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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
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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 $r=r_{\text{s}}$.