Does mass increase in SR mean higher gravity in GR?

[I812]

It has been years since I have thought of relativity, but I came across a book by Jerome Drexler in which he attempts to support his view that ultra high energy protons (cosmic rays) are what dark matter is all about in the galactic halos. In spite of the fact that it seems to me these protons would interact and emit radiation (hence be detectable), he argues that the mysterious gravitational influences from these photons is due to the relativistic mass increase due to their high velocities. This doesn't make sense to me. If this were true, it seems like I could tie a ball to a string and twirl it around fast enough to create enough relativistic mass to bring the moon crashing down to Earth! Obviously this is crazy, but I think the idea is the same. Does anyone have any explanations or thoughts as to what gravitational effects fast moving objects would have. It does not make sense to me that they would have any at all.

 

[Andrew Mason]

If this were true, it seems like I could tie a ball to a string and twirl it around fast enough to create enough relativistic mass to bring the moon crashing down to Earth!

You would need a very strong string. And a lot of energy. But in theory, you are correct.

Does anyone have any explanations or thoughts as to what gravitational effects fast moving objects would have. It does not make sense to me that they would have any at all.

According to SR a mass that experiences a gain of kinetic energy (eg. by absorption of a photon) experiences an increase in its mass (relativistic mass). GR says that relativistic mass contributes to gravitation the same as rest mass. I am not adept enough in GR mathematics to elaborate, but I think that is essentially the principle.

AM²

 

[Chronos]

Incorrect. The energy used to accelerate an object must be subtracted from the apparent relativistic mass gain. The net effect is zero.

 

[pervect]

It has been years since I have thought of relativity, but I came across a book by Jerome Drexler in which he attempts to support his view that ultra high energy protons (cosmic rays) are what dark matter is all about in the galactic halos. In spite of the fact that it seems to me these protons would interact and emit radiation (hence be detectable), he argues that the mysterious gravitational influences from these photons is due to the relativistic mass increase due to their high velocities. This doesn't make sense to me. If this were true, it seems like I could tie a ball to a string and twirl it around fast enough to create enough relativistic mass to bring the moon crashing down to Earth! Obviously this is crazy, but I think the idea is the same. Does anyone have any explanations or thoughts as to what gravitational effects fast moving objects would have. It does not make sense to me that they would have any at all.

Well, I've tried to work that particular problem (the gravitational field of a fast moving body) out two different ways - and I've gotten two different answers!

They agreed qualitatively, but not quantitatively :-(. Somoene else here has worked on the problem, too, but I don't really believe his answer either.

I do have a paper which works out the results for low velocity, though.

Qualtiatively, though, the field will increase in the transverse direction, i.e if X marks the "spot" (particle).


X----> direction of motion
|
|
\/ max field

There will also be a "magnetic" like field that will affect moving particles (but it won't matter to the force on a stationary particle) this is known as "frame dragging", or also as gravitomagnetism.

For the electric field of a relativistically moving charge (which will be generally similar, but not exactly the same as the gravitational result) see


java applet

for this java applet, note that the field strength is equal to the density of the field lines.

and if you'd rather see the equations, this link has them

equation form

 

[Andrew Mason]

Incorrect. The energy used to accelerate an object must be subtracted from the apparent relativistic mass gain. The net effect is zero.

Not if you add energy to the system - which you would obviously have to do in order to give the rotating mass such a huge relativistic mass that it causes the moon to fall out of its orbit. The energy added would have to be much greater than the energy contained in the mass of the Earth (ie. Earth mass all converted into energy). As I say, you would need a strong string.

AM²

 

[pmb_phy]

Incorrect. The energy used to accelerate an object must be subtracted from the apparent relativistic mass gain. The net effect is zero.

Nobody said the object was being accelerated. The question was about the mass and gravity as a function of speed. One can simply change their frame of reference to observer the effects. That way even a black hole can be observered from two different frames of references with merely a change in speed of the observer ... which requires much less energy than trying to accelerate a black hole,

Pete

 

[Chronos]

No object can change speed without being accelerated. The only relevant factor is the reference frame from which you measure it.

 

[pmb_phy]

No object can change speed without being accelerated. The only relevant factor is the reference frame from which you measure it.

