2 added 1 character in body edited Aug 30 '18 at 7:05 safesphere 7,62611644 Your solution is $$dE=c^2 dm$$ (from $$E=mc^2$$) combined with $$dE=m\,d\Phi$$ (from $$E=\Phi m$$) giving $$c^2 dm=m\,d\Phi$$ Or $$\dfrac{dm}{m}=\dfrac{d\Phi}{c^2}$$ That solves $$m = m_0 e^{\frac{\Phi}{c^2}}\tag{1}$$ However, you have overlooked the fact that $$m=m(\Phi)$$, thus $$dE=m\,d\Phi{\color{red}{+\Phi\,dm}}$$. This gives $$c^2 dm=m\,d\Phi+\Phi\,dm$$ Or $$\dfrac{dm}{m}=\dfrac{d\Phi}{c^2-\Phi}$$ Solving $$m=\dfrac{m_o}{1-\dfrac{\Phi}{c^2}}\tag{2}$$ Note that the first two terms of the Taylor series are the same for $$(1)$$ and $$(2)$$ referring to the Newtonian gravity, but not General Relativity. None of this has to do with the existence of the event horizon, because $$\Phi(r)$$ is not defined above. Its definition in relativity comes from time dilation. For example, in the Schwarzchild solution with no motion $$\dfrac{d\tau}{dt}=\sqrt{1-\dfrac{r_{\text{s}}}{r}}$$ Where $$r_{\text{s}}=\dfrac{2GM}{c^2}$$ is the radius of the event horizon. Accordingly $$\dfrac{\Phi}{c^2}=-\dfrac{1}{\sqrt{1-\dfrac{r_{\text{s}}}{r}}}$$$$\dfrac{\Phi}{c^2}=1-\dfrac{1}{\sqrt{1-\dfrac{r_{\text{s}}}{r}}}$$ Note that even if your solution $$(1)$$ were correct, $$\Phi(r)$$ would still make $$m=0$$ at the event horizon $$r=r_{\text{s}}$$. Your solution is $$dE=c^2 dm$$ (from $$E=mc^2$$) combined with $$dE=m\,d\Phi$$ (from $$E=\Phi m$$) giving $$c^2 dm=m\,d\Phi$$ Or $$\dfrac{dm}{m}=\dfrac{d\Phi}{c^2}$$ That solves $$m = m_0 e^{\frac{\Phi}{c^2}}\tag{1}$$ However, you have overlooked the fact that $$m=m(\Phi)$$, thus $$dE=m\,d\Phi{\color{red}{+\Phi\,dm}}$$. This gives $$c^2 dm=m\,d\Phi+\Phi\,dm$$ Or $$\dfrac{dm}{m}=\dfrac{d\Phi}{c^2-\Phi}$$ Solving $$m=\dfrac{m_o}{1-\dfrac{\Phi}{c^2}}\tag{2}$$ Note that the first two terms of the Taylor series are the same for $$(1)$$ and $$(2)$$ referring to the Newtonian gravity, but not General Relativity. None of this has to do with the existence of the event horizon, because $$\Phi(r)$$ is not defined above. Its definition in relativity comes from time dilation. For example, in the Schwarzchild solution with no motion $$\dfrac{d\tau}{dt}=\sqrt{1-\dfrac{r_{\text{s}}}{r}}$$ Where $$r_{\text{s}}=\dfrac{2GM}{c^2}$$ is the radius of the event horizon. Accordingly $$\dfrac{\Phi}{c^2}=-\dfrac{1}{\sqrt{1-\dfrac{r_{\text{s}}}{r}}}$$ Note that even if your solution $$(1)$$ were correct, $$\Phi(r)$$ would still make $$m=0$$ at the event horizon $$r=r_{\text{s}}$$. Your solution is $$dE=c^2 dm$$ (from $$E=mc^2$$) combined with $$dE=m\,d\Phi$$ (from $$E=\Phi m$$) giving $$c^2 dm=m\,d\Phi$$ Or $$\dfrac{dm}{m}=\dfrac{d\Phi}{c^2}$$ That solves $$m = m_0 e^{\frac{\Phi}{c^2}}\tag{1}$$ However, you have overlooked the fact that $$m=m(\Phi)$$, thus $$dE=m\,d\Phi{\color{red}{+\Phi\,dm}}$$. This gives $$c^2 dm=m\,d\Phi+\Phi\,dm$$ Or $$\dfrac{dm}{m}=\dfrac{d\Phi}{c^2-\Phi}$$ Solving $$m=\dfrac{m_o}{1-\dfrac{\Phi}{c^2}}\tag{2}$$ Note that the first two terms of the Taylor series are the same for $$(1)$$ and $$(2)$$ referring to the Newtonian gravity, but not General Relativity. None of this has to do with the existence of the event horizon, because $$\Phi(r)$$ is not defined above. Its definition in relativity comes from time dilation. For example, in the Schwarzchild solution with no motion $$\dfrac{d\tau}{dt}=\sqrt{1-\dfrac{r_{\text{s}}}{r}}$$ Where $$r_{\text{s}}=\dfrac{2GM}{c^2}$$ is the radius of the event horizon. Accordingly $$\dfrac{\Phi}{c^2}=1-\dfrac{1}{\sqrt{1-\dfrac{r_{\text{s}}}{r}}}$$ Note that even if your solution $$(1)$$ were correct, $$\Phi(r)$$ would still make $$m=0$$ at the event horizon $$r=r_{\text{s}}$$. 1 answered Aug 21 '18 at 8:37 safesphere 7,62611644 Your solution is $$dE=c^2 dm$$ (from $$E=mc^2$$) combined with $$dE=m\,d\Phi$$ (from $$E=\Phi m$$) giving $$c^2 dm=m\,d\Phi$$ Or $$\dfrac{dm}{m}=\dfrac{d\Phi}{c^2}$$ That solves $$m = m_0 e^{\frac{\Phi}{c^2}}\tag{1}$$ However, you have overlooked the fact that $$m=m(\Phi)$$, thus $$dE=m\,d\Phi{\color{red}{+\Phi\,dm}}$$. This gives $$c^2 dm=m\,d\Phi+\Phi\,dm$$ Or $$\dfrac{dm}{m}=\dfrac{d\Phi}{c^2-\Phi}$$ Solving $$m=\dfrac{m_o}{1-\dfrac{\Phi}{c^2}}\tag{2}$$ Note that the first two terms of the Taylor series are the same for $$(1)$$ and $$(2)$$ referring to the Newtonian gravity, but not General Relativity. None of this has to do with the existence of the event horizon, because $$\Phi(r)$$ is not defined above. Its definition in relativity comes from time dilation. For example, in the Schwarzchild solution with no motion $$\dfrac{d\tau}{dt}=\sqrt{1-\dfrac{r_{\text{s}}}{r}}$$ Where $$r_{\text{s}}=\dfrac{2GM}{c^2}$$ is the radius of the event horizon. Accordingly $$\dfrac{\Phi}{c^2}=-\dfrac{1}{\sqrt{1-\dfrac{r_{\text{s}}}{r}}}$$ Note that even if your solution $$(1)$$ were correct, $$\Phi(r)$$ would still make $$m=0$$ at the event horizon $$r=r_{\text{s}}$$.