<p>The two timescales are different because they are relative to two different physical situations.</p>
<p>When you talk about the free-fall timescale, you have in mind a <em>pressure-less</em> cloud of dust/gas distributed in a more or less spherical shape with average density <span class="math-container">$\rho$</span> that is collapsing under its own gravitational pull. From dimensional analysis it can be concluded that the time it takes for the distribution to collapse completely is <span class="math-container">$t_{ff} \approx {1 \over \sqrt{G\rho}}$</span>. Actually it is more like <span class="math-container">$t_{ff} \approx {1 \over 2}{1 \over \sqrt{G\rho}} \approx 35 \min \left(\rho \over g\ cm^{-3} \right)^{-{1 \over 2}}$</span></p>
<p>Kelvin-Helmholtz timescale comes up in a completely different situation. Imagine again a more or less spherical cloud. But this time it is supported by pressure, it is not collapsing in free fall. And the cloud is also hot, it has a certain temperature and radiates away energy with a luminosity <span class="math-container">$L$</span>. If <em>the cloud has no additional sources of energy</em> (e.g. nuclear burning) then the radiated energy must be taken from the internal energy reservoir <span class="math-container">$E_i$</span>. At this rate, it will take a time <span class="math-container">$t_{KH} \approx {E_i \over L}$</span> to drain the gas from all its energy. While loosing energy, the cloud shrinks, therefore after a time <span class="math-container">$t_{KH}$</span> it will be completely collapsed.</p>
<p>When talking about a real star, these two scenarios are just an idealization, because a real star will usually (i) be supported by pressure, (ii) produce energy through nuclear reactions. Yet, there are situations in which these two conditions do not apply and the two timescales may become important.</p>
<p>Example 1: during the formation of a star, a molecular cloud slowly collapses into itself. There is no nuclear burning yet, but the proto-star is supported by pressure, therefore the collapse will happen on a Kelvin-Helmholtz timescale.</p>
<p>Example 2: a massive star may end its life as a core collapse supernova. At the onset of collapse the star feels the pressure faltering and starts to collapse nearly in free-fall.</p>
<p>When you read that a star has a <span class="math-container">$t_{kh} &lt; t_{ff}$</span> it just means that the star is extremely luminous, so that, if the nuclear reactions happened to instantly halt and stop producing energy (magically or by other means), the star would need to collapse in a very short time, to keep up with the radiated luminosity. The collapse would be so quick that, if the material of the star were in free fall, the star would be slower to collapse.</p>

CC BY-SA 4.0
2022-02-09T08:48:04.207Z
@ user:42349 So what does tkh<tff means in the context of star formation? What does it mean when we say a high-mass protostar has tkh<tff ? Does it mean that it's accreting mass such that it's tff is bigger ? Can you please clarify ?
Rian
CC BY-SA 4.0
2022-02-10T07:33:48.650Z
@Rian I was talking about adult stars, not protostars. I think it would be unusual for a protostar to have tkh<tff, because since the protostar is not yet doing nuclear fusion, it evolves on a Kelvin-Helmholtz timescale. So tkh<tff would mean that the star is actually contracting faster than in free fall
Prallax
CC BY-SA 4.0
2022-02-10T10:08:14.087Z
@ Prallax But for high mass protostars tkh<tff
Rian
CC BY-SA 4.0
2022-02-08T23:33:32.173Z
2022-02-10T07:40:40.650Z
2022-02-10T07:40:40.650Z
Prallax
<p>The freefall timescale, <span class="math-container">$\tau_{\rm ff}$</span>, is exactly that. It assumes any collapse is unopposed by any pressure.</p>
<p>In general any contraction will be opposed by internal pressure and energy. However, that internal energy can be radiated away, allowing contraction to proceed. The Kelvin-Helmholz timescale, <span class="math-container">$\tau_{\rm KH}$</span>, is the time required for an object to radiate away its heat energy via its intrinsic luminosity.</p>
<p>If we think about a contracting/collapsing object then that evolution will take place at a rate given by the <em>larger</em> of these two timescales<span class="math-container">$^1$</span>.</p>
<p>For example, in a forming protostar, there are no nuclear reactions, the star contracts and heats up. <span class="math-container">$\tau_{\rm KH}$</span> might be something like 10 Myr for a 1 solar mass pre-main sequence star, while <span class="math-container">$\tau_{\rm ff}$</span>, which is approximately <span class="math-container">$(G \rho)^{-1/2}$</span> where <span class="math-container">$\rho$</span> is the density, might be something like 1 day. In such circumstances the contraction occurs on the K-H timescale.</p>
<p>But in other situations the order of these timescales can be reversed. That is a consequence of the very steep dependence of luminosity on mass and temperature. Thus in a high mass star the denominator (the luminosity) of <span class="math-container">$\tau_{\rm KH}$</span> increases much more than the numerator (the internal energy) and hence the K-H timescale becomes much shorter. Towards the end of its life, <span class="math-container">$\tau_{\rm KH}$</span> can become shorter than <span class="math-container">$\tau_{\rm ff}$</span> <em>in the core of the star</em>. (Not for the entire star).</p>
<p>What this means, is that once nuclear reactions cease, the contraction proceeds on a free fall timescale - leading to &quot;core collapse&quot;.</p>
<p><span class="math-container">$^1$</span> There is a third timescale to be considered - the time taken for the nuclear fuel of the star to be consumed in providing its current luminosity. This is known as the &quot;nuclear timescale&quot;, <span class="math-container">$\tau_{\rm nuc}$</span>.</p>
<p>So now you have three timescales to consider, but the rule is the same - evolution takes place at a rate governed by the longer timescale. In a main sequence star, <span class="math-container">$\tau_{\rm nuc}$</span> is much longer than either of the other two timescales and so the star evolves slowly (e.g. <span class="math-container">$\tau_{\rm nuc}$</span> is about 10 billion years for a newly formed 1 solar mass star). As the source of nuclear fuel runs out, or if the luminosity increases, then <span class="math-container">$\tau_{\rm nuc}$</span> gets much shorter and eventually one of the other timescales can take over and govern to evolution.</p>

