Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mass dilation

  1. Apr 9, 2010 #1


    User Avatar

    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?
  2. jcsd
  3. Apr 9, 2010 #2
    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.
  4. Apr 10, 2010 #3


    User Avatar

    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.

    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.
  5. Apr 10, 2010 #4
    This is not difficult:


    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.
  6. Apr 10, 2010 #5


    User Avatar

    Thanks for the response but unfortunately, being mathematically ignorant to the nth.degree (to use an algebraic factor), I do not possess 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.
  7. Apr 11, 2010 #6
    A quick Google search unearthed this comment in a LHC chat group http://www.lhcportal.com/Forum/viewtopic.php?f=4&t=71
    and this from an LHC outreach FAQ http://lhc-machine-outreach.web.cern.ch/lhc-machine-outreach/lhc-machine-outreach-faq.htm
    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.
    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.
    Last edited: Apr 11, 2010
  8. Apr 11, 2010 #7


    User Avatar

    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.

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

    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.
  9. Apr 11, 2010 #8
    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.

    No, again. Relativistic mass does not affect the gravitational attraction of a body. This is a recurring mistake in people's minds.
  10. Apr 11, 2010 #9
    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.
  11. Apr 11, 2010 #10
    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.

    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.
    Last edited: Apr 11, 2010
  12. Apr 11, 2010 #11


    User Avatar

    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.
  13. Apr 11, 2010 #12


    User Avatar

    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?
  14. Apr 11, 2010 #13
    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.
  15. Apr 11, 2010 #14
    This is exactly what you were claiming at the end of your post #7.
  16. Apr 11, 2010 #15


    User Avatar

    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.

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

    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.
  17. Apr 11, 2010 #16


    User Avatar

    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.
  18. Apr 12, 2010 #17


    User Avatar

    The end of my post #7 was -

    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!
  19. Apr 12, 2010 #18
    No, there wouldn't.
  20. Apr 12, 2010 #19
    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

    (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
    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:

    [tex]F = \frac{GMm_0}{r^2}[/tex]


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

    Seems a simple enough question.
    Last edited: Apr 12, 2010
  21. Apr 12, 2010 #20
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook