Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: PmbPhy on 22/07/2016 13:05:19

I came across a special relativity text which says A particle does not become heavier with increasing speed. Do you believe the author is correct? What would you expect would happen to the magnitude of the gravitational field if the source of the field was moving?
Authors who don't use and don't like the concept of relativistic mass make arguments such as the following which are from Appendix F in the text Special Relativity by T.M. Helliwell page 259. I'm going to paraphrase a bit to save myself some typing but the essence of what I'm quoting will not be changed. The author uses the symbol m_{R} to represent relativistic mass.
(i) By hiding √{1  v^{2}} in the mass, we may forget it is there. ...
(ii) One may get the mistaken impression that to go from classical to relativistic mechanics its only necessary to replay all masses by m_{R}. This certainly works for the momentum p = m_{R}v, but it does not work, for example, for kinetic energy: The relativistic kinetic energy is not (1/2)m_{R}v^{2}. And it works in Newton's second law F = ma only in the very special case where the force exerted on a particle is perpendicular to its velocity. In all other cases F != ma.
(iii) Relativity fundamentally serves to correct our notions about time and space. That is, it is really the dynamical equations dealing with motion, like energy and momentum, that ought to be changed, and not the properties of individual particles, like mass.
(iii) When relativity is cast in fourdimensional spacetime form, ...., the idea of relativistic mass is out of place and clumsy.
Of course I have my own opinions about each item but first I'm curious as to what others think about these arguments, so therefore please let me know what you think of each one.
I'd like to thank all of you for giving me your opinion on this.

I came across a special relativity text which says A particle does not become heavier with increasing speed. Do you believe the author is correct? What would you expect would happen to the magnitude of the gravitational field if the source of the field was moving?
This is a very interesting post. Define heavier. Does an object become heavier simply because it crashes into another object at speed? How is relativistic mass interpreted? Is it an increase only in internal kinetic energy or some orther form?
Authors who don't use and don't like the concept of relativistic mass make arguments such as the following which are from Appendix F in the text Special Relativity by T.M. Helliwell page 259. I'm going to paraphrase a bit to save myself some typing but the essence of what I'm quoting will not be changed. The author uses the symbol m_{R} to represent relativistic mass.
(i) By hiding √{1  v^{2}} in the mass, we may forget it is there. ...
(ii) One may get the mistaken impression that to go from classical to relativistic mechanics its only necessary to replay all masses by m_{R}. This certainly works for the momentum p = m_{R}v, but it does not work, for example, for kinetic energy: The relativistic kinetic energy is not (1/2)m_{R}v^{2}. And it works in Newton's second law F = ma only in the very special case where the force exerted on a particle is perpendicular to its velocity. In all other cases F != ma.
(iii) Relativity fundamentally serves to correct our notions about time and space. That is, it is really the dynamical equations dealing with motion, like energy and momentum, that ought to be changed, and not the properties of individual particles, like mass.
(iii) When relativity is cast in fourdimensional spacetime form, ...., the idea of relativistic mass is out of place and clumsy.
Of course I have my own opinions about each item but first I'm curious as to what others think about these arguments, so therefore please let me know what you think of each one.
I'd like to thank all of you for giving me your opinion on this.
I am still thinking about the 3 points above.

Define heavier.
My understanding is that heavier = having more weight, and that weight is the result of gravitational attraction between one massive (=mass/energy) object and another.
Does the author actually mean “heavier” or “more massive”?
I too an thinking about the three points, but it may take a while.

This is a very interesting post. Define heavier.
A good definition of the term weight for purposes such as this is best defined in the article
The equivalence principle and the question of weight by Kenneth Nordtvedt, Jr., Am. J. Phys., 43(3), Mar. (1975)
You can download this article from the following URL: http://booksc.org/book/20829048
The weight of a body is meant to be the force (e.g., the compression of a spring scale) required to either support the body in a gravitational field (gravitational weight) or to accelerate the body relative to an inertial space (inertial weight).
That definition means that the term heavier means that a greater force is required to support the body than that required when the body is at rest.
Does an object become heavier simply because it crashes into another object at speed?
Absolutely not.
How is relativistic mass interpreted? Is it an increase only in internal kinetic energy or some orther form?
I don't know what you're asking when you write How is relativistic mass interpreted? I'm not even sure that its meaningful to ask how any physical quantity is interpreted. To interpret means to explain the meaning. I really don't know what you're asking for here Jeff.
The relativistic mass, aka inertial mass, M, of a body of proper mass, m, is defined as M = m/√(1  v^{2}/c^{2}). This expression is determined by requiring that the quantity p = Mv is conserved in completely elastic collisions. The derivation is on my website at: http://www.newenglandphysics.org/physics_world/sr/inertial_mass.htm