CC BY-SA 4.0
2022-02-10T10:09:11.987Z
@ ProfRob So what does tkh<tff means in the context of star formation? What does it mean when we say a high-mass protostar has tkh<tff ? Does it mean that it's accreting mass such that its density decreases and tff gets bigger ? Can you please clarify ?
Rian
CC BY-SA 4.0
2022-02-10T13:30:56.823Z
@Rian I would not have thought that $t_{\rm KH}$ is less than $t_{\rm ff}$ during star formation. Could you provide a reference for where you have seen this and I can investigate. The implication would simply be that the contraction is proceeding on the free fall timescale in those circumstances.
ProfRob
CC BY-SA 4.0
2022-02-11T12:15:20.867Z
Here is the link https://www.eas-journal.org/articles/eas/abs/2011/06/eas1151012/eas1151012.html
Rian
CC BY-SA 4.0
2022-02-11T12:32:38.490Z
@Rian, it's paywalled so I can't see it. You will need to quite the relevant passage in your question.
ProfRob
CC BY-SA 4.0
2022-02-11T12:36:19.167Z
"In the case of massive stars, the free-fall time is longer than the Kelvin–Helmholtz timescale, so that the massive stars in formation reach thermal equilibrium before the accretion is completed. This is why the history of the accretion rates for massive stars is so critical. "
Rian
CC BY-SA 4.0
2022-02-10T08:39:38.400Z
2022-02-10T08:48:03.100Z
2022-02-10T08:48:03.100Z
ProfRob
<p>What is the actual difference between these two timescales? Both of these timescales define the time for collapse when there is only gravity. Also, what does it mean by saying Kelvin-Helmholtz timescale is smaller than free-fall timescale for high-mass protstars? When are these timescales important in star formation/evolution ?</p>

<star><astrophysics><stellar-evolution><star-formation><stellar-astrophysics>
What is the relation between Kelvin-Helmholtz timescale and free-fall timescale?
Rian
# What is the relation between Kelvin-Helmholtz timescale and free-fall timescale?

(Asked by: username_0 on 2022-02-08)

What is the actual difference between these two timescales? Both of these timescales define the time for collapse when there is only gravity. Also, what does it mean by saying Kelvin-Helmholtz timescale is smaller than free-fall timescale for high-mass protstars? When are these timescales important in star formation/evolution ?

**Tags:** star, astrophysics, stellar-evolution, star-formation, stellar-astrophysics

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## 2 answers

### Answer by username_1 on 2022-02-08. Score: 5

The two timescales are different because they are relative to two different physical situations.

When you talk about the free-fall timescale, you have in mind a *pressure-less* cloud of dust/gas distributed in a more or less spherical shape with average density $\rho$ that is collapsing under its own gravitational pull. From dimensional analysis it can be concluded that the time it takes for the distribution to collapse completely is $t\_{ff} \approx {1 \over \sqrt{G\rho}}$. Actually it is more like $t\_{ff} \approx {1 \over 2}{1 \over \sqrt{G\rho}} \approx 35 \min \left(\rho \over g\ cm^{-3} \right)^{-{1 \over 2}}$

Kelvin-Helmholtz timescale comes up in a completely different situation. Imagine again a more or less spherical cloud. But this time it is supported by pressure, it is not collapsing in free fall. And the cloud is also hot, it has a certain temperature and radiates away energy with a luminosity $L$. If *the cloud has no additional sources of energy* (e.g. nuclear burning) then the radiated energy must be taken from the internal energy reservoir $E\_i$. At this rate, it will take a time $t\_{KH} \approx {E\_i \over L}$ to drain the gas from all its energy. While loosing energy, the cloud shrinks, therefore after a time $t\_{KH}$ it will be completely collapsed.

