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Special Relativity and Mass Variation

  1. Jul 20, 2005 #1
    Please correct me if I am wrong, but it is my understanding of Special Relativity that the faster a particle moves the more force it requires to accellorate - laymans terms: becomes more massive. I had a post earlier where I tried to present this in the framework of an "atomic pop gun". I was told that I did not understand the conservation of energy, so I will ask my question and its basic level to try to find out where my understanding derails.

    If a particle is pushed in one direction to the a velocity close to the speed of light, the object that is accelorating the particle will be pushed back as if the particle was more massive than it's at rest mass. This would be the conservation of energy, wouldn't it?

    If both the particle and the object were allowed to travel for a period of time and then the particle was stoped by the object so that both the particle and the object were at rest relative to each other, where would be the center of mass for the particle/object? Since the particle was traveling at relativenistic speeds and became more "massive" and the conservation of energy would state that the center of mass would remain the same during this event. However, once the particle/object stopped, the center of mass would move towards the object since the particle is nolonger as massive.

    This is my thinking...as the particle and the object are traveling in different directions, the object is traveling as if the particle was more massive. When the object "stops" the particle, the particle will pull the object back as if it were more massive until it come to rest. For example: accelleration takes 1 sec, both the particle and object travel in opposite directions for 10 secs and then the object exserts a force on the particle to stop it which takes 1 second. The deacceloration of 1 sec will move the object back the distance the acceloration for 1 second moved it away. However, the 10 seconds of travel with no forces being applied would have moved the object farther away than what would have been expected if the particle did not travel at relativnistic speeds. So when the particle/object are at rest following this process, the center of mass is different then when the process began.

    At the conclusion of this process, it appears to violate the conservation of energy, but it also appears to me with my understanding that this is the only way to follow the law of conservation of energy when the forces are applied while using Special Relativity in this scenerio. Any suggestions on where my thought process/understanding is going astray? Thank you!
     
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  3. Jul 20, 2005 #2

    pervect

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    This isn't exactly wrong, but it isn't exactly right, either. Take a look at the sci.physics.faq "Does mass change with velocity" for instance.

    http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

    No. Try to recall, research, or read the threads about how one calculates the momentum and energy of a moving point mass, i.e.

    p = momentum = mv/sqrt(1-(v/c)^2)
    E = energy = m/sqrt(1-(v/c)^2))

    What you need most to solve your atomic "pop-gun" problem is the conservation of momentum, not the conservation of energy.

    Can you see why you need to solve the equation

    m1*v1/sqrt(1-(v1/c1)^2) + m2*v2/sqrt(1-(v2/c2)^2) = 0, i.e. the equation that the total momentum is zero before and after the pop-gun fires?

    Note that I am using m1 and m2 above as the invariant mass, not the relativistic mass! (See again the link

    http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

    so that we can get the terminology issues straightened out!)

    You have stumbled on a problem where the concept of relativistic mass is actually useful, but to get the right answers you need to understand the distinction between invariant mass and relativistic mass.
     
    Last edited: Jul 20, 2005
  4. Jul 20, 2005 #3
    This cannot be true. The law of conservation of the center of mass is not violated in special relativity. A full analysis would make this clear.
    Yes. You're neglecting the mechanism of how things are slowing down or speeding up. For example: suppose there was a spring involved in the acceleration. Then you'd have to take into account the increased mass of the compressed spring since when you compress the spring there is an increase in rest mass of the spring due to the increase of potential energy and the corresponding increase in the rest mass.

    Pete
     
  5. Jul 21, 2005 #4
    Thank you for the post. I guess I should have stated that I am a novice on the mathematics side of this topic. Awhile ago when I viewed a TV show concerning relativity on the discovery channel. I became interested in this theory so I read the SLAC web page concerning special relativity: http://www2.slac.stanford.edu/vvc/theory/relativity.html along with several other websites. I quickly learned that there are alot of websites out there that shouldn't be trusted as absolute fact on this subject. That is why I thought I would try to get a better understanding by posting my question on this site.

    Anyway, I have also read some of the work of Dr. James F. Woodward, Professor at Cal State Fullerton. He had a thought experiment where a person on a skateboard had a brick tied to a string. He was investigating if it was possible to vary the mass of the brick so that it was heavier when it was thrown and lighter when it was pulled back (my interpetation of his thought experiment).

