Can kinetic energy lead to black hole?

[Stephanus]

Dear PF Forum,
I have a question in mind. But I'm not sure if this belong to SR forum, cosmology or classical physics.
So I post it here.
And perhaps as some of you might have known before or thought it over. It's about kinetic energy.
Supposed this...
A rocket, with a rest mass 1 ton.
And the rocket can somehow grab matter and anti matter in interstellar medium. I know that anti matter just can't scattered freely in interstellar medium, but let's just dispense with the technical difficulties.
So in its journey the rocket always accelerates 1 g. And in doing so it has consumed, say..., 10 solar mass of matter and anti matter annihilation as its energy source.
Now after some times its speed is 99.9999% the speed of light (or should I add more nines?)
Then this 1 ton rocket hits a neutron star, say 1.5 solar mass. Can this neutron star become a black hole if it is hit by this rocket, considering this 1 ton rocket relativistic mass must be about, say 3 solar mass (or more/less?), after spending much of its energy in the form of momentum to eject its propelant?
Thanks for any clarification.

 

[Doc Al]

See: If you go too fast, do you become a black hole?

 

[Stephanus]

See: If you go too fast, do you become a black hole?

Yes, thanks Doc Al for the link. I forgot I have read that link moments ago. And I read it again. But the situation is different here.
The rocket stop now with all its potential energy. It is stopped by a neutron star. And I read that the maximum size of a neutron star is about 2 or 3 solar mass before it become a black hole. So even the mass of the rocket is 1 ton, it does spend a lot of energy to accelerate it near the speed of light wrt the neutron star right?
I can visualize if a rocket travels near the speed of light wrt me, than I can't perceive it as a black hole. But, what if the rocket is stopped by other object with all of its potential kinetic energy? Is the situation the same as the link you give me?

 

[Doc Al]

But the situation is different here.

Yes, your situation is a bit different. I'll let others chime in.

 

[pervect]

I believe the basic answer is yes. As discussed in http://link.springer.com/article/10.1134/S2070046612020069#page-1, if you have in theory, two hyper-relativistic particles colliding, they can form a black hole due to their kinetic energy even though neither one of the particles is itself a black hole.

We can then imagine an appropriate transform to a frame where one of the particles is at rest to get something reasonably similar to the situation in your question, a collision between a moving object and a stationary one creating a black hole.

The gravitational field solution of a hyper-relativistic particle in the extreme ultra-relativistic limit is called the Aichelburg-sexl ultraboost. https://en.wikipedia.org/w/index.php?title=Aichelburg–Sexl_ultraboost&oldid=551268593. So the process of colliding two hyper-relativistic particles can be "conveniently" modeled as the collision of two such "ultraboost" gravitational plane waves.

I've never seen the detailed calculations of such a collision, but my recollection and the paper I cited above support the idea that such collisions are known to form singularities and black holes.

 

[PeterDonis]

Can this neutron star become a black hole if it is hit by this rocket, considering this 1 ton rocket relativistic mass must be about, say 3 solar mass (or more/less?), after spending much of its energy in the form of momentum to eject its propellant?

You've stated the situation vaguely, so I'm not sure if your intent was that the rocket, in its travels, had collected 3 solar masses' worth of rest mass, or just that its relativistic mass (i.e., total energy) was 3 solar masses, prior to colliding with the neutron star.

In the first case (3 solar masses' worth of rest mass prior to collision), the rocket might well have become a black hole purely by absorbing enough matter, even before colliding with the neutron star. The only way it could have avoided doing that would have been to turn into something like a neutron star itself (or else igniting into an ordinary star).

In the second case (3 solar masses' worth of total energy prior to collision), the rocket's rest mass could still be much smaller than that of a star (or even much smaller than a planet, if it's moving fast enough). But what happens in the collision doesn't depend on the rocket's rest mass; it depends on its total energy and momentum. It isn't what you're imagining; see below.

The rocket stop now with all its potential energy.

No, it won't. We'll stick to the second case above, where we have a neutron star at rest with a rest mass 3 solar masses, being struck by a rocket (plus a bunch of accumulated debris) moving so fast that its total energy is 3 solar masses. We assume that after the collision, we end up with a single final object.

The first thing we need to figure is the rocket's momentum. We don't know, from your statement of the problem, exactly how much rest mass the rocket has, but since it is moving at a highly relativistic velocity, we can assume that it's much, much less than its total energy. So to a first approximation, its momentum will be very close to its total energy, i.e., about 3 solar masses' worth of momentum.

Since momentum is conserved, we must also have 3 solar masses' worth of momentum after the collision. Assuming, again, that the rocket's rest mass is negligible, the final object will have a rest mass approximately equal to that of the original neutron star, 3 solar masses. If it also has a total momentum of 3 solar masses, then its total energy is $\sqrt{m^2 + p^2} = 3 \sqrt{2}$ solar masses. In other words, the final object ends up with a relativistic $\gamma$ factor of $\sqrt{2}$, which corresponds to a speed $v$ of $1 / \sqrt{2} \approx 0.7$ times the speed of light. This certainly can't be described as the rocket being "stopped" by the neutron star.

