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General idea: The air in the car accelerates with it. Air is compressible, so, during acceleration, it's denser on the rear than on the front. Similarly, in a rotating cylinder, there is a radial density gradient (see this answer). Assume constant acceleration: this is indistinguishable from a "horizontal" gravitational field (see the Equivalence Principle) that creates a stratification in density, analogous to what happens in our atmosphere (if you go high air is less dense, as given by the hydrostatic equation).  The mosquito (or, for simplicity a "dust particle"), will move because the air must find a new hydrostatic equilibrium that is not just uniform density. Since air must change its density configuration, this gives rise to a transient current because of the continuity equation. The dust particle will be advected by such a current. This effect is too small to be observed in a car but gets proportionally bigger by increasing the acceleration and/or the length of the vehicle.

Rough estimate of the effect: Assume, for mere convenience, that your vehicle acceleration is $g\approx 9.8 m/s$. We know that (approximatively) air density changes by 10% in 1 km of elevation, see the Pressure-Elevation graph. Assuming ideal gas law for air (and constant temperature everywhere), the same is true also for the density (i.e. density is proportional to the pressure).

Therefore: if your vehicle is undergoing constant $1g$ acceleration and is long $1km$, the air in the front is 10% less dense than air in the rear. This assumes hydrostatic equilibrium, which is reached after an initial transient is damped (air is a bit viscous). The fine details of this transient depend on the exact dynamics: you have to solve the hydrodynamic equations for a gas in an accelerating box, for a given prescribed acceleration and shape of the box. Therefore, the transient phase depends on the jerk, namely on how you go from zero acceleration to constant finite acceleration and finally reach a new hydrostatic equilibrium.

Of course, your car is not 1km long and can not accelerate at $g$. However, you can rescale the result. For example, let's say it's a bus and it's 10m (again, let's keel the unrealistic acceleration of $g$). Hence, from the front to the rear density changes by $10\% \times 10m/1km = 10\%/100=0.1\%$: nearly 1 out of 1000 "air particles" that were on the front now it's on the rear (very roughly). There is a net air displacement from the initial hydrostatic equilibrium (uniform air density) to the "accelerating" hydrostatic equilibrium. Assuming that your mosquito is not a mosquito but a "dust particle", we can conclude that it has been displaced accordingly. This displacement is not noticeable because the dust particle will undergo random Brownian motion. However, the nature of such tiny displacement is not stochastic (like Brownian motion) but systematic, due to the fact that acceleration modifies the hydrostatic solution of the air in the car, which is not just uniform density. Now, by changing the acceleration to a more reasonable one, the result scales proportionally and becomes even smaller.

General idea: The air in the car accelerates with it. Air is compressible, so, during acceleration, it's denser on the rear than on the front (similarly, in a rotating cylinder, there is a radial density gradient, see this answer). Assume constant acceleration: this is indistinguishable from a "horizontal" gravitational field (see the Equivalence Principle) that creates a stratification in density, analogous to what happens in our atmosphere, as given by hydrostatics.  The mosquito (or, for simplicity a "dust particle"), will move because the air must find a new hydrostatic equilibrium that is not just uniform density. Since air must change its density configuration, this gives rise to a transient current because of the continuity equation. The dust particle will be advected by such a current. This effect is too small to be observed in a car but gets proportionally bigger by increasing the acceleration and/or the length of the vehicle. Details on the "accelerating" equilibrium (i.e. equilibrium in a gravitational field) are discussed here.

Rough estimate of the effect: Assume, for mere convenience, that your vehicle acceleration is $g\approx 9.8 m/s$. We know that (approximatively) air density changes by 10% in 1 km of elevation, see the Pressure-Elevation graph. Assuming ideal gas law for air (and constant temperature everywhere), the same is true also for the density (i.e. density is proportional to the pressure).

Therefore: if your vehicle is undergoing constant $1g$ acceleration and is long $1km$, the air in the front is 10% less dense than air in the rear. This assumes hydrostatic equilibrium, which is reached after an initial transient is damped (air is a bit viscous). The fine details of this transient depend on the exact dynamics: you have to solve the hydrodynamic equations for a gas in an accelerating box, for a given prescribed acceleration and shape of the box. Therefore, the transient phase depends on the jerk, namely on how you go from zero acceleration to constant finite acceleration and finally reach a new hydrostatic equilibrium.

Of course, your car is not 1km long and can not accelerate at $g$. However, you can rescale the result. For example, let's say it's a bus and it's 10m (again, let's keel the unrealistic acceleration of $g$). Hence, from the front to the rear density changes by $10\% \times 10m/1km = 10\%/100=0.1\%$: nearly 1 out of 1000 "air particles" that were on the front now it's on the rear (very roughly). There is a net air displacement from the initial equilibrium (uniform air density) to the "accelerating" equilibrium. Assuming that your mosquito is not a mosquito but a "dust particle", we can conclude that it has been displaced accordingly. This displacement is not noticeable because the dust particle will undergo random Brownian motion. However, the nature of such tiny displacement is not stochastic (like Brownian motion) but systematic, due to the fact that acceleration modifies the hydrostatic solution of the air in the car, which is not just uniform density. Now, by changing the acceleration to a more reasonable one, the result scales proportionally and becomes even smaller.

