[PmbPhy (OP)]

If you accelerate proton to really really high velocity,
and hit it with other stationary proton,
there will be created proton-antiproton pair:

Off-topic. 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*gamma-m0*c^2
where gamma=1/sqrt(1-v^2/c^2)
Basically, subtract relativistic-mass from rest-mass, 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)mc²

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 free-particle energy) of a particle as the sum of kinetic energy + rest energy or in symbols -> E = K + E₀. Therefore

K = E - E₀ = γmc² -  mc²

Or factoring out  mc² we get our final expression

K = (γ - 1)mc²

I'm curious. Where did you get the idea that

E.K.=m0*c^2*gamma-m0*c^2

Also, on what basis do you justify the assertion that when you subtract relativistic-mass from rest-mass, 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 relativistic-mass from rest-mass you'll get a negative number since relativistic-mass 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. 

[syhprum]

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 ?

[PmbPhy (OP)]

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.

[jeffreyH]

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 off-topic. 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.

[syhprum]

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.

[PmbPhy (OP)]

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 4-momentum cannot in general be used to define the mass of the body. The magnitude of 4-momentum found as follows:

E² - (pc)² = (mc²)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 x-axis in S as follows; Let

• represent a small sphere. Then

+y
|
|
|
|
--------

• =====
• =====
• ------> +x

|
|


At t = 0 each body emits a two photons of equal energy in opposite directions parallel to the y-axis. 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 4-momentum can't be defined and have a meaningful magnitude. However the relativistic mass will always be meaningful.

[JohnDuffield]

Where's my post? Who moved it to new theories?

I said energy gravitates. See what Einstein said: "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.

[CPT ArkAngel]

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 non-relativistic doppler shift). But this is where gravity and inertia have a clear distinction. The local character of inertia and the non-local 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. 

[PmbPhy (OP)]

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/8-962-general-relativity-spring-2006/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.

[CPT ArkAngel]

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.