Spaghettification near event horizon

[Stephanus]

Still wrong. If you're getting any answer other than the one I gave, you're doing it wrong. This calculation has been done thousands of times by physicists and students in GR. Please consult a textbook.
[..]
$$
a = \frac{GM}{r^2 \sqrt{1 - \frac{2GM}{c^2 r}}} = \frac{GM}{r \sqrt{r^2 - \frac{2GM}{c^2} r}} = \frac{G M c}{r \sqrt{c^2 r^2 - 2 G M r}}
$$

Plugging in $r = 2GM/c^2$ then gives

$$
a = \frac{G M c}{\left( \frac{2GM}{c^2} \right) \sqrt{c^2 \left( \frac{2GM}{c^2} \right)^2 - 2 G M \left( \frac{2GM}{c^2} \right)}} = \frac{G M c^3}{2 G M \sqrt{ \frac{4 G^2 M^2}{c^2} - \frac{4 G^2 M^2}{c^2} }} = \frac{c^4}{4 G M \sqrt{1 - 1}}
$$

Note that if we choose units so $c = 1$ and $G = 1$, this is identical to what I gave before. Also, of course, it is undefined.

Come on PeterDonis.
Is this wrong?
$\frac{c^4}{4GM\sqrt{1-c^2}}$
and yours
$\frac{c^4}{4GM\sqrt{1-1}}$
But I didn't change c to 1. Still needs time to understand why we can substitute c with 1 (or even G with 1).
I made some error in calculating 2GM^2.
$2GM^2 = 2G^2M^2$ it should be $4G^2M^2$ that's why I make a wrong calculation.
And it's not $2GM^2$ it's $(2GM)^2$.

 

[A.T.]

Still needs time to understand why we can substitute c with 1 (or even G with 1).

https://en.wikipedia.org/wiki/Geometrized_unit_system

 

[Stephanus]

https://en.wikipedia.org/wiki/Geometrized_unit_system

Wow, thanks for the link

 

[Stephanus]

[..]Because that just amounts to a choice of units. You already understand what it means to set $c = 1$; it just means we're measuring speeds in units of $c$, so the speed of light is $1$. Setting $G = 1$ means we are measuring mass in units of length; $G / c^2$ is just a conversion factor from ordinary units of mass to mass in length units. So the mass of the Earth, for example, instead of being about $6 \times 10^{24}$ kg, is about 4.5 millimeters in these units (called "geometric units").[..]

Okay, let's see if I can change G =1 and c = 1
$G = 6.674×10−11 N⋅m2/kg2$
in unit
$G = (kg⋅\frac{m}{s^2}) ⋅m2/kg2$
$G = \frac{m^3}{s^2⋅kg}$

$c = 300,000,000m/s$
in unit
$c = m/s$
$\frac{G}{c^2} = \frac{6.674⋅10^{-11}\frac{m^3}{s^2⋅kg}}{\frac{9⋅10^16m^2}{s^2}}$

$\frac{G}{c^2} = \frac{6.674⋅10^{-11}m}{9⋅10^{16}⋅kg}$

$\frac{G}{c^2} = \frac{7.41556⋅10^{-28}m}{kg}$

Earth mass = 6 x 10²⁴ Kg
I don't even know why we should multiply or divide Earth mass with G/c2
Okay, I choose multiply
$6 * 10^{24} Kg * \frac{7.41556⋅10^{-28}m}{kg}$
$6 * 10^{24} * 7.41556⋅10^{-28}m$
$44.49 * 10^{-4}m = 4.449 mm$ Okay...
I imagine we divide mass of the Earth with this unit, not multiply it.
Perhaps it's the same if we have a car that can runs 100km/s
And we say, what is 20 seconds according to you.
Then we multiply 20 seconds with Car_speed, then we'll say 20 seconds for my car is 200 km.
Or perhaps
There are two towns, 500 km apart. What is the distance of the towns.
Then we divide 500 km with Car_speed, then we'll say the distance of the towns is 5 seconds for my car.
I have to think it over, still don't get it fully.

 

[Stephanus]

I think G/c^2 is half of Schwarzschild radius.
If we multiply the mass of an (planetary) object with this unit, we'll get acceleration at that distance. But distance must in the format of
Distance/300000000
if We multiply the mass of that object with twice of this unit, we'll get its schwarzschild radius
I'll try to understand it deeper.

 

[PeterDonis]

Is this wrong?

Yes. Look at it; it's not the same as what I wrote, is it? That means it's wrong.

But I didn't change c to 1.

Neither did I in the equation you referred to here. In the post you quoted, I did the entire derivation without changing $c$ to 1, leaving it as $c$ in all the formulas. And I got $c^4 / 4 G M \sqrt{1 - 1}$; not $c^4 / 4 G M \sqrt{1 - c^2}$. So what you wrote is wrong. Go back and check your algebra again. Or look carefully at what I posted and see how it goes, step by step.

I imagine we divide mass of the Earth with this unit, not multiply it.

Why would you imagine that, since you just got the right answer by multiplying? You just figured out the right answer, and now you're throwing it away?

I think G/c^2 is half of Schwarzschild radius.

No, it isn't. It is a conversion factor from ordinary units to geometric units. Think of it as $G / c^2$ meters per kilogram. You have a number in kilograms, like the mass of the Earth; you want to know what it is in meters; so you multiply the number in kilograms by the conversion factor, $G / c^2$ meters per kilogram, to get the number in meters.

If you want to say that, for a given mass $M$, the distance $GM / c^2$ is half the Schwarzschild radius of a black hole of mass $M$, that would be correct. But the distance $GM / c^2$ can be used much more generally than just in a scenario with black holes. In GR, that distance is the mass in geometric units, the same way the distance $ct$ is the time $t$ in natural relativistic units.

If we multiply the mass of an (planetary) object with this unit, we'll get acceleration at that distance.

I have no idea where you're getting this from. It's wrong. You don't get an acceleration, you get a distance. See above.