rest mass/relativistic mass question
Chalky
post it here and hopefully someone more knowledgeable than me can clear
the confusion.
As a thought experiment, suppose, we are accelerating a proton in a
large accelerator (as large as you need it to be). It will continuously
gain mass/energy. Will a time come when it has gained sufficient mass
energy to form a black hole? If not why not?
If yes, then what happens to an observer who is always at rest relative
to the proton? To that observer nothing is happening but then suddenly
the proton turns into a hole? Isn't physics seemingly violated for that
observer?
thanks for any clarification.
Jon Bell
Chalky <utpalcha...@gmail.com> wrote:
>
>As a thought experiment, suppose, we are accelerating a proton in a
>large accelerator (as large as you need it to be). It will continuously
>gain mass/energy. Will a time come when it has gained sufficient mass
>energy to form a black hole? If not why not?
This question is discussed in the Usenet Physics FAQ:
http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.html
--
Jon Bell <jtb...@presby.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA
Jonathan Thornburg -- remove -animal to reply
> As a thought experiment, suppose, we are accelerating a proton in a
> large accelerator (as large as you need it to be). It will continuously
> gain mass/energy. Will a time come when it has gained sufficient mass
> energy to form a black hole? If not why not?
>
> If yes, then what happens to an observer who is always at rest relative
> to the proton? To that observer nothing is happening but then suddenly
> the proton turns into a hole? Isn't physics seemingly violated for that
> observer?
This is a Frequently Asked Question. Fortunately, there are informative
answers to be had in the Physics FAQ (written by one of our esteemed
moderators, no less!). Check out
http://math.ucr.edu/home/baez/physics/
particularly the "Black Holes" section.
ciao,
--
-- "Jonathan Thornburg -- remove -animal to reply" <jth...@aei.mpg-zebra.de>
Max-Planck-Institut fuer Gravitationsphysik (Albert-Einstein-Institut),
Golm, Germany, "Old Europe" http://www.aei.mpg.de/~jthorn/home.html
"Washing one's hands of the conflict between the powerful and the
powerless means to side with the powerful, not to be neutral."
-- quote by Freire / poster by Oxfam
Greg Egan
> I was thinking about something and got confused so I thought I will
> post it here and hopefully someone more knowledgeable than me can clear
> the confusion.
>
> As a thought experiment, suppose, we are accelerating a proton in a
> large accelerator (as large as you need it to be). It will continuously
> gain mass/energy. Will a time come when it has gained sufficient mass
> energy to form a black hole? If not why not?
>
> If yes, then what happens to an observer who is always at rest relative
> to the proton? To that observer nothing is happening but then suddenly
> the proton turns into a hole? Isn't physics seemingly violated for that
> observer?
>
> thanks for any clarification.
This kind of thing has been discussed before on this newsgroup, and
probably others as well, so you might be able to find some useful
commentary via Google Groups.
If you're willing to change the thought experiment to something a little
simpler so we can forget about such complications as electromagnetic
effects and gravitational waves, consider this: according to an observer
flying in a straight line past the Earth at close to the speed of light,
the Earth's total mass-energy is M=M_0/sqrt(1-v^2), where M_0 is the
Earth's rest mass and v is the relative velocity in units where c=1.
Suppose the closest the observer comes to the Earth is r, which will be
the same in his/her frame as in the Earth's frame, because it's measured
transversely to the motion. With v high enough, can't M be made so high
that r will fall inside the Schwarzschild radius and the observer will
necessarily be captured?
