Does converting rest mass to energy instantly change the gravitational attraction it exerts?












10












$begingroup$


Let there be two hydrogen atoms having 1.1 mass each. We use solar fusion to convert the two hydrogen plasma to helium with a mass of 2 and 0.2 gamma radiation.



Before the fusion process gravity was acting on the system's mass plus the two hydrogen atoms. Afterwards it is acting on the system plus the helium. There is less mass and hence the gravitational interaction is less (negligible but still).



My question: is the process gradual? Instantaneous? Or is there a lag associated with this process until the photon is able to free itself from the star?






gravity relativity mass-energy fusion







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Gravity couples to energy not mass.
    $endgroup$
    – Qmechanic♦
    Dec 11 '18 at 1:53
















10












$begingroup$


Let there be two hydrogen atoms having 1.1 mass each. We use solar fusion to convert the two hydrogen plasma to helium with a mass of 2 and 0.2 gamma radiation.



Before the fusion process gravity was acting on the system's mass plus the two hydrogen atoms. Afterwards it is acting on the system plus the helium. There is less mass and hence the gravitational interaction is less (negligible but still).



My question: is the process gradual? Instantaneous? Or is there a lag associated with this process until the photon is able to free itself from the star?






gravity relativity mass-energy fusion







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Gravity couples to energy not mass.
    $endgroup$
    – Qmechanic♦
    Dec 11 '18 at 1:53














10












10








10


1



$begingroup$


Let there be two hydrogen atoms having 1.1 mass each. We use solar fusion to convert the two hydrogen plasma to helium with a mass of 2 and 0.2 gamma radiation.



Before the fusion process gravity was acting on the system's mass plus the two hydrogen atoms. Afterwards it is acting on the system plus the helium. There is less mass and hence the gravitational interaction is less (negligible but still).



My question: is the process gradual? Instantaneous? Or is there a lag associated with this process until the photon is able to free itself from the star?






gravity relativity mass-energy fusion







share|cite|improve this question











$endgroup$




Let there be two hydrogen atoms having 1.1 mass each. We use solar fusion to convert the two hydrogen plasma to helium with a mass of 2 and 0.2 gamma radiation.



Before the fusion process gravity was acting on the system's mass plus the two hydrogen atoms. Afterwards it is acting on the system plus the helium. There is less mass and hence the gravitational interaction is less (negligible but still).



My question: is the process gradual? Instantaneous? Or is there a lag associated with this process until the photon is able to free itself from the star?






gravity relativity mass-energy fusion




gravity relativity mass-energy fusion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 16:19









knzhou

43.6k11118208




43.6k11118208










asked Dec 10 '18 at 21:15









SigexSigex

1535




1535








  • 4




    $begingroup$
    Gravity couples to energy not mass.
    $endgroup$
    – Qmechanic♦
    Dec 11 '18 at 1:53














  • 4




    $begingroup$
    Gravity couples to energy not mass.
    $endgroup$
    – Qmechanic♦
    Dec 11 '18 at 1:53








4




4




$begingroup$
Gravity couples to energy not mass.
$endgroup$
– Qmechanic♦
Dec 11 '18 at 1:53




$begingroup$
Gravity couples to energy not mass.
$endgroup$
– Qmechanic♦
Dec 11 '18 at 1:53










2 Answers
2






active

oldest

votes


















20












$begingroup$


Before the fusion process gravity was acting on the systems mass plus the two hydrogen atoms. After it is acting on the system plus the helium. There is less mass and hence the gravitational Interact is less (negligible but still).




One subtle thing about gravity that only arises in situations like this is that the source of gravity is actually something called the stress energy tensor. This includes energy, momentum, pressure, and stress. So even though light does not have mass it still gravitates since it has energy and momentum and pressure.