As for whether the final object would be a black hole, it's hard to say. Some energy would probably be radiated away in the collision, so it could even be that the final object's rest mass would be smaller than that of the original neutron star. Even if it turned out to be larger, whether a black hole was formed or not would depend on how close the original neutron star was to the maximum neutron star mass. It would also depend on other details like how much the neutron star's structure was compressed by the collision process; it's possible that that could form a black hole even if the final object's mass was below the maximum mass limit for a neutron star.

In short, it is certainly possible for a black hole to be formed in this type of process, but I don't think we can say very much about the detailed conditions that would determine whether one does in a specific case.

 

[PAllen]

If you make the problem extreme enough, you don't need to worry about the dynamics Peter pointed to, if one is willing to assume the (unproven but plausible) hoop conjecture. Then, if in the COM frame, the total energy of neutron star + rocket would correspond to a Schwarzschild radius larger than the neutron star + rocket, then a black hole will form. Note, this is a substantially higher threshold than the Chandrasekhar limit.

 

[Vanadium 50]

Stephanus, as usual you have way, way, way overcomplicated the situation with all this talk of antimatter and the interstellar medium.

Your question is whether an inelastic collision between an object of mass m1 at rest and one of mass m2 with momentum p can produce an object of invariant mass m3. That's a straightforward question in relativistic mechanics, and depending on m1, m2 and p, it could end up either above or below m3.

 

[Stephanus]

Thanks PeterDonis for your answer.

You've stated the situation vaguely, so I'm not sure if your intent was that the rocket, in its travels, had collected 3 solar masses' worth of rest mass, or just that its relativistic mass (i.e., total energy) was 3 solar masses, prior to colliding with the neutron star.

What I mean is this. Supposed there are anti matters and matters enough in interstellar medium. And every time the rocket "grabs" say 1 mili gram (or gram?) matter/anti matter and annihilate them to propel the rocket. After some times, the rocket has "grabbed" perhaps tens of solar mass matter and anti matter.
Now, I was just thinking. To accelerate the rocket to that velocity, the rocket has consumed tremendous amount of energies.
Its kinetic energy $E_k = \frac{1}{2}mv^2$ is bigger than $E = mc^2$?? So, the energy of this 1 ton rocket is bigger than $1 ton * c^2$ I guess. That's what I mean. That's my answer to your above paragraph. I have read your next paragraphs, there are good answers there.

The first thing we need to figure is the rocket's momentum. We don't know, from your statement of the problem, exactly how much rest mass the rocket has,

Come on...

Dear PF Forum,
I have a question in mind. [..]
Supposed this...
A rocket, with a rest mass 1 ton.
And the rocket can somehow grab matter and anti matter in interstellar medium. [..]

[..]but since it is moving at a highly relativistic velocity, we can assume that it's much, much less than its total energy. So to a first approximation, its momentum will be very close to its total energy, i.e., about 3 solar masses' worth of momentum.

Can kinetic energy be bigger than E=mc2?

Since momentum is conserved, [..]then its total energy is $\sqrt{m^2 + p^2} = 3 \sqrt{2}$ solar masses.

What is "p" here? "mv"?

In other words, [..]This certainly can't be described as the rocket being "stopped" by the neutron star.

Yeah, there's no irresistible force and there's no immovable object either. I think if we fire a bullet right in the heart of a rigid body like neutron star, the neutron star will move at least 1 micrometer or 1 nanometer, or less?

As for whether the final object would be a black hole, it's hard to say. [..]
In short, it is certainly possible for a black hole to be formed in this type of process, but I don't think we can say very much about the detailed conditions that would determine whether one does in a specific case.

Thanks for the answer. This post is just my reply to your post PeterDonis. But I have to (and am) read(ing) your post all over again.

 

[Stephanus]

I believe the basic answer is yes. As discussed in http://link.springer.com/article/10.1134/S2070046612020069#page-1, if you have in theory, two hyper-relativistic particles colliding, they can form a black hole due to their kinetic energy even though neither one of the particles is itself a black hole.

Particle?? Ahh, I read that CERN is creating (tiny) black hole on daily basis, by colliding particles. Thanks for your reply pervect. I'll reread your post again. And thanks for the links.

 

[Stephanus]

If you make the problem extreme enough, you don't need to worry about the dynamics Peter pointed to, if one is willing to assume the (unproven but plausible) hoop conjecture. Then, if in the COM frame, the total energy of neutron star + rocket would correspond to a Schwarzschild radius larger than the neutron star + rocket, then a black hole will form. Note, this is a substantially higher threshold than the Chandrasekhar limit.

Chandrasekhar? It's 1.4 solar mass, right.

