Does relativistic mass have a gravitational component?

[j8hart]

Can anyone help me with something I have been puzzling over on and off for 15 years?

I studied Physics at university, but went on to become a computer consultant, and I have forgotten quite a bit.

Moreover I have not been able to find anyone willing to take the time to explain it.

The puzzle came to me whilst I was reading "A brief history of time". At one point in the book Hawkin talks about the possibility of manufacturing a Black Hole by accelerating a particle until it becomes massive enough.

I thought, OK, so we know that relativistic mass effects the inertia of a particle, but that does not necessarily bring gravity into it.

In short, does relativistic mass have a gravitational component? And if so, does it increase in proportion to the inertia?

My university course obviously covered Special Relativity, but we did not really cover General Relativity in depth.

So I devised a couple of thought experiments which I hopped did not require General Relativity.

First, imagine two spheres in empty space. Each has a mass of 1Kg, and they are 1m apart.

In this situation, we do not have to worry about General Relativity, we can calculate the force exerted by one sphere on the other using Newton’s law of Universal Gravitation.

F = GM2/r2

M = 1Kg and r = 1m, therefore F = G Newtons.

Lets say for the moment that this force is exactly counterbalanced by an electrostatic force of repulsion between the two spheres (i.e. also G Newtons).

The spheres therefore remain at the same relative distance apart (1m).

Now imagine an observer "o" at exactly the mid point between the two spheres. Also imagine that o observes the spheres as moving with a velocity close to the speed of light along a course exactly perpendicular to a line drawn between the two spheres and through o.

The diagram below shows the situation, o in the middle, the two spheres are represented by *, and the arrows show the direction on motion.

*-------------->

o

*-------------->

If the velocity of the spheres is chosen so that the relativistic mass is 10 times the rest mass, then we should still be able to use Newton’s law of Universal Gravitation to calculate the force exerted by one sphere on the other as observed by o.

To observer o, M' = 10Kg. r is still 1m (obviously no contraction, since it is perpendicular to the line of motion).

Thus theoretically the gravitational force (F') as observed by o is 100xG Newtons.

In other words F' = 100xF.

If an observer on the spheres sees no acceleration towards the other sphere because of the electrostatic force, then nether can o.

If therefore the Gravitational force increases 100 fold, then so must the electrostatic force.

This is probably explainable by Lorentzian contraction, since the field lines will be closer together, but I have never had enough time to think this through.

If we take away the electrostatic force, and instead assume that the spheres just start to accelerate towards one another at the moment they pass o, then an observer on one of the spheres would calculate the acceleration as:

a = F/M i.e. G Newtons/1Kg, or G m/s2.

The observer o would calculate:

a' = F'/M'

F' = 100 x F, and M' = 10 x M therefore:

a' = (100 x F)/(10 x M) = 10 x (F/m)

or

a' = 10 x a

This is not the result I would expect. Time for the observer o is running 10 times faster than on the spheres.

Since acceleration is dD/dT2 from a time dilation point of view I would expect:

a' = 100 x a

In other words, velocity along the line between the two spheres should be 10 times faster as seen by o, acceleration, being the rate of change of velocity with time should therefore be 100 times faster.

If I work the other way around and assume that I do not know how the gravitational force is affected by the relative velocities, however I assume that:

1) a' = p2 x a where p is the ratio of rest mass to relativistic mass (10 in this case) is correct.

2) F' = q x F where q is some factor relating the gravitational force at rest with the gravitational force in motion.

In 1) we can substitute a' = F'/M' = (q x F)/(p x M) (in this case M=Inertia, so M' = p x M). This gives:

(q x F)/(p x M) = p2 x a = p2 x (F/M)

If we divide both sides by F/M we get:

q/p = p2

or

q = p3

In this case p=10, so F' = 1000 x F, not 100 as I calculated earlier.

This does not look right either.

Can anyone shed some light on this?

Thanks in advance,

Jonathan

 

[Entropy]

Yes relativistic mass has gravity.

