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Qmechanic ♦
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Let us assume that we have a large gravitational field, then the gravitational redshift can be expressed as,

$$\frac {v_{\infty}} {v_e} = (1-r_s/R_e)^{1/2}$$

In this equation $v_{\infty}$ represents the frequency of the light measured by an observer at infinity, $v_e$ is the frequency of the emitted wavelength, $r_s$ is the schwarzschild radius, $r_S=2GM/c^2$, and finally $R_e$ is the radius which photon is emitted.

And I argue that when we  $R_e$ get closer to $r_s$ the frequency of photon will decrease. And so does the energy. And at $r_s=R_e$ the energy of the photon will be zero, at the surface of the event horizon.

"When the photon is emitted at a distance equal to the Schwarzschild radius, the redshift will be infinitely large, and it will not escape to any finite distance from the Schwarzschild sphere." Reference to this quote

At this point I argue that the energy of the photon also decreases. I find an experiment that explains the energy decrease, here and the math of it here

Again, at this point, can someone else argue that "Light doesn't lose energy as it ascends. It was emitted with less energy at a lower elevation."  ? I didnt understand what it means and also how it explains the experiment that I shared.

Are both of the arguments true at certain conditions or only one of them is true  ?

Thanks

Let us assume that we have a large gravitational field, then the gravitational redshift can be expressed as,

$$\frac {v_{\infty}} {v_e} = (1-r_s/R_e)^{1/2}$$

In this equation $v_{\infty}$ represents the frequency of the light measured by an observer at infinity, $v_e$ is the frequency of the emitted wavelength, $r_s$ is the schwarzschild radius, $r_S=2GM/c^2$, and finally $R_e$ is the radius which photon is emitted.

And I argue that when we  $R_e$ get closer to $r_s$ the frequency of photon will decrease. And so does the energy. And at $r_s=R_e$ the energy of the photon will be zero, at the surface of the event horizon.

"When the photon is emitted at a distance equal to the Schwarzschild radius, the redshift will be infinitely large, and it will not escape to any finite distance from the Schwarzschild sphere." Reference to this quote

At this point I argue that the energy of the photon also decreases. I find an experiment that explains the energy decrease, here and the math of it here

Again, at this point, can someone else argue that "Light doesn't lose energy as it ascends. It was emitted with less energy at a lower elevation."  ? I didnt understand what it means and also how it explains the experiment that I shared.

Are both of the arguments true at certain conditions or only one of them is true  ?

Thanks

Let us assume that we have a large gravitational field, then the gravitational redshift can be expressed as,

$$\frac {v_{\infty}} {v_e} = (1-r_s/R_e)^{1/2}$$

In this equation $v_{\infty}$ represents the frequency of the light measured by an observer at infinity, $v_e$ is the frequency of the emitted wavelength, $r_s$ is the schwarzschild radius, $r_S=2GM/c^2$, and finally $R_e$ is the radius which photon is emitted.

And I argue that when we  $R_e$ get closer to $r_s$ the frequency of photon will decrease. And so does the energy. And at $r_s=R_e$ the energy of the photon will be zero, at the surface of the event horizon.

"When the photon is emitted at a distance equal to the Schwarzschild radius, the redshift will be infinitely large, and it will not escape to any finite distance from the Schwarzschild sphere." Reference to this quote

At this point I argue that the energy of the photon also decreases. I find an experiment that explains the energy decrease, here and the math of it here

Again, at this point, can someone else argue that "Light doesn't lose energy as it ascends. It was emitted with less energy at a lower elevation."? I didnt understand what it means and also how it explains the experiment that I shared.

Are both of the arguments true at certain conditions or only one of them is true?

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safesphere
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Let us assume that we have a large gravitational field, then the gravitational redshift can be expressed as,

$$\frac {v_{\infty}} {v_e} = (1-r_s/R_e)^{1/2}$$

In this equation $v_{\infty}$ represents the frequency of the light measured by an observer at infinity, $v_e$ is the frequency of the emitted wavelength, $r_s$ is the schwarzschild radius, $r_S=2GM/c^2$, and finally $R_e$ is the radius which photon is emitted.

And I argue that when we  $R_e$ get closer to $r_s$ the wavelengthfrequency of photon will decrease. And so does the energy. And at $r_s=R_e$ the energy of the photon will be zero, at the surface of the event horizon.

