How Does Acceleration Affect the Relativistic Mass of Particles?

[cos]

In his book 'Deep Space' Colin A Ronan showed (136, Pan, 1983) that when an electron is accelerated in a cathode-ray tube it will be 'pulled downwards' by gravity but that this deviation can be overcome by the application of an electric charge via deflecting plates.

This presumably complies with Kaufmann's circa 1901 cathode-ray experiments showing that the mass of an electron is subject to change and that the change depends on its velocity (161, 'Fiction Stranger Than Truth', 1981, Nikolai Rudakov).

I understand that gamma factors in excess of 400,000 times a particle's (proton's?) rest mass have been generated by the LHC.

My specific question is - how long does it take to accelerate a particle from rest to a velocity whereby its relativistic mass has increased to 400,000 times its rest mass?

 

[IttyBittyBit]

In his book 'Deep Space' Colin A Ronan showed (136, Pan, 1983) that when an electron is accelerated in a cathode-ray tube it will be 'pulled downwards' by gravity but that this deviation can be overcome by the application of an electric charge via deflecting plates.

Why do you think this has anything to do with mass change?

To answer your question, it depends on your equipment. Given extreme enough conditions, I see no reason why particles couldn't be accelerated to 3.5 TeV (current LHC level) in as short a time interval as desired.

Of course, it's much more practical to make a machine that builds up the momentum gradually, like the LHC does.

 

[cos]

Why do you think this has anything to do with mass change?

Ronan continues - "A faster moving electron is more massive and curves further down." (which requires a stronger force applied beneath the particle to restore its required horizontal trajectory).

I read that as indicating that the faster moving particle will incur a greater mass change in excess of the mass change incurred by the slower moving particle.

Rudakov's comment that Kaufmann's experiments showed "...that the mass of an electron is subject to change and that the change depends on its velocity.." also indicates to me that they are talking about mass change.

On the basis that we are talking about an accelerated particle it should be obvious that the mass referred to is not its rest mass.

To answer your question, it depends on your equipment. Given extreme enough conditions, I see no reason why particles couldn't be accelerated to 3.5 TeV (current LHC level) in as short a time interval as desired.

Of course, it's much more practical to make a machine that builds up the momentum gradually, like the LHC does.

On the basis that my OP referred to a gamma factor generated by the LHC it should be obvious that I am referring to those experiments not to experiments involving other equipment.

To what sort of time interval do you refer with your comment "..as short a time interval as possible." bearing in mind that I am talking about extant LHC tests?

I see no reason to 'make a machine that builds up the momentum gradually' on the basis that, apparently, the LHC already does this.

 

[starthaus]

In his book 'Deep Space' Colin A Ronan showed (136, Pan, 1983) that when an electron is accelerated in a cathode-ray tube it will be 'pulled downwards' by gravity but that this deviation can be overcome by the application of an electric charge via deflecting plates.

This presumably complies with Kaufmann's circa 1901 cathode-ray experiments showing that the mass of an electron is subject to change and that the change depends on its velocity (161, 'Fiction Stranger Than Truth', 1981, Nikolai Rudakov).

I understand that gamma factors in excess of 400,000 times a particle's (proton's?) rest mass have been generated by the LHC.

My specific question is - how long does it take to accelerate a particle from rest to a velocity whereby its relativistic mass has increased to 400,000 times its rest mass?

This is not difficult:

v=at/sqrt(1+(at/c)^2)

where a=F/m0. Generally, F=qE for particle accelerators.

m0 is the rest mass of the electron.

To the above, add m=m0/sqrt(1-(v/c)^2)

You know that m/m0=400,000 in the case of your exercise. You now have all the equations to solve the problem.

 

[cos]

You now have all the equations to solve the problem.

Thanks for the response but unfortunately, being mathematically ignorant to the nth.degree (to use an algebraic factor), I do not possesses the expertise.

I coincidentally quit high school math at the age of 14 - as did Einstein - and in the month he died.

On the basis of your mathematical ability and in accordance with the information you provided - could you tell me if it would take a fraction of a second or several seconds to accelerate a proton from rest to a gamma factor in excess of 400,000 via the LHC?

I fully appreciate that it is not your responsibility to do the work for me but I feel that this is just as much of a scientific question as many others in this group.