 

[Njorl]

The problem is more difficult due to the point masses/charges. I would suggest changing it to two arbitrarily long wires, held together by gravity, pushed apart by electrostatic forces. Then, the observer speeds down the center, between the wires. This gives you symmetry in time.

Njorl

 

[Hurkyl]

General relativistically speaking, yes, the relativistic mass has gravity, but so does its velocity and its acceleration (the latter aren't attractive), so a fast moving particle doesn't gravitate the same as a stationary particle with the same relativistic mass.

 

[Alkatran]

General relativistically speaking, yes, the relativistic mass has gravity, but so does its velocity and its acceleration (the latter aren't attractive), so a fast moving particle doesn't gravitate the same as a stationary particle with the same relativistic mass.

So particles have different masses and thus exert different forces in different frames? I'm hoping this is all somehow counterbalanced by time dilation and length contraction? (Obviously in a flurry of very complicated equations )

 

[j8hart]

Entropy, Hyrkyl,

You both answer my original question in the affirmative, but could you add some more detail?

In particular is the increased gravitational force proportional to the square or the cube of the relativistic mass, or is there some other relationship?

Do you have any sources you could quote for me?

In a sense it is the first part of my problem which worries me the most, since if the electro-static force also increases why are the particle Physicists spending so much money on accelerators? If I was right, wouldn't the increase in the force on a particle balance out the increase in inertia, thus making it much easier to accelerate the particle to a high fraction of the speed of light?

Njorl: Thanks for the interesting comment. However since I can reduce the problem to only considering the instant that the spheres pass the observer I am not too worried about the time symmetry. To me, your problem sounds more difficult! I don't think I want to go there, but if you come up with something interesting I'd be delighted to know.

 

[chroot]

Moving this one out of TD, too.

- Warren

 

[turin]

For two lines of charge, you usually (in undergrad physics) call the effect on the electric force "magnetic force." The two infinite lines of charge are definitely simpler. Another thing that would (counterintuitively) make this simpler is to simply accept the issue of general relativity. In general relativity, mass-energy gravitates, not just (rest) mass. Kinetic energy is energy, so moving particles gravity by virtue of their motion as well as their rest mass.

To put is simply, yes, this is all compensated by length contraction and time dilation, though that is a rather simplified way to put it.

 

[DW]

No, relativistic mass is not what is used in Newton's law of gravitation. Relativistic mass has no place in relativity anymore at all. A black hole can not be formed fromed by accelerating a particle any more than changing frames turns a particle at rest into a black hole in motion. In relativity it is the stress energy tensor that is the source of gravitation. Energy density is an incomplete contribution to that source. As with the can's shadows analogy the complete physical property described by that stress energy tensor does not depend on frame. Physics does not depend on frame.

 

[LURCH]

I have to agree with D.W. on this one. I too have been pondering this question for some time, and when answering in the affirmative have come up with paradoxical situations. For example, if relativistic mass has a gravitational consequence, we can construct a black hole from which objects escape with comparative ease.

Simply accelerate a neutron star until its mass is increased several thousand times, generating a black hole with a particular diameter of event horizon. The diameter of this event horizon should be sufficient to include the position of a particular small satellite with a highly elliptical orbit when it reaches parahelium. Of course, from the satellites frame of reference, the neutron star is not traveling at such high speed. So the neutron star is not a black hole from the satellites frame of reference, and it continues to orbit in its highly elliptical orbit.

If the neutron star is a black hole from our from reference, then the satellite repeatedly dipps into and then emerges from the event horizon.

 

[yogi]

But viewing what is transpiring as "increasing mass" may be the wrong way to look at what is going on - we are really taking about an equation that says a greater force is necessary to change the velocity of a given M due to its velocity with respect another frame in which it is observed -what is increasing is the effective inertia - but this can be due to the difference in the time increment in the other frame dv/dt - so the effect of time retardation (different time increment) in the other frame leads to the same formula for the increased inertia - in other words, you get to the same result if you consider M constant and time retarded in the frame of the moving M

 

[Hurkyl]

For example, if relativistic mass has a gravitational consequence, we can construct a black hole from which objects escape with comparative ease.

