1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Energy dependence on observer framework

  1. Jun 13, 2014 #1
    Does mechanical energy of a system depend on the framework of an observer (neglecting a constant)?
  2. jcsd
  3. Jun 13, 2014 #2


    User Avatar

    Staff: Mentor

    Yes. The kinetic energy of a bullet is zero in the frame of an observer who is at rest relative to the bullet, non-zero for an observer who is at rest relative to the target of the bullet.
  4. Jun 13, 2014 #3


    User Avatar
    Science Advisor

    I agree with Nugatory but I can't help but wonder what you mean by "neglecting a constant".
  5. Jun 15, 2014 #4
    Ok, Right. The statement "neglecting a constant" is my mistake.
    I clarify my purpose of the question:
    Newton's laws are only valid in inertial framework. I like to know whether energy formalism is valid in non-inertial framework or not? In other words, can one solve the problems exactly, using conservation of energy in non-inertial framework?
  6. Jun 15, 2014 #5

    [itex]\int_{t_0}^{t_1}\vec{F}(t)\cdot\vec{v}(t)dt = \frac{1}{2}m v^2(t_1) - \frac{1}{2}m v^2(t_0)[/itex] is valid in frames where [itex]\vec{F}(t) = m \frac{d\vec{v}(t)}{dt}[/itex]

    That is, in inertial frames.

    You still can use it in non-inertial frames IF you add "inertial forces".

    [itex]\int_{t_0}^{t_1}\vec{F}(t)\cdot\vec{v}(t)dt = U(x(t_0),y(t_0),z(t_0))- U(x(t_1),y(t_1),z(t_1))[/itex] is valid in any frame where [itex]\vec{F}(x,y,z) = -\nabla U(x,y,z)[/itex]

    where [itex]U(x,y,z)[/itex] does not vary with time in this frame.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook