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1

So I think $\partial_\mu F$ should also be 4 vectors, each being the directional derivative along a coordinate axis. It's a single covector, not a vector and not a collection of vectors. A gradient of a scalar field is a classic example of how we obtain a covector. In index notation, a free index such as your $\mu$ is interpreted as having the potential ...


0

The covariant derivative should introduce an extra index, which I presume its rho in your case. If you apply the general formula defining the covariant derivative of a tensor you will get your result.


2

There are already many good answers. Besides the fact that the standard definition of work directly relates to the work-energy theorem and the notion of potential energy, here is a geometric argument. I) The force $F_i(x,v,t)$, $i\in\{1,2,3\},$ transforms as $(0,1)$ co-vector $$\tag{1} F_i ~=~\sum_{j=1}^3F^{\prime}_j \frac{\partial x^{\prime j}}{\partial ...


3

For starters, these are not the same thing. The integration by parts rule makes this fairly obvious: $$\int_i^f y\,\mathrm{d}x = y_f x_f - y_i x_i - \int_i^f x\,\mathrm{d}y$$ But then you might be wondering what makes $\int \vec{F}\cdot\mathrm{d}\vec{s}$ the "right" definition for work while $\int \vec{s}\cdot\mathrm{d}\vec{F}$ is the "wrong" one. In a ...


6

The reason the relationship $$ W=\int\mathbf s\cdot d\mathbf F $$ doesn't work is because Work is defined as the result of a force $\mathbf F$ on a point that moves along a distance. The point follows a curve $\mathbf s$ with a velocity $\mathbf v$. The small amount of work, $\delta W$, that occurs of the instant of time $dt$ is $$ \delta W=\mathbf F(\mathbf ...


4

Because, according to your definitions, if I strain a rubber bar with constant force until it rips apart, I haven't done one joule of work to it.


1

Actually it was that maxwellian electromagnetism had no problems, in contrast to the newtonian classical mechanics framework. Theory of Relativity alters the Newtonian framework not the Maxwellian framework. i would say that even if Einstein hadn't invented SR, someone else would (as indeed many others notably Poincare, Lorentz et al) were alredy on the ...


0

The Lagrangian must be a gauge invariant and Lorentz invariant object that can be integrated over the entire spacetime $\Sigma$. So, we must first obtain an $n$-form (for $n$ the dimension of spacetime), and all that we have for that is the gauge field $A$, which itself transforms in an ugly way under gauge transformation. The only object we can build out ...


3

In general, the statement that $\nabla_\mu V^\mu$ transforms as a scalar does not quite fix the transformation properties of $V^\mu$. Rather, the most general such transformation would be $$V^\mu \mapsto V'^\mu + C^\mu,$$ where $V'^\mu = \frac{\partial x'^\mu}{\partial x^\nu} V^\nu$ is the ordinary vector transformation law, and $C^\mu$ is any quantity ...


1

When you write $V^\mu$ you mean that $V$ is a vector. Next $\nabla_\mu V^\mu$ is called divergence of a vector. Finally answering your question, a vector field with constant zero divergence is called incompressible or solenoidal – in this case, no net flow can occur across any closed surface (according to the Gauss law). If it is a constant, but not zero, ...


5

The statement from the Wikipedia articles is, as written, wrong. The EM field tensor - as a tensor - does change under change of reference frames. It is covariant, but not invariant under the Lorentz group, while the electric and magnetic field are neither, but they are covariant under the rotation group. The electric and magnetic fields are ordinary, ...



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