I came across a special relativity text which says A particle does not become heavier with increasing speed. Do you believe the author is correct?
I think it's wrong, on the particle there should be a greater force, proportional to the gamma factor. But maybe the author intended to refer to its mass and not to its weight (which kind of book is it? Is it a universitary text?)What would you expect would happen to the magnitude of the gravitational field if the source of the field was moving?
Don't know, but is it really the same question as the particle's weight?Authors who don't use and don't like the concept of relativistic mass make arguments such as the following which are from Appendix F in the text Special Relativity by T.M. Helliwell page 259. I'm going to paraphrase a bit to save myself some typing but the essence of what I'm quoting will not be changed. The author uses the symbol m_{R} to represent relativistic mass.
(i) By hiding √{1  v^{2}} in the mass, we may forget it is there. ...
(ii) One may get the mistaken impression that to go from classical to relativistic mechanics its only necessary to replay all masses by m_{R}. This certainly works for the momentum p = m_{R}v, but it does not work, for example, for kinetic energy: The relativistic kinetic energy is not (1/2)m_{R}v^{2}. And it works in Newton's second law F = ma only in the very special case where the force exerted on a particle is perpendicular to its velocity. In all other cases F != ma.
(iii) Relativity fundamentally serves to correct our notions about time and space. That is, it is really the dynamical equations dealing with motion, like energy and momentum, that ought to be changed, and not the properties of individual particles, like mass.
(iii) When relativity is cast in fourdimensional spacetime form, ...., the idea of relativistic mass is out of place and clumsy.
I agree with these three points and I would even add others, but I've already discussed about them in many other threads. All of this in SR only, however, even because I don't know anything of GR.

lightarrow

Is there a definite answer, when two particles are moving on parallel tracks at a velocity near c is there a greater gravitational attraction between them than when they are moving at a more modest velocity.
My own humble opinion is that there is not.

Kinetic energy; 1/2MV2, cannot reach or exceed 1/2MC2, since rest mass cannot move at the speed of light. If we tried to induce a mass to approach the speed of light, we would need to add infinite energy, yet kinetic energy is only able to reach the low finite limit of 1/2MC2. Relativistic mass accounts for the energy difference; E=MrC2, so energy conservation applies.
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I think it's wrong, on the particle there should be a greater force, proportional to the gamma factor. But maybe the author intended to refer to its mass and not to its weight (which kind of book is it? Is it a universitary text?)
No. He really meant weight. No physicist would ever make that kind off mistake. The text is a textbook on relativity. Although its as mathematically sophisticated as any text used in a university course, I doubt that this text would ever be used in a course on relativity
Don't know, but is it really the same question as the particle's weight?
No, of course not. This is an entirely different question.
I agree with these three points and I would even add others, but I've already discussed about them in many other threads.
Let's take each point at a time:
(i) (i) By hiding √{1  v^{2}} in the mass, we may forget it is there.
You mean to tell me that just because γ is not explicitly written down you'd forget that it was there? If you wouldn't forget its there then who is it that you think would? If you have in mind a student then all that means is that the student doesn't know the physics that well and needs to study it more. I can think of no actual equation or scenario in which someone would forget that it's there. When making a calculation its always clear what's being calculated so that nobody could reasonably emit it. E.g. please provide a reasonable scenario in which someone could forget γ.
(ii) One may get the mistaken impression that to go from classical to relativistic mechanics its only necessary to replay all masses by m_{R}.
Not something that could reasonably happen. After all, who do you think makes such calculations? Only someone who knows special relativity. After all, only a very poor student would attempt to write down an equation and use it when they've never learned how to calculate kinetic energy. No relativist could ever make such a mistake.
(iii) Relativity fundamentally serves to correct our notions about time and space. That is, it is really the dynamical equations dealing with motion, like energy and momentum, that ought to be changed, and not the properties of individual particles, like mass.
Also not a reasonable argument. Such an argument would also have to apply to everything else in relativity including time and length. A proper time interval Δτ is the magnitude of a timelike displacement 4vector while a coordinate time interval Δt is the time component of the same displacement 4vector. Proper distance Δλ is the magnitude of a spacelike displacement 4vector while distance ΔL is the magnitude of Δr which is the distance between two points in space represented by a spacelike displacement 4vector.

