# Gravitational field of a moving object

1. Sep 8, 2005

### Ahmes

Hello,
Two physics professors said two different things about the invariance of mass.
One said that "mass" is a quantity which is always measured in the object rest frame - and therefore invariant to the Lorentz transformation. In additions, laws of motion in "real life" (relativistic motion) aren't the same as in high school. For example, Newton's second law would be $$F=\gamma m a$$. Where "m" is what he called "mass" and $$\gamma$$ is the velocity-related constant.
The other professor said that mass is not invariant, and is given by $$m'=\gamma m_0$$ (will always be bigger than the self-mass).

And now for the gravity thing:
An object at rest will produce a field of gravity given by $$Gm/r^2$$.
Now I'm at rest and some object (say a spaceship) is moving near me, and I want to measure its mass by the gravitational force it applies on me.
I know that my own mass, at my own frame, is $$m_{me}$$ so the force between us (if the spaceship were at rest) would be $$F=Gm_{me} m/r^2$$. The mass of the spaceship, as I see it, would be $$m=F r^2/G m_{me}$$.

Which 'm' did I find? Is it the same mass I would measure if I took the spaceship and weighed it in my frame, or is it $$\gamma$$ times that mass?

Thank you.

2. Sep 8, 2005

### pervect

Staff Emeritus
The gravitational field will be given by neither of the above expressions.

Much like the electric field of a moving charge, the gravitational field of a moving object, to the extent it can be defined, will vary with angle.

The definitional problem is that there is no such thing as a gravitationally neutral test particle. So while the electric field at a point can be directly measured by comparing the motion of a charged and an uncharged particle, the gravitational field at a point cannot be directly measured.

What can be directly measured is not the gravitational field itself, but its rate of change, i.e. the tidal force on an object.

I go through some detailed calculations of the tidal force in this thread

but I'm afraid they aren't as clearly edited as they could be.

To give a subset of the results, consider two co-moving particles A and B, where particle B has a mass M

A-------B

The stretching tidal force on particle A is given by 2 G M L/R^3, it is oriented in a line joining A to B. Here L is the length of the tidal force measuring device (for instance a Forward mass detector), M is the mass of particle B, and R is the distance between A and B.

 Actually, R is the Schwarzschild coordinate of the particle at A with respect to the large mass M located at B - this is almost the same thing, but not quite.
[end edit]

There are also compressive tidal forces, but I won't get into those, see the original posts.

Consider a particle C located at the exact same position as A.

(A,C)-------B

If C moves directly towards B, there is no change in the tidal force.

If C moves "up the page", the stretching tidal force increases.

Rather than 2 G M L/R^3, one gets

(2 G M L / R^3) * (1 + (v/c)^2/2) / (1 - (v/c)^2)

Last edited: Sep 8, 2005
3. Sep 9, 2005

### janusz

:rofl: I am enthusiastic to the field -this is sort of delicate warning so do not kill me straightaway.I am under influence of popular science books and I have some important for me questions.Typical illustration of the gravitation in these sort of books is acceleration of the lift or rockets where direction of the force is down on the object.I wonder why always the gravitation of the object is negligible.What I know in the close approach of the speed of the light the mass of the object can increase to infinity? Lets stop when reach the mass of the Jupiter ( the bag of 1 kg of sugar for instance) and do not accelerate any more.Is it existing real mowing mass with distorted symmetry of spacetime? Can be detectible?

4. Sep 9, 2005

### Ahmes

Thank you pervect, This is more complicated than I thought. My original intention though, was that the spaceship was traveling towards me in a straight line. I wasn't particularly worried about the non-existence of a neutral test particle since I already now what forces I feel when the spaceship is very far away. Do I miss the point?

I almost fell off my seat when I first read now "longitudinal mass" Google from some reason didn't find any impressive explanations. Mass is a scalar of course and I still didn't quite figure out what "longitudinal mass" means. I'll thank you if you can tell me where to find more informations about the subject (ie. website, books).

5. Sep 9, 2005

Staff Emeritus

The Lorentz increase in mass only occurs in one formalism of special relativity. In the other formalism, favored by particle physicists, the nergy increases with relative speed but the mass is invariant. If I am not mistaken, the increasing mass formalism breaks equivalence, so the inertial increased mass is not the gravitational mass, and thus your argument is moot.

6. Sep 9, 2005

### pervect

Staff Emeritus
The problem is definitely more complicated than it looks. You know what the forces are on a far away object only after you define a coordinate system.

