How integral and gradient cancels?

[Istiak]

Homework Statement

How integral and gradient cancels?

Relevant Equations

Integration, Differentiation

I know that gradient is multi-variable derivatives. But, here line integration (one dimensional integral) had canceled gradient. How?

 

[anuttarasammyak]

Let me know in your formula which appears $V_i$ or potential V ?

 

[Istiak]

Let me know in your formula which appears $V_i$ or potential V ?

$V_i$

 

[Orodruin]

https://en.wikipedia.org/wiki/Gradient_theorem

 

[anuttarasammyak]

I know
$$\nabla V \cdot d\mathbf{s}=\sum_i (\nabla V)_i ds_i$$
but am puzzled by your formula containing three i's
$$\sum_i (\nabla V_i)_i ds_i$$

 

[Istiak]

I know
$$\nabla V \cdot d\mathbf{s}=\sum_i (\nabla V)_i ds_i$$
but am puzzled by your formula
$$\sum_i (\nabla V_i)_i ds_i$$

first one is correct. When did I say the second one?

 

[anuttarasammyak]

I found three i's in LHS.

 

[Istiak]

$$\nabla V \cdot d\mathbf{s}=\sum_i (\nabla V)_i ds_i$$

That's what you wrote either. $$\sum_i (\nabla V)_i ds_i=\sum_i \nabla_i V_i ds_i$$

 

[Orodruin]

The $i$ in your expressions does not seem to be a spatial index. The sum seems to be a sum over several contributing potentials.

 

[anuttarasammyak]

That's what you wrote either. $$\sum_i (\nabla V)_i ds_i=\sum_i \nabla_i V_i ds_i$$

I can hardly find a meaning in RHS. What is $V_i$ ? I see
$$\sum_i \int_1^2 (\nabla V)_i ds_i = [V]_1^2$$
where $i=\{x,y,z\}$ or $\{1,2,3\}$ and V is potential function. I hope it is informative for your problem.

 

[Orodruin]

I can hardly find a meaning in RHS. What is $V_i$ ? I see
$$\sum_i \int_1^2 (\nabla V)_i ds_i = [V]_1^2$$
where $i=\{x,y,z\}$ or $\{1,2,3\}$ and V is potential function. I hope it is informative for your problem.

Again, the index $i$ here does not seem to be a spatial index but rather a counting index for several different contributions.

 

[Steve4Physics]

Homework Statement:: How integral and gradient cancels?
Relevant Equations:: Integration, Differentiation

View attachment 287953I know that gradient is multi-variable derivatives. But, here line integration (one dimensional integral) had canceled gradient. How?

Hi. What you have written is not recognisable notation (to me, anyway) and you have provided virtually no explanation/background information. Some problems that immediately stand out are:

1. What is the meaning of ‘∇ᵢ’? If ∇ is the gradient operator then the subscript (i) is meaningless.

2. What does ‘dsᵢ’ mean? Is ‘dsᵢ’ simply notation for dx, dy and dz? Or have you got multiple paths and dsᵢ is a line-element along the i-th path?

3. What is i? An index (i=1, 2 or 3) corresponding to the 3 spatial dimensions, or an arbitrary number corresponding to how many V’s there are?

4, Why have you got minus signs on both sides of the equation (when they can be removed)?

5. Why is the vertical bar on the right hand side used? This symbol is typically used for the end values of a definite integral, but there is no integral on the right hand side

What is the source of your equation?

 

[Orodruin]

1. What is the meaning of ‘∇ᵢ’? If ∇ is the gradient operator then the subscript (i) is meaningless.

The natural interpretation to me is a set of particles, indexed by $i$ moving in an external potential and each with position $\vec s_i$ and potential $V_i = V(\vec s_i)$. The gradient $\nabla_i$ would be the gradient acting on the $\vec s_i$ coordinates.

2. What does ‘dsᵢ’ mean? Is ‘dsᵢ’ simply notation for dx, dy and dz? Or have you got multiple paths and dsᵢ is a line-element along the i-th path?

See above. The $d\vec s_i$ would be the differential of the position of particle $i$.

5. Why is the vertical bar on the right hand side used? This symbol is typically used for the end values of a definite integral, but there is no integral on the right hand side

The integral is on the LHS. The evaluation of the integral at the endpoints are therefore on the RHS.

 

[Steve4Physics]

The natural interpretation to me is a set of particles, indexed by $i$ moving in an external potential and each with position $\vec s_i$ and potential $V_i = V(\vec s_i)$. The gradient $\nabla_i$ would be the gradient acting on the $\vec s_i$ coordinates.See above. The $d\vec s_i$ would be the differential of the position of particle $i$.The integral is on the LHS. The evaluation of the integral at the endpoints are therefore on the RHS.

Aha. Thanks @Orodruin. That’s very helpful. I was being a bit slow.

$\nabla_i$ is the directional derivative in the direction $\vec {s_i}$, not a simple gradient operator. For neatness, ‘$i$’ has been used as the subscript rather than (the more usual) $\vec {s_i}$.

As for my query on the use of the vertical bar, I was being silly.

And for anyone else as confused as I was initially, I’ll add that the ‘1’ and ‘2’ in the given equation aren’t actual values but are shorthand for the start-coordinates (1) and the end-coordinate (2) of each particle’s path. It's briefer (though less clear) than writing something like $r_{i_1}$ and $r_{i_2}$ for the start and end points of the i-th particle.

In a physical context, the equation is simply expressing the fact that the work done moving a set of particles (with independent paths) in a conservative force field equals the total change in potential energy.