Direction of Work: Electron-Electron Repulsion

[nouveau_riche]

how can we account for the direction in which the work is being done by a particle while it is interacting?
say as an example,electron -electron repulsion,we know that there is a repulsive force but in which direction is the electron working(in direction of motion of force/any other)?
justify your answer

 

[LostConjugate]

I would expect

The direction along the vector between the two electrons.

The force is in all directions but the force acting on another electron is along this vector. The sign depends on if your taking the electron to work on the other electron or itself.

 

[magpie5]

Hi, when work is done, there must be some kind of enregy transfer or transformation eg from kinetic to kintec or potential to kinetic etc.
For two electrons interracting, say colliding, then one electron will gain an amount of energy due to the interraction and the other will loose an equal amount.
Because work deals with energy which is a scalar, then there is no direction associated with this. One electron does work on the other and vis versa. It just depends on the question.

 

[nouveau_riche]

I would expect

The direction along the vector between the two electrons.

The force is in all directions but the force acting on another electron is along this vector. The sign depends on if your taking the electron to work on the other electron or itself.

which direction are you talking about,there will be many in between?

Hi, when work is done, there must be some kind of enregy transfer or transformation eg from kinetic to kintec or potential to kinetic etc.
For two electrons interracting, say colliding, then one electron will gain an amount of energy due to the interraction and the other will loose an equal amount.
Because work deals with energy which is a scalar, then there is no direction associated with this. One electron does work on the other and vis versa. It just depends on the question.

but you missed the nature of interaction ,repulsion/attraction
with collision an energy is gained/lost but the collision can occur from various direction,say head on or at 90 degree
what i mean to say is even if it is a head on,can u say it will always be repulsion?

 

[Dale]

Work does not have a direction. It is a scalar, not a vector.

 

[nouveau_riche]

Work does not have a direction. It is a scalar, not a vector.

okay,let us talk about force in the same context,i mean electron electron interaction?

 

[sophiecentaur]

The force will be along a line joining the electrons. The work will correspond to the integral of that force times distance (∫fdx)along that line.

 

[Dale]

okay,let us talk about force in the same context,i mean electron electron interaction?

The force between two stationary electrons is given by Coulomb's law and is always repulsive. The force between two moving electons is given by the Lienard Wiechert potentials and will also always be approximately repulsive, but due to the finite speed of light it may not always be pointing exactly directly away from the current position of the other electron.

 

[sophiecentaur]

Yes, that must be right but will not each electron still 'see' a force which is radial? You would need to do a slightly modified line integral, possibly, but each electron would be accelerated along the line that it 'sees' the other one.

 

[Dale]

Yes, that must be right but will not each electron still 'see' a force which is radial?

How would you would define a "radial force" in the case of an arbitrarily moving charge?

 

[sophiecentaur]

It would be the same direction that the electron would look in order to see the other. That wouldn't be difficult to calculate, given the inclination, using a bit of SR. (Would it?)

 

[A.T.]

but each electron would be accelerated along the line that it 'sees' the other one.

Electrons are bad example for seeing each other. Take two negatively charged space ships at high relative speed.

 

[Dale]

It would be the same direction that the electron would look in order to see the other. That wouldn't be difficult to calculate, given the inclination, using a bit of SR. (Would it?)

I would have to work out the math to be sure, but I don't think that the force necessarily comes from that direction. For example, I think that the force from a charge moving at constant velocity comes from its actual position, but due to abberation you would look at a different position to see it. I am not certain about this, however, so don't rely too strongly on it.

 

[A.T.]

I would have to work out the math to be sure, but I don't think that the force necessarily comes from that direction. For example, I think that the force from a charge moving at constant velocity comes from its actual position, but due to abberation you would look at a different position to see it.

This is correct. For inertial movement the field itself in the rest frame of the charge is static and radial, so the force is towards/from the charge. But light is a disturbance of the field that travels c. So the moving charge sees aberration.

In the rest frame of the "watching" charge the directions are different too, because the light comes from an old position of the light emitting charge. The field is Lorentz contracted, but moves with the charge and the force still points directly to the moving charge.

 

[nouveau_riche]

The force will be along a line joining the electrons. The work will correspond to the integral of that force times distance (∫fdx)along that line.

how can you say that the force will be along the line joining the electrons?,i guess the direction of motion of electron is not the answer

The force between two stationary electrons is given by Coulomb's law and is always repulsive. The force between two moving electons is given by the Lienard Wiechert potentials and will also always be approximately repulsive, but due to the finite speed of light it may not always be pointing exactly directly away from the current position of the other electron.

how can you say that the force will be along the line joining the electrons?,i guess the direction of motion of electron is not the answer

 

[Dale]

how can you say that the force will be along the line joining the electrons?

I didn't say that. I said the force would be given by the Lienard Wiechert potential.

 

[nouveau_riche]

I didn't say that. I said the force would be given by the Lienard Wiechert potential.

can u explain that a bit?,and work will be a vector if the force is non conservative

 

[sophiecentaur]

I can't see how a scalar can suddenly become a vector.??

 

[Dale]

I didn't say that. I said the force would be given by the Lienard Wiechert potential.

can u explain that a bit?

