Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The total energy of an object comes from the time part of the four-momentum, and so isn't a Lorentz invariant. On the other hand, is the potential energy of a compressed spring a Lorentz invariant?

share|improve this question
1  
When a spring is compressed its mass increases by a tiny amount; so as far as energy is concerned a compressed spring is equivalent to an uncompressed but a little heavier spring. –  user10001 Jul 15 '12 at 2:40
    
The density of potential energy is frequently a Lorentz scalar even though this is not a necessary condition for a relativistic theory. –  drake Sep 13 '12 at 4:17

2 Answers 2

It depends of what potential energy you are using. If you refer to a potential energy that depends only on position variables, $V=V(x)$, this is not Lorentz invariant. If you refer to potential energy that depends both on position and time variables, this can be Lorentz invariant. Some examples of Lorentz invariant potential energies are given in the textbook by Schieve [1]

[1] Classical Relativistic Many-Body Dynamics; M. A. Trump and W. C. Schieve; 1999; Kluwer Academic Publishers

share|improve this answer
    
Thanks, I'll check out the the reference. –  Larry Harson Oct 13 '12 at 14:16
    
Check "4.2.4 The Invariant Potential" for general thoughts and "5.6.1 The Harmonic Oscillator" for application to springs –  juanrga Oct 15 '12 at 10:03

The potential energy is not a Lorentz invariant, it is just a field energy for the spring parts, including the most significant contribution, the increased electronic energy from the compression of the electronic wavefunction, and it transforms along with the field momentum as a four-vector. In general, it isn't useful to separate energy into potential energy and kinetic energy in special relativity, the energy must have local flow, and is described by field energy, a stress-energy tensor, and it is due to a combination of local fields.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.