- Quote:
*Ironically, most particle physicists don’t use either of these two simple interpretations of the relativity formula. We use a third one!* - Endnote:
*Taking E to be total energy and m to be rest mass, we view***E = m[c**as true only for stationary objects. Otherwise^{2}]**E > m[c**; in words, the total energy of a moving object always exceeds its internal energy (by an amount that can be easily expressed in terms of the quantity called momentum.)^{2}]

We particle physicists write

where ** p** is momentum. In terms of the object’s speed

**, we don’t use Newton’s formula for momentum,**

*v***=**

*p***, but instead we use an Einsteinian version,**

*mv*Note: this is not the same as Newton’s formula ** p** =

**, because we particle physicists use equation (1), which says that for a moving object**

*mv***is**

*E***NOT**equal to

**! [Got to keep on your toes, here!]**

*mc*^{2}Instead, combining equations (1) and (2), we find

Bringing the second term to the left-hand side and factoring.

Solving for ** E** gives

a formula which appears in endnote 7 of this chapter and the ensuing discussion.

Let’s return now to equation (1). For a moving object, ** v** is non-zero, so

**is non-zero too, and**

*p***is positive. Therefore**

*p*^{2}and thus, taking the square root of both sides *(and remembering both motion energy and mass are always positive)*, we find that

for a moving object. This can also be seen from equation (3); if ** v** is non-zero, then (

*v/c*)^{2}is positive but less than 1, so the square root is less than 1, and one over the square root is greater than 1.

Meanwhile for a stationary object, ** v** and

**are zero, so returning to equation (1) and taking the square root, or simply using equation (3), we find**

*p**[See also this article in which I discuss and give intuition as to how Einstein’s formulas were a sensible generalization of Newton’s formulas.]*