Direction of Work: Electron-Electron Repulsion

[Dale]

i can prove this -- if a force is non conservative a line integral between two points will be different for different paths,so i will have to specify the directions along which a test particle moves under the influence of source

Path dependence is not the same as being a vector.

http://en.wikipedia.org/wiki/Path_dependence_(physics )
http://en.wikipedia.org/wiki/Euclidean_vector

 

[nouveau_riche]

Just because work done is path dependent in some cases does not make it a vector.

so what makes vector, a vector?

because in the case of non conservative force i will need both direction and magnitude to signify the work

 

[nouveau_riche]

Work is the integral of F dot dr. That's the definition, plain and simple. What the dot product means is that the answer is the scalar magnitude of the force vector F projected along the length vector r. Because of this definition, the work done becomes a vector with one dimension of direction, + or -. So we make the important distinction that Energy is a scalar while Work is a vector. I just finished my AP Physics class not 3 months ago, so this is pretty fresh in memory. Work can either add or subtract from KE, PE, etc. In the case of a repulsive force, negative work is done, decreasing the KE and increasing PE. I believe what you're thinking of is the direction in which KE(Kinetic Energy) or the like is changing. If the electron is "climbing" a potential spike, then work done subtracts from KE, and if the electron is "falling" into a well, then the opposite is true. The justification is the definition of work and the nature of the force.

Also related is momentum, which is a vector quantity, and may be more what you're looking for. Momentum is the product of the scalar mass and vector velocity, which yields a second vector representing momentum. Force acting over time changes momentum.

i was thinking upon work done as change in energy given to a particle,call it as K.E or P.E

 

[nouveau_riche]

Do you have a published reference that states Work = vector? I'd love to see it!

i haven't seen it published as yet but that doesn't mean i could stop calling it as a vector until i am proved wrong

 

[Hootenanny]

Suppose I am sliding a box of mass 1 kg across a flat, horizontal, rough surface. The coefficient of kinetic friction is 0.4. Suppose I move the box through 2 meters. The work done is then

W = mg\mu\ell = 1\times9.81\times0.4\times2 = 7.848\;J

How is that a vector?

 

[FireStorm000]

nouveau, you quote me in support of your argument that work is a vector, but if you continue reading I later realized and admitted that I was wrong, thus using my post in support of your argument is pointless. I already said it, but I'll say it again: the only direction work has is + or -; as per the prevailing definition, that makes it a scalar quantity.

This is because the dot product is simply a projection: just as a shadow is a 2D flattening of a 3D object, so the dot product is a scalar projection of one vector onto another. You're simply measuring the size of the shadow it casts. The fact you use the cosine operator reduces the domain of direction from all real numbers to a domain of n * pi. In short, you can only call it a vector if the domain of direction includes all real numbers.

 

[FireStorm000]

W=mgμℓ=1×9.81×0.4×2=7.848J

Wouldn't that come out to -7.848J since you are doing work against friction? The vector of friction points exactly opposite the vector of displacement. Cos pi = -1. Got do remember that cosine ;)

 

[Hootenanny]

Wouldn't that come out to -7.848J since you are doing work against friction? The vector of friction points exactly opposite the vector of displacement. Cos pi = -1. Got do remember that cosine ;)

The mechanical work done by force \boldsymbol{F} on an object displaced by \boldsymbol{\ell} is \boldsymbol{F}\cdot\boldsymbol{\ell} = F\ell\cos\theta, where \theta is the angle between the vectors.

So although I am doing work against friction, the direction of "my" force is in the same direction as the displacement , hence the work done by "me" is positive. The work done by friction, on the other hand, would be negative.

 

[Doc Al]

so what makes vector, a vector?

For one thing, a vector has a direction.

because in the case of non conservative force i will need both direction and magnitude to signify the work

No you don't. You need the specific path to calculate the work, but the work itself has no direction.

Example: Say you move an object from A to B against a non-conservative force. One path requires 8 J of work, another path requires 10 J. Neither of those results have a direction. If you think they do, how would you specify it?

 

[Dale]

i haven't seen it published as yet but that doesn't mean i could stop calling it as a vector until i am proved wrong

No, that is not the way that this forum works. The rules of this forum are that we are not allowed to post non-mainstream material. That means that you must be able to furnish references in the mainstream scientific literature (textbooks or peer-reviewed papers). This response is inappropriate for the rules of this forum.

