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Relativistic momentum

  1. Jan 19, 2006 #1
    1. Why do we have to assume mass doesn't change? And always use [tex]m_0[/tex]?
    2. OK, let's assume we always use [tex]m_0[/tex]. Then why is momentum is defined as [tex]\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]. If measuring in object's frame of reference, shouldn't we use his distance and time, where they are [tex]\frac{x_0}{\gamma(v)}[/tex] and [tex]t_0 \gamma(v)[/tex], and v would be [tex]\frac{x_0}{t_0 \gamma(v)^2}[/tex]. It seems that we're using [tex]x_0[/tex] either [tex]t_0[/tex] and not both, nor none. Isn't this inconsistent?
     
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  3. Jan 20, 2006 #2

    jtbell

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    It's basically a matter of taste whether one defines relativistic momentum with one formula using rest mass, or a different formula using relativistic mass. They give the same numeric value for momentum in either case. We use one of these definitions because that is the quantity whose sum turns out to be conserved in collisions and other processes involving two or more particles. The way to derive it is to analyze a collision carefully and look for a quantity whose sum is conserved.

    See http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000054000009000804000001&idtype=cvips&gifs=yes [Broken] for one derivation. This is not the one that is usually given in textbooks, but the paper lists some textbooks that do contain the usual derivation. It might be on the net somewhere, but I'm not going to spend a half hour looking for it. You're welcome to do that if you want. :wink:
     
    Last edited by a moderator: May 2, 2017
  4. Jan 20, 2006 #3
    please have a look at
    http://arxiv.org./abs/physics/0505025 where a derivation of the transformation equations for mass, momentum and energy does not use conservation laws. A post use review is appreciated.
     
  5. Jan 29, 2006 #4
    jtbell, that link you gave seems not to be available freely (username+password, or $$$). Having not paid to AIP because I think physics is a science and not a subject of commerce, I couldn't see that paper. Thanks anyway.

    Bernhard, thanks for that paper. However, according to that paper, mass change seems to be necessary. So, that would quite disagree with usenet??
     
    Last edited: Jan 29, 2006
  6. Jan 29, 2006 #5
    You don't have to assume that. It follows from the definition of mass(speed) = |momentum|/speed. Those who chose rest mass over relativistic mass do something similar but defined "mass" m to be mass(0) = m. However that is a limited definition which does not hold in general. m defined in this way is not always the intrinsic property of an object as some might believe. The mass(0) = m definition holds only for isolated (aka 'closed') systems or for point particles. When it comes to fields or extended objects under stress
    (or for mass density) then the definition where "m" is an invariant makes no sense.

    Because those who use the term "mass" to mean "proper mass" and label it "m" express it as "m" rather than m0.
    If you're measuring in the objects frame of reference then [itex]\gamma[/itex] = 1. I don't understand what you mean by the quantities [tex]\frac{x_0}{\gamma(v)}[/tex] and [tex]t_0 \gamma(v)[/tex]. What is x0? Do you mean the x-coordinate as measured in the rest frame of the object? If so then since [itex]\gamma[/itex] = 1 you have not really done anything. Please clarify for me.

    Note: (ct, x, y, z) is the coordinate of an event while distances and time intervals are the components of the a spacetime displacement.

    Pete
     
    Last edited: Jan 29, 2006
  7. Jan 29, 2006 #6

    jtbell

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    Sorry about that. We must have an institutional account here which gives access to everyone who connects from my college's network.

    Anyway, I'm sure you can find the "standard" derivation that uses a glancing collision, in many second-year level "introdution to modern physics" textbooks. I know it's in Beiser and in Taylor/Zafiriatos/Dubson. It might even be on a Web page somewhere if you plow through the hits that Google gives you for "derivation of relativistic momentum".
     
  8. Jan 29, 2006 #7
    pmb_phy, I'm pretty certain I'm wrong at somewhere, I just don't know where.
    I assume a coordinate system, moving with -v according to the mass. So, according to mass, there should be a length contraction in coordinate system right? My question is, what if mass thinks "oh, I'm moving at v, and coordinate system stands still, from this POV, what is my momentum? my speed is: in a time interval I measure the distance I've taken (which was contracted according to and outside observer, standing still to the coordinate system). and my m is m_0. so my momentum should be m_0 (dr/gamma(dr/dt))/dt"?

    What I mean by t_0 and l_0 are proper time and proper length. Okay, it seems his time is proper time, but there should be some contraction in length, right?

    jtbell, in Taylor/Zafaritos, it first assumes that momentum is mdr/dt_0, then tests it in a thought experiment (well, actually it doesn't do it even, full solution it's given as an exercise!). Good assumption I say, but why to assume that.

    Or is this the whole story: "find an expression that is consistent with conservation of momentum, oh, and it should include gamma somewhere".
     
