Define the  to be   $$J(f)={{d^3 f} \over {dt^3}}$$  Why don't we use the concept of jolt more often considering that the change in acceleration is most definitely an important characteristic in the universe. Is there some mathematical proof that essentially kicks jolt out of the useful units club? However, it seems like initial conditions relying on jerk/jolt should be the norm rather than the exception. I mean, the universe doesn't care about what's easy to measure...  In addition, what about the other neglected members of the rate of change community. Why isn't   $${{d^4 f} \over {dt^4}}$$  or  $${{d^{55} f} \over {dt^{55}}}$$  Used in physics? In fact wouldn't having a function that just gives all order derivatives of a function useful? You could presumably study the rate of change of the spread rates of change!  $$S(f(x),n)={{d^nx} \over {dt^n}}$$  In fact, on a more math intensive note, using fractional calculus, you could treat this Spread function as a real continuous function, since fractional calculus defines fractional order derivatives.  Of course, lets not leave out the integral. Shouldn't an expression that integrates the position with respect to time have physical meaning?  $$\int x \ dt$$  what about $$\iint x \ dt$$ or even $$\iiint x \ dt$$  In fact, it seems you could derive a function very similar to the one mentioned above.  Questions: So why don't we use these other forms of rate of change? In addition, isn't the integral of position important. I mean, it has units of $L \cdot M$ which seems to me at least to be very important. In addition, what about this Spread function, what physical significance does it have. In addition aren't initial conditions subject to the effects of jerk/jolt etc...!!
62154
Why is the Lagrangian a function of the position and velocity (possibly also of time) and why are dependences on higher order derivatives (acceleration, jerk,...) excluded?  Is there a good reason for this or is it simply "because it works".
71708
I ask because, as a first-year calculus student, I am running into the fact that I didn't quite get this down when understanding the derivative:  So, a derivative is the rate of change of a function with respect to changes in its variable, this much I get.  Thing is, definitions of 'differential' tend to be in the form of defining the derivative and calling the differential 'an infinitesimally small change in x', which is fine as far it goes, but then why bother even defining it formally outside of needing it for derivatives?  And THEN, the bloody differential starts showing up as a function in integrals, where it appears to be ignored part of the time, then functioning as a variable the rest.  Why do I say 'practical'? Because when I asked for an explanation from other mathematician parties, I got one involving the graph of the function and how, given a right-angle triangle, a derivative is one of the other angles, where the differential is the line opposite the angle.   I'm sure that explanation is correct as far it goes, but it doesn't tell me what the differential DOES, or why it's useful, which are the two facts I need in order to really understand it.  Any assistance?
38589
I'll be generous and say it might be reasonable to assume that nature would tend to minimize, or maybe even maximize, the integral over time of $T-V$.  Okay, fine.  You write down the action functional, require that it be a minimum (or maximum), and arrive at the Euler-Lagrange equations.  Great.  But now you want these Euler-Lagrange equations to not just be derivable from the Principle of Least Action, but you want it to be equivalent to the Principle of Least Action.  After thinking about it for awhile, you realize that this implies that the Principle of Least Action isn't really the Principle of Least Action at all:  it's the "Principle of Stationary Action".  Maybe this is just me, but as generous as I may be, I will not grant you that it is "natural" to assume that nature tends to choose the path that is stationary point of the action functional.  Not to mention, it isn't even obvious that there is such a path, or if there is one, that it is unique.  But the problems don't stop there.  Even if you grant the "Principle of Stationary Action" as fundamentally and universally true, you realize that not all the equations of motions that you would like to have are derivable from this if you restrict yourself to a Lagrangian of the form $T-V$.  As far as I can tell, from here it's a matter of playing around until you get a Lagrangian that produces the equations of motion you want.  From my (perhaps naive point of view), there is nothing at all particularly natural (although I will admit, it is quite useful) about the formulation of classical mechanics this way.  Of course, this wouldn't be such a big deal if these classical ideas stayed with the classical physics, but these ideas are absolutely fundamental to how we think about things as modern as quantum field theory.  Could someone please convince me that there is something natural about the choice of the Lagrangian formulation of classical mechanics (I don't mean in comparison with the Hamiltonian formulation; I mean period), and in fact, that it is so natural that we would not even dare abandon these ideas?
