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Many quantities in physics are defined as ratio of infinitesimal quantities. For example: $$\rho(x)=\frac{dm}{dx}$$ or $$P(t)=\frac{dW}{dt}$$

Are these quantities actually derivatives? I mean if we want to calculate density it wouldn't make sense to select a infinitesimal length element $dx$ and then measure the mass difference $dm$ (same for work).

Also are $dm$ and $dW$ infinitesimal? I can't understand why the change of a function $df=f(x+dx)-f(x)$ must be infinitesimal. Or because we measure them in infinitesimal intervals $dx$ and $dt$ we use the differential notation? Another example is the first law of thermodynamics where: $$dU=dq+dw$$ Why again we use the differential notation for change in internal energy? Is the sum of two infinitesimals an infinitesimal?

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You could think of the functions $m(x)$ or $W(t)$ as functions that represent some sort of accumulation. For example with power, you could define $W(t)$ as the work done since some time $t=t_0$. Then the instantaneous power is just $P=\text dW/\text dt$. For linear density you could think of $m(x)$ as the the mass you have "counted" starting at the edge of the body. Then the linear density is just $\rho=\text dm/\text dx$.

In any case, we usually express these equations a little differently $$\text dm=\rho\,\text dx$$ $$\text dW=P\,\text dt$$ and these are better at getting at the heart of what we mean. The infinitesimal amount of mass $\text dm$ contained at some location $x$ is $\rho(x)\, \text dx$. The infinitesimal amount of work $\text dW$ performed at time $t$ is $P(t)\,\text dt$.

This is also reflected in the first law of thermodynamics $\text dU=\text dq+\text dw$. The infinitesimal change in the internal energy $\text dU$ of the system is accounted for by the infinitesimal amount of energy that enters/leaves due to heat $\text dq$ and the infinitesimal amount of energy that enters/leaves due to work $\text dw$. Note that sometimes you might instead see $\delta q$ and $\delta w$ to denote a path deoendency. Yes, the sum of two infinitesimal values is also infinitesimal in general.

It does not have to be the case in general that $\text df=f(x+\text dx)-f(x)$, this is because in general $f$ does not need to be continuous. I would think in that case then $\text df$ would not be defined since the limit $f(x+\text dx$ as $\text dx\to 0$ would not be defined, but I am not a mathematician so I might be off here. In any case, we usually work with continuous functions with continuous derivatives, so this usually isn't an issue.

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  • $\begingroup$ Thanks for the answer. The accumulation concept solved some of my misunderstanding. What that I still don't understand is why we assume infinitesimal amount of mass at some location $x+dx$ and not a finite? $\endgroup$ – Antonios Sarikas Jul 4 at 16:34
  • $\begingroup$ @AntoniosSarikas You can do finite (which you have to do in reality), but then you get errors, as with any numeric method. $\endgroup$ – BioPhysicist Jul 4 at 17:44
  • $\begingroup$ Why $dm$ must be so small? We can make as small as we want the interval ($dx$) but why the mass element at that interval must be so small and not $5$ kg for example? $\endgroup$ – Antonios Sarikas Jul 4 at 18:05
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    $\begingroup$ @AntoniosSarikas Like I said, that would mean the function is not continuous. $\endgroup$ – BioPhysicist Jul 4 at 18:50
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    $\begingroup$ @AntoniosSarikas: $\Delta m$ could be 5kg for a small $\Delta x$. But it will be even smaller for smaller $\Delta x$, and (assuming the function is reasonably well-behaved, ie. differentiable) the ratio $\frac{\Delta m}{\Delta x}$ will tend to some value. $\frac{dm}{dx}$ represents that value. $\endgroup$ – BlueRaja - Danny Pflughoeft Jul 4 at 20:54
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If your are interested in a bit more mathematical rigor:

You are absolutely right, $\rho(x)$, mathematically speaking, is the derivative. It is nothing but $m^{'}(x)$. If people pull tricks like multiplying by $\mathrm{d}x$ or something similar, is just a notational trick (although it works every (?) time for "well-behaved" functions).

The total differential $\mathrm{d}U = \mathrm{d}q + \mathrm{d}w$ is another beast, so to say, it is called a "one-form" and - mathematically speaking - to be understood in terms of integration, such that:

$\int \mathrm{d}U = \int \mathrm{d}q + \int \mathrm{d}w$

The talk about "infinitessimal quantities" is a nice model of thought and often used for practical calculations but ultimately you have to look at the mathematical definitions to "know what the objects really are".

Hope this helps.

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The mass density $\rho(x)$, and power $P(t)$ are defined as derivatives. This is analog to the (one-dimensional) velocity, $v = dx/dt$. At first glance it does not make sense to define the velocity as a derivative, but seams to be more intuitive to define it as the ratio of finite differences, $\Delta x/\Delta t$. However, one we think of each quantity as a function the derivative definition becomes clearer.

So let's consider a particular case $x(t) = t^2$ and ask: "What is the velocity at time $t_1=2s$?" Taking the concept of finite difference, we could e.g. take the position at times $t_0=1.5s$ and $t_2=2.5s$ and calculate $v(t_1) \approx \frac{x(t_2)-x(t_0)}{t_2 - t_1}$. However, plotting the position as a function of time in blue, and the calculated velocity in red, we see that the result is not perfect,

velo1

Hence, it seams natural to reduce the time difference $\Delta t = t_2 - t_1$. Doing so, we will observe that the red line approaches "the local shape" of the blue line closer and closer. In the limit $\Delta t \to 0$, which is merely the derivative, we obtain the closest match.

The same argument is true for your examples. E.g. in the case of the mass density, we ought to think in terms of functions. Hence, the mass $m(x)$ is a function, and therefore the mass density $\rho(x)$ becomes a function.

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We do not deal in infinitesimal quantities in standard analysis. Even in non-standard analysis in which infinitesimal quantities are permissible, we cannot mathematically construct a single instance of an infinitesimal quantity.

The derivative is not a ratio of infinitesimal quantities, it is the limit of the ratio of small quantities

$$ y' = \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\delta y}{\delta x} $$

where the limit means that for any $\epsilon >0$ there is a $\delta >0$ such that if $0 < \delta x < \delta$ then $$\bigg|y' - \frac{\delta y}{\delta x} \bigg| < \epsilon .$$

For the purpose of physics, we are not interested in exact numerical results, but in results to the accuracy of measurement. In practice we just need the measured result to be within $\epsilon$ of the notional value, where $\epsilon$ refers to measurement accuracy.

It is common therefore to treat very small quantities $\delta x$ as though they were infinitesimal quantities $dx$ and also to use the notation $dx$ and refer to an infinitesimal quantity when strictly we are talking of quantities sufficiently small that any error is less than required measurement accuracy.

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