This is important here because not only does the fusion produce helium, it also produces neutrinos and light. The sun doesn’t lose gravitation until the neutrinos and light actually leave the sun. For the neutrinos that happens quite quickly, but for the light it can take a long time as the light is repeatedly scattered, absorbed, and re-emitted.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I suspected this was the case "The sun doesn’t lose gravitation until the neutrinos and light actually leave the sun". I also did some calculations to see how much mass the Sun has used up. My original thought was could the mass conversion account for the missing mass observed in the Coma cluster due to "dark matter". I calculated 10^26kg used up in the life time of the Sun. While its mass is 10^30kg so it appears to be a big NO. I am interested on getting others feedback on this still.
    $endgroup$
    – Sigex
    Dec 13 '18 at 9:26



















2












$begingroup$

There isn't actually less mass after the fusion reaction.



In relativity, the mass is defined by $$m^2 = E^2 - mathbf{p}^2$$ where $E$ is the energy and $mathbf p$ is the three-momentum. These should be taken as the sums over all particles in the system, and they are always conserved. Therefore, the mass is also always conserved.



Since the equation is non-linear, mass doesn't just add. While a single photon is massless, the system of a a hydrogen atom and a photon has more mass than just the hydrogen atom. Indeed, a gas of many photons can have mass, even though it's made up entirely of massless particles.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How can a gas of many photons have mass, if all photons are massless? Do you may have link for further reading or a term I can use to google this? Thanks in advance
    $endgroup$
    – undefined
    Dec 12 '18 at 14:13






  • 2




    $begingroup$
    @undefined because in relativity mass isn't additive. You just need two photons with opposite momenta. Then the total energy is $2E$ but the total momentum is 0, so $m^2 = 4E^2 - 0 > 0$.
    $endgroup$
    – Robin Ekman
    Dec 12 '18 at 14:19






  • 1




    $begingroup$
    I believe what you are referring to here is the effective mass and not the physical mass its self?
    $endgroup$
    – Sigex
    Dec 13 '18 at 9:23










  • $begingroup$
    It is not an effective mass. It is the actual physical invariant mass of the system. Note that it is the mass of the system as a whole
    $endgroup$
    – Dale
    Dec 13 '18 at 23:00











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









20












$begingroup$


Before the fusion process gravity was acting on the systems mass plus the two hydrogen atoms. After it is acting on the system plus the helium. There is less mass and hence the gravitational Interact is less (negligible but still).




One subtle thing about gravity that only arises in situations like this is that the source of gravity is actually something called the stress energy tensor. This includes energy, momentum, pressure, and stress. So even though light does not have mass it still gravitates since it has energy and momentum and pressure.



This is important here because not only does the fusion produce helium, it also produces neutrinos and light. The sun doesn’t lose gravitation until the neutrinos and light actually leave the sun. For the neutrinos that happens quite quickly, but for the light it can take a long time as the light is repeatedly scattered, absorbed, and re-emitted.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I suspected this was the case "The sun doesn’t lose gravitation until the neutrinos and light actually leave the sun". I also did some calculations to see how much mass the Sun has used up. My original thought was could the mass conversion account for the missing mass observed in the Coma cluster due to "dark matter". I calculated 10^26kg used up in the life time of the Sun. While its mass is 10^30kg so it appears to be a big NO. I am interested on getting others feedback on this still.
    $endgroup$
    – Sigex
    Dec 13 '18 at 9:26
















20












$begingroup$


Before the fusion process gravity was acting on the systems mass plus the two hydrogen atoms. After it is acting on the system plus the helium. There is less mass and hence the gravitational Interact is less (negligible but still).




One subtle thing about gravity that only arises in situations like this is that the source of gravity is actually something called the stress energy tensor. This includes energy, momentum, pressure, and stress. So even though light does not have mass it still gravitates since it has energy and momentum and pressure.