As I specifically pointed out, the objects velocity and acceleration (ok, ok, momentum and momentum flow) also have a part in gravitation; you can't simply ignore them.

 

[j8hart]

Simply accelerate a neutron star until its mass is increased several thousand times, generating a black hole with a particular diameter of event horizon.

Thanks for this one Lurch I really like it. Especially since you do not of course have to accelerate the neutron star, you simply have to be traveling towards/away from it close to the speed of light, preferably along a line perpendicular to the orbit of the satellite.

The orbit must then have the same size and shape in our frame of reference as it does in the satellites frame of reference.

Perhaps someone should start scanning the sky for very red shifted black holes .

Seriously though, I am also convinced (as it looks like Hurkyl and others are) that the acceleration, (rate of change of momentum or whatever) is different in the two frames of reference. Also, as I said what started me thinking about this originally was a line from "A brief history of time" where Steven Hawking talks about doing exactly what you and DW say is impossible, namely creating a black hole by accelerating a particle (I did write to him at the time to ask if he was sure this was possible, but of course got a standard "he gets many letters" response ).

Does this not mean we have an interesting set of paradoxes, and if they are that interesting someone must have worked on them a long time ago?

 

[LURCH]

Yes, it is a very interesting set of questions, and we've hashed it out a tme or two here in the Forums before. I myself onxce started a thread called "Relative Black Holes?" based on the afforementioned model. Don't know as we've ever reached a consensus, but I can't get passed the idea that we'd have an object that's a BH to me, but not to someone else.

 

[turin]

When did a BH become an absolute invariant object?

I want to point out that I'm not jibing; I just don't ever remember encountering this distinction explicated for a BH.

 

[Hurkyl]

It's quite simple. If, to me, nothing can escape a given region of space-time, then, for you, nothing can escape that region of space-time.

 

[mijoon]

This thread (and many others) shows why Einstein referred to the idea of a "relativistic mass" as "dangerously misleading".

 

[yogi]

If two equal masses M1 and M2 are moving parallel separated by a distance d, and they have a gravitational force between them given by Newton = F1 .. if they are both accelerated equally to some new velocity v (relative to a second frame) where their effective inertia is doubled relative to said second frame - then what is the new G force between them and between either of them and a third equal mass M3 in the second frame. Does the observer in the second frame calculate the force between M1 and M2 as being greater than F1? If at some point M2 is midway between M1 and M3 as M1 and M2 pass by - will M2 be attracted to M3 with a greater force than M1?

 

[pmb_phy]

The puzzle came to me whilst I was reading "A brief history of time". At one point in the book Hawkin talks about the possibility of manufacturing a Black Hole by accelerating a particle until it becomes massive enough.

I find it hard to believe that Hawking said that since, as you've stated it, that is incorrect. The gravitational field of a moving body is a function of the body's speed, so yes. Relativistic mass plays a very important part in gravity and general relativity. In fact that is precisely what the Misner, Thorne and Wheeler (MTW) mean when, in their text Gravitation, they write on page 404 of that text in chapter 17 How Mass-Energy Generates Curvature (Note: when MTW use the term "mass" on that page they mean what you call "relativistic mass". They just use the term "mass".)

Mass is the source of gravity. The density of mass-energy as measured by any observer with 4-velocity u is

rho = u*T*u

Therefore the stress-energy tensor T is the frame-independant "geometric object" that must act as the source of gravity.

What that means is this - There is a mathematical object which fully describes the source of gravity. That source is mass. Mass in one frame is current density and current flux in another frame.

To understand what this means consider first what the source of the EM field is in electrodynamics: Consider a static lump of charge (i.e. finite charge density in its rest frame but no current in that rest frame). In that lump's frame, call it S, the charge density is non-zero but the current density is zero. This lump will give rise to an electric field (no magnetic field in this frame). Now change to a frame, call it S', in which the lump is moving. There is now a non-zero current density. The charge density is now different than it was in S. There may be a magnetic field in S'. The EM field itself can be described by one mathematical object, the Faraday tensor (aka the EM field tensor). The mathematical quantity which plays the role of source in the 4-tensor version of Maxwell's equations is the 4-current. The time component of the 4-current is the charge density. The spatial portion is the current density.

