---
title: "From a wind pulse to a restoring force"
description: "Derive ram acceleration from a gas column, compare impulse with vertical gravity, and see when a static pressure threshold is useful."
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language: "en"
page_type: "story"
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  zh-CN: "https://sandboxphysics.com/zh/stories/ram-pressure-stripping/from-a-wind-pulse-to-restoring-force/"
last_modified: "2026-11-11"
---

[Ram pressure and returning gas](https://sandboxphysics.com/stories/ram-pressure-stripping/)

Ram pressure and returning gas · L2

# From a wind pulse to a restoring force

Sandbox Physics · 2026-11-11

A wind must act long enough to move gas. A slowly varying pressure can be compared with the strongest restoring force above a disk; a brief gust is better described by the momentum it transfers. Their time scales decide which comparison helps.

## Before you begin

Read why a tail can fall back. You will use force equals mass times acceleration, energy and a simple time integral; the physical meaning of the integral is explained below.

## Choose a gas sheet and its coordinates

Take one small patch of a gas disk. Its area is $A$ and its mass is $m$. Define its gas column as mass divided by area, so column has units of kilograms per square metre. A large column gives the wind more mass to accelerate behind the same exposed area.

Use the galaxy rest frame. The $z$ axis points along the disk normal and the incoming wind; $R$ is cylindrical distance from the centre in the disk plane. The incoming speed $V$ is positive, while the patch has its own normal velocity. With a fixed area, the force from relative normal motion follows the signed momentum flux.

Writing the relative-speed factor with an absolute value matters. When a patch moves faster than the incident flow, the direction of the drag reverses. Replacing it everywhere with the incident pressure is a useful scale estimate only when the patch speed is small compared with the wind.

The path offset is $q=b/r_c$: closest-approach distance $b$ divided by the atmosphere core scale $r_c$. The five computed values are 0, 0.37, 1.5, 2.72 and 4. The galaxy speed stays fixed; changing the offset selects a different complete pressure history.

$$
\Sigma_g=\frac mA,\qquad a_{\rm ram}=\frac{\rho\lvert V-v_z\rvert(V-v_z)}{\Sigma_g}
$$

Here $\rho$ is the prescribed incident density, $v_z$ is the gas normal velocity, and acceleration has units of length divided by time squared. The full calculation keeps the gas mass changing through star formation. This teaching sheet law sets the normal momentum-coupling coefficient to one.

Reading: Original fixed-slice and static-restoring-force materials

## Why vertical gravity has a maximum

The force toward a smooth midplane is zero exactly at the plane, rises above it and eventually weakens far from the galaxy. A gentle steady push can shift a gas sheet to a new height where upward pressure balances downward gravity. A stronger push can exceed the largest available restoring acceleration at that fixed radius.

The local static reference multiplies that maximum acceleration by the gas column. It is closely related to the Gunn–Gott pressure comparison. Its meaning is a force balance for a slowly changing configuration. It does not by itself describe how far a patch moves during a short pulse.

Our independent reference uses the declared three-component potential and searches positive heights at each initial radius. It shows one curve with the initial column and another after the specified consumption up to nominal closest approach. Those columns differ because gas mass is gradually converted into stars while the sheet area stays fixed.

$$
P_{\rm restore}(R)=\Sigma_g(R)\max_{z>0}\left|\frac{\partial\Phi(R,z)}{\partial z}\right|
$$

$\Phi$ is potential energy per unit mass, chosen to approach zero at infinity. The radius is held fixed in this reference; actual moving sheets can change radius.

![Static fixed-radius restoring pressure for initial and nominal closest-approach gas columns](https://sandboxphysics.com/_astro/force.C0w_4Omt.svg)

Independent numerical maxima of the original declared potential at $64$ initial radii. Blue uses initial columns; gold multiplies them by $\exp(-500/2000)$. The horizontal line is the central incident peak. These are local quasi-static references, not a finite-pulse escape prediction. [Open the full figure](https://sandboxphysics.com/_astro/force.C0w_4Omt.svg)

Reading: Köppen, Jáchym, Taylor & Palouš, Ram Pressure Stripping Made Easy, arXiv:1806.05887v1 (2018) · Original fixed-slice and static-restoring-force materials

## A short gust delivers an impulse

If the pulse is much shorter than the time the gas needs to move up and down in the potential, the gas barely changes position while the strongest push occurs. Add up pressure multiplied by each tiny time interval. The result is impulse per unit area: pressure integrated over time.

Divide that impulse by the column and you obtain a velocity change in the simplest short-pulse approximation. A pulse of twice the height and half the duration has the same impulse. That statement assumes the column stays nearly constant and ignores gravity and the changing relative speed during the short push.