The answer is no. In the Earth's frame, the observer is moving by at a
very high speed, well outside the Schwarzschild radius 2 M_0 (in units
where G=c=1). And if you do a general-relativistic calculation (treating
the observer as a test mass, and assuming a Schwarzschild geometry
unperturbed by his/her presence), then the acceleration required for the
observer to travel in a straight line at the point of closest approach is:
a = (2v^2+1)/(1-v^2) M_0/r^2 1/sqrt(1-2M_0/r)
(G=c=1) which is obviously finite when v<1. Specifically, at the value
of v where r=2 M, we have:
v^2 = 1 - 4 M_0^2 / r^2
and
a = (3/(4 M_0) - 2 M_0 / r^2) / sqrt(1-2M_0/r)
This is actually quite high for "small" masses like the Earth, about
1.5*10^18 gees! (Remember we're using units with G=c=1, so you have to
convert things back and forth to get that answer.) So the relativistic
change in the mass does have a signigicant effect. But the acceleration
is still finite, whereas the acceleration required to stay outside an
event horizon is infinite.
As for the question "why doesn't the relativistic mass turn the Earth
into a black hole", one very rough short answer is that an event horizon
is a global property of space-time that depends on the matter responsible
for it sticking around, not flying off into the distance at close to the
speed of light. If your proton is in a linear accelerator, I'm sure it
will never become a black hole, because that's basically just changing
the frame of reference.
The kinetic energy of objects *confined* inside a bounded region would
contribute to the total mass-energy in that region -- e.g. a very, very
fast spin could turn the Earth into a black hole (if it was made of
something strong enough) and I guess even a proton moving in a suitably
implausible cyclotron could (along with the energy density of the
magnetic field needed to confine it) turn the cyclotron into a black hole.
Greg Egan
Email address (remove name of animal and add standard punctuation):
gregegan netspace zebra net au
Chalky
i was pretty sure that a mass moving inertially will not turn into a
black hole (as described in the Baez faq). I understood that much.
However, as I said in my first post the proton is not moving
inertially. It is continuously accelerating.
The acceleration does not make any difference?
Greg Egan Write:
-------------------------------------------------------------------------------------
f your proton is in a linear accelerator, I'm sure it
will never become a black hole, because that's basically just changing
the frame of reference.
The kinetic energy of objects *confined* inside a bounded region would
contribute to the total mass-energy in that region -- e.g. a very, very
fast spin could turn the Earth into a black hole (if it was made of
something strong enough) and I guess even a proton moving in a suitably
implausible cyclotron could (along with the energy density of the
magnetic field needed to confine it) turn the cyclotron into a black
hole.
-------------------------------------------------------------------------------------
So a linear acceleration of a proton does not turn it into a hole, but
an angular acceleration might?
Pmb
"Chalky" <utpalcha...@gmail.com> wrote in message
news:1140127497.9...@f14g2000cwb.googlegroups.com...
If a object is not a black hole in one frame of reference then it won't be a
black hole in any other frame even though the grazvitational field will
increase with speed. Most people believe an object is a black hole *because*
it has more than a certain amount of mass. But that is not the case. The
mass must be confined within a certain sperical region of space. Take Mount
Everest as an example. We all know that Mt. Everest is not a black hole. But
its theoretically possible to have a black hole with the same mass of Mt.
Everest. Black holes around this size are refered to as mini-black holes.
Pete
Greg Egan
Chalky <utpalcha...@gmail.com> wrote:
> thanks everyone for taking the time to reply.
>
> i was pretty sure that a mass moving inertially will not turn into a
> black hole (as described in the Baez faq). I understood that much.
>
> However, as I said in my first post the proton is not moving
> inertially. It is continuously accelerating.
>
> The acceleration does not make any difference?
>
>
> Greg Egan Write:
>
-------------------------------------------------------------------------------------
> If your proton is in a linear accelerator, I'm sure it
> will never become a black hole, because that's basically just changing
> the frame of reference.
>
> The kinetic energy of objects *confined* inside a bounded region would
> contribute to the total mass-energy in that region -- e.g. a very, very
> fast spin could turn the Earth into a black hole (if it was made of
> something strong enough) and I guess even a proton moving in a suitably
> implausible cyclotron could (along with the energy density of the
> magnetic field needed to confine it) turn the cyclotron into a black
> hole.
>
-------------------------------------------------------------------------------------
>
> So a linear acceleration of a proton does not turn it into a hole, but
> an angular acceleration might?