Here is a good place to start: http://tinyurl.com/3aovkjz

and work will be a vector if the force is non conservative

No.

 

[ZealScience]

I don't quite understand your question. What do you meant by direction of work? You can consider displacement as its direction, or you can consider the direction of the force as its direction. I think there is no difference between two vectors of the dot product or dot product integrated.

 

[nouveau_riche]

I don't quite understand your question. What do you meant by direction of work? You can consider displacement as its direction, or you can consider the direction of the force as its direction. I think there is no difference between two vectors of the dot product or dot product integrated.

well direction of work mean the path that will be followed by the test particle under the influence of source particle

Here is a good place to start: http://tinyurl.com/3aovkjz

No.

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

I can't see how a scalar can suddenly become a vector.??

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

 

[Doc Al]

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

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

 

[FireStorm000]

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.

 

[FireStorm000]

http://physics.info/vector-multiplication/"

Curiously, they list the dot product as producing a scalar. I guess what I'm getting from that is that there is some confusion/disagreement as to what constitutes a vector or a scalar. When I refer to a vector, it is any number with an attached direction, regardless of how many components that direction has. A scalar is any number that is independent of direction; a magnitude alone, without and direction, even + or -. Thus as I was taught, Work == Vector, Energy == Scalar. That might not be true if you choose to define vector/scalar quantities differently. If someone could clarify that inconsistency, I'd be much obliged.

 

[Doc Al]

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.

Yikes!

Note that another name for the 'dot' product is the scalar product of two vectors. Quoting from hyperphysics: "The scalar product of two vectors can be constructed by taking the component of one vector in the direction of the other and multiplying it times the magnitude of the other vector." (See: http://hyperphysics.phy-astr.gsu.edu/hbase/vsca.html" )

 

[FireStorm000]

Yikes! ...

Yeah, read the post after that. What I was taught is not being consistent with what the INTERNET is saying. Could you clarify?

 

[Doc Al]

Curiously, they list the dot product as producing a scalar.

Of course they do! As will any legit reference.

I guess what I'm getting from that is that there is some confusion/disagreement as to what constitutes a vector or a scalar.

Not really.

When I refer to a vector, it is any number with an attached direction, regardless of how many components that direction has. A scalar is any number that is independent of direction; a magnitude alone, without and direction, even + or -. Thus as I was taught, Work == Vector, Energy == Scalar. That might not be true if you choose to define vector/scalar quantities differently. If someone could clarify that inconsistency, I'd be much obliged.

A scalar can have a sign, just not a direction.

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

 

[Doc Al]

Yeah, read the post after that. What I was taught is not being consistent with what the INTERNET is saying. Could you clarify?

It's not just 'the internet' saying that. What reference do you have?

 

[WannabeNewton]

Curiously, they list the dot product as producing a scalar. I guess what I'm getting from that is that there is some confusion/disagreement as to what constitutes a vector or a scalar.

Not at all. An inner product space over \mathbb{R} has a map <.,.>: V\times V \mapsto \mathbb{R}. The inner product over this field will always map vectors to the reals so there is no way work is a vector. You can't define it that way, it makes no sense to.

 

[FireStorm000]

A scalar can have a sign, just not a direction.

You actually clarified everything right there. Under that definition, work is a scalar with a domain from + to - infinity. Energy has a domain of 0 - +infinity. The definition of a scalar I was using in my AP physics class was that a scalar had domain 0 to +infinity, and everything else was a vector, because everything else would constitute a direction. Under this definition, which would appear to be the prevailing one, then work is a scalar. It just has a +/- sign where energy does not.

I went through the http://asd20.org/education/components/docmgr/default.php?sectiondetailid=37896&fileitem=43607&catfilter=5055" again, and the instructor specifically states that the answer is a scalar because the dot product can be simplified as [scalar * scalar * cos direction = scalar].

Generally speaking, I think my logic for coming to the conclusion I did is straightforward though: "A vector is any quantity that has direction"; few people would disagree with that statement, though apparently it isn't quite right. "magnitude is the same as the absolute value." In other words, the magnitude does not include a +/-. We use the same symbolism for the two operations. --> "Anything where |V| != V is a vector" --> "|cos pi| != cos pi." --> "Work is not a scalar, it must be a vector".

If all of those statements hold, then the logic is sound, so somewhere in there a distinction must be made that makes a quantity with 1 dimension of direction a scalar. In the end, I guess it makes sense though. You can't really consider a number line as having a direction, as there are only two choices of angle, 0 and pi. A plane has one choice of direction, that is, an angle that can take on any (real) value. A 3D coordinate system has 2 free directions along which we must define an angle. Thus a better way to define a scalar is by the number of free axises of direction it has. A vector is any value more than 0, a scalar has 0.

Curiously, we end up then with two classes of scalars, those that must be 0 or above, and those that can take on any real value. Temperature, Energy, Counts, Speed and some others fall into this first category. Things like Work, Rates of changes of the first class, and some others fall into this second class. So I guess my misconception was to group this second class in with vectors when it's really just another type of scalar.

 

[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.