 

[yuiop]

...
I understand that gamma factors in excess of 400,000 times a particle's (proton's?) rest mass have been generated by the LHC.

My specific question is - how long does it take to accelerate a particle from rest to a velocity whereby its relativistic mass has increased to 400,000 times its rest mass?

A quick Google search unearthed this comment in a LHC chat group http://www.lhcportal.com/Forum/viewtopic.php?f=4&t=71

1. The brief story of a proton accelerated through the accelerator complex at CERN is as follows:
- Hydrogen atoms are taken from a bottle containing hydrogen. We get protons by stripping orbiting electrons from hydrogen atoms.
- Protons are injected into the PS Booster (PSB) at an energy of 50 MeV from Linac2.
The booster accelerates them to 1.4 GeV.
- The beam is then fed to the Proton Synchrotron (PS) where it is accelerated to 25 GeV.
- Protons are then sent to the Super Proton Synchrotron (SPS) where they are accelerated to 450 GeV.
- They are finally transferred to the LHC (both in a clockwise and an anticlockwise direction, the filling time is 4’20’’ per LHC ring) where they are accelerated for 20 minutes to their nominal energy of 7 TeV. Beams will circulate for many hours inside the LHC beam pipes under normal operating conditions.
- Protons arrive at the LHC in bunches, which are prepared in the smaller machines.

and this from an LHC outreach FAQ http://lhc-machine-outreach.web.cern.ch/lhc-machine-outreach/lhc-machine-outreach-faq.htm

How long does it take for a proton to go from zero to 14 TeV ?

When a proton leaves the source, it crosses the linac and reaches the PSB in a few microseconds. In the PSB it is accelerated from 50 MeV to 1.4 GeV in 530 ms,
then after less than a microsecond it is injected in the PS where it can either:
- be accelerated/manipulated/extracted in 1025 ms
- or wait for 1.2 more seconds before being accelerated, if it's part of the first PSB batch to the PS.

Then it is sent to the SPS where it waits for 10.8, 7.2, 3.6, or zero seconds whether it's part of the first, second, third, or fourth PS batch to the SPS.
The SPS accelerates it to 450 GeV in 4.3 seconds, and sends it to the LHC.

So the time it takes from the source to the exit of the SPS is between
0.53 + 1.025 + 4.3 = 5.86 seconds
and
0.53 + 1.2 + 1.025 + 10.8 + 4.3 = 17.86 seconds

Then our proton has to wait up to 20 minutes on the LHC 450 GeV injection plateau before the 25 minutes ramp to high energy, and these 45 minutes dominates the transit time.

This indicates the acceleration time from rest is in the region of 25 minutes plus a few seconds.

The figure of 14 Tev in the second quote is, I think, the combined energy of two counter rotating beams, each with an individual energy of 7 Tev and the first quote is for two counter rotating beams of 3.5 Tev each.

The FAQ suggests the energy of a proton at 7 TeV is "only" 7,460 times that of its rest mass, rather than the 300,000 that you mention. I am not sure where that discrepancy comes from. THe 20 minute holding time at the injection plateau is the delay while batches or bunches of protons are injected into the the LHC from the SPS prior to the final acceleration of all the batches in the LHC. As I understan it the beam is not continuous, but made up of bunches with gaps between, which makes it easier to switch (kick) a bunch of protons out of circulation into sidestream processes.

What is interesting is that the first quote mentions that the beam can remain circulating at the final velocity for many hours and gravity would definitely have to be taken into account to keep the beam horizontal over that sort of time period.

Ronan continues - "A faster moving electron is more massive and curves further down." (which requires a stronger force applied beneath the particle to restore its required horizontal trajectory)...

I think this is the more interesting question. The "curves further down" statement is misleading. A fast moving horizontal beam curves less than a slow moving beam. Both the slow moving beam and the fast moving beam hit the floor at the same time, because the downward acceleration due to gravity is the same for both beams. The combination of the horizontal and vertical components means the faster beam curves less and not more as implied by Ronan.

