escape velocity

[RichIsnang]

What is the acceleration at the schwarzschild radius of a black hole?

is it 3x10^8? ©?

if it is, what is the acceleration at 1m below the radius?

more than the c?

of course gravity obeys a 1/r^2 law

 

[mathematic]

What units are you using for 3 x 10^8? The speed of light is 3x10^8 meters/sec. - this a speed not an acceleration.

[RichIsnang]

The units for acceleration are m/s^2

That is the unit I use

[timo]

In that case, the acceleration is neither c, not more or less than c. 13 m/s cannot be compared to 8 m/s^2 in a similar sense as five elephants not being comparable to 25 Kelvin. An escape velocity is conceptually not the same as an acceleration - and despite popular belief, thinking about black holes, photons, curved spacetime, and extra dimensions is no help for understanding concepts of school physics (but apparently more interesting than the school physics).

 

Is "of course gravity obeys a 1/r^2 law" a claim or a question? The gravitational force of a point source (and sources well-approximated as a point) obeys a 1/r^2 law in Newtonian gravity. But considering that black holes are an effect of General relativity that is not present in Newtonian gravity, it's quite an extrapolation to claim that this property of Newtonian gravity necessarily holds true. I think you are underestimating how different General relativity is from school physics.

[granpa]

the acceleration at the event horizon depends on the mass of the black hole.

[RichIsnang]

i see your point

i understand the difference between acceleration and velocity.

the acceleration due to gravity on the earth is 9.81m/s^2 is it not?

and the escape velocity is measured in m/s is it not? (for earth 11000m/s)

and that was a claim i thought. you say a point source obeys newtonian gravity, i would say that a black hole is the closest thing we are ever going to get to a point source

seeing as all its mass in concentrated in a single point?

i thought black holes were predicted by newtonian gravity?

how do they differ if i am wrong?

 

What is the acceleration at the schwarzschild radius of a black hole?

[timo]

And that was a claim i thought. you say a point source obeys newtonian gravity, i would say that a black hole is the closest thing we are ever going to get to a point source seeing as all its mass in concentrated in a single point?

That's probably true from a perspective of Newtonian gravity, yes.

 

I thought black holes were predicted by newtonian gravity?

Nope.

 

how do they differ if i am wrong?

In Newtonian gravity, the (negative) potential energy E of an object of mass m that is a distance d away from a point-like gravitational source of mass M is given by [math] E = G Mm / d [/math], where G is the gravitational constant. This implies that an object with a kinetic energy larger than this E can get an arbitrary distance away from the gravitational source - get free from the attracting source. If you evaluate the velocity that would result in this kinetic energy you get the escape velocity.

 

The equation for the potential energy I gave directly implies that for every pairs of masses m, M and for every non-zero distance between them, there is some (possibly high) kinetic energy that would allow the particle to break free of the attraction. Black holes are characterized by the fact that there are regions for which an escape from the attracting source is impossible irrespective of the kinetic energy - a feature that qualitatively does not exist in Newtonian Gravity (as I just explained). The boundary separating those regions from the normal regions is called the event horizon. In standard coordinates is forms a sphere with a radius that is called Schwarzschild radius.

 

For people who never heard about relativity it seems suggestive to define a classical Schwarzschild radius at the distance at which the escape velocity calculated via E=mv^2/2 equals c, since no particle's velocity can ever exceed c. However, in relativity E=mv^2/2 is not the correct expression to calculate the kinetic energy as a function of velocity. In relativity, a particle of mass m can reach an arbitrarily high kinetic energy even for velocities <c.

 

What is the acceleration at the schwarzschild radius of a black hole?

General relativity foots on a different concept of space and time than other physics (or chemistry, biology, engineering, ...). As a result, I don't think there is a standard interpretation of what acceleration is supposed to be in the context of your question. It's intriguing to answer your question with "infinite". Sounds cool, after all. But "not obvious what's meant with it" really is the better answer - despite sounding like a weak excuse.

 

Btw.: Please use proper capitalization - and possibly also punctation.

Edited March 26, 2012 by timo

[RichIsnang]

Ah, i see, thanks timo

How do we work out the kinetic energy of a particle in relativity?

And how does a particle have higher kinetic energy than its velocity would suggest?

[granpa]

http://www.physicsforums.com/showthread.php?p=3787339#post3787339

 

classically' date=' an electron accelerating from rest in a uniform electric field will have a kinetic energy proportional to the distance 'd' from its point of origin.

 

will this continue to hold even when the electron is moving at relativistic velocity?

 

I understand that the formula for relativistic kinetic energy is

 

[img']http://upload.wikimedia.org/wikipedia/en/math/6/5/6/656314a4a1ad9593e71227d9c2184c57.png[/img]

 

so basically kinetic energy is proportional to gamma - 1

 

so gamma(d) ≡ d + 1?

[timo]

How do we work out the kinetic energy of a particle in relativity?

As a function of velocity the kinetic energy is [math]E_{\rm kin} = \left( \frac{1}{\sqrt{ 1 - v^2/c^2}} -1 \right) mc^2[/math]. But there's little information to be gained from this relation other than that

a) for 0<=v<c any possible positive energy can be achieved, and

b) the reverse statement: For any finite kinetic energy, the corresponding speed of the object is always smaller than that of the speed of light (which in turn is a popular argument for why nothing can exceed the speed of light).

 

And how does a particle have higher kinetic energy than its velocity would suggest?

It merely has a higher kinetic energy than you would expect from non-relativistic physics. The reason is that non-relativistic physics is wrong for high speeds.

[RichIsnang]

I see,

the only thing I don't get is how 'In relativity, a particle of mass m can reach an arbitrarily high kinetic energy even for velocities <c.'

[mathematic]

I see,

the only thing I don't get is how 'In relativity, a particle of mass m can reach an arbitrarily high kinetic energy even for velocities <c.'

The additional energy ends up as an increase in mass due to Lorentz transformation.