Mixing
Based on black hole thermodynamics, shouldn't empty space contain infinite energy?
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
According to the Wikipedia page on Hawking Radiation:
"In SI units, the radiation from a Schwarzschild black hole is blackbody radiation with temperature
$$T=hbar c^3over8pi G k_bM$$
where $hbar$ is the reduced Planck constant, $c$ is the speed of light, $k_B$ is the Boltzmann constant, $G$ is the gravitational constant, and M is the mass of the black hole."
By taking the limit of $T$ as $M$ goes to zero, the following is found:
$$lim_Mto0^+ T=lim_Mto0^+hbar c^3over8pi G k_bM=+infty$$
Wouldn't this mean that empty space would have infinite energy? As when $M=0$ the Schwarzschild radius is also $0$, so every point in space would be paradoxically hot. I know I'm probably wrong, I just don't know why I'm wrong.
thermodynamics black-holes hawking-radiation
edited 1 hour ago
Avantgarde
8631312
asked 2 hours ago
user189728
3959
add a comment |Â
up vote
4
down vote
favorite
According to the Wikipedia page on Hawking Radiation:
"In SI units, the radiation from a Schwarzschild black hole is blackbody radiation with temperature
$$T=hbar c^3over8pi G k_bM$$
where $hbar$ is the reduced Planck constant, $c$ is the speed of light, $k_B$ is the Boltzmann constant, $G$ is the gravitational constant, and M is the mass of the black hole."
By taking the limit of $T$ as $M$ goes to zero, the following is found:
$$lim_Mto0^+ T=lim_Mto0^+hbar c^3over8pi G k_bM=+infty$$
Wouldn't this mean that empty space would have infinite energy? As when $M=0$ the Schwarzschild radius is also $0$, so every point in space would be paradoxically hot. I know I'm probably wrong, I just don't know why I'm wrong.
thermodynamics black-holes hawking-radiation
edited 1 hour ago
Avantgarde
8631312
asked 2 hours ago
user189728
3959
1
If $M$ is zero what is supposed to be radiating (nothing ?) and where does the energy for this radiation come from (nowhere ?) ?
– StephenG
1 hour ago
@StephenG I share your confusion, but the math doesn’t lie
– user189728
1 hour ago
From your link " Unlike most objects, a black hole's temperature increases as it radiates away mass. The rate of temperature increase is exponential, with the most likely endpoint being the dissolution of the black hole in a violent burst of gamma rays. A complete description of this dissolution requires a model of quantum gravity, however, as it occurs when the black hole approaches Planck mass and Planck radius."
– Bruce Greetham
1 hour ago
Hawking's original computation involves quantum fields in curved spacetime, where curvatures are small compared to the Planck length. I believe that in order to take a limit all the way to $M=0$, one must know how Planck scale quantum gravity works. Maybe there's something Hawking says in his article "Black hole explosions?" but I can't open it at the moment: nature.com/articles/248030a0
– Avantgarde
51 mins ago
Newtonian 'black holes' ($c to infty$) also have infinite temperature. But this would be wrong since we derived $T$ using relativistic field theory.
– Avantgarde
29 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
According to the Wikipedia page on Hawking Radiation:
"In SI units, the radiation from a Schwarzschild black hole is blackbody radiation with temperature
$$T=hbar c^3over8pi G k_bM$$
where $hbar$ is the reduced Planck constant, $c$ is the speed of light, $k_B$ is the Boltzmann constant, $G$ is the gravitational constant, and M is the mass of the black hole."
By taking the limit of $T$ as $M$ goes to zero, the following is found:
$$lim_Mto0^+ T=lim_Mto0^+hbar c^3over8pi G k_bM=+infty$$
Wouldn't this mean that empty space would have infinite energy? As when $M=0$ the Schwarzschild radius is also $0$, so every point in space would be paradoxically hot. I know I'm probably wrong, I just don't know why I'm wrong.
thermodynamics black-holes hawking-radiation
edited 1 hour ago
Avantgarde
8631312
asked 2 hours ago
user189728
3959
According to the Wikipedia page on Hawking Radiation:
"In SI units, the radiation from a Schwarzschild black hole is blackbody radiation with temperature
$$T=hbar c^3over8pi G k_bM$$
where $hbar$ is the reduced Planck constant, $c$ is the speed of light, $k_B$ is the Boltzmann constant, $G$ is the gravitational constant, and M is the mass of the black hole."
