2
$\begingroup$

What does it actually mean "this concept is defined as ..."? for example when someone tells me "Work is defined as $W=\int \vec{F}\cdot d\vec{x}$".

What I understand:

  • I understand the need for the integral sign, the need for summing up all the small work's that make up the total work done.

What I don't understand:

  • why does it have to be this way? Why does $W= \int\vec{F}\cdot d\vec{x}$ have to be the definition for work? why does work done have to be defined as "The product of the force and distance moved in the direction of the force"? Does it have a mathematical derivation or is it just defined empirically (Through experiments)?

On the contrary the concept of electric flux $\phi=\vec{E}\cdot\vec{A}$
I do understand why it has to be this way. I understand that electric flux is proportional to the number of field lines that pass through a surface and that $\mid\vec{E}\mid$ is proportional to the electric flux density-$\dfrac{Number OfFieldLines}{Area}$ thus it makes sense the definition $\phi=\vec{E}\cdot\vec{A}$

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ You seem to be mixing up definitions and derivations. Definitions don't have to be anything. You define something first, and then you can see if it has any useful properties using derivations. $\endgroup$
    – BioPhysicist
    Commented Apr 26, 2019 at 22:30
  • 1
    $\begingroup$ Gaspard-Gustave Coriolis determined empirically that work was the "weight lifted through a height", which was based on the use of early steam engines to lift buckets of water out of flooded ore mines. The steam engine has contributed more to science than science has to the steam engine. $\endgroup$
    – Cinaed Simson
    Commented Apr 27, 2019 at 2:48

5 Answers 5

Reset to default
3
$\begingroup$

why does work done have to be defined as "The product of the force and distance moved in the direction of the force"? Does it have a mathematical derivation or is it just defined empirically (Through experiments)?

Neither. It's a definition. Definitions are not derived. Definitions are anything you want them to be. I can define the "kinetic momentum" of an object to be the product of the object's momentum and kinetic energy: $$p_K=pK=\frac12m^2v^3$$

And it's done. I defined this value. Nothing else needs to be said about it. I didn't do any derivations, proofs, etc. That's how definitions work. You define something, and that's what it is.

The real question is, "Is this definition useful in describing the world around us?" My new definition probably is not. But work is definitely useful. You can show that is value defined by $\text dW=\mathbf F\cdot\text d\mathbf x$ relates to another defined quantity, the kinetic energy. You can show that the work done by conservative forces relates to another defined quantity called potential energy. And I could go on and on.

So, work is defined the way it is because it's useful.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Oh so you mean we defined some quantities like kinetic energy, potential energy etc. And then we also defined Work done to be a quantity that if it is non zero then that means it changes the energy of a system. And it just seems so that this arbitrary definition links very well with the other definitions we created thus it is useful? $\endgroup$
    – Luca Ion
    Commented Apr 27, 2019 at 7:48
  • $\begingroup$ @LucaIon kind of, yes. Although these definitions probably had some motivation behind them than just making up random things out of thin air and seeing what works. The other answers go into some motivations behind the definitions. I really just wanted to point out that you don't derive definitions. You seemed to be hung up on this idea $\endgroup$
    – BioPhysicist
    Commented Apr 27, 2019 at 10:32
  • $\begingroup$ Allright I think I get it now thank you. But how do I know, when I get introduced to a new concept that the equations are definition and are not relations that can be derived from a more general definition? $\endgroup$
    – Luca Ion
    Commented Apr 27, 2019 at 10:44
  • $\begingroup$ @LucaIon You can tell from the context it is given in. Usually a book will use the word "defined", "given by", etc. Things with names are definitions, velocity, momentum, work, electric field, etc. Then you can derive relations between these from either math or experiments. Newton's laws, work-energy theorem, Gauss's law, Snell's law, etc. Or "derivations" could be something like "the momentum of a photon is given by $p=E/c$" where even though momentum is a definition you can derive what the momentum of something is in terms of other definitions depending on the system. $\endgroup$
    – BioPhysicist
    Commented Apr 27, 2019 at 12:25
  • $\begingroup$ @LucaIon You just need to read things carefully and think about what the equations presented to you really represent. $\endgroup$
    – BioPhysicist
    Commented Apr 27, 2019 at 12:27
3
$\begingroup$

The definition of electric flux that you quote "has to be" that way because it is a special case of the notion of flux in general. On the other hand, the definition of work that you quote does not have reference to a more general notion: that is just what work is. The motivation for giving the quantity $\int\vec{F}\cdot d\vec{x}$ a name is that it turns out to be a very useful quantity: the work-energy principle provides the motivation for defining the quantity $W$, but does not itself constitute a definition.

