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10

To make progress we need to be clear what we mean by the laws of physics and observer. A law of physics is just some set of equations that we use to predict what happens. So if for example we're trying to describe how charges interact with light our set of equations, i.e. our law of physics, would be Maxwell's equations. But to write down Maxwell's ...


9

Actually you're quite correct, though possibly not in the way you expected. Ordinary velocity isn't an invariant because obviously different observers moving at different speeds will measure different velocities. However there is an invariant form of velocity called the four velocity that is an invariant under special relativistic (i.e. Lorentz) ...


5

Let's start at the beginning: The setting for relativity - be it special or general - is that spacetime is a manifold $\mathcal{M}$, i.e. something that is locally homeomorphic to Cartesian space $\mathbb{R}^n$ ($n = 4$ in the case of relativity), but not globally. Such manifolds possess a tangent space $T_p\mathcal{M}$ at every point, which is where the ...


4

The thrower's height doesn't change i.e. it is the same in both the reference frames of the thrower and the bug. That's because distances normal to the direction of motion are not changed by Lorentz transformations. In the bug's frame the thickness of the thrower decreases, so the thrower is flattened in the direction of motion, but the height is unchanged. ...


3

Just to add to John Rennie's answer, the objects where we expect to see the largest frame dragging effects are spinning black holes. There, there is actually a surface called the ergosphere (outside of the event horizon), where it is impossible for observers to stay stationary with respect to observers far from the black hole. In a sense, their reference ...


3

I will expand my comment above into an answer, but I will not comment further on it to avoid the usual very long discussions of your posts. In my opinion, you are trying to argue on a logical level, but it is not clear if you have enough knowledge of logical theories to do so on a mathematical/physical level. Without entering too much into details, a ...


3

The spacetime outside a spinning mass is described by the Kerr metric. To explain how the Kerr metric produces frame dragging is hard, because it's not something for which there's an easy intuitive model. Frame dragging arises because the spacetime geometry links the angle measured around the spinning object to time, and this means the angle changes with ...


2

First of all, light waves and matter waves may be treated together, using the same maths, because the waves associated with light and the waves associated with matter are fundamentally the same thing. Second, all the waves before they interfere and after they interfere may be written in terms of the probability current $j^\mu (x,y,z,t)$, and its ...


2

No, it doesn't mean that. One must distinguish two things: "laws of physics that apply to an object" and "laws of physics formulated from an object's viewpoint". These are two different things. Laws of physics apply to all objects. And the behavior of the objects may be described relatively to many coordinate systems or "frames of reference". The special ...


2

Almost none. Let's be much more generous than your idea of human-carrying craft. Let's just use the fastest probe. The Helios II craft, after nearing the sun, reached a heliocentric speed somewhere near 70 km/s. Obviously, its speed was more due to the gravitational influence of the sun than its engines. $$t = \frac{t_o}{\sqrt{1 - \frac{v^2}{c^2}}} $$ ...


2

So let's just say that the spacecraft can accelerate until it's moving away from the Earth at the speed of the fastest currently-existing spacecraft First, note that the fastest speed, relative to Earth, that a spacecraft has obtained is an exceedingly small fraction of the $c$ and, thus, one should not expect significant time dilation. For ...


2

This is taking too far the concept of relativistic mass. Einstein himself was not fond of this concept, according to his quote on this wikipedia article. If I understand your question correctly, you find an apparent paradox pursuing the consequences of plugging the relativistic correction factors ($\gamma$) in both sides of Kepler's third law. On one side, ...


2

Well, the definition of the causality relations are on Wikipedia. But they do not do what you think it does: Events that causally precede another event are merely the events that could have influenced the event they precede (assuming that no causal influence is possible that is faster than light) - the causal relation does not make any statement about "how ...


2

In what sense does the answer make sense? Does it describe some physical phenomenon? You've done the algebra right, and gotten a result, but I don't think your inputs (imaginary velocity) mean anything physically. So the answer can't be expected to mean anything physically.


2

According to Einstein's theory of mass-energy equivalence, if the photon is a particle of pure energy, and if $E=mc^2$, then the photon is theoretically traveling at $c^2$; not $c$ You have neglected dimensional analysis: E $\to$ Joule = $\frac{\rm kg\,m^2}{\rm s^2}$ m $\to$ kilogram = $\rm kg$ which means that $c^2$ has units of m$^2$/s$^2$, not ...


2

If two events both have a spacetime interval of zero, can they both be said to be happening “now”? There is an interval associated with any two events but there is not an interval associated with an event. From the Wikipedia article "Spacetime": In spacetime, the separation between two events is measured by the invariant interval between the ...


1

The question is phrased in terms of dynamical concepts like force and mass, but there's a more fundamental kinematical answer that trumps these issues. If an object is moving with speed $u$, and you then apply a boost $v$, the object's new speed is not $u+v$ but rather $(u+v)/(1+uv/c^2)$. This is always less than $c$. Therefore it's not possible to ...


1

First, from the point of view of $O$. The lightning strikes at points $A$ and $B$ happen simultaneously. Light propagates away from those points, and since $O$ is halfway between $A$ and $B$, the light fronts reach him at the same moment (equal distances and equal velocities gives equal times). Now, for things as $O'$ sees them. The important thing to ...


1

Consider $$M_{31} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \text{ and } P_0 = -i \begin{pmatrix} 0 & 0 & 0 & 0 & 1 ...


1

I'm not sure this is a complete answer to your question, but thinking about special relativity that way will get you into trouble. Essentially, that way of interpreting special relativity attributes all of its weirdness to signal delay. Here's how I think you're interpreting the barn door experiment: The ladder is put stationary in the barn and is found to ...


1

how is the detailed connection between the statement that the length measurement has to be simulanous and the quoted derivation? Sally didn't measure a length, she measured a time and, from that, calculated a length. Both Sally and Sam agree that their relative speed is $v$ so Sally can calculate the distance between the ends of the platform by ...


1

This is my first problem, as the modulus of a vector shouldn't be negative. First, while there are many useful properties of introductory linear algebra you should keep in mind with GR, thinking in Cartesian terms with positive definite matrices simply has to go. Vectors in relativity can very much have negative norm. Even though it's not often done in ...


1

In special relativity, in the rest-frame of the proton, the moving magnet m appears as a magnet m’ and an electric dipole p’. The electrostatic E field created by the proton makes rotate this electric dipole, actually the magnet. And the E’ created by the electric dipole is the responsible for the force the proton experiences. Remark that you will not found ...


1

You vastly overestimate the meaning of frame. A frame is a (local) choice of coordinates on the spacetime manifold $\mathcal{M}$. All physical laws can be directly formulated on the manifold itself, without referring to frames at all. That is at the heart of relativity, and that is what Lorentz invariance means. Let's go through your numbered points one by ...



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