About the mass-energy relation

  • #1
716
9

Main Question or Discussion Point

I see how the premises

[tex]

p = \gamma m v

[/tex]

[tex]

F = \frac {dp}{dt}

[/tex]

and

[tex]


W= \int F dx

[/tex]

lead to

[tex]

dW = mc^2 d \gamma

[/tex]

and therefore

[tex]

W = \gamma mc^2 + k

[/tex]

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

Answers and Replies

  • #2
Doc Al
Mentor
44,882
1,129
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
716
9
I get

[tex]

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


[/tex]

so

[tex]

k = -mc^2 \neq 0

[/tex]

Since W(v=0) = 0 this is indeed the kinetic energy and not the total energy. Thanks, Doc Al.
 
  • #4
716
9
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?
 
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