Kinetic energy; 1/2MV2, cannot reach or exceed 1/2MC2, since rest mass cannot move at the speed of light.
What is the "M" that you used in that expression? Is it relativistic mass or proper mass? In either case that's not the expression for kinetic energy. Kinetic energy, K, is given by (letting m = proper mass aka rest mass)
K = (γ  1)mc^{2}
If we tried to induce a mass to approach the speed of light, we would need to add infinite energy, yet kinetic energy is only able to reach the low finite limit of 1/2MC2.
That's not true at all. Kinetic energy can have any value whatsoever. I.e. it can get as large as one would like. The value of K is the work done on the particle and you can do as much work as you'd like on a particle.
Relativistic mass accounts for the energy difference; E=MrC2, so energy conservation applies.
That expression is wrong. The energy associated with relativistic mass, M, is the energy E = Mc^{2}. It's equal to the sum of the particles kinetic energy and rest energy, i.e. E = Mc^{2} = K + E_{0} = K + mc^{2}

Is there a definite answer, when two particles are moving on parallel tracks at a velocity near c is there a greater gravitational attraction between them than when they are moving at a more modest velocity.
My own humble opinion is that there is not.
Firstly your post should be given particular attention since it raises a very important point which should be discussed further. In reply to a post of Pete's above I agree that weight would have to depend upon speed. So then what syhprum says is of particular significance. I haven't had time to think about other aspects. The internal kinetic energy remarks with respect to relativistic mass I may write up later. It is to do with an actual theory I am working on. Yes I actually have a theory!

No particle's inertial mass or gravitational mass does not increase with speed. The issue is when someone applies an absolute frame of reference of speed assuming that the force necessary to accelerate it 1mph faster gets harder and harder to do the closer you are to the speed of light. Velocity isn't absolute, it approaches light speed and the momentum approaches infinity. The issue is that F=ma considers acceleration with absolute frame of reference. The idea of relativistic mass is only necessary in order to make sense of relativity in absolute Newtonian terms. It does not exist as an actual phenomenon. The issue is the Newtonian conception of inertia breaks down.

No particle's inertial mass or gravitational mass does not increase with speed. The issue is when someone applies an absolute frame of reference of speed assuming that the force necessary to accelerate it 1mph faster gets harder and harder to do the closer you are to the speed of light. Velocity isn't absolute, it approaches light speed and the momentum approaches infinity. The issue is that F=ma considers acceleration with absolute frame of reference. The idea of relativistic mass is only necessary in order to make sense of relativity in absolute Newtonian terms. It does not exist as an actual phenomenon. The issue is the Newtonian conception of inertia breaks down.
What you appearing to be talking about is a coordinate value for inertia. So that it only appears to change with a change in reference frame. Correct me if I am wrong. I think input from a particle accelerator engineer would be useful.

