If a balloon is filled with hot air, it is rising due to buoyancy: the mass of the hot air inside the balloon is lower than the mass of the same volume of the cold air outside the balloon cavity.

However, in absence of a balloon, how and where the pressures producing the buoyancy force are applied to? - the hot air simply diffuses into colder regions and vice versa, but the collisions between the molecules seem too rare to talk about pressure.

In terms of continuum hydrodynamics we are talking about the difference between the heat conduction and convection. The question is then: when can a gas can be approximated by a continuous liquid, so that talking about convection makes sense? Is this description applicable to air?

I will particularly appreciate the answer with actual estimates (in terms of air density, mean free path, etc.).

The question boils down to how we calculate pressure on microscopic level in absence of a rigid boundary. In statistical physics pressure by a gas on a boundary is calculated from the change of momentum of atoms/molecules reflected by the boundary. In absence of a boundary one could suggest scattering against other molecules - but then, what is the difference between mascroscopic (bouyancy, convection) microscopic (diffusion, heat conduction) behavior of a gas?

  • $\begingroup$ "but the collisions between the molecules seem too rare to talk about pressure." How you've came to this conclusion? Seems to me that the higher kinetic energy hot air molecules displacing cooler air molecules being the central mechanism behind the buoyancy phenomenon and actual reason why hot air rises up for a normal atmosphere pressure gradient of one At at sea level. $\endgroup$
    – Markoul11
    Nov 14, 2021 at 16:36
  • $\begingroup$ A question: isn't convection a form of buoyancy anyway? $\endgroup$
    – Alchimista
    Nov 14, 2021 at 16:54
  • $\begingroup$ @Alchimista convection is bouyancy, but not diffusion. $\endgroup$
    – Roger V.
    Nov 14, 2021 at 17:32
  • $\begingroup$ @Markoul11 think of how pressure is calculated in statistical mechanics - it has nothing to do with intermolecular collisions, which are rare in gas (as opposed to a liquid). $\endgroup$
    – Roger V.
    Nov 14, 2021 at 17:35
  • $\begingroup$ Does this answer your question? Is rising air high pressure or low pressure? $\endgroup$ Nov 18, 2021 at 8:17

4 Answers 4


The reason for this is actually the temperature-dependent distribution of the absolute value of particle velocities (higher temperature results in higher average velocities as well as greater variance of the underlying distribution) and the impact forces will have on particles with different velocities. The buoyancy is not a result of a pressure force (it does not emerge from a pressure term) but instead it emerges directly from gravity acting upon the gas molecules moving at lower or higher speed. On a macroscopic level this turns out to be correlated to the local density. I will try to illustrate this in the following section by adapting the macroscopic (Archimedes principle and Navier-Stokes-Fourier equations) and the microscopic view (Boltzmann equation) of physics.

My post turned out quite lengthy. I am pretty sure one could give a more concise answer but after having put a lot of time into writing it I do not feel like deleting most of it again.

You have probably already heard the saying (commonly attributed to the Brit George Box) "All models are wrong, but some are useful": Models fall short of the complexity of real physics but some can be used to make valuable predictions if somebody is aware of the limits and pitfalls. Maybe some things cannot be described by a model because it is too coarse to capture the phenomena (it does not resolve the scale of interest) and some models emerge as special cases from other models.

In the case of macroscopic fluid dynamics of gases (as a matter of fact most common fluids) the macroscopic description is given by a set of partial differential equations in these macroscopic variables, the Navier-Stokes-Fourier equations. The description assumes that a gas is so dense that one can find limit values for macroscopic properties like density, stresses or specific forces

$$\rho := \lim\limits_{\Delta V \rightarrow 0} \frac{\Delta m}{\Delta V}, \phantom{abcdefg} \sigma_{ij} := \lim\limits_{\Delta A_i \rightarrow 0} \frac{\Delta F_j}{\Delta A_i}, \phantom{abcdefg} g_i := \lim\limits_{\Delta m \rightarrow 0} \frac{\Delta G_i}{\Delta m}$$

as well as temperature $T$, an assumption commonly referred the continuum assumption. This breaks down for very dilute gases, where diluteness (generally specified by $Kn \gtrsim 0.01$) is a question of the scale of interest, generally characterised by the Knudsen number $Kn$ which defines that ratio between the mean free path $\lambda$ and the characteristic length scale of the problem $L$:

$$ Kn := \frac{\lambda}{L} \phantom{spacespace} \frac{\text{mean free path}}{\text{representative physical length scale}}$$

This means that something that might seem dense on a larger scale might be too dilute on a smaller scale to be considered a continuum after all.

