1. Mar 31, 2010

### snoopies622

I see how the premises

$$p = \gamma m v$$

$$F = \frac {dp}{dt}$$

and

$$W= \int F dx$$

$$dW = mc^2 d \gamma$$

and therefore

$$W = \gamma mc^2 + k$$

where m is the rest mass and k is a constant of integration. But why do we conclude that k=0?

2. Mar 31, 2010

### Staff: Mentor

That constant won't equal 0. (The work done equals the KE, not the total energy.) Assume you start from rest and integrate to speed v.

3. Mar 31, 2010

### snoopies622

I get

$$W = \gamma mc^2 - mc ^2 = ( \gamma - 1 ) mc^2$$

so

$$k = -mc^2 \neq 0$$

Since W(v=0) = 0 this is indeed the kinetic energy and not the total energy. Thanks, Doc Al.

4. Apr 1, 2010

### snoopies622

Follow-up: Am I to conclude that the constant of integration here represents the (negative) "rest energy" of the object, or is there a better way to arrive at that relationship?