If there is a star at rest in my frame of reference and I move to a new frame of reference which is moving relative to the old one with a speed v = 0.9c. I will now measure the speed of the star to be v = 0.9c. How much work did I do on the star? Where did that energy come from?

However if the person swinging the ball got energy from outside the Earth to swing the ball then the gravitational field of the Earth/ball system would increase.

Pete

 

[gonzo]

Actually, I would be really happy if someone could give me a good explanation of how kinetic energy works in special relativity.

I mean, if a 1 kg rock hits the Earth at 100 m/s, you usually calculate the energy of that collision by using the mass of the rock. Why don't you use the mass of the Earth? Why isn't the Earth moving towards the rock instead?

 

[yogi]

Einstein concluded in one of his papers that a particle's increase in inertial mass due to motion would result in an increase in its gravitational mass As Andrew Mason has said. I don't know of any actual reported experiments however - anyone have a reference on this. It does seem to raise some paradoxical questions- but it also seems that it must follow from the equivalence principle.

 

[Andrew Mason]

If there is a star at rest in my frame of reference and I move to a new frame of reference which is moving relative to the old one with a speed v = 0.9c. I will now measure the speed of the star to be v = 0.9c. How much work did I do on the star? Where did that energy come from?

Of course, the measure of kinetic energy is frame dependent. I can give the entire Earth more rotational energy in my frame of reference by traveling west in my car along the equator at a very high speed. But I can only collide with a tree once and that energy is lost.

The gravitational effect, however, is the same in both reference frames. Since gravity is the product of both masses, it doesn't matter which mass the \gamma factor applies to. It is still F \propto \gamma m M/R^2

AM

 

[pervect]

Of course, the measure of kinetic energy is frame dependent. I can give the entire Earth more rotational energy in my frame of reference by traveling west in my car along the equator at a very high speed. But I can only collide with a tree once and that energy is lost.

The gravitational effect, however, is the same in both reference frames. Since gravity is the product of both masses, it doesn't matter which mass the \gamma factor applies to. It is still F \propto \gamma m M/R^2

AM

See either of the two links I posted (the java applet, or the equations) for why the electric field of a charge is NOT just qE1 E2 / R^2 when the charge is moving at a relativistic velocity.

Columb's law does not work for relativistically moving charges. Similarly, Newton's law of gravitation will not work for relativistically moving masses.

 

[pervect]

Actually, I would be really happy if someone could give me a good explanation of how kinetic energy works in special relativity.

I mean, if a 1 kg rock hits the Earth at 100 m/s, you usually calculate the energy of that collision by using the mass of the rock. Why don't you use the mass of the Earth? Why isn't the Earth moving towards the rock instead?

OK, here's the good concise explanation of the problem you asked about.

The physically significant quantity in a collision between two masses is the energy in the center-of-mass frame.

If a 1kg rock hits the Earth at 100m/s, the center of mass frame is essentially that of the Earth.

You can acutally work the problem of the rock hitting the Earth out in any frame you desire, but the center of mass frame is the simplest, and the most standard.

Note that this is why particle accelerators go to the bother of creating anti-protons to slam into protons with an equal and opposite velocity. The energy that counts is the energy in the center-of-mass frame, and much of this energy is lost when a hyper-relativistic particle strikes a stationary target. You get well over double the energy in a "head-on" collision.

 

[pmb_phy]

The gravitational effect, however, is the same in both reference frames. Since gravity is the product of both masses, it doesn't matter which mass the \gamma factor applies to. It is still F \propto \gamma m M/R^2

AM

That is an incorrect expression for the gravitational force in GR.

The active gravitational mass of a moving body is not \gamma M, its \gamma M(1+\beta^2).

See - http://www.geocities.com/physics_world/gr/moving_body.htm

Pete

 

[pervect]

I still don't see where you get eq 3a) in your derivation - I was hoping that the paper you cited by Harris would answer this question, but it does not.

Harris ("Analogy between general relativity and electromagnetism ...") gives the formula

<br /> m\frac{d^2 x^u}{d \tau^2} + m\Gamma^u{}_{ab}(\frac{dx^a}{d \tau})(\frac{dx^b}{d \tau})<br />

his eq 1a), as the formula for deriving "force", which I do believe, at least as long as one is in a coordinate basis (the last point may seem picky, but it was the source of at least some of my earlier difficulties).

