B Speed of light and this Accelerating Rocket scenario

[John SpaceY]

I get $v=0.91 \ c$

Yes sorry I have made an error in my calculation ...
I find now 0,91c
We see here that the speed of the spacecraft will increase thanks to the external gravity MG to m
And so, the acceleration is not zero ...
I agree that the acceleration will be zero when v will tend towards c and so the speed will never be higher than c but before reaching c the external gravity will increase the speed up to c.
Now the question is : if the spacecraft continu to go closer to MG, the speed is already to its maximum c, and what will be the effect of the external gravity in this case ? the external gravity reduces the time inside the spacecraft but the time is already at zero and so what is the next step ?
The equation before is only OK when r > rs but when r = 10 rs we are higher than rs and we are also higher than the radius of Sagittarius A for example (the radius of Sagittarius A is 12 millions of km and its rs value is around 3 millions of km).

 

[Ibix]

And so, the acceleration is not zero

The coordinate acceleration is non-zero given the particular coordinate choice I made. It would be zero in other coordinate systems. The proper acceleration is zero independent of any choice I make. You will remain confused about what we are saying until you understand that there are two separate concepts here.

 

[Dale]

And so, the acceleration is not zero ...

The coordinate acceleration is not 0 in the Schwarzschild coordinates. The proper acceleration is still 0, and the coordinate acceleration still must be specified relative to given coordinates.

Now the question is : if the spacecraft continu to go closer to MG, the speed is already to its maximum c, and what will be the effect of the external gravity in this case ? the external gravity reduces the time inside the spacecraft but the time is already at zero and so what is the next step ?

The next step (and the previous several steps) is to learn chapters 1-3 of the lecture notes I posted for you earlier.

Please do not skip that this time.

 

[John SpaceY]

The coordinate acceleration is not 0 in the Schwarzschild coordinates. The proper acceleration is still 0, and the coordinate acceleration still must be specified relative to given coordinates.The next step (and the previous several steps) is to learn chapters 1-3 of the lecture notes I posted for you earlier.

Please do not skip that this time.

Your reasoning makes no sense because you are still trying to treat General Relativity as if it were Newtonian gravity with a few tweaks, rather than the reality that it is a completely different model with a completely different approach to most things. In particular, it doesn't model gravity as a force at all and you have no justification for your claim that $dr$ goes to zero as $v$ goes to $c$ (I suspect you've done some horrible twisting of length contraction, which doesn't apply here). However, $v<c$ is the correct conclusion.

The correct calculation yields that if the ship has initial velocity far from the black hole $v_i$ and you are hovering at radial coordinate $r$ above the black hole, the ship will pass you at $c\sqrt{(c^2r_s-(r_s-r)v_i^2)/(c^2r)}$, where $r_s$ is the Schwarzschild radius. Note that the calculation used to get there is only valid for $r>r_s$, and that the square root is less than 1 in that range. Note also that I'm being very careful to say that this is not a general expression for the speed of the ship (there isn't any such thing), but specifying that it's the speed measured at close range by a specific observer.

Hello Ibix,
I am French and I don't understand all the words and I have a question here :
when you write "you are hovering at radial coordinate r above the black hole" you want to say that the speed of the spacecraft is in the direction of the center of the black hole or the spacecraft is turning around the black hole at a distance r ?
For me I understand that the initial speed vi is in the direction of the center of the black hole and the new speed v is also in the same direction.

If the initial speed vi is in the direction of the center of the black hole and the new speed v is tangential to the black hole (the space craft is turning around the black hole) there is a change in the direction and I would like to understand why there is this change in the direction ? and what force has created this ?

 

[John SpaceY]

And also the equation is only OK for r > rs
Do you know what will happen for r < rs when the speed is near to c when r will arrive to rs ?
I understand it is not possible to go to rs, because there is some matter at r = rs, but it is for a theoretical calculation
Thanks in advance

 

[Nugatory]

when you write "you are hovering at radial coordinate r above the black hole" you want to say that the speed of the spacecraft is in the direction of the center of the black hole or the spacecraft is turning around the black hole at a distance r ?

Neither. A hovering object is one that that is firing a rocket straight down with enough force to keep $r$ (which must be greater than $R_S$ for hovering to be possible) fixed so neither climbing up out of the gravity well nor falling deeper in.
If we are dealing with a solid planet that is not a black hole, then the solid surface will be at some $r$ greater than $R_S$ so it is possible to "hover" just by standing on the surface; the upwards force of the ground on our feet works just as well as a rocket to keep us in the same place, neither falling inwards nor climbing up into the sky.