The macroscopic description can be derived as the limit case for dense gases from a microscopic (or rather mesoscopic) description from kinetic theory of gases, the Boltzmann equation, in the limit of dense gases by a perturbation analysis, the Chapman-Enskog expansion. One might argue that the Navier-Stokes equations emerge as a special from the more general Boltzmann equation and that the Boltzmann equation is more general. They are two different levels of description and can be used to explain the same phenomena, while the reason for certain phenomena cannot be fully explained on the coarse macroscopic level. I would argue that advection, diffusion and heat conduction are all actually macroscopic phenomena: They are caused by microscopic mechanisms but are actually only visible on a much larger scale.

Now we are going to investigate the buoyancy on these two levels of description for your two different cases and see how they two are different.

1. Buoyancy on the macroscopic level

1.1 With the balloon

Already 200 B.C. Archimedes, a greek mathematician and physicist, found out that static buoyancy is connected to a difference in density (Archimedes' principle):

Any object, totally or partially immersed in a fluid or liquid, is buoyed up by a force equal to the weight of the fluid displaced by the object.

A difference in density and the resulting displacement will result in a force: A balloon filled with hot air having a lower density than the surrounding cold air will result in a net force upwards. This means neglecting friction (viscous effects) one could actually easily calculate the resulting force. This can be expressed in mathematical more rigorous terms by integration of stresses:

$$ F = - \int\limits_{0}^{\pi} \sigma_{r \theta} |_{r=R} \sin(\theta) dA + \int\limits_{0}^{\pi} \sigma_{r r} |_{r=R} \cos(\theta) dA $$

If you are able to determine the distribution of stresses $\sigma_{r \theta}$ and $\sigma_{r r}$ acting inside the balloon and outside of it, you can determine the resulting force.

For the static case this is pretty straight forward while for any dynamic case this can be pretty complicated as one would have to consider the full Navier-Stokes-Fourier equations (continuity equation \eqref{1}, momentum equation \eqref{2}, energy equation \eqref{3}, Fourier's law \eqref{4} and the constitutive equation \eqref{5} as well as the equation of state \eqref{6}) and can only be done for very simple symmetric cases in laminar flows.

$$\frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\rho u_j )}{\partial x_j }=0 \tag{1}\label{1}$$

$$\frac{\partial (\rho u_i )}{\partial t} + \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_i u_j )}{\partial x_j} = \sum\limits_{j \in \mathcal{D}} \frac{\partial \sigma_{ij}}{\partial x_j } + \rho g_i \tag{2}\label{2}$$

$$\frac{\partial (\rho e)}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\rho u_j e)}{\partial x_j} = - \sum\limits_{j \in \mathcal{D}} \frac{\partial q_j}{\partial x_j} + \sum\limits_{(i, j) \in \mathcal{D}} \frac{\partial (\sigma _{ij} u_i)}{\partial x_j} + \sum\limits_{j \in \mathcal{D}} \rho u_j g_j \tag{3}\label{3}$$

$$q_i = - k \frac{\partial T}{\partial x_i}. \tag{4}\label{4}$$

$$\sigma_{ij} = - p \delta_{ij} + 2 \mu S_{ij} - \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} S_{kk} \delta_{ij}. \tag{5}\label{5}$$

$$p \, v = \frac{p}{\rho} = R_m T \tag{6}\label{6}$$

where $S_{ij}$ is the rate of strain tensor

$$S_{ij} := \frac{1}{2} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right).$$

If you are assuming a sperical body that is only falling or rising slowly you can actually neglect most of the terms and simplify the equations to determine the terminal sinking speed analytically by extending the derivation here with a simple force balance. The most important equation in this context is the momentum equation \eqref{2} which is a continuum equivalent of Newton's second law.

