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41

The question is a little hard to define because, as pointed out in the comments, it's not clear what the "outside volume" is, or how to define it in curved space to compare it to regular space. Still, I think there is a sense in which the answer is "yes". A regular sphere has area $A = 4 \pi R^2$ and volume $V = 4 \pi R^3/3$, giving a relation $V = (1/6)\...


27

Surprisingly, the answer is yes. Or at least, yes, if you rephrase the question a little bit. As you might know, gravity can deform spacetime. (actually this is not completely true, it is actually the deformation of spacetime, which we call gravity) This means that around heavy objects, distances or timescales, can become longer or shorter. This will form ...


25

In a question like this you need to ask what does the volume change relative to. So it's a little bit ambiguous. However, the answer to your question is "yes" in the following restricted sense. Imagine having a "swarm" of test objects, with mass so small that their effect on the spacetime around them is negligible. Assume that they are in freefall, i.e. ...


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 ...


16

Squeezing the bottle does decrease its volume. Rather than a bottle, it may be more helpful to think of a full toothpaste tube; the mechanics will be the same. If you squeeze the middle of the tube, the middle will collapse, the back will expand, and the front will expand and squirt out some toothpaste. Treating the toothpaste in the tube (or the water in ...


16

In dimension $2n>2$ they are not equivalent since (for time-independent transformations) canonical is equivalent to $$\sum_{k=1}^n dq^k\wedge dp_k = \sum_{k=1}^n dQ^k\wedge dP_k\tag{1}$$ whereas conservation of oriented volume means $$dq^1\wedge \cdots \wedge dq^n \wedge dp_1\cdots \wedge dp_n = dQ^1\wedge \cdots \wedge dQ^n \wedge dP_1\cdots \wedge dP_n\:...


16

I'd like to add another option to the answers provided by Javier and Crimson. There is a spacetime geometry popularised by Wheeler (I'm not sure it originated with him) called the bag of gold spacetime. This doesn't have a Wikipedia page I can link to but I have mentioned it in my answer to Creation of a miniature universe? There are a few other related ...


16

The physical concept most closely mimicking the experience of fantasy items like bag of holding is a traversable wormhole. Essentially, it is a connection between two different regions of space, a portal (yes, like in a videogame), or maybe even bridge between different universes. In the former case the inside volume of a car's trunk would be located ...


14

The event horizon is a lightlike surface, and so its area is coordinate-invariant. For a Schwarzschild black hole, $$ds^2 = -\left(1-\frac{2m}{r}\right)dt^2 + \left(1-\frac{2m}{r}\right)^{-1}dr^2 + r^2(d\theta^2+\sin^2\theta\,d\phi^2)$$ The horizon suface is at $r = 2m$ of Schwarzschild radial coordinate, and so at any particular Schwarzschild time ($dt = 0$)...


14

The surface area of the bottle is conserved, but the volume is not. Squeezing the bottle deforms it into a shape whose volume to surface area ratio is lower than it was previously. As an example consider a bottle whose cross-section is initially a circle. The volume of the bottle will be $V_0=\pi r^2h$ where $h$ is the height of the bottle, and the ...


14

There's actually not one simple answer to your question, which is why you are a bit confused. To specify your problem fully, you must specify exactly how and whether the gas swaps heat with its surroundings and how or even whether it is compressed. You should always refer to the full gas law $P\,V=n\,R\,T$ when reasoning. Common situations that are ...


9

Like all good physicists we'll start by assuming the animal is spherical. Actually the calculation I'm going to describe is basically the same whatever shape the animal is, but choosing a sphere means I can write down some simple formulae. If the radius of the spherical animal is $r$, then the total area of its skin is the surface area of the sphere: $$ A =...


9

Counterexample: The transformation $$Q^1~=~2q^1 ,\qquad P_1~=~p_1,\qquad Q^2~=~\frac{1}{2}q^2 ,\qquad P_2~=~p_2 $$ preserves phase space volume & orientation, but is not a symplectomorphism.$^1$ For 2D phase space, the canonical phase space volume form $$\Omega~=~\frac{1}{n!}\omega^{\wedge n}$$ is the symplectic 2-form $\omega$ itself, so that the ...


8

Double the ammount of water does not need doulbe the ammount of time to heat, since while the energy needed is doubled indeed, losses due to vaporization and radiation from the kettle should be approximately constant. You can plot the time needed for a given ammount of water to boil and try to fit a function into that. With two data points you can manage to ...


