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3

To add to the answer of @ACuriousMind, it may be remarked that the determinant of a square matrix is equal to the (signed) volume of the parallelepiped spanned by its columns. Since this parallelepiped is exactly the image of the unit cube, the intuitive meaning of the substitution theorem (the name under which I know it) is that you can change 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 ...

3

Landau and Lifshitz apparently like to "cancel as many $ds$ in the numerator and the denominator as possible", perhaps for aesthetic reasons. However, to make sense of the integrals, it seems more pedagogical to express$^1$ everything as functions (and derivative of functions) of the parameter $s$, i.e., there should only appear a single $ds$ in an integral ...

2

You can't calculate the displacement like this. The application note clearly says " When implementing positioning in 3 axes, extra processing is required to null the earth's gravity effect.", and that's the understatement of the month! Nulling gravity is an enormous problem, unless your accelerometer is perfectly perpendicular to Earth's gravity vector! The ...

2

So I think you should check out this problem for some useful commentary. Problem understanding sign of volume integral in Minkowski space Essentially if you look at the integral on the LHS we see it takes the form explained in the problem above $$I^{\mu\nu}[f] = \int d^4k\, f(k^2)k^\mu k^\nu$$ Where f is just 1 The integral is clearly Lorentz invariant ...

3

The question is not what is possible, but what is useful to obtain analytic results. In multi-loop integrals one is often interested in analytic results, because it can be very hard to confirm numerical results to a level where you trust them completely. Now the two-loop integrals usually involve propagators of the type $$F(q_1, q_2) = \frac{i}{(p - q_1 - ... 3 You are massively overthinking the problem. The question was "find the maximum speed". If you know the greatest depth of the tunnel, then compute the difference in potential energy between the surface and that point. Now convert that to kinetic energy, and then to velocity. Evaluating$$U = \frac{mgr^2}{2R} at both $r=R$ and $r=h$, and setting the ...

0

Two dimensional case Let's look at Gauss's law. The law says that if you have a surface $S$ which encloses a charge $Q_\textrm{enc}$, and there is an electric field $\mathbf{E}(\mathbf{r})$ on the surface $S$, then the electric field and the charge enclosed are related by $Q_\textrm{enc} / \epsilon_0 = \iint_S \mathbf{E}(\mathbf{r}) \cdot d\mathbf{A}$. Now ...

3

These quantities are all infinitesimals. $dx$ and $dl$ are often used to denote infinitesimal line elements, while $dS$ and $dA$ are conventionally used for infinitesimal surface (area) elements. The list is completed with $dV$, the infinitesimal volume element. On the application of these infinitesimals: It is often simple and intuitive to think about ...

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