MadSci Network: Physics

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

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 
terms of things you already know about.  Since:
   E^2 – p^2  =  (E+p)(E-p)  = (E+p) k
And we already have
   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 
particularly fundamental about it.

  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 

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

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