How Does Acceleration Affect the Relativistic Mass of Particles?

[starthaus]

This is fascinating. You said h = (angular momentum/rest mass). Are you seriously suggesting momentum is independent of v ??

Yes, one more time, h does not depend on v, it is a constant. You need to start reading here.

 

[yuiop]

Sure it does, both your equations are not applicable. The correct equation of motion can be obtained through the application of the Euler-Lagrange formalism:

d^2u/dphi^2+u=m/h^2

where m=MG/c^2 from the Schwarzschild metric : ds^2=(1-2m/r)dt^2-...-r^2*(dphi)^2
h=specific angular momentum (angular momentum/particle rest mass)

The relativistic mass plays NO role.

Neither side of the ODE is in units of force..

I am aware that none of the equations you gave are not in terms of force. I was wondering why you said the conclusions I have drawn from my equations of force are wrong, when you have clearly not bothered to work out what the equation for force is in your terms.

This paper by Matsas http://arxiv.org/PS_cache/gr-qc/pdf/0305/0305106v1.pdf about the relativistic submarine paradox is well known and not considered crank or controvertial. In it he concludes

Thus according to observers at rest with the fluid, the gravitational field on the moving submarine increases effectively by a factor of \gamma

which is clearly in agreement with the last equation I gave in #19.It is also in agreement with the claim that the force of gravity acting on a particle is velocity dependent.

He also gives this equation for the forces acting on a relativistic submarine as:

F_{TOT} = -mg(\gamma -1/\gamma)

which expands to:

F_{TOT} = -\gamma m g +mg/\gamma

where the first term on the right is the gravitational force acting downwards on the submarine and the second term is the buoyancy force acting upwards on the submarine. Both forces are velocity dependent.

This paper http://arxiv.org/PS_cache/arxiv/pdf/0708/0708.2488v1.pdf reaches similar conclusions. It states

The force required of a man on the train to keep an apple at a fixed height is higher when the train moves than when it is at rest relative to the platform.

. Clearly the author thinks gravitational force is velocity dependent. He later quantifies that as:

The force required to keep the apple of rest mass m at a fixed height relative to the train is thus given by F = m\gamma^2a.

This is the proper force (or received force) measured by an observer on the train.

Hence the given force is smaller than the received force by a factor of \gamma. Because the received force required to keep an object (like an entire train) moving along a straight horizontal line relative to a vertically accelerating reference frame is proportional to \gamma^2 (as discussed in Sec. II), it follows that the force required by the rail to support the train scales with a factor of \gamma

This clearly answers this question I posed earlier:

Imagine a train on a long straight horizontal monorail that is suspended from springs that directly measure the weight of the train and the rail. When the train is accelerated to the same velocity as the bullet in the OP will the track scales directly measure the train to be 7 time heavier?

in the afirmative and agrees with the assertion that force of gravity acting on a moving body is velocity dependent which agrees with what Ben and cos said earlier. If you are asserting that the force of gravity acting on a particle is not velocity dependent it seems you find yourself in a minority.


.

 

[starthaus]

I am aware that none of the equations you gave are not in terms of force. I was wondering why you said the conclusions I have drawn from my equations of force are wrong, .

There is no "force" in the formalism presented by Rindler because there is no such thing as gravitational force in GR. Trying to hack the relativistic mass into the Newton formula is definitely the wrong way to go.

when you have clearly not bothered to work out what the equation for force is in your terms

I tried to help you learn the correct approach. I am sorry I failed. Perhaps you should invest in buying the Rindler book.

 

[utesfan100]

Yes, one more time, h does not depend on v, it is a constant. You need to start reading here.

So, because angular momentum in quantum mechanics comes in discrete steps, angular momentum in GR is not dependent on v?

I refer you to http://en.wikipedia.org/wiki/Angular_momentum#Angular_momentum_in_relativistic_mechanics".

The basic definition of angular momentum is the product of radius with linear momentum.

Now the relativistic dependence of the radius with v depends on the angle of the radius with the velocity vector. Linear momentum clearly has a factor of gamma relative to the velocity.

