Istiak
- 158
- 12
- Homework Statement
- How integral and gradient cancels?
- Relevant Equations
- Integration, Differentiation
##V_i##anuttarasammyak said:Let me know in your formula which appears ##V_i## or potential V ?
first one is correct. When did I say the second one?anuttarasammyak said: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
That's what you wrote either. $$\sum_i (\nabla V)_i ds_i=\sum_i \nabla_i V_i ds_i$$anuttarasammyak said:\nabla V \cdot d\mathbf{s}=\sum_i (\nabla V)_i ds_i
I can hardly find a meaning in RHS. What is ##V_i## ? I seeIstiakshovon said:That's what you wrote either. $$\sum_i (\nabla V)_i ds_i=\sum_i \nabla_i V_i ds_i$$![]()
Again, the index ##i## here does not seem to be a spatial index but rather a counting index for several different contributions.anuttarasammyak said: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.
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:Istiakshovon said: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?
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.Steve4Physics said:1. What is the meaning of ‘∇ᵢ’? If ∇ is the gradient operator then the subscript (i) is meaningless.
See above. The ##d\vec s_i## would be the differential of the position of particle ##i##.Steve4Physics said: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?
The integral is on the LHS. The evaluation of the integral at the endpoints are therefore on the RHS.Steve4Physics said: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
Aha. Thanks @Orodruin. That’s very helpful. I was being a bit slow.Orodruin said: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.