# Does relativistic mass have a gravitational component?

Zanket said:
I don’t see how it’s two different frames in the “case I’ve just explained.” It seems to me that it’s all happening in my frame. Things can move in my frame right?
I made a mistake. I meant that the situation of having one lump of clay (which isn't a black hole) at rest in a frame is different than having two lumps of clay approaching each other.
Say I’m floating in deep space, watching a light-year-in-diameter sphere of space. Within that sphere are the two lumps of clay (let’s say boxcar-sized) approaching each other. They collide in the middle of the sphere, releasing enough heat energy to create a black hole with an event horizon that encloses the sphere. Didn’t I witness a black hole spontaneously form in a region of space to which no mass was added? All in one frame?
It takes more than energy to create a black hole. There is enough mass in my body to create a black hole. All that needs to be done to create a black hole is to have all my matter within a much smaller region of space.

In what you've just described, that energy must be compressed into a region of space so small as to have all the matter all within the Schwarzchild radius. However the mass remained constant in the situation you just described. Because mass is a conserved quantity the total mass in the end was identical to the total mass in the beginning. You forgot that there is mass just due to the motion of the clay lumps, i.e. the fast a lump of clay, the greater its mass is.

Pete

Last edited:
pmb_phy said:
In what you've just described, that energy must be compressed into a region of space so small as to have all the matter all within the Schwarzchild radius. However the mass remained constant in the situation you just described. Because mass is a conserved quantity the total mass in the end was identical to the total mass in the beginning. You forgot that there is mass just due to the motion of the clay lumps, i.e. the fast a lump of clay, the greater its mass is.
By "spontaneous" I mean "unconnected to a cause," not "immediately."

Yes, the mass remained constant (I didn’t forget; in the example I’m using that fact to make a point). In my frame, this constant mass was compressed to within a Schwarzschild radius when the lumps of clay were one light year apart. But not until they collided at least six months later did the black hole form (with an event horizon diameter of one light year).

Unlike the case of an imploding star which surface falls through its Schwarzschild radius, in which case a black hole exists immediately, in the lumps example the black hole did not form until at least six months after all the mass fell through the Schwarzschild radius. The six-month delay seems strange.

All that needs to be done to create a black hole is to have all my matter within a much smaller region of space.
It took more than that in the case of the lumps. The lumps' relativistic mass had to be converted to rest mass (heat energy). That’s what’s strange.

Last edited:
j8hart
Chronos said:
I do not interpret this as meaning you can accelerate a particle less massive than the Planck energy into a black hole. Rather, that if you could accelerate and collide less massive particles at high enough energies, a Planck energy size particle could be created [i.e., a tiny black hole result]. This is the same process CERN used to discover the W+, W- and Zo particles. They did not convert less massive particles to these highly massive particles through acceleration, it was through the energy-matter conversion resulting from collisions between highly accelerated less massive particles
Thanks for that Chronos. I think that was the quote. And I am sure you are right, so now I finally understand it

So he is talking about creating theoretical particle whose rest mass is more than 1022 times greater than an electron?

Sounds like a “don’t try this at home kids” kind of experiment!

But wouldn’t the particle also have to be contained within a radius of about 10-35 of a meter?

j8hart said:
In short, does relativistic mass have a gravitational component?
Yes.
And if so, does it increase in proportion to the inertia?
No. The gravitational force, G on a moving particle has componentsGk is given by

$$G_k = m\Gamma^{\alpha}_{k\beta}v_{\alpha}v^{\beta}$$

where the relativitic mass m is given by

$$m = \frac{dt}{d\tau}$$

which is a function of both speed and gravitational potential. For a derivation of this result, and the definition of the terms used, please see

http://www.geocities.com/physics_world/gr/grav_force.htm

So while you've included the speed factor of the mass you've neglected the functional dependance of m on gravitational potential. You've also incorrectly assumed that the velocity dependance of the gravitational force depends only on m.

F = GM2/r2
From this relationship I assume that you're refering to the gravitational force on one particle due to the mass of the other body.

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.
Does "o" remain stationary while the two paricles move?
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.
Newton's Law is inadequate here since Newton's Law is not a function of the speed of the particle etc.
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.
I don't follow. The gravitational force on what? And where did the factor of 100 come in? I thought you said gamma = 10.

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.
The electric field does increase by a factor of gamma. I'm not clear on the specifics of your question though

More later

Pete

Last edited:
Pmb_phy (or anyone),

Any comment on my post above? Can you confirm whether or not it is widely accepted (among experts in general relativity) that the theory supports the spontaneous formation of a black hole an arbitrarily long time after all the mass fell through the Schwarzschild radius? (Where the time is measured by a distant observer at rest with respect to the region in which the hole eventually forms.)

DW
pmb_phy said:
Yes.

No. The gravitational force, G on a moving particle has componentsGk is given by

$$G_k = m\Gamma^{\alpha}_{k\beta}v_{\alpha}v^{\beta}$$

where the relativitic mass m is given by

$$m = \frac{dt}{d\tau}$$
...
Now you're just getting rediculous.
The electric field does increase by a factor of gamma. I'm not clear on the specifics though
Don't tell me you are about to propose un-invariant charge now?!
Mass is invariant the same way that charge is. Factors of $$\gamma$$ in the dynamics equations come from time dilation in proper time derivatives, not from the mass, and just the same not from the charge.

