Speeding black hole?

1. Sep 8, 2009

Mentallic

As matter tends towards c, its mass increases.
At some point, if a particle's density is too great, it will collapse into a blackhole.
Now, since this matter doesn't increase in size as it moves faster, doesn't that mean its density will be increasing? Couldn't its density be so great at some point it will become a moving blackhole?
What is also confusing me is that this matter's mass is relative. i.e. the mass will be very great compared to an observer moving near c, but the mass of the matter from an observer not moving relative to it, will not have a very large blackhole-worthy density.

2. Sep 9, 2009

DaveC426913

An object moving at relativistic speeds experiences no mass increase in its own frame of reference. It is only from an external frame of reference where the mass increase is measureable.

3. Sep 9, 2009

4. Sep 9, 2009

Mentallic

Yes I realize that, but can't comprehend what it's supposed to mean.
The blackhole won't be created because for those in their frame of reference, the mass is very slight? How about for those in the other frame of reference that is moving at large relativistic speeds? Do they witness a very large mass which would be perfectly capable of creating a blackhole if and only it was rest mass being observed?

5. Sep 9, 2009

DaveC426913

In the FoR of the particle, its mass has not changed at all.

Yes. Fire a particle at a target at relativistic velocities and it will impact with more mass than it had when it was stationary. But good luck getting the partcile to such a high veloicty that its mass approaches that of a BH. It'll take a vast amount of energy. In fact ... E=mc^2...

6. Sep 9, 2009

JesseM

Spacetime curvature (which determines gravitational effects in general relativity) is not determined by "relativistic mass" (even in special relativity the concept of 'relativistic mass' is pretty misleading, since mass is supposed to refer to inertia--resistance to acceleration--but the resistance to acceleration of a moving object actually varies depending on which direction you're trying to accelerate it). Most physicists these days avoid using the concept of "relativistic mass" altogether, and just talk about things like energy and momentum.

7. Sep 9, 2009

Mentallic

Isn't the requisite for a blackhole only high density? Mass increases and then due to length contraction... yeah

Ahh this is very interesting! I never thought mass had another definition besides a product that is affected by a force such as gravity. i.e. creating weight.

8. Sep 9, 2009

Staff: Mentor

The FAQ yenchin linked to says it all. The formation of a black hole is a result of a specific solution to the Einstein field equations, and that solution is for a stationary mass so it does not even apply. Also, mass by itself is not the source of gravity, but rather the stress-energy tensor which includes momentum, pressure, energy and momentum flux, etc. For a relativistically moving particle those other terms become large and cannot be ignored.

9. Sep 9, 2009

stevebd1

Regarding a high velocity object in flat space, one possible way to look at it is that the gain in kinetic energy is proportional to the lose in potential energy (and somewhere in this, the quantity of 'work done' would fit in). So it's not simply a case of an object will increase in mass as it increases in speed, it's more of a case of an object will require more energy in order to increase in speed. An object wouldn't simply increase to near light speed and therefore increase in mass, the energy to obtain that speed has to come from somewhere i.e. the propulsion system and fuel, the fuel being considered potential energy which would be used up and converted to kinetic energy.

Last edited: Sep 9, 2009
10. Sep 10, 2009

stevebd1

In addition to my post above, the following equation demonstrates how much fuel (PE) is required to attain a specific velocity (based on the fuel being turned completely into energy i.e. 100% annihilation), the equation is based on conservation of energy and conservation of momentum (source of equation- http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html" [Broken])

$$\frac{M}{m}=\gamma\left(1+\frac{v}{c}\right)-1$$

where M represents the fuel required to accelerate the payload m and $\gamma$ is the Lorentz factor.

Say for a ship accelerating to 0.5c, M/m would equal ~0.73 so for every 1 kg of payload, 0.73 kg of fuel would be required. According to the web page, '..This equation is true irrespective of how the ship accelerates to velocity v..' and I'm making the assumption that once the required velocity has been attained, the fuel would have been used up and only the payload mass would apply when calculating the resulting kinetic energy. At 0.5c, the kinetic energy (for every 1 kg of payload) would be 0.16 kg (using the relativistic equation to calculate KE) which is considerably less than the fuel that was used up to attain this speed.

As the required velocity increases (say to within 12 decimal places of c) the resulting KE tends towards half of the initial mass of the fuel required. At 0.999c, M/m=43.71 and KE=21.37 so it appears that PE would always be larger than the resulting KE.

How this might apply to high energy muons and particles in an accelerator I don't know but I'm guessing the principle might be the same.

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11. Sep 10, 2009

Staff: Mentor

If you want to calculate how a moving object acts as a source of gravity (which is what would determine whether it can form a black hole), you have to look at its stress energy tensor, not just its kinetic energy (which is only a portion of its total energy anyway--it leaves out the energy contained in its rest mass) or its momentum.http://en.wikipedia.org/wiki/Stress-energy_tensor" [Broken] gives an equation for the stress-energy tensor of an isolated, non-interacting particle; if we simplify that expression so the particle is moving only in the x-direction, we find that the only non-zero components of the stress-energy tensor are:

$$T^{tt} = \gamma m c^2$$

$$T^{tx} = \gamma m c v$$

$$T^{xx} = \gamma m v^2$$

where v is the velocity in the x-direction and $$\gamma$$ is the relativistic time dilation/length contraction factor.