However, the particles with higher horizontal energy have effectively more inertia and the force of gravity acting upon them is effectively greater (in order that the vertical acceleration of gravity should be constant). The higher energy particles moving horizontally, require a greater electromagnetic force to compensate for the effect of gravity in order to maintain a horizontal trajectory. In modern interpretations, rest mass is always constant and the term "relativistic mass" is deprecated. It is thought of in terms of the relativistic equations for force, acceleration and inertia being different from those of Newtonian mechanics. Another way to thionk of it, is that applying a force from a stationary device to a moving particle (such as using the compensting electromagnetic field in the LHC) is not the the same as applying a force from a stationary device to a stationary particle.

Although the inertial behaviour of a moving particle might be regarded as an indication of relativistic mass, this use of the term is generally avoided because it requires the awkward concept that the transverse inertial mass of a given particle is different from the parallel inertial mass of the same particle and so the modern use is that the inertial mass, gravitational mass and rest mass of a particle are all equivalent and constant and just called "mass" without any need for qualification.

P.S> Non of the above is intended to be authoritive. I am just mulling over my own understanding, in the hope that any misconceptions I have, will be corrected by the more knowledgeable people in this forum.

 

[cos]

The FAQ suggests the energy of a proton at 7 TeV is "only" 7,460 times that of its rest mass, rather than the [400,000] that you mention. I am not sure where that discrepancy comes from.

That GF was provided by jtbell #2 Feb27-09 in a response to my posting 'Mass dilation determination' who wrote that the acceleration of electrons and positrons to 209 GeV corresponds to a Lorentz gamma factor of about 409000.

What is interesting is that the first quote mentions that the beam can remain circulating at the final velocity for many hours and gravity would definitely have to be taken into account to keep the beam horizontal over that sort of time period.

I assume that gravity would also have to be taken into account to keep the beam horizontal over a period of a few seconds.

I think this is the more interesting question. The "curves further down" statement is misleading. A fast moving horizontal beam curves less than a slow moving beam. Both the slow moving beam and the fast moving beam hit the floor at the same time, because the downward acceleration due to gravity is the same for both beams. The combination of the horizontal and vertical components means the faster beam curves less and not more as implied by Ronan.

However, the particles with higher horizontal energy have effectively more inertia and the force of gravity acting upon them is effectively greater (in order that the vertical acceleration of gravity should be constant). The higher energy particles moving horizontally, require a greater electromagnetic force to compensate for the effect of gravity in order to maintain a horizontal trajectory.

My specific interest is in the fact that a particle 'at rest' in the laboratory requires an electromagnetic force of 1 'unit' in order to overcome gravity whereas, presumably, a particle that has been accelerated to 209 GeV would require an em force of around 409000 'units' in order to maintain a required horizontal trajectory.

If there is 1 'unit' of gravity acting upon the 'at rest' particle then, when it has been accelerated to 209 GeV, there would, presumably, be 409000 'units' of gravity acting upon that particle.

 

[starthaus]

, a particle that has been accelerated to 209 GeV would require an em force of around 409000 'units' in order to maintain a required horizontal trajectory.

No. If you completed the calculations I showed you you would have found that the force is necessary in order to reach gamma=400,000.

If there is 1 'unit' of gravity acting upon the 'at rest' particle then, when it has been accelerated to 209 GeV, there would, presumably, be 409000 'units' of gravity acting upon that particle.

No, again. Relativistic mass does not affect the gravitational attraction of a body. This is a recurring mistake in people's minds.

 

[starthaus]

.

I coincidentally quit high school math at the age of 14 - as did Einstein

Einstein did not quit high school. In fact, he went on to college and got a doctoral degree. Your comparing yourself to him is offensive.

 

[IttyBittyBit]

I repeat my question: What the hell does that (larger field required for faster electron) have to do with mass change?

As kev and starthaus explained, it doesn't.

My specific question is - how long does it take to accelerate a particle from rest to a velocity whereby its relativistic mass has increased to 400,000 times its rest mass?

On the basis that my OP referred to a gamma factor generated by the LHC it should be obvious that I am referring to those experiments not to experiments involving other equipment.

To what sort of time interval do you refer with your comment "..as short a time interval as possible." bearing in mind that I am talking about extant LHC tests?

I see no reason to 'make a machine that builds up the momentum gradually' on the basis that, apparently, the LHC already does this.

What is obvious is that you don't understand how to be polite. Usually when people ask 'how long does it take for xxx' to happen, they mean independent of any particular setup. If you had a particular setup in mind, you should have explicitly stated it. Don't want to? Fine, but don't go blaming others for lack of ability to read your mind.

