Gravitational field of a moving object

In summary, the two professors said that mass is not invariant, and is given by m'=\gamma m_0 (will always be bigger than the self-mass).
  • #1
Ahmes
78
1
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 [tex]F=\gamma m a[/tex]. Where "m" is what he called "mass" and [tex]\gamma[/tex] is the velocity-related constant.
The other professor said that mass is not invariant, and is given by [tex]m'=\gamma m_0[/tex] (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 [tex]Gm/r^2[/tex].
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 [tex]m_{me}[/tex] so the force between us (if the spaceship were at rest) would be [tex]F=Gm_{me} m/r^2[/tex]. The mass of the spaceship, as I see it, would be [tex]m=F r^2/G m_{me}[/tex].

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 [tex]\gamma[/tex] times that mass?

Thank you.
 
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  • #2
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.

[edit] 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)
 
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  • #3
:smile: 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? Let's 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
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 followed the link you borough and links inside it and will read http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html soon.
I almost fell off my seat when I first read now "longitudinal mass" :biggrin: 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
janusz said:
:smile: 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? Let's 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?


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
Ahmes said:
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?

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).


I followed the link you borough and links inside it and will read http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html soon.
I almost fell off my seat when I first read now "longitudinal mass" :biggrin: 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).

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

To find the real origin of the concept of relativistic mass, you have to look back to the earlier papers of Lorentz. In 1904 Lorentz wrote a paper Electromagnetic Phenomena in a System Moving With Any Velocity Less Than That of Light. There he introduced the "longitudinal" and "transverse" electromagnetic masses of the electron. With these he could write the equations of motion for an electron in an electromagnetic field in the Newtonian form F = ma, where m increases with the electron's speed. Between 1905 and 1909 Planck, Lewis and Tolman developed the relativistic theory of force, momentum and energy. A single mass dependence could be used for any acceleration if F = d/dt(mv) is used instead of F = ma. This introduced the concept of relativistic mass which can be used in the equation E = mc2 even for moving objects. It seems to have been Lewis who introduced the appropriate velocity dependence of mass in 1908, but the term "relativistic mass" appeared later. [Gilbert Lewis was a chemist whose other claim to fame in physics was naming the photon in 1926.]

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

[tex]
p = \frac{mv}{\sqrt{1-\left(\frac{v}{c}\right)^2}}
[/tex]

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

then

[tex]
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}}}
[/tex]

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

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

[tex]
\gamma = \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}
[/tex]
 
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  • #7
selfAdjoint said:
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.
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
Ahmes said:
Hello,
Two physics professors said two different things about the invariance of mass.
For some reason that doesn't surprise me. :smile:
One said that "mass" is a quantity which is always measured in the object rest frame - and therefore invariant to the Lorentz transformation.
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?
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 [tex]F=\gamma m a[/tex]. Where "m" is what he called "mass" and [tex]\gamma[/tex] is the velocity-related constant.
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 high school. This is at a time where most students have not be introduced to calculus. Therefore the real definition of force is

[tex]F = \frac{dp}{dt}[/tex]

Feynman made this quite clear in The Feynman Lectures but this good stuff is always ignored for some dumb reason.
The other professor said that mass is not invariant, and is given by [tex]m'=\gamma m_0[/tex] (will always be bigger than the self-mass).
He's a smart man. :biggrin:

I'm sure he'd enjoy reading my first cut of this article
http://www.geocities.com/physics_world/mass_paper.pdf
And now for the gravity thing:
An object at rest will produce a field of gravity given by [tex]Gm/r^2[/tex].
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 [tex]m_{me}[/tex] so the force between us (if the spaceship were at rest) would be [tex]F=Gm_{me} m/r^2[/tex]. The mass of the spaceship, as I see it, would be [tex]m=F r^2/G m_{me}[/tex].

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 [tex]\gamma[/tex] 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
pmb_phy said:

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

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 [itex] \nabla \times E \neq 0 [/itex]
and the line integral of E is not zero around every closed loop.

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).

Measuring the active gravitational mass of a moving object

D. W. Olson and R. C. Guarino
Department of Physics, Southwest Texas State University, San Marcos, Texas 78666

(Received 27 October 1983; accepted 10 July 1984)

If a heavy object with rest mass M moves past you with a velocity comparable to the speed of light, you will be attracted gravitationally towards its path as though it had an increased mass. If the relativistic increase in active gravitational mass is measured by the transverse (and longitudinal) velocities which such a moving mass induces in test particles initially at rest near its path, then we find, with this definition, that Mrel=gamma(1+beta2)M. Therefore, in the ultrarelativistic limit, the active gravitational mass of a moving body, measured in this way, is not gamma M but is approximately 2gamma M.

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.
 