The above just says that, in the lab frame, a moving object appears to have a total energy increased by $$\gamma$$ (this includes its "rest energy" and its kinetic energy), a linear momentum in the x-direction of $$\gamma m v$$ (the extra factor of c in the t-x term of the tensor is to correct the units from momentum to energy), and a pressure along the x-axis proportional to $$\gamma m v^2$$ (to really make this a pressure we would have to divide by an invariant volume element). This pressure can be thought of as the object's "resistance to being Lorentz contracted" along the x-direction.

If we want to look at how this moving object will act as a source of gravity, the key quantity is the trace of the stress-energy tensor, which is (for those who know about index gymnastics, I'm glossing over some technicalities in how this is calculated):

$$T^{tt} - T^{xx} = \gamma m c^2 \left( 1 - \frac{v^2}{c^2} \right) = m c^2$$

which is just the object's original rest energy. In other words, considered as a source of gravity, the motion of the object makes no difference.

Last edited by a moderator: May 4, 2017
12. Sep 10, 2009

Staff: Mentor

Oops, made a mistake in the above post. The factor in front of all the terms should be $$\gamma^2$$, not $$\gamma$$; the details of how this arises from the Lorentz transformation of the stress-energy tensor are in my posts at the bottom of https://www.physicsforums.com/showthread.php?t=334974&page=2". That means that we have:

$$T^{tt} = \gamma^2 m c^2$$

$$T^{tx} = \gamma^2 m c v$$

$$T^{xx} = \gamma^2 m v^2$$

and the trace should be

$$T^{tt} - T^{xx} = \gamma^2 m c^2 \left( 1 - \frac{v^2}{c^2} \right) = m c^2$$

which ends up being the same answer I got before, just arrived at correctly this time. :-)

Btw, this makes me think that the formula in the Wikipedia page I linked to is not correct (it has $$\gamma$$ in front of all the terms, as I wrote before, not $$\gamma^2$$). If I have time in the near future, I'll add a note to the discussion for that Wikipedia page suggesting that the formula should be corrected.

Last edited by a moderator: Apr 24, 2017
13. Sep 12, 2009

stevebd1

Hello Peter

While I understand that the conventional trace for the stress-energy tensor for an object of mass is $tr(T^{\alpha \beta})=\rho+3p$ and where $\rho$ and $p$ fit in the 4x4 matrix (as shown http://en.wikipedia.org/wiki/Stress-energy_tensor#Stress-energy_of_a_fluid_in_equilibrium"), could you shed some light on where $T^{tx}$ and $T^{xx}$ fit in the 4x4 matrix and how they were derived (assuming that $T^{tt}$ is simply synonymous with $\rho$ (or $T^{00}$) allowing for the Lorentz factor). Also how you derived $T^{tt} - T^{xx}$ and how it results in the equation above.

regards
Steve

Last edited by a moderator: Apr 24, 2017
14. Sep 12, 2009

Staff: Mentor

stevebd1:

The stress-energy tensor you gave is for a perfect fluid in equilibrium, in the rest frame of the fluid. The one I gave is for an isolated free particle of mass m, in a frame in which that particle is moving with velocity v in the x-direction. There are some similarities between the two, but they're not the same.

The "cross" term $$T^{tx}$$ in my tensor comes from the motion of the particle (it's simply the particle's momentum in the x-direction, with a factor of c stuck in to make the units the same as for the other tensor components). If you take the stress-energy tensor you gave, for a perfect fluid, and transform it into a frame in which the fluid is moving, you'll see "cross" terms there too. (In the thread I linked to in my last post, I worked through more details of how the transformation works.)

The trace formula that I used for the tensor I wrote down, for an isolated particle, is simply:

$$tr \left( T^{ab} \right) = T^{ab} g_{ab}$$

where $$g_{ab}$$ is the metric tensor and we're using the Einstein summation convention (so the above is really a sum over all possible pairs of indices a, b).

The metric tensor I used is just the Minkowski metric, which in the units I was using (and with the "timelike" sign convention, where the t-t component is positive) is:

$$g_{tt} = 1$$

$$g_{xx} = g_{yy} = g_{zz} = -1$$

(Note: the Wikipedia page you cited uses an opposite "spacelike" sign convention, where the sign of the t-t component of the metric is negative. They also seem to use a somewhat different convention for units, with factors of c or $$c^2$$ appearing in different places than I put them.)