 

[Stephanus]

Stephanus, as usual you have way, way, way overcomplicated the situation with all this talk of antimatter and the interstellar medium.

Your question is whether an inelastic collision between an object of mass m1 at rest and one of mass m2 with momentum p can produce an object of invariant mass m3. That's a straightforward question in relativistic mechanics, and depending on m1, m2 and p, it could end up either above or below m3.

Could you elaborate more? What is m3? Is it m1+m2, or below it?

 

[pervect]

Let's clear up a few things that other posters have mentioned

Dear PF Forum,
I have a question in mind. But I'm not sure if this belong to SR forum, cosmology or classical physics.
So I post it here.
And perhaps as some of you might have known before or thought it over. It's about kinetic energy.
Supposed this...
A rocket, with a rest mass 1 ton.
And the rocket can somehow grab matter and anti matter in interstellar medium. I know that anti matter just can't scattered freely in interstellar medium, but let's just dispense with the technical difficulties.
So in its journey the rocket always accelerates 1 g. And in doing so it has consumed, say..., 10 solar mass of matter and anti matter annihilation as its energy source.

You didn't specify what happens to the extra mass that the rocket "picks up". I assumed that the rocket "burned" this extra mass, in the form of matter-antimatter fuel, expelling it out the rear of the rocket, and that we are to assume the rocket does all this with sufficient efficiency that the thrust generated by burning this fuel more than compensates for any momentum penalty involved in the "pick up" process. Furthermore I assumed that the rocket burned and expelled all the fuel, so that the rocket's final rest mass was the same as its initial rest mass, 1.e. 1 ton.

To specify the end result (rather than the process of obtaining it), my assumption was that we are to consider the rest mass of the rocket as still 1 ton, the same as it started out as, and the kinetic energy of the rocket (relative to its starting frame) is about 3 solar masses * c^2.

Can you confirm if this was your intent?

Going from the broad overview of the title
Now after some times its speed is 99.9999% the speed of light (or should I add more nines?)
Then this 1 ton rocket hits a neutron star, say 1.5 solar mass. Can this neutron star become a black hole if it is hit by this rocket, considering this 1 ton rocket relativistic mass must be about, say 3 solar mass (or more/less?), after spending much of its energy in the form of momentum to eject its propelant?
Thanks for any clarification.

This particular form of the problem is unnecessarily difficult. Would you be satisfied with considering a simpler problem, two such rockets, colliding head on?

With this modification, the center-of-mass of the collision is stationary, and the center-of-mass energy density is what is immportant.

Since 1 solar mass is about 1.5 km, a rocket smaller than 1km would easily produce the energy densities required to create a black hole.

While you didn't ask this question, I don't believe it would even be necessary for the two rockets to actually collide, passing within, say, the previoiusly mentioned kilometer of each other should be easily good enough to create a black hole.

The p-p wave approach I mentioned previously is probably a bit too advanced, I hope this simpler scenario is easier to grasp.

 

[PeterDonis]

upposed there are anti matters and matters enough in interstellar medium. And every time the rocket "grabs" say 1 mili gram (or gram?) matter/anti matter and annihilate them to propel the rocket. After some times, the rocket has "grabbed" perhaps tens of solar mass matter and anti matter.

But none of this increases the rest mass of the rocket; the matter and antimatter is annihilated and ejected out the back of the rocket. So the total mass that it has "grabbed" is not relevant except that it contributes to the rocket's total energy in the frame of the neutron star.

Can kinetic energy be bigger than E=mc2?

Assuming $m$ is rest mass, certainly it can. There is no limit in principle to how much kinetic energy an object can have. In this case, if the rest mass $m$ of the rocket is 1 ton all through the scenario, and its total energy is 3 solar masses, that equates to kinetic energy being about $10^{27}$ times $m$.

What is "p" here? "mv"?

$p$ is momentum. The correct relativistic formula is not $p = mv$. It's $p = \gamma m v$, where $\gamma = 1 / \sqrt{1 - v^2}$ (in units where $c = 1$).

I think if we fire a bullet right in the heart of a rigid body like neutron star, the neutron star will move at least 1 micrometer or 1 nanometer, or less?

Yes, in principle the neutron star will always move by some amount--more precisely, it will gain some nonzero velocity in the frame in which it was originally at rest. But for an ordinary object like a bullet hitting it, that velocity will be way too small to detect--many orders of magnitude too small. The reason I pointed it out in the case you described was that the neutron star's speed after the collision is certainly not too small to detect--it's going to be a substantial fraction of the speed of light.