 

[pmb_phy]

I still don't see where you get eq 3a) in your derivation

See Eq. (8) in - http://www.geocities.com/physics_world/gr/grav_force.htm

Pete

 

[pervect]

I should add, that by applying the above definition, for the metric for Newtonian gravity (U = - \Phi)

(1+2U)dx^2 + (1+2U)dy^2 + (1+2U)dz^2 - (1-2U) dt^2

with a 4-velocity
<br /> (t,x,y,z) = (\gamma, \beta \gamma, 0, 0)<br />

\beta = v/c, \gamma = 1/\sqrt{1-\beta^2}

I'm getting

<br /> \frac{d^2 x}{d \tau^2} = \frac{\frac{\partial U}{\partial x} - 2 \gamma^2 \frac{\partial U}{\partial t}}{1+2U}<br />

the main concern is the term in the partial of U with repsect to t (as U is not necessarily indpedent of time depending on the coordinate system used).

[add]

also
<br /> \frac{d^2 y}{d \tau^2} = \frac{1+\beta^2}{(1+2U)(1-\beta^2)} \,\frac{\partial U}{\partial y}<br />

which also didn't seem compatible

 

[pervect]

See Eq. (8) in - http://www.geocities.com/physics_world/gr/grav_force.htm

Pete

http://www.geocities.com/physics_world/gr/eq_gif/gr01-eq-08.gif

conflicts with Harris's article (Anology between General Relativity and electromagnetism for slowly moving paticles in weak gravitational fields) and MTW, which give

<br /> m\frac{d^2 x^u}{d \tau^2} + m\Gamma^u{}_{ab}(\frac{dx^a}{d \tau})(\frac{dx^b}{d \tau})<br />

instead of what you wrote, so that the equivalent 4-force is

-m\Gamma^u{}_{ab}(\frac{dx^a}{d \tau})(\frac{dx^b}{d \tau})

If we use the above formula to calculate the 'x' component of the 4-acceleration with a 4-velocity \beta in the x direction, we get

<br /> \frac{d^2x}{d \tau^2} = -\frac{\partial U}{\partial x} / (1 + 2U)<br />

which reduces to

mx / R^3 in the limit where R >> m, R being of course x^2+y^2+z^2

(This is because the metric at this point is static, we consider the mass stationary and the particle to be the only thing that is moving).

This compares very well with my "method two" which was based on consideration of the formula (for a Schwarzschild geometry)

rdot = \frac{dr}{d \tau} = \sqrt{E^2-(1-2*M/r)*(1+L^2/r^2)}

We can use this to calculate \frac{d^2 r}{d \tau}^2 as follows

\frac{d^2 r}{d \tau^2} = \frac{d rdot}{d r} \frac{dr}{d \tau} = \frac{d rdot}{dr} rdot

and when L = 0, for an radially infalling/outgoing particle, we get \frac{d^2 r}{d \tau^2} = -M/r^2 regardless of the value of rdot, which is the same as the previous result.

Furthermore, I realized the "simple" way to work this problem is to work entirely within the simple static metric, then convert the 4-acceleration to a new set of moving coordinates after the 4-acceleration is computed in the "mass centered" coordinates via a final Lorentz boost.

I still get rather strange-looking results with a factor of (1-3v^2) in them for the x-acceleration after doing the final boost, but I'm beginning to believe them, though my explanation for the change in sign of the force at v^2 = 1/3 (!) is a bit strained at this point.

 

[pmb_phy]

http://www.geocities.com/physics_world/gr/eq_gif/gr01-eq-08.gif

conflicts with Harris's article (Anology between General Relativity and electromagnetism for slowly moving paticles in weak gravitational fields) and MTW, which give

<br /> m\frac{d^2 x^u}{d \tau^2} + m\Gamma^u{}_{ab}(\frac{dx^a}{d \tau})(\frac{dx^b}{d \tau})<br />

I don't recall MTW using a definition of force and I lost my copy of Harris.