So given the above, the trace of the tensor I gave is (giving only the nonzero components):

$$tr \left( T^{ab} \right) =T^{tt} g_{tt} + T^{xx} g_{xx} = T^{tt} - T^{xx}$$

15. Sep 12, 2009

Staff: Mentor

I realized on reading over this thread again that I should also comment on the above statement. As you've stated it, it's incorrect: the trace of the perfect fluid stress-energy tensor is *not* $\rho + 3p$. However, that does not mean that that expression doesn't have a key role to play in how a fluid acts as a gravitational source; it does. Let me run through how that comes about.

(Btw, what I'm going to say is based on John Baez' excellent tutorial on http://math.ucr.edu/home/baez/einstein/" [Broken]. He gives a lot more details than I'm giving here.)

First of all, I was a bit cavalier in an earlier post when I said that the trace of the stress-energy tensor is the key quantity in determining how an object acts as a gravitational source. It is, for the particular example I was considering (the isolated moving free particle), but that's a special case (I'll show below how that special case arises from the more general analysis we're about to do). In the general case, the quantity you need to look at is the 0-0 (or t-t) component of the Ricci tensor, in the rest frame of the object (or fluid, or whatever) whose behavior as a "source" of gravity you want to investigate. (Baez' tutorial explains why this is true; I won't go into the details here.)

The Ricci tensor is related to the stress-energy tensor via the Einstein field equation, in the following form:

$$R^{ab} = \frac{8 \pi G}{c^4} \left( T^{ab} - \frac{1}{2} g^{ab} T \right)$$

where G is Newton's gravitational constant and c is the speed of light. (Those constants have to appear as they do on the RHS of the above in order for us to use ordinary units for both the Ricci and stress-energy tensors; the Ricci tensor is a "geometric" object, meaning its normal units are units of length, but the stress-energy tensor's normal units are energy density, or pressure--the two are really the same. The $\frac{G}{c^4}$ coefficient in front of the RHS is the "conversion factor" between energy density and length. In many textbooks on general relativity that part of the RHS is left out, and "geometric units" are used for all quantities--meaning everything is written in terms of length, even things like energy and pressure.)

So if we have an expression for the stress-energy tensor in the rest frame of the "source" object, we can write down the Ricci tensor in that same frame. The 0-0, or t-t, component is the one we want; it is:

$$R^{tt} = \frac{8 \pi G}{c^4} \left( T^{tt} - \frac{1}{2} g^{tt} T \right)$$

With the sign conventions I've been using, $$g^{tt} = 1$$, and the trace of the stress-energy tensor for a perfect fluid is

$$tr \left( T^{ab} \right) = T = T^{tt} - T^{xx} - T^{yy} - T^{zz} = \rho c^2 - 3 p$$

where I've used $\rho$ for the *mass* density of the fluid, so its energy density (in its rest frame) is $\rho c^2$. Now we have all we need to find the expression for the t-t component of the Ricci tensor:

$$R^{tt} = \frac{8 \pi G}{c^4} \left[ \rho c^2 - \frac{1}{2} \left( \rho c^2 - 3 p \right) \right] = \frac{4 \pi G}{c^4} \left( \rho c^2 + 3 p \right)$$

That's where the $\rho c^2 + 3 p$ comes from, and as you can see, it *is* the appropriate expression to describe the "strength of the source of gravity" for a perfect fluid.

Now, let's suppose we had a fluid that had *zero* pressure. Then we would have simply:

$$R^{tt} = \frac{4 \pi G}{c^4} \left( \rho c^2 \right) = \frac{1}{2} T^{tt} = \frac{1}{2} T$$

In other words, in this special case (where the only non-zero component of the stress-energy tensor in the rest frame of the object is the t-t component), the quantity we need for "strength of source", the t-t component of the Ricci tensor, is equal to half the trace of the stress-energy tensor. Of course, it's also equal to half the t-t component of the stress-energy tensor. But it's nicer to express it as the trace because the trace is an invariant quantity; it's the same in any frame of reference.

So, for example, if we happen to have an expression for the stress-energy tensor in a frame in which the object is in motion (aha!), instead of having to go to all the trouble of transforming the whole tensor back into the object's rest frame, we just take its trace and we're done--which is, of course, what I did with the isolated free particle example above. (I'll leave it as an exercise to show that, in the rest frame of the isolated free particle, the only non-zero stress-energy tensor component *is* the t-t one, so that the special case applies.)

(I should also note that the expression given on the Wikipedia page for the isolated free particle stress-energy tensor, which I used above, sweeps some technicalities under the rug. As I noted above, the "normal" units for the stress-energy tensor are energy density units--but as the isolated free particle tensor is given above, its units are just energy-- $m c^2$ --not energy density. To be rigorously correct, the m should be a $\rho$, for mass density, but you can't really define the "mass density" of an isolated point particle. These technicalities don't affect the main point I was making, so I didn't mention them in my earlier post. The perfect fluid stress-energy tensor expression avoids the problem by just assuming that the fluid is a continuous substance, not a bunch of isolated points; as long as the distance scales you're working on are much larger than the actual size of the particles, that assumption works fine.)

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