Why is work a scalar?

  • Thread starter Forrest T
  • Start date
  • #1
Forrest T
23
0
I am trying to understand why work is a scalar, without knowing ahead of time that work is defined as:

[itex]W_{ab} = \int ^{\vec{r_{b}}}_{\vec{r_{a}}} \vec{F} \cdot d{\vec{r}}[/itex]

Essentially, I am trying to understand how this definition was derived (based on the one-dimensional work-energy theorem, which is easy to figure out).
 
Last edited:

Answers and Replies

  • #2
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
29,905
15,565
What else would it be? It can't be a vector, since it has no direction.
 
  • #3
Forrest T
23
0
Why doesn't it have a direction?
 
  • #4
DaveC426913
Gold Member
21,445
4,925
Why doesn't it have a direction?
Moving an object North and moving an object West still accomplish the same amount of work. Work is only a magnitude.
 
  • #5
Forrest T
23
0
How do you know that?
 
  • #6
DaveC426913
Gold Member
21,445
4,925
How do you know that?
Know what? :grumpy:


That moving north is the same amount of work as moving west? Because I spun your laboratory 90 degrees before letting you take your measurements. And you didn't notice.


That work is a magnitude? That's the definition. If it had a direction it would be ... a displacement.
 
  • #7
Forrest T
23
0
Here's a better way to phrase my question:

How can you show that

[itex]\sum W[/itex] = [itex]W_{x} + W_{y} + W_{z}[/itex]

using the work-energy theorem in one dimension

[itex]\int^{b}_{a} F_{x}dx[/itex] = [itex]\frac{1}{2}mv^{2}_{b} - \frac{1}{2}mv^{2}_{a}[/itex]

and basically anything other than the three-dimensional definition of work?
 
  • #8
Pengwuino
Gold Member
5,124
17
Work can't be broken down into components, your first line makes no sense.
 
  • #9
33,853
11,544
I am trying to understand how this definition was derived
This approach doesn't make sense to me. Definitions are not derived, they are just defined. Then you go on from the definitions and derive other properties, such as conservation principles. You don't derive definitions.
 
  • #10
Pengwuino
Gold Member
5,124
17
This approach doesn't make sense to me. Definitions are not derived, they are just defined. Then you go on from the definitions and derive other properties, such as conservation principles.

To add to this, it is defined in such a manner so that certain conservation laws, that we know are valid in reality, are valid mathematically and in theory.

If we defined work as something different such as

[itex]W = \int {\bf{F}} \cdot d{\bf{v}}[/itex]

we wouldn't be able to define work as part of any sort of energy conservation that we're accustomed to.
 
  • #11
Forrest T
23
0
I am trying to derive the equation for work in three dimensions. Here is what I have so far:

In one dimension:

[itex]\int ^{b}_{a} F_{x}dx = \int ^{b}_{a} m\frac{v_{x}}{dt}dx = \int ^{b}_{a} m\frac{v_{x}}{dt}v_{x}dt = \int ^{b}_{a} mv_{x}dv_{x} = \frac{1}{2}mv^{2}_{b} - \frac{1}{2}mv^{2}_{a}[/itex]

[itex] W_{ab} = K_{b} - K_{a}[/itex]

In three dimensions:

[itex]W_{ab} = \sum _{i} (W_{i})_{ab} = W_{x} + W_{y} + W_{z}[/itex]

[itex]W_{ab} = \int ^{x_{b}}_{x_{a}} F_{x}dx +\int ^{y_{b}}_{y_{a}} F_{y}dy + \int ^{z_{b}}_{z_{a}} F_{z}dz [/itex]

[itex]W_{ab} = \int ^{\vec{r_{b}}}_{\vec{r_{a}}} \vec{F} \cdot d{\vec{r}}[/itex]

[itex]W_{ab} = (\frac{1}{2}mv^{2}_{x_{b}} - \frac{1}{2}mv^{2}_{x_{a}}) + (\frac{1}{2}mv^{2}_{y_{b}} - \frac{1}{2}mv^{2}_{y_{a}}) + (\frac{1}{2}mv^{2}_{z_{b}} - \frac{1}{2}mv^{2}_{z_{a}}) [/itex]

