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1

If we measure length by a scale there is no change of energy. So it depends on subjects to be measured.


3

Susskind's original argument doesn't work. Alice just needs Bob to send a message to her saying "I'm still alive!" She doesn't have to illuminate Bob. Of course, it's hard to get a message out from the near-horizon region of a black hole because of the redshift, but there's no theoretical reason that this shouldn't work. Suppose you don't have a ...


2

Conjecture: The distance to the moon was approximately known by astronomers in Newton's days. Once you know the orbital radius and period, a simple diagram shows you how much the moon "falls". When angle $q$ is small, the distance $d$ is calculated from $$\frac{d}{v~t}= \frac{v~t}{R}\\ d=\frac{v^2 t^2}{R}$$ Note - interesting details on this ...


0

Good point, What Collier should be saying is: By assuming his own law to be true, and given what we know today, Newtons might could have reckoned that the Moon actually ‘falls’ not 5 m, but 1.37 mm (0.00137 m or about one sixteenth of an inch) below a straight line trajectory in 1 second. It wasn't until centuries later than anyone verified this ...


1

Advantages Very accurate ( In fact its accuracy allows it to be utilised to calibrate other thermometers) Wide Range Independent of gas used Disadvantages Large and bulky(inconvenient to carry and handke) Slow to Respond (due high heat capacity) Expensive to manufacture and keep I'm not sure about its use in any specific industries.


3

The most famous "terrestrial" experiment that I know of was the measurement of the speed of sound in Nevile's Court in Trinity College, Cambridge. As the tour guides will tell you, Newton stomped his foot and listened for the echo from the wall on the opposite end of the North Cloister. He timed the echo by making a (short) pendulum, and adjusting the length ...


0

$\chi_I$ is a number, not an operator, do not mix up $\chi_I$ with the projection operator. This whole formalism is actually very simple in this case. $P_I$ is a projection operator onto the space of states which correspond to a particle being inside the interval $I$. Suppose we live in discrete space for the time being so we can work with explicit ...


1

Notice that the probability of measuring say the position of a particle whose wavefuction is $\psi(x)$ in the interval $I=(a,b)$ is $$\int_a^b \left|\psi(x)\right|^2 \mathrm{d}x.$$ We can define a multiplication operator on the state space much like the position operator $\hat{X}\psi(x)=x\psi(x)$ as follows. $P_I(\psi)=\chi_I(x)\psi(x)$. It is a projection ...


0

For any self-adjoint operator $A$ (with continuous or discrete spectrum), the so-called spectral theorem tells you there is unique a family $(P_{\lambda}(A))_{\lambda\in\mathbb{R}}$ of orthogonal projections (called spectral family) such that $P_{\lambda_2}(A)-P_{\lambda_1}(A)$ gives you the projection on the subspace with spectrum in the interval $[\...


0

The concept of 'independence' of units is difficult to maintain. For large distances for example, we tend to use light-hours or light-years, effectively merging length and time. As Wrzlprmft already pointed out, you could reduce all units to numbers introducing similar concepts. I tend to treat units as a concept to make things easier – both in everyday ...


0

You're measuring the wrong thing. Measure the momentum of a punch, not the "force". The force will be different if you hit something soft or hard, so you can't depend on that. The momentum is just the mass of hand+glove+forearm times velocity of punch. How to measure that? Hit the punching bag and see how far it swings.


0

Considering the stormy nature of Jupiter's atmosphere, and the planet's vast size, no I don't think that kind of shrinkage rate is significant, and therefore not measurable. Moreover, it is certainly not the kind of useful information which it is worth the effort trying to obtain.


1

You've incorrectly found that the matrix for $H$ is given by $$ \left(\begin{array}{cc} 1&0\\0&H \end{array}\right) $$ where $H$ and $1$ are $2\times 2$ matrices, and we're using the following ordering of basis elements: $(|00\rangle,|01\rangle,|10\rangle,|11\rangle)$. This isn't true. An easy way to see that is to compute the matrix element $\...


1

First an important clarification about simultaneity you need to be aware of: In special relativity we learn that there is no such thing as two spatially separated events A and B happening 'at the same time', at least not in any absolute sense. If one inertial observer sees the events as simultaneous, another perfectly legitimate inertial observer sees A ...


0

The speed of light is invariable but the number or ratio of the frames of reference are variable as determined by the amount of gravity, the result of which is gravitational lensing.



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