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34

For a packing of grains to stay wet up to a height $h$, the gravitational pressure $\rho g h$ needs to be balanced by the capillary pressure $\sigma cos(\theta)/r$. Here, $r$ represents the effective pore radius of the packing, $\theta$ the wetting angle (angle at which the air-water interface meets the sand grains), $\rho$ the water density, $g$ the ...


22

There is nothing wrong with the argument. The mathematics are quite simple and the conclusion is sound - scale cancels out. Let's consider the essence of the question; How does the scale size of an animal affect the absolute height it can jump? Let's assume an on-the-spot spring jump so we exclude a run-up. Now consider an arbitrary animal (let's call it a ...


22

The bigger (and longer) the object, more will be the torque experienced by it. Let's say the length of the chalk we have is $\frac{L}{2}$ (Chalk 1) and $L$ (Chalk 2). When the chalk falls on the floor, it's most likely to hit on one of its edges. Given that it is dropped from the same height, the force on the heavier mass (Chalk 2) will be more than the one ...


18

What's wrong is : For a jump of height h one needs energy proportional to $L^3/h$ Taking L as a measure of animal size then we should actually have $$E \approx Mgh \propto L^3h $$ So not divided by $h$ but multiplied by $h$ ! And a little thought would show that dividing by $h$ would make no sense, as it implies you need less energy to jump higher. ...


17

It is mainly a mathematical reason. Extensive quantities grow with system size. If two quantities scale in the same way with a variable (in this case system size), it cancels out in the division. Mini-example: $A$ and $B$ are extensive physical quantities both dependent on $n$. Their ratio is called $C = A / B$. If you scale the system up, $A$ and $B$ grow ...


15

Big objects break because they are heavier than small objects, so they hit the ground harder. You might think that a big object is also stronger than a small object. That's true, but it's not enough to compensate for the heaviness. To see why, imagine two objects of the same shape, one twice as long as the other. Since the big object has twice the height, ...


13

Here is a heuristic. The actual details will depend on the details of what type of rock it is, and materials science and chemistry beyond my pay grade, but this gives what I think should be the general idea. All rocks get wet when you put water on them, the surface gets slick, and the like. When this happens, what you get is the water adhering to the ...


11

The fact that animals are all very different aside, it ignores some important facts. The first big problem is that it ignores the fact that an animal scaled up might not be able to stand at all. By the original argument, $F\propto x^2$ and $mg\propto x^3$, where $x$ is some scaling factor. So it should be obvious that, at some point, $mg>F$ and the ...


8

Let's assume our gravitational potential is zero at our center of mass just before the jump. Our initial mechanical energy is zero. We do nonconservative work to increase our mechanical energy. Then our feet leave the floor and our kinetic energy diminishes until we reach height $h$. We have $$W_{\mathrm{nc}} = F d = \frac{1}{2}m v^2 + m g d = m g h.$$ ...


8

You are referring to scaling laws for the energy confinement time ($\tau_{E}$), which is a key performance parameter for a fusion reactor. For example, a stellarator currently has \begin{equation} \tau_{E} \propto \, a^{2.33} B^{0.85}, \end{equation} where $a$ is the minor radius and $B$ is the toroidal magnetic field. This particular scaling is of the Bohm ...


8

One of the reasons stems from extreme value statistics. Objects break at their least resistant (call it softest) spot. The probability of having a softer spot is larger in a larger object. You could think of a chain with $N$ links. Each link has a maximum force it can bear, $F$. Since links are not all the same, $F$ comes from a probability distribution, $P(...


7

From "Perturbative quantum field theory" Edward Witten (page 446 in volume 1 of "Quantum fields and strings : A course for mathematicians"):


7

The moore's (empirical) "law" states that the number of transistors in a chip increases exponentially (doubles every 2 years). So the question is : is there a hard limit in the number of transistors in a chip? Or, in other words : Are there limits on the size of a chip and on the size of transistors? Indeed there are (almost). The matter is made of atoms, ...


7

I basically agree with the answer of GiorgioP and only have a little more to add to it. First, statements in words can be just as rigorous as statements in mathematical symbols, because in the end they are two ways of presenting precisely the same assertion. The assertion in terms of maths is that a quantity $A$ is extensive if it is homogeneous of degree ...


6

It's because fractal systems are, pretty much by definition, self-similar which means that there is no preferred length scale. If something else depends on the length scale $L$ as a function $f(L)$, the argument $L$ must have units – and no unit is better than any other – so it is "dimensionful". On the other hand, $f(L)$ is a quantity that must have well-...


6

What you have shown is that Newton's law is not scale-invariant for a force $F(x,\dot{x},t)$ that is scale-invariant, since you implicitly assumed that $F$ transforms as a scalar under the dilation1. This is kind of a trivial statement: If the l.h.s. of an equation transforms non-trivially and you assume that the r.h.s. transforms trivially, the equation as ...


