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This question already has an answer here:

My university lecturer told me that:

If $$F=m\dfrac{dv}{dt}$$ it's incorrect to write $$F\,dt=m\,dv\tag{1}$$ but it is okay to write $$\int F\,dt=\int m\,dv$$ for Newtons' second law.

But never explained why $(1)$ is mathematically incorrect.

My high school teacher told me that:

Derivatives with respect to one independent variable can be treated as fractions.

So this implies that $(1)$ is valid.

This is clearly a contradiction as my high school teacher and university lecturer cannot both be correct. Or can they?

Another example of this misuse of derivatives uses the specific heat capacity $c$ which is defined to be $$c=\frac{1}{m}\frac{\delta Q}{dT}\tag{2}$$

Now in the same vain another lecturer wrote that $$\delta Q=mc\,dT$$ by rearranging $(2)$.

Another contraction to the first lecturer. I this really allowed or if it's invalid then which mathematical 'rule' has been violated here?


EDIT:

In my question here I have used formulae that belong to Physics but these were just simple examples to illustrate the point. My question is much more general and applies to any differential equation in mathematics involving the treatment of derivatives with respect to one independent variable as fractions.

Specifically; Why is it 'strictly' incorrect to rearrange them without taking the integral of both sides?

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marked as duplicate by Community Oct 30 '16 at 14:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This is a question about useful intuition vs formal correctness. I agree with the second lecturer. Related: math.stackexchange.com/questions/1812327/…, math.stackexchange.com/questions/1977134/most-useful-heuristic/… $\endgroup$ – Ethan Bolker Oct 30 '16 at 14:03
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    $\begingroup$ It's usually okay to write $Fdt=mdv$. But your university lecturer may have some reasons behind by saying it's incorrect, so you should ask him/her for details. $\endgroup$ – Cave Johnson Oct 30 '16 at 14:04
  • $\begingroup$ You can write $F\,dt=m\,dv$ if you mean differential forms. Note, that $\delta Q$ isn't exact differential form, so you can't write it as $dQ$. $\endgroup$ – Canis Lupus Oct 30 '16 at 14:41
  • $\begingroup$ The fact that it can be written as $F\,dt=m\,dv$ without anything going wrong is the beauty of the Leibniz notation of differentials. There are plenty of questions here (apart from the marked duplicate) regarding this only. This] post has the links of the related posts as well. $\endgroup$ – StubbornAtom Oct 30 '16 at 17:17
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It is possible that your lecturer is telling you that, on its own, the expression $dt$ is meaningless, whereas $\int...dt$ does mean something quite specific, i.e. an operator or instruction to integrate with respect to $t$.

In contrast, $\delta t$ does mean something specific, i.e. a small increment in the value of $t$.

However, most people are fairly casual about this sort of thing.

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The difference is not so much between university lecturers and highschool teachers as between mathematicians and physicists. Some mathematicians tend to frown on certain procedures that are perfectly acceptable to physicists. I was careful to write "some" because mathematicians familiar with Robinson's framework with infinitesimals do assign a perfectly rigorous meaning to formulas like $F\, dt = m\, dv$; see Keisler's beautiful textbook Elementary Calculus for details.

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More precisely,

$$F=\frac{dp}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}$$

$$\int F\, dt=\Delta p$$

If $\dfrac{dm}{dt}=0$, then $$\int F\, dt=m\int dv$$

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  • $\begingroup$ In classical Mechanics $\frac{dm}{dt}=0$ $\endgroup$ – Ganesh Oct 30 '16 at 14:32
  • $\begingroup$ @RAM, That's not true for rockets. $\endgroup$ – Ng Chung Tak Oct 30 '16 at 14:36
  • $\begingroup$ Yes,In theory of relativity that is not true. $\endgroup$ – Ganesh Oct 30 '16 at 14:41
  • $\begingroup$ @RAM, See rocket principle $\endgroup$ – Ng Chung Tak Oct 30 '16 at 14:46
  • $\begingroup$ Yes i agree with you. I said dm/dt=0 is not true $\endgroup$ – Ganesh Oct 30 '16 at 14:47

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