 

[DrGreg]

Its kinetic energy $E_k = \frac{1}{2}mv^2$

By the way. that's the wrong formula. In relativity it's $E_k = (\gamma - 1) mc^2$

 

[Stephanus]

Thanks pervect for your post. I use matter/anti matter in this example not chemical because I want to know if $E_k = \frac{1}{c}mv^2$ can be bigger than $E = mc^2$
If I use chemical reaction, then we'are talking with joulse not mass/energy equivalent. So, let's say that the rocket grabs 1 gram matter and 1 gram anti matter and annihilate it. The mass of the rocket? Not affected, still 1 ton. Why I don't use joule? Because it will be difficult to convert it to mass, with matter/anti matter, we just can say that the rocket consumes energy equivalent to say... 3 solar mass

Let's clear up a few things that other posters have mentioned
You didn't specify what happens to the extra mass that the rocket "picks up". I assumed that the rocket "burned" this extra mass, in the form of matter-antimatter fuel, expelling it out the rear of the rocket, and that we are to assume the rocket does all this with sufficient efficiency that the thrust generated by burning this fuel more than compensates for any momentum penalty involved in the "pick up" process.

Yes.

Furthermore I assumed that the rocket burned and expelled all the fuel, so that the rocket's final rest mass was the same as its initial rest mass, 1.e. 1 ton.

Yes

To specify the end result (rather than the process of obtaining it), my assumption was that we are to consider the rest mass of the rocket as still 1 ton, the same as it started out as, and the kinetic energy of the rocket (relative to its starting frame) is about 3 solar masses * c^2.

Yes
Can you confirm if this was your intent? I did

This particular form of the problem is unnecessarily difficult. Would you be satisfied with considering a simpler problem, two such rockets, colliding head on?[..]

Yeah, should have thought it over. Remember particle-particle collision in LHC, or accelerator?

While you didn't ask this question, I don't believe it would even be necessary for the two rockets to actually collide, passing within, say, the previoiusly mentioned kilometer of each other should be easily good enough to create a black hole.

Yes, I didn't ask that. But I think it over lunch just now. I was just thinking, still with neutron star. If the neutron star is 3 solar mass, and the rocket with its 3 solar mass potential energy pass by the rocket less then 10 km should be enough. Then I go back to my computer and see your quote. Doesn't actually collide, for 3 solar mass it will be... 3 * 3 * 3 = 81 km?

 

[Stephanus]

But none of this increases the rest mass of the rocket; the matter and antimatter is annihilated and ejected out the back of the rocket. So the total mass that it has "grabbed" is not relevant except that it contributes to the rocket's total energy in the frame of the neutron star.

Yes, it is ejected. The mass of the rocket still 1 ton when it hit the neutron star. I only want to know, in the small compact radius, smaller than Schwarzschild Radius, with only 3 solar mass, what if we put additional energy in that radius, not mass, mass is neglible 1 ton compared to 3 solar mass, right.. Can it become a black hole? Because the potential kinetic energy of the rocket is 3 solar mass.

Assuming $m$ is rest mass, certainly it can. There is no limit in principle to how much kinetic energy an object can have. In this case, if the rest mass $m$ of the rocket is 1 ton all through the scenario, and its total energy is 3 solar masses, that equates to kinetic energy being about $10^{27}$ times $m$.

Oh

$p$ is momentum. The correct relativistic formula is not $p = mv$. It's $p = \gamma m v$, where $\gamma = 1 / \sqrt{1 - v^2}$ (in units where $c = 1$).

Hmm..., okay. DrGreg post below states it. I'll read it.

Yes, in principle the neutron star will always move by some amount--more precisely, it will gain some nonzero velocity in the frame in which it was originally at rest.
Nonzero, yes.
But for an ordinary object like a bullet hitting it, that velocity will be way too small to detect--many orders of magnitude too small.
Too small, I agree. How small? Have to calculate it for solar mass vs 10 grams
The reason I pointed it out in the case you described was that the neutron star's speed after the collision is certainly not too small to detect--it's going to be a substantial fraction of the speed of light.
Yes, that's right. I already suspect that. The neutron star will move a substantial fraction the speed of light. But over lunch, I thought. The rocket doesn't necessarily HIT the neutron star. Just when the rocket arrives at the vicinity of the neutron star and their combined mass is below the schwarzschild radius, it can become a black hole. Then I read pervect post.

While you didn't ask this question, I don't believe it would even be necessary for the two rockets to actually collide, passing within, say, the previoiusly mentioned kilometer of each other should be easily good enough to create a black hole.

The p-p wave approach I mentioned previously is probably a bit too advanced, I hope this simpler scenario is easier to grasp.

Thanks for your confirmation PeterDonis

 

[Stephanus]

By the way. that's the wrong formula. In relativity it's $E_k = (\gamma - 1) mc^2$

So, what is $\frac{1}{2}mv^2$ This is what you get when you accelerate an object with Newtonian law right?
Suppose I push an object 1 kg for 1 n for 50 meters, its velocity will be 10 m/s, etc, etc. I don't have to tell you this DrGreg . You must have known this inside out!
Now, if the object suddenly stops. Then it's kinetic energy is $\frac{1}{2}mv^2$, right. Because it's the amount of energy spent to move that object.
Hmmm, I'll study your formula later.

