15.25: Addition of Kinetic Energies
( \newcommand{\kernel}{\mathrm{null}\,}\)
I want now to consider two particles moving at nonrelativistic speeds – by which I mean that the kinetic energy is given to a sufficient approximation by the expression 12mu2 and so that parallel velocities add linearly.
Consider the particles in figure XV.37, in which the velocities are shown relative to laboratory space.
Referred to laboratory space, the kinetic energy is 12m1u21+12m2u22. However, the centre of mass is moving to the right with speed V=(m1u1+m2u2)(m1+m2), and, referred to centre of mass space, the kinetic energy is 12m1(u1−V)2+12m(u2+V)2. On the other hand, if we refer the situation to a frame in which m1 is at rest, the kinetic energy is 12m2(u1+u2)2, and, if we refer the situation to a frame in which m2 is at rest, the kinetic energy is 12m1(u1+u2)2.
All we are saying is that the kinetic energy depends on the frame to which speed are referred – and this is not something that crops up only for relativistic speeds.
Let us put some numbers in. Let us suppose, for example that
m1=3kg u1=4m s-1
m2=2kg u3=4m s-1
so that
V=1.2m s-1.
In that case, the kinetic energy
referred to laboratory space is 33 J,
referred to centre of mass space is 29.4 J,
referred to m1 is 49 J,
referred to m2 is 73.5 J.
In this case the kinetic energy is least when referred to centre of mass space, and is greatest when referred to the lesser mass.
Exercise. Is this always so, whatever the values of m1, m2 , u1 and u2?
It may be worthwhile to look at the special case in which the two masses are equal (m) and the two speeds(whether in laboratory or centre of mass space) are equal (u).
In that case the kinetic energy in laboratory or centre of mass space is mu2, while referred to either of the masses it is 2mu2.
We shall now look at the same problem for particles travelling at relativistic speeds, and we shall see that the kinetic energy referred to a frame in which one of the particles is at rest is very much greater than (not merely twice) the energy referred to a centre of mass frame.
If two particles are moving towards each other with “speeds” given by g1 and g2 in centre of mass space, the g of one relative to the other is given by equation 15.16.14, and, since K = g - 1, it follows that if the two particles have kinetic energies K1 and K2 in centre of mass space (in units of the m0c2 of each), then the kinetic energy of one relative to the other is
K=K1⊕K2=K1+K2+K1K2+√K1K2(K1+2)(K2+2).
If two identical particles, each of kinetic energy K1 times m0c2, approach each other, the kinetic energy of one relative to the other is
K=2K1(K1+2).
For nonrelativistic speeds as K1→0, this tends to K=4K1, as expected.
Let us suppose that two protons are approaching each other at 99% of the speed of light in centre of mass space (K1 = 6.08881). Referred to a frame in which one proton is at rest, the kinetic energy of the other will be K = 98.5025, the relative speeds being 0.99995 times the speed of light. Thus K=16K1 rather than merely 4K1 as in the nonrelativistic calculation. For more energetic particles, the ratio KK1 is even more. These calculations are greatly facilitated if you wrote, as suggested in Section 15.3, a program that instantly connects all the relativity factors given there.
Two protons approach each other, each having a kinetic energy of 500 GeV in laboratory or centre of mass space. (Since the two rest masses are equal, these TWO spaces are identical.) What is the kinetic energy of one proton in a frame in which the other is at rest?
(Answer: I make it 535 TeV.)
The factor K (the kinetic energy in units of m0c2) is the last of several factors used in this chapter to describe the speed at which a particle is moving, and I take the opportunity here of summarising the formulas that have been derived in the chapter for combining these several measures of speed. These are
β1⊕β2=β1+β21+β1β2.
γ1⊕γ2=γ1γ2+√(γ21−1)(γ22−1).
k1⊕k2=k1k2
z1⊕z2=z1z2+z1+z2.
K=K1⊕K2=K1+K2+K1K2+√K1K2(K1+2)(K2+2).
ϕ1ϕ2=ϕ1+ϕ2
If the two speeds to be combined are equal, these become
β1⊕β1=2β11+β21.
γ1⊕γ1=2γ21−1
k1k1=k21
z1⊕z1=z1(z1+2)
K1⊕K1=2K1(K1+2).
ϕ1⊕ϕ1=2ϕ.
These formulas are useful, but for numerical examples, if you already have a program for interconverting between all of these factors, the easiest and quickest way of combinng two “speeds” is to convert them to ϕ. We have seen examples of how this works in Sections 15.16 and 15.18. We can do the same thing with our example from the present section when combining two kinetic energies. Thus we were combining two kinetic energies in laboratory space, each of magnitude K1 = 6.08881 (ϕ1 = 2.64665). From this, ϕ = 5.29330, which corresponds to K = 98.5025.