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14

The way atomic clocks work is to produce a microwave signal with exactly the frequency of the atomic transition being used. So for caesium this would be 9,192,631,770Hz. Then we can count the oscillations of our microwave generator to measure the time. Practically you do this by tuning your microwave signal to maximise its absorption by the caesium atoms. ...

5

A tensile pulse travels through the rope at the speed of sound. This speed depends on the density and the bulk modulus of the material - a rope strong enough to support its own weight would probably have a very high bulk modulus. The equation is $$v = \sqrt{\frac{K}{\rho}}$$ Where $K$ is the bulk modulus and $\rho$ the density.

3

The black axes give the frame of the meter stick. The black vertical axis is the worldline of the left end of the stick and the parallel black line is the worldline of the right end of the stick. Your condition 1) says that the spatial distance from A to B, measured in the black frame, is shorter than the spatial distance from A to B, measured in the ...

2

Consider this nomenclature for simplicity: Photon 1 is the photon emitted by the spaceship when it was at rest with respect to the outside observer and Photon 2 is the photon emitted by the spaceship when it is moving at a uniform speed with respect to the outside observer. Since the mechanism through which the photon is emitted from the spaceship is not ...

2

Addressing the 3rd question, the clocks' frequencies are the same when each is viewed in its own rest frame. When a clock is viewed in a moving frame, that's when its frequency is changed. You record the ticks and their locations (which are different since the clock is moving) and you discover that the time between ticks is now longer.

2

There is absolutely a gravitational radiation reaction and solving for it is one of the very active fields in classical relativity theory at present. Basically, particles with nontrivial masses distort the spacetime around them; this causes them to not move on geodesics of the "background" spacetime (the spacetime that one would have found had the secondary ...

2

Physics is an observational science. It measures and observes the way nature behaves, and ever since Newton, models these measurements and observations with mathematics. Mathematics is a discipline where one starts with axioms uses logic and arrives to theorems and expressions that can be proven using the axioms and other theorems. One ends a proof in ...

2

Excellent question. And well put. You're quite correct in saying that any obseverver can claim to be moving at a constant speed. However, doesn't George also have the right to claim/perceive himself as the one who is in motion, and if so wouldn't he perceive Gracie's clock to tick faster than his? No, the clock George looks at (on his wrist or ...

1

No. Cooling down the atom will not alter it's frequency. What the scientist meant by saying that the clocks will be better is that they will be much more accurate. This comes from the principle of Quantum mechanics, in fact one of its most beautiful consequence, Heisenberg's uncertainty principle. Which says (in one of it's variety) that the uncertainty in ...

1

No. It is not about any measurement error or the wrong measurements being carried out by some observer or something. Relativity is about the actual changes that come when you change the frame of reference. It is the heart of relativity that all the observers are equivalent and thus surely it doesn't suggest that the 'moving' observers carry out 'wrong' ...

1

yes, time dilation still occurs. The reason is that the mirrors are there to provide an intuitive view of how/why time dilation occurs, not to to create it. This time dilations still occurs regardless of the existence of the mirrors or the direction of motion. But the drawing will no longer have explanatory power.

1

In that case, time dilation still occur, of course. In order to show this using t=d/v, you'd have to take into account the space contraction in the direction of motion. Mathematically, if d is the height of the clock, then the time taken from a photon at the bottom to reach the top of the clock isn't $\frac{d+vt}{c}$ but $\frac{d/\gamma+vt}{ c}$. When you ...

1

Comments to the question (v8): Let us here for simplicity consider point mechanics. The generalization to field theory is straightforward and left to the reader. Given an (off-shell) action functional $$\tag{1} I[q]~=~\int_{t_i}^{t_f} \! dt~L,$$ it seems that OP in the first half of his post mainly confirms that the functional/variational derivative ...

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