 

[Drakkith]

Now, if the object suddenly stops. Then it's kinetic energy is 12mv2\frac{1}{2}mv^2, right. Because it's the amount of energy spent to move that object.

No, that equation is only useful when speeds are very low when compared to the speed of light. The work done to accelerate an object actually follows the equation that DrGreg gave. If you were to accelerate an object to 0.9c, you would find that 1/2 mv² is incorrect and gives a kinetic energy of less than 1/3 the actual value.

 

[DrGreg]

$$
E_k = (\gamma - 1)mc^2 = \left( \frac{1}{\sqrt{1 - v^2 / c^2}} - 1 \right) mc^2
$$
When $v$ is very small compared with $c$ and you do the maths, it turns out that $E_k$ is approximately $\tfrac{1}{2}mv^2$.

If you want to know why the relativistic formula is exactly correct and the Newtonian formula is only an approximation, you'll need to study a textbook as there's no simple one-paragraph explanation. But as a plausibility argument you should see that $E_k$ increases towards $\infty$ as $v$ approaches $c$, so no matter how much energy you supply to a mass, it never reaches the speed of light. If the Newtonian formula were correct, you'd only need to supply $\tfrac{1}{2}mc^2$.to reach the speed of light.

 

[PAllen]

Chandrasekhar? It's 1.4 solar mass, right.

Actually, I meant the Tolman–Oppenheimer–Volkoff limit which is not well known, but is believed to be around 3 solar masses.

 

[Stephanus]

No, that equation is only useful when speeds are very low when compared to the speed of light. The work done to accelerate an object actually follows the equation that DrGreg gave. If you were to accelerate an object to 0.9c, you would find that 1/2 mv² is incorrect and gives a kinetic energy of less than 1/3 the actual value.

Actually I was trying to do the algebra for DrGreg equation before your post comes up.

$$
E_k = (\gamma - 1)mc^2 = \left( \frac{1}{\sqrt{1 - v^2 / c^2}} - 1 \right) mc^2
$$
When $v$ is very small compared with $c$ and you do the maths, it turns out that $E_k$ is approximately $\tfrac{1}{2}mv^2$.

But if v is very small then $\sqrt{1-\frac{v^2}{c^2}}$ is almost 1 right. So, if gamma is 1, then $E_k = 0$? If $V = \sqrt{0.75}$ then $E_k = mc^2$.

Can this formula lead to $E_k = \frac{1}{2}mv^2$?

$E_k = (\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1)mc^2$

$E_k = (\frac{c}{\sqrt{c^2-v^2}}-1)mc^2$

$E_k = (\frac{c\sqrt{c^2-v^2}}{c^2-v^2}-1)mc^2$

$E_k = \frac{c\sqrt{c^2-v^2}-c^2+v^2}{c^2-v^2}mc^2$

$E_k = m\frac{c\sqrt{c^2-v^2}-c^2+v^2}{c^2-v^2}c^2$

I give up!. I think $(\gamma-1)mc^2$ can't never reach $\frac{1}{2}mv^2$. But for very small v...?
Perhaps I should use limit?
$\displaystyle\lim_{v \rightarrow 0} (1+\frac{1}{2}v)(1+\frac{1}{2}v)-1 = v$
I see that for a very small v, then $E_k ≈ \frac{1}{2}mv^2$. I can't do the algebra, I can't do the limit. But, I somewhat can see where this equation lead to.

 

[PAllen]

But if v is very small then $\sqrt{1-\frac{v^2}{c^2}}$ is almost 1 right. So, if gamma is 1, then $E_k = 0$? If $V = \sqrt{0.75}$ then $E_k = mc^2$.

Can this formula lead to $E_k = \frac{1}{2}mv^2$?

$E_k = (\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1)mc^2$

$E_k = (\frac{c}{\sqrt{c^2-v^2}}-1)mc^2$

$E_k = (\frac{c\sqrt{c^2-v^2}}{c^2-v^2}-1)mc^2$

$E_k = \frac{c\sqrt{c^2-v^2}-c^2+v^2}{c^2-v^2}mc^2$

$E_k = m\frac{c\sqrt{c^2-v^2}-c^2+v^2}{c^2-v^2}c^2$

I give up!. I think $(\gamma-1)mc^2$ can't never reach $\frac{1}{2}mv^2$. But for very small v...?
Perhaps I should use limit?
$\displaystyle\lim_{v \rightarrow 0} (1+\frac{1}{2}v)(1+\frac{1}{2}v)-1 = v$
I see that for a very small v, then $E_k ≈ \frac{1}{2}mv^2$. I can't do the algebra, I can't do the limit. But, I somewhat can see where this equation lead to.

Do you know Taylor expansion? It falls out immediately that when v/c >> (v/c)², the Newtonian formula is 'correct' to any achievable precision. Key results of Taylor series are:

√(1+x) ≈ 1 + x/2 when x << 1
and
1/(1+x) ≈ 1 -x when x << 1.