[itex] = \frac{1}{2}m(v^{2}_{x_{b}} + v^{2}_{y_{b}} + v^{2}_{z_{b}}) - \frac{1}{2}m(v^{2}_{x_{a}} + v^{2}_{y_{a}} + v^{2}_{z_{a}}) [/itex]

[itex] = \frac{1}{2}m(\vec{v_{b}} \cdot \vec{v_{b}}) - \frac{1}{2}m(\vec{v_{a}} \cdot \vec{v_{a}}) [/itex]

[itex] = \frac{1}{2}mv^{2}_{b} - \frac{1}{2}mv^{2}_{a} [/itex]

Therefore,

[itex] P = \frac{dW}{dt} = \frac{d}{dt}(\int ^{x}_{x_{a}} F_{x}dx +\int ^{y}_{y_{a}} F_{y}dy + \int ^{z}_{z_{a}} F_{z}dz) [/itex]

Using the fundamental theorem of calculus,

[itex] P = F(x)\frac{dx}{dt} + F(y)\frac{dy}{dt} + F(z)\frac{dz}{dt} [/itex]
[itex] P = F(x)v_{x} + F(y)v_{y} + F(z)v_{z} = \vec{F} \cdot \vec{v} [/itex]

The equations

[itex]W_{ab} = \int ^{x_{b}}_{x_{a}} F_{x}dx +\int ^{y_{b}}_{y_{a}} F_{y}dy + \int ^{z_{b}}_{z_{a}} F_{z}dz [/itex]
[itex] F(x)v_{x} + F(y)v_{y} + F(z)v_{z} [/itex]

provide motivation for the definition of the dot product as

[itex] \vec{A} \cdot \vec{B} = A_{x}B_{x} + A_{y}B_{y} + A_{y}B_{y} [/itex]

which can be used to show that

[itex] \vec{A} \cdot \vec{B} = AB [/itex]cos[itex] \theta [/itex]

where [itex] \theta [/itex] is the angle between vectors [itex] \vec{A} [/itex] and [itex] \vec{B} [/itex].


The thing is, I don't know how to substantiate my claim that

[itex]W_{ab} = \sum _{i} (W_{i})_{ab} = W_{x} + W_{y} + W_{z}[/itex]

Does that make sense?
 
Last edited:
  • #12
Forrest T
23
0
I guess I am trying to justify the definition rather than derive it.
 
  • #13
Pengwuino
Gold Member
5,124
17
I don't know how to substantiate my claim that

[itex]W_{ab} = \sum _{i} (W_{i})_{ab} = W_{x} + W_{y} + W_{z}[/itex]

Does that make sense?

You're labeling work as components of a vector but they're not. You could say [itex]W_{x} = \int F_x dx[/itex] and say that [itex]W_{x}[/itex] is the amount of work done by movement in the x-direction, but it certainly doesn't promote work to something that can be defined as [itex]{\bf{W}} = (W_x, W_y, W_z)[/itex]
 
  • #14
Forrest T
23
0
The point of that equation is that the addition of work follows the rules of scalar addition rather than vector addition. For example,

[itex] \vec{F} \neq F_{x} + F_{y} + F_{z} [/itex]

but

[itex] W = W_{x} + W_{y} + W_{z} [/itex]
 
  • #15
JeffKoch
400
1
Because it is simply a scalar, you defined it by the equation in your first post and could have saved the effort of typing the next dozen equations after that. If that does not suffice, consider that work is energy expended, and energy is a scalar.
 
  • #16
Forrest T
23
0
I did not want to go off of what I wrote in my first post because I wanted to understand the motivation for the definition of the dot product.

Is it possible to answer my question without presuming that I know anything about energy? Why is energy a scalar anyway?
 
  • #17
khemist
258
0
Because energy doesn't have a direction...?
 
  • #18
Forrest T
23
0
I am assuming that I don't know anything about energy, if that is possible.
 