By taking the limit of $T$ as $M$ goes to zero, the following is found:
$$lim_Mto0^+ T=lim_Mto0^+hbar c^3over8pi G k_bM=+infty$$
Wouldn't this mean that empty space would have infinite energy? As when $M=0$ the Schwarzschild radius is also $0$, so every point in space would be paradoxically hot. I know I'm probably wrong, I just don't know why I'm wrong.
thermodynamics black-holes hawking-radiation
thermodynamics black-holes hawking-radiation
edited 1 hour ago
Avantgarde
8631312
asked 2 hours ago
user189728
3959
edited 1 hour ago
Avantgarde
8631312
asked 2 hours ago
user189728
3959
edited 1 hour ago
Avantgarde
8631312
edited 1 hour ago
Avantgarde
8631312
edited 1 hour ago
Avantgarde
8631312
8631312
asked 2 hours ago
user189728
3959
asked 2 hours ago
user189728
3959
asked 2 hours ago
user189728
3959
3959
1
If $M$ is zero what is supposed to be radiating (nothing ?) and where does the energy for this radiation come from (nowhere ?) ?
– StephenG
1 hour ago
@StephenG I share your confusion, but the math doesn’t lie
– user189728
1 hour ago
From your link " Unlike most objects, a black hole's temperature increases as it radiates away mass. The rate of temperature increase is exponential, with the most likely endpoint being the dissolution of the black hole in a violent burst of gamma rays. A complete description of this dissolution requires a model of quantum gravity, however, as it occurs when the black hole approaches Planck mass and Planck radius."
– Bruce Greetham
1 hour ago
Hawking's original computation involves quantum fields in curved spacetime, where curvatures are small compared to the Planck length. I believe that in order to take a limit all the way to $M=0$, one must know how Planck scale quantum gravity works. Maybe there's something Hawking says in his article "Black hole explosions?" but I can't open it at the moment: nature.com/articles/248030a0
– Avantgarde
51 mins ago
Newtonian 'black holes' ($c to infty$) also have infinite temperature. But this would be wrong since we derived $T$ using relativistic field theory.
– Avantgarde
29 mins ago
add a comment |Â
1
If $M$ is zero what is supposed to be radiating (nothing ?) and where does the energy for this radiation come from (nowhere ?) ?
– StephenG
1 hour ago
@StephenG I share your confusion, but the math doesn’t lie
– user189728
1 hour ago
From your link " Unlike most objects, a black hole's temperature increases as it radiates away mass. The rate of temperature increase is exponential, with the most likely endpoint being the dissolution of the black hole in a violent burst of gamma rays. A complete description of this dissolution requires a model of quantum gravity, however, as it occurs when the black hole approaches Planck mass and Planck radius."
– Bruce Greetham
1 hour ago
Hawking's original computation involves quantum fields in curved spacetime, where curvatures are small compared to the Planck length. I believe that in order to take a limit all the way to $M=0$, one must know how Planck scale quantum gravity works. Maybe there's something Hawking says in his article "Black hole explosions?" but I can't open it at the moment: nature.com/articles/248030a0
– Avantgarde
51 mins ago
Newtonian 'black holes' ($c to infty$) also have infinite temperature. But this would be wrong since we derived $T$ using relativistic field theory.
– Avantgarde
29 mins ago
1
1
If $M$ is zero what is supposed to be radiating (nothing ?) and where does the energy for this radiation come from (nowhere ?) ?
– StephenG
1 hour ago
If $M$ is zero what is supposed to be radiating (nothing ?) and where does the energy for this radiation come from (nowhere ?) ?
– StephenG
1 hour ago
@StephenG I share your confusion, but the math doesn’t lie
– user189728
1 hour ago
@StephenG I share your confusion, but the math doesn’t lie
– user189728
1 hour ago
From your link " Unlike most objects, a black hole's temperature increases as it radiates away mass. The rate of temperature increase is exponential, with the most likely endpoint being the dissolution of the black hole in a violent burst of gamma rays. A complete description of this dissolution requires a model of quantum gravity, however, as it occurs when the black hole approaches Planck mass and Planck radius."