In your case with the hot and cold air and the elastic membrane the two would push the membrane such that there is an equilibrium between the two with equal pressure. If we assume that the membran of the balloon is adiabatic (and further assuming density differences that are sufficiently small such that the terminal speed is still laminar in the creeping regime) then the formulas mentioned in the post above would hold.

1.2 Without the balloon

Without the balloon other terms of the Navier-Stokes-Fourier equation system will gain importance. The conservation equations in the end are nothing but advection-diffusion equations with different diffusivities of the following form:

$$\underbrace{\frac{\partial \Phi_i}{\partial t}}_{\text{temporal change}} + \underbrace{\sum\limits_{j \in \mathcal{D}} \frac{\partial (\Phi_i u_j )}{\partial x_j }}_{\text{change due to advection}} = \underbrace{ \sum\limits_{j \in \mathcal{D}} \frac{\partial D_i}{\partial x_j } }_{\text{diffusion}} + \underbrace{S_i}_{\text{source}}$$

As you can see from \eqref{2} and the equation above the gravity force term is not part of the pressure term - instead it is a source term.

Now if there is no balloon the air of different temperature can mix (clearly if they were two kinds of different fluids and the coherence between the fluid of the same kind would too large the two would not). This means that the differences between the hot and the cold air will even out over time due to the diffusive term. You would still have the buoyancy but it would interfere with diffusion of momentum and energy. If you wait sufficiently enough the air would have the same temperature and density. Which mechanism dominates depends on the difference in temperature and the corresponding diffusivities.

I want to make a brief remark at this point on the post that sparked your question: It is true that in natural convection the pressure of the two hot and the cold air is actually identical. Pressure disturbances propagate with the speed of sound

$$ c_s := \sqrt{\left( \frac{\partial p}{\partial \rho} \right)_S } = \sqrt{\gamma R_m T} $$

This means that for phenomena that involve fluid flow at lower speeds (characterised by a Mach number $Ma := U / c_s$ lower than 1) the flow will be aware of the increase in pressure, resulting in a smooth change in pressure between two reservoires of hot and cold air. For supersonic flow ($Ma > 1$, e.g. a balloon with a sufficiently high pressure difference to the surrounding medium and a de Laval nozzle) this is different and there would be a difference in pressure between the two streams.

2. Kinetic theory of gases

Now if we want to know what happens on a level of gas molecules we will have to turn to the kinetic theory of gases. For this we assume a simple model gas made of billiard-ball-like molecules that interact with each other in elastic collisions. These collisions preserve mass, momentum and energy. Contrary to your assumptions and as already outlined in my previous answer, collisions in gases are not rare but actually pretty common due to the immense number of particles. While in a real gas you will have far-field interactions due to repulsive forces, it allows us to get quite a good approximation.

2.1 Boltzmann equation

Properties like the density $\rho ( \vec x, t )$ are a measure for the average state of particles. When considering dilute gases a density cannot be found but instead one may find particles at a certain point in space $\vec x$ at a certain time $t$ with a corresponding velocity $\vec \xi$ with a corresponding probability. The basis of kinetic theory is the introduction of a particle distribution function (which can be seen as a generalisation of the density $\rho$, which is only a function of time and space $\rho(\vec x,t)$)

$$f(\vec x, \vec \xi, t) := \frac{dN}{d \vec x \, d \vec \xi}$$

where $N$ denotes the number of molecules.

Now all relevant macroscopic variables can be determined as moments of the distribution function through integration over the velocity space. The zeroth moment equals to the density

$$ \rho(\vec x, t) = m_P \int f(\vec x, \vec \xi, t) d \vec \xi, $$ the first moment to the impuse

$$\rho(\vec x, t) \vec u(\vec x, t) = m_P \int \vec \xi f(\vec x, \vec \xi, t) d \vec \xi, $$

where $\vec u$ is the mean macroscopic velocity. The second moment delivers the total energy density

$$ \rho(\vec x, t) e(\vec x, t) = \frac{m_P}{2} \int \vec \xi^2 f(\vec x, \vec \xi, t) d \vec \xi, $$