5

There isn't a simple answer to your question. The scaling will be different in different situations. Let's take your example of gravity. The acceleration is given by: $$ a = G \frac{M}{r^2} $$ so $a$ scales as mass$^1$ and distance$^{-2}$. But consider some other quantity like the orbital period, which is given by: $$ T = 2\pi \sqrt{\frac{r^3}{GM}} $$ ...


5

Suppose that, for a temperature $T_1$, you know $$ \rho(\lambda,T_1) = \lambda^{-5}f(\lambda T_1) $$ for every value of $\lambda$. Now, for a temperature $T_2$, let's introduce a variable $$\bar{\lambda} = \lambda T_2/T_1. $$ Then $$ \begin{align} \rho(\lambda,T_2) &= \lambda^{-5}f(\lambda T_2)\\ &= (T_2/T_1)^5 \,\bar{\lambda}^{-5}f(\bar{\lambda} ...


5

Wetting here is most likely a capillary effect: Your question is about the size of the air gaps between grains (or rocks) of sand, not the size of the grains or rocks. In practice, except perhaps for very peculiarly shaped objects, these will be of similar magnitude to the smallest gap-filling grains. What happens is that the energy required to create the ...


4

One could make an argument that we are just about the size we need to be. There is a fascinating paper from 1980 by William H. Press: Man's size in terms of fundamental constants, where he argues that intelligent beings have to have a scale of $$ L_H \sim \left( \frac{\hbar^2}{m_e e^2} \right) \left( \frac{ e^2 }{ G m_p^2 } \right)^{1/4} \sim a_0 10^9 \...


4

Think about it like this: in order to get all lengths (i.e. your own length as well as the height of your jump) to scale down by a factor $\alpha$, while keeping the contraction velocity of your muscles the same, you have to rescale all lengths and all times occurring in the problem. That means that you have to scale down the gravity (length over time ...


4

It is the timescale required by viscosity $\nu$ to diffuse momentum significantly over a characteristic length scale $L$. Analogous to how mass and heat can be transfered by molecular diffusion through collisions between particles, momentum can also be transfered through similar mechanisms. For mass and heat transfer the respective time scales are $L^2/\...


4

Assume for now that both P,V are extensive quantities. By the definition of an extensive quantity if the size of system is increased by a factor of λ the extensive quantity is multiplied by λ but the intensive quantities remain the same. So P->(λP) and V->(λV) therefore $$(PV)\rightarrow λ^{2}(PV) $$ This λ square might be the non-linearity that the ...


4

The correct requirement of homogeneity of degree one is a little more precise than what you have cited: Extensive thermodynamic quantities are homogeneous functions of degree one only with respect to their extensive variables. This solves the problem with your example of $U(V,T)$, but clearly reintroduces the problem of what an extensive variable is. ...


3

Many simplifying assumptions of our reality must be made to arrive at the results you're looking for (and that the other answers gave). I'll try to give my own 2 cents. I hope that anything incorrect in here will be corrected by other users - I'm mostly just trying to address some problems and confusion with these assumptions. When you jump, you lower your ...


3

I will give some hints: It is anomalous scaling dimension. scaling dimension is defined as $$x \rightarrow \lambda x,\\ \phi(x) \rightarrow \lambda^\Delta \phi(\lambda x) $$ From the formula (3.45) in the reference (maybe it is better to be $\phi(x) \rightarrow \Lambda^\frac{d-2}{2}\phi(\Lambda^{-1 }x)$), we know that the classical dimension of $\phi$ is $\...


3

Exponentials come as solutions of this differential equation: dx/dt=c*x where c is a constant and x and t variables. The solution is of the form: x(t)=e^(c*t) A great number of measurements and observations we make can be approximated by this equation. Once the solution is exponential it is logical to take the log since the numbers become large fast ...


3

In your definition of the dimensionless time you have assumed that the characteristic scale for time is $\frac{L^2}{\nu}$. If you instead assume that the characteristic scale is the inverse of the vortex-shedding frequency $f^{-1}$ and redo the analysis you will retrieve the Strouhal number. You will need to rescale the characteristic scale for the pressure ...


3

Hint: The gravitational force between cubes with side length $a$ is $${\bf F}(a) ~=~G\rho^2 \iiint_{[-a,0]\times[0,a]^2}\! d^3{\bf r} \iiint_{[0,a]^3}\! d^3{\bf r}^{\prime} \frac{{\bf r}-{\bf r}^{\prime}}{|{\bf r}-{\bf r}^{\prime}|^3}. $$ Now replace $a\to ka$. Try to look for a substitution of the integration variables.


3

Both force and velocity are vector quantities. This means that they can be split up into components. Here you have been given two components of a velocity and need to use trigonometry to calculate the "resultant velocity". I.e. reconstruct the velocity components back to the "total", true velocity. This is analogous to saying that I am walking East at $...


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