In GR the mathematical object which describes the source of gravity can be stress-energy-momentum tensor T (you can divide T by c² to get a different tensor which works just as well. I fondly call the resultant tensor the mass tensor for reasons of illustration such as this). T⁰⁰ = c²(mass density), T^0k = current density, etc. Mass in one frame is momentum in another frame etc.

So you can really say that (relativistic) mass is gravitational charge and relativistic momentum is a gravitational current (is the source of the gravitomagnetic field)

For details from a modern cosmology text see an online sample at
http://assets.cambridge.org/0521422701/sample/0521422701WS.pdf
(This is the cosmology text that they use at MIT)
See especially page 17 to 18

In short, does relativistic mass have a gravitational component? And if so, does it increase in proportion to the inertia?

Yes, if I understood you correctly that is.

(more later - I can't sit that long due to back injury)

This thread (and many others) shows why Einstein referred to the idea of a "relativistic mass" as "dangerously misleading".

That is not what this thread shows. It shows that someone has a question on a subject and got their guess wrong. That happens all the time in all areas of physics. That does not mean that all areas of physics are a a bad idea.

By the way. Einstein never said that relativistic mass as "dangerously misleading". He only said that

It is not good to introduce the concept of the mass M = gamma*m for which no clear definition can be given. Instead of introducing M it is better to mention the expression for the momentum and energy of a body in motion.

People are always misquoting Einstein and when doing so they always don't understand what he meant when he said that. Relativistic mass is not defined as M = gamma*m. Relativistic mass is defined as the m such that p = mv is a conserved quantity for free particles (or, rather, particle's which only interact by contact forces) in an inertial frame. m defined as such is what Einstein used in GR in gravitational fields for slowly moving particles. That m is not rest mass m₀ but is a function of the metric. In fact it is for that reason that Einstein, in his text The Meaning of Relativity wrote on page 100

The inertia of a body must increase when ponderable masses are piled up in its neighborhood.

Pete

 

[turin]

If two masses move along side each other, then of course their gravitational influence does not depend on the frame. But, I thought the issue was, what happends to the gravitational influence of a source that is moving with respect to some other object that it influences gravitationally. The 00 component of the curvature caused by a moving object is different than the 00 component of the curvature caused by a stationary object, no? (the 00 component basically being the Newtonian part of it)

 

[pmb_phy]

If two masses move along side each other, then of course their gravitational influence does not depend on the frame. But, I thought the issue was, what happends to the gravitational influence of a source that is moving with respect to some other object that it influences gravitationally. The 00 component of the curvature caused by a moving object is different than the 00 component of the curvature caused by a stationary object, no? (the 00 component basically being the Newtonian part of it)

00 component of curvature? Don't you mean the 00 component of T? i.e. T = c^2(mass ensity) where T = energy-momentum tensor.

By the way, I think its incorrect to say that if two masses move along side of each other that the gravitational inluence is frame independant. Why would you say that? Please clarify.

E.g. Do you think that If two charges move along side each other, then their electromagnetic influence does not depend on the frame? Please describe what that means? Thanks

Pete

 

[turin]

00 component of curvature? Don't you mean the 00 component of T?

No, I mean curvature, taking for granted that T₀₀ will increase:

R₀₀ = -κ (T₀₀ - (1/2)g₀₀T),

if I remember correctly. I admit it is slightly recursive, but, that's the nature of the GR beast.

... I think its incorrect to say that if two masses move along side of each other that the gravitational inluence is frame independant. Why would you say that?

I am in agreement with a few that have brought up the point that you can change the relative velocity between yourself and another phyical system by chaning your own velocity without doing anything whatsoever to the other system. Then, whatever happens in that other system should not be affected by this "rotation" of perspective. I should have been more specific to say that the net gravitational attraction between two bodies perpendicular to the boost should not be affected by the boost.