This is important here because not only does the fusion produce helium, it also produces neutrinos and light. The sun doesn’t lose gravitation until the neutrinos and light actually leave the sun. For the neutrinos that happens quite quickly, but for the light it can take a long time as the light is repeatedly scattered, absorbed, and re-emitted.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I suspected this was the case "The sun doesn’t lose gravitation until the neutrinos and light actually leave the sun". I also did some calculations to see how much mass the Sun has used up. My original thought was could the mass conversion account for the missing mass observed in the Coma cluster due to "dark matter". I calculated 10^26kg used up in the life time of the Sun. While its mass is 10^30kg so it appears to be a big NO. I am interested on getting others feedback on this still.
    $endgroup$
    – Sigex
    Dec 13 '18 at 9:26














20












20








20





$begingroup$


Before the fusion process gravity was acting on the systems mass plus the two hydrogen atoms. After it is acting on the system plus the helium. There is less mass and hence the gravitational Interact is less (negligible but still).




One subtle thing about gravity that only arises in situations like this is that the source of gravity is actually something called the stress energy tensor. This includes energy, momentum, pressure, and stress. So even though light does not have mass it still gravitates since it has energy and momentum and pressure.



This is important here because not only does the fusion produce helium, it also produces neutrinos and light. The sun doesn’t lose gravitation until the neutrinos and light actually leave the sun. For the neutrinos that happens quite quickly, but for the light it can take a long time as the light is repeatedly scattered, absorbed, and re-emitted.






share|cite|improve this answer











$endgroup$




Before the fusion process gravity was acting on the systems mass plus the two hydrogen atoms. After it is acting on the system plus the helium. There is less mass and hence the gravitational Interact is less (negligible but still).




One subtle thing about gravity that only arises in situations like this is that the source of gravity is actually something called the stress energy tensor. This includes energy, momentum, pressure, and stress. So even though light does not have mass it still gravitates since it has energy and momentum and pressure.



This is important here because not only does the fusion produce helium, it also produces neutrinos and light. The sun doesn’t lose gravitation until the neutrinos and light actually leave the sun. For the neutrinos that happens quite quickly, but for the light it can take a long time as the light is repeatedly scattered, absorbed, and re-emitted.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 11 '18 at 3:23

























answered Dec 10 '18 at 22:09









DaleDale

5,4721826




5,4721826








  • 1




    $begingroup$
    I suspected this was the case "The sun doesn’t lose gravitation until the neutrinos and light actually leave the sun". I also did some calculations to see how much mass the Sun has used up. My original thought was could the mass conversion account for the missing mass observed in the Coma cluster due to "dark matter". I calculated 10^26kg used up in the life time of the Sun. While its mass is 10^30kg so it appears to be a big NO. I am interested on getting others feedback on this still.
    $endgroup$
    – Sigex
    Dec 13 '18 at 9:26














  • 1




    $begingroup$
    I suspected this was the case "The sun doesn’t lose gravitation until the neutrinos and light actually leave the sun". I also did some calculations to see how much mass the Sun has used up. My original thought was could the mass conversion account for the missing mass observed in the Coma cluster due to "dark matter". I calculated 10^26kg used up in the life time of the Sun. While its mass is 10^30kg so it appears to be a big NO. I am interested on getting others feedback on this still.
    $endgroup$
    – Sigex
    Dec 13 '18 at 9:26








1




1




$begingroup$
I suspected this was the case "The sun doesn’t lose gravitation until the neutrinos and light actually leave the sun". I also did some calculations to see how much mass the Sun has used up. My original thought was could the mass conversion account for the missing mass observed in the Coma cluster due to "dark matter". I calculated 10^26kg used up in the life time of the Sun. While its mass is 10^30kg so it appears to be a big NO. I am interested on getting others feedback on this still.
$endgroup$
– Sigex
Dec 13 '18 at 9:26




$begingroup$
I suspected this was the case "The sun doesn’t lose gravitation until the neutrinos and light actually leave the sun". I also did some calculations to see how much mass the Sun has used up. My original thought was could the mass conversion account for the missing mass observed in the Coma cluster due to "dark matter". I calculated 10^26kg used up in the life time of the Sun. While its mass is 10^30kg so it appears to be a big NO. I am interested on getting others feedback on this still.
$endgroup$
– Sigex
Dec 13 '18 at 9:26











2












$begingroup$

There isn't actually less mass after the fusion reaction.