To put it another way, if you can force an object to maintain a constant velocity and direction with respect to some coordinate system, i.e. make the object move in a "straight line", you can use an accelerometer to measure the gravitational force on the moving object.

Unfortunately, space itself is curved near a large mass, so it's not terribly clear what a "straight line" really means when velocities approach 'c'. Various terms due to the curvature of space which one wants to neglect become non-negligible. You want to neglect these terms, because you want to assume that space is flat. But when you do the math, the terms aren't negligible :-(.

One way around this difficulty is to compute the answer in terms of tidal forces. The answer in terms of tidal forces is physically meaningful and simpler to compute and also doesn't suffer this definitional problem.

The question of what the electric field of a moving charge is like is also quite interesting and much easier to solve, and helps provide much insight into what the gravitational field of a moving mass is, though the two questions are NOT equivalent.

Take a look at the following links:

http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_15.pdf
http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_14.pdf

When you move straight towards a charge, two things happen:

1) the electric field does not change
2) the measured distance to the charge goes down due to Lorentz contraction.

The combination of effects 1 and 2 can be interpreted as an _weakening_ of the electric field in the direction of motion.

However, the electric field in the _transverse_ direction is strengthened.

Very similar things happen with gravity, though the numerical factors by which the field is weakened and strengthened are NOT exactly the same.

(Note that the "weakened" field is physically the same for two co-located particles, it's just that the moving particle measures a shorter distance).

From the sci.physics.faq I quoted, here's the part that talks about longitudinal mass.

With calculus, it's not too hard to compute the value of the longitudinal mass.
The point is that given that the momentum p is given by the formula

$$p = \frac{mv}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$$

where m is the "invariant" mass aka "rest mass"

then

$$F = \frac{dp}{dt} = \frac{dp}{dv} \frac{dv}{dt} = \frac{m a}{\left(1-\left(\frac{v}{c}\right)^2\right)^{\frac{3}{2}}}$$

where F is the force (dp/dt), and a is the acceleration (dv/dt).

This gives a ratio F/a, the "longitudinal mass" of $\gamma^3 m$ where

$$\gamma = \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$$

Last edited by a moderator: Apr 21, 2017
7. Sep 11, 2005

### janusz

What I understood my 1 kg bag of sugar stay the same even the energy been pumped to it to reach the speed closed to C? How come the same 1kg bag gain the gravitational mass when is placed to Jupiter surface? My understanding of relativity was maybe wrong - I thought the gravity of the object is strictly tied to influence of spacetime symmetry ,so 1 kg bag weights more on Jupiter,time goes slower and the planet is bigger then normally due to relativity interaction.

8. Sep 11, 2005

### pmb_phy

For some reason that doesn't surprise me. :rofl:
More correctlyh referred to as proper mass. You can't really measure the mass of a photon in that way. What you failed to state was how they were doing the measuring?
That is a widely spread misunderstanding of the definition of force. Your professors have failed you in this respect. Most people first learned about force = ma in highschool. This is at a time where most students have not be introduced to calculus. Therefore the real definition of force is

$$F = \frac{dp}{dt}$$

Feynman made this quite clear in The Feynman Lectures but this good stuff is always ignored for some dumb reason.
He's a smart man.

http://www.geocities.com/physics_world/mass_paper.pdf
That depends on the particular gravitational field you're speaking of. It does not hold in general and that is a Newtonian expression. Not a relativistic one. For an example of a Newtonian g-field which can be produced which is not of that nature note that if you have a spherical body with uniform mass density and you cut out a cavity which is small enough to lie totally within the sphere but not concentric to it then you'll have a uniform field which, when measured in the rest frame of the body, will be f = mg where g is a constant which is a function of the distance the center of the cavity is from the center of the sphere.

Now I'm at rest and some object (say a spaceship) is moving near me, and I want to measure its mass by the gravitational force it applies on me.
I know that my own mass, at my own frame, is $$m_{me}$$ so the force between us (if the spaceship were at rest) would be $$F=Gm_{me} m/r^2$$. The mass of the spaceship, as I see it, would be $$m=F r^2/G m_{me}$$.

Which 'm' did I find? Is it the same mass I would measure if I took the spaceship and weighed it in my frame, or is it $$\gamma$$ times that mass?

Piece of cake - Please see -
http://www.geocities.com/physics_world/gr/moving_body.htm

Also read the following and show it to these two profs.

Measuring the active gravitational mass of a moving object, D.W. Olson, R.C. Guarino, Am. J. Phys. 53(7), July 1985

Pete

9. Sep 11, 2005

### pervect

Staff Emeritus
Sorry, but that write-up is seriously flawed.