Now we want to determine how this distribution evolves in time, the so called Boltzmann transport equation. In a real system an ensemble contains theoretically information about its state at any time but due to the enormous number of interactions over time this is converted into subtle correlations that appear chaotic (The time over which the system may be predicted is very limited.) and almost random. When deriving this equation Boltzmann therefore assumed that the particles are un-correlated before collisions (one-sided molecular chaos). In dilute gasses it can be further assumed that the particles ($i=1, \ldots , N$) spend most of their lifespan on free trajectories apart from collisions involving only two particles (Collisions involving multiple particles are so rare they can be neglected.) at a time for which in the case of mono-atomic (billiard balls) gas classical Newtonian physics can be assumed

$$\frac{d x_i}{dt}= \xi_i = \frac{P_i}{m_P}, \phantom{space} \frac{d P_i}{dt} = F_i.$$

where $\vec P$ stands for the transferred momentum. In such a model the particle distribution would follow a simple transport equation, the Boltzmann transport equation. The total derivative of a distribution function $f$ given by

$$\left. \frac{Df}{Dt} \right\vert_{transport} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x_i} \underbrace{ \frac{\partial x_i}{\partial t} }_{\xi_i} + \frac{\partial f}{\partial \xi_i} \underbrace{ \frac{\partial \xi_i}{\partial t} }_{\frac{F_i}{m_P}}. $$

must be equal to the changes caused by the collision. This can be rewritten using $\partial x_i/\partial t=\xi_i$ and $\partial \xi_i/\partial t=F_i/m$ as well as introducing the notation $\Omega(f)=Df/Dt$ for the not yet defined collision integral to

$$\underbrace{ \frac{\partial f}{\partial t} + \vec \xi \boldsymbol{\cdot} \vec \nabla f + \frac{\vec F}{m_P} \boldsymbol{\cdot} \vec \nabla_{\vec \xi} f }_\text{Propagation} = \underbrace{ \Omega(f) }_\text{Collision}$$

Under certain assumptions one can derive such a collision operator and simplified versions of it, such as the BGK, and interestingly by considering symmetries we can find a peculiar distribution that describes $f$ in the equilibrium state $f$ as sketched out over here.

Again you can see there is an explicit body force term, as we will see this is again the source of the body force on a macroscopic level and therefore of our buoyancy.

Now we could ask ourselves what happens on the level of density, momentum and energy on a much larger scale. This can be done by calculating the aforementioned moments of the Boltzmann equation. Now certainly we do not know what the distribution $f$ looks like. The easiest way of doing so is by assuming an equilibrium distribution. In equilibrium regardless of the precise form of the collision integral it has to be orthogonal to any collision invariant, meaning it has to preserve mass, momentum and energy. When doing so, you end up with the Euler equations for inviscid flow.

When assuming a perturbation from the equilibrium, with a perturbation series of the form

$$f = \sum_{n=0}^{\infty} \epsilon^n f^{(n)}$$

one has to evaluate the integrals of the collision operator and the whole things becomes quite complicated, a procedure known as the Chapman-Enskog expansion, and one ends up with the full Navier Stokes equations in the limit of small Knudsen numbers.

Anyways the interesting term here for us is the force term in the momentum equation. Let us take the first moment (which results in the momentum equation) of the force term

$$ m_P \int \xi_i \, \underbrace{\frac{\vec F}{m_P}}_{\vec g} \boldsymbol{\cdot} \vec \nabla_{\vec \xi} f \, d \vec \xi $$

alone. With integration by parts and using $\int \frac{\partial f}{\partial \xi_i} d \vec \xi= 0$ we get

$$ m_P \int \xi_j \frac{\partial f}{\partial \xi_i} d \vec \xi= - m_P \int \frac{\partial \xi_j}{\partial \xi_i} f d \vec \xi = - \rho \delta_{i j} $$

and as a result first moment of the forcing term on the left-hand side of the Boltzmann equation $m_P \int \xi_i \, \vec g \boldsymbol{\cdot} \vec \nabla_{\vec \xi} f \, d \vec \xi$ becomes the gravity term $\rho g_i$ on the right-hand-side of the macroscopic momentum equation.

The force term in the Navier Stokes equation completely emerges from the force term in the Boltzmann equation, it does not contribute to pressure there either. Basically when moving faster on a macroscopic level (as the faster molecules of the hotter gas do) the effects of gravity are less apparent and somehow the resulting term is proportional to the density on a macroscopic level.

