Relativity and collapsing stars

[Identity]

Sorry if this is a newby question, but it was something I was tinkering with a bit last night

"As an object increases its speed relative to you, its relative mass increases. Therefore, if a neutron star passes by you at a certain speed, it should turn into a black hole."

The schwartzchild radius is $$R = \frac{2MG}{c^2}$$ (which can be derived by setting $$v=c$$ in the escape velocity equation)

So a mass of $$M \geq \frac{Rc^2}{2G}$$ will be a black hole.

Using special relativity,

$$M = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

So we can solve the equation

$$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} \geq \frac{Rc^2}{2G}$$ for v.

$$\therefore v \geq \sqrt{c^2-\left(\frac{2Gm_0}{Rc}\right)^2}$$

If we plug in the data for a neutron star,
$$\rho = 4.9 \times 10^{17}\ \mbox{kg/m}^3$$
$$r = 12 \times 10^3\ \mbox{m}$$

We get $$v > 0.9c$$, which is certainly achievable.

In your frame of reference, you would see the star collapse into a black hole. In the neutron star's frame of reference, nothing would happen.


Does that work?

 

[Dale]

In GR gravity does not just couple to energy (mass) but also to pressure, momentum, stress, etc. Such an object, moving at relativistic velocity, would have an enormous amount of momentum which would prevent the black hole from forming.

 

[Gorn]

Hello..

The second line of your text is what disturbs me. As an objects speed increases an increase in energy is needed. With an increase in energy..a resultant mass increase takes place as well.

A neutron star has a 'rest mass'. If its speed increases..so will its mass to the point of becoming a black hole. If its speed increases relative to you..it will become a black hole.

But what you are saying..is just a certain speed relative to you NOT an increase in speed
Bye
Gorn

 

[Dale]

A neutron star has a 'rest mass'. If its speed increases..so will its mass to the point of becoming a black hole. If its speed increases relative to you..it will become a black hole.

No, it won't. See my answer to the OP above.

 

[jtbell]

The schwartzchild radius is $$R = \frac{2MG}{c^2}$$ (which can be derived by setting $$v=c$$ in the escape velocity equation)

As I undertsand it, the proper derivation of the Schwartzschild radius comes from solving Einstein's field equations of general relativity for a spherically symmetric, non-rotating object at rest. The fact that the "classical" derivation gives the same result is basically a lucky accident.

For a moving object, you have to set up the Einstein field equations differently. They use the stress-energy tensor whose components relate to the object's energy and momentum, more specifically the energy density and momentum density at each point inside the object. I don't know if anyone has actually solved the equations for this case, but I would not expect the results to correspond to simply plugging the object's "relativistic mass" into the solution for a stationary object, because of the extra terms introduced by the non-zero momentum.

 

[yuiop]

Try this thought experiment. An object is dropped a from a height of 9.8 meters at the North pole. It takes about 1 second to fall. An space traveller passing the Earth parallel to the equator at 0.8c measures the fall time to be 1.666 seconds. According to relativity the space traveller can consider himself to be stationary and the Earth is passing by him at 0.8c. From his point of view, the falling object takes longer to fall when the Earth is moving relative to him, so the acceleration due to the gravity of the Earth appears to be less when it is moving and not more as would be implied by assuming the gravitational mass to increase in the same way as the relativistic mass is supposed to increase.

 

[Gorn]

Hello..
I thot I would just throw this in for conversational purposes.

I light photon is supposed to be a rest massless particle...but it is altered in its trajectory in space by gravitiational fields..so...?
Bye
Gorn

 

[jtbell]

An object doesn't need to have mass in order to follow the curvature of spacetime.

Mass isn't even absolutely needed in order to produce spacetime curvature. Curvature is related to the stress-energy tensor whose components are basically energy density and components of momentum density. It's been calculated that two light beams can attract or repel each other, even when describing them as classical electromagnetic radiation (no photons involved).

 

[Gorn]

Hello..

A response to the previous thought experiement. Because I am somewhat limited in time..I will use 'old' relativity speak.

The preceding example has 'frame of reference' problems. From the 'f of R' of the ball.. no matter what speed the Earth is traveling thru space (and as far as it is concerned) the Earth can be considered as not moving at all..the experimenter who is with the ball experiences no changes in the 'gravitational' mass of the Earth or anything else.

But as far as the neutron star is concerned..the best word to use when it comes to 'its' relationship to the Earth..if they 'interact' would be "KABOOM".

The reason that the person on the neutron star sees a 1.66s elapsed time taking place in his calculations is because the ball "does not" take a straight line path to hit the ground as an experimenter in the f of R of the ball will measure.

With relation to the neutron star (or from its f of R or from its 'reality') the ball takes a 'parabolic' trajectory to get to the ground..that is why he measures a greater time passage.

* I would like to charge that gravitational mass and relativistic mass are 'equivalent'. Oops...mistake.. (RM) is larger.

Whether a black hole can form because of other issues..I am not sure why 'yet'.
Also, I believe that photons 'do' have grav/mass..but only when they get to relativistic velocities will it become apparent thru experience.

If someone has the time or desire..because 'I' am somewhat limited..I would not mind having the stress-energy tensor thing articulated (or maybe not) so I can think about other issues.

Bye.

 

[atyy]

I would like to charge that gravitational mass and relativistic mass are 'equivalent'.

Yes, but only heuristically. Newton's law of gravitation says "derivative of the gravitational field = gravitational mass". If gravitation is the curvature of spacetime, then the gravitational field is the spacetime metric, which is a tensor. Unfortunately, the relativistic mass is not a tensor, so we can't write some equation in which "derivative of spacetime metric = relativistic mass". Luckily, the relativistic mass is a component of the stress-energy tensor, and we can write an equation "derivative of spacetime metric = stress-energy tensor", which turns out to be correct.

As DaleSpam said, the stress-energy tensor contains not just relativistic mass, but also momentum, pressure etc. So you cannot make more than heuristic conclusions about gravity with only the relativistic mass. To get correct conclusions, you must also use momentum , pressure etc.