General idea: The air in the car accelerates with it. Air is compressible, so, during acceleration, it's denser on the rear than on the front (but just a very tiny bit, it's impossible to notice). Similarly, in a rotating cylinder, there is a radial density gradient (see this answer). Assume constant acceleration: this is indistinguishable from a "horizontal" gravitational field (see the Equivalence Principle) that creates a stratification in density, analogous to what happens in our atmosphere (if you go high air is less dense, as given by the hydrostatic equation).

Rough estimate of the effect: Assume, for mere convenience, that your vehicle acceleration is $g\approx 9.8 m/s$. We know that (approximatively) air density changes by 10% in 1 km of elevation, see the Pressure-Elevation graph. Assuming ideal gas law for air (and constant temperature everywhere), the same is true also for the density (i.e. density is proportional to the pressure).

Therefore: if your vehicle is undergoing constant $1g$ acceleration and is long $1km$, the air in the front is 10% less dense than air in the rear. This assumes hydrostatic equilibrium, which is reached after an initial transient is damped (air is a bit viscous). The details of this transient depend on the exact dynamics: you have to solve the hydrodynamic equations for a gas in an accelerating box, for a given prescribed acceleration and shape of the box. Therefore, the transient phase depends on the jerk, namely on how you go from zero acceleration to constant finite acceleration and finally reach a new hydrostatic equilibrium.

Of course, your car is not 1km long and can not accelerate at $g$. However, you can rescale the result. For example, let's say it's a bus and it's 10m (again, let's keel the unrealistic acceleration of $g$). Hence, from the front to the rear density changes by $10\% \times 10m/1km = 10\%/100=0.1\%$: nearly 1 out of 1000 "air particles" that were on the front now it's on the rear (very roughly). There is a net air displacement from the initial hydrostatic equilibrium (uniform air density) to the "accelerating" hydrostatic equilibrium. Assuming that your mosquito is not a mosquito but a "dust particle", we can conclude that it has been displaced accordingly. This displacement is not noticeable because the dust particle will undergo random Brownian motion. However, the nature of such tiny displacement is not stochastic (like Brownian motion) but systematic, due to the fact that acceleration modifies the hydrostatic solution of the air in the car, which is not just uniform density. Now, by changing the acceleration to a more reasonable one, the result scales proportionally and becomes even smaller.

General idea: The air in the car accelerates with it. Air is compressible, so, during acceleration, it's denser on the rear than on the front. Similarly, in a rotating cylinder, there is a radial density gradient (see this answer). Assume constant acceleration: this is indistinguishable from a "horizontal" gravitational field (see the Equivalence Principle) that creates a stratification in density, analogous to what happens in our atmosphere (if you go high air is less dense, as given by the hydrostatic equation). The mosquito (or, for simplicity a "dust particle"), will move because the air must find a new hydrostatic equilibrium that is not just uniform density. Since air must change its density configuration, this gives rise to a transient current because of the continuity equation. The dust particle will be advected by such a current. This effect is too small to be observed in a car but gets proportionally bigger by increasing the acceleration and/or the length of the vehicle.

Rough estimate of the effect: Assume, for mere convenience, that your vehicle acceleration is $g\approx 9.8 m/s$. We know that (approximatively) air density changes by 10% in 1 km of elevation, see the Pressure-Elevation graph. Assuming ideal gas law for air (and constant temperature everywhere), the same is true also for the density (i.e. density is proportional to the pressure).

Therefore: if your vehicle is undergoing constant $1g$ acceleration and is long $1km$, the air in the front is 10% less dense than air in the rear. This assumes hydrostatic equilibrium, which is reached after an initial transient is damped (air is a bit viscous). The fine details of this transient depend on the exact dynamics: you have to solve the hydrodynamic equations for a gas in an accelerating box, for a given prescribed acceleration and shape of the box. Therefore, the transient phase depends on the jerk, namely on how you go from zero acceleration to constant finite acceleration and finally reach a new hydrostatic equilibrium.