 

[Stephanus]

Do you know Taylor expansion? It falls out immediately that when v/c >> (v/c)², the Newtonian formula is 'correct' to any achievable precision. Key results of Taylor series are:

√(1+x) ≈ 1 + x/2 when x << 1
and
1/(1+x) ≈ 1 -x when x << 1.

No, I'm afraid not. This is math, right? My math is weak. But I already have the answers from the mentors/advisors gave me.

[..]In short, it is certainly possible for a black hole to be formed in this type of process, but I don't think we can say very much about the detailed conditions that would determine whether one does in a specific case.

While you didn't ask this question, I don't believe it would even be necessary for the two rockets to actually collide, passing within, say, the previoiusly mentioned kilometer of each other should be easily good enough to create a black hole.

Thanks everybody.

 

[Stephanus]

Do you know Taylor expansion? It falls out immediately that when v/c >> (v/c)², the Newtonian formula is 'correct' to any achievable precision. Key results of Taylor series are:

√(1+x) ≈ 1 + x/2 when x << 1
and
1/(1+x) ≈ 1 -x when x << 1.

Frankly I don't know Taylor expansion. But I can see something like this.
$1.1^2 = 1.21$
$1.01^2 = 1.0201$
$1.001^2 = 1.002001$
$1.0000000000001^2 = 1.00000000000020000000000001$
It looks like for a very small x then $\sqrt{1+x} = 1+\frac{x}{2}$ Like you write.

 

[DrGreg]

It looks like for a very small x then $\sqrt{1+x} = 1+\frac{x}{2}$ Like you write.

Note that $\left(1+\frac{x}{2}\right)^2 = 1 + x + \frac{x^2}{4}$.

So if $x^2$ is small enough to be negligible, but $x$ is not, you can say $\left(1+\frac{x}{2}\right)^2 \approx 1 + x$, i.e. $1+\frac{x}{2} \approx \sqrt{1 + x}$,

What I've said is an algebraic version of the calculations you did.

 

[Drakkith]

For the OP, here's an online calculator from Hyperphysics that will calculate the relativistic kinetic energy of an object and also tell you what the Newtonian kinetic energy would be at that velocity: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/releng.html#c5

 

[Stephanus]

For the OP, here's an online calculator from Hyperphysics that will calculate the relativistic kinetic energy of an object and also tell you what the Newtonian kinetic energy would be at that velocity: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/releng.html#c5

Wow, thanks for the calculator Drakkith. I have had Schwarzschild calculator from this site, too.

 

[Stephanus]

Btw, what is v<<1? V much smaller than 1?

 

[DrGreg]

Btw, what is v<<1? V much smaller than 1?

Yes

 

[anorlunda]

I thought that the OP was seeking confirmation that total energy, not just rest mass, gravitates and thus should be included in calculating the Chandrasekhar limit. (Even if his example and expression of the question was flawed.)

A recent PF Insights, https://www.physicsforums.com/insights/do-photons-have-mass/, used the definition "The invariant mass of a particle is defined as the total energy of the particle measured in the particle’s rest frame divided by the speed of light squared." The definition says particle, but I take it to apply anybody or system including a black hole.

Physicists have stopped using the term relativistic mass because it causes confusion. Isn't the situation the same with the term rest mass? Why not abandon the word mass in all contexts and talk only about energy instead? If we did so, then then OP might not have needed to ask his question in the first place.

I see that particle physics with Feynman diagrams already uses elecron-volts for the energy of particles entering and leaving, eschewing use of mass.

 

[PeterDonis]

I thought that the OP was seeking confirmation that total energy, not just rest mass, gravitates

But that is not correct. What gravitates is the stress-energy tensor; "total energy" is not a good heuristic because it is frame-dependent. Failure to recognize that is what leads to the common misconception that you can turn yourself into a black hole if you move fast enough.

Physicists have stopped using the term relativistic mass because it causes confusion. Isn't the situation the same with the term rest mass?

No, because rest mass is an invariant. (Many relativity texts now use the term "invariant mass" instead of "rest mass" to underscore this point.) If you want to do a quick estimate of how much a given system will gravitate, the best quantity to use is the system's invariant mass. If the system is composed of multiple objects that are moving with respect to each other, the kinetic energy of those objects in the system's center of mass frame will contribute to the invariant mass of the system, and therefore will affect how much it gravitates. But you can't increase how much the system gravitates by transforming to a frame in which the system's center of mass is moving at relativistic speed and looking at the total energy in that frame.

I see that particle physics with Feynman diagrams already uses elecron-volts for the energy of particles entering and leaving, eschewing use of mass.

This has nothing to do with using energy instead of mass; it has to do with using energy units instead of mass units, because energy units are easier to compare to actual measurements.

 

[anorlunda]

What gravitates is the stress-energy tensor; "total energy" is not a good heuristic because it is frame-dependent.

