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38

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! :)


22

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 ...


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

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

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

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 ...


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, $4\pi$.


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

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

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

-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 ...


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

This is more like a maths question to me. This is just an identity, which is true and facilitates the calculation and it is valid for any vector field. The proof, using Einstein summation convention would be something like: $$ (\nabla \times \vec u )\times \vec u = \epsilon_{ijk}(\nabla \times u)_j u_k = \\ \epsilon_{ijk}\epsilon_{jlm}\partial_l (u_m) u_k ...


4

1.) The differentiation operator acting will give rise to Kronecker-Deltas since $\frac{\partial x_a}{\partial x_b}=\delta_{ab}$ This will kill one summation. More specifially: $\frac{\partial U}{\partial x_a}=-1/2 \sum_{ij}b_{ij}(\delta_{ai}x_j+\delta_{aj}x_i)=-1/2( \sum_{j}b_{aj}x_j+\sum_{i}b_{ia}x_i)=-\sum_{j}b_{aj}x_j$. Rename j to be i and you're ...


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 ...


4

There are three cases here: The acceleration is a function of time $a(t)$. Then the velocity is $$v(t)=\int a(t)\,{\rm d}t$$ and the position as a function of time $$x(t)=\int v(t)\,{\rm d}t$$ The distance is calculated from $x(t)$. The acceleration is function of position $a(x)$. Then the velocity as a function of position is $$ \frac{1}{2}v(x)^2 = \int ...


4

Nice question! The answer to this depends on the version of Newton's first law you use. In the Principia, the statement of the first law, as translated by Machin, is: Every body perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed thereon. This is immediately followed by ...


4

We take: $$x=r\cos\theta\cos\phi$$ $$y=r\cos\theta\sin\phi$$ $$z=r\cos\theta$$ Now, you know the definition of the gradient in spherical coordinates: $\vec{\nabla}=\frac{\partial}{\partial x}\hat{x}+\frac{\partial}{\partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z}$ Now, we use the chain rule or each component. For instance, $$\frac{\partial}{\partial ...


4

More mathematically, it comes from the change in volume element when making a change of variable. I will give here some intuitive arguments in 1D and 2D and give the general formula then: In 1D, if you integrate along the real line and change from a variable $x$ to $X = f(x)$, you know that the measure element $dX = f'(x) dx$ which implies that $dx = ...


4

(I'm addressing this from the point of view of standard analysis) I don't think you will have a satisfactory understanding of this until you go to multivariable calculus, because in calculus 2 it's easy to think that $\frac{d}{dx}$ is all you need and that there's no need for $\frac{\partial}{\partial x}$ (This is false and it has to do with why in general ...


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 ...


3

One way to see that considering the dependence of $\dot{x}$ on $x$ is problematic is as follows: $x(t)$ maps a real number $t$ to another real number $x$. So $\dot{x}=dx/dt$ is the derivative of that map, meaning we take $$\lim_{\Delta t \to 0} \frac{x(t+\Delta t) - x(t)}{\Delta t}$$ So we can see that $dx/dt$ is itself another map from a real number $t$ ...


3

Your premise appears to be that imperfections in the physical world can invalidate precise mathematical statements. However, there is a difference between a statement not applying in detail and a statement being entirely invalid. Sure, you can always look closely enough and find forces acting on an object. Similarly, if you have a gas someone claims is at ...


2

I have the same book, so I take it you are referring to Problem 1.16, which wants to find the divergence of $\frac{\hat{r}}{r^2}$. If you look at the front of the book. There is an equation chart, following spherical coordinates, you get $\nabla\cdot\vec{v} = \frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d}r}\left (r^2 v_r\right) + \text{ extra terms}$. Since the ...


2

I find that it helps a great deal to understand the fundamental phenomenon. You have your equation correct, but consider it's derivation: We start with Newton's second law, ${\bf F} = \dot{\bf p}$ where ${\bf F}$ is the force vector and $\dot{\bf p}$ is the derivative with respect to time of the momentum. The equation you gave is obtained by assuming a ...


2

There isn't a simple answer to your question. The sort of basic calculus you need to understand the equations of motion isn't especially hard maths, but it takes a while to get comfortable with it. Your profile says you're 16, so I'd guess that if you do Physics and Maths at school you'll soon be learning calculus. In the UK you learn it as part of your ...


2

$$\text d f(x)=f'(x)\text d x\equiv\frac{\partial f}{\partial x}\text d x.$$ The second equation you posted is of course only a component representation, the vectors in the first equation are most definately higher dimensional (3 dimensional). And so the derivative in the second equation is chosen such that it goes in the direction of the path, denoted y s. ...


2

Using $v^2 = \vec v \cdot \vec v$ and $\dfrac{d}{dt} (f g) = f \dfrac{dg}{dt} + \dfrac{df}{dt} g $ write: $\dfrac{d}{dt}(v^2) = \dfrac{d}{dt}(\vec v \cdot \vec v) = \vec v \cdot \dfrac{d\vec v}{dt} + \dfrac{d\vec v}{dt} \cdot \vec v = 2 (\dfrac{d\vec v}{dt} \cdot \vec v)$ The fact that the $dt$ and $1/dt$ "cancel" in the RHS integral means that you're ...



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