Search Results

Usually you model an observer by a worldline. You then associate at every event a orthogonal basis (with respect to the Minkowski metric) such that the time-like 4-vector is tangent to the worldline. …

The formulas are equivalent to your paragraph. $E\times B$ is in the direction of the normal of the plane spanned by $E,B$. Since $E,B$ are perpendicular, $$ |E\times B|=|E||B| $$ So $\frac{E\times B} …

If you are talking about beams, then you can assimilate them as uniformly charged lines moving at their respective velocities. For simplicity, I’ll assume the beams to be parallel. You won’t need rela …

No you cannot switch the derivatives in general. This is not like Schwartz' theorem for partial derivatives. In general: $$ \begin{align} \frac{d}{dx}\frac{d}{dy} &= \frac{d}{dx}\frac{dx}{dy}\frac{d}{ …

The issue when measuring lengths is that you need to define a line (hyperplane of spacetime in general, but in your case this is a line) of simultaneity that is space-like. In this plane, you can then …

No need for 1+3D, just focus on 1+1D. The whole point is that Rindler coordinates are not obtained by a Lorentz transformation (or their inverse, which is also a Lorentz transformation since it is gro …

it is simply a question of terminology, the relative velocity of two observers is the velocity of one with respect to the other. This way, you do not need to appeal to a third frame that is not well-d …

It's always tricky to gain physical understanding using time dilation/ length contraction without talking about relativity of simultaneity. It is actually this last phenomena that truly helps preserve …

In general, there are two classic ways to represent the Poincaré group. The first one comes from its definition. They are the "rigid" motions for a 4D Minkowski spacetime. Identifying space-time with …

It's simply the converse of length contraction. In $S$, the spacing between the electrons was contracted compared to its rest frame $S'$, so when you switch to $S'$, you have an effective length dilat …

I’ll just formalize my previous comments. Let me restrict to 2D flat spacetime with a certain inertial frame $t,x$ ($c=1$ and the metric signature is $(+,—)$ like in particle physics). Then hyperbolic …

I will use the more standard Rindler coordinates ($a=1$): $$ \begin{align} t &= X\sinh T & x &= X\cosh T \end{align} $$ with $t,x$ inertial coordinates and $T,X$ your accelerated frame and $a$ the pro …

The natural way is to specify proper acceleration with respect to proper time. This uniquely specifies your trajectory, given that you give its initial spacetime event and corresponding normalised 4- …

Your calculation is consistent, a bit misleading. You should view it as a constraint on the possibilities of $\Phi$ since the force $f$ must be spacelike. This is the case here since $\Phi$ has only s …

If I understand correctly, let $(t,r,\theta,\phi)$ be the spherical coordinates entered about an arbitrary origin of an inertial frame, you want a new coordinate system $(\tau,\rho,\theta,\phi)$ such …