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The only reason one would ever define something like $A^{\mu}$ is "an effort to go and pull electrodynamics into a four-dimensional framework. Now, in order to do that, one can notice that you have t …

You're appealing to the distant stars, but that is picking out the reference frame of "the distant stars", at which point you're not talking about special relativity, you're talking about Robertson-wa …

In special relativity, you think of a 4-dimensional space-time. The key point here is that two events, 1, and 2 happening at $t_{1}, x_{1}, y_{1}, z_{1}$ and $t_{2},x_{2},y_{2},z_{2}$ have a distance …

I'm guessing that the OP is uncomfortable with the idea of a universal time existing in cosmology interfering with the spirit of the principle of relativity. The missing piece in seeing why there isn' …

It's probably worth adding to the conversation that "how do I measure the interval" depends on whether it is timelike, spacelike, or null. For timelike intervals, you find an inertial frame where the …

The simplest way to check to see if a frame is inertial is to check whether momentum and angular momentum are conserved in it. To check to see whether non-conservation of momentum is equivalent to non …

If "first principles" means "The Minkowski Metric", then note that the spacetime interval (I use units where $c=1$ for this whole discussion. Feel free to replace $v$'s with $v/c$'s if you like): $$d …

The answer to this question depends on the nature of the object and its motion. The standard Kerr solution to Einstein's equation, which describes a spinning black hole, can be expressed in the form …

if it's a tensor, it transforms like a tensor, which means, the lorentz transformation matrices can act on it. If nothing else, the levi-civita symbol appears in things like the definition of the det …

If you allow for superluminal travel, it is relatively easy to construct curves where you can go to earlier points on your own timeline, or send signals back to a younger version of yourself, or whate …

four-momentum is literally the first (proper) time derivative of position, multiplied by mass, so it is a vector with the same transformation rule.

Let's just justify this physically without making an explicit appeal to the Lagrangian math. Assume the Minkowski metric and $c=1$ units. We already know, if the theory is going to make any sense at …

Just to add to the other answers, one might ask, "why the square root"? The heart of it is the underlying geometry. When we ask "what is the diagonal of a square with sides $a$ and $b$, Euclid's rule …

If you've written down your equation correctly, then it's simply a matter of: $$F_{ab}\partial^{a}F^{bc}$$ has two dummy indices, $a$ and $b$. You can freely swap their names, because they are summed …

Better yet, you can derive this by starting with the Lagrangian: $$L = \sqrt{|g|}\left(\frac{1}{16\pi}R + \frac{1}{4}F^{ab}F_{ab}\right)$$ And just finding the equations of motion (I recommend cheatin …