<p>What you are missing is that the <a href="https://en.wikipedia.org/wiki/Shell_theorem" rel="nofollow">shell theorem</a>, which says that for a spherically symmetric mass distribution (we can count the pre- and post-supernova states as this for the sake of argument), that the gravitational field at some distance $d$ from the mass is the same as if all the mass were concentrated at the centre of the distribution.</p>

<p>Therefore the gravitational field due to the pre-supernova star is unchanged after the supernova at a distance $d$ until some of the mass (or equivalently in General Relativity, energy) as travelled <em>beyond</em> a distance $d$.</p>

<p>After a core-collapse supernova, if the pre-supernova star was say $15 M_{\odot}$, then what might happen is that $1.4 M_{\odot}$ is left as a neutron star remnant, whilst $10^{46}$ J of energy is released - mostly in the form of neutrinos travelling at almost the speed of light and the envelope of the star expanding outwards at tens of thousands of km/s and a kinetic energy of about $10^{44}$ J.</p>

<p>The neutrinos carry an equivalent rest mass of a mere $0.05M_{\odot}$, so if you were on a (indestructable) planet in orbit around the supernova, then yes, after the main neutrino pulse has passed there would be a small decrease in the gravitational force felt towards the centre of the planet's orbit (not instantaneously, the neutrino pulse lasts some tens of seconds) that would result in an outward acceleration and  the orbit widening slightly.</p>

<p>Sometime later (the orbital radius divided by the supernova ejecta speed) the main mass loss from the supernova would pass by and this would result in a drastic decrease in the gravitational force and a drastic widening of the orbit.</p>

CC BY-SA 3.0
2016-03-28T13:10:06.490Z
2016-03-28T13:10:06.490Z
ProfRob
<p>As a star collapses, there is a massive release of energy. A supernova releases $10^{44}$ Joules, equivalent to a mass of $10^{27}kg$ which is a lot. But compared to the mass of the sun ($2\times10^{30}$kg) it is not so much: 0.0005 solar masses. So there is a change in the gravitational field, but this is not the cause or a major contribution to the supernova. The field doesn't change "in an instant" Most of the energy is carried away by neutrinos at just under the speed of light, and changes to the gravitational fields propagate at the speed of light (there is some gravitational radiation emitted too)</p>

<p>Supernovae are powered principally by the release of gravitational potential energy as the core falls in on itself. The explosion we see is only a small fraction of the energy released, most is in the form of neutrinos. Supernovae are so ridiculously energetic that you can't trust your intuition, but computer models are becoming better at <a href="https://www.youtube.com/watch?v=2RxIwtxdEnQ" rel="nofollow">predicting the dynamics</a></p>

CC BY-SA 3.0
2016-03-28T09:29:15.060Z
2016-03-28T09:29:15.060Z
James K
<p>During the collapse of a star at the commencement of supernovae, the intense pressures force fusion and fission to create the heavier elements and release vast amounts of energy. As this energy/mass is released - the gravitational field must also decrease in an instant, and as this "Warping of space time" is instantly released and returns to flat space (Or flatter space) - does this create or contribute to the explosion of the supernovae itself? I have always been puzzled by the concept of the collapsed outer layers "Bouncing off of an in-compressible core" as the source of the explosion. Isn't this instantaneous "Relaxation of the curvature of the space" not what we see as the explosion?</p>

<general-relativity><supernova><gravitational-waves>
Gravitational field released during supernovae
Steve Feldman
# Gravitational field released during supernovae

(Asked by: username_0 on 2016-03-28)

During the collapse of a star at the commencement of supernovae, the intense pressures force fusion and fission to create the heavier elements and release vast amounts of energy. As this energy/mass is released - the gravitational field must also decrease in an instant, and as this "Warping of space time" is instantly released and returns to flat space (Or flatter space) - does this create or contribute to the explosion of the supernovae itself? I have always been puzzled by the concept of the collapsed outer layers "Bouncing off of an in-compressible core" as the source of the explosion. Isn't this instantaneous "Relaxation of the curvature of the space" not what we see as the explosion?

**Tags:** general-relativity, supernova, gravitational-waves


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## 2 answers

### Answer by username_1 on 2016-03-28. Score: 3

What you are missing is that the [shell theorem](https://en.wikipedia.org/wiki/Shell_theorem), which says that for a spherically symmetric mass distribution (we can count the pre- and post-supernova states as this for the sake of argument), that the gravitational field at some distance $d$ from the mass is the same as if all the mass were concentrated at the centre of the distribution.

Therefore the gravitational field due to the pre-supernova star is unchanged after the supernova at a distance $d$ until some of the mass (or equivalently in General Relativity, energy) as travelled *beyond* a distance $d$.

After a core-collapse supernova, if the pre-supernova star was say $15 M\_{\odot}$, then what might happen is that $1.4 M\_{\odot}$ is left as a neutron star remnant, whilst $10^{46}$ J of energy is released - mostly in the form of neutrinos travelling at almost the speed of light and the envelope of the star expanding outwards at tens of thousands of km/s and a kinetic energy of about $10^{44}$ J.

The neutrinos carry an equivalent rest mass of a mere $0.05M\_{\odot}$, so if you were on a (indestructable) planet in orbit around the supernova, then yes, after the main neutrino pulse has passed there would be a small decrease in the gravitational force felt towards the centre of the planet's orbit (not instantaneously, the neutrino pulse lasts some tens of seconds) that would result in an outward acceleration and the orbit widening slightly.

Sometime later (the orbital radius divided by the supernova ejecta speed) the main mass loss from the supernova would pass by and this would result in a drastic decrease in the gravitational force and a drastic widening of the orbit.


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### Answer by username_2 on 2016-03-28. Score: 1

As a star collapses, there is a massive release of energy. A supernova releases $10^{44}$ Joules, equivalent to a mass of $10^{27}kg$ which is a lot. But compared to the mass of the sun ($2\times10^{30}$kg) it is not so much: 0.0005 solar masses. So there is a change in the gravitational field, but this is not the cause or a major contribution to the supernova. The field doesn't change "in an instant" Most of the energy is carried away by neutrinos at just under the speed of light, and changes to the gravitational fields propagate at the speed of light (there is some gravitational radiation emitted too)

Supernovae are powered principally by the release of gravitational potential energy as the core falls in on itself. The explosion we see is only a small fraction of the energy released, most is in the form of neutrinos. Supernovae are so ridiculously energetic that you can't trust your intuition, but computer models are becoming better at [predicting the dynamics](https://www.youtube.com/watch?v=2RxIwtxdEnQ)


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