Do you think that If two charges move along side each other, then their electromagnetic influence does not depend on the frame?

Yes.

Please describe what that means?

I mean that, if, for instance, two like charges are held together by a spring so that they are in equilibrium, then you will not observe the spring to stretch nor contract regardless of the amount of boost (if the boost is perpendicular to the spring). This is the case even though the electric part of the electromagnetic influence increases. One resolution is offered by considering the magnetic influence that results from the fact that the two charges are moving. Another resolution is to realize that the physical situation is the same, even though the components of the Faraday tensor are seen to "rotate" according to the boost (so that the purely electric components "rotate" somewhat into the magnetic components).

 

[pmb_phy]

No, I mean curvature, taking for granted that T₀₀ will increase:

R₀₀ - (1/2)g₀₀R = κT₀₀,

No?

Okay. I see what you're saying. Note that R₀₀ is not the curvature tensor. R₀₀ can be zero in regions of spacetime where there is non-zero curvature. In fact R₀₀ = 0 at all points in spacetime where there is no matter, i.e. R₀₀ = 0 at all points in spacetime where T = 0. This follows from a way that you can recast Einstein's field equation, i.e. the EFE can be written as

R^uv = -(8*pi*G/c⁴)(T^uv - (1/2)g^uvT)

But it would be incorrect to say that spacetime is not curved where there is no matter.

I am in agreement with a few that have brought up the point that you can change the relative velocity between yourself and another phyical system by chaning your own velocity and not really doing anything whatsoever to the other system.

By changing your relative velocity between you and the other system you are not changing the other system. What you're doing is changing to a system in which the field generated by that system is different. E.g. consider a large sheet of charge in the xy-plane. Let the charge distribution in S be such that in S the charge density is constant everywhere and there are no moving charges. Then in frame S there will be an electric field E which is proportional to the charge density of the sheet. Now move to a frame S' which is moving in the +x-direction with respect to S. In S' the charge density is larger than the charge density in S. The electric field in S' is therefore larger than it is in S (This can be obtained straight from the transformation equations of the fields too). In S' there is now a magnetic field. So while I'm not affecting the sheet by me changing my frame of reference, I am moving to a new frame of reference in which the fields are different. There are now velocity dependant forces in this new frame etc. The only thing that is really "being done" is that the sheet is now Lorentz contracted giving a higher charge density. Anymore than this is getting into semantics though. However a very similar thing happens with GR

I should have been more specific to say that the net gravitational attraction between two bodies perpendicular to the boost should not be affected by the boost.

Attraction is measured in terms of things like coordinate acceleration and such quantities change upon a change in frame.

Pete

 

[kurious]

Philip Gibbs writes on Usenet FAQ:

In fact objects do not have any increased tendency to form black holes due to their extra energy of motion. In a frame of reference stationary with respect to the object, it has only rest mass energy and will not form a black hole unless its rest mass is sufficient. If it is not a black hole in one reference frame, then it cannot be a black hole in any other reference frame.

 

[pmb_phy]

Philip Gibbs writes on Usenet FAQ:

In fact objects do not have any increased tendency to form black holes due to their extra energy of motion. In a frame of reference stationary with respect to the object, it has only rest mass energy and will not form a black hole unless its rest mass is sufficient. If it is not a black hole in one reference frame, then it cannot be a black hole in any other reference frame.

That entire page is here

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.html

Pete

 

[j8hart]

Thanks to all especially Pete, I think with a fair amount of study I might possibly understand. I read though some of it, but I still have a long way to go.

I find it hard to believe that Hawking said that since, as you've stated it, that is incorrect.

I'd assume I misunderstood him if I were you, it seams highly likely to me! I don't have the book on me, but I will look at it again when I can and see if what you have said helps me to a new interpretation of the passage.