Of course, your car is not 1km long and can not accelerate at $g$. However, you can rescale the result. For example, let's say it's a bus and it's 10m (again, let's keel the unrealistic acceleration of $g$). Hence, from the front to the rear density changes by $10\% \times 10m/1km = 10\%/100=0.1\%$: nearly 1 out of 1000 "air particles" that were on the front now it's on the rear (very roughly). There is a net air displacement from the initial hydrostatic equilibrium (uniform air density) to the "accelerating" hydrostatic equilibrium. Assuming that your mosquito is not a mosquito but a "dust particle", we can conclude that it has been displaced accordingly. This displacement is not noticeable because the dust particle will undergo random Brownian motion. However, the nature of such tiny displacement is not stochastic (like Brownian motion) but systematic, due to the fact that acceleration modifies the hydrostatic solution of the air in the car, which is not just uniform density. Now, by changing the acceleration to a more reasonable one, the result scales proportionally and becomes even smaller.

The air in the car accelerates with it. Air is compressible, so, during acceleration, it's denser on the rear than on the front (but just a very tiny bit, it's impossible to notice). Assume constant acceleration: this is indistinguishable from a "horizontal gravitational field" (Einstein said so, see the Equivalence Principle) that creates a stratification in density, analogous to what happens in our atmosphere (if you go high air is less dense, as given by the hydrostatic equation).

Let's play this game: assume that you have a really powerful car and that your acceleration is $g\approx 9.8 m/s$. We know that (approximatively) air density changes by 10% in 1 km of elevation, see the Pressure-Elevation graph. Assuming ideal gas law for air (and constant temperature everywhere), the same is true also for the density (i.e. density is proportional to the pressure).

Therefore: if your car is undergoing constant $1g$ acceleration and is long $1km$, the air in the front is 10% less dense than air in the rear. This assumes hydrostatic equilibrium, which is reached after an initial transient is damped (air is a bit viscous). The details of this transient phase depend on the exact dynamics: you have to solve the hydrodynamic equations for a gas in an accelerating box, for a given prescribed acceleration and shape of the box. Therefore, the transient phase depends on the jerk, namely on how you go from zero acceleration to constant finite acceleration and finally reach a new hydrostatic equilibrium.

Of course, your car is not 1km long, so let's say it's a bus and it's 10m. Hence, from the front to the rear density changes by $10\% \times 10m/1km = 10\%/100=0.1\%$: nearly 1 out of 1000 "air particles" that were on the front now it's on the rear (very roughly). There is a net air displacement from the initial hydrostatic equilibrium (uniform air density) to the "accelerating" hydrostatic equilibrium. Assuming that your mosquito is not a mosquito but a "dust particle", we can conclude that it has been displaced accordingly. This displacement is not noticeable because the dust particle will undergo random Brownian motion. However, the nature of such tiny displacement is not stochastic (like Brownian motion) but systematic, due to the fact that acceleration modifies the hydrostatic solution of the air in the car, which is not just uniform density.

General idea: The air in the car accelerates with it. Air is compressible, so, during acceleration, it's denser on the rear than on the front (but just a very tiny bit, it's impossible to notice). Similarly, in a rotating cylinder, there is a radial density gradient (see this answer). Assume constant acceleration: this is indistinguishable from a "horizontal" gravitational field (see the Equivalence Principle) that creates a stratification in density, analogous to what happens in our atmosphere (if you go high air is less dense, as given by the hydrostatic equation).

Rough estimate of the effect: Assume, for mere convenience, that your vehicle acceleration is $g\approx 9.8 m/s$. We know that (approximatively) air density changes by 10% in 1 km of elevation, see the Pressure-Elevation graph. Assuming ideal gas law for air (and constant temperature everywhere), the same is true also for the density (i.e. density is proportional to the pressure).

Therefore: if your vehicle is undergoing constant $1g$ acceleration and is long $1km$, the air in the front is 10% less dense than air in the rear. This assumes hydrostatic equilibrium, which is reached after an initial transient is damped (air is a bit viscous). The details of this transient depend on the exact dynamics: you have to solve the hydrodynamic equations for a gas in an accelerating box, for a given prescribed acceleration and shape of the box. Therefore, the transient phase depends on the jerk, namely on how you go from zero acceleration to constant finite acceleration and finally reach a new hydrostatic equilibrium.

Of course, your car is not 1km long and can not accelerate at $g$. However, you can rescale the result. For example, let's say it's a bus and it's 10m (again, let's keel the unrealistic acceleration of $g$). Hence, from the front to the rear density changes by $10\% \times 10m/1km = 10\%/100=0.1\%$: nearly 1 out of 1000 "air particles" that were on the front now it's on the rear (very roughly). There is a net air displacement from the initial hydrostatic equilibrium (uniform air density) to the "accelerating" hydrostatic equilibrium. Assuming that your mosquito is not a mosquito but a "dust particle", we can conclude that it has been displaced accordingly. This displacement is not noticeable because the dust particle will undergo random Brownian motion. However, the nature of such tiny displacement is not stochastic (like Brownian motion) but systematic, due to the fact that acceleration modifies the hydrostatic solution of the air in the car, which is not just uniform density. Now, by changing the acceleration to a more reasonable one, the result scales proportionally and becomes even smaller.