– Bruce Greetham
1 hour ago
From your link " Unlike most objects, a black hole's temperature increases as it radiates away mass. The rate of temperature increase is exponential, with the most likely endpoint being the dissolution of the black hole in a violent burst of gamma rays. A complete description of this dissolution requires a model of quantum gravity, however, as it occurs when the black hole approaches Planck mass and Planck radius."
– Bruce Greetham
1 hour ago
Hawking's original computation involves quantum fields in curved spacetime, where curvatures are small compared to the Planck length. I believe that in order to take a limit all the way to $M=0$, one must know how Planck scale quantum gravity works. Maybe there's something Hawking says in his article "Black hole explosions?" but I can't open it at the moment: nature.com/articles/248030a0
– Avantgarde
51 mins ago
Hawking's original computation involves quantum fields in curved spacetime, where curvatures are small compared to the Planck length. I believe that in order to take a limit all the way to $M=0$, one must know how Planck scale quantum gravity works. Maybe there's something Hawking says in his article "Black hole explosions?" but I can't open it at the moment: nature.com/articles/248030a0
– Avantgarde
51 mins ago
Newtonian 'black holes' ($c to infty$) also have infinite temperature. But this would be wrong since we derived $T$ using relativistic field theory.
– Avantgarde
29 mins ago
Newtonian 'black holes' ($c to infty$) also have infinite temperature. But this would be wrong since we derived $T$ using relativistic field theory.
– Avantgarde
29 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
Cute question. I don't think it's possible to give a definite answer unless we decide on a definite description of the manner in which the limit is to be taken.
If we're just considering the limit of an asymptotically flat spacetime consisting of a single black hole, as the black hole's mass goes to zero, then I think the answer is pretty straightforward. The radiated power is $Ppropto M^-2propto r_s^-2$, where $r_s$ is the Schwarzschild radius. If this power is considered to be emitted from the surface of a sphere with radius equal to $r_s$, then the power per unit area at $r>r_s$ is $Ipropto Pr^-2 propto r_s^-2 r^-2$. The average distance of a randomly chosen observer from the black hole is infinite, so we should consider the limit $rrightarrowinfty$. So now the question arises as to whether we should compute
$$lim_rrightarrowinftylim_r_srightarrow0 I$$
or
$$lim_r_srightarrow0lim_rrightarrowinfty I.$$
These are both indeterminate forms, so it sort of doesn't matter which one we evaluate -- neither one gives a meaningful answer without some physics input.
I think the additional physics input comes from the fact that $r_s$ probably can't be any smaller than the Planck length $ell$. Therefore we really shouldn't be taking the limit as $r_srightarrow0$ but as $r_srightarrowell$. Then the double limit is zero.
Another way of approaching the whole thing is to imagine a theory of quantum gravity in which the vacuum has virtual black holes popping in and out of existence. Then presumably one of the things you want from such a theory (which we don't yet possess) is that it doesn't predict infinite density of photons for the vacuum. I would imagine that this would happen because the density of virtual Planck-scale black holes would be small.
answered 1 hour ago
Ben Crowell
44.9k3147274
add a comment |Â
up vote
1
down vote
Wouldn't this mean that empty space would have infinite energy?
So ignoring quantum issues (and in the absence of a complete theory of quantum gravity we have no choice) and staying strictly with the classical approach let's consider the problem.
Regardless of what amount of power is radiated by the black hole, that power is removed from the energy of the black hole. But the black holes you are talking about have zero energy and so there is no way for them to power hawking radiation.
The mistake you are making is ignoring that nature balances it's books and in this case the balance is that you can't reduce the mass below zero.
There's another reason why your logic is failing.
The entire idea of Hawking radiation depends on the exist of a curved space time and an event horizon. But when $M=0$ we just get a flat spacetime. There is no event horizon. And note that an $R=0$ event horizon would mean there was nothing inside the black hole - no volume, nothing.
