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.For instance, if we want to find the vector which represents the“state in which the object is moving at velocity v = 12c, we must Lorentz transform the “at rest” vector to a frame moving at velocity −12c.The result is shown in the figure opposite, top-right.Notice that, unlike the Galilei transformed case, not only the space-component but also the time-component of the vector has changed.What stays invariant is the area of the diamond with the vector as one of its sides, and the diagonals at 45◦ angles from the horizontal.(See figure.) When the object is at rest, this diamond becomes a square with the length of its sides proportional to the mass of the object, so its area is proportional to the mass squared.In a similar fashion, we can determine the vectors that represent the motion of the object at all other velocites.When we do this, the tip of the vector will move along the curve shown in the bottom figure.This curve is of a kind known as a hyperbola, and its asymptotes are the light-cone of the spacetime origin.The distance between the spacetime origin and the hyperbola is proportional to the object’s mass.13.1 The energy–momentum vector233ct (m)ct (m)xx(m)(m)ct (m)This distance isproportional tothe massx(m)234Relativistic dynamicsNow, as in the Newtonian case, it is convenient to draw a graph for the“state of motion” vectors separately from the spacetime diagram.For the Newtonian case, we used a graph with mass notched on the vertical axis, and momentum = mass × velocity notched on the horizontal axis.In the relativistic case, however, because the vertical axis of the spacetime diagram is ct and not t, the corresponding “state of motion” graph will also have the vertical axis multiplied by c.So the quantity notched on the vertical axis will be mass × (speed of light), while that on the horizontal axis will be mass × velocity as before.The units for both axes will be kg · m/s.The vector representing the “state of the object of mass m at rest”on this graph is then one with time-component equal to mc, and space-component equal to zero.The vector representing the “state of the object of mass m moving at velocity v” will be the Lorentz transform of the“at rest” vector, as shown in the figure.As we noted above, the time-component is longer than what it was at rest.We express the factor by which the time-component has lengthened with the greek letter γ, that is:(time-component at velocity v) = γ(time-component at rest) = γmc.Then, since the ratio of the time-component and the space-component is determined by the velocity v, we must have (space-component at velocity v) = γmv.Note that the factor γ is a number that depends on v and changes with it.When v = 0, there is no lengthening of the time-component, so γ = 1.However, as v grows toward c, the diamond shown in the figure will get flatter and flatter and collapse onto the light cone, while still maintaining a constant area.For this to happen γ must diverge to infinity as v approaches c.1Now, the time-component of our vector represents the “tenacity of the motion” in the time-direction, while its space-component represents the“tenacity of the motion” in the space-direction.The space-component γmv is called the momentum, just as in the Newtonian case (though its numerical value differs by the factor γ), and is represented by the symbolp = γmv.The time-component γmc represents the inertia of the object, so you may be inclined to call it the mass (times c)2.However, we would like to reserve the term for m, which is the amount of “stuff” contained in the 13.1 The energy–momentum vector235object, and independent of the frame.So we will call the time-component of our vector by another name: the energy.To be more precise, the energy E is defined as the time-component of the vector times c: E = γmc2.Thus, the vector we have defined will be called the “energy–momentum vector.”3Notice that when the object is at rest, v = 0, the γ-factor is equal to one, and the above equation reduces to E = mc2.E = mc2 is the energy that the object has when it is at rest, so it is called the rest energy.Now, this is Einstein’s famous formula, but at this point, we have just taken the mass, multiplied it by c2, and given it the alias rest energy, so it does not have any physical meaning yet.We will revisit this equation later.E/c␥ mcmc␥ myp (kg.m/s)236Relativistic dynamics13.2 The energy–momentum vector of a photon The photon is the particle associated with light, so it travels at the speed of light c [ Pobierz całość w formacie PDF ]