Note that a hovering observer will have non-zero proper acceleration.

 

[Ibix]

Do you know what will happen for r < rs when the speed is near to c when r will arrive to rs ?

This analysis isn't possible at or below $r_s$. You can do a different analysis and get any speed you like (as you can outside the event horizon - this is the whole point).

I understand it is not possible to go to rs,

There's no problem going to the Schwarzschild radius. You just can't stay there, so my analysis (which is based on the view of a hovering observer) isn't valid there because the hovering observer can't exist, and assuming it does leads to a number of contradictory in the maths.

 

[John SpaceY]

The coordinate acceleration is not 0 in the Schwarzschild coordinates. The proper acceleration is still 0, and the coordinate acceleration still must be specified relative to given coordinates.The next step (and the previous several steps) is to learn chapters 1-3 of the lecture notes I posted for you earlier.

Please do not skip that this time.

Hello Dale,
Thanks for the information
I have tried several times to read those chapters but I am always blocked by several points : the 2 main points are the following :
I am not familiar with the symbols used there (tensors, metric tensor, stress energy tensor, ...) and so I don't understand the main points. And there is no simple examples with numerical applications : I understand things only when I see numerical applications and it is the reason why I ask you all these questions with numerical applications.
The other point is that equations are only valid for r > rs and so for me there is something which is missing and so probably there is something missing in these equations but I don't know how to explain this.
Maybe I should not have read that Einstein himself has said that he has some doubt about what he has added on Gravity on the GR compared to the SR but I don't understand what he was talking about.
But I will try again to try to understand the chapters 1-3 by first trying to understand what is beyond the symbols
But thanks Dale again for the information

 

[John SpaceY]

Neither. A hovering object is one that that is firing a rocket straight down with enough force to keep $r$ (which must be greater than $R_S$ for hovering to be possible) fixed so neither climbing up out of the gravity well nor falling deeper in.
If we are dealing with a solid planet that is not a black hole, then the solid surface will be at some $r$ greater than $R_S$ so it is possible to "hover" just by standing on the surface; the upwards force of the ground on our feet works just as well as a rocket to keep us in the same place, neither falling inwards nor climbing up into the sky.

Note that a hovering observer will have non-zero proper acceleration.

Thanks Nugatory for the information
What I understand now is that the observer is hovering and the spacecraft will continue to go in the direction of the external gravity MG and this observer will see a speed of the spacecraft equal to 0,91c

 

[Dale]

I am not familiar with the symbols used there (tensors, metric tensor, stress energy tensor, ...) and so I don't understand the main points. And there is no simple examples with numerical applications : I understand things only when I see numerical applications and it is the reason why I ask you all these questions with numerical applications.

Then please ask questions about those. You will need to understand those concepts in order to understand GR and even properly formulate the questions you are asking so that they can be answered.

You need to not try to skip this effort if you are actually interested in the questions you have posed here.

The other point is that equations are only valid for r > rs

That is not true. The equations in those chapters 1-3 are valid everywhere.

But I will try again to try to understand the chapters 1-3 by first trying to understand what is beyond the symbols

Please ask for help on that as needed here. But do that prior to proceeding on the topics of this thread.

 

[Ibix]

The other point is that equations are only valid for r > rs and so for me there is something which is missing and so probably there is something missing in these equations but I don't know how to explain this.

The analysis I did cannot be extended below the horizon because it uses hovering observers. You can't hover at or below the horizon, so the analysis doesn't work. Plenty of other things can be considered below the horizon, but not this. Basically, that's because it's something like the Newtonian-style anaysis you were attempting to do, and reality is so different from Newton's conception of it once you get inside the black hole that there is no way to make it look Newtonian.

You need to learn how GR works if you want to understand black holes. Bodging something kind-of-Newtonian and crossing your fingers just won't work.

 

[John SpaceY]

This analysis isn't possible at or below $r_s$. You can do a different analysis and get any speed you like (as you can outside the event horizon - this is the whole point).

There's no problem going to the Schwarzschild radius. You just can't stay there, so my analysis (which is based on the view of a hovering observer) isn't valid there because the hovering observer can't exist, and assuming it does leads to a number of contradictory in the maths.