No particle's inertial mass or gravitational mass does not increase with speed.
That's incorrect. It's a well known fact that inertial mass increases with speed. Engineers have to take this fact into account when designing particle accelerators. The educational material at CERN explains this. See:
http://lhcmachineoutreach.web.cern.ch/lhcmachineoutreach/lhcmachineoutreachfaq.htm
Basically the relativistic mass of a particle increases with velocity and tends to infinity as the velocity approaches the speed of light.
where the term relativistic mass is just another name for inertial mass. Gravitational mass also increases with speed.
See also: https://cerntruth.wordpress.com/
In the fall of 2015 CERN will begin colliding groups of 70 million lead hadrons at 287 tev, unpacking millions of quarks in each collision. Those quarks will be first accelerated at light speed, acquiring relativistic mass, becoming heavier strange quarks, the substance of a strange quarkgluon soup called a ‘strangelet‘. The strange liquid has the potential to become stable and start an ‘ice9′ bigbang reaction. If that happens that effectively transforms the Earth into a pulsar.
The issue is when someone applies an absolute frame of reference of speed assuming that the force necessary to accelerate it 1mpfaster gets harder and harder to do the closer you are to the speed of light.
There's no such thing as an absolute frame of reference so what exactly are you claiming here?
Velocity isn't absolute, it approaches light speed and the momentum approaches infinity. The issue is that F=ma considers acceleration with absolute frame of reference.
Force isn't defined as F =ma. It's defined as F = dp/dt.
The idea of relativistic mass is only necessary in order to make sense of relativity in absolute Newtonian terms. It does not exist as an actual phenomenon. The issue is the Newtonian conception of inertia breaks down.
You're quite wrong. Mass is defined as the m in p = mv. Your assertion about Newtonian terms is incorrect. Your mistake is based on the erroneous assumption that there is a difference in the laws of mechanics between Newtonian mechanics and relativity. There isn't and there never have been. In fact Newton defined force as the time rate of change of momentum. It was Euler who wrote force as F = ma, not Newton. In both Newtonian mechanics and relativity force is given by F = dp/dt. Newton's three laws also hold in relativity where the only difference being that in Newtonian mechanics Galilean transformations are used where in relativity Lorentz transformations are used to transform between inertial frames. Newton's third law applies only between objects in contact.

What you appearing to be talking about is a coordinate value for inertia. So that it only appears to change with a change in reference frame. Correct me if I am wrong. I think input from a particle accelerator engineer would be useful.
Increase in inertial mass is a fact that can be found in any text on accelerator physics. It's in the new text I bought recently.

What you appearing to be talking about is a coordinate value for inertia. So that it only appears to change with a change in reference frame. Correct me if I am wrong. I think input from a particle accelerator engineer would be useful.
Increase in inertial mass is a fact that can be found in any text on accelerator physics. It's in the new text I bought recently.
Frazier seems to have an alternate view.

Is there a definite answer, when two particles are moving on parallel tracks at a velocity near c is there a greater gravitational attraction between them than when they are moving at a more modest velocity.
My own humble opinion is that there is not.
For a measurement made in the particles referential, the gravitational atraction is the same. Nothing should change.
For a measurement made from the referential from which they have such a high speed, I'm not sure.
If they are really heavier, the aceleration due to the bigger gravitational field should make them colide in a shorter time.
But I'm not sure if relativistic mass can be used straightforwardly for a=GM/r˛.
It is not valid for F=ma, (we could naively suppose it was), because dp/dt produces another 1/(1v˛/c˛) term (when the force has the same direction of velocity).

The gamma function includes a value for velocity that relates to the frame of the particle under consideration. Not a value as measured from a remote frame. In which case the particle has to become heavier in its own frame of reference. This will be undetectable in that frame since all objects will become heavier in propotion to anything under observation. Including measuring equipment. So that the laws of physics are the same as in a rest frame. As long as the velocity is constant and no external forces are applied to the frame. This is an important consideration. The question is then as syhprum stated above. Will gravitation draw particles together when they are both moving at relativistic speed and on parallel paths. Since both are becoming heavier then the inertia of both particles is increasing in direct proportion. This cannot have any effect to counter the increased attraction since acceleration due to gavity is independent of mass. All objects fall at the same rate. So we have a conundrum.