The proton itself would not turn into a black hole, but it seems possible
to me (in principle, though not with any realistic machine) that the
whole system that was both accelerating and confining the proton -- into
which energy has been poured, via an electric current, and stored, as
kinetic energy and in the electromagnetic field -- might collapse into a
black hole if you could make the energy content high enough. The proton
would be radiating both electromagnetic and gravitational waves, so the
system would also be losing energy, and at the very least the real-world
parameters of the materials used to construct the magnets would surely
make the whole thing vaporise and/or explode long before this point was
reached.
As the FAQ notes, this is a tricky issue, and I certainly can't prove my
hunch. But I'd argue as follows. Suppose we power the cyclotron by
shipping lots of hydrogen and antihydrogen into its neighbourhood. These
are then combined to liberate energy which is somehow captured and used
to generate electricity. From far enough away, the gravitational effects
would be blind to all this interconversion of energy between different
forms; they would just depend on the amount of mass/energy that has
entered the neighbourhood of the cyclotron. The detailed behaviour of
the spacetime curvature close to the cyclotron *would* depend on all
these details, so it might turn out to be much, much harder to form an
event horizon by piping electricity into a cyclotron than by accumulating
a large cloud of cold hydrogen. But it's hard for me to believe that the
difficulty would be *infinitely insurmountable*, that is, no matter how
much electromagnetic and kinetic energy you managed to get into the
cyclotron, a horizon still wouldn't be created. (Of course, the
real-world difficulties would include confining the energy in these forms
in the first place.)
Blackbird
> thanks everyone for taking the time to reply.
>
> i was pretty sure that a mass moving inertially will not turn into a
> black hole (as described in the Baez faq). I understood that much.
>
> However, as I said in my first post the proton is not moving
> inertially. It is continuously accelerating.
>
> The acceleration does not make any difference?
Ok. It will be easier to answer that question if you tell us why you think
acceleration should make a difference.
In special relativity, acceleration is nothing but "inertial motion"
increasing with time. In any inertial frame, even an accelerating object
can be regarded as moving with constant speed if you only make your time
interval short enough (technically speaking, acceleration is a second order
concept). So applying special relativity to accelerating objects in simply
an exercise in integration.
Blackbird
Chalky
i was pretty sure that a mass moving inertially will not turn into a
black hole (as described in the Baez faq). I understood that much.
However, as I said in my first post the proton is not moving
inertially. It is continuously accelerating getting as close to c as
possible in a limiting sense.
The acceleration does not make any difference?
Greg Egan Write:
-----------------------------------------------------------------------
if your proton is in a linear accelerator, I'm sure it
will never become a black hole, because that's basically just changing
the frame of reference.
The kinetic energy of objects *confined* inside a bounded region would
contribute to the total mass-energy in that region -- e.g. a very, very
fast spin could turn the Earth into a black hole (if it was made of
something strong enough) and I guess even a proton moving in a suitably
implausible cyclotron could (along with the energy density of the
magnetic field needed to confine it) turn the cyclotron into a black
hole.
John Bell
> As a thought experiment, suppose, we are accelerating a proton in a
> large accelerator (as large as you need it to be). It will continuously
> gain mass/energy. Will a time come when it has gained sufficient mass
> energy to form a black hole? If not why not?
I am in general agreement with the other respondents on this matter.
However, even though I don't know quite where this is taking me, there
is something about physics in the reference frame of the linearly
accelerating proton, which could be very loosely described as vaguely
analogous to black hole physics (despite also being very clearly
different). I am referring to the well-known fact that an observer
linearly accelerating at 1g, will outrun a photon if given a head start
of about a year. Thus, for the linearly accelerating proton, its source
will have an effective 'temporal event horizon' at about t = c/g.