Zanket said:
Pmb_phy (or anyone),

Any comment on my post above? Can you confirm whether or not it is widely accepted (among experts in general relativity) that the theory supports the spontaneous formation of a black hole an arbitrarily long time after all the mass fell through the Schwarzschild radius? (Where the time is measured by a distant observer at rest with respect to the region in which the hole eventually forms.)
Yes. That's correct.

Pete

pervect
Staff Emeritus
DW said:
Now you're just getting rediculous.

Don't tell me you are about to propose un-invariant charge now?!
First I want to point out that I am _not_ a fan of relativistic mass. I prefer to use the term energy, it leads to a lot less arguments and confusion.

Secondly, I want to point out that the transverse electric field of a moving charge *DOES* increase by a factor of gamma.

The easiest way to show this is to boost the Faraday tensor

F_{ab}= \left[ \begin {array}{cccc} 0&-{\it Ex} &-{\it Ey}&-{\it Ez}\\\noalign{\medskip}{\it Ex}&0&{\it Bz}&-{\it By} \\\noalign{\medskip}{\it Ey}&-{\it Bz}&0&{\it Bx}\\\noalign{\medskip}{ \it Ez}&{\it By}&-{\it Bx}&0\end {array} \right]

if you boost around the z axis you should get

F_{a'b'} = \left[ \begin {array}{cccc} 0&{\frac {-{\it Ex}+{\it By}\,v}{\sqrt {1 -{v}^{2}}}}&-{\frac {{\it Ey}+{\it Bx}\,v}{\sqrt {1-{v}^{2}}}}&-{\it Ez}\\\noalign{\medskip}-{\frac {-{\it Ex}+{\it By}\,v}{\sqrt {1-{v}^{2 }}}}&0&{\it Bz}&{\frac {{\it Ex}\,v-{\it By}}{\sqrt {1-{v}^{2}}}} \\\noalign{\medskip}{\frac {{\it Ey}+{\it Bx}\,v}{\sqrt {1-{v}^{2}}}}& -{\it Bz}&0&{\frac {{\it Ey}\,v+{\it Bx}}{\sqrt {1-{v}^{2}}}} \\\noalign{\medskip}{\it Ez}&-{\frac {{\it Ex}\,v-{\it By}}{\sqrt {1-{ v}^{2}}}}&-{\frac {{\it Ey}\,v+{\it Bx}}{\sqrt {1-{v}^{2}}}}&0 \end {array} \right]

My favorite link for the relativistic E-field of a moving charge is at

http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_14.pdf

you can of course do gauss's intergal on the resulting E-field in accordance with gauss' law to get a conserved charge.

Last edited by a moderator:
DW
pervect said:
First I want to point out that I am _not_ a fan of relativistic mass. I prefer to use the term energy, it leads to a lot less arguments and confusion.

Secondly, I want to point out that the transverse electric field of a moving charge *DOES* increase by a factor of gamma.

The easiest way to show this is to boost the Faraday tensor

F_{ab}= \left[ \begin {array}{cccc} 0&-{\it Ex} &-{\it Ey}&-{\it Ez}\\\noalign{\medskip}{\it Ex}&0&{\it Bz}&-{\it By} \\\noalign{\medskip}{\it Ey}&-{\it Bz}&0&{\it Bx}\\\noalign{\medskip}{ \it Ez}&{\it By}&-{\it Bx}&0\end {array} \right]

if you boost around the z axis you should get

F_{a'b'} = \left[ \begin {array}{cccc} 0&{\frac {-{\it Ex}+{\it By}\,v}{\sqrt {1 -{v}^{2}}}}&-{\frac {{\it Ey}+{\it Bx}\,v}{\sqrt {1-{v}^{2}}}}&-{\it Ez}\\\noalign{\medskip}-{\frac {-{\it Ex}+{\it By}\,v}{\sqrt {1-{v}^{2 }}}}&0&{\it Bz}&{\frac {{\it Ex}\,v-{\it By}}{\sqrt {1-{v}^{2}}}} \\\noalign{\medskip}{\frac {{\it Ey}+{\it Bx}\,v}{\sqrt {1-{v}^{2}}}}& -{\it Bz}&0&{\frac {{\it Ey}\,v+{\it Bx}}{\sqrt {1-{v}^{2}}}} \\\noalign{\medskip}{\it Ez}&-{\frac {{\it Ex}\,v-{\it By}}{\sqrt {1-{ v}^{2}}}}&-{\frac {{\it Ey}\,v+{\it Bx}}{\sqrt {1-{v}^{2}}}}&0 \end {array} \right]

My favorite link for the relativistic E-field of a moving charge is at

http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_14.pdf

you can of course do gauss's intergal on the resulting E-field in accordance with gauss' law to get a conserved charge.
Just to be clear, I was not saying that it didn't. I was just saying that the factor of $$\gamma$$ does not come from the charge which like mass is invariant. It comes instead ultimately from time dilation.

Last edited by a moderator:
pmb_phy said:
Yes. That's correct.
Seems strange! Thanks much for the confirmation.

Does relavistic mass have a gravitational component?

Certainly.

Gravity may be defined as an acceleration d/sq(t)
and E = m * sq(c)

compnent sq(c) = (d*d) / sq(t) = d * (d/sq(t))

Thus d/sq(t) is a component of m * sq(c)

Hope this helps ;)