What does this equation for density mean? $$\rho = \lim_{\Delta V\to\varepsilon^3} \ \frac{\Delta m}{\Delta V}$$
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2$\begingroup$ This seems like a perfectly reasonable question, if someone has not been exposed to calculus and it's application to physics. $\endgroup$– StephenG - Help UkraineCommented Apr 22 at 19:53
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1$\begingroup$ Where is this equation from? Were $\Delta m$, $\Delta V$, and $\epsilon$ defined? Is $\rho$ being defined at a point $x$? Did the source specify that "$\epsilon\rightarrow 0$? Related: "Density definition implies total mass integral". Also related: "Mass-density functions: how is there mass-density at points?" and "How do we define density (real one, not mathematical)?"?. $\endgroup$– David BaileyCommented Apr 23 at 6:56
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In the equation for density:
$$\rho = \lim_{\Delta V\to\varepsilon^3} \ \frac{\Delta m}{\Delta V}$$
$\rho$ is density.
$\Delta V$ represents a small change in volume.
$\Delta m$ represents a small change in mass, corresponding to the change in volume $\Delta V$.
The symbol $\varepsilon^3$ signifies the volume of an infinitesimally small region, and taking the limit as $\Delta V$ approaches $\varepsilon^3$ allows us to consider the density at a single point in space. This concept is often used in calculus and mathematical analysis to define density at a point.
answered Apr 22 at 11:23
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1$\begingroup$ Thank you so much. This helps but why is it the symbol E^3. And why would the similar equation for pressure be E^2? $\endgroup$– sebbbbCommented Apr 22 at 11:44
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$\begingroup$ $\varepsilon^3$ I think represents an incrementally small 1 dimension cubed, i.e. 1d to the power of 3 is volume (3d), or an incrementally small volume. Not zero volume, but near to zero volume. $\endgroup$ Commented Apr 22 at 11:57
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2$\begingroup$ I would have thought $\Delta V \rightarrow 0$ would be a more standard and less confusing notation. $\endgroup$ Commented Apr 23 at 6:34
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1$\begingroup$ The SI units for pressure are kg * m^-1 * s^-2. This is because pressure is a force (kg * m) divided by an area (m^2). To find the pressure at an infinitesimally small point, you'd take the limit as the area the force is divided over approaches 0, ie E^2. $\endgroup$– elfeiinCommented Apr 23 at 8:02