Only in the case where the radius is aligned with the velocity do they cancel, but then wait for the object to rotate any amount in its plane of rotation and this will not be true any more.

 

[yuiop]

Yes, one more time, h does not depend on v, it is a constant. You need to start reading here.

The link you gave talks about spin angular momentum and the Planck constant \hbar which is certainly a constant, but I have a strong hunch that the h used by Rindler is not the Planck constant or the quantum mechanical spin. Could you check that?

 

[starthaus]

So, because angular momentum in quantum mechanics comes in discrete steps, angular momentum in GR is not dependent on v?

"Furthermore, experiments show that most subatomic particles have a permanent, built-in angular momentum, which is not due to their motion through space. This spin angular momentum comes in units of h_bar. For example, an electron standing at rest has an angular momentum of h_bar/2."


If you have further issues with the above, I suggest that you look at the Rindler derivation.

 

[starthaus]

The link you gave talks about spin angular momentum and the Planck constant \hbar which is certainly a constant, but I have a strong hunch that the h used by Rindler is not the Planck constant or the quantum mechanical spin. Could you check that?

Yes, I checked. I suggest that you buy the book.

 

[yuiop]

There is no "force" in the formalism presented by Rindler because there is no such thing as gravitational force in GR. Trying to hack the relativistic mass into the Newton formula is definitely the wrong way to go.

The authors of the papers (Matsas and Jonsson) I quoted are happy to hack the way I did for a local first order aproximation. You are right that there is no such thing as gravitational force for a free falling object in GR, but there is a force when a horizontally moving object has zero vertical velocity and is therefore not free falling.

 

[starthaus]

The authors of the papers (Matsas and Jonsson) I quoted are happy to hack the way I did for a local first order aproximation.

Yes, I saw that. Two wrongs don't make a right. I would always take Rindler over both of the others. I just started looking at the Jonsson paper, it is especially bad, especially when you look at his derivation for equations (1)-(3). As Pauli would have said, "not even wrong".

 

[utesfan100]

There is no "force" in the formalism presented by Rindler because there is no such thing as gravitational force in GR. Trying to hack the relativistic mass into the Newton formula is definitely the wrong way to go.

Does GR fail to have momentum? I would think that force could still be defined in terms of dP/dt. Is the momentum along a geodesic constant? I think the Earth's momentum as it follows its geodesic around the Sun demonstrates otherwise.

I will concede that such might not be a tensor, as gravity itself cannot be a tensor due to its intrinsic second order effects.

If the energy in the stress energy tensor includes kinetic energy, gravitation must be velocity independent.

I am not certain that the formula:

F=G/c⁴ E₁E₂/r² might not be better than Newton's law of gravitation. It certainly works better for photons than using rest mass.

 

[cos]

Kev, in another thread you wrote -

Imagine a train on a long straight horizontal monorail that is suspended from springs that directly measure the weight of the train and the rail. When the train is accelerated to the same velocity as the bullet in the OP will the track scales directly measure the train to be 7 time heavier?

I, too, was disappointed that you received no response. I submitted no comment due to my inadequacy.

Please excuse my ignorance and possibly incorrect terminology but in accordance with the principle of equivalence your train, initially at rest, is analogous to an apple on a tree which is being acted upon by a gravitational field. When the stem gives way the apple starts moving (as does your train) and because it is then moving it incurs a relativistic mass increase as a result of which the Earth applies a stronger force of gravity to the apple and the apple's velocity accordingly increases (ad infinitum).

If the apple lands on a set of scales wouldn't that device initially, momentarily, determine a greater mass for the apple than its rest mass (analogous to your track scales determining that the train has become 'heavier')?

 

[cos]

...there is no such thing as gravitational force for a free falling object in GR.

Isn't it more correct to state that a free falling object feels no force acting upon it?

A free falling object accelerates because there is a gravitational field acting upon it.

If an observer is falling in a gravitational field he is presumably entitled to realize that the cliff face is accelerating past him because an (albeit otherwise indeterminable from his point of view) gravitational field is acting upon him.