  9. Jan 30, 2006 #8

    pervect

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    Momentum is defined in this manner so that it remains a conserved quantity when frames of reference are changed via the Lorentz transform.

    The Newtonian defintion of momentum, p=mv, is compatible with the Gallilean transform - the relativistic deveition of momentum, p = mv/sqrt(1-(v/c)^2) is compatible with the Lorentz transform.

    There is a detailed proof in many physics textbooks, Goldstein's "Classical mechanics" comes to mind.
     
  10. Jan 30, 2006 #9
    Have a look please at a paper
    why p=gamamv? because of the principle pf relativity
    http:arxiv.org/abs/physics/0404059
     
  11. Jan 30, 2006 #10

    DrGreg

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  12. Jan 31, 2006 #11

    DrGreg

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    I am having some problems understanding this paper. I don't see how equation (11) was obtained, and equation (14) doesn't even make sense, as it equates a velocity to a momentum!
     
  13. Jan 31, 2006 #12
    Please move (m) from the right side to the left one in (14).
    With a little perseverence you can derive (11) using the addition law of relativistic velocities.
     
  14. Feb 3, 2006 #13
    Then perhaps what you're looking for is in this paper -

    http://www.geocities.com/physics_world/mass_paper.pdf

    I wrote it specifically for people who have questions regarding mass in relativity, just like yourself. I also shined some light on well-established but little known facts of mass in relativity. I used one in a paradox regarding the mass density of a magnetic field.

    Let me know what you think of it. I've been having a hard time getting feedback on it. What little I got was very good though.

    Pete
     
    Last edited: Feb 3, 2006
  15. Feb 9, 2007 #14
    I enjoyed your paper.

    Let me describe what I think I understand and you can point out the errors.

    Mass A is near several masses B, C, and D which have random velocities.

    Each of the masses B, C, and D see a distortion of mass A depending on their velocity with respect to A.

    Each of the masses B, C, and D have a distortion component and the “rest energy” component.

    The energy required to accelerate A in a particular direction as perceived by B will depend on the “rest energy” of A and the velocity A has with respect to B in the vector direction A is to be accelerated.

    Mass C and D do not experience the velocity mass B has attributed to A but only the “rest energy” of A and then of course C and D each have a perceived mass distortion depending on their differential velocity with respect to A.

    I also have a question.

    If A is accelerating with respect to B will B perceive A to have an apparent mass distortion from the “rest energy” mass of A?

    Duane
     
  16. Feb 10, 2007 #15
    I'm glad to hear that Duane. Thanks for letting me know.
    I'll do my best.
    So particles A, B, C and D for a system of particles with random velocities..
    What kind of distortion? Gravitational?
    I don't understand what this means. Particles B, C and D have a "rest energy" component? Do you mean that there is a rest mass associated with the system of the B,C.D particles???
    If you're asking whether the gravitational force is a velocity dependant one then yes, it is.
    Sorry. You lost me. There is nothing about which indicates that C and D do not interact with A and B.
    What kind of distortion?

    Pete
     
  17. Feb 10, 2007 #16
    Sorry for the terminology.

    Mass A has a rest mass.

    Mass B has a velocity with respect to mass A.

    Mass B has a gravitational force toward A and the mass which is used in the gravitational equation to determine the gravitational force is the relativistic mass of A as perceived by B.

    So mass A would "appear" to have more mass then its rest mass.

    Is this correct?
     
  18. Feb 10, 2007 #17

    pervect

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    If you intend to use the Newtonian force equation GmM/r^2 with m being the relativistic mass of the moving mass, this is most definitely not correct.

    To get a really correct answer, one needs to think about the gravitational field as being "tidal forces", because this is what one can actually measure at a point. One can also describe these tidal forces as "curved space-time" - these two descriptions are equivalent.

    To get an approximately correct answer, one can think of the gravitational field as being not spherically symmetrical, i.e. one can think of the field as being "squished" so that the transverse field is stronger, much like the electric field of a moving charge.

    However, if one does not take into account the issues raised with regards to how one defines the gravitational field, the above description will still not be exactly correct, though it does give one a rough idea of what happens.

    Using the spherically symmetrical Newtonian formula is "right out".
     
  19. Feb 10, 2007 #18
    I understand that curved space has complicated equations.

    Does it simplify in the case I just described if masses A and B are moving directly away from each other and what would the true equation be for the force of gravity on B which is considered the stationary objcect toward A with is considered to be moving?

    If it is to difficult to explain don’t bother posting it.

    I don’t know if I can handle 10 pages of equations.

    Duane
     
  20. Feb 10, 2007 #19
    I have a second question.

    Mass A and B are as before.

    A is seen by B as being flattened along the line of site between A and B.