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  • #10
pervect said:
Sorry, but that write-up is seriously flawed.
Please provide proof.
Your equation 3a) appears mysteriously from nowhere and defines an ambiguous quantity.
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.

The final result is suspect on its mere appearance. The electrostatic force on a moving charge is not the gradient of *any* scalar potential
Don't confuse analogy with identity. I.e. one shouldn't expect the answer to be identical to that of the EM analogy.

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.
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. :smile:
 
  • #11
pmb_phy said:
Please provide proof.

The biggest thing that makes me think your analysis is wrong are the statements 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.

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. :smile:

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
selfAdjoint said:
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.
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.
Please be patient again.
Let's say that in very remote future humanity will built the big spaceship to reach another side of galaxy.For them like to humans doesn't matter where they go just to achieve the goal.They do not care about planetary enviriment and adopt the Moon to trip the Jupiter is going to be the fuel tank.This very important day for humanity just come and big spaceship started.Slowly it reach speed as close possible to the speed of light.Meantime humankind die off because of new holy war but just before that they observe for almost 5 years the ship slowly reaching Alpha Centauri.The other side of galaxy is 100 000 light years away,and that much time humans on the Earth had to wait to get results,but thanks to relativity travel time on the ship is going to be only one month.
The daytime on the Moon ship start with raising of the gravity it is summary of gravity of the surface plus acceleration,on night time the hair raising because of the fuel tank ,it is located in the safe distance say 200000 km away.Stars start moving ,in the 100 000 years they move quiet a lot.but they still stay under influence of Milky Way and its central bulge.They has to spend huge energy to clean up the path and finally spaceship reached the destination.It was not enough fuel to stop the ship,but humans did not care about it.Space ship slowdown like using aerobreaking, slow down to the point of inertia,certain point alowwed by spacetime for this quantity of mass.
Spaceship should do straight line course a,a1,a2 but due to expansion of the space straight line change to cone of possibilities abcd..
No illusions in the physics and no negligible rest in the math?
From the point of view of the traveler,his length of the galaxy is one month of flight,his body been streched to millions of kilometers ,been exited in different slowdown universe.
And now I come to the point.
I imagine the time like sheet of paper ,membrane penetrating through ashes of energy.,no straight line where you can ask for past and future,so imagine that every moment everywhere in the universe curl up POINT of expansion with 3 spatial dimension and membrane of time.Here is our spaceship because of the speed he get caught up by some and skipped some exploding POINTS in his movement.When he tunneled through the spacetime it changed parameters of spacetime according to Einstein the other one .mass has to occur too in the direction of the movement - that's how I Understand longtudinal mass of Lorentz.In every second one atom was present in the line of the points .Movement close to speed of the light is like quantum mechanics with the only probability of egsistens of the particle on the line of the move.Expansion is irregular in the no presence of matter,how this can influence the movement in the empty pockets of galaxy,would be harder to accelerate? So mass increase?Everything moving,is no rest point for matter,for energy.Why Expansion of the universe in every its point is not taken into account?It happen even in within ourself.In the latest book of Brian Greene I have red that expansion is slower in the presence of matter,so close to conclusion ,our world is deforming ,and another one if matter has natural tendence to collapse and Points of space to expansion why not to interconnect this two forces into force of every force - fields and put funny gene graviton into the bin.In this instance inertial mass is fully explainable.Thank you for beeing patient and please for answer...
 

FAQ: Gravitational field of a moving object

What is the gravitational field of a moving object?

The gravitational field of a moving object refers to the force exerted by the object on other objects around it, due to its mass and motion. This force is described by Newton's Law of Universal Gravitation, which states that the force is directly proportional to the mass of the object and inversely proportional to the square of the distance between the objects.

Does the gravitational field of a moving object change?

Yes, the gravitational field of a moving object changes as the object's mass and velocity change. As an object gains or loses mass, its gravitational pull on other objects will also change. Additionally, as the object's velocity changes, the strength and direction of its gravitational field will also change.

How is the gravitational field of a moving object calculated?

The gravitational field of a moving object can be calculated using the formula F = (GmM)/r^2, where F is the force of gravity, G is the universal gravitational constant, m and M are the masses of the two objects, and r is the distance between them. This formula takes into account both the mass and motion of the objects involved.

Can the gravitational field of a moving object be measured?

Yes, the gravitational field of a moving object can be measured using various instruments such as a gravimeter. This device measures the acceleration due to gravity, which is directly related to the strength of the gravitational field. By measuring the acceleration of a falling object, the gravitational field can be calculated.

How does the gravitational field of a moving object affect other objects?

The gravitational field of a moving object affects other objects by exerting a force on them. This force can cause objects to accelerate towards or away from the moving object, depending on their relative positions and masses. The gravitational field also plays a crucial role in the motion of celestial bodies, such as planets and satellites, in our solar system.

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