 

[yuiop]

Please excuse my ignorance and possibly incorrect terminology but in accordance with the principle of equivalence your train, initially at rest, is analogous to an apple on a tree which is being acted upon by a gravitational field. When the stem gives way the apple starts moving (as does your train) and because it is then moving it incurs a relativistic mass increase as a result of which the Earth applies a stronger force of gravity to the apple and the apple's velocity accordingly increases (ad infinitum).

The case for vertical velocity is a little tricky and if I am honest I would admit I not certain of the answer to this. (It would be nice to have a definitive answer to the horizontal case first). It is worth considering that that the inertial resistance of the apple to being accelerated downwards also increases and this might cancel out the apparent increased force of gravity acting on it so that appears to accelerate downwards normally. It is also worth noting that in coordinate terms a particle dropped from infinty towards a Schwarzschild black hole appears to slow down as it gets near the event horizon while locally it appears to be speeding up. Tricky eh! Sorry, I can not be more helpful here.

If an observer is falling in a gravitational field he is presumably entitled to realize that the cliff face is accelerating past him because an (albeit otherwise indeterminable from his point of view) gravitational field is acting upon him.

In relativity he is also entitled to consider himself stationary and "something" is accelerating the cliff upwards.

 

[utesfan100]

"Furthermore, experiments show that most subatomic particles have a permanent, built-in angular momentum, which is not due to their motion through space. This spin angular momentum comes in units of h_bar. For example, an electron standing at rest has an angular momentum of h_bar/2."


If you have further issues with the above, I suggest that you look at the Rindler derivation.

If you can link to Rindler's derivation that would be useful, as I don't have it available.

I am not arguing that subatomic particles do not have spin 1/2 values that contribute to angular momentum. That would be silly, as the splitting of electron rays in a magnetic field shows otherwise.

It is only the intrinsic angular momentum that is quantized and constant. The dynamic angular momentum is very velocity dependent. In most systems (including electrons in the outer shells of the heavier atoms) the dynamic angular momentum of the particles greatly exceeds their intrinsic momentum.

Introducing planks constant into a relativity discussion is futile until a quantum theory of gravity is achieved. At least wait until we can explain the http://www.hindawi.com/journals/aa/2009/931920.html" .

 

[starthaus]

If you can link to Rindler's derivation that would be useful, as I don't have it available.

It covers several pages, so, I suggest you invest in the book.

 

[utesfan100]

It covers several pages, so, I suggest you invest in the book.

You obviously don't appreciate the financial constraints of community college support staff.

 

[yuiop]

Yes, I saw that. Two wrongs don't make a right. I would always take Rindler over both of the others. I just started looking at the Jonsson paper, it is especially bad, especially when you look at his derivation for equations (1)-(3).

I would tend to agree that Rindler is more authorative, but he is also harder to follow and understand. I managed to find a google preview of the Rindler book http://books.google.co.uk/books?id=...&resnum=3&ved=0CA4Q6AEwAg#v=onepage&q&f=false but unforunately the page with equations you mention is not available.

The preceding pages are visible and I can see that on page 238 he states h is a constant and on page 239 he defines h as angular momentum per unit rest mass. This suggests to me he is not referring to h as quantum mechanical spin angular momentum, but rather just regular angular momentum being used as a sort of affine parameter. In other words he is taking the initial angular momentum at a radial coordinate very far from the gravitational body and using that as a constant parameter which is valid. This means the quantity on the right is mass times a constant. Now I will have to try and figure out what the rest of the equation is saying.

 

[utesfan100]

I would tend to agree that Rindler is more authorative, but he is also harder to follow and understand. I managed to find a google preview of the Rindler book http://books.google.co.uk/books?id=...&resnum=3&ved=0CA4Q6AEwAg#v=onepage&q&f=false but unforunately the page with equations you mention is not available.