The formula you are using for temperature is based on a model which starts out with a non-zero positive mass and then makes a first order approximation close to the event horizon (Wikipedia has a description of this). But that formula does not apply when you're using $Mto 0$. Again, you're using an approximation based on the assumption of a curved spacetime ($M>0$) and applying it outside it's "designed purpose" as an approximation.
answered 19 mins ago
StephenG
5,01821323
add a comment |Â
up vote
0
down vote
When the mass drops down to 228 metric tonnes, that's the signal that
exactly one second remains. The event horizon size at the time will be
340 yoctometers, or 3.4 × 10^-22 meters: the size of one wavelength of
a photon with an energy greater than any particle the LHC has ever
produced. But in that final second, a total of 2.05 × 10^22 Joules of
energy, the equivalent of five million megatons of TNT, will be
released. It's as though a million nuclear fusion bombs went off all
at once in a tiny region of space; that's the final stage of black
hole evaporation.
https://www.forbes.com/sites/startswithabang/2017/05/20/ask-ethan-what-happens-when-a-black-holes-singularity-evaporates/#3d39e14b7c8c
The point is, in reality, M never hits zero.
Theory is great but we can't take it literally.
answered 22 mins ago
CramerTV
20228
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Cute question. I don't think it's possible to give a definite answer unless we decide on a definite description of the manner in which the limit is to be taken.
If we're just considering the limit of an asymptotically flat spacetime consisting of a single black hole, as the black hole's mass goes to zero, then I think the answer is pretty straightforward. The radiated power is $Ppropto M^-2propto r_s^-2$, where $r_s$ is the Schwarzschild radius. If this power is considered to be emitted from the surface of a sphere with radius equal to $r_s$, then the power per unit area at $r>r_s$ is $Ipropto Pr^-2 propto r_s^-2 r^-2$. The average distance of a randomly chosen observer from the black hole is infinite, so we should consider the limit $rrightarrowinfty$. So now the question arises as to whether we should compute
$$lim_rrightarrowinftylim_r_srightarrow0 I$$
or
$$lim_r_srightarrow0lim_rrightarrowinfty I.$$
These are both indeterminate forms, so it sort of doesn't matter which one we evaluate -- neither one gives a meaningful answer without some physics input.
I think the additional physics input comes from the fact that $r_s$ probably can't be any smaller than the Planck length $ell$. Therefore we really shouldn't be taking the limit as $r_srightarrow0$ but as $r_srightarrowell$. Then the double limit is zero.
Another way of approaching the whole thing is to imagine a theory of quantum gravity in which the vacuum has virtual black holes popping in and out of existence. Then presumably one of the things you want from such a theory (which we don't yet possess) is that it doesn't predict infinite density of photons for the vacuum. I would imagine that this would happen because the density of virtual Planck-scale black holes would be small.
answered 1 hour ago
Ben Crowell
44.9k3147274
add a comment |Â
up vote
1
down vote
Cute question. I don't think it's possible to give a definite answer unless we decide on a definite description of the manner in which the limit is to be taken.
If we're just considering the limit of an asymptotically flat spacetime consisting of a single black hole, as the black hole's mass goes to zero, then I think the answer is pretty straightforward. The radiated power is $Ppropto M^-2propto r_s^-2$, where $r_s$ is the Schwarzschild radius. If this power is considered to be emitted from the surface of a sphere with radius equal to $r_s$, then the power per unit area at $r>r_s$ is $Ipropto Pr^-2 propto r_s^-2 r^-2$. The average distance of a randomly chosen observer from the black hole is infinite, so we should consider the limit $rrightarrowinfty$. So now the question arises as to whether we should compute
$$lim_rrightarrowinftylim_r_srightarrow0 I$$
or
$$lim_r_srightarrow0lim_rrightarrowinfty I.$$
These are both indeterminate forms, so it sort of doesn't matter which one we evaluate -- neither one gives a meaningful answer without some physics input.
I think the additional physics input comes from the fact that $r_s$ probably can't be any smaller than the Planck length $ell$. Therefore we really shouldn't be taking the limit as $r_srightarrow0$ but as $r_srightarrowell$. Then the double limit is zero.
Another way of approaching the whole thing is to imagine a theory of quantum gravity in which the vacuum has virtual black holes popping in and out of existence. Then presumably one of the things you want from such a theory (which we don't yet possess) is that it doesn't predict infinite density of photons for the vacuum. I would imagine that this would happen because the density of virtual Planck-scale black holes would be small.
answered 1 hour ago
Ben Crowell
44.9k3147274
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Cute question. I don't think it's possible to give a definite answer unless we decide on a definite description of the manner in which the limit is to be taken.