What you appearing to be talking about is a coordinate value for inertia. So that it only appears to change with a change in reference frame. Correct me if I am wrong. I think input from a particle accelerator engineer would be useful.
Increase in inertial mass is a fact that can be found in any text on accelerator physics. It's in the new text I bought recently.
Frazier seems to have an alternate view.
I guess I should have made my question clearer. Obviously, anybody who knows me knows very well that I know everything that there is to know about relativistic mass, rest mass, proper mass, the energymomentum tensor, fourmomentum, all aspects of the debate regarding relativistic mass vs rest mass, etc. I don't need Frazier to repeat what I know so well. In all likelihood I know more about this subject than most physicists do.
But let me make something very clear: I do not wish to debate relativistic mass or debate whether mass should be defined in a particular way and whether it's better to do so. I already know this issues and have a stance on it. In fact I wrote an article on the subject located at: http://arxiv.org/abs/0709.0687
The purpose of this thread is for me to understand who it is that people think are making such terrible mistakes that I quoted in the three points in the OP. I also wanted to know whether people assume that weight depends on speed or not. By the way, the weight as measured in the rest frame of the body is a function of the velocity of the source. Here's two examples for those who can understand the math:
http://www.newenglandphysics.org/physics_world/gr/grav_field_sheet.htm
http://www.newenglandphysics.org/physics_world/gr/grav_field_rod.htm
As far as syphrums question goes; I'd have to actually sit down and calculate it. But it should be simple just to think about it. In the rest frame of one of the bodies the other body is also at rest. So you can easily calculate the force between the two bodies; believe it or not it has the same value as in Newtonian gravity. Now just transform your observations to a moving frame and calculate the quantities that you want to.
See also: Measuring the active gravitational mass of a moving object by D.W. Olson and R.C. Guarino, Am. J. Phys., 53(7), Jul. (1985). You can download it from here: http://booksc.org/book/20873864
The abstract reads as follows.
From: http://scitation.aip.org/content/aapt/journal/ajp/53/7/10.1119/1.14280
Abstract: 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 M rel=γ(1+β^2)M. Therefore, in the ultrarelativistic limit, the active gravitational mass of a moving body, measured in this way, is not γM but is approximately 2γM.
That's why I'm all for relativistic mass. In all aspects, relativistic mass behaves as we expect mass should behave. I suspect that's why Misner, Thorne and Wheeler use it in their text Gravitation. Where the say that Mass is the source of gravity. they're referring to the energymomentum tensor, which really should be called the mass tensor since it fully explains/describes mass.

What you appearing to be talking about is a coordinate value for inertia. So that it only appears to change with a change in reference frame. Correct me if I am wrong. I think input from a particle accelerator engineer would be useful.
I don't know any particle accelerator engineers but I do have the book Was Einstein Right by Clifford M. Will. Will is a renown experimental physicist. His area of expertise is relativity.
All that I'm doing in this post is responding to Jeff's remark about input from an engineer. I'm assuming that he's okay with an experimental physicist in relativity. But I'm not doing this to start a debate on relativistic mass. We've beat that horse silly in this forum and other forums and I'd have an anxiety attack if I had to see people repeating those same old arguments in this thread. That would be offtopic. It was my intention when I created this thread that we discuss relativistic mass and weight and who those people are that make those mistakes. That's all.
On page 262 Clifford Will writes
The U.S. National Budget. On 1983, particle physicists proposed that the United States build a gigantic 85 kilometer circumference particle accelerator called the superconducting super collider, costing over 6 billion dollars. One reason for the enormous size and cost is the special relativistic increase in the inertia of a particle moving near the speed of light that makes it harder and harder to accelerate it to higher speeds.
On page 273 Will says just about the same thing, i.e.
The principle of relativity can also be applied to more complicated situations, such as the collision between two bodies, or the motion of a charged body in an electric field. In order for the outcomes of experiments like these to be independent of the inertial frame, the effective mass, or inertia, of a moving particle must increase. The relativistic increase of inertia is what prevents particles from being accelerated up to and beyond the speed of light, because the inertia of the particle increases without bound as it approaches c. This has been observed countless times in particle accelerators, and must be figured into all the engineering specifications and cost estimates for more powerful accelerators.
Think how difficult it would have been to explain that without invoking an increase in inertia.

If you accelerate proton to really really high velocity,
and hit it with other stationary proton,
there will be created protonantiproton pair:
p+ + p+ > p+ + p+ + p+ + p
Input two particles are creating four particles on output.
Proton has restmass 938.272 MeV/c^2
Antiproton (also) have restmass 938.272 MeV/c^2
So basically from kinetic energy of incoming particles there are created two new particles (proton and antiproton)..
Baryon Number conservation:
prior event: +1 +1
after event: +1 +1 +1 1
+1+1=+1+1+11
BTW, kinetic energy in Special Relativity is not E.K.=1/2*m*v^2 but
E.K.=m0*c^2*gammam0*c^2
where gamma=1/sqrt(1v^2/c^2)
Basically, subtract relativisticmass from restmass, and multiply by c^2.