In relativity, the mass is defined by $$m^2 = E^2 - mathbf{p}^2$$ where $E$ is the energy and $mathbf p$ is the three-momentum. These should be taken as the sums over all particles in the system, and they are always conserved. Therefore, the mass is also always conserved.



Since the equation is non-linear, mass doesn't just add. While a single photon is massless, the system of a a hydrogen atom and a photon has more mass than just the hydrogen atom. Indeed, a gas of many photons can have mass, even though it's made up entirely of massless particles.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How can a gas of many photons have mass, if all photons are massless? Do you may have link for further reading or a term I can use to google this? Thanks in advance
    $endgroup$
    – undefined
    Dec 12 '18 at 14:13






  • 2




    $begingroup$
    @undefined because in relativity mass isn't additive. You just need two photons with opposite momenta. Then the total energy is $2E$ but the total momentum is 0, so $m^2 = 4E^2 - 0 > 0$.
    $endgroup$
    – Robin Ekman
    Dec 12 '18 at 14:19






  • 1




    $begingroup$
    I believe what you are referring to here is the effective mass and not the physical mass its self?
    $endgroup$
    – Sigex
    Dec 13 '18 at 9:23










  • $begingroup$
    It is not an effective mass. It is the actual physical invariant mass of the system. Note that it is the mass of the system as a whole
    $endgroup$
    – Dale
    Dec 13 '18 at 23:00
















2












$begingroup$

There isn't actually less mass after the fusion reaction.



In relativity, the mass is defined by $$m^2 = E^2 - mathbf{p}^2$$ where $E$ is the energy and $mathbf p$ is the three-momentum. These should be taken as the sums over all particles in the system, and they are always conserved. Therefore, the mass is also always conserved.



Since the equation is non-linear, mass doesn't just add. While a single photon is massless, the system of a a hydrogen atom and a photon has more mass than just the hydrogen atom. Indeed, a gas of many photons can have mass, even though it's made up entirely of massless particles.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How can a gas of many photons have mass, if all photons are massless? Do you may have link for further reading or a term I can use to google this? Thanks in advance
    $endgroup$
    – undefined
    Dec 12 '18 at 14:13






  • 2




    $begingroup$
    @undefined because in relativity mass isn't additive. You just need two photons with opposite momenta. Then the total energy is $2E$ but the total momentum is 0, so $m^2 = 4E^2 - 0 > 0$.
    $endgroup$
    – Robin Ekman
    Dec 12 '18 at 14:19






  • 1




    $begingroup$
    I believe what you are referring to here is the effective mass and not the physical mass its self?
    $endgroup$
    – Sigex
    Dec 13 '18 at 9:23










  • $begingroup$
    It is not an effective mass. It is the actual physical invariant mass of the system. Note that it is the mass of the system as a whole
    $endgroup$
    – Dale
    Dec 13 '18 at 23:00














2












2








2





$begingroup$

There isn't actually less mass after the fusion reaction.



In relativity, the mass is defined by $$m^2 = E^2 - mathbf{p}^2$$ where $E$ is the energy and $mathbf p$ is the three-momentum. These should be taken as the sums over all particles in the system, and they are always conserved. Therefore, the mass is also always conserved.



Since the equation is non-linear, mass doesn't just add. While a single photon is massless, the system of a a hydrogen atom and a photon has more mass than just the hydrogen atom. Indeed, a gas of many photons can have mass, even though it's made up entirely of massless particles.






share|cite|improve this answer











$endgroup$



There isn't actually less mass after the fusion reaction.



In relativity, the mass is defined by $$m^2 = E^2 - mathbf{p}^2$$ where $E$ is the energy and $mathbf p$ is the three-momentum. These should be taken as the sums over all particles in the system, and they are always conserved. Therefore, the mass is also always conserved.



