Basic exercise of path integral

In summary, the line integral would be:$$\int \vec{F}\cdot \vec{dr} = A \int_{1}^{0}xy dx = A\int_1^0 x dx = -\dfrac{A}{2}$$
  • #1
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Homework Statement
Hi everyone.

I was studying an example of Kleppner's book and I have a question.

In this example, the path integral is calculated, starting from the point (0,0) and going to the point (0,1). The force is $$F = A(xy \hat{i}+ y^2 \hat{j})$$. The path followed is denoted by (a,b,c) (see figure).
Relevant Equations
$$W = \vec{F}\cdot \vec{dr}$$
In the book it is mentioned that, in path c, the line integral would be:

$$\int \vec{F}\cdot \vec{dr} = A \int_{1}^{0}xy dx = A\int_1^0 x dx = -\dfrac{A}{2}$$.

but I think that dx is negative in that case, the result would be positive, right?
 

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  • #2
Exactly because dx is negative (because the lower limit of integration is bigger than the higher limit of integration, for this look up the definition of integral as the limit of Riemann Sum) the result would be negative (if A is taken positive).
In order to be more accurate in what is exactly happening, when we take the dot product of ##\vec{F}## and ##d\vec{r}## those two are in opposite directions :##\vec{F}## points (we can neglect the ##\hat j## component since we take the dot product with path c which is perpendicular to j) in the direction of ##\hat i## since ##Axy## is positive on path c , and ##\vec{dr}=dx\hat i## but dx is negative on path c, so ##\vec{dr}## points in opposite direction to ##\hat i##.

Since they point in opposite directions, the dot product is negative.
 
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  • #3
To add to that, ##dx## is negative in that integral already because you are integrating from 1 to 0.
 
  • #4
Thanks for the answers. It is clear to me that the integral must be negative since the force is opposite to the displacement.

However, the path integral is defined as

$$ W = \int_{a}^{b} \vec{F}\cdot d\vec{r} $$

where a and b refer to the start and end points, respectively.

In this example a = 1 and b = 0, ##dr = -dx \hat{i}##. So:

$$ W = \int_{1}^{0} \vec{F}\cdot d\vec{r} =\int_{1}^{0} (xy \hat{i} +y^2\hat{j })\cdot (-dx \hat{i}) = -\int_{1}^{0} xy dx $$

The results, in that case, is positive. According to your answers, my error is that the sign of dx is contain in the integration limit. In that case, I need to consider a positive dx.
 
  • #5
No I disagree, it is always ##\vec{dr}=dx \hat i +dy \hat j +dz\hat k##, it is that dx , dy, dz can be negative or positive (or zero) depending on the path direction.
 
  • #6
So how would you set up the integral of this path?,

$$ W = \int_{1}^{0} \vec{F}\cdot d\vec{r} =\int_{1}^{0} (xy \hat{i} +y^2\hat{j })\cdot (dx \hat{i}) = \int_{1}^{0} xy dx $$

For me it is not totally clear yet the reason.
 
  • #7
Ark236 said:
So how would you set up the integral of this path?,

$$ W = \int_{1}^{0} \vec{F}\cdot d\vec{r} =\int_{1}^{0} (xy \hat{i} +y^2\hat{j })\cdot (dx \hat{i}) = \int_{1}^{0} xy dx $$

For me it is not totally clear yet the reason.
What is not clear? Now you doing it correctly, you just have to put y=1 cause y remains constant and equal to 1 in path c.
 
  • #8
Because from the definition, the integration limits are the initial and the final points. From path c the particle starts from the point (0,1) and ends at the point(1,0). Also, the ##\vec{dr}## is the direction of the displacement, in that case, would be negative (##\vec{dr} = -dx \hat{i}##).
 
  • #9
Ark236 said:
Because from the definition, the integration limits are the initial and the final points. Also, the ##\vec{dr}## is the direction of the displacement, in that case, would be negative (##\vec{dr} = -dx \hat{i}##).
It should be ##d\vec{r} = -\lvert dx \rvert\,\hat i##. If you want to put the minus sign in explicitly, you have to use the absolute value of ##dx## to ensure you're using its magnitude.
 
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  • #10
Ark236 said:
Because from the definition, the integration limits are the initial and the final points. From path c the particle starts from the point (0,1) and ends at the point(1,0). Also, the ##\vec{dr}## is the direction of the displacement, in that case, would be negative (##\vec{dr} = -dx \hat{i}##).
I don't know why you insist on this, the vector calculus book I 've read (ok its a Greek book written by Greek authors and unknown to the many), defines ##d\vec{r}=dx\hat i +...## and this definition holds regardless of what is the path. It is that depending on path the dx,dy,dz,..., can become negative or positive or zero. In path c it is dy=0 and dx=negative.
 
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  • #11
Thanks so much for the answer. Now for me is clear :D .
 
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1. What is the purpose of the basic exercise of path integral?

The basic exercise of path integral is a mathematical tool used to calculate the probability of a particle moving from one point to another. It is commonly used in quantum mechanics and statistical mechanics to analyze the behavior of particles.

2. How does the basic exercise of path integral work?

The basic exercise of path integral involves breaking down a path into small segments and calculating the probability of the particle moving along each segment. These probabilities are then summed together to determine the overall probability of the particle's path.

3. What are the limitations of the basic exercise of path integral?

The basic exercise of path integral is limited to systems with a finite number of states and does not take into account interactions between particles. It also assumes a continuous and deterministic path, which may not always be the case in quantum mechanics.

4. How is the basic exercise of path integral applied in real-world situations?

The basic exercise of path integral is commonly used in fields such as quantum mechanics, statistical mechanics, and field theory to analyze the behavior of particles and systems. It has also been applied in other areas such as finance and economics to model the behavior of stock prices and market trends.

5. Are there alternative methods to the basic exercise of path integral?

Yes, there are alternative methods to the basic exercise of path integral, such as the Feynman path integral and the Wiener path integral. These methods may provide more accurate results in certain situations, but they are based on similar principles and still involve breaking down a path into smaller segments.

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