = 1 2 m d d τ μ , ν η μν u μ u ν + dm d τ μ , ν η μν u μ u ν = − c 2 dm d τ . (8.38) Thus, if ∑ μ , ν η μν u μ f ν = 0, then the rest mass of the particle does not change as the particle moves along its worldline. The example of special relativistic particle mechanics also provides a good illus-tration of the manner in which spacetime tensors combine together—into a single quantity—quantities that would have been viewed as entirely distinct in pre-relativity physics. (This occurs whether or not the dynamical laws themselves need to be modified to be manifestly special relativistically covariant.) In pre-relativity physics, the energy of a particle, E = 1 2 m v 2 , is a scalar quantity, and one has the freedom to add an arbi-trary constant to it. It is entirely distinct from the momentum of the particle, p = m v . However, in special relativity, these quantities are identified as the components of the 4-momentum p μ of the particle: ( E / c , p ) = p μ = ( mu 0 , m u ) = (γ mc , γ m v ) . (8.39) This gives rise to new formulas for momentum and energy, namely, p = γ m v , and E = γ mc 2 . Most importantly, since E is now the component of a spacetime vector rather than a scalar quantity, there is no longer any freedom to modify its definition by the addition of a constant. Thus, a particle at rest must be assigned the energy E = mc 2 .

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  Physics, Volume 1

  • Robert Resnick, David Halliday, Kenneth S. Krane(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  20-23 is nearly equal to 1; at low speeds Eq. 20- 23 reduces to the familiar classical formula Equa- tion 20-23 thus also satisfies this necessary criterion of relativistic formulas. Of course, the ultimate test is agreement with experi- ment. Figure 20-21 shows a collection of data, based on in- dependent determinations of the momentum and velocity of electrons. The data are plotted as p/mv, which should have the constant value 1 according to classical physics. The re- sults agree with the relativistic equation and not with the classical one. Note that the classical and relativistic predic- tions agree for low speeds, and in fact the difference between the two is not at all apparent until the speed exceeds 0.1c, which accounts for our failure to observe the relativistic cor- rections in experiments with ordinary laboratory objects. p B  m v B . P yi  P yf  0. P xi  P xf  2mv 1  v 2 /c 2 , p x  mv x √1  v 2 /c 2 and p y  mv y √1  v 2 /c 2 . p B  m v B √1  v 2 /c 2 . v B p B  m v B , 466 Chapter 20 / The Special Theory of Relativity y x y' x' y' x' y x S S S 1 1 2 1 1 2 2 2 v v –v –v Frame S' Frame Before the collision After the collision S' S' u = –v u = –v 2v 1+ v 2 /c 2 v 2 – v 2 /c 2 √ v 2 – v 2 /c 2 √ –v 1 – v 2 /c 2 √ v 1 – v 2 /c 2 √ (a) (b) (c) (d) v v Figure 20-20. A collision between two particles of the same mass is shown (a) before the collision in the reference frame of S, (b) after the collision in the reference frame of S, (c) before the collision in the reference frame of S, and (d ) after the collision in the reference frame of S. Sample Problem 20-6. What is the momentum of a proton moving at a speed of v  0.86c? Solution Using Eq. 20-23, we obtain The units of kg  m /s are generally not convenient in solving prob- lems of this type. Instead, we evaluate the quantity pc: where we have used the conversion factor 1 MeV  1.60  10 13 J.

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  Forces in Physics

  A Historical Perspective

  • Steven N. Shore(Author)
  • 2008(Publication Date)
  • Greenwood

    (Publisher)

  8 THE RELATIVITY OF MOTION In the last few days I have completed one of the finest papers of my life. When you are older, I will tell you about it. —Albert Einstein to his son Hans Albert, November 4, 1915 Electromagnetic theory achieved a unification of two forces but at a price. It requi- red a medium, the ether, to support the waves that transmit the force. But this takes time and there is a delay in arrival of a signal from a source when something is changing or moving during this time interval. How do you know something is mov- ing? Remember that Aristotle used the notion of place within space or memory to recognize first change and then motion. So there the motion is known by compari- son to a previous state. But this isn’t the same question that recurs in the seven- teenth century. According to Descartes, we have to look instead at the disposition of bodies relative to which motion is known. These aren’t the same statement, although that would again divert us into metaphysics. Instead, we can look at what this means for the force concept. As Newton was at pains to explain, inertia requires two quantities, velocity and mass, to specify the quantity, momentum that is conserved. While mass is both a scalar quantity and a primitive in the dynamical principles, the momentum is not. It requires a direction and therefore, if we identify the inertial state as one moving at constant velocity, we need to know relative to what we take this motion. This is where we begin our discussion of the revolutionary developments at the start of the twentieth century: time, not only space and motion, is relative to the observer and dependent on the choice of reference frame and, consequently, has the same status as a coordinate because of its dependence on the motion of the observers. It follows that the notion of force as defined to this point in our discussions must be completely re-examined.

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