If you accelerate proton to really really high velocity,
and hit it with other stationary proton,
there will be created protonantiproton pair:
Offtopic. That has nothing to do with the subject of this thread. Please stay on topic.
BTW, kinetic energy in Special Relativity is not E.K.=1/2*m*v^2.....
Um..... everyone reading this thread already knows that. In fact it's one part of the subject of this thread. And in case they didn't know that I made if very clear in Reply #8.
E.K.=m0*c^2*gammam0*c^2
where gamma=1/sqrt(1v^2/c^2)
Basically, subtract relativisticmass from restmass, and multiply by c^2.
That expression and your comment following "Basically" is incorrect. Your expression is dimensionally incorrect. Kinetic energy has the units of energy (as it must of course) and the units of energy in terms of kilograms (kg), meters (m) and seconds (s) is Newton*meter which is the same as (kg)*[(m)/(s)]^2. What you have is (K) = (kg)[(m)/(d)]^4. The correct expression as I explained above is in Reply #8 is
K = (γ  1)mc^{2}
where I used m for proper mass. The derivation is on my webpage at:
http://www.newenglandphysics.org/physics_world/sr/work_energy.htm
The value for kinetic energy is in Eq. (12). That page is wicked messy. I don't know what the heck I was thinking when I wrote that. When I rewrite it, it will be much more efficient, i.e. it will be straight to the point next time. I.e. by definition; Kinetic energy = Work done on particle to change its speed from 0 to v.
It's also easily found by writing down the identity for the total inertial energy (aka freeparticle energy) of a particle as the sum of kinetic energy + rest energy or in symbols > E = K + E_{0}. Therefore
K = E  E_{0} = γmc^{2}  mc^{2}
Or factoring out mc^{2} we get our final expression
K = (γ  1)mc^{2}
I'm curious. Where did you get the idea that
E.K.=m0*c^2*gammam0*c^2
Also, on what basis do you justify the assertion that when you subtract relativisticmass from restmass, and multiply by c^2 you'd get kinetic energy? Could it be that perhaps you had in right in your mind but when you wrote it down it came out backwards? After all, if you subtract relativisticmass from restmass you'll get a negative number since relativisticmass is always greater than rest mass.
By the way. If you'd like to use Greek symbols then all you have to do is place your mouse pointer over the Greek symbol that you want and do a "copy shortcut". Then paste the result in the message dialog box where you want it to go and do a paste. Then cut away all the extraneous stuff and you're all set. If it works you can pay me back by buying me a cyber beer. :)

Although it is not practical to measure whether two passengers sitting side by side in a spaceship approaching c experience an increased gravitational attraction it is possible to observe a bundle of Quarks called a Proton at high speed in the LHC where we are told it losses its spherical shape and becomes more like a pancake.
Is this due to increased gravity between its parts or is there another explanation ?

Although it is not practical to measure whether two passengers sitting side by side in a spaceship approaching c experience an increased gravitational attraction it is possible to observe a bundle of Quarks called a Proton at high speed in the LHC where we are told it losses its spherical shape and becomes more like a pancake.
Is this due to increased gravity between its parts or is there another explanation ?
It's a due to Lorentz contraction. Regarding using a particle accelerator to measure the gravitational attraction between protons, it's impractical to do something like that. Since the gravitational attraction between two protons is so small its impossible to measure it in an accelerator. Not to mention the fact that the electric force dominates over the gravitational force by many orders of magnitude. Also, we can't really observe the shape of a proton directly. Especially one moving near the speed of light. Sorry my friend but you're idea won't work.