When talking about a real star, these two scenarios are just an idealization, because a real star will usually (i) be supported by pressure, (ii) produce energy through nuclear reactions. Yet, there are situations in which these two conditions do not apply and the two timescales may become important.

Example 1: during the formation of a star, a molecular cloud slowly collapses into itself. There is no nuclear burning yet, but the proto-star is supported by pressure, therefore the collapse will happen on a Kelvin-Helmholtz timescale.

Example 2: a massive star may end its life as a core collapse supernova. At the onset of collapse the star feels the pressure faltering and starts to collapse nearly in free-fall.

When you read that a star has a $t\_{kh} < t\_{ff}$ it just means that the star is extremely luminous, so that, if the nuclear reactions happened to instantly halt and stop producing energy (magically or by other means), the star would need to collapse in a very short time, to keep up with the radiated luminosity. The collapse would be so quick that, if the material of the star were in free fall, the star would be slower to collapse.

> **Comments:**
> **username_0**: @ user:42349 So what does tkh<tff means in the context of star formation? What does it mean when we say a high-mass protostar has tkh<tff ? Does it mean that it's accreting mass such that it's tff is bigger ? Can you please clarify ?  
> **username_1**: @Rian I was talking about adult stars, not protostars. I think it would be unusual for a protostar to have tkh<tff, because since the protostar is not yet doing nuclear fusion, it evolves on a Kelvin-Helmholtz timescale. So tkh<tff would mean that the star is actually contracting faster than in free fall  
> **username_0**: @ Prallax But for high mass protostars tkh<tff  

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### Answer by username_2 on 2022-02-10. Score: 2

The freefall timescale, $\tau\_{\rm ff}$, is exactly that. It assumes any collapse is unopposed by any pressure.

In general any contraction will be opposed by internal pressure and energy. However, that internal energy can be radiated away, allowing contraction to proceed. The Kelvin-Helmholz timescale, $\tau\_{\rm KH}$, is the time required for an object to radiate away its heat energy via its intrinsic luminosity.

If we think about a contracting/collapsing object then that evolution will take place at a rate given by the *larger* of these two timescales$^1$.

For example, in a forming protostar, there are no nuclear reactions, the star contracts and heats up. $\tau\_{\rm KH}$ might be something like 10 Myr for a 1 solar mass pre-main sequence star, while $\tau\_{\rm ff}$, which is approximately $(G \rho)^{-1/2}$ where $\rho$ is the density, might be something like 1 day. In such circumstances the contraction occurs on the K-H timescale.

But in other situations the order of these timescales can be reversed. That is a consequence of the very steep dependence of luminosity on mass and temperature. Thus in a high mass star the denominator (the luminosity) of $\tau\_{\rm KH}$ increases much more than the numerator (the internal energy) and hence the K-H timescale becomes much shorter. Towards the end of its life, $\tau\_{\rm KH}$ can become shorter than $\tau\_{\rm ff}$ *in the core of the star*. (Not for the entire star).

What this means, is that once nuclear reactions cease, the contraction proceeds on a free fall timescale - leading to "core collapse".

$^1$ There is a third timescale to be considered - the time taken for the nuclear fuel of the star to be consumed in providing its current luminosity. This is known as the "nuclear timescale", $\tau\_{\rm nuc}$.

So now you have three timescales to consider, but the rule is the same - evolution takes place at a rate governed by the longer timescale. In a main sequence star, $\tau\_{\rm nuc}$ is much longer than either of the other two timescales and so the star evolves slowly (e.g. $\tau\_{\rm nuc}$ is about 10 billion years for a newly formed 1 solar mass star). As the source of nuclear fuel runs out, or if the luminosity increases, then $\tau\_{\rm nuc}$ gets much shorter and eventually one of the other timescales can take over and govern to evolution.

> **Comments:**
> **username_0**: @ ProfRob So what does tkh<tff means in the context of star formation? What does it mean when we say a high-mass protostar has tkh<tff ? Does it mean that it's accreting mass such that its density decreases and tff gets bigger ? Can you please clarify ?  
> **username_2**: @Rian I would not have thought that $t_{\rm KH}$ is less than $t_{\rm ff}$ during star formation. Could you provide a reference for where you have seen this and I can investigate. The implication would simply be that the contraction is proceeding on the free fall timescale in those circumstances.  
> **username_0**: Here is the link https://www.eas-journal.org/articles/eas/abs/2011/06/eas1151012/eas1151012.html  
> **username_2**: @Rian, it's paywalled so I can't see it. You will need to quite the relevant passage in your question.  
> **username_0**: "In the case of massive stars, the free-fall time is longer than the Kelvin–Helmholtz timescale, so that the massive stars in formation reach thermal equilibrium before the accretion is completed. This is why the history of the accretion rates for massive stars is so critical. "  

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