However, this horizon is never observed from the accelerating reference
frame, because, as that horizon is approached, light from the source
continues to become progressively more red shifted. In other words,
light from the source between time 0 and time t = c/g is observed
progressively more stretched out, from time 0 to infinity, in the
reference frame of the constantly accelerating observer. Clearly,
something very similar happens when we look at light from the
continuously accelerating observer, when viewed in the reference frame
of the source. This consideration alone should help to confirm that the
linearly accelerated proton should never become a black hole relative
to the source, and vice versa.
However, if a cyclotron accelerates the proton, the situation could
become significantly different, because the mean velocity of the proton
remains zero, relative to the earth. I thus agree that it then seems
theoretically possible to 'pump up' the total rest mass of the
system indefinitely, provided sufficient energy can be found to
maintain adequate strength in the containment field, as well as to
maintain acceleration of the particle. So, yes, the system might then
eventually turn into a black hole (theoretically).
> If yes, then what happens to an observer who is always at rest relative
> to the proton?
The simple answer is that he will probably have been killed by huge
differential centrifugal forces, long before that system becomes a
black hole. In fact, it is difficult to imagine that those centrifugal
forces will not cause the whole system to explode long before that time
too.
>To that observer nothing is happening
Not so. Massive experienced forces will become progressively vaster.
> but then suddenly
> the proton turns into a hole
I don't think so. If he could survive the forces, that observer should
eventually find himself in the same hole as the proton, and nothing is
particularly strange about that.
Hope this helps.
John Bell
Chalky
continuously linearly accelerating mass. John and Greg's reasoning
seemed particularly elucidating.
Blackbird wrote:
--------------------------------------------------------------------------------
Ok. It will be easier to answer that question if you tell us why you
think
acceleration should make a difference.
--------------------------------------------------------------------------------
Correct I should have clarified that at first. That is where the
confusion started. Basically, it seemed to me that as and when I added
mass/energy to any object it contributed to its stress-energy tensor.
You have a static uncharged spehere. You spin it up, it adds to its
stress-energy. You add charges to it, it adds to it stress-energy. In
each case as energy is spent on it some of it is converted into some
form of energy that contributes to the stress-energy tensor and hence
there seems to be no theoritical reason on why that tensor cannot be
made to form a black hole given enough energy is added to it.
Now, with a linearly accelerating mass, energy is definitely being
spent on it. That energy is being transformed into kinetic energy
(which is the other confusing concept because this meaure of energy
seems to be frame dependant) which is as "real" as any other energy
because it can be converted back if necessary to any other form of
energy. However, this kinetic energy for some reason does not
contribute to the stress-energy tensor.
So all kinds of mass-energy do not add to stress-energy tensor? So in
some sort of weird sense energy is not conserved in GR?
Blackbird
> ok, i think i can understand why the black hole will not form with a
> continuously linearly accelerating mass. John and Greg's reasoning
> seemed particularly elucidating.
>
> Blackbird wrote:
> --------------------------------------------------------------------------
------
> Ok. It will be easier to answer that question if you tell us why you
> think
> acceleration should make a difference.
> --------------------------------------------------------------------------
------
>
> Correct I should have clarified that at first. That is where the
> confusion started. Basically, it seemed to me that as and when I added
> mass/energy to any object it contributed to its stress-energy tensor.
The big "curver of space" is the energy-momentum vector. For an isolated
particle, the length of this vector is just its rest mass. For a system of
non-interacting particles, the energy-momentum of the system is simply the
sum of the energy-momentums of its constituents, i.e., the sum of their rest
masses. So the kinetic energy of a system's parts does not per se add to
the system's mass. That is why the statement "energy is mass" is somewhat
misleading, since Einstein only meant E=mc^2 to be applied to a particle in
its rest frame.
Now with the cyclotron someone mentioned in another post, it is a different
matter. Here, the particle interacts with the cyclotron (it has to, since
it's supposed to follow a non-geodesic, circular orbit). This interaction
manifests itself as potential energy (you have to apply a constant inwards
force to keep the particle in orbit). This energy will literally
materialize itself as an increase in the systems mass.