But you can't increase how much the system gravitates by transforming to a frame in which the system's center of mass is moving at relativistic speed and looking at the total energy in that frame.

Isn't that handled in ZapperZ's definition by referring to the object's rest frame? That eliminates kinetic energy seen by other frames. Once again, "The invariant mass of a particle is defined as the total energy of the particle measured in the particle’s rest frame divided by the speed of light squared."

We don't know what happens inside a BH. All the original mass could be converted to massless particles or not. The point should be that we don't need to know. It is irrelevant with respect to the gravity external to the event horizon. E/c^2 gravitates the same as the m. I say that without reference to GR.

Doesn't the same thing arise in the "radiation dominated" version of the FRW equation in cosmology? It describes the gravitational potential energy in a universe with no matter. Again without reference to GR.

 

[Stephanus]

I thought that the OP was seeking confirmation that total energy, not just rest mass, gravitates and thus should be included in calculating the Chandrasekhar limit. (Even if his example and expression of the question was flawed.)

"Total energy"? is that the term. Yes, I think it's simpler. I should have rephrased my question. But I didn't think of that until your post came up.
Chandrasekhar limit?

Actually, I meant the Tolman–Oppenheimer–Volkoff limit which is not well known, but is believed to be around 3 solar masses.

PAllen post stated another limit. Tolman-Oppenheimer-Volkoff. I'll look it up in wiki.

[..]Failure to recognize that is what leads to the common misconception that you can turn yourself into a black hole if you move fast enough.

I often read that phrase.
"If you move fast enough, your clock slows down."
But I think the correct statement is, "If you move fast enough then your clock looks slower wrt a rest observer", is this right?
Should the statement be changed to this?
"You look like black hole according to the rest observer if you move fast enough according to the rest observer"?
Because as some mentors/advisors wrote in previous post.
"We are moving at 99.99%c according to particle at LHC. Do we see our clock slow down?"
Thanks for the responds guys.

 

[Nugatory]

But I think the correct statement is, "If you move fast enough then your clock looks slower wrt a rest observer", is this right?

Although that statement can be read in a way that makes it not flat-out wrong, it is so easily misunderstood and so misleading that I cannot bring myself to say that it is "right". The problem is that to speak in terms of you "moving fast enough" as opposed to being a "rest observer" is to invite the demon of absolute motion back into the discussion.

You look like black hole according to the rest observer if you move fast enough according to the rest observer

And that's an example of the sort of misunderstanding that results from accepting the previous statement.

Perhaps "Clocks moving relative to you read slower than clocks at rest relative to you" would be better? However, there's only so precise a natural-language statement can be.

 

[PeterDonis]

"Total energy"? is that the term.

No. See my post #32.

Chandrasekhar limit?

That limit is not a limit on total energy, which, as I pointed out, is frame-dependent. It is best viewed as a limit on invariant mass.

Should the statement be changed to this?
"You look like black hole according to the rest observer if you move fast enough according to the rest observer"?

No. Whether or not an object is a black hole is an invariant; it doesn't change if you change frames. An object cannot look like a black hole to some observers but not others. This is exactly the sort of confusion I was warning against in post #32.

 

[PeterDonis]

Isn't that handled in ZapperZ's definition by referring to the object's rest frame?

Obviously not, since your phrasing has led Stephanus into exactly the sort of confusion I warned against. The term "total energy" suggests something frame-dependent; the term "invariant mass" does not. That's why you use the latter when you are talking about the conditions for an object becoming a black hole, since being a black hole is an invariant.

We don't know what happens inside a BH.

We can't observe what happens inside from the outside, true. But whether or not an object is a black hole is still an invariant, so we should use invariant language when talking about it.

Doesn't the same thing arise in the "radiation dominated" version of the FRW equation in cosmology? It describes the gravitational potential energy in a universe with no matter.

No, it doesn't. The concept of gravitational potential energy is not well-defined in an expanding universe. It's only well-defined in stationary spacetimes.

Also, the radiation dominated cosmology has no "matter" in the sense of "stuff moving at non-relativistic speeds". But it certainly has a nonzero stress-energy tensor, and, as I've already pointed out, the stress-energy tensor is the source of gravity.

Again without reference to GR.

Um, what? You do know that the FRW spacetimes (all of them, not just the radiation-dominated one) are GR solutions, right?

 

[Stephanus]

"Total energy"
What if a 3 solar mass neutron star collides with an 3 solar mass anti matter neutron star?
(I know this is very unlikely. Even for a non scientist, I have read about baryongenesis and something like baryon asymetry. And again, let's dispense with technical difficulties, that there is somehow an anti matter neutron star)
As you know, this will create enormouse energy. What is the result? There's just pure energy, no mass in that so packed diameter.
Also I have read a "wrong question". What if a black hole from matter collide with anti matter black hole. And not to be confused with our discussion, there's no matter (or anti matter, for that matter ) black hole.
So if a huge energy is packed in a small radius, smaller than the Schwarzschild radius for matter equivalent, will it become a black hole?
@PeterDonis
I see that you edit your post before I reply.
Yes, I knew, it's in your latter post rather in your first post.