That is not what this thread shows. It shows that someone has a question on a subject and got their guess wrong. That happens all the time in all areas of physics. That does not mean that all areas of physics are a a bad idea.

A very polite and kind way of looking at it, thanks! However in a sense I agree with the sprit of mijoon's post (whether the quote was correct or not) in that I've always seen this problem as a dangerous mix of SR and Newtonian mechanics concocted by me in a lazy way so as (I hoped) to avoid having to study GR to understand it.

I'm not bothered BTW if the original guesses were wrong, understanding is my goal.

I guess I was still hoping for a words of one syllable none GR explanation when I posted it. Just to be absolutely sure, and please remember that not only have I not studied GR, the physics I did study has been gathering dust in my brain for 20 years, I can re-work the original problem like this...

If the two spheres remain at one location, and the observe o travels away and turns round timing the trip so as to arrive back just as the two balls meet (I'll give them a non zero radius, say 5cm) his time for the 45cm journey is 1/10th the time of another observer who stayed with the spheres.

Now suppose we take one of the spheres away and put a (not particularly powerful) rocket on the remaining sphere. Whist o travels away and comes back the rocket is used to take the sphere through the same 45cm journey as before (the amount of fuel burnt being gradually increased to simulate the inverse square law).

If the two observers have never heard of GR they can still explain this easily enough.

The observer o sees the fuel expelled by the rocket as having 10 times the momentum as the observer who stays behind, expelled over 1/10th of the time. Therefore at all times he would calculate the force of the rocket as being 100 times and the acceleration as being 10 times what the observer who stays behind calculates. When o calculates the distance the sphere will have traveled by the time he returns he gets 45cm, as does the observer who stays behind.

If they now bring the second sphere back, and fit it with an identical rocket. If both rockets are pointed towards each other and used at the lower 1G Newtons setting, the spheres remain at there original 1m (center to center) separation during o's entire trip.

o reports to the observer who stayed behind that the 100 fold increase in the force of the rockets was exactly balanced by a 100 fold increase in the force of gravity.

Is this in any way sound, or do the two of them just have to get down onto their lazy buts and invent GR?

 

[kurious]

j8Hart:
I'm not bothered BTW if the original guesses were wrong, understanding is my goal.

Kurious:

Apparently, Einstein guessed that the stress energy tensor of relativity theory
was related to the curvature of space-time, so you are in good company.

 

[j8hart]

The observer o sees the fuel expelled by the rocket as having 10 times the momentum as the observer who stays behind, expelled over 1/10th of the time. Therefore at all times he would calculate the force of the rocket as being 100 times and the acceleration as being 10 times what the observer who stays behind calculates. When o calculates the distance the sphere will have traveled by the time he returns he gets 45cm, as does the observer who stays behind.

OK, so I posted in hast, left the internet café and did not have a chance to get back and correct it.

I still have not had a chance to think it through properly, but I now believe I should have written:

The observer o sees the fuel expelled by the rocket as having 100 times the momentum as the observer who stays behind, expelled over 1/10th of the time. Therefore at all times he would calculate the force of the rocket as being 1000 times and the acceleration as being 100 times what the observer who stays behind calculates. When o calculates the distance the sphere will have traveled by the time he returns he gets 45cm, as does the observer who stays behind.

So then when I said:

If they now bring the second sphere back, and fit it with an identical rocket. If both rockets are pointed towards each other and used at the lower 1G Newtons setting, the spheres remain at there original 1m (centre to centre) separation during o's entire trip.

o reports to the observer who stayed behind that the 100 fold increase in the force of the rockets was exactly balanced by a 100 fold increase in the force of gravity.

Is this in any way sound, or do the two of them just have to get down onto their lazy buts and invent GR?

I should have said that if they do that they will not be able to explain why o does not see the spheres move apart. Since in the frame of o a crude application of Newton’s universal law of gravitation suggests a 100 fold increase in the force of gravity. Not enough to balance the increased force of the rockets, and therefore presumably to answer my own question they do indeed have to go away and invert GR.

Always assuming I got it right this time round.

