I have come across the following lines in "Introduction to Mechanics" by Kleppner and Kolenkow.

A peculiar property of energy is that the value of mechanical energy $E$ is arbitrary; only changes in $E$ have physical significance. This comes about because the equation $$U_b - U_a = -\int_{a}^{b} \vec F \cdot \vec {dr}$$ defines only the difference in potential energy between $a$ and $b$ and not the potential energy itself. We could add an arbitrary constant to $U_b$ and the same constant to $U_a$ and still satisfy the defining equation. However, since $E = K +U$, adding an arbitrary constant to $U$ increases $E$ by the same amount.

What does he mean by "$E$ is arbitrary; only changes in $E$ have physical significance."? What is this physical significance he is talking about?

The concept of potential energy is still unknown to me. The author has introduced this form of energy by mathematically restating the equation of work done by a conservative force. Potential energy will be discussed a little later.


1 Answer 1


In this case, the "physical significance" is referred to the evolution of the system: it will be defined only by the differences between certain values of $E$, then $E$ is arbitrary since whatever constant you add to $E$ (i.e. you change $E=K+U$ with $E'=K+U+c$) it will not be relevant since the constant will cancel when you evaluate the difference between two value of $E'$.

Think about a problem involving the gravitational potential energy $U=mgh$ of a body of mass $m$ suspended on an height $h$: whatever problem will you think about, the solution will ever depend on the difference between potentials evaluated on different height, and it will never depend on a single potential evaluated in a point. Then, if we redefine $U$ as $U'=mgh+c$ (where $c$ is an arbitrary constant) the physical situation of your problem will never change, since evaluating the difference between $U'$ in two different point, the constant $c$ will cancel and the result will be identical to the one you can obtain with the original potential.

In this sense, "no physical significance" means that the constant $c$ is not relevant at all in the evolution of your system. Being $E=K+U$, the reasoning done on $U$ can be extended on $E$, since whatever constant $c$ you add to $E$, it can be considered, for example, as a constant added to $U$.

  • $\begingroup$ Changes in $U$ are quite evident from the equation of the work done by a conservative force. However, what do we mean by saying "changes in E"? $\endgroup$
    – R004
    Commented Jan 29, 2018 at 11:31
  • $\begingroup$ Just like the change in $U$, the change in $E$ is the difference of the values of $E$ in two different time, or in two different point of the motion. The changes of $U$ are directly correlated to the work done by a conservative force, while the changes in $E$ should be equal to zero in an insulated system while in presence of conservative-only forces. $\endgroup$
    – Alessandro
    Commented Jan 29, 2018 at 14:15
  • $\begingroup$ If I understand well( correct me if I am wrong ), changes in $E$ occur only when there are external forces( conservative or non-conservative ) acting on the system. This tells us that changes in $E$ have a physical meaning. $E$ is arbitrary because it can take any value without altering the physical situation of the problem( adding constants to potential energy functions does not alter the physical meaning of the problem because potential energy at a point has no physical meaning whatsoever ). $\endgroup$
    – R004
    Commented Jan 30, 2018 at 3:10
  • $\begingroup$ I think you are right, since $E=K+U$, changes of $E$ can occur if $U$ is changed (e.g. the body moves to a position to which corresponds a different potential energy) or if $K$ is changed (e.g. a force acts on the body). Despite this, if the system is closed and only conservative forces are acting, then the change of $E$ should be zero, i.e. the potential energy increase when the kinetic energy decrease, and vice versa. Finally, adding a constant value to $E$ in all the calculation in your problem will not change the results, since, as you say, the constant can be added to the potential. $\endgroup$
    – Alessandro
    Commented Jan 30, 2018 at 8:26
  • $\begingroup$ I think I was wrong stating that changes in $E$ occur when there are external forces acting on the system. If I take a system of mass $m$ and I observe that it is moving in a conservative field like gravitational field, I witness that the work done by gravity( external ) does not alter the total energy of the system at any time. The correct statement is "Non-zero work done by non-conservative forces( internal or external ) changes the energy of the system." $\endgroup$
    – R004
    Commented Jan 30, 2018 at 12:51

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