Also, why do you repeat the same question over and over again? kev and starthaus are genuinely trying to help you. The least you can do is be polite and actually read what they are saying.

 

[cos]

Einstein did not quit high school. In fact, he went on to college and got a doctoral degree. Your comparing yourself to him is offensive.

Amongst my several reference books the one that comes more readily to mind and to hand is 'Einstein for Beginners' in which the authors - Schwartz and McGuiness (34, Writers and Readers, 1979) - wrote :-

"After two months on his own [after the rest of the family departed to Milan in 1894], Albert obtains a doctor's certificate saying that he is suffering a nervous breakdown. The school authorities dismiss him."

You will most likely find it outrageously offensive that I compare myself to Einstein inasmuch as I, too, was castigated by my teachers on the basis that my presence in class was disruptive and affected other students (32, 'Einstein for Beginners') but even more egregious is my comment that, like Einstein, I too am of the male gender and have two legs and a moustache - vilification of his memory, I know.

 

[cos]

If there is 1 'unit' of gravity acting upon the 'at rest' particle then, when it has been accelerated to 209 GeV, there would, presumably, be 409000 'units' of gravity acting upon that particle.

No, again. Relativistic mass does not affect the gravitational attraction of a body. This is a recurring mistake in people's minds.

Why, then, does a falling object accelerate?

I trust that you are not implying that I was suggesting that an increase in mass of an accelerated object increases the gravitational field strength of the object i.e. its gravitational attraction?

 

[starthaus]

Amongst my several reference books the one that comes more readily to mind and to hand is 'Einstein for Beginners' in which the authors - Schwartz and McGuiness (34, Writers and Readers, 1979) - wrote :-

"After two months on his own [after the rest of the family departed to Milan in 1894], Albert obtains a doctor's certificate saying that he is suffering a nervous breakdown. The school authorities dismiss him."

You claimed that Einstein quit high school at the age of 14 (just like you). When challenged, you changed your tune and you now claim something totally different.

 

[starthaus]

I trust that you are not implying that I was suggesting that an increase in mass of an accelerated object increases the gravitational field strength of the object i.e. its gravitational attraction?

This is exactly what you were claiming at the end of your post #7.

 

[cos]

I repeat my question: What the hell does that (larger field required for faster electron) have to do with mass change?

It is my understanding that the relativistic mass increase (change) of an accelerated particle requires the application of a proportionally increased em field beneath the particle in order to maintain its horizontal trajectory.

What is obvious is that you don't understand how to be polite.

The tone of my responses is directly in accordance with that of the authors.

Usually when people ask 'how long does it take for xxx' to happen, they mean independent of any particular setup. If you had a particular setup in mind, you should have explicitly stated it. Don't want to? Fine, but don't go blaming others for lack of ability to read your mind.

I had a particular setup in mind and specified same inthat my posting directly referred to LHC experiments that have generated gamma factors in excess of 400000.

 

[cos]

You claimed that Einstein quit high school at the age of 14 (just like you). When challenged, you changed your tune and you now claim something totally different.

According to Schwartz and McGuiness (34, 'Einstein for Beginners' Writers and Readers, 1979) Einstein obtained a doctor's certificate knowing that it would result in his dismissal from school which, according to those authors, was his intention!

Your insulting attitude is unwarranted, inappropriate and terminating.

 

[cos]

This is exactly what you were claiming at the end of your post #7.

The end of my post #7 was -

My specific interest is in the fact that a particle 'at rest' in the laboratory requires an electromagnetic force of 1 'unit' in order to overcome gravity whereas, presumably, a particle that has been accelerated to 209 GeV would require an em force of around 409000 'units' in order to maintain a required horizontal trajectory.

If there is 1 'unit' of gravity acting upon the 'at rest' particle then, when it has been accelerated to 209 GeV, there would, presumably, be 409000 'units' of gravity acting upon that particle.

There is nothing in those comments that implies an increase in the particle's gravitational field strength!

The gravitational field applicable to the 'at rest' particle in the first paragraph, above, was the planet's gravitational field NOT the particle's comparably miniscule gravitational field strength!

The 1 'unit' of gravity acting upon the at rest particle in the second paragraph was a reference to the planet's gravitational field not that of the particle.

