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I know that work = force times displacement with vectors ($W = \vec{F}\cdot\vec{D}$) can be calculated in the same way as a dot product, but it seems to me that it isn't really a dot product. The reason being is that they seem to be vectors in different coordinate systems. One of the vectors has a unit of length of "force size", and the other vector has a unit of length of "distance travelled". I'm not arguing with the process of calculation, but it seems to me that calling it a dot product is actually a confusion of terms. Can someone explain to me how to make it make sense as a dot product since the units are different?

The way that I would solve it from first principles is to convert the force into the direction of the distance using trig (which would get the force multiplied by $\cos(\theta)$), and then multiply it by the scalar distance that it travels in that direction.

Note - I am aware of this question, which covers the calculation piece. Again, my question is to whether this is truly a dot product or just calculated with the same steps.

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    $\begingroup$ Given the components of the vectors you can always factor out the dimensions and work with dimensionless vector components. For example $W=(1\;\mathrm{N},1\;\mathrm{N},1\;\mathrm{N})\cdot (1\,\mathrm{m},0,0) = (1,1,1)\cdot(1,0,0)\;\mathrm{Nm}$. $\endgroup$
    – mb28025
    Commented Jan 22 at 20:06
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    $\begingroup$ Why do you think the two vectors of a dot product need to have the same unit? $\endgroup$
    – Thomas Fritsch
    Commented Jan 22 at 20:06
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    $\begingroup$ You are right, work is better defined as the integral of a 1-form (the force) and so in this context the force is in fact a different type of vector than displacement ("vectors in different coordinate systems" is more accurately "vectors in different vector spaces"). Vectors are for things like describing trajectories, while 1-forms are for eating vectors and spitting out numbers like work done. They both can be treated like "arrows in space" but how they are used is pretty distinct geometrically $\endgroup$
    – Er Jio
    Commented Jan 22 at 20:36
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    $\begingroup$ You can multiply numbers with different units. You can do the same with vectors. If you add numbers, they must have the same units. Vectors are the same. $\endgroup$
    – mmesser314
    Commented Jan 22 at 21:10
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    $\begingroup$ When you multiply speed and time to find the corresponding displacement, you multiply two quantities of different units, and get a third quantity of still different units. The same happens with force and displacement to get the work. Quantities, including vectors, do not need to have the same units for their multiplication to be meaningful. $\endgroup$
    – Ján Lalinský
    Commented Jan 22 at 23:37

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Technically, the $\vec F$ term in $W = \vec F \cdot \vec D$ isn't a vector - it is a covector. It is a linear function that takes a distance vector $\vec D$ as input, and outputs a scalar $W$. There is no problem (mathematically) with a covector and a vector living in different vector spaces.

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    $\begingroup$ This is not how classical physics understands dot product(scalar product) of two vectors in 3D space, when using it to define work. Both $\vec{F}$ and $\vec{D}$ here are vectors. A covector is a linear map that takes in a vector and outputs a number; a function that produces dot product of the input vector with $\vec{F}$ is one such possible covector, and thus vector $\vec{F}$ can be used to define this special covector. This does not mean that $\vec{F}$ is a covector in the context of definition of work. $\endgroup$
    – Ján Lalinský
    Commented Jan 22 at 23:35
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    $\begingroup$ However, the classical physics interpretation of force as a vector seems to be the incorrect notion to begin with. Force is most naturally the time rate of change of momentum, momentum is certainly a covector (as it is the derivative of the Lagrangian, scalar, wrt a vector). Also eg look at the Lorentz force law where the force is the electromagnetic 2 form contracted with the velocity vector, yielding a 1 form. No matter which direction you approach force from a theoretical standpoint, it's always a covector. The interpretation is it inputs a distance (vector) and gives work (scalar) $\endgroup$
    – Toyesh Jayaswal
    Commented Jan 23 at 2:19
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It being a dot product is alright, because a dot product is meant to handle different values.

Otherwise, you'd be adding or subtracting vectors. It is during addition or subtraction that vectors should be the same. This is the same for scalar mathematics, but vectors just have the additional component of direction to go with.

Are you confused about scalar products similarly? I guess not. If so, simply extend the intuition to dot and vector products.

For dot products, for example with $W=\vec F \cdot \vec d$, we are only concerned with the direction of d, and the angle of F only affects the result because of the share of it's component in the direction of the displacement, which is what contributes to work.

Or more accurately, a dot product for two vectors of fixed sizes is maximum when they are aligned, and minimum when they are perpendicular. It measures how collinear two vectors are.

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  • $\begingroup$ I'm probably going to accept this answer as it is the more intuitive one. I'm still thinking on it, though. The more I think about it the "angle" part is unified between the two vectors, so it isn't like they are completely different coordinate systems. So the dot product is taking the angle into account prior to multiplication. That seems to make sense to me. $\endgroup$
    – johnnyb
    Commented Jan 22 at 21:52
  • $\begingroup$ I also appreciate @mb28025's comment on the OP as it helped me conceptualize it. $\endgroup$
    – johnnyb
    Commented Jan 22 at 21:54

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