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14

Maxwell's equation can be given in the form $$\text dF = 0$$ $$\text d\star F + J = 0$$ where $F$ is a 2-form and $J$ an $n-1$-form (a current density) which in principle can be generalised to any manifold (for physical reasons one might want to consider pseudo-Riemannian manifolds with signature $(+,-,\cdots,-)$). In the four dimensional theory one usually ...


7

You can generalize Maxwell's equations to an arbitrary number of dimensions by using either the tensor or differential form version, as the vector formalism does not help too much (For instance, in two dimensions, the magnetic field is a (pseudo) scalar field, not a vector field). The equations are then : $\partial_\alpha F^{\alpha\beta} = \mu_0 J^\beta$ ...


1

John's answer is correct. However, to get an intuitive - and I think not too far wrong - picture of why we call the singularity one-dimensional and not zero-dimensional, imagine the analogous situation without general relativity. Instead of a black hole, we'd just have the singularity - still a point of infinite density, just one that doesn't mess with ...


5

There are a few complications with your question. For example you need to distinguish between coordinate singularities and curvature singularities. A coordinate singularity is a place where the coordinate system becomes singular. A good example of this is a black hole event horizon. A curvature singularity is a place where the spacetime curvature becomes ...


4

The concept of a singularity is a mathematical concept, not a physics concept, and you have to get that clear in your understanding. As simple as the function 1/r has a singularity at r=0, i.e the number becomes infinite. If one variable is used the singularity is one dimensional, if r is the length of a vector the singularity comes when x,y,z are zero. ...


0

Comment to the question (v6): It seems that OP is basically seeing the effect that Gauss's law forces Coulomb's law to be $$\tag{1} F~=~ k_e \frac{Q_1Q_2}{r^{D-1}} $$ in $D$ spatial dimensions. If we choose Lorentz–Heaviside/CGS/Gaussian units with $c=1=\hbar$, then Coulomb's constant $k_e$ becomes dimensionless. It then follows from Coulomb's law (1) ...


1

I think Gennaro covered, this, I'll give a layman's explanation. In spacetime there are four general dimensions, three of space and one of time. Why is it that other dimensioned qualities seem to be rarely considered as part of spacetime? For example, why isn't speed part of spacetime, forming a five-dimension spacetime-speed manifold? Objects in ...


2

We need to clarify what we mean by dimensions of a mechanical system. They refer, in the standard terminology, to the number of different degrees of freedom we need to describe the kinematics of a point particle (in this case). To this extend, given any reference frame $S$, an event in the space-time is identified by its position $(x,y,z)$ and the time $t$ ...


-3

A spacial dimension is a direction in which a body can move. This line of direction must lay 90° to other dimensions. The 3 dimensions commonly known to us are forward-back, up-down and front-back. A moving body in our universe used one or more of these directions when it moves. So far we have not detected any 4th spacial direction/dimension in our universe. ...


1

In 2D a circunference is $$(x-x_c)^2 + (y-y_c)^2 = r^2$$ A circle is: $$(x-x_c)^2 + (y-y_c)^2 \leq r^2$$ In 3D sphere is: $$(x-x_c)^2 + (y-y_c)^2 + (z-z_c)^2 = r^2$$ A ball is: $$(x-x_c)^2 + (y-y_c)^2 + (z-z_c)^2 \leq r^2$$ You can even have a "1D ball/circle": $$(x-x_c)^2 \leq r^2$$ which is just a line segment. A 1D circunference/sphere is ...



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