Notes on Relativity and Cosmology - Physics Department, UCSB

154 CHAPTER 6. DYNAMICS: ENERGY AND ...
You may recall seeing an expression like (6.6) in some of your homework. It
came up there because it gives the first few terms in the Taylor’s series expansion
of the time-dilation factor,
1
√
1 − v2 /c = 1 + 1 v 2
+ small corrections, (6.7)
2 2 c2 a factor which has appeared in almost every equation we have due to its connection
to the interval and Minkowskian geometry.
It is therefore natural to guess that the correct relativistic formula for the total
energy of a moving object is
E =
m 0 c 2
√
1 − v2 /c 2 = m 0c 2 coshθ. (6.8)
This is exactly the formula 9 that will be derived in section 6.6.
6.4.2 Momentum and Mass
Momentum is a little trickier, since we only have one term in the expansion
so far: p = mv + small corrections. Based on the analogy with energy, we
expect that this is the expansion of something native to Minkowskian geometry
– probably a hyperbolic trig function of the boost parameter θ. Unfortunately
there are at least two natural candidates, m 0 c sinhθ and m 0 c tanhθ (which is
of course just m 0 v). The detailed derivation is given in 6.6, but it should come
as no surprise that the answer is the sinhθ one that is simpler from the point
of Minkowskian geometry and which is not the Newtonian answer. Thus the
relativistic formula for momentum is:
p =
m 0 v
√
1 − v2 /c 2 = m 0c sinhθ. (6.9)
If you don’t really know what momentum is, don’t worry too much about it.
We will only touch on momentum briefly and the brief introduction in section
6.2.2 should suffice. I should mention, however, that the relativistic formulas for
energy and momentum are very important for things you encounter everyday –
like high resolution computer graphics! The light from your computer monitor 10
is generated by electrons traveling at 10 − 20% of the speed of light and then
hitting the screen. This is fast enough that, if engineers did not take into
account the relativistic formula for momentum and tried to use just p = mv,
the electrons would not land at the right places on the screen and the image
would be all screwed up. There are some calculations about this in homework
problem (5).
9 The old convention that you will see in some books (but not in my class!!) is to define
m
a ‘velocity-dependent mass’ m(v) through m(v) = √ 0
= 1−v 2 /c 2 E/c2 . This is an outdated
convention and does not conform to the modern use of the term ‘mass.’
10 So long as it is not an LCD display.