The preceding pages are visible and I can see that on page 238 he states h is a constant and on page 239 he defines h as angular momentum per unit rest mass. This suggests to me he is not referring to h as quantum mechanical spin angular momentum, but rather just regular angular momentum being used as a sort of affine parameter. In other words he is taking the initial angular momentum at a radial coordinate very far from the gravitational body and using that as a constant parameter which is valid. This means the quantity on the right is mass times a constant. Now I will have to try and figure out what the rest of the equation is saying.

This sounds like the angular momentum relative to the center of mass of a rotating object, Iw. In this case I would expect the moment of inertia to increase with the relativistic mass and w to slow down canceling the effect.

The angular momentum of the object about its own axis is constant. Its angular momentum about some distant object remains very velocity dependent.

 

[yuiop]

This sounds like the angular momentum relative to the center of mass of a rotating object, Iw. In this case I would expect the moment of inertia to increase with the relativistic mass and w to slow down canceling the effect.

The angular momentum of the object about its own axis is constant. Its angular momentum about some distant object remains very velocity dependent.

I think Rindler is effectively saying that as the particle descends and its inertial mass increases, it's angular (orbital) velocity decreases (in coordinate terms), such that its angular momentum is a conserved quantity and remains constant.

This fourmilab page http://www.fourmilab.ch/gravitation/orbits/ seems to be using similar equations to Rindler, except they use L with a tilde over it instead of h.

 

[cos]

It is worth considering that that the inertial resistance of the apple to being accelerated downwards also increases and this might cancel out the apparent increased force of gravity acting on it so that appears to accelerate downwards normally.

The apple does not simply appear to accelerate it does accelerate and because it accelerates then it seems obvious that the increasing force of gravity acting upon it is more than sufficient to overcome the apple's increasing inertial resistance.

It is also worth noting that in coordinate terms a particle dropped from infinty towards a Schwarzschild black hole appears to slow down as it gets near the event horizon while locally it appears to be speeding up.

As you effectively point out - the particle appears, to a distant observer, to be slowing down as it approaches the event horizon but assuming that he is aware of the fact that as the particle moves closer to the black 'hole' the light that the particle emits reaches him at progressively slower velocities he should be capable of realizing that what appears to be taking place is nothing more than a visual illusion.

In relativity he is also entitled to consider himself stationary and "something" is accelerating the cliff upwards.

He falls off a cliff and at that very instant 'is' of the opinion that not only the cliff but also the entire universe is accelerating in a particular direction!

Presumably being scientifically inclined he might perhaps ask himself what (greater-than-infinite and increasing) force could cause an infinite universe to accelerate.

 

[Frame Dragger]

The apple does not simply appear to accelerate it does accelerate and because it accelerates then it seems obvious that the increasing force of gravity acting upon it is more than sufficient to overcome the apple's increasing inertial resistance.



As you effectively point out - the particle appears, to a distant observer, to be slowing down as it approaches the event horizon but assuming that he is aware of the fact that as the particle moves closer to the black 'hole' the light that the particle emits reaches him at progressively slower velocities he should be capable of realizing that what appears to be taking place is nothing more than a visual illusion.



He falls off a cliff and at that very instant 'is' of the opinion that not only the cliff but also the entire universe is accelerating in a particular direction!

Presumably being scientifically inclined he might perhaps ask himself what (greater-than-infinite and increasing) force could cause an infinite universe to accelerate.

Uhhh... doppler shifting isn't "illusory", it's a real effect, and it's also very real that the observer will NEVER be able to see the 'astronaut' falling past the horizon. In the same way, the issue of the scientist you propose requires that he be aware of initial conditions to make his judgement, otherwise... still relative. The fact that he'll fall to his death doesn't change the nature of relative motion or Relativity.

 

[utesfan100]

I think I found Starthaus' objection to the long monorail example. Compare the velocity described for the train to the escape velocity of the Earth and consider which way the force must go.

Now, suppose a thin ring is supported by a stand and then spun to a speed where its mass is moving at v=0.6c along the ring. Would the stand feel more weight to support the ring?

 

[yuiop]

I think I found Starthaus' objection to the long monorail example. Compare the velocity described for the train to the escape velocity of the Earth and consider which way the force must go.

