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




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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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迪纳利

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迪纳利峰 位于迪纳利国家公园和保留区的迪纳利峰（旧名麦金利峰） 最高点 海拔 20,237英尺（6,168米）   NGVD 29 . [1] [2] 相對高度 20,073英尺（6,118米） 列表 各洲最高峰列表 特独立山峰 各国最高点列表 美国最高点  排名第一 [3] 坐标 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley ) 坐标： 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley )   地理 位置   美國阿拉斯加州迪纳利自治市镇 迪纳利国家公园和保留区 山脈 阿拉斯加山脉 攀山 首次登頂 （ 英语 ： First ascent ） 1913年6月7日 哈德逊·史塔克 哈里·卡斯坦斯 沃尔特·哈珀 罗伯特·塔特姆 最簡路線 西坡路线（攀登冰川/雪） 迪纳利 （北阿萨巴斯卡语支、 英语： Denali ， / d ᵻ ˈ n ɑː l i / ，直译“高山”），1917年至2015年官方名称为 麦金利山 （另译作 麦金利峰 ， 英语： Mount McKinley ），位于阿拉斯加州东南部、阿拉斯加山脉中段，海拔6194米，是北美洲最高峰，也是美国的最高峰。它的俄语名称是 Большая Гора ， 罗马化：Bolshaya Gora ，意思是“大山”。 迪纳利山地区拥有变幻莫测的高山风、典型的北极植被以及野生动植物。这里大部分地区终年积雪，山间经常浓雾不断，雾气在皑皑白雪中缭绕弥漫时，几百米之外的景物便不可见。夏季，迪纳利山的青青山坡上鲜花盛开，紫色的杜鹃和精巧的铃状石南花随处可见。迪纳利山还是世界登山爱好者的汇集之地。每年5月到7月有数百人登山，只有一半不到的人能登顶成功。一次登山约花3个星期。 2015年8月31日，美國政府宣佈將山名由“麦金利山”改回原名“迪纳利山”。美國內政部發表聲
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南乌拉尔铁路局

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南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
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