Does mass increase in SR mean higher gravity in GR?

pervect

http://www.geocities.com/physics_wor...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

[tex]
m\frac{d^2 x^u}{d \tau^2} + m\Gamma^u{}_{ab}(\frac{dx^a}{d \tau})(\frac{dx^b}{d \tau})
[/tex]

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

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

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

[tex]
\frac{d^2x}{d \tau^2} = -\frac{\partial U}{\partial x} / (1 + 2U)
[/tex]

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)

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

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

[tex]\frac{d^2 r}{d \tau^2} = \frac{d rdot}{d r} \frac{dr}{d \tau} = \frac{d rdot}{dr} rdot [/tex]

and when L = 0, for an radially infalling/outgoing particle, we get [tex] \frac{d^2 r}{d \tau^2} = -M/r^2 [/tex] 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

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

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 [tex]\mbox{{d^2 x}/{d \tau^2}}[/tex] 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/PHY2...ativity_15.pdf

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 equasions F=G(Mm/r^2), the M and m variables can use relavistic mass. In an Email to a good freind 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/physic...y/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

Well, I just heard that from a physist, 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 equasion 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

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

[tex]
\frac{d^2 x^a}{d \tau^2} + \Gamma^a{}_{uv} (\frac{dx^u}{d\tau})(\frac{dx^v}{d\tau}) = 0
[/tex]

""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

[tex]
m \Gamma^a{}_{uv}(\frac{dx^u}{d\tau})(\frac{dx^v}{d\tau})
[/tex]

as a sort of force, as long as the spacial 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 [tex]\Gamma^a{}_{uv}[/tex] in any coordinate basis from the metric via the formulas

[tex]
\Gamma_{auv} = \frac{1}{2} (\frac{\partial g_{au}}{\partial v} + \frac{\partial g_{av}}{\partial u} - \frac{\partial g_{uv}}{\partial_a})
[/tex]

and

[tex]
\Gamma^a{}_{uv} = \sum_x g^{ax} \Gamma_{xuv}
[/tex]

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/...0-5/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


a diagram

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

[tex]\beta \equiv \frac{v}{c}[/tex]

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

- Warren

pervect

What makes you say this? No disrespect, but I think this is incorrect. Certainly, F=K q1 q2 / r^2 does not work for the force between a relativistically moving charge and a stationary one (see my previous post) - why would one think that the analogous equation works for gravity under the same conditions?

Furthermore, it is difficult if not impossible to explain how pressure causes gravity if velocity does not change the gravitational force between particles. If we use the "swarm of particles" model for a perfect fluid, the difference between a fluid without pressure and a fluid with pressure is just the fact that the particles are moving. If the motion of the particles had no effect, the gravitational field of a pressureless fluid would be the same as that of a fluid with pressure - but it is not, the pressure of the fluid contributes to the stress energy tensor.

Creator

Thanks for your response, Pervect, to my Aikelburg-Sexl comments, and for the reference.
Before we go on to address the origin of your differences with that of Pete's derivation (the discussion of which I find very interesting), let me first ask you a question:

1st. Aikelburg & Saxl (whose authority you seem to accept) state that (as determined in both the linearized and exact solutions) ..."Physically the gravitational field of a rapidly moving particle shows the same characteristic behavior as its electromagnetic field: it is...compressed in the direction of motion".

2ndly, Your own link (http://www.phys.ufl.edu/~rfield/PHY2...ativity_15.pdf )
verifies the validity of this statement and in the diagram given clearly shows shortened field lines in the direction of motion; (look at it closely if you missed it).

Why then do you insist on saying, " ..... 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."??
Respectfully, on this point then, I think you are possibly in error.
If it is your derivation that has led you to such a conclusion, then it is not only contrary to that of Pete's but also to the above referenced items.

Creator

pervect

I probably should have posted a link to the predecessor article

http://www.phys.ufl.edu/~rfield/PHY2...ativity_14.pdf

Note that at the same location in space-time Ex, measured in the O frame, is the same as Ex', measured in the O' frame.

This is a consequence of how the electromagnetic field transforms under a boost

this link

might be useful in confirming that fact (or it might be an unneeded complication at this point, I dunno).

So why is the field clearly shorter in the diagram? It's shorter not because the field at the same point in space-time is different because of the boost - it's because the field is expressed in different variables. x' is different from x by a factor of gamma, and when one is moving towards the source, the distance r' is diffrent from the distance r by a factor of gamma.

As far as the non-electrostatic problem goes, the summary of my calculations are at

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

I feel fairly happy about the results in 3), but not very happy about what happens when I "boost" back to the coordinate system in 5). Interestingly, the steps needed to go from 3 to 5 are fairly simple. The results in 5) are the ones the most comparable to Pete's results, and the pictures I've presented, and what I want to find out - the fields that a stationary observer would see as a large mass "whizzes by" him.

I don't see at this point how to reconcile the results with Pete's. I have some ideas on things I want to try and look at, but I haven't gotten around to doing any more serious work, in part because of a nasty flu that I've come down with.

I'm also still suspicious of the fact that Pete's results are the gradient of a scalar function.

Chronos

Respects to pervect and creator. Good technical points all. Anyways, in defense of Chroot [like he needs it] he is correct in the real universe. The mass of an accelerated object does increase as it gains speed, but the system mass does not. [a self propelled object actually bleeds off mass entropically due to efficiency losses in the propulsion system]. This is a classic fallacy in SR. While it is true a mass possessing body accelerated to the speed of light acquires infinite mass, it requires an infinite amount of energy to accelerate it to that speed. If you dutifully obey the laws of thermodynamics and do the math [e=mc^2], you will find the net effect is zero in a self propelled system. Paradoxes only occurs when you import 'free energy' from nowhere to power the acceleration. I suggest that a similar energy cutoff [albeit gigantic] should be considered in any real universe solutions to field equations. The universe may be infinite, but, the causally connected fragment we reside in, is not.