Sure, if the train is exceeding orbital velocity on a spherical massive body then it will take off, but I would like to consider a very localised version on a very large flat earth, to give us an idea of what happens when considering situations like the one you describe below.

By the orbital argument, a mass moving with orbital velocity parallel to the surface of the Earth experiences zero or negative gravity. Now consider a helicopter with masses on the end poles instead of lifting blades on a gravitational body with no atmosphere. What will happen if the rotor rotates with relativistic tip velocities that exceed the escape velocity of the massive body? Will the helicopter be able to hover or ascend in no atmosphere just using the rotational velocity of its rotor tip weights? I think the answer is no.

Now, suppose a thin ring is supported by a stand and then spun to a speed where its mass is moving at v=0.6c along the ring. Would the stand feel more weight to support the ring?

Most of the things I have read would say the answer is yes.

 

[utesfan100]

By the orbital argument, a mass moving with orbital velocity parallel to the surface of the Earth experiences zero or negative gravity. Now consider a helicopter with masses on the end poles instead of lifting blades on a gravitational body with no atmosphere. What will happen if the rotor rotates with relativistic tip velocities that exceed the escape velocity of the massive body? Will the helicopter be able to hover or ascend in no atmosphere just using the rotational velocity of its rotor tip weights?

In this case the rotor provided the force to keep the tips in place, circling above to no effect, other than to keep the helicopter grounded by the increase in mass caused by their rotation.

 

[yuiop]

In this case the rotor provided the force to keep the tips in place, circling above to no effect, other than to keep the helicopter grounded by the increase in mass caused by their rotation.

OK, we are in agreement here.

 

[cos]

Uhhh... doppler shifting isn't "illusory", it's a real effect,

My comment had absolutely nothing whatsoever to do with doppler shifting!

If a far distant observer is looking at a beam of light that is heading directly toward a black 'hole' that beam will accelerate! (He cannot, of course, actually see that beam - it is only a hypothetical situation.)

Similarly if he is looking at a beam that is moving directly away from that object the velocity of that beam, relative to him, will be dependent upon the distance of its source from that black star (assuming that the source is external to the event horizon).

and it's also very real that the observer will NEVER be able to see the 'astronaut' falling past the horizon.

Irrelevant. I made no reference to an [object] falling past the horizon.

In the same way, the issue of the scientist you propose requires that he be aware of initial conditions to make his judgement...

I specifically wrote that the scientist falls off a cliff! He cannot fall off a cliff unless he is initially located on that cliff ergo is presumably aware of the initial conditions - assuming mental competence/awareness of course.

He could, of course, have been rendered unconscious and thrown off the cliff but I prefer to stick with relevance.

 

[yuiop]

I think I found Starthaus' objection to the long monorail example. Compare the velocity described for the train to the escape velocity of the Earth and consider which way the force must go.

I found another article http://www2.warwick.ac.uk/fac/sci/physics/teach/module_home/px436/notes/lecture16.pdf that has equations similar to those of Rindler. Starthaus is obviously talking about "effective potential" where the effect of centrifugal potential due to orbital velocity around the massive body is subtracted from the gravitational potential. For a circular orbit the potential gradient is zero. In the case of the weight of a small spinning ring, orbital velocities are not relevant.

 

[starthaus]

I found another article http://www2.warwick.ac.uk/fac/sci/physics/teach/module_home/px436/notes/lecture16.pdf that has equations similar to those of Rindler.

Yes, this is good but not as good as Rindler. In essence it is an abstract of Rindler's chapter 11, minus the proofs.

Starthaus is obviously talking about "effective potential" where the effect of centrifugal potential due to orbital velocity around the massive body is subtracted from the gravitational potential.

Rindler is talking about particle orbits in Schwarzschild space. This describes exactly the motion of particles at LHC, i.e. it answers the OP.
As an aside, the Jonsson paper is pure nonsense, AmJPhys publishes a lot of bad papers when it comes to relativity. The other paper (Matsas?) is not even peer-reviewed.