> You have a static uncharged spehere. You spin it up, it adds to its
> stress-energy. You add charges to it, it adds to it stress-energy. In
> each case as energy is spent on it some of it is converted into some
> form of energy that contributes to the stress-energy tensor and hence
> there seems to be no theoritical reason on why that tensor cannot be
> made to form a black hole given enough energy is added to it.
You are right on the target. Sooner or later, this sphere will form a black
hole, unless it explodes first.
> Now, with a linearly accelerating mass, energy is definitely being
> spent on it. That energy is being transformed into kinetic energy
> (which is the other confusing concept because this meaure of energy
> seems to be frame dependant)
Yes it is, that's why energy-momentum, which is frame invariant, and other
frame invariant concepts, are the preferred notions in relativity. I think
Greg Egans introduction is very readable:
http://gregegan.customer.netspace.net.au/FOUNDATIONS/03/found03.html#s3
Greg Egan
"Chalky" <utpalcha...@gmail.com> wrote:
[snip]
> Now, with a linearly accelerating mass, energy is definitely being
> spent on it. That energy is being transformed into kinetic energy
> (which is the other confusing concept because this meaure of energy
> seems to be frame dependant) which is as "real" as any other energy
> because it can be converted back if necessary to any other form of
> energy. However, this kinetic energy for some reason does not
> contribute to the stress-energy tensor.
Measures of energy certainly *are* frame-dependent! Even in Newtonian
physics, the kinetic energy of a body will depend on what frame you
choose to measure it in. In relativistic physics, although a body will
have a frame-independent rest mass, aka proper mass, which is its mass
measured in a frame in which it is stationary, an observer in another
frame will consider the body's total energy (kinetic plus rest mass) to
be the time coordinate (in the observer's frame, rather than the body's
rest frame) of the body's 4-momentum vector.
Similarly, all the individual components of the stress-energy tensor will
depend on the frame they are measured in. In that sense, kinetic energy
certainly *does* "contribute to the stress-energy tensor".
> So all kinds of mass-energy do not add to stress-energy tensor? So in
> some sort of weird sense energy is not conserved in GR?
There are some tricky technical problems with energy conservation in GR,
but I don't believe they are the source of your confusion here. Those
problems are absent in special relativity, and if you can get a solid
understanding of the 4-momentum vector and the stress-energy tensor in SR
I think that will help clear up some of your confusion.
The 4-momentum vector, p, of a body is equal to its rest mass times its
4-velocity, and tells you what direction through spacetime its
mass/energy is moving in. If you choose a frame in which the body is
stationary, you get coordinates:
(m,0,0,0),
and the mass/energy of the body is just moving in what you consider to be
the time direction. If you choose a frame in which the body is moving
with an ordinary velocity v in the x direction, you get coordinates:
(m,mv,0,0)/sqrt(1-v^2)
and the flow of mass-energy now consists of m/sqrt(1-v^2) moving in your
time direction, and mv/sqrt(1-v^2) moving in your x-direction.
However, note that the magnitude of the 4-vector p is *invariant*:
|p|^2 = p.p = -p_t*p_t + p_x*p_x + p_y*p_y + p_z*p_z
In the first frame, this is just:
|p|^2 = -m^2
In the second frame, this is:
|p|^2 = -m^2/(1-v^2) + m^2 v^2/(1-v^2)
= -m^2
It's also important to understand that p exists as a geometric object
independent of any reference frame. That 4-vector p "is what it is",
regardless of the numerical values of its coordinates.
The stress-energy tensor T tracks the flow of the density of
energy-momentum through spacetime. If you choose a frame in which a body
of density rho is stationary, the coordinates of T (inside the body) are
just:
| rho 0 0 0 |
| 0 0 0 0 |
| 0 0 0 0 |
| 0 0 0 0 |
whereas if you choose a frame in which it is moving with an ordinary
velocity of v in the x direction, the coordinates become:
| rho/(1-v^2) rho v/(1-v^2) 0 0 |
| rho v/(1-v^2) rho v^2/(1-v^2) 0 0 |
| 0 0 0 0 |
| 0 0 0 0 |
Like a vector taking on different coordinates when measured in different
frames, a tensor also takes on different coordinates in different frames.