The 409000 units of gravity referred to in that same paragraph was in relation to the Earth's gravitational field NOT that of the particle!

 

[starthaus]

If there is 1 'unit' of gravity acting upon the 'at rest' particle then, when it has been accelerated to 209 GeV, there would, presumably, be 409000 'units' of gravity acting upon that particle.

No, there wouldn't.

 

[yuiop]

I think cos has an interesting point that has not really been addressed. Here is a thought experiment that I hope makes the issue clearer. Let us say we have two cylinders and the curved walls of the cylinders are perfectly vertical and perfectly frinctionless. In each cylinder a charged particle is circulating horizontally. Now let us say one particle is circulating at 0.8666c (gamma=2) and the other is circulating at 0.9860c (gamma=6) relative to the lab in which the cylinders are at rest. To stop the particles falling at the usual 9.8 m/s/s, a compensating vertical electromagnetic force is required. Will they require exactly the same compensating electromagnetic force so that they remain at constant height? The answer to this this question must be know in scientific circles, because this has to be taken into account when designing particle accelerators like the LHC. In another thread we seemed to conclude that the faster particle would require 3 times the amount of compensating force, than the slower particle to maintain constant altitude. Agree or disagree?

P.S. Here is what Ben had to say in post #5 of the other thread https://www.physicsforums.com/showthread.php?t=387599&highlight=springs

... The equivalence principle requires that everything have the same downward acceleration. However, that doesn't mean that everything has to feel the same force. An object moving horizontally with \gamma=7 has 7 times its normal inertia, but it also has 7 times the normal amount of gravitational force acting on it, so it has the same acceleration it normally would have.

(My bold). This is I think the crux of the matter that cos is getting at and most people in this current thread seem to be contradicting Ben's statement in the other thread.

In the other thread I also put forward this straightforward question

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?

but no one responded. Are we avoiding a thorny issue here? Is the force of gravity (according to a local observer at rest with the massive body) acting on a particle with a horizontal velocity of v relative to the massive body (and a vertical velocity of zero) proportional to:

F = \frac{GMm_0}{r^2}

or

F = \frac{GMm_0}{r^2\sqrt{1-v^2/c^2}} ?

Seems a simple enough question.

 

[starthaus]

In the other thread I also put forward this straightforward question but no one responded. Are we avoiding a thorny issue here? Is the force of gravity (according to a local observer at rest with the massive body) acting on a particle with a horizontal velocity of v relative to the massive body (and a vertical velocity of zero) proportional to:

F = \frac{GMm_0}{r^2}

or

F = \frac{GMm_0}{r^2\sqrt{1-v^2/c^2}} ?

Seems a simple enough question.

Neither answer is correct, the correct answer is given in chapter 11.9, equation 11.45 (and its derivation) of Rindler's book "Relativity, Special , General and Cosmological". In a different thread, bcrowell and I had had a long debate on this subject as applied to the trajectories of "massive" photons. This thread is just a simpler case of the other thread.

 

[yuiop]

Neither answer is correct, the correct answer is given in chapter 11.9, equation 11.45 (and its derivation) of Rindler's book "Relativity, Special , General and Cosmological". In a different thread, bcrowell and I had had a long debate on this subject as applied to the trajectories of "massive" photons. This thread is just a simpler case of the other thread.

The equations I gave are to first order, locally and in the weak field limit, so that the spherical geometry and gravitational radiation can be ignored. Seeing as we we will probably never agree on the exact equations, maybe you can answer the simple yes/no question that was contained in the same post:

Let us say we have two cylinders and the curved walls of the cylinders are perfectly vertical and perfectly frinctionless. In each cylinder a charged particle is circulating horizontally. Now let us say one particle is circulating at 0.8666c (gamma=2) and the other is circulating at 0.9860c (gamma=6) relative to the lab in which the cylinders are at rest. To stop the particles falling at the usual 9.8 m/s/s, a compensating vertical electromagnetic force is required. Will they require exactly the same compensating electromagnetic force so that they remain at constant height?