    One day I wondered why these to ideas could not be put togther: the increase of mass/gamma based on an objects velocity and the brick in Dr. Woodward's presentation. I understand that the conservation of momentum (I used energy before - sorry) states that state1 must equal state2. But I am having a hard time picturing this in my head. It seems like if the particle requires 7 time more force to move it at 99% of C, then it will require the same amount of force to bring it to a stop relative to the object. If the object is more massive than the particle, this force will not push the object at relativenistic speeds so its mass/gamma will not increase at the same ratio as the particle. This means that as the particle and the object travel away from each other, the object is moving as if the particle's mass was 7 times its at-rest mass. If this separation lasts for 10 secs, 10 minutes or 10 years, the object would be considerably further away from where it would have been if it only accellorated the particle to 98% of C. The only way I am aware that when the object stopped the particle that momentum would = 0 would be if the particle pulled on the object with a greater force that would pull it all the way back to the place where it would have been if Special Relativity did not exsist.

    I am hoping that someone would be able to explain my error in laymans terms. Something like: because the particle is traveling so fast, the object will need to apply a force to stop the particle right after acceloration thus the object is immediately deaccelorating so there will be no coasting time.

    Am I a lost cause or can someone help me?
     
  6. Jul 21, 2005 #5

    Aer

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    No postulate of special relativity requires mass to increase with velocity. Until pmb_phy or someone else points to any experimental proof that relativisitic mass exists then you need not concern yourself with such notions. I'll point out that if you search the site you linked above for "relativistic mass" you'll come up empty.
     
  7. Jul 22, 2005 #6

    pervect

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    Dunno yet :-)

    This seems reasonable so far.

    This seems to me to be wrong. The easiest way to check whether or not it's right or wrong is to actually do the math. The math is not terribly hard. Doing the math is easier (at least for me) than following your verbal statements - much easier to follow, as a matter of fact, English is terribly imprecise. Math is logic, written in symbols. Doing the math is therfore an easy way of checking your logic.

    The math says:

    the initial momentum of the system is zero
    the final momentum of the system is zero
    the energy of the system is conserved

    IF you think of your high speed particle as being like a alpha or beta particle being emitted from an atomic nucleus, the conservation of energy says that the mass of the nucleus decreases by more than the invariant mass of the emitted particle (you could say that the mass of the nucleus decreased by the relativistic mass of the particle).

    You can also say that the center of mass of the system is located not by taking the average position of the invariant masses, but by taking the average position of the relativistic masses. (That seems to be the same point your rather imprecise verbiage is trying to make).

    You can even say that momentum is relativistic mass * velocity, rather than invariant mass * velocity * gamma.

    So the mathematical equations are EXTREMELY simple in this particular case when you use relativistic mass. This is actually somewhat rare in my opinion, so congratulations for finding a real-world problem where the idea of relativistic mass is actually useful :-).
     
  8. Jul 22, 2005 #7

    Aer

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    It would be much more useful to get rid of relativistic mass altogether. It has no fundamental use and only leads to confusion, ignorance, and violence.
     
  9. Jul 22, 2005 #8
    In some textbooks[1] I read, author rather uses term "center-of-momentum" (in short, its again CM) for a frame of reference in which (three) momentum of system is zero. So "relativistic mass" is simply avoided.


    [1] D. Griffiths, "Introduction to elementary particles", J. Wiley & sons (1987.)
     
  10. Jul 22, 2005 #9

    pervect

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    I don't use "center of mass" much in relativity either, though it is common to pick a frame where the total momentum is zero, it's less common to insist that the origin of said frame be located any place in particular. Thus "center of momentum" usually sufficies.

    The fact that the center of mass does not move when relativistic mass is used is fairly easy to demonstrate in the case of two particles and no fields. The result should generalize to any number of particles, but the presence of any fields which carry momentum or energy would be expected to invalidate it unless the energy or momentum of the fields is also included.

    The math is fairly easy. Let the "relativistic mass" of masses 1 and 2 be called m1_r and m2_r. (I'm not sure how standard this notation is, I don't use relativistic masses much).

    Then m1_r * v1 + m2_r * v2 = 0, by the conservation of momentum - the totla momentum must be zero. (v1 and v2 must have opposite signs, a positve velocity is when x increases with time, and a negative velocity is when x decreases with time).

    This implies that m1_r * dx1/dt + m2_r * dx2/dt = 0

    Integrating gives that m1_r * x1 + m2_r * x2 = C

    If m1 and m2 both start at x=0 (the trivial center of mass), the value of C must initally be zero. Since it is a constant, it cannot change, and must always be zero.
     
  11. Jul 22, 2005 #10

    EnumaElish

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    How can learning something new (properly explained) lead to ignorance? This reminds me of the Internet myth that drinking coffee actually decreases one's body fluid volume, or that eating celery decreases one's total calorie intake at the end of the day.

    And under what circumstances exactly has this concept led to violence, Aer? What else have I been missing in these forums?
     
  12. Jul 22, 2005 #11

    Aer

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    Sarcasm.




    This message is not too short!
     