So now we have all the tools to explain the effects on a microscopic level, which is quite similar to the macroscopic case.

2.2 With the balloon

Our molecules to both sides of the membrane, the ones belonging to the hot and the cold air, will collide with it. They will displace the membrane until there is an equilibrium in pressure to both sides. The density on the hot side will be smaller than on the cold side as the equilibrium means that on that side atoms are more apart and less densely packed. This can also be seen from the mean free path which can be calculated in kinetic theory (again with the same assumptions as above) to

$$\lambda = \frac{m_P}{\sqrt{2} \pi d^2 \rho}.$$

2.3 Without the balloon

Without the balloon we will start with two distributions that are different between hot and cold air. They will not be an equilibrium distribution in each case but something in between the two distributions. Every collision between molecules of the two distributions will even out this difference and drive the gas locally to equilibrium. This equilibrated molecules will collide with their neighbours resulting in an diffusive layer that spreads out and blurs the interface between the two. The diffusivity for the momentum is the viscosity which we can estimate to

$$ \mu = \frac{ 5\sqrt{\pi}}{16} \frac{\sqrt{k_B m_P T}}{\pi d^2} $$

As pointed out before the gravity will have less of an effect of the molecules travelling on higher speeds, while so to speak a more prominent effect on slower moving molecules, which happens to be proportional to the local density. These two effects are acting on the same time resulting in a blurring of the interface between the hot and cold air due to diffusion and to a rising motion of the hot air due to the density differences.

  • $\begingroup$ Thanks a lot! Very well written. $\endgroup$
    – Roger V.
    Dec 7, 2021 at 8:31
  • $\begingroup$ @RogerVadim You are welcome and sorry for taking that long. Lately time has become a bit of a comodity. I had written a first draft the same day you asked but did not find the time to finish it until yesterday. $\endgroup$
    – 2b-t
    Dec 7, 2021 at 18:34
  • $\begingroup$ Well written and great stuff. definitely a upvote; Now, can you tell us why Cumulus clouds has a straight lower edge? As it must rise from some clear physical cause. My answer here is related; physics.stackexchange.com/questions/679359/… $\endgroup$
    – Jokela
    Dec 8, 2021 at 20:11
  • $\begingroup$ @Jokela I will try to have a look at it on the weekend! $\endgroup$
    – 2b-t
    Dec 10, 2021 at 0:45

You are confusing several concepts and mechanisms here

  1. Diffusion: this is a random microscopic process associated with large scale physical motion of molecules. Since in air a molecule collides about every $10^{-5} cm$ with another molecule, it takes a long time (minutes or even hours) to mover over a distance of 1 m or so, despite moving at several hundred meters per second (since the motion is random, the number of scattering events needed increases with the square of the distance). In any case, an atom would be likely to have lost its original energy after that amount of time (distributed it to the colder air)

  2. Heat conduction: this happens faster than diffusion as molecules don't have to move physically over longer distance but just transfer their kinetic energy to neighboring molecules. In this case, the time required is only linear with distance, but still it takes a relatively long time as the process also depends on random collisions.

  1. Convection: this is essentially buoyancy without the balloon skin and is effectively what a 'convection heater' in your home uses. The absence of the balloon skin makes actually not that much of a difference as the action of the pressure force on the heated air volume happens with the speed of sound, so a heated air volume of a size of 1 m or so will start rising within milliseconds, way shorter than the time needed for the volume to disintegrate due to diffusion or heat conduction.

For the last 2 points see the Hyperphysics website for some more info.

The crucial point here is that, even if there is no solid container around the air volume, the pressure is still transmitted to it via inter-molecular collisions. As mentioned above already, the average mean free path for an air molecule is about $10^{-5} cm$, so on average a molecule will bounce back from other molecules after about this distance. In an isothermal atmosphere this will not result in a net pressure force though as the thermal pressure force from below exactly neutralizes the sum of the thermal and gravitational pressure (i.e. weight) forces from above. However, if the pocket of air has, let's say, half the density but twice the temperature compared to the surrounding medium, then, even though the dynamic pressure is unchanged (as per the ideal gas law $p=nkT$) the gravitational pressure force from above is now less due to the lower density/mass. This means the heated air volume will be pushed up (buoyancy). Of course, without a containing shell, the volume will not be stable in the long term as some molecules will diffuse from the hot diluted volume into the cold denser volume and vice versa, thus not only causing a progressive fizzling out of the volume boundaries but also leading to an equalization of the densities and temperatures and the eventual disappearance of the buoyancy effect (the latter will also happen with a hot air balloon though if you leave it to itself and don't re-heat (and thus re-dilute) the air inside its shell).