"When the photon is emitted at a distance equal to the Schwarzschild radius, the redshift will be infinitely large, and it will not escape to any finite distance from the Schwarzschild sphere." Reference to this quote

At this point I argue that the energy of the photon also decreases. I find an experiment that explains the energy decrease, here and the math of it here

Again, at this point, can someone else argue that "Light doesn't lose energy as it ascends. It was emitted with less energy at a lower elevation." ? I didnt understand what it means and also how it explains the experiment that I shared.

Are both of the arguments true at certain conditions or only one of them is true ?

Thanks

Let us assume that we have a large gravitational field, then the gravitational redshift can be expressed as,

$$\frac {v_{\infty}} {v_e} = (1-r_s/R_e)^{1/2}$$

In this equation $v_{\infty}$ represents the frequency of the light measured by an observer at infinity, $v_e$ is the frequency of the emitted wavelength, $r_s$ is the schwarzschild radius, $r_S=2GM/c^2$, and finally $R_e$ is the radius which photon is emitted.

And I argue that when we  $R_e$ get closer to $r_s$ the wavelength of photon will decrease. And so does the energy. And at $r_s=R_e$ the energy of the photon will be zero, at the surface of the event horizon.

"When the photon is emitted at a distance equal to the Schwarzschild radius, the redshift will be infinitely large, and it will not escape to any finite distance from the Schwarzschild sphere." Reference to this quote

At this point I argue that the energy of the photon also decreases. I find an experiment that explains the energy decrease, here and the math of it here

Again, at this point, can someone else argue that "Light doesn't lose energy as it ascends. It was emitted with less energy at a lower elevation." ? I didnt understand what it means and also how it explains the experiment that I shared.

Are both of the arguments true at certain conditions or only one of them is true ?

Thanks

Let us assume that we have a large gravitational field, then the gravitational redshift can be expressed as,

$$\frac {v_{\infty}} {v_e} = (1-r_s/R_e)^{1/2}$$

In this equation $v_{\infty}$ represents the frequency of the light measured by an observer at infinity, $v_e$ is the frequency of the emitted wavelength, $r_s$ is the schwarzschild radius, $r_S=2GM/c^2$, and finally $R_e$ is the radius which photon is emitted.

And I argue that when we  $R_e$ get closer to $r_s$ the frequency of photon will decrease. And so does the energy. And at $r_s=R_e$ the energy of the photon will be zero, at the surface of the event horizon.

"When the photon is emitted at a distance equal to the Schwarzschild radius, the redshift will be infinitely large, and it will not escape to any finite distance from the Schwarzschild sphere." Reference to this quote

At this point I argue that the energy of the photon also decreases. I find an experiment that explains the energy decrease, here and the math of it here

Again, at this point, can someone else argue that "Light doesn't lose energy as it ascends. It was emitted with less energy at a lower elevation." ? I didnt understand what it means and also how it explains the experiment that I shared.

Are both of the arguments true at certain conditions or only one of them is true ?

Thanks

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seVenVo1d
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Gravitational redshift and Energy of a photon

Let us assume that we have a large gravitational field, then the gravitational redshift can be expressed as,

$$\frac {v_{\infty}} {v_e} = (1-r_s/R_e)^{1/2}$$

In this equation $v_{\infty}$ represents the frequency of the light measured by an observer at infinity, $v_e$ is the frequency of the emitted wavelength, $r_s$ is the schwarzschild radius, $r_S=2GM/c^2$, and finally $R_e$ is the radius which photon is emitted.

And I argue that when we  $R_e$ get closer to $r_s$ the wavelength of photon will decrease. And so does the energy. And at $r_s=R_e$ the energy of the photon will be zero, at the surface of the event horizon.

"When the photon is emitted at a distance equal to the Schwarzschild radius, the redshift will be infinitely large, and it will not escape to any finite distance from the Schwarzschild sphere." Reference to this quote

At this point I argue that the energy of the photon also decreases. I find an experiment that explains the energy decrease, here and the math of it here

Again, at this point, can someone else argue that "Light doesn't lose energy as it ascends. It was emitted with less energy at a lower elevation." ? I didnt understand what it means and also how it explains the experiment that I shared.

Are both of the arguments true at certain conditions or only one of them is true ?

Thanks