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Why is work a scalar?

  1. Oct 16, 2011 #1
    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: Oct 16, 2011
  2. jcsd
  3. Oct 16, 2011 #2

    Vanadium 50

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    What else would it be? It can't be a vector, since it has no direction.
     
  4. Oct 16, 2011 #3
    Why doesn't it have a direction?
     
  5. Oct 16, 2011 #4

    DaveC426913

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    Moving an object North and moving an object West still accomplish the same amount of work. Work is only a magnitude.
     
  6. Oct 16, 2011 #5
    How do you know that?
     
  7. Oct 16, 2011 #6

    DaveC426913

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    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.
     
  8. Oct 16, 2011 #7
    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?
     
  9. Oct 16, 2011 #8

    Pengwuino

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    Work can't be broken down into components, your first line makes no sense.
     
  10. Oct 16, 2011 #9

    Dale

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    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.
     
  11. Oct 16, 2011 #10

    Pengwuino

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    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.
     
  12. Oct 16, 2011 #11
    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: Oct 16, 2011
  13. Oct 16, 2011 #12
    I guess I am trying to justify the definition rather than derive it.
     
  14. Oct 16, 2011 #13

    Pengwuino

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    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]
     
  15. Oct 16, 2011 #14
    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]
     
  16. Oct 16, 2011 #15
    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.
     
  17. Oct 16, 2011 #16
    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?
     
  18. Oct 16, 2011 #17
    Because energy doesnt have a direction...?
     
  19. Oct 16, 2011 #18
    I am assuming that I don't know anything about energy, if that is possible.
     
  20. Oct 16, 2011 #19
    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.
     
  21. Oct 16, 2011 #20
    Wait, what? Wikipedia is a good place to start.
     
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