The definition I used is dP_u/dt where u is a covariant index and let u = k = 1,2,3. Had I not used this definition the gravitational force would not reduce Newton's expression in the Newtonian limit. Had I use proper time rather than coordinate time then a moving body would not weigh more than the same body at rest, which it must. Both Moller and Mould use this definition. Ohanian uses dP_u/d(tau)

Pete

 

[pervect]

I've uploaded my latest derivation to

http://www.geocities.com/pervect303/movingmass.pdf

I don't see how to reconcile the results at this point.

I previously attempted to upload the file as an attachment, but it was too big

 

[Creator]

I came across a book by Jerome Drexler in which he attempts to support his view that ultra high energy protons (cosmic rays) are what dark matter is all about in the galactic halos. ... he argues that the mysterious gravitational influences from these photons is due to the relativistic mass increase due to their high velocities.

I812...it may interest you to know that...
Generally, the results of most determinations of the gravitational field of a massless particle show the grav. field to be non-vanishing in a plane containing the particle and which is orthogonal to its direction of motion.
(see ref#1 below & the like).

This is the extreme case of that presented by Pete for relativistic massive bodies which shows the gravitational field compressed in the direction of, and dilated normal to, the direction of motion; (see the diagram at the end of his link to help you visulaize).

Thus a massless particle in effect is predicted to have grav. field which is a plane fronted gravity wave...

1.) 'Gravitational Field of a Massless Particle', P. Aichelburg & R. Sexl, Gen. Relativity & Gravitation, vol.2, no.4 (1971).

Creator

 

[pervect]

http://arxiv.org/abs/gr-qc/0110032 also has a good article on the Aichelberg-Sexl boost. (This link points to the abstract, to get the full paper click on the format desired, i.e. 'pdf').

I agree with Pete's results qualitatively, and my results also predict an increase in the transverse gravitational field of a moving particle. However, my independent derivation doesn't seem to match his on some important points, one of the key issues is the issue of the field when one is moving directly towards a gravitating object at a high velocity. I find that \mbox{{d^2 x}/{d \tau^2}} is the same for a particle that's stationary and one that's moving directly towards or away from the gravitating object. (Here x is measured in the frame in which the gravitating body is stationary). This result is similar to the behavior of an electric field - the component of the field E in the direction of a boost is unchanged by the boost, i.e. Ex is unchanged by a boost (velocity change) in the x direction.

I also am a bit skeptical that the force is the gradient of a scalar potential function - I'm fairly sure that this is not true for electromagnetism.

See http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_15.pdf

The E-field of a moving charge is a
remarkable electric field. It is a field that
no stationary charge distribution, in
whatever form, can produce. For this
electric field has the property that
[it has a nonzero curl]
and the line integral of E is not zero
around every closed loop.

Thus the electric field of a moving charge is not the gradient of a scalar potential function, because the curl of a gradient is zero, and the curl of this field is not zero.

 

[Gamish]

I was just woundering today of relavistic mass counts towered gravitational atraction. My asumption from reading this thread is yes, when using the equations F=G(Mm/r^2), the M and m variables can use relavistic mass. In an Email to a good friend of mine (a physicist), he said that there is NO such thing as non-relavistic mass.

 

[pervect]

Physics definiteily has a concept called "invariant mass", that does not depend on the velocity of an object. It's the invariant length of the energy-momentum 4-vector. See for instance the sci.physics faq "does mass increase with velocity"

http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

It is sheer folly to say that "there is no such thing as invariant mass".

Your assumption that you can plug "relaitivistic mass" into the equations F=GmM/r^2 (a nonrelativistic formula) and get sensible answers is almost surely wrong, and is an example of the common MISSUSE of the term relativistic mass. This sort of missuse is why I think the concept is not particularly useful.

 

[Gamish]

mass

Well, I just heard that from a physicist, I tried to tell him that there is a "relavent mass" and an "invarient mass", he did not believe me. So, if relavent mass cannot be plugged into the gravitatiotional force, but relavent mass does create a gravitational atracction, then give me the equation to calculate it. And if you can, explaine it.



I'm the master at time!

 

[chroot]

The gravitational attraction of a body with a given (rest-, or invariant-) mass is always the same, no matter how fast it's moving with respect to anything else.