76398
I'm very confused about the definition of the inverse Fourier transform between time and frequency spaces. In many places, including , the Fourier transform pairs are defined by:  \begin{align} f(t)=&amp;\int_{-\infty}^\infty\tilde{f}(\omega)e^{i\omega t}\frac{d\omega}{2\pi} \\ \tilde{f}(\omega)=&amp;\int_{-\infty}^\infty f(t)e^{-i\omega t}dt \, . \end{align}  In other sources, such as "Many-body quantum theory in condensed matter physics" by Bruus and Flensberg, or , they are instead defined by:  \begin{align} f(t)=&amp;\int_{-\infty}^{\infty}\tilde{f}(\omega)e^{-i\omega t}\frac{d\omega}{2\pi} \\ \tilde{f}(\omega)=&amp;\int_{-\infty}^\infty f(t)e^{i\omega t}dt \, . \end{align}  Why are there these two different definitions? I thought maybe they mean the same thing, but this would only seem to be the case for $\tilde{f}(\omega)=\tilde{f}(-\omega)$ and I don't see why this should be true in general.  Indeed, I originally noticed this discrepancy when I tried to calculate the impedance of a capacitor using the second definition above, and obtained: $$Z=-\frac{1}{i\omega C} \, ,$$ and when I looked up the solution, found that it is generally accepted to be: $$Z=\frac{1}{i\omega C} \, ,$$ which is what I obtain using the second definition above. I don't understand how these two functions could have the same physical interpretation. That's why I asked here instead of the math stackexchange, because I'm sure they can be defined either way formally, but it seems like if we want to think physically about time and frequency, one must be correct.  Can anybody tell me which one is correct, or explain why they are equivalent?
47569
I am working on a problem and am a little bit confused.  I need to find the distribution that corresponds to the MGF:  $2e^t\over3-e^t$  Do we need to separate this into something like:  2e$^t$ and $1\over3-e^t$  It seems like we might be able to separate this into X ~ Exp() or Poi()
3973
I do not know much about fractional calculus, except what I have read in a few short posts at MSE and .  I know that order of a derivative can be extended from rational values to real values, but all I know is what is written .  So my question is, what is the simplest way to understand and/or define $\dfrac{d^n}{dx^n}f(x),$ for $n\in\mathbb{R}$?  For example what is $\dfrac{d^\pi}{dx^\pi}f(x)$?  Also, what about complex values?  What might we say about $\dfrac{d^{s}}{dz^s}f(z)$ for $s\in\mathbb{C}$?  And what are some uses for such derivatives?
84747
I have read the most basic and important parts of vector calculus are gradient, divergence and curl. These three things are too important to analyse a vector field and I have gone through the physical meaning of gradient, divergence, and curl.  Gradient indicates the direction of maximum rate of change of a function, and it is defined as $\nabla u(\mathbf{x})\equiv\mathrm{grad} \, u(\mathbf{x})$ where $u(\mathbf{x})$ is a scalar function of the position vector $\mathbf{x}$.   Whereas divergence represents the volume density of the outward flux from an infinitesimal volume around a point and is defined as $\boldsymbol{\nabla} \boldsymbol{\cdot} \mathbf{A}(\mathbf{x})\equiv\mathrm{div} \, \mathbf{A}(\mathbf{x})$  And the curl signifies the infinitesimal rotation around a point and is defined as $\boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}(\mathbf{x})\equiv \mathrm{curl} \, \mathbf{A}(\mathbf{x})$  And now I have study and also mathematically formulate that $\boldsymbol{\nabla} \boldsymbol{\cdot} \left[\boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}(\mathbf{x})\right]=\mathrm{div} \left[\mathrm{curl} \mathbf{A}(\mathbf{x})\right]=0$.  And $\boldsymbol{\nabla} \boldsymbol{\times}\left[\nabla u(\mathbf{x})\right] =\mathrm{curl}\left[ \mathrm{grad} \, u(\mathbf{x})\right]= \boldsymbol{0}$ where u is a scalar function  Now my question is how can I interpret those two expressions? I mean, I want the physical significance of these two expressions in the sense of definition of gradient, divergence, curl, and also using graphical representation.
24037
According to , the derivatives of position with respect to time are position, velocity, acceleration, jerk, snap, crackle, pop.  Who came up with snap, crackle, and pop?  What about higher order derivatives?
45340
I am having trouble understanding the concept of convolution. I get the mathematical idea behind it. I am able to evaluate that the convolution of the following problem  is $$f*g(t) = {0,\ \ t &lt;= -1,\ \ t &gt;= 1;\\ t+1, -1 &lt;= t &lt;= 0;\\ 1 - t,\ \ 0 &lt;= t &lt;=1}$$ However, what exactly is happening to the signal on more physical level?
