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











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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







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  • 4




    Gravity couples to energy not mass.
    – Qmechanic♦
    Dec 11 at 1:53















up vote
10
down vote

favorite
1












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









New contributor




Sigex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 4




    Gravity couples to energy not mass.
    – Qmechanic♦
    Dec 11 at 1:53













up vote
10
down vote

favorite
1









up vote
10
down vote

favorite
1






1





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









New contributor




Sigex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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









New contributor




Sigex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Sigex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 11 hours ago









knzhou

40.1k11113194




40.1k11113194






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asked Dec 10 at 21:15









Sigex

1535




1535




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Sigex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Sigex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.








  • 4




    Gravity couples to energy not mass.
    – Qmechanic♦
    Dec 11 at 1:53














  • 4




    Gravity couples to energy not mass.
    – Qmechanic♦
    Dec 11 at 1:53








4




4




Gravity couples to energy not mass.
– Qmechanic♦
Dec 11 at 1:53




Gravity couples to energy not mass.
– Qmechanic♦
Dec 11 at 1:53










2 Answers
2






active

oldest

votes

















up vote
20
down vote



accepted











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



















  • 1




    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.
    – Sigex
    Dec 13 at 9:26


















up vote
2
down vote













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























  • 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
    – undefined
    Dec 12 at 14:13






  • 2




    @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 $m^2 = 4E^2 - 0 > 0$.
    – Robin Ekman
    Dec 12 at 14:19






  • 1




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










  • 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
    – Dale
    Dec 13 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








up vote
20
down vote



accepted











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



















  • 1




    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.
    – Sigex
    Dec 13 at 9:26















up vote
20
down vote



accepted











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



















  • 1




    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.
    – Sigex
    Dec 13 at 9:26













up vote
20
down vote



accepted







up vote
20
down vote



accepted







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















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 at 3:23

























answered Dec 10 at 22:09









Dale

4,4191623




4,4191623








  • 1




    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.
    – Sigex
    Dec 13 at 9:26














  • 1




    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.
    – Sigex
    Dec 13 at 9:26








1




1




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.
– Sigex
Dec 13 at 9:26




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.
– Sigex
Dec 13 at 9:26










up vote
2
down vote













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























  • 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
    – undefined
    Dec 12 at 14:13






  • 2




    @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 $m^2 = 4E^2 - 0 > 0$.
    – Robin Ekman
    Dec 12 at 14:19






  • 1




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










  • 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
    – Dale
    Dec 13 at 23:00















up vote
2
down vote













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























  • 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
    – undefined
    Dec 12 at 14:13






  • 2




    @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 $m^2 = 4E^2 - 0 > 0$.
    – Robin Ekman
    Dec 12 at 14:19






  • 1




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










  • 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
    – Dale
    Dec 13 at 23:00













up vote
2
down vote










up vote
2
down vote









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














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 at 23:46

























answered Dec 11 at 12:36









Robin Ekman

12k12144




12k12144












  • 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
    – undefined
    Dec 12 at 14:13






  • 2




    @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 $m^2 = 4E^2 - 0 > 0$.
    – Robin Ekman
    Dec 12 at 14:19






  • 1




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










  • 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
    – Dale
    Dec 13 at 23:00


















  • 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
    – undefined
    Dec 12 at 14:13






  • 2




    @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 $m^2 = 4E^2 - 0 > 0$.
    – Robin Ekman
    Dec 12 at 14:19






  • 1




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










  • 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
    – Dale
    Dec 13 at 23:00
















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
– undefined
Dec 12 at 14:13




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
– undefined
Dec 12 at 14:13




2




2




@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 $m^2 = 4E^2 - 0 > 0$.
– Robin Ekman
Dec 12 at 14:19




@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 $m^2 = 4E^2 - 0 > 0$.
– Robin Ekman
Dec 12 at 14:19




1




1




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




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












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
– Dale
Dec 13 at 23:00




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
– Dale
Dec 13 at 23:00










Sigex is a new contributor. Be nice, and check out our Code of Conduct.










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格利澤436b

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格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP
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When can things happen in Etherscan, such as the picture below?

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2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  | 
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