Recently, I'm doing my personal task which is to formalize every definition and concept in physics, by means of formal language and of course with intuitional notes. Because I found myself that the most of the concepts I'm possesing were quite informal and unclear, for instance, energy. I think I found and read many definitions about energy, but it's still a hard concept to grab.

I know what the work is. My understanding is, roughly speaking, a force $F$ is said to do work on an object $a$ when the displacement $s$ of $a$ is changed in the direction of $F$. And the amount of work $W$ is defined by $\int_C F\cdot\mathrm{d}s$ where $C$ is the path of $a$ in which $F$ is exerting. It's unit called Joule, denoted by $\mathrm{J}$. (Actually, I defined this sorely in formal language, but it is really long to write all of them and hard to read.)

And my understanding of the term energy is that the energy of an object is the highest amount of work which the object can provide to other objects.

My question is, is there any problem to understand the concept of energy like the above? Is the definition inconsistent?

  • $\begingroup$ If I may, let me add a quick very intuitive view on work: Work is, exactly as it sounds from the word itself, useful energy spent to to something. If you spent a lot of force, but didn't actually do anything (no displacement - like pushing hard on a wall), then you didn't do any useful work. If you move something far but it wasn't very hard to do (didn't require any significant force - like pushing a balloon) then you also didn't do much work. To do work requires a useful effort. - That's a non-mathematical explanation at least. $\endgroup$
    – Steeven
    Oct 12, 2015 at 15:19
  • $\begingroup$ @Steeven Yes. I thought so too. When I first encountered the concept of work, I was thinking, "If I think work as some sort of an amount of the exerted force, then multiplying with time will also work. But why displacement?". $\endgroup$
    – Shin Kim
    Oct 12, 2015 at 15:27
  • $\begingroup$ You might get a different answer if you are discussing Newtonian Physics, Special Relativity, General Relativity, or Quantum Theory. $\endgroup$
    – Timaeus
    Oct 12, 2015 at 16:57
  • $\begingroup$ Flanders and Swann even sang a song about this called First and Second Law $\endgroup$
    – Henry
    Aug 31, 2018 at 6:59

2 Answers 2


This kind of definition is actually very conventional at the introductory level. You can find versions of it in many first-year textbooks. Indeed I used my version of this recipe in a recent answer.

You can probably get through an entire undergraduate education in physics on the basis of that kind of definition, but it you delve deeply enough you will eventually find it supplanted by the notion that energy is the Noetherian current associated with invariance in time.


As of such it is a fine view in my opinion. (Remember though that it is not possible still to convert all this energy into work in any case.)

But you might miss out on heat which is energy in transit (actually just like work).

Also, the formulation is not precise, as "the energy of an object" does not cover energy which is not stored within the object. Like potential energies. The best example might be gravitational potential energy, which is "stored" in the system, not in the object or in the Earth alone; it is only present because those objects are at a certain distance of each other.

  • $\begingroup$ I don't think the OP's definition excludes potential energy. (for the record, I have no opinion on his idea) $\endgroup$
    – garyp
    Oct 12, 2015 at 17:19
  • $\begingroup$ @garyp Potential energy is by definition not a property of the object itself. the energy of an object is the highest amount of work which the object can provide to other objects. This sentence thus needs rewriting to make sense, if potential energy must be included. $\endgroup$
    – Steeven
    Oct 12, 2015 at 17:28
  • $\begingroup$ Ok. It would be better to replace the word "object" with "system". Agreed. To some, "object" might refer to a point particle. $\endgroup$
    – garyp
    Oct 12, 2015 at 18:12

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