<p>Interesting question! The existence of quark stars is hypothetical and they are often the <a href="https://physics.stackexchange.com/questions/524720/can-quark-stars-form-under-an-event-horizon">target</a> of woo-hoo.</p>
<p>Even if one computed the escape velocity at the surface of a quark star (which would be a complicated calculation that would take me a while to research how to do correctly, which is probably an open question anyway), there would still be differences from an (absolute) event horizon. That is, there are aspects of an event horizon that a quark star with superluminal escape velocity does not have.</p>
<p>For example, although there are no timelike outgoing geodesics from this quark star, as with an event horizon, there are spacelike outgoing geodesics, unlike an event horizon.</p>
<p>Also, if we consider such a quark star to be spinning, so that we compare it to a Kerr black hole, then the bulk motion of the quark star material will induce gravity gradients in the spacetime which, I think, in principle could be measured and compared to the Kerr spacetime. It's not clear whether the spacetime of such a spinning quark star is indistinguishable from the Kerr spacetime.</p>

CC BY-SA 4.0
2021-05-25T14:24:09.197Z
2021-05-25T14:24:09.197Z
Daddy Kropotkin
<p>The escape velocity (in terms of Schwarzschild coordinates) from the surface of a quark star will be given by
<span class="math-container">$$ v = c\left(\frac{r_s}{r_q}\right)^{1/2}, $$</span>
where <span class="math-container">$r_s$</span> is the Schwarzschild radius, <span class="math-container">$2GM/c^2$</span>, and <span class="math-container">$r_q$</span> is the &quot;radius&quot; of the quark star. Strictly speaking, <span class="math-container">$2\pi r_q$</span> is the circumference of the quark star.</p>
<p>Since <span class="math-container">$r_s &lt; r_q$</span>, this does not exceed the speed of light.</p>
<p>However, this applies to objects launched radially outwards. The escape velocity is higher at other angles. Not even light can escape if launched tangentially at <span class="math-container">$r = 1.5 r_s$</span> and quark stars could be smaller than this.</p>
<p>If the quark star is spinning (and it would be) the escape velocity becomes even more complex.</p>

CC BY-SA 4.0
2021-05-25T17:13:48.960Z
2021-05-25T17:13:48.960Z
ProfRob
<p>Adding to the other answers and in particular ProfRob's answer: <a href="https://en.wikipedia.org/wiki/Buchdahl%27s_theorem" rel="nofollow noreferrer">Buchdahl's theorem</a> states that the minimal radius of a static, spherically symmetric matter configuration that acts as a perfect fluid is <span class="math-container">$$r_{B}=\frac{9GM}{4c^2}=\frac{9}{8}r_S.$$</span>
The reason is that otherwise the pressure at the centre will become infinite, implying a collapse.</p>
<p>This means that the escape velocity from the surface of a &quot;Buchdahl star&quot; is <span class="math-container">$v_{esc}=c\sqrt{8/9}\approx 0.94c$</span>. A stable quark star will hence have a subluminal escape velocity.</p>
<p>As noted in the linked Wikipedia article, this limit constrains alternatives to black holes as compact objects and <a href="https://link.springer.com/article/10.1007%2Fs41114-019-0020-4" rel="nofollow noreferrer">one can use what assumptions they breach in the theorem to categorize the proposals</a>.</p>

CC BY-SA 4.0
2021-05-25T21:20:04.967Z
2021-05-25T21:20:04.967Z
Anders Sandberg
<p>Quark stars are hypothetical compact stars that are denser than neutron stars and maybe the last stage of upholding matter before stars that collapse into a singularity. Neutron stars have escape velocities up to half the speed of light, and the escape velocity of quark stars must be even closer to c. Could it be that some or all black holes are quark stars whose escape velocity is just above 1 c, rather than singularities?</p>

CC BY-SA 4.0
2021-05-25T14:15:05.823Z
Related https://physics.stackexchange.com/questions/524720/can-quark-stars-form-under-an-event-horizon
Daddy Kropotkin
<light><neutron-star><degenerate-matter><escape-velocity>
What escape velocity would quark stars have?
Giovanni
# What escape velocity would quark stars have?

(Asked by: username_0 on 2021-05-25)

Quark stars are hypothetical compact stars that are denser than neutron stars and maybe the last stage of upholding matter before stars that collapse into a singularity. Neutron stars have escape velocities up to half the speed of light, and the escape velocity of quark stars must be even closer to c. Could it be that some or all black holes are quark stars whose escape velocity is just above 1 c, rather than singularities?

**Tags:** light, neutron-star, degenerate-matter, escape-velocity
> **Comments:**
> **username_1**: Related https://physics.stackexchange.com/questions/524720/can-quark-stars-form-under-an-event-horizon  

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## 3 answers

### Answer by username_1 on 2021-05-25. Score: 1 | Accepted answer

Interesting question! The existence of quark stars is hypothetical and they are often the [target](https://physics.stackexchange.com/questions/524720/can-quark-stars-form-under-an-event-horizon) of woo-hoo.

Even if one computed the escape velocity at the surface of a quark star (which would be a complicated calculation that would take me a while to research how to do correctly, which is probably an open question anyway), there would still be differences from an (absolute) event horizon. That is, there are aspects of an event horizon that a quark star with superluminal escape velocity does not have.

For example, although there are no timelike outgoing geodesics from this quark star, as with an event horizon, there are spacelike outgoing geodesics, unlike an event horizon.

Also, if we consider such a quark star to be spinning, so that we compare it to a Kerr black hole, then the bulk motion of the quark star material will induce gravity gradients in the spacetime which, I think, in principle could be measured and compared to the Kerr spacetime. It's not clear whether the spacetime of such a spinning quark star is indistinguishable from the Kerr spacetime.


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### Answer by username_2 on 2021-05-25. Score: 4

The escape velocity (in terms of Schwarzschild coordinates) from the surface of a quark star will be given by
$$ v = c\left(\frac{r\_s}{r\_q}\right)^{1/2}, $$
where $r\_s$ is the Schwarzschild radius, $2GM/c^2$, and $r\_q$ is the "radius" of the quark star. Strictly speaking, $2\pi r\_q$ is the circumference of the quark star.

Since $r\_s < r\_q$, this does not exceed the speed of light.

However, this applies to objects launched radially outwards. The escape velocity is higher at other angles. Not even light can escape if launched tangentially at $r = 1.5 r\_s$ and quark stars could be smaller than this.

If the quark star is spinning (and it would be) the escape velocity becomes even more complex.


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### Answer by username_3 on 2021-05-25. Score: 3

Adding to the other answers and in particular ProfRob's answer: [Buchdahl's theorem](https://en.wikipedia.org/wiki/Buchdahl%27s_theorem) states that the minimal radius of a static, spherically symmetric matter configuration that acts as a perfect fluid is $$r\_{B}=\frac{9GM}{4c^2}=\frac{9}{8}r\_S.$$
The reason is that otherwise the pressure at the centre will become infinite, implying a collapse.

This means that the escape velocity from the surface of a "Buchdahl star" is $v\_{esc}=c\sqrt{8/9}\approx 0.94c$. A stable quark star will hence have a subluminal escape velocity.

As noted in the linked Wikipedia article, this limit constrains alternatives to black holes as compact objects and [one can use what assumptions they breach in the theorem to categorize the proposals](https://link.springer.com/article/10.1007%2Fs41114-019-0020-4).


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