What you appearing to be talking about is a coordinate value for inertia. So that it only appears to change with a change in reference frame. Correct me if I am wrong. I think input from a particle accelerator engineer would be useful.
I don't know any particle accelerator engineers but I do have the book Was Einstein Right by Clifford M. Will. Will is a renown experimental physicist. His area of expertise is relativity.
All that I'm doing in this post is responding to Jeff's remark about input from an engineer. I'm assuming that he's okay with an experimental physicist in relativity. But I'm not doing this to start a debate on relativistic mass. We've beat that horse silly in this forum and other forums and I'd have an anxiety attack if I had to see people repeating those same old arguments in this thread. That would be offtopic. It was my intention when I created this thread that we discuss relativistic mass and weight and who those people are that make those mistakes. That's all.
On page 262 Clifford Will writes
The U.S. National Budget. On 1983, particle physicists proposed that the United States build a gigantic 85 kilometer circumference particle accelerator called the superconducting super collider, costing over 6 billion dollars. One reason for the enormous size and cost is the special relativistic increase in the inertia of a particle moving near the speed of light that makes it harder and harder to accelerate it to higher speeds.
On page 273 Will says just about the same thing, i.e.
The principle of relativity can also be applied to more complicated situations, such as the collision between two bodies, or the motion of a charged body in an electric field. In order for the outcomes of experiments like these to be independent of the inertial frame, the effective mass, or inertia, of a moving particle must increase. The relativistic increase of inertia is what prevents particles from being accelerated up to and beyond the speed of light, because the inertia of the particle increases without bound as it approaches c. This has been observed countless times in particle accelerators, and must be figured into all the engineering specifications and cost estimates for more powerful accelerators.
Think how difficult it would have been to explain that without invoking an increase in inertia.
Thanks Pete for the various replies. I will be considering all the points you mentioned. When you weren't here the answers were missing! I think your knowledge helps quite a few here.

I have an unfortunate tendency to act as the devils advocate and put forward unlikely scenarios just to see how thoroughly they will be demolished I know all abought Lorenz transforms and how weak gravity is and also about the word not being flat despite what I was taught at school about parabolic moving missiles.

I have an unfortunate tendency to act as the devils advocate and put forward unlikely scenarios just to see how thoroughly they will be demolished I know all abought Lorenz transforms and how weak gravity is and also about the word not being flat despite what I was taught at school about parabolic moving missiles.
I'm sorry, but I don't understand what the point is that you're trying to make here.
I forgot to mention the reason why the magnitude of a body's 4momentum cannot in general be used to define the mass of the body. The magnitude of 4momentum found as follows:
E^{2}  (pc)^{2} = (mc^{2})^{2}
This expression is the magnitude. Solving for m gives the proper mass. The problem with this is that in general its only valid for a single point particle or a body who can be treated as a point.
Consider a rod lying at rest on the xaxis in S as follows; Let  represent a small sphere. Then
+y







At t = 0 each body emits a two photons of equal energy in opposite directions parallel to the yaxis. In S' the sphere's won't emit radiation at the same time but in sequence. Then in S the object "Rod with spheres" will have only two values of energy and momentum but in S' it will have three different values. Therefore the magnitude of a 4momentum can't be defined and have a meaningful magnitude. However the relativistic mass will always be meaningful.

Where's my post? Who moved it to new theories (http://www.thenakedscientists.com/forum/index.php?topic=67757.0)?
I said energy gravitates. See what Einstein said (http://einsteinpapers.press.princeton.edu/vol6trans/197?highlightText=gravitatively%20): "the energy of the gravitational field shall act gravitatively in the same way as any other kind of energy". That's no new theory. Nor is the wave nature of matter.

Does a particle's weight increase with speed?
No, because gravity moves at the speed of light. In any cases, the gravity wave is relativistically redshifted by exactly the same amount as the increase in mass (taking off the nonrelativistic doppler shift). But this is where gravity and inertia have a clear distinction. The local character of inertia and the nonlocal character of gravity. I think this is why it is important to keep the concept of proper mass alive.
If you add a relativistic mass and a fixed velocity of propagation for gravity to Newton's point of view, it is comprehensible. Einstein started from Newton's point of view and modified it according to thoughts and real experimental results.