 

[anorlunda]

Obviously not, since your phrasing has led Stephanus into exactly the sort of confusion I warned against. The term "total energy" suggests something frame-dependent; the term "invariant mass" does not.

The prasing was not mine. It is ZapperZ's phrasing in the PF Insights article linked above. The phrasing does not dodge use of invariant mass it is the definition of invariant mass.

Um, what? You do know that the FRW spacetimes (all of them, not just the radiation-dominated one) are GR solutions, right?

See the Sussikind video below. He derives the FRW equations, including radiation dominated, from Newtonian mechanics. FRW may be a solution of GR, but GR is not necessary to derive it.

 

[PeterDonis]

The prasing was not mine.

I realize you didn't write the article, but you linked to it in this thread, so you're the one claiming its phrasing is relevant to this discussion.

He derives the FRW equations, including radiation dominated, from Newtonian mechanics.

No, he shows how the FRW equations can make sense if you use an interpretation based on Newtonian mechanics. This is not news. But it is not the same as an actual derivation.

 

[Dale]

This particular form of the problem is unnecessarily difficult. Would you be satisfied with considering a simpler problem, two such rockets, colliding head on?

With this modification, the center-of-mass of the collision is stationary, and the center-of-mass energy density is what is immportant.

Since 1 solar mass is about 1.5 km, a rocket smaller than 1km would easily produce the energy densities required to create a black hole.

While you didn't ask this question, I don't believe it would even be necessary for the two rockets to actually collide, passing within, say, the previoiusly mentioned kilometer of each other should be easily good enough to create a black hole.

Don't you need spherical symmetry also? I believe that you are basically referencing Birkhoff's theorem, but I am not sure that there is an equivalent theorem for axisymmetric scenarios instead of spherically symmetric scenarios.

 

[PeterDonis]

Don't you need spherical symmetry also? I believe that you are basically referencing Birkhoff's theorem

I think he's actually referring to the hoop conjecture, which doesn't assume spherical symmetry.

 

[Stephanus]

Dear Mentors/Advisors,
I should have stopped watching this thread , once I got the answer from @PeterDonis that a one ton rocket hitting a neutron star at 99.9999...% the speed of light or just passing by inside the supposed Schwarzschild radius as stated by @pervect is enough to create a black hole. It's all the answer that fulfill my curiousity.
But then again I remember DrGreg post.

By the way. that's the wrong formula. In relativity it's $E_k = (\gamma - 1) mc^2$

Now I have another question confusion.
Motion is relative, right. What if we apply this equation to the neutron star. $E_k = (\gamma - 1)m_{star}c^2$, not $E_k = (\gamma - 1)m_{rocket}c^2$
The neutron star shouldn't move at 99.9999% the speed of light. Just 99% speed of light will do, then gamma is 7. And if the neutron star hits the rocket the neutron star and the rocket meet then could this process create a black hole?
What if we don't have to move the neutron star, it will require tremendous energy, right. What if we push the rocket just 99% the speed of light. Perhaps this process will take, say.. 10 years. And in 10 years + 1 day, all the other observer away from the star and the rocket will only see that these two objects are approaching each other.
And what should the observer calculate the process?
A: $E_k = (\gamma - 1) m_{star}c^2$, or
B: $E_k = (\gamma -1) m_{rocket}c^2$?
Perhaps the answer is a technical one, not mathematic. Still, I hope this can clear some confusion.

 

[Dale]

I think he's actually referring to the hoop conjecture, which doesn't assume spherical symmetry.

Oh, cool. I didn't know about that.

 

[pervect]

I think he's actually referring to the hoop conjecture, which doesn't assume spherical symmetry.

Yes, the hoop conjecture, mentioned by other posters, is mainly what I was thinking of. Using Google, there is also some other theoretical work that has been studying gravity wave collisions in higher-dimensional ADS space-times, see for instance http://arxiv.org/pdf/0902.4062.pdf, http://link.springer.com/article/10.1007/s11232-009-0152-x#page-1. These may not be strictly relevant to the results of near-collision of ultra-relativistic masse in Minkowskii space-time, though I find them highly suggestive, especially combined with the hoop conjecture. I am also assuming that the formation of an apparent horizon (a trapped null surface) would imply the formation of an absolute horizon in the classical context where there aren't any issues with black hole evaporation. I could be wrong on this point, too.

 

[PeterDonis]

I am also assuming that the formation of an apparent horizon (a trapped null surface) would imply the formation of an absolute horizon in the classical context where there aren't any issues with black hole evaporation.

This is correct; in fact, it's basically the content of the singularity theorems of Hawking and Penrose.