  13. Jul 22, 2005 #12

    EnumaElish

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    There is no way I could have missed a lot of sarcasm. Just curious about the violence bit.
     
  14. Jul 22, 2005 #13
    Thank you to all who have posted to this!

    Well I will try to show my work on how I understand it using math, but please be patient with me. It has been almost 20 years since I had a math class.

    A particle and an object are floating in space with 0 velocity relative to each other. The object’s mass (M1) is 71,000 times the particles mass (M2).

    M1 = 71,000*M2

    The object exerts a force onto the particle to move it away from the object at 99% the speed of light. It is my understanding of Special Relativity that the faster an object moves, the more force it take to push it faster (appears to become more massive). This equation is:

    Gamma = 1/sqrt(1-v2/c2)

    Solving this for the particle (M2):

    Gamma = 1/sqrt(1 – (0.99* c) 2/c2) = 7.1 (approximately)

    So if the object pushed the particle so that the particle was traveling at 99% of the speed of light relative to the object, the object would be traveling in the opposite direction at:

    V1 = V2 * (Gamma * M2) / M1 = 99%C * (7.1 * 1) / 71,000 = 0.0099%C

    If the object traveled for 1 second and then exerted a force on the particle to bring it to a complete stop relative to the object, the object’s distance traveled (OD) and the particle’s distance traveled (PD) would be:

    OD = 0.0099% * (C/1 second) = 0.0099% * 186,300 miles = 18.4437 miles
    PD = 99% * (C/1 second) = 99% * 186,300 miles = 184,437 miles

    However, if the object pulled the particle back so that the particle was traveling only 1% of the speed of light, the gamma would be extremely low:

    Gamma = 1/sqrt(1 – (0.01* c) 2/c2) = 1.00005 (basically no additional force)

    The object would then be traveling only:

    V1 = V2 * (Gamma * M2) / M1 = 1%C * (1.00005 * 1) / 71,000 = 0.000014%C

    Traveling at 1% of C, it would take the following amount of time for the particle to travel back to its starting position:


    Time = Original distance/(1% of C) = 184,437 miles / (.01 * 186,300 m/s) = 99 seconds

    If the object exerted a force on the particle to pull it back to the particle’s original starting position (184,437 miles) then the object would travel:

    OD = 99 seconds * .000014% of C = 99s *.000014%*186,300m/s = 2.58 miles

    Conclusion (at least the way I understand this subject): If the object pushes the particle to 99% the speed of light, it will travel approx. 18.4437 miles in 1 sec and the particle will travel 184,437 miles. If the object pushes the particle at only 1% of the speed of light, then it will take the particle 99 seconds to travel the same distance and the object will only travel 2.58 miles. If this is the case, then when the object stops the particle from traveling at 99% the speed of light, then either the object would have traveled an additional 15.86 miles or the force needed to stop the particle would result in pulling the object back 15.86 miles. The later would mean that it would require more force to stop the particle then was applied to move the particle and this additional force is dependant on the time the object and particle are traveling in opposite directions. If the first is correct and the object will travel 15.86 miles further than if the particle traveled at non-relativenistic speeds, then by applying energy to push the particle at relativenistic speeds and pulling it back at non-relativenistic speeds would result in moving the object thru space.

    What am I missing? Any help would be greatly appreciated! Thanks!
     
  15. Jul 24, 2005 #14

    pervect

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    (Snip wrong equations)

    If m1 exerts a force on m2, m2 should exert a force on m1.

    Using the invariant masses m1 and m2 for this go around (check my previous post for the equations with relativistic masses)

    we can say that

    F = dp1/dt which is gamma^3 * m1 * a1

    See for instance, Doc Al's post about Longitudinal mass about why gamma gets cubed

    https://www.physicsforums.com/showpost.php?p=687759&postcount=71

    (or, if your math is up to it, you could do the derivative). But you don't really need to be able to do calculus to find the velocities! (Do I hear a sigh of relief :-)).

    Write down the equation for the total momentum of the system. The force acting on m1 and m2 will always act in such a way that momentum remains conserved. Thus, if at t=0, v1=0 and v2=0, the total momentum is zero. We can then write:


    gamma1*m1*v1 + gamma2*m2*v2 = 0

    or m1*v1/sqrt(1-v^2) + m2*v2/sqrt(1-v^2) = 0, if we for convenience take velocities as a number between 0 and 1, 1 being equal to the speed of light (i.e. we set c=1, or use relativistic units).

    given m1,v1, and m2, this equation directly allows one to find v2 with no mucking about with calculus.

    Therefore I would suggest using this equation for v2, and proceeding on with the rest of your analysis, which I more or less skimmed over.

