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Notes on Relativity and Cosmology - Physics Department, UCSB

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156 CHAPTER 6. DYNAMICS: ENERGY AND ...<br />

Furthermore, we see that this ‘size’ does not depend <strong>on</strong> the frame of reference<br />

<strong>and</strong> so does not depend <strong>on</strong> how fast the object is moving. However, for a<br />

rapidly moving object, both the time part (the energy) <strong>and</strong> the space part (the<br />

momentum) are large – it’s just that the Minkowskian noti<strong>on</strong> of the size of a<br />

vector involves a minus sign, <strong>and</strong> these two parts largely cancel against each<br />

other.<br />

6.4.3 How about an example?<br />

As with many topics, a c<strong>on</strong>crete example is useful to underst<strong>and</strong> certain details<br />

of what is going <strong>on</strong>. In this case, I would like to illustrate the point that while<br />

energy <strong>and</strong> momentum are both c<strong>on</strong>served, mass is not c<strong>on</strong>served.<br />

Let’s suppose we take two electr<strong>on</strong>s <strong>and</strong> places them in a box. Suppose that<br />

both electr<strong>on</strong>s are moving at 4/5c, but in opposite directi<strong>on</strong>s. If m e is the rest<br />

mass of an electr<strong>on</strong>, each particle<br />

|p| =<br />

m e v<br />

√<br />

1 − v2 /c 2 = 4 3 m ec, (6.12)<br />

<strong>and</strong> an energy<br />

E =<br />

m e c 2<br />

√<br />

1 − v2 /c 2 = 5 3 m ec 2 . (6.13)<br />

We also need to c<strong>on</strong>sider the box. For simplicity, let us suppose that the box also<br />

has mass m e . But the box is not moving, so it has p Box = 0 <strong>and</strong> E Box = m e c 2 .<br />

Now, what is the energy <strong>and</strong> momentum of the system as a whole? Well, the<br />

two electr<strong>on</strong> momenta are of the same size, but they are in opposite directi<strong>on</strong>s.<br />

So, they cancel out. Since p Box = 0, the total momentum is also zero. However,<br />

the energies are all positive (energy doesn’t care about the directi<strong>on</strong> of moti<strong>on</strong>),<br />

so they add together. We find:<br />

So, what is the rest mass of the systemas a whole?<br />

p system = 0,<br />

E system = 13<br />

3 m ec 2 . (6.14)<br />

E 2 − p 2 c 2 = 169<br />

9 m2 e c4 . (6.15)<br />

So, the rest mass of the positr<strong>on</strong>ium system is given by dividing the right h<strong>and</strong><br />

side by c 4 . The result is 13<br />

3 m e, which is significantly greater than the rest mass<br />

of the Box plus twice the rest mass of the electr<strong>on</strong>!<br />

Similarly, two massless particles can in fact combine to make an object with a<br />

finite n<strong>on</strong>-zero mass. For example, placing phot<strong>on</strong>s in a box adds to the mass<br />

of the box. We’ll talk more about massless particles (<strong>and</strong> phot<strong>on</strong>s in particular)<br />

below.

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