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Work (in physics) is a scalar. Why is it not a vector?

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    $\begingroup$ What direction would you like it to point in? $\endgroup$ – Michael Brown Oct 25 '13 at 11:20
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It's defined as a dot-product (or scalar product) of force and displacement, both of which are vectors.

A scalar product of two vectors gives a scalar result (aptly named!).

$$dW = \vec{F}\cdot\vec{S} = {\|F\|}{\|S\|}\cos\theta$$ ($\theta$ being the angle between the vectors).

No direction, only magnitude.

Thinking logically, what would be the direction of work, anyway? You may say, "In the direction of displacement!", but then why not in the direction of force? And if you say the direction of both, well then, it isn't always the same! A force can do work on a body even displacing at an angle to the direction of force ($\theta$!).

=>Note that when $\theta$ is $90^\circ$, the result will be zero ($\cos 90^\circ = 0$). When force and displacement are perpendicular, the force does no work on the body!


Edit: As said by @anna: Please also note that work is part of the energy in a system (work and energy) and energy is a scalar. If it were not so we would not be talking of "conservation of energy" as an experimental observation. Energy is a scalar.

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Another way to see this is to test how it transforms under rotation of coordinate axes. Vectors and scalars have distinct transformation patterns. For simplicity if we assume a three dimensional Cartesian coordinate system then knowing that both force and displacement are vectors, i.e., their components transform under same rotation as: $$A_i \rightarrow A_i^{\prime}= \sum_{j=1}^{3}a_{ij}A_j$$ where the $a_{ij}$'s are elements of an orthogonal matrix with determinant=+1, one can check that work done $W \rightarrow W^{\prime}=W$, i.e., work done remains invariant under rotation of coordinate axes. In other words, work done due to displacement caused by a force is a scalar quantity.

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The dot product of vector quantities is always scalar which means it is has only magnitude and no direction.

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  • $\begingroup$ You are correct, but to clarify work is defined as (in case some might argue that work does not follow from the dot product of two vectors): $$ W=\int_C{\vec{F}\cdot d\vec{x}} $$ $\endgroup$ – fibonatic Feb 5 '14 at 12:33
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In addition to the scalar product reason mentioned above ,I would go a step back and explain it based on the why we need to define vector quantity. Some quantity could be measured in + and -ve like temperature , distance etch because we can represent it in terms of positive and negative value and it give us full information. But quantity like displacement need us to define direction as displacement of 20m doesn't specify which direction the 20m is , which is necessary as displacement is shortest distance between two points and it couldn't be in any direction. For Work , for example, if we do 10 J work when we push object to 10m in east and then when we push 10 m in West with say 15 Joule, the total work would be 25 Joule here we are not bothered here about direction as it doesn't add any extra information. (Please note if force is in opposite direction to displacement then work is negative) .

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protected by Qmechanic May 21 '14 at 15:17

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