 

[Frame Dragger]

My comment had absolutely nothing whatsoever to do with doppler shifting!

If a far distant observer is looking at a beam of light that is heading directly toward a black 'hole' that beam will accelerate! (He cannot, of course, actually see that beam - it is only a hypothetical situation.)

I either stand corrected, or deeply confused. Actually, the latter regardless of the former.

Similarly if he is looking at a beam that is moving directly away from that object the velocity of that beam, relative to him, will be dependent upon the distance of its source from that black star (assuming that the source is external to the event horizon).

That I understand, but I'm not grasping why this is significant in this situation. I don't mean that sarcastically, I just (clearly) don't get it.

I specifically wrote that the scientist falls off a cliff! He cannot fall off a cliff unless he is initially located on that cliff ergo is presumably aware of the initial conditions - assuming mental competence/awareness of course.

Agreed, but given all of that, how is this a question which touches on Relativity at all? You have your initial and final velocities of the hapless scientist (the final being most upsetting to him), and all other conditions set; you have your IRF of the scientist established.

He could, of course, have been rendered unconscious and thrown off the cliff but I prefer to stick with relevance.

Well, relevant, but genuinely cruel,

If there is a single source of my confusion here, it would be that what you're ascribing to an illusory effect, seems more to do with gravity as fictitious force, than relative motion. I wouldn't argue that we can establish IRFs, or that at the end of the fall distant observers will agree on the end result (thud), but I don't see what it is you're illustrating.

 

[cos]

If a far distant observer is looking at a beam of light that is heading directly toward a black 'hole' that beam will accelerate! (He cannot, of course, actually see that beam - it is only a hypothetical situation.)

I either stand corrected, or deeply confused. Actually, the latter regardless of the former.

You wrote that you understand that if a far distant observer is looking at a beam that is moving directly away from that object the velocity of that beam, relative to him, will be dependent upon the distance of its source from that black star.

In accordance with that concept - the light emanating from a source that is at a (hypothetically) fixed location relative to a black star will travel toward the distant observer at a slower rate than the light from an object that is located further away from the star.

Similarly, the light emitted by both sources toward the star will accelerate.

Imagine that you are at a vast distance from a light source that emits beams of light to your right hand side and to your left. Obviously (although you cannot actually see those beams) they will both be moving at identical speeds away from the source however if the source starts accelerating in the same direction as one of those beams that beam will be moving away from the source at a slower speed than the other beam (but only whilst the source is accelerating).

In accordance with the principle of equivalence - if a source is at a fixed distance from a black star the beam that is projected toward the star will accelerate whilst the other beam will travel away from the source at a (slower) velocity that is dependent upon the source's distance from the star (however only from your, far distant, point of view not from that of an observer located alongside the source).

I specifically wrote that the scientist falls off a cliff! He cannot fall off a cliff unless he is initially located on that cliff ergo is presumably aware of the initial conditions - assuming mental competence/awareness of course.

Agreed, but given all of that, how is this a question which touches on Relativity at all?

If you fail to see how this touches on relativity and mass dilation I cannot clarify the situation.

You have your initial and final velocities of the hapless scientist (the final being most upsetting to him), and all other conditions set; you have your IRF of the scientist established.

We do not have our IRF of the scientist!

He is accelerating hence is not in an inertial reference frame!

If there is a single source of my confusion here, it would be that what you're ascribing to an illusory effect, seems more to do with gravity as fictitious force, than relative motion. I wouldn't argue that we can establish IRFs...but I don't see what it is you're illustrating.

My description of an illusory effect is that it is the fact that the speed of the light traveling toward a distant observer from an object that is falling into a black star slows down as the object enters stronger gravitational tidal areas giving the distant observer the impression that the object's rate of travel toward the star is decreasing but in reality (and the distant observer should be fully aware of this fact) the object is accelerating.

The illusory effect is analogous to when a person sees a mirage; the scene is not actually where it appears to be in the same way that if we see a star that seems to have changed locations as its light bypasses a massive body (e.g. during an eclipse) that star has not physically moved; this is only a visual illusion.