I asked many people why work is scalar. But the questions and the answers just cycles.

My question : Why is work a scalar quantity?

Their answer : Because it is the dot product of Force and Displacement. So it is a scalar quantity.

So I asked,

My question : Why is work the dot product of force and displacement and not the cross product?

Their answer : Because it is a scalar quantity and not a vector.

So can anyone please tell me why work is scalar apart from this cycle of questions and answers? Is there any other logic which can answer both questions?

Edit : $W=Fs\cos\theta$. So $Fs\cos0 = -Fs\cos180$. So shouldn't work be a vector? This confusess me.

  • $\begingroup$ If you are happy with my answer, you can accept it. Otherwise let me know in a comment under my answer if you still have doubts and I can address them. $\endgroup$
    – Andrea
    Aug 1, 2018 at 11:10

1 Answer 1


Well, say that you are looking at a man lifting boxes. Each box weighs 10kg. At first you look at him standing, but since just looking at him made you tired, you decide to lie down. Now, from your horizontal position, the scene looks different, but is the man doing more, less, or the same amount of work per box?

The word “scalar” is often used to just mean “number”, but it actually has a technical meaning: a scalar is a quantity that does not change when the system is transformed following one of the symmetries of the theory.

In this case, I think you are talking about classical mechanics, and the transformations involved are rotations in 3D space. As you know, the dot product of two vectors is invariant under rotation, that is why it's called “the scalar product”. The cross product, also called the vector product, transforms as a vector.

Now there is a final bit of the puzzle. The magnitude of a vector is also a scalar. Rotating the vector does not change its length. So in fact we have two ways to obtain a scalar from two vectors $\vec F$ and $\vec s$: taking the scalar product $\vec F \cdot \vec s$ or the magnitude of the vector product $|\vec F \times \vec s|$. How do we chose one?

Well, we know that when the force and the displacement are aligned, the work is maximum, and when they are perpendicular the work is minimum. This allows us to chose one alternative.

  • 1
    $\begingroup$ Oh sorry. I totally forgot about that. $\endgroup$ Aug 1, 2018 at 11:31
  • $\begingroup$ @AsifIqubal no worries ;) $\endgroup$
    – Andrea
    Aug 2, 2018 at 7:06
  • $\begingroup$ So the new definition would be that scalar is not frame dependant and Vector is frame dependant?. Please answer $\endgroup$
    – user253164
    Mar 22, 2020 at 11:32
  • 1
    $\begingroup$ @user253164 Yes. This is not "the new" definition. This the most common definition that is most often used in physics: a scalar quantity is a quantity that does not change when you change frames. A vector quantity is one that changes in a certain way. $\endgroup$
    – Andrea
    Mar 25, 2020 at 13:22
  • $\begingroup$ @Andrea oh okay. This was actually never mentioned in my physics book, so i thought it was a new definition... $\endgroup$
    – user253164
    Mar 25, 2020 at 13:36

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