Relativistic Definition of Energy: E=Fd or E=md2/t2?

[mrspeedybob]

In classical physics E=Fd and F=ma so E=mad.
a=d/t² so E=md²/t²
Measurements of d and t will get complicated by Lorentz transformation, so, is E=Fd still a correct definition of energy, or is it a Newtonian approximation which is not accurate at relativistic velocities?

 

[PeterDonis]

is E=Fd still a correct definition of energy, or is it a Newtonian approximation which is not accurate at relativistic velocities?

The basic definition E = Fd is still valid in relativity; it just says that a change in energy is produced by exerting a force through a distance (i.e., doing work). (More precisely, this is a change in kinetic energy.) However, the other formulas you give are no longer correct in relativity; they are just Newtonian approximations. For more on the correct definition of force in relativity, see here.

 

[bcrowell]

E=Fd is not a definition of energy even in nonrelativistic mechanics. It's an expression for mechanical work, which is one type of energy transfer.

Fundamental conserved quantities like energy do not have single-equation definitions. There is a nice analogy in section 4-1 of the Feynman lectures, where Feynman compares a conservation law to the observation that a bishop on a chessboard is always observed to end up on a square of the same color.

 

[Nugatory]

In classical physics E=Fd and F=ma so E=mad.
a=d/t² so E=md²/t²
Measurements of d and t will get complicated by Lorentz transformation, so, is E=Fd still a correct definition of energy, or is it a Newtonian approximation which is not accurate at relativistic velocities?

The $F$ in $F=ma$ is the net force on an accelerating object, and plugging it into $W=Fd$ will give you the work done by the net force, which is the kinetic energy not the total energy. You can see this difference if you imagine a mass of one kg being moved one meter by a force of 10 Newtons in one direction while a 9 Newton frictional force is working in against the applied force. The net force will be one Newton and the total kinetic energy will be one joule. However, the total work done by all the forces, and hence the total energy expended, will be 19 joules; the missing 18 joules will show up as heat.

Add in the special relativity stuff, and $W=Fd$ still works as long as you're careful to pick one frame and work in that frame (but you have to be careful to avoid the pitfall of measuring the force in the frame of the accelerated object and the distance in the frame of an observer watching the acceleration). Kinetic energy is $(\gamma-1)m_0c^2$ where $m_0$ is the rest mass of the object. Total energy is given by the relationship $E^2=(m_0c^2)^2+(pc)^2$ where $p=\gamma{m_0}v$ is the momentum.