Does a particle's weight increase with speed?
No, because gravity moves at the speed of light.
That's incorrect. Einstein proved this in 1905 although you may not recognize it as such. The derivation of the proof is on my website at: http://www.newenglandphysics.org/physics_world/gr/weight_moving_body.htm
As far as redshift goes, I don't see how that has anything to do with this problem. I assume that what you really have in mind is gravitational time dilation. Gravitational redshift only appears if and when the light is moving through a gravitational field. In this experiment there's no light. I assume that you're referring to gravitational time dilation. However that's irrelevant. All that could effect is the amount that the weight will change. It has nothing to do with whether the weight is a function of speed. That's a fact that is observer independent since all observers can determine whether the spring balance will depend on speed or not.
There's one very important thing that you neglected to take into account and that's the equivalence principle of general relativity which states that a uniform gravitational field is equivalent to a uniformly accelerating frame of reference. This means that you can test this result by producing a uniform gravitational field by changing your frame of reference to a uniformly accelerating one. In that frame there are no time delays having to do with the finite speed of the propagation of changes in the gravitational field. Now read the article On the Electrodynamics of Moving Bodies by Albert Einstein, Annalen der Physik, 17lk, (1905). This paper is online on Fermi lab's website at: https://www.fourmilab.ch/etexts/einstein/specrel/www/
According to general relativity the weight of a body equals the transverse mass times the component of acceleration transverse to the bodies velocity. Transverse mass equals relativistic mass. This is the only mistake in Einstein's paper, i.e. he calculated the wrong value for the transverse mass (TR), m_{t}. TR was defined and I believe measured before Einstein wrote this SR paper. Ohanian explains this in his book Einstein's Mistakes. I placed that portion on my website for people to read.
It's at http://www.newenglandphysics.org/other/Ohanians_Third_Mistake.pdf
As Ohanian explains, when Einstein calculated the transverse mass using the expression F = ma he used the force as measured in one frame but used the acceleration as measured in another frame. This is a very odd thing to do. When physicists measure quantities and then use the results of those measurements to calculate things like force and acceleration to get mass one must use the same frame of reference for all the measurements. That is to say, if you wish to predict what the particle will weigh when its moving on a spring balance then you must measure both the acceleration and force in the same frame of reference. In this case TM was already defined. Why he chose to redefine it is confusing unless he was unaware of the results already obtained by others. In this case Max Abraham had already defined these terms three years earlier in 1902. See:
https://en.wikipedia.org/wiki/Mass_in_special_relativity
The author of that Wikipedia page refers to it as an unfortunate definition.
The speed of gravity has nothing to do with this. The gravitational field is static and therefore the field isn't propagating. The only time that you have to think about the speed of gravity is when the source of gravity, i.e. matter, has a time dependence of the distribution.
There's a a bit of a story behind why I'm interested in this subject if you're interested. One of the reasons that I decided to study general relativity (GR) was to learn how to calculate the weight of an object moving at speeds close to the speed of light. I was trying to determine whether the concept of relativistic mass was equal to passive gravitational mass. I found out that it is by.
I have a friend at MIT who I was discussing this with him. The professor, Dr. Edmund Bertschinger, liked this problem so much that he's made it a homework problem and its been a homework problem every since I raised the subject over 16 years ago.
If you want to read it please see: http://ocw.mit.edu/courses/physics/8962generalrelativityspring2006/assignments/pset06.pdf
Although he gets the same result that I do his solutions aren't available online. I can send it to you in a PM if you'd like to see how the professor solves the problem and gets the same result that I do. I also posted above a reference to an article in the American Journal of Physics which demonstrates that the moving body generates a larger gravitational field as it moves.
I think this is why it is important to keep the concept of proper mass alive.
I'm quite happy to see you use the term proper mass rather than rest mass. I think that it'd be a serious mistake to think that by preferring relativistic mass over proper mass means abandoning the use of proper mass. That's be like abandoning proper time for coordinate time, which would also a be a serious mistake.

Forget that, I contradicted myself and everything I wrote before.
I was referring to the mass of a photon in its direction of motion and I extended it inappropriately.
A photon is spontaneously emitted on the moon in your direction on earth. Everything else being constant, the mass of the moon remains constant until the photon reaches you. Thus the photon had no gravitational interaction with you along its path. From your point of view, the photon energy was in the moon all along. I know that mathematically it is a matter of reference frame.
Your web site is very nice!
Thanks for the references.