Since the equation is non-linear, mass doesn't just add. While a single photon is massless, the system of a a hydrogen atom and a photon has more mass than just the hydrogen atom. Indeed, a gas of many photons can have mass, even though it's made up entirely of massless particles.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 11 '18 at 23:46

























answered Dec 11 '18 at 12:36









Robin EkmanRobin Ekman

12.1k12244




12.1k12244












  • $begingroup$
    How can a gas of many photons have mass, if all photons are massless? Do you may have link for further reading or a term I can use to google this? Thanks in advance
    $endgroup$
    – undefined
    Dec 12 '18 at 14:13






  • 2




    $begingroup$
    @undefined because in relativity mass isn't additive. You just need two photons with opposite momenta. Then the total energy is $2E$ but the total momentum is 0, so $m^2 = 4E^2 - 0 > 0$.
    $endgroup$
    – Robin Ekman
    Dec 12 '18 at 14:19






  • 1




    $begingroup$
    I believe what you are referring to here is the effective mass and not the physical mass its self?
    $endgroup$
    – Sigex
    Dec 13 '18 at 9:23










  • $begingroup$
    It is not an effective mass. It is the actual physical invariant mass of the system. Note that it is the mass of the system as a whole
    $endgroup$
    – Dale
    Dec 13 '18 at 23:00


















  • $begingroup$
    How can a gas of many photons have mass, if all photons are massless? Do you may have link for further reading or a term I can use to google this? Thanks in advance
    $endgroup$
    – undefined
    Dec 12 '18 at 14:13






  • 2




    $begingroup$
    @undefined because in relativity mass isn't additive. You just need two photons with opposite momenta. Then the total energy is $2E$ but the total momentum is 0, so $m^2 = 4E^2 - 0 > 0$.
    $endgroup$
    – Robin Ekman
    Dec 12 '18 at 14:19






  • 1




    $begingroup$
    I believe what you are referring to here is the effective mass and not the physical mass its self?
    $endgroup$
    – Sigex
    Dec 13 '18 at 9:23










  • $begingroup$
    It is not an effective mass. It is the actual physical invariant mass of the system. Note that it is the mass of the system as a whole
    $endgroup$
    – Dale
    Dec 13 '18 at 23:00
















$begingroup$
How can a gas of many photons have mass, if all photons are massless? Do you may have link for further reading or a term I can use to google this? Thanks in advance
$endgroup$
– undefined
Dec 12 '18 at 14:13




$begingroup$
How can a gas of many photons have mass, if all photons are massless? Do you may have link for further reading or a term I can use to google this? Thanks in advance
$endgroup$
– undefined
Dec 12 '18 at 14:13




2




2




$begingroup$
@undefined because in relativity mass isn't additive. You just need two photons with opposite momenta. Then the total energy is $2E$ but the total momentum is 0, so $m^2 = 4E^2 - 0 > 0$.
$endgroup$
– Robin Ekman
Dec 12 '18 at 14:19




$begingroup$
@undefined because in relativity mass isn't additive. You just need two photons with opposite momenta. Then the total energy is $2E$ but the total momentum is 0, so $m^2 = 4E^2 - 0 > 0$.
$endgroup$
– Robin Ekman
Dec 12 '18 at 14:19




1




1




$begingroup$
I believe what you are referring to here is the effective mass and not the physical mass its self?
$endgroup$
– Sigex
Dec 13 '18 at 9:23




$begingroup$
I believe what you are referring to here is the effective mass and not the physical mass its self?
$endgroup$
– Sigex
Dec 13 '18 at 9:23












$begingroup$
It is not an effective mass. It is the actual physical invariant mass of the system. Note that it is the mass of the system as a whole
$endgroup$
– Dale
Dec 13 '18 at 23:00




$begingroup$
It is not an effective mass. It is the actual physical invariant mass of the system. Note that it is the mass of the system as a whole
$endgroup$
– Dale
Dec 13 '18 at 23:00


















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Dieringhausen

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Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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