If we're just considering the limit of an asymptotically flat spacetime consisting of a single black hole, as the black hole's mass goes to zero, then I think the answer is pretty straightforward. The radiated power is $Ppropto M^-2propto r_s^-2$, where $r_s$ is the Schwarzschild radius. If this power is considered to be emitted from the surface of a sphere with radius equal to $r_s$, then the power per unit area at $r>r_s$ is $Ipropto Pr^-2 propto r_s^-2 r^-2$. The average distance of a randomly chosen observer from the black hole is infinite, so we should consider the limit $rrightarrowinfty$. So now the question arises as to whether we should compute
$$lim_rrightarrowinftylim_r_srightarrow0 I$$
or
$$lim_r_srightarrow0lim_rrightarrowinfty I.$$
These are both indeterminate forms, so it sort of doesn't matter which one we evaluate -- neither one gives a meaningful answer without some physics input.
I think the additional physics input comes from the fact that $r_s$ probably can't be any smaller than the Planck length $ell$. Therefore we really shouldn't be taking the limit as $r_srightarrow0$ but as $r_srightarrowell$. Then the double limit is zero.
Another way of approaching the whole thing is to imagine a theory of quantum gravity in which the vacuum has virtual black holes popping in and out of existence. Then presumably one of the things you want from such a theory (which we don't yet possess) is that it doesn't predict infinite density of photons for the vacuum. I would imagine that this would happen because the density of virtual Planck-scale black holes would be small.
answered 1 hour ago
Ben Crowell
44.9k3147274
Cute question. I don't think it's possible to give a definite answer unless we decide on a definite description of the manner in which the limit is to be taken.
If we're just considering the limit of an asymptotically flat spacetime consisting of a single black hole, as the black hole's mass goes to zero, then I think the answer is pretty straightforward. The radiated power is $Ppropto M^-2propto r_s^-2$, where $r_s$ is the Schwarzschild radius. If this power is considered to be emitted from the surface of a sphere with radius equal to $r_s$, then the power per unit area at $r>r_s$ is $Ipropto Pr^-2 propto r_s^-2 r^-2$. The average distance of a randomly chosen observer from the black hole is infinite, so we should consider the limit $rrightarrowinfty$. So now the question arises as to whether we should compute
$$lim_rrightarrowinftylim_r_srightarrow0 I$$
or
$$lim_r_srightarrow0lim_rrightarrowinfty I.$$
These are both indeterminate forms, so it sort of doesn't matter which one we evaluate -- neither one gives a meaningful answer without some physics input.
I think the additional physics input comes from the fact that $r_s$ probably can't be any smaller than the Planck length $ell$. Therefore we really shouldn't be taking the limit as $r_srightarrow0$ but as $r_srightarrowell$. Then the double limit is zero.
Another way of approaching the whole thing is to imagine a theory of quantum gravity in which the vacuum has virtual black holes popping in and out of existence. Then presumably one of the things you want from such a theory (which we don't yet possess) is that it doesn't predict infinite density of photons for the vacuum. I would imagine that this would happen because the density of virtual Planck-scale black holes would be small.
answered 1 hour ago
Ben Crowell
44.9k3147274
answered 1 hour ago
Ben Crowell
44.9k3147274
answered 1 hour ago
Ben Crowell
44.9k3147274
answered 1 hour ago
Ben Crowell
44.9k3147274
44.9k3147274
add a comment |Â
add a comment |Â
up vote
1
down vote
Wouldn't this mean that empty space would have infinite energy?
So ignoring quantum issues (and in the absence of a complete theory of quantum gravity we have no choice) and staying strictly with the classical approach let's consider the problem.
Regardless of what amount of power is radiated by the black hole, that power is removed from the energy of the black hole. But the black holes you are talking about have zero energy and so there is no way for them to power hawking radiation.
The mistake you are making is ignoring that nature balances it's books and in this case the balance is that you can't reduce the mass below zero.
There's another reason why your logic is failing.
The entire idea of Hawking radiation depends on the exist of a curved space time and an event horizon. But when $M=0$ we just get a flat spacetime. There is no event horizon. And note that an $R=0$ event horizon would mean there was nothing inside the black hole - no volume, nothing.
