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46

It means don't be a jerk. The third derivative of position (i.e. the change in acceleration) is called "jerk", though it's a little used quantity. It's called jerk because a changing acceleration is felt as a "jerk" in that direction.


41

It is exactly because we have a factor of $\frac 1 2$ in the area formula of a triangle. To understand what I'm saying, consider what is the $v(t)$ graph of a particle under constant acceleration. Some say, a good plot is worth a million words! :)


30

In physics, sometimes the third derivative of position with respect to time is called jerk.


26

When I asked my undergrad analytic mechanics professor "what does it mean for a rotation to be infinitesimal?" after he hand-wavily presented this topic in class, he answered "it means it's really small." At that point, I just walked away. Later that day I emailed my TA who set me straight by pointing me to a book on Lie theory. Fortunately, I don't ...


23

There is an old tradition, going back all the way to Leibniz himself and carried on a lot in physics departments, to think of differentials intuitively as "infinitesimal numbers". Through the course of history, big minds have criticized Leibniz for this (for instance the otherwise great Bertrand Russell in Chapter XXXI of "A History of Western Philosophy" ...


18

The result you've got would be better known as this: $$\int_0^t\biggl(\int_0^{t'} a\mathrm{d}t''\biggr)\mathrm{d}t' = \frac{1}{2}at^2$$ In other words, it's a derivation of the formula for uniformly accelerated motion. This derivation, or something like it, is one of the first things students in a good calculus-based introductory physics class learn. The ...


11

We have also the same notions of derivation, curl, etc... for functions that are less regular. When you write Maxwell's equations, you are writing a system of partial differential equations. To investigate them, you have to specify the type of solution you look for (in the language of PDEs: classic, mild, weak...) and the functional space you set your ...


11

In German, this property is known as the Transformationssatz, but I do not know any appropriate translation for it. This is, however, a special case of coordinate tranformations changing the measure by the determinant of their Jacobian, since obviously $\frac{\partial y_i}{\partial x_j} = A_{ij}$. That it is the determinant that plays a role in the ...


10

Here is a brief historical ideosyncratic intro to calculus. Calculus of finite differences Consider this problem from a typical IQ test: 2 5 10 17 26 ? What's the next number you expect in the sequence (this is not hard, you should do it). The n-th term in the sequence is given by: $$ n^2 + 1 $$ as you can see by substituting n=1,2,3,4,5, so the next ...


8

$a_x \Delta t = \Delta v_x = v_{xf} - v_{xi}$ $\Delta x = v_{x,average}\Delta t = v_{xi}\Delta t + \dfrac{1}{2}a_x (\Delta t)^2$ $\Rightarrow v_{x,average} = v_{xi} + \dfrac{1}{2}a_x \Delta t = v_{xi} + \dfrac{1}{2}(v_{xf} - v_{xi}) = \dfrac{v_{xf}+ v_{xi}}{2}$ Is there a geometric interpretation or does it just work out mathematically?


8

I think your math teacher is right. One way to see that differentials are not normal numbers is to look at their relation to so called 1-forms. I do not know if you already have had forms in calculus 2, but it is easy to look up on the internet. Since you chose a tag "integrals" in your question, let me give you an example based on an integral. Let's say ...


7

Technically, the equation $$d = \frac{\mathrm{d}x}{\mathrm{d}t}t + \frac{\mathrm{d}^2x}{\mathrm{d}t^2}\frac{t^2}{2}$$ is not right. Instead, for constant acceleration, you need $$d = \left(\left.\frac{\mathrm{d}x}{\mathrm{d}t}\right|_0\right) t + \left(\left.\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\right|_0\right) \frac{t^2}{2}$$ In other words, a quantity ...


7

The result is sometimes called Flanders' lemma. The remarkable point is that it does not need that $f$ is analytic, but just that it is $C^\infty$. So it does not relies upon the Taylor series as it could seem at first glance, since that series may not converge. It works in any open star-shaped neighborhood of points in $\mathbb R^n$. A set $A\subset ...


6

Since the force is a function of distance, you need to integrate: $$F = kx\\ W = \int F\ dx\\ W = \int k\ x\ dx\\ W = \frac12kx^2$$ Add signs as needed... Your work considered the force to be constant - and that's not how springs work.


6

-What is an infinitesimal quantity like $\delta$ to the physicist? To most physicists, it means the same thing it meant to Newton, Leibniz, and Euler. It means something that's small enough that we can apply a certain informally defined body of techniques to it and get correct answers. To physicists who know more about math after 1960, it means the ...


6

This is not an equality, strictly speaking. Looks like your lecturer used spherical coordinates. If the integrand is spherically symmetric, i.e. it only depends on the magnitude of $\mathbf{p}$, then the integration over the angular coordinates is trivial and just gives you the solid angle subtended by a sphere, $4\pi$.