For a numerical scale, take a pressure of one ten-trillionth pascal acting for 100 million years on a column of 0.002 kilograms per square metre. With the declared time-unit convention, the simple impulse estimate gives about 158 kilometres per second. Stretching the same pressure across a much longer interval eventually invalidates the short-pulse approximation because gravity and motion have time to respond. Use this estimate only when the vertical response time is much longer than the stated pulse duration.

$$
\Delta v_z\simeq\frac{1}{\Sigma_g}\int P_{\rm inc}(t)\,dt
$$

The area under the pressure–time graph is the impulse per unit area. This estimate describes the brief impulse; the complete trajectory follows the changing force throughout.

$$
\begin{aligned}\Delta t&=100\,\mathrm{Myr}\\&=3.156\times10^{15}\,\mathrm s\\\Delta v_z&\simeq\frac{10^{-13}\,\mathrm{Pa}\,(3.156\times10^{15}\,\mathrm s)}{2\times10^{-3}\,\mathrm{kg}\,\mathrm m^{-2}}\\&=1.578\times10^5\,\mathrm m\,\mathrm s^{-1}\\&\simeq158\,\mathrm{km}\,\mathrm s^{-1}\end{aligned}
$$

The declared numerical time unit is $1\,\mathrm{Myr}=3.156\times10^{13}\,\mathrm s$. Pressure is force per area, so pressure times time divided by mass per area has velocity units.

Reading: Köppen, Jáchym, Taylor & Palouš, Ram Pressure Stripping Made Easy, arXiv:1806.05887v1 (2018) · Original fixed-slice and static-restoring-force materials

## Let the vertical motion set the comparison

Near the midplane, a smooth potential often looks approximately like a spring: restoring acceleration grows roughly in proportion to height. The gas then has a natural vertical oscillation period. A pulse lasting much longer can be followed almost statically; a much shorter pulse resembles a kick.

Between these limits the gas moves while the pressure changes. Its velocity changes the drag, its column evolves, and gravitational work matters throughout the encounter. Integrating the actual equation of motion is how to follow that middle region.

The pressure curves in the reference figure therefore help form a prediction, then the saved trajectory tests its consequence. A local static threshold, a maximum tail height and the eventual bound mass describe related parts of the same event. They are useful together when their definitions remain explicit.

Reading: Köppen, Jáchym, Taylor & Palouš, Ram Pressure Stripping Made Easy, arXiv:1806.05887v1 (2018) · Original fixed-slice and static-restoring-force materials

## Exercises and explained answers

1. Two pulses have equal peak pressure, but one lasts ten times longer. Must they remove the same gas?

   Read the explanation

   No. A short pulse may deliver much less impulse. If both are slow compared with the vertical response, a static comparison can be useful; otherwise duration changes the motion.

2. If the gas column halves while area and gravity stay fixed, what happens to the static restoring pressure?

   Read the explanation

   It halves because pressure is column times maximum restoring acceleration. At the same incident pressure the gas also accelerates more readily.

## Try the prediction

### [Follow gas lifted by a cluster wind](https://sandboxphysics.com/labs/ram-pressure-stripping/)

1. Compare two supported paths at the same native time. Use the pressure history to explain why the gas has reached its present height.
2. Predict whether a high displaced fraction must equal the positive-energy fraction, then inspect both readouts.

[Previous: Why a galaxy tail can fall back](https://sandboxphysics.com/stories/ram-pressure-stripping/why-a-galaxy-tail-can-fall-back/) [Next: Follow fuel across the disk boundary](https://sandboxphysics.com/stories/ram-pressure-stripping/follow-fuel-across-the-disk-boundary/)

## Sources and calculation scope

1. [Köppen, Jáchym, Taylor & Palouš, Ram Pressure Stripping Made Easy, arXiv:1806.05887v1 (2018)](https://arxiv.org/abs/1806.05887v1)

   Sections2.1–2.2 and5.4 explain the slow-pressure/impulse limits and re-accretable gas. Our fixed three-component potential, equal-mass sheets and consuming gas are specified independently; the own static comparison does not reproduce every published simulation or observational sample.

2. [Original fixed-slice and static-restoring-force materials](https://sandboxphysics.com/stories/ram-pressure-stripping/materials/ram-pressure-science-v1.zip)

   Our five exact synthetic slices with sixty-four sheets and thirteen fields, initial common arrays, two global-light histories/three screens/four-age band subset, independent Python static maxima and mask/energy/colour diagnostics. Original trajectory and native SSP production are distinct earlier calculations; this offline package repeats diagnostics and band ratios. No browser Engine, original FITS/SED, paper, author code or observed-galaxy data is included.