The formula you are using for temperature is based on a model which starts out with a non-zero positive mass and then makes a first order approximation close to the event horizon (Wikipedia has a description of this). But that formula does not apply when you're using $Mto 0$. Again, you're using an approximation based on the assumption of a curved spacetime ($M>0$) and applying it outside it's "designed purpose" as an approximation.
answered 19 mins ago
StephenG
5,01821323
add a comment |Â
up vote
1
down vote
Wouldn't this mean that empty space would have infinite energy?
So ignoring quantum issues (and in the absence of a complete theory of quantum gravity we have no choice) and staying strictly with the classical approach let's consider the problem.
Regardless of what amount of power is radiated by the black hole, that power is removed from the energy of the black hole. But the black holes you are talking about have zero energy and so there is no way for them to power hawking radiation.
The mistake you are making is ignoring that nature balances it's books and in this case the balance is that you can't reduce the mass below zero.
There's another reason why your logic is failing.
The entire idea of Hawking radiation depends on the exist of a curved space time and an event horizon. But when $M=0$ we just get a flat spacetime. There is no event horizon. And note that an $R=0$ event horizon would mean there was nothing inside the black hole - no volume, nothing.
The formula you are using for temperature is based on a model which starts out with a non-zero positive mass and then makes a first order approximation close to the event horizon (Wikipedia has a description of this). But that formula does not apply when you're using $Mto 0$. Again, you're using an approximation based on the assumption of a curved spacetime ($M>0$) and applying it outside it's "designed purpose" as an approximation.
answered 19 mins ago
StephenG
5,01821323
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Wouldn't this mean that empty space would have infinite energy?
So ignoring quantum issues (and in the absence of a complete theory of quantum gravity we have no choice) and staying strictly with the classical approach let's consider the problem.
Regardless of what amount of power is radiated by the black hole, that power is removed from the energy of the black hole. But the black holes you are talking about have zero energy and so there is no way for them to power hawking radiation.
The mistake you are making is ignoring that nature balances it's books and in this case the balance is that you can't reduce the mass below zero.
There's another reason why your logic is failing.
The entire idea of Hawking radiation depends on the exist of a curved space time and an event horizon. But when $M=0$ we just get a flat spacetime. There is no event horizon. And note that an $R=0$ event horizon would mean there was nothing inside the black hole - no volume, nothing.
The formula you are using for temperature is based on a model which starts out with a non-zero positive mass and then makes a first order approximation close to the event horizon (Wikipedia has a description of this). But that formula does not apply when you're using $Mto 0$. Again, you're using an approximation based on the assumption of a curved spacetime ($M>0$) and applying it outside it's "designed purpose" as an approximation.
answered 19 mins ago
StephenG
5,01821323
Wouldn't this mean that empty space would have infinite energy?
So ignoring quantum issues (and in the absence of a complete theory of quantum gravity we have no choice) and staying strictly with the classical approach let's consider the problem.
Regardless of what amount of power is radiated by the black hole, that power is removed from the energy of the black hole. But the black holes you are talking about have zero energy and so there is no way for them to power hawking radiation.
The mistake you are making is ignoring that nature balances it's books and in this case the balance is that you can't reduce the mass below zero.
There's another reason why your logic is failing.
The entire idea of Hawking radiation depends on the exist of a curved space time and an event horizon. But when $M=0$ we just get a flat spacetime. There is no event horizon. And note that an $R=0$ event horizon would mean there was nothing inside the black hole - no volume, nothing.
The formula you are using for temperature is based on a model which starts out with a non-zero positive mass and then makes a first order approximation close to the event horizon (Wikipedia has a description of this). But that formula does not apply when you're using $Mto 0$. Again, you're using an approximation based on the assumption of a curved spacetime ($M>0$) and applying it outside it's "designed purpose" as an approximation.
answered 19 mins ago
StephenG
5,01821323
answered 19 mins ago
StephenG
5,01821323
answered 19 mins ago
StephenG
5,01821323
answered 19 mins ago
StephenG
5,01821323
5,01821323
add a comment |Â
add a comment |Â
up vote
0
down vote
When the mass drops down to 228 metric tonnes, that's the signal that
exactly one second remains. The event horizon size at the time will be
340 yoctometers, or 3.4 × 10^-22 meters: the size of one wavelength of
a photon with an energy greater than any particle the LHC has ever
produced. But in that final second, a total of 2.05 × 10^22 Joules of
energy, the equivalent of five million megatons of TNT, will be
released. It's as though a million nuclear fusion bombs went off all
at once in a tiny region of space; that's the final stage of black
hole evaporation.
https://www.forbes.com/sites/startswithabang/2017/05/20/ask-ethan-what-happens-when-a-black-holes-singularity-evaporates/#3d39e14b7c8c
The point is, in reality, M never hits zero.