6

I can give you an intuitive view from a physicist. Charges are the sources and sinks for the electrical field. Consider the extreme case where the volume enclosed by the surface is empty space, so no charges. Then any field line that enters the volume must exit the volume somewhere else. Thus, the integral of the field over the entire surface is 0. If ...


5

While it is true that some parts of physics are discrete; matter, energy, charge, etc. There are many others that are continuous. Distances, time, temperature, probability, and angles are just a few examples of continuous quantities. In addition, even though things such as Energy are "discrete" for one system, the smallest unit of Energy is not the same for ...


5

As user BebopButUnsteady mentions in a comment, this is essentially an exercise in Gaussian integration. With the caveat that the integration variables take values in a Lie algebra representation. (Warning: We will ignore factors of 2 and $\pi$ in what follows, and sometimes use Einstein summation convention.) The three bosonic fields $X_1\equiv X$, ...


5

Consider Tsiolkovsky's rocket equation $ \Delta v = v_e \ln \left( m_0/m_f \right) $ with $\Delta V$ the total change in velocity, $v_e$ the exhaust speed of the reaction products, $m_0$ the initial mass (structure+payload+propellant) and $m_f$ the final mass (structure+payload). If you ignore the atmosphere and other such "nuisances", it should be ...


5

Pretty sure the question is about $\frac{\hat{r}}{r^2}$, i.e. the electric field around a point charge. Naively the divergence is zero, but properly taking into account the singularity at the origin gives a delta-distribution.


5

You must first rewrite the old partial derivatives in terms of the new ones. A priori, they're some linear combinations with coefficients that could depend on the spacetime coordinates in general but here they don't depend because the transformation is linear. The rules $$ t'=t, \quad x'=x-Vt,\quad y'=y $$ get translated to $$ \frac{\partial}{\partial t} = ...


5

You may be imagining that if you push with constant force $F$, the spring will compress until the spring has such a resistive force. But since the spring was not counteracting that force, your constant force $F$ was accelerating the mass. Upon reaching the point where the spring has force $F$ as well, the mass does not stop but has a speed such that $KE = ...


5

$$A=\pi r^2$$ $$\frac{dA}{dr}=\pi\cdot2r$$ $$dA=2\pi rdr$$ Alternatively, you can write : $\lim_{\Delta r\to 0}\frac{\Delta A}{\Delta r}=\lim_{\Delta r\to 0}\frac{\pi\{(r+\Delta r)^2-r^2\}}{\Delta r}=\lim_{\Delta r\to 0}\frac{2\pi r\Delta r+\Delta r^2}{\Delta r}=2\pi r+0$ You have to ignore $(dr)^2$ as it is very small. Why? Because you took the limit ...


5

1. Since $x\gg p$, we see that $\sin(px)$ is highly oscillatory. In fact, the integral becomes $$\int_0 ^\infty \mathrm{d}p\ p \sin px \ e^{-it\sqrt{p^2 +m^2}}\sim \int_{-\infty} ^\infty \mathrm{d}p\ p\ e^{ipx-it\sqrt{p^2 +m^2}}$$ modulo some factor of $\pm2/i$. Observe now this integral resembles $\int f(p)\exp(g(p))\,\mathrm{d}p$. We find the point ...


4

In mathematics the notation $\def\d{\mathrm d}\d x$ is actually a linear form, this means that $\d x$ is a linear function taking a vector a giving a scalar. Let us take a differentiable function $f$ defined over $\def\R{\mathbf R}\R$ and consider it at point $a$. The tangent to the curve of $f$ at the point $a$ has a slope $f'(a)$. The point on this ...


4

The integral $$I(k) = \int_{-\infty}^\infty \frac{s e^{isr}}{(s-k)(s+k)} ds \tag{1}$$ where $k$ is real and the integration is for real $s$, is not really well-defined. This is precisely because the integrand has singularities on the integration domain. However consider if $k$ is a complex number $k = k_r + ik_i$ with $k_i >0$. Then the integrand is ...


4

The integral evaluates to $$ \int_0^{2\pi}\cos^2(\theta)\mathrm{d}\theta = \pi$$ and they use it there exactly that way (note the sneaky $\pi$ in $(101)$). No approximation involved. The talk of average values there is due to the fact that the average value is defined by $$ \frac{1}{2\pi - 0}\int_0^{2\pi}\cos^2(\theta)\mathrm{d}\theta = \frac{1}{2}$$ ...


4

Divergence is an operation that maps a vector field $\vec D(x,y,z)$ to a scalar field ${\rm div}\,\vec D(x,y,z)$. How do you calculate ${\rm div}\,\vec D(x,y,z)$? Either you follow the definitions using derivatives, which you can't if you don't know what a derivative is. Or you imagine the following: the vector field $\vec D(x,y,z)$ tells you about the ...


4

You can keep on adding higher order derivatives until they become vanishingly small. A convenient point of entry to this topic would be the Wikipedia article Jerk (physics). Bear in mind that when you're in a car, jerk is only of relevance during the time when the accelerator pedal is actually moving, to a first-order approximation. Update: It seems a ...



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