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Posted

Thank you for working out the math.

(acceleration proper makes you into a "pancake", not a "spaghetti")

Why would a large acceleration turn you into a pancake? Wouldn't it accelerate each part of you equally, preserving your shape?
Posted

Thank you for working out the math.

Why would a large acceleration turn you into a pancake? Wouldn't it accelerate each part of you equally, preserving your shape?

Because you would collapse under your own weight. Imagine that you weighed 10 tonnes. Your skieleton would not be able to support you.

Posted

Only if you are on something to support you, which is not the case when falling into a black hole.

True for freefall. Not true for the realistic case of approaching the BH in aby type of spaceship. The electromagnetic force of the ship will work against your enormous weight.

Posted

We don't know what happens inside the EH. We will NEVER know (because no information escapes from beyond the EH).

 

You could. All you need is a little push in the right direction.

 

Don't worry about getting the information out. We'll send in the directions once you're inside... >:D

Posted

Math is fine ZZ, but the mathematical model has to be judiciously applied to the situation, otherwise it leads to non-sensical results like dimensionless points of infinite density.

 

So while its true no outside observer will ever know Bob's fate once he reaches the event horizon, Bob will definitely know.

Posted (edited)

Math is fine ZZ, but the mathematical model has to be judiciously applied to the situation, otherwise it leads to non-sensical results like dimensionless points of infinite density.

What's your point? The math was applied correctly and it reflects the mainstream literature I quoted.

 

 

 

So while its true no outside observer will ever know Bob's fate once he reaches the event horizon, Bob will definitely know.

 

I never claimed otherwise, nor is the discussion about that. It is about where the "spaghettification" occurs: outside the EH vs. inside.

Edited by zztop
Posted

Sure, and your claim was that it occurs OUTSIDE the event horizon, while everyone else said it depends on the mass of the BH.

Or do I need to quote you ?

Posted

Sure, and your claim was that it occurs OUTSIDE the event horizon, while everyone else said it depends on the mass of the BH.

Or do I need to quote you ?

Can you follow the math that compares [math]r[/math] where the spaghettification occurs with the location of the EH? I think I had two very detailed posts on the subject. I also posted the wiki references that show the case where spaghettification occurs way outside the EH. I suggest that you re-read them.

Posted (edited)

Yes, I just re-read your post #6, and again in #8.

No reference to size of the EH at all.

Do you have selective memory ???

Edited by MigL
Posted

Yes, I just re-read your post #6, and again in #8.

No reference to size of the EH at all.

Do you have selective memory ???

[math]r_s[/math] is the radius of the EH.

Posted (edited)

True for freefall. Not true for the realistic case of approaching the BH in aby type of spaceship. The electromagnetic force of the ship will work against your enormous weight.

In case of the spaceship, the force compressing you depends entirely on the spaceship, and not on the bh. Any realistic spaceship would be freefalling with you, or perhaps resisting slightly.

 

More on topic: I think zztop was very clear about calculating for which size of bh the spaghettification happens inside or outside the eh.

Edited by Bender
Posted

In case of the spaceship, the force compressing you depends entirely on the spaceship, and not on the bh. Any realistic spaceship would be freefalling with you, or perhaps resisting slightly.

.

The "resisting slightly" is what determines the amount of "pancaking".

 

 

 

More on topic: I think zztop was very clear about calculating for which size of bh the spaghettification happens inside or outside the eh

Yes, I did. This is why the language of physics is math.

Posted

The "resisting slightly" is what determines the amount of "pancaking".

 

Yes, I did. This is why the language of physics is math.

Yes, but would not the resistance depend on the thrust of the ship, in which case you'd get the same pancaking effect regardless of whether there is a black hole involved or not?

Posted

Yes, but would not the resistance depend on the thrust of the ship, in which case you'd get the same pancaking effect regardless of whether there is a black hole involved or not?

Sure, if you could find a spaceship capable of a 1000g+ thrust.

Posted

The point was generating the effect in the absence of the BH.

There is also no pancaking in the presence of the BH.

Posted (edited)

There is also no pancaking in the presence of the BH.

Incorrect, the EEP teaches you that one can get 1000g from the combined effects of a BH with 800g and a spaceship of 200g (opposite sense of vectors). In the absence of the BH the gravitational effect is only 200g. This thread is degenerating from science into trolling.

Edited by zztop
Posted

What is EEP? I'm not a native English speaker and I'm unfamiliar with that acronym.

 

If the spaceship can only produce 200g, the force pancaking you can never exceed your mass, multiplied with 200g. The acceleration of the BH simply has no effect, since it applies to both you as to the spaceship. (of course, 200g is enough to do a decent amount of pancaking, but I don't think that is the point here.)

Posted

True for freefall. Not true for the realistic case of approaching the BH in aby type of spaceship. The electromagnetic force of the ship will work against your enormous weight.

Um, what? The ship will also be in freefall.

Posted (edited)

Um, what? The ship will also be in freefall.

The ship as a 200g acceleration wrt to some frame of reference, in the absence of any gravitating body.

According to the Equivalence Principle, a body inside the spaceship is subjected to a hravitational acceleration of 200g.

You now add in a gravitating body that generates a gravitational field of 800g. The Equivalence Principle tells you that the effective acceleration exerted on the ship is now 1000g.

Edited by zztop
Posted

The ship as a 200g acceleration wrt to some frame of reference, in the absence of any gravitating body.

According to the Equivalence Principle, a body inside the spaceship is subjected to a hravitational acceleration of 200g.

You now add in a gravitating body that generates a gravitational field of 800g. The Equivalence Principle tells you that the effective acceleration exerted on the ship is now 1000g.

200g is a condition you put in later, not part of the discuscussion in the post I responded to.

 

So absent this newly positioned goalpost, there is no effect from the acceleration.

Posted

The ship as a 200g acceleration wrt to some frame of reference, in the absence of any gravitating body.

According to the Equivalence Principle, a body inside the spaceship is subjected to a hravitational acceleration of 200g.

You now add in a gravitating body that generates a gravitational field of 800g. The Equivalence Principle tells you that the effective acceleration exerted on the ship is now 1000g.

Only if the ship is accelerating away from the gravitating body with the equivalent of 200g. And in an 800g gravitational field, that requires... a 1000g acceleration.

Posted (edited)

200g is a condition you put in later, not part of the discuscussion in the post I responded to.

 

So absent this newly positioned goalpost, there is no effect from the acceleration.

The discussion is about the contribution of the BH gravitational field, not of the spaceship acceleration.

Since several of you seem not to understand the concept, I can give a rigorous proof that a rod falling radially into a BH gets stretched due to the gravitational field of the BH.

Only if the ship is accelerating away from the gravitating body with the equivalent of 200g.

That is precisely what I said.

 

 

 

And in an 800g gravitational field, that requires... a 1000g acceleration.

 

(Proper) acceleration is absolute, meaning that it isn't relative to the gravitating body.

Looks like (at least) three of you have some severe misconceptions about the fact that not only the acceleration gradient but the acceleration proper causes objects' deformation during radial freefall. A short mathematical explanation might help. Would you be interested in learning something new?

Edited by zztop

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