"If a black hole has more mass then the event horizon shrinks" ???

[martix]

The title is a direct quote of this video by Dr. Becky Smethurst, an astrophysicist specializing in black hole research.

This is a mistake, right?

Supermassive black holes, for example, don't have tiny radii, compared to stellar mass BHs.

Then there's the equation she presents seconds later.

The Kerr solution has an extra term in the form of 1 + sqrt(...). The square root is always positive, which means the extra term is ≥1, meaning the result (i.e. event horizon radius) is always greater than the Schwarzschild radius.

Furthermore, she says "when a equals zero [...] that whole equation reduces back to Schwarzschild's radius", which if you substitute a=0, is clearly not the case.

But there are no pinned comments or the description ackowledging any kind of error. Yet she says is so emphatically, and multiple times and it makes me wonder if I missed some detail?

 

[Ibix]

Furthermore, she says "when a equals zero [...] that whole equation reduces back to Schwarzschild's radius", which if you substitute a=0, is clearly not the case.

This is correct, although depending on the coordinate choice it may not be obvious.

Don't know about the rest and I'm on a train with a patchy internet connection, so not watching videos at the moment. It would be helpful to state the maths if you can.

 

[PeroK]

The title is a direct quote of this video by Dr. Becky Smethurst, an astrophysicist specializing in black hole research.

This is a mistake, right?

Supermassive black holes, for example, don't have tiny radii, compared to stellar mass BHs.

Correct. Never trust anyone who says "times by" instead of "times" or "multiplied by"!

Then there's the equation she presents seconds later.

The Kerr solution has an extra term in the form of 1 + sqrt(...). The square root is always positive, which means the extra term is ≥1, meaning the result (i.e. event horizon radius) is always greater than the Schwarzschild radius.

Furthermore, she says "when a equals zero [...] that whole equation reduces back to Schwarzschild's radius", which if you substitute a=0, is clearly not the case.

This video is correct. The Schwarzschild radius is $R_e = \frac{2GM}{c^2}$.

 

[martix]

Correct. Never trust anyone who says "times by" instead of "times" or "multiplied by"!

This video is correct. The Schwarzschild radius is $R_e = \frac{2GM}{c^2}$.

I'm a bit confused as to which statements you're adressing.
For the first response it seems fairly clear that you are agreeing the title statement is a mistake.

However for the second, you say the video is correct, without saying which part.
Here are the things that confuse me:

The Kerr solution equation that comes up on screen
1. contradicts the title statement (honest mistake in the script, fair enough)
2. contradicts the "when a equals zero [...]" statement later (plain contradiction with the equation - but unclear which is wrong - the statement or equation)
3. kind of doesn't make sense to me? a looks like the fraction contribution of momentum to a black hole's mass. So it kind of makes sense for it to be $R_e = \frac{2GM}{c^2}(1+\sqrt{a^2}) = \frac{2GM}{c^2}(1+a)$, in which case a=0 means non-rotating black hole, reducing to Schwarzschild's equation, and when a>0, we have a positive contribution of momentum to the black hole's mass.

 

[Ibix]

The Kerr metric reduces to the Schwarzschild metric in the $a=0$ case, as I said before. This is easy enough to confirm in Boyer-Lindquist coordinates - the wikipedia page on the Kerr metric has the maths if you wish to confirm.

If the video uses other coordinates the result may not be so obvious. Or the maths may be wrong. The statement isn't wrong, though.

 

[PeroK]

I'm a bit confused as to which statements you're adressing.
For the first response it seems fairly clear that you are agreeing the title statement is a mistake.

Yes. It's difficult to know whether it was a slip of the tongue. That said, her obvious hand gestures as she says the word "shrink" are difficult to explain. You should get in touch with her.

 

[martix]

The Kerr equation, from the video is as follows:

$R_e = \frac{2GM}{c^2}(1+\sqrt{1 - a^2})$

and I could not find it in that form on wikipedia.

In the above equation, when a=0 you get $R_e = \frac{2GM}{c^2}*2$, which is clearly not the same as the Schwarzschild radius. At the same time, on wikipedia there is something that looks similar: $r_{\rm H}^{\pm} = 1 \pm \sqrt{1-a^2}$, but I lack the background necessary to parse the math.

Yes. It's difficult to know whether it was a slip of the tongue. That said, her obvious hand gestures as she says the word "shrink" are difficult to explain. You should get in touch with her.

I suppose there are contact details on her youtube channel. I could try, but I'm not sure what kind of response I ought to expect regarding questions about a 3 month old video from some random person on the internet.

 

[Ibix]

That appears to be a typo, then. The correct form is$$r_\pm=\frac{GM}{c^2}\pm\sqrt{\left(\frac{GM}{c^2}\right)^2-a^2}$$The $r_H^\pm$ formula you quote appears to be the same but expressed in geometric units, where $G=M=c=1$ and $a$ is dimensionless.

The positive version reduces to the Schwarzschild radius in the case $a=0$. The negative version reduces to zero, which tells you that the inner horizon isn't a feature of Schwarzschild spacetime.

 

[PeroK]

The Kerr equation, from the video is as follows:

$R_e = \frac{2GM}{c^2}(1+\sqrt{1 - a^2})$

No, it's not. In the video she gives $R_e = \frac{GM}{c^2}(1+\sqrt{1 - a^2})$

 

[PeroK]

I suppose there are contact details on her youtube channel. I could try, but I'm not sure what kind of response I ought to expect regarding questions about a 3 month old video from some random person on the internet.

Someone already made that comment in any case.

 

[martix]

No, it's not. In the video she gives $R_e = \frac{GM}{c^2}(1+\sqrt{1 - a^2})$

Oh crud, I did not see the 2 in there and blindly copied your latex... nor in the original equation. Things suddenly make a lot more sense now. So the Kerr equation is probably correct and I've been misreading the Schwarzschild variant this whole time.

Tho the "shrinks" part still seems a mistake.

 

[Ibix]

Tho the "shrinks" part still seems a mistake.

It decreases for increasing $a$ and fixed $M$ (although do note that you can't increase $a$ without increasing $M$, so this is kind of apples and oranges).