Your equation 3a) appears mysteriously from nowhere and defines an ambiguous quantity.

The final result is suspect on its mere appearance. The electrostatic force on a moving charge is not the gradient of *any* scalar potential

See the large print in:

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

Thus it is definitely wrong for you to say (as you do) that the E-field of a moving charge is the gradient of some potential function, and it is highly suspicious that the gravitational field of a moving mass would be the gradient of a scalar potential.

You appear to conflate gravity and electromagnetism quite frequently in your writeup - it's not entirely clear whether you are or are not claiming that they transform in an identical manner.

Note that I don't have any argument with Olsen's paper at this point (I don't have the full paper, but the abstract appears quite sensible).

Note that Olsen is not attempting to define the ambiguous notion of gravitational force in a curved space-time, but instead looks at the total velocity imparted when a large mass passes close by, something that can be defined unambiguously because space-time will be flat after the close encounter with the large mass.

Last edited by a moderator: Apr 21, 2017
10. Sep 11, 2005

### pmb_phy

You asked me about that a long time ago and I gave you the reference to the equation. Its found in all general relativity texts which define the gravitational force. All it is is the simple equation F = dp/dt when the 4-force on the particle is zero and t = coordinate time. For the derivation please see -- http://www.geocities.com/physics_world/gr/grav_force.htm

You can find this same equation in Basic Relativity, Richard A. Mould, Springer Verlag Press. Its most likely in Moller too.

Don't confuse analogy with identity. I.e. one shouldn't expect the answer to be identical to that of the EM analogy.

I don't recall either saying that or thinking that.

Pete

ps - For the next month or two you'll have me at a disadvantange since I'll be hitting major surgery soon and won't feel much like posting anything, never mind arguing. But please don't make claims of so-called "ambigious" anything since this is a standard formula in GR in those GR texts which calculate gravitational force. You need simply ask and I'd provide as I did when you first asked me. But I've been in pain for over a year and a half and am not in the mood for correcting ommisions of references in my website yet. But one need only ask.

11. Sep 13, 2005

### pervect

Staff Emeritus
The biggest thing that makes me think your analysis is wrong are the statments you make about the electric field of a moving charge being the gradient of a scalar field, which is clearly wrong (see the web reference I posted a while back).

There are other problems, too.

You are taking a fundamentally ambiguous coordinate dependent concept (gravitational force), and asserting it has a definite value.

This is something that you should prove, and discuss in detail, not something that I should have to disprove. MTW and Carlip both talk about the ambiguity of gravity as a force, (though I can't seem to find the specific paper by Carlip which talks about the problem offhand, alas). MTW carefully avoids defining a "gravitational field" in any specific terms, other than to say that any of the various instruments used to study gravity (such as the Riemann tensor) can be losely called a "gravitational field". They definitely do not use the equation that you wrote down. (They do of course use the geodesic equation.)

My own comments about measuring the force requiring one to define a straight line in curved space should also illustrate the difficulty, however - you don't seem to "get it", and I'm not too sure how to explain it more clearly.

I very much suspect you've taken an approxmiation, which works at low velocities and nearly Newtonian fields, and used the approximation where it just doesn't apply.

Another test is if your results give the correct transformation of the tidal forces. We know how the Riemann should transform, in fact that's how I got the answers I got - so it's basically just a matter of re-computing your answers and comparing them to mine by computing the tidal tensor from your results. I think I'd have to extend my solution to cover the general case rather than just the "head on" and "tangential" cases to make the comparision, though.

I've been meaning to look at this more closely, but I haven't been able to get motivated enough.

I think there is more going on here than just the ambiguity I talked about, though - because of the clearly wrong results you get for the electric field.

Well good luck with your surgery - I can see that now is not the time to do a lot of work to find your error, cause you won't be listening anyway. So this whole thread will just re-occur the next time someone asks the same question, cause I just don't believe your results.

12. Sep 13, 2005

### janusz

Thank you very much for your response.Unfortunately I represent real bottom of sciencitific community and I can find handful number of people willing talk with me about cosmology or general relativity,in workers envirement .I have some basic knowledge and imagination doors in my brain are really wide open.I love general relativity theory is magnificent and unbeatable achievement of xx century.I'm its propagator as much I can grasp of it with my mind.What I would like to ask is your help with finishing my quest to find the answer to my question about it.I want to know the whole true,to build the sensible and easy explainable picture.