# 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?

Jonathan

Related Special and General Relativity News on Phys.org
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
Staff Emeritus
Gold Member
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
Homework Helper
Hurkyl said:
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
Staff Emeritus
Gold Member
Moving this one out of TD, too.

- Warren

turin
Homework Helper
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.

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
Staff Emeritus
Gold Member
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
LURCH said:
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 travelling 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
Homework Helper
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
Staff Emeritus
Gold Member
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.

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

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?

j8hart said:
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), T0k = 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)

mijoon said:
This thread (and many others) shows why Einstein refered 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 Eistein 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

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turin
Homework Helper
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)

turin said:
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
Homework Helper
pmb_phy said:
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.

pmb_phy said:
... 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.

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

pmb_phy said:
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).

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turin said:
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

Ruv = -(8*pi*G/c4)(Tuv - (1/2)guvT)

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

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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.

kurious said:
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