So I have been trying to understand why there is a negative sign in the following$\Delta U= -\int\vec{F}\cdot d\vec{l}$

I wanted to try to understand it by trying to derive the gravitational potential energy $E_p=-\dfrac{GMm}{r}$ I start by defining potential energy to be $0$ at $\infty$ and see what I get if I let gravity do work on the mass until a distance r away from a planetenter image description here

$W_G=\int_{\infty}^{r} \vec{G}\cdot{d\vec{l}} = \int_{\infty}^{r} G\cos{\theta}dl = \int_{\infty}^{r} G dx$ G is the gravitational force so: $W_G= \int_{\infty}^{r} -\dfrac{GMm}{x^2}dx = \Bigg[\dfrac{GMm}{x}\Bigg]$ applying limits of integration gives $W_G = \dfrac{GMm}{r}$

Thus I am confused why the minus sign, from the derivation I wrote it seems to have to be $\Delta U= \int\vec{F}\cdot d\vec{l}$ at least for gravity. I also did the same derivation for a mass on earth ($W_G = -mgh$ Where $\vec{G}$ is considered to be constant) and the result indeed is the correct one $\Delta U= -\int\vec{F}\cdot d\vec{l}$ so my confusion arises, why don't i get the same result for both cases.

I am also wondering if I carry out the same derivation but for elastic potential energy, what would I get... Thank you for any help!


2 Answers 2


The negative sign is there due to convention. We like the idea that conservative forces cause things to move from higher to lower energy, and this also allows us to define the total mechanical energy as $E= K + U$, where $K$ is kinetic energy and $U$ is potential energy, rather than $E=K-U$. However, I think you are really asking about the consistency of this sign choice in how we calculate the work done by conservative forces, which I will go into now.

The definition of work done by a force is $$W=\int \mathbf F\cdot\text d\mathbf x$$

The potential energy of a conservative force relates to the force by the definition of potential energy $$\mathbf F=-\nabla U$$

Putting these two things together, the work done by a conservative force is $$W=\int -\nabla U\cdot\text d\mathbf x=-\Delta U$$ by the fundamental theorem of calculus$^*$. i.e. work done by a conservative force can easily be determined using the (negative) change in potential energy along the path we are calculating the work on.

Moving the negative sign, for a conservative force we then have: $$\Delta U=-W=-\int \mathbf F\cdot\text d\mathbf x$$ the above discussion is true for any conservative force, so you can use it for a Hookean force, $F=-kx$, as well since it is conservative.

In your first example $W_G$ is positive because your gravitational force does positive work as the object is moved to smaller $r$. This means that the potential energy must decrease, which you can easily verify that it does. I think your error is assuming $W_G=\Delta U$, which is not correct.

Because of our discussion in the comments, it looks like you are unclear on how to set this integral up for the work done by gravity. I will show you the general set up. You can then apply it to your specific scenario.

When using this definition you should always think about what $\mathbf F$ and $\text d\mathbf x$ are before trying to set the integral up. So we need to specify the force, which points radially inward (I will use your convention of $x$ being the radial variable): $$\mathbf F=-\frac{GMm}{x^2}\hat x$$ and your infinitesimal displacement$^{**}$ $$\text d\mathbf x=\text dx\ \hat x$$

Therefore, if the object starts at a distance $x=a$ from the center of the field, and ends at a distance $x=b$ from the center of the field, we have $$W=\int_a^b\left(-\frac{GMm}{x^2}\hat x\right)\cdot(\text d x\ \hat x)=\int_a^b-\frac{GMm}{x^2}\text d x=GMm\left(\frac1b-\frac1a\right)$$

From the discussion above then $$\Delta U=-W=GMm\left(\frac1a-\frac1b\right)$$ which is what we could have gotten by using the potential energy in the first place: $$\Delta U=U(b)-U(a)=\left(-\frac{GMm}{b}\right)-\left(-\frac{GMm}{a}\right)=GMm\left(\frac1a-\frac1b\right)$$ (if these two lines seem redundant, that is because they are, since we have shown everything ends up giving the same result above. I am just trying to drive the point home here).

Notice how when $b>a$ (raising the object) we have negative work done by gravity and an increase in potential energy, and when $b<a$ (falling object) we have positive work done by gravity and a decrease in potential energy. So, it all checks out.

$^*$If you are unfamiliar with vector calculus, just do the 1D analog. $F=-\frac{\text d U}{\text d x}$ and $W=\int F\text dx$. You will find the same thing to be true as explained above.

$^{**}$ Notice that this encodes with it the direction of the displacement based on your limits of integration. Therefore, for example, if the displacement was towards smaller $x$ values, you would not say $\text d\mathbf x=-\text dx\ \hat x$ if your starting and ending locations were your lower and upper limits of integration respectively. The sign of $\text dx$ is determined by our limits of integration already.

  • $\begingroup$ Hmm, I have a question. now looking at what I did better, I spotted a mistake in my math. $W_G= \int_{\infty}^{r} -\dfrac{GMm}{x^2}dx = \Bigg[\dfrac{GMm}{x}\Bigg]$ should be $W_G= \int_{\infty}^{r} \dfrac{GMm}{x^2}dx = - \Bigg[\dfrac{GMm}{x}\Bigg]$ because when doing the scalar product, i should only take the magnitude of G ie $|\vec{G}|$ so this gets rid of the minus sign in the integral and from there it should all work fine. Is this a sensible reason for my mistake? $\endgroup$
    – Luca Ion
    Commented Apr 29, 2019 at 14:49
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    $\begingroup$ @LucaIon No, the work done by gravity is positive in your example. You are making a sign error. $\endgroup$ Commented Apr 29, 2019 at 14:58
  • $\begingroup$ yes, and in my example, I took the magnitude of the force by gravity with a minus sign which is wrong because it is the magnitude thus it is unsigned, if I make that correction then it ends up the right formula $\endgroup$
    – Luca Ion
    Commented Apr 29, 2019 at 15:11
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    $\begingroup$ @LucaIon What do you mean by "right formula". If $W_G<0$ then you have made a mistake. If $\Delta U>0$ then you have made a mistake. $\endgroup$ Commented Apr 29, 2019 at 17:08
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    $\begingroup$ @LucaIon Please see my edit. It deals with the correct way to set up the integral. I am unsure why you say we only consider the magnitude for the scalar product, when the scalar product involves two vector quantities. $\endgroup$ Commented Apr 29, 2019 at 17:27

It is just a matter of definition because Change in potential energy of system of particles is defined as the negative of work done by internal conservative forces of the system.

It is defined in this way so we are always used to put a negative sign.


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