 

[quartodeciman]

from Baez link: Philip Gibbs 1996 - If you go too fast do you become a black hole?

"M is contained within a sphere of radius 2GM/c2 (the Schwarzschild radius)"
Spherical symmetry is pretty much lost when the star is transformed by a very large boost.

"One way to avoid this is to not speak about relativistic mass and think only in terms of invariant rest mass"
How can one avoid something by merely not talking about it?

 

[turin]

Note that R₀₀ is not the curvature tensor. R₀₀ can be zero in regions of spacetime where there is non-zero curvature.

Yes, I agree. I was only pointing out the 00 component because that is the closest thing to the non-relativistic Newtonian gravitation component in Einstein's Equation. (It is actually more like the Laplacian of the Newtonian gravitational potential). Most of the other components of the Ricci tensor do not have a non-relativistic analogue (unless you want to count gravitomagnetism as non-relativistic). It wasn't the curvature so much as that particular component of the Ricci tensor that I wanted to emphasise.

... the EFE can be written as

R^uv = -(8*pi*G/c⁴)(T^uv - (1/2)g^uvT)

Yeah, I editted my post to express the EFE in this more relevant form. You must have replied before I had gotten done editting. It would have been rather sloppy to use the EFE in the other form to explain how the Ricci tensor depends on the stress energy.

But it would be incorrect to say that spacetime is not curved where there is no matter.

Of course. That's not what I was trying to say at all. Even in the Newtonian theory, there can be (there is) a tidal field.

By changing your relative velocity between you and the other system you are not changing the other system.

Yes, that's what I was trying to say. Sorry for the confusion.

 

[pmb_phy]

from Baez link: Philip Gibbs 1996 - If you go too fast do you become a black hole?

"M is contained within a sphere of radius 2GM/c2 (the Schwarzschild radius)"
Spherical symmetry is pretty much lost when the star is transformed by a very large boost.

"One way to avoid this is to not speak about relativistic mass and think only in terms of invariant rest mass"
How can one avoid something by merely not talking about it?

Don't ask me. I think some of those FAQs have a lot to be desired. That one is an example. It basically says "Don't think so much and don't ask probing questions unless you know as fact that you're right the first time." Dumb. Really dumb.

Whether you call relativistic mass the source of gravity or you call energy the source of gravity. Either way the same question arises. i.e. "The faster a particle goes the greater its energy. Energy is the source of gravity (one component of energy-momentum tensor, energyu in one frame is momentum and momentum flux in another etc). So why doesn't the partilce become a black hole?" - The answer is this - An object's rest mass confined within a given region of space is what determines ife a body is a black hole or not.

Pete

 

[pmb_phy]

from Baez link: Philip Gibbs 1996 - If you go too fast do you become a black hole?

"M is contained within a sphere of radius 2GM/c2 (the Schwarzschild radius)"
Spherical symmetry is pretty much lost when the star is transformed by a very large boost.

"One way to avoid this is to not speak about relativistic mass and think only in terms of invariant rest mass"
How can one avoid something by merely not talking about it?

Don't ask me. I think some of those FAQs have a lot to be desired. That one is an example. It basically says "Don't think so much and don't ask probing questions unless you know as fact that you're right the first time." Dumb. Really dumb.

Whether you call relativistic mass the source of gravity or you call energy the source of gravity. Either way the same question arises. i.e. "The faster a particle goes the greater its energy. Energy is the source of gravity (one component of energy-momentum tensor, energy in one frame is momentum and momentum flux in another etc). So why doesn't the partilce become a black hole?" - The answer is this - An object's rest energy confined within a given region of space is what determines ife a body is a black hole or not.

Pete

 

[j8hart]

from Baez link: Philip Gibbs 1996 - If you go too fast do you become a black hole?

"M is contained within a sphere of radius 2GM/c2 (the Schwarzschild radius)"
Spherical symmetry is pretty much lost when the star is transformed by a very large boost.

"One way to avoid this is to not speak about relativistic mass and think only in terms of invariant rest mass"
How can one avoid something by merely not talking about it?

