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Ignore the diagram. It's complex because it includes all the various bits of electronics required for the experiment and that confuses the issue. The experiment is just a variant of Mössbauer spectroscopy. The experiment uses a $^{57}$Fe source that emits a gamma ray with an energy of 14.4 keV. Since this energy matches the spacing in the energy levels of ...
As you correctly note, you need to prove $t_B > t_A$ is preserved by Lorentz transformations. It's not clear to me whether your final answer demonstrates this (since the sign of $x_b - x_a$ isn't totally obvious), but I think you're on the right track. I might have responded with the following argument. The invariant interval $$ds^2 = -dt^2 + dx^2 ... 2 No your argument is not correct. Firstly, velocities do not add linearly like 3-vectors in Euclidean space: the relativistic sum of two velocities always has a speed of less than c if both the velocities' magnitudes are less than c (no matter what their direction). Secondly, photons have no rest frame: that's a basic property of things that have zero ... 2 You can think of this question as someone a distance 2d ahead of you releasing a pulse of light (the man in the mirror). In that way, the problem simplifies to "how far can light travel in 0.80 \mu s?" Solving this will give you the value of 2d (2d = ct). So divide by 2 to get d, and the calculation is simply d = ct/2 = 120m. EDIT: I should ... 2 What you have done here is a Galilean transform, that is a non-relativistic transformation. Take your final result (which is quite correct):$$ t' = \frac{\sqrt{\beta^2 + \alpha^2}}{\sqrt{\eta^2 + \mu^2}} \tag{1} $$We know that the vertical velocity is \eta, so the vertical distance moved in our time t is given by:$$ \beta = \eta t $$We also know ... 2 You can determine the charge of an electron from a static measurement in one frame. Another frame could determine the charge of an electron from a static measurement in their frame. And they might agree or disagree. We postulate they agree, but we had three options: We could postulate that whether or not something is an electron depends on your frame ... 1 Let us say you have two frames of reference; frame F and frame F' such that F' is moving at velocity v in the positive x direction of F. Given a space time event that occurs at (ct,x,y,z) in frame F the Lorentz transform helps us to find the space-time coordinates (ct',x',y',z') of that event in frame F'. If, however, you know the event ... 1 Relativity just requires "constant speed of light in vacuum". It makes no claims about the speed of light in a medium. When you are moving relative to water, you will observe a different speed of light depending on your relative velocity. But you will still have all the other effects of relativity at work - such as time dilation. 1 The necessity of anti-particles was first noticed when trying to construct quantum mechanical descriptions of particles that obey the relativistic energy-momemntum-mass m^2c^4 = E^2 - (\mathbf{p}c)^2 relationship. The Schrödinger equation is intuited from a combination of de Broglie's rules E = hf and p = h/\lambda and the classical Hamiltonian E = ... 1 I don't know the "formal" proof, but here is my proof: Time dilation and length contractions are given to us by the Lorentz transformations by: t’ = t/(1-v2/C2)1/2 and d’ = d/(1-v2/C2)1/2 (in other words “same” or proportional to each other) where: t = distance/length traveled through the T dimension in observers own frame of ... 1 The equation you quote:$$ t' = t\sqrt{1-\frac{3GM}{rc^2}} \tag{1} $$gives the time relative to an observer at infinity. You want the time relative to an observer on the Earth's surface. You need to calculate:$$ t_\text{satellite} = t\sqrt{1-\frac{3GM}{r_\text{satellite}c^2}} $$and:$$ t_\text{Earth} = t\sqrt{1-\frac{2GM}{r_\text{Earth}c^2}}  ...