Theory is great but we can't take it literally.
answered 22 mins ago
CramerTV
20228
add a comment |Â
up vote
0
down vote
When the mass drops down to 228 metric tonnes, that's the signal that
exactly one second remains. The event horizon size at the time will be
340 yoctometers, or 3.4 × 10^-22 meters: the size of one wavelength of
a photon with an energy greater than any particle the LHC has ever
produced. But in that final second, a total of 2.05 × 10^22 Joules of
energy, the equivalent of five million megatons of TNT, will be
released. It's as though a million nuclear fusion bombs went off all
at once in a tiny region of space; that's the final stage of black
hole evaporation.
https://www.forbes.com/sites/startswithabang/2017/05/20/ask-ethan-what-happens-when-a-black-holes-singularity-evaporates/#3d39e14b7c8c
The point is, in reality, M never hits zero.
Theory is great but we can't take it literally.
answered 22 mins ago
CramerTV
20228
add a comment |Â
up vote
0
down vote
up vote
0
down vote
When the mass drops down to 228 metric tonnes, that's the signal that
exactly one second remains. The event horizon size at the time will be
340 yoctometers, or 3.4 × 10^-22 meters: the size of one wavelength of
a photon with an energy greater than any particle the LHC has ever
produced. But in that final second, a total of 2.05 × 10^22 Joules of
energy, the equivalent of five million megatons of TNT, will be
released. It's as though a million nuclear fusion bombs went off all
at once in a tiny region of space; that's the final stage of black
hole evaporation.
https://www.forbes.com/sites/startswithabang/2017/05/20/ask-ethan-what-happens-when-a-black-holes-singularity-evaporates/#3d39e14b7c8c
The point is, in reality, M never hits zero.
Theory is great but we can't take it literally.
answered 22 mins ago
CramerTV
20228
When the mass drops down to 228 metric tonnes, that's the signal that
exactly one second remains. The event horizon size at the time will be
340 yoctometers, or 3.4 × 10^-22 meters: the size of one wavelength of
a photon with an energy greater than any particle the LHC has ever
produced. But in that final second, a total of 2.05 × 10^22 Joules of
energy, the equivalent of five million megatons of TNT, will be
released. It's as though a million nuclear fusion bombs went off all
at once in a tiny region of space; that's the final stage of black
hole evaporation.
https://www.forbes.com/sites/startswithabang/2017/05/20/ask-ethan-what-happens-when-a-black-holes-singularity-evaporates/#3d39e14b7c8c
The point is, in reality, M never hits zero.
Theory is great but we can't take it literally.
answered 22 mins ago
CramerTV
20228
answered 22 mins ago
CramerTV
20228
answered 22 mins ago
CramerTV
20228
answered 22 mins ago
CramerTV
20228
20228
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f431935%2fbased-on-black-hole-thermodynamics-shouldnt-empty-space-contain-infinite-energ%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
If $M$ is zero what is supposed to be radiating (nothing ?) and where does the energy for this radiation come from (nowhere ?) ?
– StephenG
1 hour ago
@StephenG I share your confusion, but the math doesn’t lie
– user189728
1 hour ago
From your link " Unlike most objects, a black hole's temperature increases as it radiates away mass. The rate of temperature increase is exponential, with the most likely endpoint being the dissolution of the black hole in a violent burst of gamma rays. A complete description of this dissolution requires a model of quantum gravity, however, as it occurs when the black hole approaches Planck mass and Planck radius."
– Bruce Greetham
1 hour ago
Hawking's original computation involves quantum fields in curved spacetime, where curvatures are small compared to the Planck length. I believe that in order to take a limit all the way to $M=0$, one must know how Planck scale quantum gravity works. Maybe there's something Hawking says in his article "Black hole explosions?" but I can't open it at the moment: nature.com/articles/248030a0
– Avantgarde
51 mins ago
Newtonian 'black holes' ($c to infty$) also have infinite temperature. But this would be wrong since we derived $T$ using relativistic field theory.
– Avantgarde
29 mins ago