    Actually there is one more thing to beware of. The way I've written this, it looks like m2 is a constant. But it is not - it is a variable. Imagine that m2 is a nucleus, and that m1 is an emitted particle (a beta particle, and alpha particle, whatever). When a heavy nucleus emits a particle, its mass is lowered by more than the rest mass of the particle.

    In terms of forces, the energy needed to accelerate m1 comes from somewhere, and we can presume in this scenario that it's being generated by m2, which is a system (like the heavy nucleus) that is converting part of its mass to energy to cause m1 to accelerate.

    We can idealize the situation by saying that the total energy is constant, thus


    This gives us one more equation, the conservation of enregy

    gamma1*m1 + gamma2*m2 = constant

    These two equations allow us to calculate both m2 and v2 given m1 and v1 - with no calculus required.

    Hopefully you can proceed again with your analysis with the corrected equations (cross fingers!).

    Your analysis will be VERY MUCH simplfied if you assume that the acceleration of m1 to it's final velocity happens very quickly. Otherwise you'll have to take into account the fields that do the acceleration.

    So, you need to pick m1,v1, and the initial value of m2.

    You then calculate m1,v1, the value of m2 after the emission of m1 (call it m2'), and then v2. You have two equations to do this, and two unknowns.
     
    Last edited: Jul 24, 2005
  16. Jul 24, 2005 #15
    My appologies...I originally wrote my last post in MS Word then pasted it into this site. When I did so, my superscript/subscripts were converted to regular text. I will repost below to show my superscripts as c^2 instead of just c2, etc. I will also go thru the suggestions on the last posting and make the corrections suggested when I get time later this week. Thank you for your help!

    +++++++++++++++++++++++++++
    Repost:

    Well I will try to show my work on how I understand it using math, but please be patient with me. It has been almost 20 years since I had a math class.

    A particle and an object are floating in space with 0 velocity relative to each other. The object’s mass (M1) is 71,000 times the particles mass (M2).

    M1 = 71,000*M2

    The object exerts a force onto the particle to move it away from the object at 99% the speed of light. It is my understanding of Special Relativity that the faster an object moves, the more force it take to push it faster (appears to become more massive). This equation is:

    Gamma = 1/sqrt(1-v^2/c^2)

    Solving this for the particle (M2):

    Gamma = 1/sqrt(1 – (0.99* c)^2/c^2) = 7.1 (approximately)

    So if the object pushed the particle so that the particle was traveling at 99% of the speed of light relative to the object, the object would be traveling in the opposite direction at:

    V1 = V2 * (Gamma * M2) / M1 = 99%C * (7.1 * 1) / 71,000 = 0.0099%C

    If the object traveled for 1 second and then exerted a force on the particle to bring it to a complete stop relative to the object, the object’s distance traveled (OD) and the particle’s distance traveled (PD) would be:

    OD = 0.0099% * (C/1 second) = 0.0099% * 186,300 miles = 18.4437 miles
    PD = 99% * (C/1 second) = 99% * 186,300 miles = 184,437 miles

    However, if the object pulled the particle back so that the particle was traveling only 1% of the speed of light, the gamma would be extremely low:

    Gamma = 1/sqrt(1 – (0.01* c)^2/c^2) = 1.00005 (basically no additional force)

    The object would then be traveling only:

    V1 = V2 * (Gamma * M2) / M1 = 1%C * (1.00005 * 1) / 71,000 = 0.000014%C

    Traveling at 1% of C, it would take the following amount of time for the particle to travel back to its starting position:


    Time = Original distance/(1% of C) = 184,437 miles / (.01 * 186,300 m/s) = 99 seconds

    If the object exerted a force on the particle to pull it back to the particle’s original starting position (184,437 miles) then the object would travel:

    OD = 99 seconds * .000014% of C = 99s *.000014%*186,300m/s = 2.58 miles

    Conclusion (at least the way I understand this subject): If the object pushes the particle to 99% the speed of light, it will travel approx. 18.4437 miles in 1 sec and the particle will travel 184,437 miles. If the object pushes the particle at only 1% of the speed of light, then it will take the particle 99 seconds to travel the same distance and the object will only travel 2.58 miles. If this is the case, then when the object stops the particle from traveling at 99% the speed of light, then either the object would have traveled an additional 15.86 miles or the force needed to stop the particle would result in pulling the object back 15.86 miles. The later would mean that it would require more force to stop the particle then was applied to move the particle and this additional force is dependant on the time the object and particle are traveling in opposite directions. If the first is correct and the object will travel 15.86 miles further than if the particle traveled at non-relativenistic speeds, then by applying energy to push the particle at relativenistic speeds and pulling it back at non-relativenistic speeds would result in moving the object thru space.

    What am I missing? Any help would be greatly appreciated! Thanks!
     
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