108663
Is there an upper limit for  jerk in physics? What about higher derivatives?   A consequence of  special relativity is that no material body can reach or exceed the speed of light in vacuum (due to the relativistic mass increase, or the Lorentz contraction).    I am not interested in obtaining these upper limits  based on postulates (and following mathematical implications), like the postulate  of the constancy of speed of light in special relativity.    I would like to know whether there are experiments that would lead to these upper limits, as a matter of consistency with empirical evidence. For the speed of light in vacuum we have Maxwell's equations. As far as I know, it was established experimentally that the constant that appears in Maxwell's equations is the speed of light in vacuum.  For acceleration,  there is a previous question on StackExchange, , where in an answer it is discussed Caianiello's maximum acceleration (the result is linked to Heisenberg's uncertainty principle, that's the type of answer that I am interested in, as an example).   So for speed and acceleration,  I understand that there is  experimental evidence that tells us that in order to have consistency (with the empirical evidence),  then we must accept that there is a maximum speed and acceleration in physics.  I am not interested in arguments involving the Planck system of units, since these arguments would be sensitive to the choice of normalization.   What about jerk in physics, and higher order derivatives. Is there any experimental data that would force us to accept upper limits for these quantities? Would it be useful to imagine experimental scenarios that would force us to accept these upper limits?  I will reformulate the question,  for clarity. I will not focus on the chronological emergence of the mathematical models that explain the experimental data.  In physics, maximum speed can be linked to Maxwell's equations. Maximum acceleration can be linked to Heisenberg's uncertainty principle. Does this trend continue,  for jerk and higher derivatives? Because  intuition tells me that in physics there must be upper limits/bounds for all of them, and there is a countable infinity of these quantities.
8237
My calculus text (Swokowski, Olnikc, Pence, 6th edition) gives the  formula for work as $W = Fd$ and then goes on to explain that if the force varies over the distance the formula becomes an integral.    As part of an example, it then shows that the work to lift a 500 lb beam 30 feet would be 500 * 30 = 15,000 ft-lb.  But don't we have to exert an upward force greater than the weight of the beam somewhere in our model in order to get the beam to move up?  Then the work would be greater than 15,000 ft-lbs according to the definition of work.  It seems that in setting up integrals or just using the formula  W = Fd examples always use the weight of the increment or object to be  lifted without taking into account the fact that more than that force  must be exerted at some point in order for the thing to move upwards. By the formula W = Fd, if we greatly accelerate an object upwards, the  work done lifting the object will be greater than if a lesser  acceleration is applied across the same distance, but examples in my  book don't seem to take this into account.  (I am assuming that  F = ma, and of course the mass stays constant.)
10960
I have read the thread regarding , but it does not answer my question.  I am confused about the notation for change in Physics. In Mathematics, $\delta$ and $\Delta$ essentially refer to the same thing, i.e., change. This means that $\Delta x = x_1 - x_2 = \delta x$. The difference between $\delta$ and $d$ is also clear and distinct in differential calculus. We know that $\frac{dy}{dx}$ is always an operator and not a fraction, whereas $\frac{\delta y}{\delta x}$ is an infinitesimal change.   In Physics, however, the distinction is not as clear. Can anyone offer a clearer picture?
24646
In this equation:  $$ \mathbf a_i\overset{\rm def}{=}\left(\frac{d^2\mathbf r}{dt^2}\right)_i=\left(\frac{d\mathbf v}{dt}\right)_i=\left[\left(\frac{d}{dt}\right)_r+\boldsymbol\Omega\times\right]\left[\left(\frac{d\mathbf r}{dt}\right)_r+\boldsymbol\Omega\times\mathbf r\right] $$ (), why is  $$\left(\frac{d}{dt}\right)_r \boldsymbol{\Omega} \times \mathbf{r}=\frac{d\boldsymbol{\Omega}}{dt}\times\bf{r}+\bf{\Omega}\times \bf{V_r}$$  In particular, I have qualms with the term  $$\frac{d\bf{\Omega}}{dt}\times\bf{r}$$  Why are we deriving the angular velocity? Why is it a derivative not of the rotational type (Namely $(\frac{d}{dt})_r$ )  Other sources do not point out that term I have problems with. In any case, I want to know how you evaluate that derivative.
