### Re: Special relativity, conceptual question on E-p

Date: Fri Apr 2 09:05:38 2004
Posted By: Guy Beadie, Staff, Optical sciences, Naval Research Lab
Area of science: Physics
ID: 1079477978.Ph
Message:
```

Hello, Ben.

The fact that you’re thinking about the meanings behind equations is
good.  Too many students do algebra without thinking about what all the
symbols mean.

However, algebra also provides clues to the relationships between
quantities.  In particular, it can help us rewrite your variable k=E-p in
E^2 – p^2  =  (E+p)(E-p)  = (E+p) k
E^2 = m^2 + p^2
We can write:
k = m^2 / (E+p)

There you have it – that’s k.  We could discuss the different k values
of a particle if we wished, but we would find that there isn’t anything

It sounds like you’re interested in how momentum is tied to energy in
reference frames moving at different speeds.

“Is k to be thought of as reflecting the system's internal structure &
internal motions in an observer's frame? I.e., does k represent
momenta "tied up" in the system due to the system's traveling slower than
light?”

Perhaps it would help to clarify what all these things mean.

First of all, be aware that the value of total energy E is NOT
independent of inertial frame.  From my point of view, the energy of the
(stationary) coffee cup next to me is equal to its rest mass.  In an
inertial frame zipping by at some fast speed, my cup no longer looks
stationary.  An observer in that frame would calculate my cup’s energy to
be more energetic than just the rest mass.

[Digression - If you’re confused by this because you remember that
energy should be conserved, then I should point out energy conservation
applies to the CHANGE in energy over time.  Energy conservation still
holds for my coffee cup, because the other observer and I agree that there
is no _change_ in the energy of my cup as it sits next to me.  This is a
trivial example of conservation of energy; the principle becomes more
useful when discussing collisions between two or more particles.]

So, if the other observer measures a higher energy for my coffee cup,
where did this ‘extra’ energy come from?  How much extra energy is there?
What is this energy?

I’ll answer those questions in reverse order.  First, the extra energy
is called the kinetic energy, T, of an object.  Second, it is the
difference between the total energy of a moving object and its rest mass:

T = sqrt(m^2 + p^2) – m

Third, it represents the amount of energy that would be dissipated if the
object were brought to a halt.  If I hold out a hand next to my coffee
cup, there’s nothing my cup can do to it.  However, if the other observer
holds out their hand as they fly by, my coffee cup could do a lot of
damage.  That’s the nature of the kinetic energy.

One final point, in case you’re still hung up on “internal motions in an
observer’s frame.”  My coffee cup is made up of many, many particles, all
moving around in small ways.  Because it’s so complicated, let’s go all
the way down to the scale of a single water molecule.

A water molecule is composed of two hydrogen atoms and an oxygen atom.
However, a water molecule does not have the same rest mass as the three
atoms by themselves.  Why?  Because a water molecule is more stable.  When
the three atoms form water, they ‘snap’ together.  In the process they
release some energy.  By conservation of energy, since the energy of the
three isolated atoms at rest is equal to the water molecule at rest PLUS
some extra energy, the molecule must have a lower rest mass.

Thus, the interactions between particles that form bigger objects count
towards computing the rest mass of the object as a whole.  Once that
bookkeeping is done, we use the new rest mass and the momentum of the
whole object to compute kinetic energy of the new particle.

Good question,

- Guy

```

Current Queue | Current Queue for Physics | Physics archives