    If A consisted of two objects, A1 and A2.
    A1 AND A2 are on a line of site from B to A.
    A2 is farther away from B then A1.
    Would the distance between the two objects A1 and A2 also seem shorter to B due to the contraction?

    Duane
     
  21. Feb 10, 2007 #20
    Yes, in the special theory of relativity it is not only objects that contract, the whole space in the direction of the relative motion contracts.


    Look at the beauty of relativity:

    When a smaller box s is situated, relatively at rest, inside the hollow space of a larger box S, then the hollow space of s is a part of the hollow space of S, and the same "space", which contains both of them, belongs to each of the boxes. When s is in motion with respect to S, however, the concept is less simple. One is then inclined to think that s encloses always the same space, but a variable part of the space S. It then becomes necessary to apportion to each box its particular space, not thought of as bounded, and assume that these two spaces are in motion with respect to each other.

    Before one has become aware of this complication, space appears as an unbounded medium or container in which material objects swim around. But it must be remembered that there is an infinite number of spaces, which are in motion with respect to each other.

    The concept of space as something existing objectively and independent of things belongs to pre-scientific thought, but not so the idea of the existence of an infinite number of spaces in motion relatively to each other. This latter idea is indeed unavoidable, but is far from having played a considerable role even in scientific thought.

    Fascinating isn't it?
     
    Last edited: Feb 10, 2007
  22. Feb 10, 2007 #21
    Reality just kind of sucks you into it doesn't it.

    I had two posts.

    I was kind of hoping there would be an easy transformation from Newton’s gravity equation to space time if certain conditions were met without eliminating the velocity difference, like that A and B are moving directly away from each other or something to make the system symmetrical.

    Did you miss it or was the relationship just to complicated to discuss?

    Also, I have always imagined that relativistic mass was just a distortion of the information received about mass A as observed by mass B due to the motions of mass A.

    Is this a dangerous view point?

    would the masses A1 and A2 appear from B to interact ( fall toward each other) as if they had relativistic mass ( or whatever space time predicted ) or would they interact as if they have their rest mass.

    I know that to an observer on A1 or A2 which are moving together they would experience rest mass attractions.

    Duane
     
  23. Feb 10, 2007 #22
    This isn't neccesarily true Duane. If body B is moving towards body A then the hte gravitational force may not neccesarily be parallel to the line connecting the two. I'd have to calculate it to see that. Also the force on body A is due to more than just the relativistic mass, the force is also due to the fact that the gravitational force is velocity dependant. There are two velocity dependancies here (1) one due to relativistic mass and (2) one due to the velocity dependant gravitational force. I worked out an example and placed it on my web site here

    http://www.geocities.com/physics_world/gr/force_falling_particle.htm

    Notice Eq. (12). There is an extra factor for the velocity dependant force.

    Actually it would be the moving (i.e. relativistic) mass which "appears" to have more mass. However (relativistic) mass gets altered just by its presence of being in the field.

    Pete
     
    Last edited: Feb 11, 2007
  24. Feb 10, 2007 #23

    pervect

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    I have some serious issues with Pete's calculations, as I've mentioned before. I think Chris Hillman does too, though he generally doesn't think it's worth his time to do "debunking".
     
  25. Feb 11, 2007 #24

    pervect

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    If you have a small test mass falling into a black hole on a radial path, one can say a couple of things:

    1) The tidal stretching force experienced by the infalling observer will be 2GM/r^3, where r is the schwarzschild coordinate. This is the same formula as the formula for the Newtonian tidal force (except for replacing the distance r by the Schwarzschild coordinate r) and is independent of the radial velocity of the falling object.

    (Interestingly enough, the non-radial velocity of the falling object does matter, it increases the tidal force, though this result is not in the textbooks but must be calculated.)

    See for instance "Gravitation" by Misner, Thorne, Wheeler for a detailed derivation. (I can look up the page # on request if anyone is really interested).

    The tidal stretching force on an observer falling through the event horizon of a black hole is thus finite and the same as the tidal stretching force due to Newtonian gravity. A naive interpretation using "relativistic mass" would give the incorrect result of an infinite tidal force for an object falling into a black hole (as the velocity of the infalling mass approaches the speed of light at the event horizon relative to a stationary observer).

    Also, there is an interesting paper which describes the total imparted velocity from a relativistic flyby. The result is also not equal to the "relativistic mass", there is an additional factor of (1+[itex]\beta^2[/itex])

    http://dx.doi.org/10.1119/1.14280
     
    Last edited: Feb 11, 2007
  26. Feb 11, 2007 #25
    Beggars can't be choosers.
    It’s my responsibility to find out if this is correct is it not?
    Besides if it is too easy I will not learn anything.

    I will see how much of it I can understand.

    I may go over you web page as it seems to have some sort of tutorials.

    Duane
     
    Last edited: Feb 11, 2007
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