Improve this answer
$\endgroup$
3
$\begingroup$

You choose two interesting examples.

The difficulty with the first might be that the mathematical object that we use to represent force, the vector, is really not up to the task. Force and electric field can be represented by vectors, but occasionally a situation arises when we have to recognize that there is something different between the two. There are names for the two types: contravarient (electric field), and covarient (force). As luck would have it, I used this very example, work, to explain why this distinction is necessary in an earlier StackExchange post

But there is another mathematical object that does a better job of representing a force than does a vector. It is the 1-form. In three-dimensional Euclidean space there isn't much of an advantage mathematically for using 1-forms vs. vectors, but in other situations there is. On the other hand, the 1-form does present a metaphor, or intuitive picture, that is similar to the picture of "number of field lines piercing the surface". A 1-form is pictured as a series of parallel sheets, whose normal vector would be perpendicular to the sheets. That takes care of the direction of the force. The magnitude is represented by the spacing between the sheets. The closer the spacing, the larger the magnitude.

With this picture of force as 1-form, we can associate work with "number of sheets pierced by the displacement of the object".

I've often wondered if it would be better to introduce 1-forms right away in "Physics 101" rather than leaving the topic to be discovered much too late to be of any real use. But even if that would be a good idea, and I'm not sure it is, it's not going to happen.

Improve this answer
$\endgroup$
2
$\begingroup$

To expand on Will's and Jitendra's answer:

We define work arbitrarily, to be the sum of the force along the path, $W=\int_L \vec{F}\cdot d\vec{l}$. This is a useful definition because of the work-energy theorem, that shows us that the work is the change in kinetic energy. So we ultimately came up with this concept, "work", and its definition because carving out reality using this concept turned out to be useful.

In the case of flux, we are guided by mathematical and geometric intuition to generically define the flux of any field through a surface, $\Phi=\iint_S \vec{E} \cdot d\vec{a}$. We then later find that this is a useful definition through laws such as Gauss' law for the electric field, $\Phi=q_{enc}/\epsilon_0$. So even if initially our motivation for the definition is mathematical, ultimately we still use this definition because we find it useful to carve-up reality in this way, because the laws of nature are simpler to consider and apply when we use this concept.

Improve this answer
$\endgroup$
2
$\begingroup$

There is no analytical derivation for any definition. But we can understand it through a simple example.

Assume a mass $m$ kept on a horizontal surface. A force $F$ starts to act at time $t=0$. At time $t=t$ the mass has moved a distance $x$ and acquired a velocity $v$. If one asks the change in energy of the mass due to the force we can say it is $\frac{1}{2}mv^2$. Now one can also ask what does it take for the force $F$ to create the change of energy of the mass. Or we can also say what is the work done by the force $F$ in order to change the energy of the mass.

We know $v^2-0^2=2ax$ which imples $\frac{1}{2}mv^2=\frac{1}{2}m(2ax)=max=Fx$ and we define work by force $F$ as $Fx$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ This doesn't make sense. You don't derive a definition $\endgroup$
    – BioPhysicist
    Commented Apr 26, 2019 at 22:31
  • $\begingroup$ I am not deriving it. Just trying to make sense of the definition through an example. $\endgroup$
    – Jitendra
    Commented Apr 27, 2019 at 0:40
  • $\begingroup$ I am not aware of any analytical derivation for the definition of work First sentence of your answer. This doesn't make sense to me. But your motivation of it with the example is fine. I'm sorry, I should have been more specific. $\endgroup$
    – BioPhysicist
    Commented Apr 27, 2019 at 1:28
  • 1
    $\begingroup$ Yes, I am not aware and also there might not be one. As you said definitions do not have derivations. However, we can still try to understand them through examples in order to realise why and how the definition makes sense. I think I should change "I am not aware of any analytical derivation for the definition of work" to "There is no derivation for a definition." Thank you for pointing it out. $\endgroup$
    – Jitendra
    Commented Apr 27, 2019 at 1:34
  • $\begingroup$ @Jitendra But haven't you just derived it by using the definition of kinetic energy? $\endgroup$
    – Luca Ion
    Commented Apr 27, 2019 at 7:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.