56652
I've been wondering what makes the  such an important model. What I came up with:   It is a (relatively) simple system, making it a perfect example for physics students to learn principles of classical and quantum mechanics. The harmonic oscillator potential can be used as a model to approximate many physical phenomena quite well.   The first point is sort of meaningless though, I think the real reason is my second point. I'm looking for some materials to read about the different applications of the HO in different areas of physics.
105993
Distance is paired with Displacement and it seems to be a bigger idea than just the magnitude of Displacement. Speed is paired with Velocity.  I have always thought that there is not such pairing with Acceleration.  I would teach: Acceleration is a vector, and we can talk about the magnitude of Acceleration, but we cannot talk about a scalar that describes the change in Speed over time.  Is there a scalar counterpart to Acceleration?  P.S. First question on physics.stackexchange.  Please excuse me if the question is inappropriate.  Edit to Question: I think I need to explain the question better.  If I walk to the store and back home, my displacement is zero and my velocity would also be zero (regardless of how long it took me).  Since the store was some distance away, my total distance traveled for this situation would be positive and so my speed would be positive.  I would then conclude that speed is not the magnitude of velocity, but something different (just like distance is a different concept than displacement).  In this way I can say distance &amp; speed are scalar (not magnitudes of vectors).  Now that I define my question a bit better, I think I see the problem.  It seems distance is a summary (i.e. sum of) of the magnitudes of a group of vectors (displacements).  Speed is the summary of the magnitudes of the first derivative (with respect to time) of the group of vectors.  I doubt there is (or is there a need for) a similar summary of the magnitudes of the second derivative of  the  group of displacements.  I welcome your thoughts.
91415
In a classical electrodynamics textbook (Griffiths), it is mentioned that even though the electric field function, $E:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$, is a (3D) vector valued function, the amount of information needed to fully describe it is equivalent to the amount needed to describe a scalar valued function, and this is because it is the gradient of an electric potential function, $U:\mathbb{R}^{3}\rightarrow \mathbb{R}$ and so is completely determined by it. The author goes on to explains that this is so because $E$ has the property that its curl is zero evevrywhere, and this is what restricts the freedom in determining $E$ (and what enables it to be the gradient of a scalar function in the first place).     This is all fine (and fun), but I find myself unable to answer this question: why does the imposition of zero curl provide an amount of information exactly equivalent to two scalar-valued functions (no more, no less)?   An example of the sort of reasoning that leads me to other conclusions: since zero curl is equivalent to equating three pairs of partial derivatives ($\partial E_{x}/\partial y = \partial E_{y}/\partial x$ and so on), leaving 6 out of 9 partial derivatives "free", it seems that "one third of the amount of information" required to describe $E$ has been taken up, as opposed "two thirds"... I am of course assuming here that the first order partial derivatives determine the behavior of the function (perhaps up to a constant as in the 1D case?), and that these partial derivatives are independent of each other and thus provide equal amounts of information. Is either of these assumptions wrong? Is my whole reasoning off?  Any insight into this question and a possible answer would be appreciated.  *The question is, of course, a general one about vector-valued functions, with $E$ just being a particular case.  **I have not mentioned all the obvious assumption of smoothness necessary for everything to be defined..  ***I realise that I'm playing fast-and-loose with the word "information", and that my whole question is very un-formal. Any reference to an area of mathematics which may perhaps put such intuitions on rigorous footing are more then welcome.   Edit: I am not asking why (or when) a vector field having zero curl is equivalent to it being a gradient of some scalar field. I am asking about the amount of information we get about a vector field when we determine its curl.
87653
I'm only 4 weeks into my first college-level physics course, so this is a very elementary question.  Also, in asking this question, I'm just playing devil's advocate in order to better understand the historical motivation for the use of vectors to describe physical phenomena. I'm not actually proposing we stop using them.  :) So, please don't bash me for asking this innocent question.  As I was reading my textbook today, I wondered why vectors were invented and are still useful today. I have a little experience with them from my Trigonometry course, but they've always seemed a bit counterintuitive to me, as a tool for modeling and calculating.  What are the benefits of representing magnitude and direction (or other related values) with a single variable, instead of treating them as separate values with distinct variables?  What are the drawbacks (aside from increased complexity), if any, of using vectors instead of performing calculations and building models in which the speed and direction of a moving object (for example) are treated as distinct values with a distinct variable for each quantity?