In high school, we are taught that work is delta energy, or force times displacement. Or if we take a more advanced course, work is calculated by the** line integral of nabla force** (the gradient of force, yes, the “upside-down triangle” *f*).

Because work is delta energy, work is not dependent on what path we take to move from one point to another, but only dependent on where we start and where we end. **But is there a more mathematical way of proving that, only using the definition of “work”? Or is work actually affected by other variables as well?**

This characteristic of not being dependent on path is called **“path independence.” **To prove whether work is path-independent or not, let us consider a 3-dimensional space, where there are two lines, C1 and C2 to reach point P1 from P0 (*fig. 1*). Then, we must prove the equation in the equation in the [ ].

*Fig 1. C1 and C2 on 3 dimentions*

But note that we can prove this claim by proving* equation (1) *when **C is any path** from C0 to C1. This equation is, in fact, the **fundamental theorem of line integrals,** a concept extremely pivotal in the field of multivariable calculus.

If *eq 1* is true, the right-hand side is a constant, meaning that the work for moving along any path C would be the same. The left-hand side of *eq 1* is equivalent to the right-hand side, giving us *eq 2*. Then, we parameterize x, y, and z using the parameter *t*. Then, we can represent the line integral in *eq 2* as *eq 3*. But note that the long-expression in the line integral is simply the chain rule of the function *f *differentiated by time. Hence, we get *eq 5 *and *6*, which proves the fundamental theorem of calculus. **Therefore, work is path-independent.**

But there is a** logical leap** that I made during the proof. For us to integrate *nabla f*, we must be **able to differentiate it first.** If *f* is not a differentiable function, we would not be able to get the gradient of *f*. Therefore, the fundamental theorem of line integrals is only true for a function that is differentiable within the space that our bounds enclose. For such a function, *nabla f* is called a **gradient field** or a **conservative field.**

Therefore, **work** is **path-independent,** but **only when the force is a conservative field.** This means that in real life, non-conservative forces such as air resistance or friction take part in affecting the movement, and hence work is **not entirely path-independent.**

The **proof** for this is that for a conservative force, moving from one place to another would require the **same amount of work.** For instance, in terms of gravitational force, going “up” is positive displacement while going “down” would be negative displacement. However, for frictional force, every time you move you cause heat, which would be positive work. Even if you come back to the same position, you would have a **different amount of delta energy, **meaning that frictional force is** non-conservative. **For the same reason, there is “elastic potential energy” or “gravitational potential energy,” but no “frictional potential energy.”

I think that the proof on the path-independence of work is a **beautiful example** of how math and physics are intricately connected, and how they can be used to represent our world. Also, it is impressive and elegant that we can **infer** that the force has to be a differentiable function by using mathematical analysis.

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