Tag Info

Hot answers tagged

85

No. To balance perfectly, the pencil would have to be perfectly upright and perfectly still. The uncertainty principle limits how well you can do both at the same time. Momentum and position form a conjugate pair. $\Delta x \Delta p \geq \hbar$. Angular momentum and angular position form one too. $\Delta L \Delta \Theta \geq \hbar$ This doesn't guarantee ...


85

TL;DR: there are many factors that prevent a pencil remaining perfectly balanced. The most important of these is the uncertainty principle that will make the pencil fall over in less than four seconds. For details, read on... Short answer: NO. The first photon of light that hits it would disturb your perfect equilibrium. The moon's tidal forces (which are ...


19

No. The weight of the pencil is roughly 1 Newton, and the area is about 500 square picometer (5 * 10-22) which means the pressure on the tip is around 2 ZettaPascal. That's quite a bit more than what graphite (or diamond) can withstand (that's masured in GigaPascal)


16

Let's forget physics for a moment and just talk about the mathematics of waves. The uncertainty principle is a property of waves. Think of a single, narrow pulse traveling along one direction. The pulse is narrow, and so the position of the pulse at any given time is easy to quantify. But this is a single, non-periodic pulse. You can build up such a ...


11

In classical physics you are supposed to be able to measure the coordinates and the velocity (really the momentum) of a mass with infinite precision at the same time. If you try this trick in the lab you notice that that's not the case. Either your position or your momentum measurement or both will always show some non-trivial statistical fluctuations when ...


5

The question is so ambiguous that it allows a resounding yes. This is because "balance" is not defined, neither are the dimensions and the material used for the pencil, nor the location of where the "balancing" is to happen. The material and the shape of the surface to balance the pencil on, are not specified, nor the length of time it should stay balanced. ...


4

No. Firstly, the point of the pencil is generally not sharp enough to have just one single atom. People attempt to make that kind of tip in STM's. Even if you somehow did manage to get it sharp enough, graphite is so soft the the weight of the pencil will crush the tip. It won't stay a single atom wide. So there is no way to balance the pencil on a single ...


2

Let me be the firs to answer 'Yes' (more or less). As the saying goes: In theory there is no difference between theory and practice. In practice there is. What I'm getting at is that there will always be differences between theory and practice, and that it is up to the physicist to decide which assumptions/simplications are suitable and which ones are ...


2

The zero-point energy is the energy eigenvalue $\langle H \rangle_\Omega$ of the lowest lying energy eigenstate $\Omega$. $\Delta H_\Omega = 0$, so this has nothing to do with the zero-point energy. This does also not contradict the uncertainty principle, since $\frac{\mathrm{d}\langle A \rangle}{\mathrm{d}t} = 0$ for a energy eigenstate (which are the ...


2

What you've fabricated is of course unrealistic in the physical world as CuriousOne stated, but not so in the virtual world of simulation. All the conditions you ask for can be arranged in a simulated universe. If perfectly balanced as its initial condition, the virtual pencil will not fall in this virtual world. It is unstable, but it will not fall until a ...


2

In the universe where this pencil is, there are no outside forces which can affect the pencil, other than gravity. But there are forces afoot inside the pencil. Unless you also chill the pencil to absolute zero and thus stop all molecular activity, the trillions of atoms in the pencil are vibrating in random directions. It won't take long for the ...


2

There is a related (and relevant) question with answer called Uncertainty principle and measurement but I want to give a specific answer for this situation and background. You will need top know the difference between precision and accuracy. A precise dart thrower throws darts that land very close to each other, this isn't a deep statement, in fact it is ...


1

Why don't we observe the infinite violations of conservation of energy The reason we don't see e.g. an atom spontaneously turning into a red giant for a fraction of a second is because of the extremely small timescales that such an energy difference would require - not even light could travel a tiny fraction of a proton radius in that time. The heavy ...


1

$$\Delta E \Delta t \sim \hbar$$ This is a version of the Heisenberg Uncertainty Principle. Instead of using momentum and position,however,the above form uses energy and time ($\Delta$ means change in). How can the uncertainty principle be used to deduce range of a force from properties of the force carrier? Let's take the simplest example: the ...


1

When someone says that spin measured about different axis can't both be known, they mean that whatever state you pick will have variability in at least one of the possible spin measurements you can do. So that is what you will get when measure the spin, you will get variable results. This happens even with entanglement with even just one particle. With ...


1

A highly simplified analogy that attempts to get thinking going in a direction that can sometimes lead to eventual understanding... Think of various photographs taken of a baseball in flight. Different photographs are taken with different exposures. They are part of an attempt to measure both speed and position at the same time. In particular, imagine that ...


1

Which way will the pencil fall, after you let go? Facetious answer: it would fall in the direction to which it was leaning when you let go. Okay, now to justify that: you cannot balance the pencil before letting go. For an object resting on a surface to be balanced, its centre of mass (a single point) must be directly above a point within the area of ...


1

Let $X$ and $Y$ be random variables with density functions $f$ and $\hat{f}$, where $\hat{f}$ is the Fourier transform of $f$. Then the uncertainty principle is a lower bound on $\sigma_X\cdot \sigma_Y$ (where $\sigma$ is the standard deviation). In particular, $\sigma_X\sigma_Y\ge 1/4\pi$. For example, take a particle moving in one dimension, in a fixed ...


1

Roughly speaking, and restricting to particles for now, a classical trajectory is a set of exact positions and corresponding velocities (or momenta) of the particle, which (usually) change over time. In quantum mechanics, the uncertainty principle says that it is not possible to know simultaneously the exact position and momentum of the particle, and so it ...


1

In classical mechanics, each body has an exact position at all times (and hence an exact speed given by its derivative, and the equivalent quantities for rotary motion). To understand the difference to quantum mechanics, think of it as wave mechanics: Each body is described by a wave(function) and properties such as position or speed are only defined to the ...


1

The short answer is in this paragraph, with more detail in other paragraphs. The particle acts a bit like a blob of mist spread out over position and momentum space. The intensity of the mist in a given region tells your the probability of finding it there. If you squash the blob too much in position space then it will be more spread out in momentum space ...



Only top voted, non community-wiki answers of a minimum length are eligible