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I have seen in many textbooks and sources which say that we can't experimentally measure potential energy but we can measure differences in potential energy.

$$\Delta U_g=-W_g$$

Choosing zero potential (reference point) at the ground.

Now if I measure change in gravitational potential energy from zero point to a point where an object thrown upwards attains zero velocity, then $U_g$ at that point would just be negative of the work done.

If potential energy at that point can be calculated then why is it said that absolute potential energy at a point can't be calculated?

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    $\begingroup$ Energy is a torsor. $\endgroup$
    – PM 2Ring
    Commented Nov 30, 2018 at 14:59

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Simply put, potential energy is the energy an object possesses because of its position. Position, or location, is always relative. Therefore there is no such thing as an exact or absolute position in space and consequently no exact potential energy.

Potential energy must be measured relative to something. Suppose a 1 Kg ball is suspended 1 meter above the surface of the earth. Relative to the surface of the earth it has a potential energy of 9.81 Joules. But suppose we put a 0.5 m high table underneath the ball. Relative to the surface of the table it has a potential energy of 4.9 Joules.

We haven't moved the ball, so which is the real potential energy?

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  • $\begingroup$ As I said in my question that it is possible to measure the difference in potential energy. What does this 'difference in potential energy' really mean . Because even if in your given case(or in any case) I choose the reference configuration say, 2 meters above the surface of the earth even then the difference in gravitational potential between the surface of the earth and ball would be the same as compared to the the case in your answer. Does it mean that for any 2 fixed points the difference in GPE is same and is hence measurable (regardless of reference config) ? $\endgroup$
    – Karan Mehta
    Commented Nov 30, 2018 at 17:30
  • $\begingroup$ @KaranMehta Yes, for any 2 fixed points in a gravitational field the difference in GPE is the same and is hence measurable, provided that the acceleration due to gravity, $g$, is the same between the fixed points. $\endgroup$
    – Bob D
    Commented Dec 1, 2018 at 14:17
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It's because the "zero point" you mentioned is arbitrary, and doesn't have to be zero. I could just as say that $GPE(y = 0) = 10 \text{J}$ or any other arbitrary number. In that case, even thought I could find $GPE$ at whatever height the ball reaches, I couldn't unique determine it, since it would depend on how I defined $GPE(y = 0)$.

Hope that helps!

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  • $\begingroup$ isn't GPE at zero point always zero? $\endgroup$
    – Karan Mehta
    Commented Nov 30, 2018 at 15:16
  • $\begingroup$ @Karan No, it isn't. It's convenient to set GPE=0 at ground level when working with systems close to the ground, where the variation in gravitational acceleration is negligible. But when working on a larger scale, eg calculating orbits, the usual convention is to set GPE=0 at infinite distance from the centre of mass. $\endgroup$
    – PM 2Ring
    Commented Nov 30, 2018 at 17:24
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We can measure potential energy. For example, we know that the electrostatic potential energy between a proton and a electron in a hydrogen atom is negative, and equal to -27.2 eV. This negative energy is what makes the mass of hydrogen atom be less than the sum of the proton and electron masses.

In special relativity, we know that mass, energy, and momentum of a system are related by $m^2=E^2-p^2$ (in units where $c=1$), and the energy $E$ in this equation includes all forms of energy, including potential energy. So, unlike in Newtonian physics, it is not true that only potential energy differences matter.

The electrostatic potential energy between two point charges is $q_1 q_2/r$ (in Gaussian units), not this plus an arbitrary constant. When the two charges are infinitely far apart, there is no potential energy.

Similarly, the gravitational potential energy between two point masses is $-G m_1 m_2/r$, not this plus an arbitrary constant. When two masses are infinitely far apart, there is no potential energy.

We know that this is true in the case of gravity because, in the post-Newtonian approximation of General Relativity, the negative gravitational potential energy affects the force of gravity. This has been tested in the dynamics of the solar system.

Only in Newtonian mechanics, which we have known is wrong for more than a century, is it true that only differences in potential energy matter. We know better now.

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    $\begingroup$ This answer is false! (sorry). The electrostatic potential energy between two point charges : $q_1 q_2/r$, is defined relative to space at infinity. The proof of that formula is done by evaluating the work done by the electrostatic force to bring the charges at distance $r$ from infinity. Nothing imposes the potential energy to be 0 at infinity, it's just an arbitrary choice. It's the same for the gravitationnal potential energy. Energy is a very subtle concept in relativity, especially in General Relativity. $\endgroup$
    – Cham
    Commented Dec 1, 2018 at 15:46
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    $\begingroup$ @Cham Explain how the invariant mass $m=\sqrt{E^2-p^2}$ can be invariant if part of $E$ — the potential energy — is arbitrary up to a constant. $\endgroup$
    – G. Smith
    Commented Dec 1, 2018 at 17:53
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    $\begingroup$ Also, take note that there is no potential energy in the relation $m = \sqrt{E^2 - p^2}$. $E$ is the rest energy plus kinetic energy (defined in a given frame). If you want to add up potential energy, it's something like this : \begin{equation}m^2 = (\, p_a - q \, A_a)(\, p^a - a \, A^a),\end{equation}where $p^a$ is the canonical momentum wich is relative, while $A^a$ is gauge dependant (also a "relative" quantity). The difference $p^a - a \, A^a$ is independant of the gauge, but still dependant on the frame. $\endgroup$
    – Cham
    Commented Dec 1, 2018 at 18:08
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    $\begingroup$ The mass of your atom is the rest energy/$c^2$, which is still defined relative to vacuum and infinity (boundary conditions). $\endgroup$
    – Cham
    Commented Dec 1, 2018 at 18:14
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    $\begingroup$ This subject is interesting and brings me a question that I may ask on the forum : what are the "things" we need in physics to define any kind of energy, including "mass". We need a reference frame. We need to define vacuum, and as I see it we also need boundary conditions (which may be arbitrary). What else? I believe there is much more subtle things that enters the energy concept, and we tend to forget or take as granted many of them. This may explain a lot of confusions in the physics community. $\endgroup$
    – Cham
    Commented Dec 1, 2018 at 18:37

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