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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 ...

6

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 ...

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 $$\frac{\partial}{\partial t} = ... 5 Consider Tsiolkovsky's rocket equation  \Delta v = v_e \ln \left( m_0/m_f \right)  with \Delta V the total change in velocity, v_e the exhaust speed of the reaction products, m_0 the initial mass (structure+payload+propellant) and m_f the final mass (structure+payload). If you ignore the atmosphere and other such "nuisances", it should be ... 5 As user BebopButUnsteady mentions in a comment, this is essentially an exercise in Gaussian integration. With the caveat that the integration variables take values in a Lie algebra representation. (Warning: We will ignore factors of 2 and \pi in what follows, and sometimes use Einstein summation convention.) The three bosonic fields X_1\equiv X, ... 4 This is more like a maths question to me. This is just an identity, which is true and facilitates the calculation and it is valid for any vector field. The proof, using Einstein summation convention would be something like:$$ (\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

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, $$\frac{\partial}{\partial ... 4 More mathematically, it comes from the change in volume element when making a change of variable. I will give here some intuitive arguments in 1D and 2D and give the general formula then: In 1D, if you integrate along the real line and change from a variable x to X = f(x), you know that the measure element dX = f'(x) dx which implies that dx = ... 3 One way to see that considering the dependence of \dot{x} on x is problematic is as follows: x(t) maps a real number t to another real number x. So \dot{x}=dx/dt is the derivative of that map, meaning we take$$\lim_{\Delta t \to 0} \frac{x(t+\Delta t) - x(t)}{\Delta t}$$So we can see that dx/dt is itself another map from a real number t ... 3 You can keep on adding higher order derivatives until they become vanishingly small. A convenient point of entry to this topic would be the Wikipedia article Jerk (physics). Bear in mind that when you're in a car, jerk is only of relevance during the time when the accelerator pedal is actually moving, to a first-order approximation. Update: It seems a ... 3 Your premise appears to be that imperfections in the physical world can invalidate precise mathematical statements. However, there is a difference between a statement not applying in detail and a statement being entirely invalid. Sure, you can always look closely enough and find forces acting on an object. Similarly, if you have a gas someone claims is at ... 2 There are three cases here: The acceleration is a function of time a(t). Then the velocity is$$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 ...

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 ...

2

Let $v(t)$ be velocity vector and define the scalar function $$f(t)~:=~\langle v(t),v(t)\rangle,$$ which is the absolute value of velocity (speed) squared. The time derivative is $${d \over dt}f(t)~=~ \left\langle {d \over dt}v(t),v(t)\right\rangle + \left\langle v(t),{d \over dt} v(t) \right\rangle ~=~ 2\ \left\langle{d \over ... 2 I don't have time for a detailed derivation, so the following can contain errors, so take it for what it's worth...In the following I assume that \mathbf{B} is constant in time. If not, the difference will just give (in the first approximation) the first term in the integral in the right-hand side. Let us consider the volume formed by \Sigma(t_0) and ... 2 I find that it helps a great deal to understand the fundamental phenomenon. You have your equation correct, but consider it's derivation: We start with Newton's second law, {\bf F} = \dot{\bf p} where {\bf F} is the force vector and \dot{\bf p} is the derivative with respect to time of the momentum. The equation you gave is obtained by assuming a ... 2 No need to make it that complicated--calculate the potential at the surface of a sphere with a uniform mass distribution with mass M. Then calculate the energy required to add a bit of mass with mass dm and radius dr to the top of this sphere. Express M and dm in terms of r and dr and integrate. 2$$ E_1 = Re\left[\left(E_{i0}e^{-j\beta z} + \Gamma E_{i0}e^{+j\beta z}\right)e^{j\omega t}\right] E_1 = E_{i0}\cos(\omega t - \beta z) + \Gamma E_{i0}\cos(\omega t +\beta z)  E_1 = E_{i0}\left(\cos(\omega t)\cos (\beta z) + \sin(\omega t)\sin(\beta z) \right) + \Gamma E_{i0}\left(\cos(\omega t)\cos (\beta z) - \sin(\omega t)\sin(\beta z) ...

2

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 ...

2

Since the acceleration is not constant, you need to start from the equation of motion and solve it directly. This is easiest if you consider the instantaneous balance of forces tangential to the circle. The weight has a component $mg\sin\theta$ in this direction, and the acceleration is $a=r\ddot\theta$. (That's not trivial to work out, by the way: have a ...

2

A vector in a (polynimial) $gl(3)$ representation is highest weight if it is annihilated by the raising root operators $a_jk = b_j^{\dagger}b_k$, $k>j$. In our case, the relevant operators are $a_{12}$, $a_{23}$, and $a_{13}$. We do not need to check the third case, because $a_{13} = [ a_{12}, a_{23}]$ is given by the commutator of the two other ...

1

-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 ...

1

Your notation is quite confusing, so let me reformulate and partly simplify your problem. If my reformulation/simplification is wrong... Well, tough luck... Let us consider the following magnitude: $E=\Re[E_i\exp(i(\delta z+\omega t))+E_r\exp(i(-\delta z+\omega t))]$, where $\delta$ is real, $E_i$ and $E_r$ are real and positive. We are trying to find out ...

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