# Indefinite integral of fields

1. Jun 15, 2014

### Jhenrique

Is possible to compute indefinite integrals of functions wrt its variables, but is possible to compute indefinite integrals of scalar fields and vector fields wrt line, area, surface and volume?

2. Jun 15, 2014

### Matterwave

Yes...I feel like you keep asking this same question, and I keep giving you the same answer. Look at Stoke's theorem:

$$\int_\Omega d\omega = \oint_{\partial \Omega} \omega$$

If we know that $\eta=d\omega$, then $\omega$ can be thought of as an "anti-derivative" of $\eta$, we can use the above theorem to generate a type of indefinite integral! Sadly, we get back an lower dimensional integral instead, but that is all that one can hope for as far as "indefinite integrals" work. Also this only works with exact forms, because non-exact forms are not necessarily derivatives of something. There's no way to define an "anti-derivative" if the base object is not the derivative of something.

The closest form of what you want is the Gradient theorem, which I showed you in a previous thread:

$$\int_a^b \vec{\nabla}\phi\cdot d\vec{r}=\phi(b)-\phi(a)$$

Last edited: Jun 15, 2014
3. Jun 15, 2014

### Jhenrique

Yeah, but I'm talking about a indefinite integral of a vector field! This can exist? This make sense?

So, I'm thinking in something like this:

$$\\ \int \vec{f} \cdot d\vec{r} = \int \vec{\nabla}\phi\cdot d\vec{r}=\phi$$

4. Jun 15, 2014

### Matterwave

You tell me. When do you think a path integral makes any sense if you don't specify a path? When do you think a surface integral makes any sense if you don't specify a surface?

5. Jun 15, 2014

### Jhenrique

No make sense think in a line integral without the line of integration like no make think in the area of integration of a function without the limits of integration. However, make sense think in the indefinite integral of a function, so why would make sense think in the indefinite integral of a field too?

6. Jun 15, 2014

### Matterwave

What's the path of integration for a single variable function? Think this one through. You might realize, that a path was chosen for you already without you even thinking about it.

7. Jun 16, 2014

### Jhenrique

But, independent of exist a geometric interpretation, this equation is true:
?

8. Jun 16, 2014

### Matterwave

No, you have to specify the endpoints or else that equation is meaningless.

What is $\phi$? It's a scalar function right? So it's really $\phi(x,y,z)$. Now what values of x, y and z are in the $\phi$ on the right? Does it make any sense?

If I change the path of integration, do you agree that the left hand side of that equation will change? Will the right hand side change?

Even in regular integrals. do we say:

$$\int \frac{d f}{dx} dx = f$$

? No we do not. We say:

$$\int \frac{d f}{dx} dx = f+C$$

Why is that C there? Is it necessary? Use your critical thinking. What is the "C" in the equation you posted?

9. Jun 16, 2014

### Jhenrique

I can't use a critical thinking over a subject that is almost completely obscure for me...

C is a initial condition, or a guaranteed that integral is covering all the set of possibles antiderivatives.

10. Jun 16, 2014

### Matterwave

If you are unable to even think critically about a question, it is best to learn a bit more about the subject first before making conjectures. To be honest Jhenrique, not to discourage you or your love of science/math, but a lot of your conjectures seem to make no sense. Perhaps they can lead somewhere if you can at least formulate the questions in a cogent manner.

There's a big book by Arfken and Weber called Mathematical Methods for Physicists (I'm a physicist, so I learn from physics textbooks) that covers basically everything you're worried about. You can use it as a study guide or a reference text. It's quite comprehensive. However, it looks at things from a physicist's perspective, so it's not always 100% rigorous like a math text would be. You might also want to pick up a pure math book on the subject and see where that takes you as well.

11. Jun 16, 2014

### micromass

Staff Emeritus
No, it doesn't. So if you want to discuss this further, you must provide a reference where this thing is defined.

12. Jun 19, 2014

### Jhenrique

When is about vector and tensor calculus my doubts are exponentially big... :(

More one thing... indefinite integrals requires (implicitly or not) path independence? In other words, only is possible to compute indefinite integrals of exact form?

13. Jun 19, 2014

### micromass

Staff Emeritus
Like I said, there is no such thing as indefinite path integrals.