7

As you suggest, you could use a large number of nested containers to gradually increase the pressure of the innermost container to very large values. Generally it's only the pressure difference that matters. These could even be weak containers, like a balloon within a balloon within a balloon within... But this can't go without limit. Once the absolute ...


7

In the quantum regime the answer to such questions really depends on what you mean by "volume". A photon has no volume in the sense that it can theoretically be confined in an arbitrarily small region of space. Although such confining would result in the photon having extremely high energy fluctuations (basically due to the uncertainty principle) so that ...


7

For a fairly exact measurement, I would take some harmless chemical that's easy to detect, and pour a certain volume in and let it dissolve/diffuse completely. Then, measure it in ppm, do the necessary calculations, and get your volume.


6

Pressure and volume have an inverse relationship when $n$ and $T$ are constant. How do you imagine the pressure in the balloon is increased? Either $n$ or $T$ must increase, or $V$ must decrease. Additionally, balloons are roughly constant-pressure systems. The rubber membrane is a very weak elastic, so the internal pressure of the balloon is at almost ...


6

Most materials contract on cooling. The notable exception to the rule are some phase transitions and water. But even ice contracts on cooling. Water expands on cooling only between $0^\circ\text{C}$ and $4^\circ\text{C}$ (including phase transition). This corresponds to the part of the graph below, in which density rises with temperature (note suppressed ...


6

My belief is that this refers to the density effect alone. Obviously you will get less gas on a hot day since it will be lower density at that temperature. This is separate from thermodynamic factors. I would venture an educated guess that ICEs will be less efficient on hot days, and I would suspect that this factor would be greater than the density ...


6

If you are a Schwarzschild observer the radial co-ordinate $r$ is defined as the circumference of a circle around the black hole divided by $2\pi$. Also the event horizon is at a Schwarzschild radius of $2M$ (in geometrised units), so for a Schwarzschild observer the area of the event horizon is simply $16\pi M^2$. But this is a somewhat trivial answer as ...


6

The key for producing a nice tone, is a mix of two facts: first, offering the air stream a somehow geometrical regular hole, where a stationary wave can be born with a certain frequency and other possible frequencies are filtered out, and second, a low enough stream velocity, so that no turbulences can happen and the flow is ordered (laminar). The whistle ...


6

First: what frequency should you hit? There are many, many different factors at play in determining the natural frequency of an object I know from experience. These are (not limited to): Thickness, density, elasticity modulus (you'll need two of those, e.g. Young's Modulus and Poisson Ratio), and of course shape. I'm not aware of any papers publishing a ...


6

I assume you ask this because you want to know if a fart's mass contribution to body weight can be offset by its lighter-than-air property, if any. According to About.com the typical composition of human flatulence is as follows. I added the molar mass of each component, as the density of a gas determines whether it's heavier or lighter than air: \begin{...


6

The event horizon of a Black Hole is a lightlike surface, and so its area is coordinate-invariant. Take as an example the Schwarzschild black hole \begin{align} ds^2 &= -\left(1-\frac{2m}{r}\right)dt^2 + \left(1-\frac{2m}{r}\right)^{-1}dr^2 + r^2(d\theta^2+\sin^2\theta\,d\phi^2) \end{align} The horizon suface is at $r = 2m$ of Schwarzschild radial ...


5

Volume increase in the system is due to work done by the system. Therefore W is negative using your notation. Think of it this way, work done on the system would push the system inwards, decreasing volume. Therefore a volume increase is work done by the system. Alternatively you could reason using the formula: $dU = dQ - dW$ (using your notation ...


5

Think about this: why is the pressure increasing? If it's because you're blowing air into the balloon (which is the usual way to increase the air pressure), then what you're actually doing is raising $n$. And it makes sense that an increase in $n$ should be correlated with an increase in $p$ (or $V$, or both). Boyle's law doesn't apply in this case because $...


5

Let me begin with the second question where you don't change the dimensionality, just the volume. The entropy never decreases when you actually compress gas. The compression means that the walls are mostly moving against the colliding molecules which means that they're recoiled backwards at higher velocities. The molecules' kinetic energy increases so they ...


5

You need a psychrometric chart. On the x-axis, input the value of a common air thermometer (it's the dry bulb temperature). Now follow a vertical line until you reach the red curve that reads 90% relative humidity. Follow the horizontal line to the y-axis on the right and that humidity ratio will be the ratio of water versus dry air (mass-wise). That is, 0....


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