  • $\begingroup$ You probably meant heat conduction. Your definitions are correct, but it still does not answer the question: how the pressure arises in absence of a surface. I am thinking along the lines of how preassure is calculated for a gas in statistical physics, where one safely neglects collisions. $\endgroup$
    – Roger V.
    Nov 14, 2021 at 19:50
  • $\begingroup$ @RogerVadim Yes, sorry, that was a typo. I corrected it already. I don't understand your issue with the absence of the surface of the balloon. Whether the pressure is exerted directly to the air volume or via the skin of the balloon does not make any difference as long as the air volume keeps its shape (which it will do for a while at least as explained in my answer). Air pockets rise and fall all the time in weather systems without a material surface containing them. $\endgroup$
    – Thomas
    Nov 14, 2021 at 20:04
  • $\begingroup$ When we have a surface, the pressure arises from the molecules being scattered backwards from this surface, trasmitting momentum to it. In absence of surface they can only scatter from other molecules - it is not clear how to differentiate this from diffusion or heat conduction (what interests me here is the distinction between microscopic and macroscopic phenomena). Obviously, I do not deny that hot air rises and that atmospheric convection exists - I am asking about how it happens on a microscopic level. $\endgroup$
    – Roger V.
    Nov 15, 2021 at 8:20
  • 1
    $\begingroup$ @RogerVadim Please see my edited answer for further explanations $\endgroup$
    – Thomas
    Nov 16, 2021 at 22:45

The self-diffusion coefficient $D_A$ at normal conditions is $0.192\,\text{cm}^2/\text s$ for oxygen and $0.155\,\text{cm}^2/\text s$ for nitrogen (Source). So the diffusion length for $t=100s$ is of the order of $\sqrt{D_a t}\sim10\,\text{cm}$. For a meter-sized volume of warm air this is not much, and the volume would rise pretty high in $100\,\text s$.

  • $\begingroup$ Thanks for the estimate. But what the preassure is applied to? $\endgroup$
    – Roger V.
    Nov 14, 2021 at 17:36
  • $\begingroup$ @RogerVadim : It is applied to the surface of the volume of the warm air. Imagine a volume of air in vacuum. It will disperse very fast. Why do you think collisions in gas are rare? They are rare compared to those in liquid, so the diffusion coefficient is higher than that in liquids, but not high enough for the volume of warm air to lose its "individuality" fast enough (compared to the volume rise due to buoyancy). $\endgroup$
    – akhmeteli
    Nov 14, 2021 at 17:56
  • $\begingroup$ but there's no surface, unless the air is in a baloon. $\endgroup$
    – Roger V.
    Nov 14, 2021 at 18:34
  • $\begingroup$ @RogerVadim : There is a surface, although it is somewhat blurred. However, I agree that a more physical answer is in order. The pressure is applied to the molecules at this surface through collisions. $\endgroup$
    – akhmeteli
    Nov 14, 2021 at 18:45
  • $\begingroup$ I agree to @akhmeteli . There is a layer of air outside the volume of hot air which is only temperory and shades off quickly and gets replaced by another one. And that's why the volume of hot air reduces as it goes above, due to diffusion. $\endgroup$
    – user316791
    Dec 7, 2021 at 8:50

If this is a bulk of hot air collected together then, the particles at the boundaries of volume act like a layer which shades off from time to time but until it shades off. It can be compared to a ballon with air of less density filled into it.

The particles at top layer of volume of hot air get lesser downward force as compared to the bottom layer of volume. Therefore there is a net upward force in that volume. That net force can be distributed unevenly in that volume but their is an overall upward movement of that hot air volume.

If the number of molecules of air is one or hundred or around this, it would easily transfer excess energy to surrounding and therefore will not seemingly move upward.


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