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The equation W = $\int$$\vec F$.d$\vec s$ defines Work as a dot product of force and distance along the direction force. But as far as I know, dot product of two vectors comes from how much of one vector is projected onto another.

So what does that dot product actually mean? Is work just an invention to describe a force in scalar terms or is there an intrinsic meaning behind work?

Qmechanic
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Sashank Sriram
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    I'd suggest you to read the answer by Ron Maimon and other answers https://physics.stackexchange.com/questions/535/why-does-kinetic-energy-increase-quadratically-not-linearly-with-speed – Abhinav Dhawan Oct 19 '17 at 09:18

2 Answers2

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Your equation is not quite correct - what you described gives the differential work $dW$. Work done on a particle is defined according to a line integral: $$ W = \int_C \vec{F}\cdot d\vec{s} $$

where $C$ is the path that the particle takes, $\vec{F}$ is the force acting on the particle (which may be a function of position), and $d\vec{s}$ is the differential line element along the path.

You are correct in that the dot product represents a projection of one vector onto another. Intuitively, this integral is the mathematical equivalent of saying "at each point along the path, calculate how much the vector field $\vec{F}$ is contributing to the motion of the particle by finding the projection of the force onto the actual trajectory, and add it all up."

Billy Kalfus
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  • Thanks for the rectification. I actually meant to put a dW there, but still, it's fine. Now, allow me to reiterate my understanding of your comment in my own way. "Force, a vector, is acting on a particle, and thus, the particle moves. The direction of the motion of that particles is inclined at a particular angle to the force applied on it. So, the projection of that force onto the actual path of motion is what the scalar 'Work' is trying to explain." Is that right? – Sashank Sriram Oct 19 '17 at 09:22
  • My pleasure! If this answer helped you, it would help me if you accept the answer. – Billy Kalfus Oct 19 '17 at 09:23
  • Yes, that's correct. – Billy Kalfus Oct 19 '17 at 16:19
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Yes, the dot product $\vec a\cdot \vec b$ mathematically symbolizes the projection of one vector onto the other. In other words: The dot product multiplies the parallel components:

A cross product $\vec a\times \vec b$ is somewhat the opposite. The (length of) a cross-product is the perpendicular components multiplied.

You will often see them mathematically in these forms:

$$\vec a\cdot \vec b=a_\parallel b=a b_\parallel=ab\cos(\theta)$$ $$|\vec a\times \vec b|=a_\perp b=a b_\perp=ab\sin(\theta)$$

The last version in each line is convenient, since cosine takes care of parallel components and disregards the perpendicular ones, while sine does the opposite. $\theta$ is the angle between the two vectors.

These are merely mathematical inventions, yes. Nothing more. And they happen to be very useful in describing many physical phenomena, because many things in physical only happen between parallel or perpendicular components.

For example:

  • To tighten a screw with a wrench, you pull perpendicular - pulling parallel in the wrench will not make it turn (it will just detach). Therefore moment is defined with a cross product: $\vec \tau = \vec F \times \vec r$.
  • Work is energy added when pushing something i.e. If you push from the side on a toy train moving in its track, then you are not adding any energy to it's motion. It doesn't move sideways (the track holds it in place) and your effort was worthless. Only if you push along with the motion will you add energy and make it speed up. Therefore work is a defined as a dot product $W=\vec F\cdot \vec s$.

In the toy train example, if you push opposite to the motion, then you take energy out of the train and slow it down. You are doing work, but that work is negative.

As you see, all parallel forces do work (negative if antiparallel), while all sideways forces do nothing at all. If you push at an angle, then only the parallel component of your force has an influence. All this is included in the mathematical cosine expression of the dot product $W=Fs\;\cos(\theta)$, because a cosine is zero, when the angle is $90^\circ$.

Steeven
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