# B Difference between W=-(ΔU) and negative U.

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1. Sep 4, 2016

### donaldparida

Is there any difference between the meaning of the equation W=-ΔU and the meaning of the statement that potential energy of system is always negative when the reference point(zero point) is taken to be located at infinity or do they both convey the same message?

2. Sep 4, 2016

### Staff: Mentor

I see no link between the two statements.

Note also that the statement that "the potential energy of system is always negative when the reference point(zero point) is taken to be located at infinity" is not absolutely true. There are systems that can have positive potential energy even in that case.

3. Sep 4, 2016

### Staff: Mentor

I should also point out that the sign in W=-ΔU is also a convention, that positive work is work done by the system, not on the system. That convention is far from universal.

4. Sep 4, 2016

### donaldparida

Doesn't W=-ΔU come from the law of conservation of energy?(when there is positive increase in K.E. there must be negative increase or simply decrease in the P.E. so that the total energy remains constant.)

5. Sep 4, 2016

### Staff: Mentor

It comes from conservation of energy when dealing with adiabatic processes. But my point was about the sign in the equation, which is somewhat arbitrary.

6. Sep 4, 2016

### DarkBabylon

There is a difference.
While W=-ΔU talks about the work done W is the difference between two states of the potential energy ΔU (if both end results are at rest is the only time it really applies, on other occasions it is W=-(ΔU+ΔE[kinetic]) ).
The other says that when you consider a two body system under the influence of a conservative force, that is there is potential energy, the reference point of infinity is taken as the point of 0 energy. However the statement is inaccurate because negative energy in such a case happens in a system with at least two bodies bound to one another.

7. Sep 4, 2016

### Staff: Mentor

$\Delta U$ is the change in internal energy, not just potential energy. It includes kinetic energy.

8. Sep 4, 2016

### DarkBabylon

For some reason our professors said otherwise. Guess either they mistook or convention is not the same in every place. Some people use j as the sqrt(-1) on single plane complex numbers, where as others use i for instance.

9. Sep 4, 2016

### Staff: Mentor

@donaldparida can you clarify the context? Is this for a thermodynamic system undergoing an adiabatic process, or something else?

10. Sep 5, 2016

### donaldparida

No, this question is not related to a thermodynamic system undergoing adiabatic process. It is related to the gravitational potential energy of a system.

11. Sep 5, 2016

### Staff: Mentor

Ah, please ignore my comments above. They were in the wrong context.

The equation has nothing whatsoever to do with setting the reference at infinity. In fact, it shows that any reference can be used since $\Delta U$ is independent of the reference.

12. Sep 5, 2016

### donaldparida

Is it correct to say that if the reference point is set at infinity only then the potential energy is negative and if we select any other reference point, the potential energy is positive?

13. Sep 5, 2016

### Staff: Mentor

if the reference point is set at infinity only then the potential energy is ALWAYS negative and if we select any other reference point, the potential energy is SOMETIMES positive

14. Sep 5, 2016

### Staff: Mentor

Then ignore my comments. For me, W = -ΔU is a thermodynamical statement.

It is always useful to state the context of the question

15. Sep 5, 2016

### donaldparida

@Dale , so in conclusion, the equation W=-ΔU (meaning that the work done is the negative of the change in potential energy) and the fact that the potential energy of a system when the reference point is taken at infinity is always negative have no relation with each other and the equation W=-ΔU comes from the work-energy theorem and the law of conservation of energy. Right? Also is there any reference point for which the potential energy of a system always comes out to be positive?

16. Sep 5, 2016

### Staff: Mentor

Correct.

R=0, but in practice that is never used because it would be positive and infinite everywhere.

17. Sep 5, 2016

### DarkBabylon

For 1/r^2 forces yes.

18. Sep 6, 2016

### donaldparida

19. Sep 6, 2016

### DarkBabylon

In forces such as coulomb's law and newton's formula for gravity, notice that r is in the denominator. r refers to the distance between the object and the source. In the case of these two forces, the two objects feel each other of course. If you try to find the limit of 1/r as r approaches 0, you will find the result gets bigger and bigger and diverges to infinity.
However do note, that is true for point objects, in reality if you know the force in every single location there are some physical systems where the force does not diverge when r=0, and again it depends on the problem. In certain systems you can define r=0 as the point of 0 potential energy and not have the result infinite everywhere.
Systems such as distributed charge in a sphere, or an object with its mass distributed in a certain configuration in space, would not necessarily yield infinity when r=0.