 

[starthaus]

The equations I gave are to first order, locally and in the weak field limit, so that the spherical geometry and gravitational radiation can be ignored. Seeing as we we will probably never agree on the exact equations, maybe you can answer the simple yes/no question that was contained in the same post:

I answered your question, neither of your two equations are correct. You can't stick relativistic mass in the Newton's formula and call this a "weak limit field". The chapter in Rindler gives you the exact answer.

 

[yuiop]

I answered your question, neither of your two equations are correct. You can't stick relativistic mass in the Newton's formula and call this a "weak limit field". The chapter in Rindler gives you the exact answer.

I do not have the Rindler book. What does Rindler's equation predict the answer to "Will they require exactly the same compensating electromagnetic force so that they remain at constant height?" is for the thought experiment I described? The answer does not depend on my equations.

 

[starthaus]

I do not have the Rindler book. What does Rindler's equation predict the answer to "Will they require exactly the same compensating electromagnetic force so that they remain at constant height?" is for the thought experiment I described? The answer does not depend on my equations.

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 +3mu^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.

 

[yuiop]

The relativistic mass plays NO role.

So can I take it, that from your understanding of Rindler's equation, the answer to

"Will (two particles with the same rest mass but different horizontal velocities) require exactly the same compensating electromagnetic force so that they remain at constant height?"

is Yes?

 

[starthaus]

So can I take it, that from your understanding of Rindler's equation, the answer to

"Will (two particles with the same rest mass but different horizontal velocities) require exactly the same compensating electromagnetic force so that they remain at constant height?"

is Yes?

You need to solve the second degree differential equation. The solution depends on the initial conditions. Since the equation is second degree , the solution definitely depends on initial position and initial velocity.

 

[yuiop]

You need to solve the second degree differential equation. The solution depends on the initial conditions. Since the equation is second degree , the solution definitely depends on initial position and initial velocity.

So you in fact agreeing, that the force of gravity acting on a horizontally moving particle is velocity dependent.

Good, we are getting somewhere. That agrees with the statement by cos that the force required to maintain the altitude of a particle in the LHC is velocity dependent, although we have not hammered out the exact details of that dependence.

 

[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)

We are talking about force which has units of kg m s^{-2} while the right hand side of your equation is in units of s^2\\ m^{-3}. Even if we re-insert the mass units that were removed by using angular momentum per unit mass, the units are s^2 m^{-3} kg^{-2}. Whatever you are talking about, it is not acceleration or force. At least my equations have the correct units.

Can you define what the variable (u) is?

The relativistic angular momentum (L) in this case is given by:

h = \frac{m r^2}{\sqrt{1-\frac{r^2d\phi^2}{dt^2 c^2}}} \frac{d\phi}{dt} = \frac{m r v}{ \sqrt{1-v^2/c^2}}

and the angular momentum per unit mass (h) is:

h = \frac{ r v}{ \sqrt{1-v^2/c^2}}

As can be seen, even in terms of unit mass, the velocity dependence is implicit in your equation.

The relativistic mass plays NO role.

Please do not get fixated on the relativistic mass concept. All we want to know is if the gravitational force acting a horizontally moving particle is velocity dependent or not. Velocity dependent gravitational force does not have to imply relativistic mass.

 

[starthaus]

We are talking about force which has units of kg m s^{-2} while the right hand side of your equation is in units of s^2\\ m^{-3}.

Neither side of the ODE is in units of force.

Even if we re-insert the mass units that were removed by using angular momentum per unit mass, the units are s^2 m^{-3} kg^{-2}. Whatever you are talking about, it is not acceleration or force. At least my equations have the correct units.

Can you define what the variable (u) is?

You are wrong on all accounts. If you want to learn, u is a notation for 1/r.

The relativistic angular momentum (L) in this case is given by:

h = \frac{m r^2}{\sqrt{1-\frac{r^2d\phi^2}{dt^2 c^2}}} \frac{d\phi}{dt} = \frac{m r v}{ \sqrt{1-v^2/c^2}}

and the angular momentum per unit mass (h) is:

h = \frac{ r v}{ \sqrt{1-v^2/c^2}}

As can be seen, even in terms of unit mass, the velocity dependence is implicit in your equation.

This is incorrect, "h" is a constant, it does not depend on v. I suggest that you try buying the Rindler book.

 

[yuiop]

This is incorrect, "h" is a constant, it does not depend on v. I suggest that you try buying the Rindler book.

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

 

[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.