In any case, here is a proof:
W=\int_P \mathbf f \cdot d\mathbf x
Where f and x are vectors. The result of the inner product of two vectors is a scalar and the integral of a scalar is a scalar. Therefore W is a scalar.

 

[nouveau_riche]

For one thing, a vector has a direction.

No you don't. You need the specific path to calculate the work, but the work itself has no direction.

Example: Say you move an object from A to B against a non-conservative force. One path requires 8 J of work, another path requires 10 J. Neither of those results have a direction. If you think they do, how would you specify it?

i would specify the direction in terms of the displacement vector and will treat the object work as sum over histories of this displacement vector

 

[Hootenanny]

i would specify the direction in terms of the displacement vector and will treat the object work as sum over histories of this displacement vector

 

[nouveau_riche]

read it again,the displacement vector will provide me the nature of forces,when i go on adding the effects of each displacement vectors between two points forming a path ,i will get the net work done when i have followed that path in directions of series of displacement vectors

 

[Hootenanny]

read it again,the displacement vector will provide me the nature of forces,when i go on adding the effects of each displacement vectors between two points forming a path ,i will get the net work done when i have followed that path in directions of series of displacement vectors

Okay. Suppose you apply a constant force \boldsymbol{F} = [0,1]^\text{T} to a particle which is constrained to move in a straight line. The initial position is \boldsymbol{x}_0 = [0,0]^\text{T} and the final position is \boldsymbol{x}_0 = [2,\sqrt{3}]^\text{T}.

Calculate for me, the work done by the force please.

 

[nouveau_riche]

Okay. Suppose you apply a constant force \boldsymbol{F} = [0,1]^\text{T} to a particle which is constrained to move in a straight line. The initial position is \boldsymbol{x}_0 = [0,0]^\text{T} and the final position is \boldsymbol{x}_0 = [2,\sqrt{3}]^\text{T}.

Calculate for me, the work done by the force please.

firstly i am not much into mathematics,secondly as i said it will depend upon the force(conservative or non conservative)

 

[Hootenanny]

firstly i am not much into mathematics,secondly as i said it will depend upon the force(conservative or non conservative)

If you're not good at mathematics, why don't you listen to what we're telling you?!

 

[Dale]

firstly i am not much into mathematics

Then why do you care if work is a vector or not? It is inherently a mathematical topic.

It's like you ask why the Prius doesn't come in a V8 version and when someone tells you about engines and fuel economy you say that you arent into automotive technology.

 

[FireStorm000]

i would specify the direction in terms of the displacement vector and will treat the object work as sum over histories of this displacement vector

OK, I think I understand what nouveau is suggesting: While unconventional, it is possible to specify work done by describing an initial point, a final point, a path between the two, and all forces acting, and whether or not they are path in/dependent. This provides all information necessary to calculate complex systems. Further, if you know position on that path as a function of time, you could simplify the givens to time dependent path, and the forces to give work at an unknown time and point P from t₀.

To make an example, you throw a ball from point A at a angle pi/4, where A = Path(t₀) and tan(pi/4) = Path'(t₀). The path is, loosely speaking, close to a parabola, diverging more as friction acts. Air resistance acts along the entire path, and in this case, influences the path. Gravity is conservative and acts constantly irrespective of position(in this case).

So by specifying the "Work" as nouveau_riche did then what we have in fact done is to specify the initial conditions in a non-standard way, but one that communicates the same information. The initial conditions will be a vector. The forces will be a vector. The answer will be a scalar, because what you are doing take the dot product of these two vectors, thus the vectors you give us BECOME scalar in nature; they loose their direction term when we find work.

 

[Born2bwire]

read it again,the displacement vector will provide me the nature of forces,when i go on adding the effects of each displacement vectors between two points forming a path ,i will get the net work done when i have followed that path in directions of series of displacement vectors

No, no you won't. That's the problem with this, if you represented work as a vector field where the vector pointed in the direction of displacement with a magnitude equal to the work at that point it would be obvious that the total work would not be the same as the sum over the work vectors of your path.

What happens if you move a box in a closed circle? If the friction is uniform then your net work "vector" would obviously be zero.

You know, it sounds very much that what you want is the force vector field. Which we already have...

I have a strong feeling that the biggest problem behind some of your recent threads is that you have a fundamentally incomplete conception of scalars, vectors and fields.