7 votes
Accepted

Vector addition for differentials in the context of electric potential

I think that your professor is showing the differential vector for infinestimal change in each coordinate component. The diagrams correspond to cartesian, cylindrical and spherical coordinate systems ...
Muhammed Çağlar TUFAN's user avatar
5 votes

What does an upside down delta mean - covariant vectors?

That is the Nabla, and it just means the covariant vector made of space derivatives. Edit: $$\vec\nabla f= \begin{pmatrix} \dfrac{\partial f}{\partial x}\\ \dfrac{\partial f}{\partial y}\\ \dfrac{\...
naturallyInconsistent's user avatar
5 votes

Partial derivatives vs Covariant derivatives in polar coordinates

Covariant derivatives take into account for both component and basis changes, thereby applicable for curved spaces - where partial derivatives only take component changes into account - is this ...
Lenard Kasselmann's user avatar
4 votes

Given a wave that follows these two conditions, is there some valid and non-trivial physical interpretation to it?

By Helmholtz's decomposition theorem if $\nabla \cdot \vec{u}=0$ and $\nabla\times\vec{u}=0$ in all space then $\vec{u}=0$.
hyportnex's user avatar
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4 votes
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How does the $\not{\partial}$ work in the Dirac Lagrangian?

The $\equiv$ symbol in eq. (2.142) allegedly means equality modulo total divergence terms (which don't change the Euler-Lagrange (EL) equations).
Qmechanic's user avatar
  • 188k
3 votes

Partial derivatives vs Covariant derivatives in polar coordinates

As OP correctly points out connections introduce a concept of differentiation of tensor fields or more in general of sections of vector bundles that takes into account how the bases of the fibers ...
Mr. Feynman's user avatar
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3 votes
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Why can the dot product of two vectors be expressed as a differential?

I would imagine that the simplest way to show this is to note that the position vector $\mathbf x$ can be expressed in either basis: $$x'^j \hat e_j' = \mathbf x = x^i \hat e_i$$ A given set of ...
J. Murray's user avatar
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3 votes

Difference and meaning of index the derivative operator

You can eventually (if you need to) learn a more rigorous treatment later, so let me instead provide a cookbook approach: An object with an open index means that its value changes when the observer ...
SonerAlbayrak's user avatar
3 votes

Vector addition for differentials in the context of electric potential

The infinitesimal displacement $d\vec{s}$ is derived from $\vec{s}$. In cartesian coordinates : $\vec{s}=x\vec{i}+y\vec{j}+z\vec{k}$ $$d\vec{s}=dx\vec{i}+xd\vec{i}+dy\vec{j}+yd\vec{j}+dz\vec{k}+zd\vec{...
M06-2x's user avatar
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3 votes
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Why the $\Delta$ in the definition of pressure? (fluid mechanics)

The $\Delta$ is there to indicate that it is the "bit" of the total force on the boundary due to the pressure pushing on the small area $\Delta S$. $P= \Delta F/\Delta S$ is better written ...
mike stone's user avatar
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2 votes

Why the $\Delta$ in the definition of pressure? (fluid mechanics)

Pressure is the amount of force applied perpendicular to the surface of an object per unit area. A problem arises when the surface is not flat ie curved. So what is done is to take an infinitesimally ...
Farcher's user avatar
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1 vote
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Invariance of d'Alembertian for Lorentz boost in arbitrary directions

I have not been able to have a thorough look at this, but maybe this could be the problem. In your first equation with $\nabla_{\mathbf x'}$ you calculate $\frac{\partial t}{\partial\mathbf x'}$. To ...
AccidentalTaylorExpansion's user avatar
1 vote
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Hermitian Conjugate Terms in Lagrangians

Under Hermitian conjugation, $$ \phi^{\dagger} \phi \mapsto \phi^{\dagger} \phi ,\implies \\ i\phi^{\dagger} \partial_\mu \phi \mapsto i\phi^{\dagger} \partial_\mu \phi -i\partial_\mu ( \phi^{\...
Cosmas Zachos's user avatar
1 vote

What do we get on differentiating the instantaneous displacement function?

Displacement is defined as difference between position vectors : $$ \vec s = \vec r - \vec r~' \tag 1$$ Differentiating that we get : $$ \frac {d\vec s}{dt} = \frac {d(\vec r - \vec r~')}{dt} = \...
Agnius Vasiliauskas's user avatar
1 vote

Does $δS = 0$ mean that "the small changes in the actions equal to zero"?

Indeed the diagram you show is not correct, since you are interpreting variations of the action along the curve that minimizes it (sorry if I misunderstood the picture), and that is not the point of ...
Gravitino's user avatar
  • 549
1 vote

Equating 2 sides of EFE

First, note that for scalar functions, the covariant derivative reduces to the partial derivative. So for scalar functions, it is true that if the covariant derivative is zero at a point, then the ...
Andrew's user avatar
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1 vote

Chain rule when the intermediary variable might be equal to zero

Recall the definition of velocity in 1D. $$v(t)=\lim_{\Delta t\to 0} \frac{\Delta x}{\Delta t}=\lim_{t\to t_0}\frac{x(t)-x(t_0)}{t-t_0}=\frac{dx}{dt}$$ From this definition, you can recall that the ...
Mario Figueroa's user avatar
1 vote

Why can the dot product of two vectors be expressed as a differential?

First observe, that for any matrix, we can pick up its $i,j$ entry by applying it first to a column vector that is zero apart from $1$ at its $j$'th position and then dotting it into a similar vector ...
Amit's user avatar
  • 1,256
1 vote

Difference and meaning of index the derivative operator

Here is a brief summary: In (relativistic) physics, it is standard to adorn a local coordinate $x^{\mu}$ with a superindex. We define a shorthand notation for the partial derivative $\partial_{\mu}:=...
Qmechanic's user avatar
  • 188k
1 vote

Vector addition for differentials in the context of electric potential

The diagrams are slightly misleading, because the infinitesimal changes in angular quantities are shown as being quite large. It may be more helpful to draw separate diagrams showing how $\vec{s}$ ...
FTT's user avatar
  • 1,217
1 vote

Covariant derivative of gauge theory in curved space

It depends on what sort of field $\phi$ is. If $\phi$ is a scalar field, you can use the ordinary derivatove, but if $\phi$ is a Dirac spinor, for instance, then you will need to include the spin ...
mike stone's user avatar
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1 vote
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Calculating the Berry potential: Questions

Yes To quote the relevant passage of Sakurai: Let us continue our discussion by taking as an example, the problem of the spin in a magnetic field mentioned earlier (spin magnitude $S$, and magnetic ...
J. Murray's user avatar
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