Thanks Ibix for the comment
What seems strange to me is that we theoretically go to rs and so we could go to r < rs but there is no Theory when r < rs
If we look at the time inside the spacecraft, if the speed of this spacecraft is 0 when r = rs, this time t inside the spacecraft will tend towards 0, only because of the gravity and its contraction factor.
But we can go to r < rs and so what will happen to the parameters inside the spacecraft ? (time, speed seen by an outside observer, ...).

I have read somewhere that when r < rs the time will become negative and the speed could be higher than c but I don't see any concrete Theory on these points and so I try to understand.
You have written also that we will find contradictory in the maths and so I think that some points are missing in the Theory today and I search to understand.

For example If I consider to be with a spacecraft of mass m which is at r higher than rs and far away the surface of an external gravity to m (but not too far : for example at 10 rs) and if this spacecraft has a speed very near of c, the time inside the spacecraft will tends towards 0 : and for me the spacecraft will be in the same situation (seen from inside) than when this spacecraft was at rs at 0 speed.
But the big difference is that at rs we can go physically to r < rs (we can pass the point) and at r higher than rs, we can not go higher than c : we cannot theoretically pass a limit
And this is strange for me : maybe something is missing in the current Theory and I don't know what and I cannot explain and so I am searching some ideas to clarify all this.

In the current Theory, by adding the effect of a very high speed and a very high external gravity to m, this mass m cannot go at a speed higher than c and nothing can help to pass the limit : there is no rs that we could pass (to go at r < rs for exmaple) and so pass the limits. And I am tryning to find an equivalent rs but far away from the rs of a planet where there is mater around and a very high gravity and a very high temperature and pressure and ...

I thought that having a very high speed (near to c) and going in the direction of an external gravity and being very near of this gravity, we could pass the limit (maybe by changing the direction of the spacecraft of mass m to accelerate it or ...) but I don't find anything that can explain this theoretically and so ?

If we find something that could pass the limit when r > rs we could understand what will be the evolution of parameters (t, speed, ...) when r < rs and also the reverse is possible ! if we can understand what will be the evolution of parameters when r < rs we could find what is missing when r > rs "to pass the limits".

And so maybe you have some information when r < rs ?
what are the contradictories in the math for example ?
Have you heard about negative time inside m and speed higher than c for m when r < rs ?
And do you have some start of theories on these points ?

Thanks in advance

 

[PeterDonis]

there is no Theory when r < rs

Yes, there is. It just doesn't work the way you are thinking.

if the speed of this spacecraft is 0 when r = rs

It isn't.

I have read somewhere

Where?

when r < rs the time will become negative

This is wrong.

and the speed could be higher than c

This is wrong.

I don't see any concrete Theory on these points

Then you need to spend time learning it. You can do so from pretty much any GR textbook. But, as @Dale has repeatedly pointed out, you can't skip steps; you need to build your understanding from the ground up, starting from the basics, otherwise you won't understand the theory. We can't help you if you are not willing to do that.

 

[Ibix]

What seems strange to me is that we theoretically go to rs and so we could go to r < rs but there is no Theory when r < rs

That would be strange if it were true...

 

[Dale]

there is no Theory when r < rs

I told you that this is not true. I gave you the theory. I offered to help you understand the theory. And I told you in no uncertain terms that you need to do that first.

Thread closed for moderator discussion of next steps.

EDIT: after a very brief discussion this thread will remain closed

 

[Nugatory]

have tried several times to read those chapters but I am always blocked by several points : the 2 main points are the following :
I am not familiar with the symbols used there (tensors, metric tensor, stress energy tensor, ...) and so I don't understand the main points. And there is no simple examples with numerical applications

In that case you may find the No-nonsense Introduction to be an easier starting point.

If you find section 2 of that introduction difficult to follow (as opposed to the hoped-for reaction: “Wow - what an excellently terse summary of the geometrical approach to special relativity!”) you may need to back up even more and develop a better understanding of special relativity before taking on the general theory. For that I recommend Taylor and Wheeler’s “Spacetime Physics” - the first edition which I learned from is free on the internet.

Starting with worked numerical examples will be ineffective because many of the quantities that you want to see calculated (“acceleration”, “gravity”, distance”, “velocity”, “force”, ….) don’t mean what you think they do. Our familiar understanding of these concepts are applicable only under the familiar special conditions of flat spacetime. Attaching a number to, for example, a “velocity” will only confuse if you have not first carefully considered what “velocity” means.