  • #19
JeffKoch
400
1
Is it possible to answer my question without presuming that I know anything about energy? Why is energy a scalar anyway?

Because it is defined that way. I think this discussion will just wrap in circles - work is by definition a scalar number. You can define some other quantity if you like that is related to energy and is a vector, for example a Poynting vector when discussing the direction of energy flux in electromagnetic waves, but the usual definition of work that everyone is familiar with is a scalar number. There's nothing deeper to uncover.
 
  • #20
khemist
258
0
I am assuming that I don't know anything about energy, if that is possible.

Wait, what? Wikipedia is a good place to start.
 
  • #21
DaveC426913
Gold Member
21,445
4,925
Wait, what? Wikipedia is a good place to start.
I think he is asking you to humour him. Pretend he knows nothing. Make no assumptions.
 
  • #22
Forrest T
23
0
Yes, please humor me. lol.
 
  • #23
Forrest T
23
0
@JeffKoch

I understand what you're saying, but...
In The Feynman Lectures on Physics, Volume I, an exploration of torque produces a motivation for the definition of the vector (cross) product. I think there is a similar procedure with work, but I don't know why work is defined as a scalar. The work-energy theorem in one dimension is not useful for this, and I don't know what else to use to show that work is a scalar.
 
  • #24
Forrest T
23
0
Anyone?
 
  • #26
JeffKoch
400
1
..and I don't know what else to use to show that work is a scalar.

Now this is going in circles, it's a scalar because it's defined to be a scalar. [gives up]
 
  • #27
johng23
294
1
Wikipedia, on the work-energy theorem:
The theorem is particularly simple to prove for a constant force acting in the direction of motion along a straight line. For more complex cases, however, it can be claimed that very concept of work is defined in such a way that the work-energy theorem remains valid.

Why insist on understanding work without any reference to energy? Energy is the more fundamental quantity.
 
  • #28
Black Integra
56
0
Ok, all I can say is

1) Dot product of two vectors is scalar, thus work is scalar.

2) Work is not vector; you could not write it as Wx+Wy+Wz because IT ISN'T VECTOR.

3) Yes...is that clear
 
  • #29
jtbell
Mentor
15,969
4,774
Note that there is a motion-related quantity that has direction: momentum, [itex]\vec p = m \vec v[/itex].

In the 1700s, there was a lot of confusion about whether the "proper" quantity that combines mass and velocity contains v (momentum) or v^2 (kinetic energy). Finally physicists realized that we need both of them, for different purposes.

We define work the way we do, so as to have a conserved quantity via the work-energy theorem:

[tex]\int{\vec F \cdot d \vec s} = W = \Delta K = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2[/tex]

When we integrate force over time (instead of over distance) we get the change in momentum, via the impuse-momentum theorem:

[tex]\int{\vec F dt} = \vec I = \Delta \vec p = m \vec p_f - m \vec p_i[/tex]
 
  • #30
Delta2
Homework Helper
Insights Author
Gold Member
5,695
2,473
@JeffKoch

I understand what you're saying, but...
In The Feynman Lectures on Physics, Volume I, an exploration of torque produces a motivation for the definition of the vector (cross) product. I think there is a similar procedure with work, but I don't know why work is defined as a scalar. The work-energy theorem in one dimension is not useful for this, and I don't know what else to use to show that work is a scalar.

So you asking why work is scalar and why dot product is a scalar? Hard to resolve this without falling back to the definition of energy and that energy is a scalar (therefore it doesn't matter if we have kinetic energy on the x,y or z axis we just add em to find the total energy). If work wasnt a scalar how it could be a vector what physical meaning could you give to the direction a work could have in order to be a vector?
 
Last edited:

Suggested for: Why is work a scalar?

Replies
15
Views
652
Replies
23
Views
2K
Replies
7
Views
388
Replies
22
Views
1K
Replies
4
Views
4K
Replies
1
Views
397
Replies
9
Views
669
  • Last Post
Replies
2
Views
414
  • Last Post
Replies
1
Views
334
  • Last Post
Replies
12
Views
354
Top