- Warren

 

[Andrew Mason]

The gravitational attraction of a body with a given (rest-, or invariant-) mass is always the same, no matter how fast it's moving with respect to anything else.- Warren

This does not seem to square with special relativity. Einstein showed that energy and mass are equivalent so that the emission of electromagnetic energy from body A which is absorbed by body B increases the rest mass of body B and reduces the rest mass of body A. Surely this means the gravity of B increases and that of A decreases.

Now if you add kinetic energy to mass B instead of electromagnetic potential energy, you will also increase the mass of B. But if, as you suggest, it does not increase B's gravity, there must be a physical reason for making a distinction between the form of energy. If so, what is it?

AM

 

[pervect]

In the standard approach, "forces" aren't used at all in GR. Instead particles with no forces acting on them are said to follow geodesics in space-time, which is considered to be curved.

The geodesic equation is

<br /> \frac{d^2 x^a}{d \tau^2} + \Gamma^a{}_{uv} (\frac{dx^u}{d\tau})(\frac{dx^v}{d\tau}) = 0<br />

""Gravitation", Misner, Thorne, Wheeler - pg 211. (This standard formula is also eq 1) in Harris, "Anaology between general realtivity and electromagnetism for slowly moving particles in weak gravitational fields")

It is possible to interpret the quantity

<br /> m \Gamma^a{}_{uv}(\frac{dx^u}{d\tau})(\frac{dx^v}{d\tau}) <br />

as a sort of force, as long as the spatial part of space-time isn't too terribly curved. In this case, one interprets the departure from geodesic motion along a straight line in space as being due to a "force" rather than a curvature. This is actually a relativistic four-force, not the usual "three-force".

You can calculate \Gamma^a{}_{uv} in any coordinate basis from the metric via the formulas

<br /> \Gamma_{auv} = \frac{1}{2} (\frac{\partial g_{au}}{\partial v} + \frac{\partial g_{av}}{\partial u} - \frac{\partial g_{uv}}{\partial_a})<br />

and

<br /> \Gamma^a{}_{uv} = \sum_x g^{ax} \Gamma_{xuv}<br />

The last formula is just the usual formula for "raising an index".

Add to this the formula for the metric in isotropic coordiantes (given below), or the easier equation of the metric of an almost Newtonian source if you're only interested in weak gravity, and you're all set!. I should point out that if you are interested in interpreting gravity as a force, you'll HAVE to restrict yourself to the almost Newtonian case, so you can save yourself a lot of work by avoiding isotropic coordinates.

http://relativity.livingreviews.org/Articles/lrr-2000-5/node9.html

Sorry this is so involved, but nobody said GR was easy. You can perhaps appreciate why there appears to be a lack of resolution on the issue of the exact result between myself and another poster considering the complexity of the problem.

However, the good news is that you can gain a considerable insight from studying the behavior of electric fields of charged particles, and how they act relativistically. The results won't be numerically exactly the same as those for gravity, but the math is a lot easier.

As I remarked to another poster, the columb force law f = K q1 q2 / r^2 does not work at relativistic velocities either. You can, however, define the electric (and, magnetic) field of a moving charged particle. Notiting that the magnetic force on a stationary charge is zero (as F=q v x B, and v=0), we can say that the force on a stationary test charge is due entirely to the electric field, and is equal to F=qE. This equation, ignoring the magnetic field, then gives the relativistically correct force between a moving particle and a _stationary_ test particle in our frame of reference.

We can then ask "what does the electric field of a moving charge look like". This will give us a lot of insight (though not the exact numbers) of what the gravitational field of a moving mass looks like.

If we were interested in gravitomagnetism, we could also look at the magnetic field of the moving charge, and compare it to the "gravitomagnetic field", but that would be a topic for another post I think.

Here are three links that compute the electric field of a moving charge
I've posted these all before, but I'm posting them again, in the hopes that interested parties will actually take the time to _look at them_ this time around


http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_15.pdf

This is probably the clearest link

just the formulas

a java applet

In interpreting the output of the java applet, it is helpful to know that the electric field strength is not directly shown, rather the electric field lines are shown. The number of field lines crossing a unit area gives the component of force in the direction perpendicular to that area.

 

[DB]

May be off subject, but what is relativistic beta?

 

[chroot]

\beta \equiv \frac{v}{c}

It's just a convenience variable -- the speed in units of c.

- Warren