My favourite quote from that page is the understated "It is actually quite difficult to define the correct conditions for a black hole to form."

But I think the point the article is making in the section you quote is that someone with my level of Physics might imagine a sphere smaller than the radius of the star 'at rest' but concentric with the star, and then say "even if you ignore the 'relativistic mass' and only consider the length contraction the mass within this sphere goes up". In this way of looking at it, the fact that the star itself is no longer a sphere does not matter, provided you can find a sphere with sufficient mass inside it.

However, the article then goes on to say "But this is based on a particular static solution to the Einstein field equations of general relativity, and ignores momentum and angular momentum as well as the dynamics of space-time itself."

In other words I guess, "it is not going to happen, but it's outside the scope of this article to explain why not in mathematical terms".

As for trying not to speak about "relativistic mass", one of the things I have learned from all this is that apparently people have been trying to avoid the term almost as long as it has existed.

The article links to another article "Does mass change with velocity" which has for example the following paragraph.

"In the final analysis the issue is a debate over whether or not relativistic mass should be used, and is a matter of semantics and teaching methods. The concept of relativistic mass is not wrong: it can have its uses in special relativity at an elementary level. This debate surfaced in Physics Today in 1989 when Lev Okun wrote an article urging that relativistic mass should no longer be taught (42 #6, June 1989, pg 31). Wolfgang Rindler responded with a letter to the editors to defend its continued use. (43 #5, May 1990, pg 13)."

When I was at university I picked up the idea from somewhere or someone I don't remember, that it was better to consider it as "relativistic inertia", which is part of what triggered my original question.

 

[pmb_phy]

"In the final analysis the issue is a debate over whether or not relativistic mass should be used, and is a matter of semantics and teaching methods. The concept of relativistic mass is not wrong: it can have its uses in special relativity at an elementary level. This debate surfaced in Physics Today in 1989 when Lev Okun wrote an article urging that relativistic mass should no longer be taught (42 #6, June 1989, pg 31). Wolfgang Rindler responded with a letter to the editors to defend its continued use. (43 #5, May 1990, pg 13)."

That article by Okun is filled with flaws. In fact each and ever argument he gave is wrong. Granted, they seem correct if you don't take the time to really look into it, but I had plenty of time and picked each one apart.

Pete

 

[quartodeciman]

...imagine a sphere smaller than the radius of the star 'at rest' but concentric with the star, and then say "even if you ignore the 'relativistic mass' and only consider the length contraction the mass within this sphere goes up". In this way of looking at it, the fact that the star itself is no longer a sphere does not matter, provided you can find a sphere with sufficient mass inside it.

I was thinking that removing spherical symmetry alters the metric and requires different solutions.

As for trying not to speak about "relativistic mass", one of the things I have learned from all this is that apparently people have been trying to avoid the term almost as long as it has existed.

That wasn't my impression. I started investigating the ideas of SR back when Einstein still lived and worked. I don't recall any complaints about relativistic mass then. People who accepted SR gladly used the Lewis-Tolman style of derivation for SR dynamical relations. I can't speak for those who objected to SR theory in the first place. And I can't relate what was complained in journal notes, because I didn't know any. For me, the energy-first (invariant-mass-only) school was just suddenly there with a vengeance. Maybe it was the mid-1960s that I first noticed it.

When I was at university I picked up the idea from somewhere or someone I don't remember, that it was better to consider it as "relativistic inertia", which is part of what triggered my original question.

My thought on that is of a strong objection to reification (treating something abstract as if it were real and substantial). Elementary masses are real existents. On the other hand, energy is an abstract concept from the beginning, so it doesn't bear that onus. The answer to "What is energy made of" is easily answered "It's not made of anything because it is a quantitative abstraction, not a substance". So, particle physicists readily convert masses to energies and deal exclusively and freely in that abstract domain. Masses are thereby filled in as final conclusions to HEP problems. That's just my supposition of the situation. The GR (what makes gravity?) situation seems to be another major battlefield for this opposition.