In a frame in which a body is moving at close to the speed of light, its
stress-energy tensor has very different coordinates than in a frame in
which it is stationary.
T itself is, however, an independent geometric object, just like the
4-momentum vector. It doesn't actually change, it just gets different
coordinates when measured by different observers.
It's worth noting that the stress-energy tensor of a gas or fluid will
have terms due to its pressure, which ultimately relate to the kinetic
energy (and momentum) of individual particles. If we're taking a
coarse-grained view of a fluid, then this kinetic energy shows up in T,
even in the fluid's rest frame, because no choice of frame can make all
of this random motion around the average go away.
Now, getting back to black holes. The existence or otherwise of a black
hole is not a frame-dependent fact. An event horizon exists when the
geometry of space-time is such that all the light cones on the horizon
point completely inwards, and that's something all observers will agree
on.
The reason, very broadly, that a horizon forms is that a sufficient
quantity of mass-energy is *confined* within a sufficiently small volume
.. when the volume and mass-energy are measured in the centre-of-mass
frame of the relevant body or system of bodies. Why do we have to
specify a particular frame like that? It's really just the easiest way
to get our hands on some invariant quantities. Just as the easiest way
to get the invariant magnitude of the 4-momentum is to choose a frame in
which the body is at rest, the easiest way to get the relevant invariants
of our would-be black hole is to look at it in its centre-of-mass frame.
Aren't Einstein's equations for GR equally valid in all frames? Yes,
they are, and if you choose a frame in which a body is moving very fast,
some coordinates of its stress-energy tensor will become very high, and
through Einstein's equations this will produce some coordinates in the
curvature tensor that are very high. But that's not what a black hole is
about. If you look at the curvature tensor close to the Earth, you can
alter the value of its coordinates just by choosing a different reference
frame, and you can give some of them extremely high values by choosing a
frame travelling very fast wrt the Earth ... but the spacetime geometry
is still the same as ever, and if you calculate certain scalar quantities
from the tensor coordinates, they will be completely independent of your
choice of frame.
Chalky
----------------------------------------------------------------------------------
http://gregegan.customer.netspace.net.au/FOUNDATIONS/03/found03.html#s3
----------------------------------------------------------------------------------
Great. That helped a lot and cleared my confusion.
mark...@yahoo.com
> If a object is not a black hole in one frame of reference then it won't be a
> black hole in any other frame even though the grazvitational field will
> increase with speed.
There are subtleties involved here that are not covered in any FAQ; in
large measure because of the vagueness of comments like those above.
As no less than Hawking, himself, has pointed out, the occurrence and
existence of causal horizons is *frame*-dependent [1] and there is, in
fact, no trapped region associated with a black hole -- that is, no
"black hole" in the usual classical sense of the term. Instead, there
is only a causal horizon, and one which is observer-dependent. Even the
event horizon associated with a black hole is frame-dependent.
So the question needs to be closely reexamined, since the FAQ and the
conventional wisdom of the previous century are both a little naive
with respect to the general issue.
This is, of course, not the only place there conventional wisdom will
go wrong and where the FAQ needs to be reexamined. Another good example
concerns the visual appearance of a moving sphere, where there is
actually a double-take end-run on conventional wisdom. The naive
assessment is that such an object appears flattened out. A closer
examination of the optics involved shows that the object would appear
spherical -- which is also naive.
The double-take occurs here in that this, too, is wrong. It actually
appears circular, since the optics is respect to *one* eye.
The actual 3-dimensional *stereographic* view of the moving object
(i.e., that obtained by receiving images in two eyes in real-time) need
not be spherical at all.
Note:
[1] "Frame" when used in the context of quantum theory in curved
spacetime does *not* mean coordinate grid, but global timelike flow.
We're not talking about coordinate dependence above, but dependence on
the selection of a timelike field.