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

Four momentum: why four?

  1. Oct 21, 2004 #1

    learningphysics

    User Avatar
    Homework Helper

    I've been trying to see how relativity shows that the quantity gamma*m*c^2 (total energy) is conserved. I assumed that this would proceed from the conservation of momentum. So I researched momentum in relativity, and noticed that it has a time-component: gamma*m*c (which is E/c). So this partially answers my question. If four-momentum is conserved so is E.

    But how and why does this fourth component get introduced? The spacelike components of momentum have been changed from m*v (classical physics) to gamma*m*v. This part I understand. But can someone show me how this fourth time component (gamma*m*c) is necessary from the first principles of relativity? Thanks a bunch!
     
  2. jcsd
  3. Oct 21, 2004 #2

    jcsd

    User Avatar
    Science Advisor
    Gold Member

    Clearly 3 vectors in genral are not Lornetz invariant so if we wish to describe quantities in a Lornetz invariant manner we need four vectors, in any frame a four vector has a time component.

    If we take an object, it's four vector velcotiy U is the vector tangent to it's worldline, it's four momentum is simply mU.
     
  4. Oct 21, 2004 #3

    learningphysics

    User Avatar
    Homework Helper

    Hi. Thanks for replying. I still have some questions. The conservation of this fourth component is troubling me. My reasoning process is like this:

    1. mv is not conserved in all frames. so we change the definition of momentum to gamma*m*v. This is conserved in all frames. So that's the first 3 components of momentum.

    2. We need a fourth component to make the norm of the momentum lorentz invariant. This is y*m*c.

    3. We guess that this fourth component of momentum is also conserved??

    Is part 3 a new hypothesis? We can verify the gamma*m*v formula by studying elastic collisions from classical physics and seeing that this new space momentum is conserved. But this fourth time component is very different. Did Einstein just guess that it is conserved also?

    Or does the fact that the norm of the momentum vector is Lorentz invariant and the first 3 components are conserved somehow prove that this fourth component is also conserved?

    I think there is something fundamental to Lorentz invariance that I'm missing.

    Thanks again!
     
  5. Oct 21, 2004 #4

    jcsd

    User Avatar
    Science Advisor
    Gold Member

    The conservation of the time component of four momentum is just the conservation of energy.
     
  6. Oct 21, 2004 #5

    learningphysics

    User Avatar
    Homework Helper

    How can one prove the conservation of energy?
     
  7. Oct 21, 2004 #6

    selfAdjoint

    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    If you step back from momentum and consider position, you find that it has four components in Minkowski spacetime, three of space and one of time. The time one has an opposite sign to the space ones, and has a factor of c to make its units the same as the space components. The components of a four-vector like this are not Lorentz covariant, but the length is because it's the inner product of the vector with itself and the inner product is covariant. So now if you differentiate this position four-vector wrt proper time you get a velocity four-vector, and you multiple that by invariant mass and you get a momentum four-vector.

    It time component is the kinetic energy of the particle, its space components span the spatial momentum of the particle mp, and its length (which is invariant, remember) is mc^2.
     
    Last edited: Oct 21, 2004
  8. Oct 21, 2004 #7

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You probably meant to say that
    "its time component is the relativistic energy of the particle"...
    The relativistic kinetic energy (i.e., the non-rest-energy) is the relativistic energy minus the rest energy: [tex]K=E-mc^2=\gamma mc^2 - mc^2 =(\gamma-1)mc^2[/tex],
    where [itex]m[/itex] is the rest mass.
     
  9. Oct 21, 2004 #8

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If you accept that total relativistic momentum [itex]\sum \gamma_{(i)} m_{(i)}v_{(i)}[/itex] is conserved in (say) an elastic collision, then I would think that applying a velocity transformation would imply the conservation of relativistic energy.
     
  10. Oct 22, 2004 #9

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    You've already noticed that the momentum 4-vector is actually the energy-momentum 4-vector. So it should not be surprising to you that there are 4 quantities in the vector, one for the enregy, and 3 for the momentum.

    But you may ask - why do energy and momentum mix in this manner? The answer is interesting, but it's unfortunately a bit advanced.

    The relevant theorem is known as Noether's theorem. It points out that the existence of a conserved momentum is a consequence of the invariance of the laws of physics to spatial translation, plus the action principle.

    THe existence of a consered energy is the result of the invariance of the laws of physics to time translation, plus the action principle.

    Since space and time get intermixed by different observers, so do energy and momentum. This is ultimately "why" the energy-momentum 4-vector transforms in the same manner as the space-time vector.

    There's a little more about Noether's theorem here:

    wikipedia

    I'll quote the points which I think are important from this article below:

    In order to understand the argument fully, though, one has to be aware of the formulation of the laws of physics in terms of Hamilton's principle, the principle of least action. This may be mentioned in undergraduate physics courses, but it's usually not taught in full until the early graduate level.

    The main point I want to make though, is that since space and time "intermix" because of relativity, so do energy and momentum, since energy is a time-translation symmetry and momentum is a space-translation symmetry.
     
  11. Oct 25, 2004 #10

    Garth

    User Avatar
    Science Advisor
    Gold Member

    The important point about using the Principle of Least Action is that it works for generalised coordinates. Hence the conserved quantity obtained from using that principle is that one appropriate for the manifold used. As energy-momentum or four-momentum is invariant in four dimensional space with a signature of -2 then the application of the principle of least action to such a manifold conserves energy-momentum of the field or the rest mass of a particle, rather than energy or three-momentum separately.

    Garth
     
  12. Oct 25, 2004 #11
    The term you speak of [itex]E = mc^2[/itex] (sometimes labeled T) is the inertial energy[/itex] of an isolated object (i.e. an object on which no force acts at all - a "free" object) and has the value

    E = Rest energy + Kinetic Energy = E0 + K

    Total energy, W (sometimes labeled E) is a term which does not have a unique meaning in the relativity literature. Sometimes it refers to inertial energy, T, and sometimes it refers to T + V where V = potential energy of position in a field. The total inertial energy of a system is conserved if no force acts on the system and if the objects of the system do not interact. If they interact then you have to take into account the energy and momentum of fields through which they act. For a free object E is proportional to the time component of its 4-momentum. W is proportional to the time component of the canonical 4-momentum.

    Consider a closed system of non-interacting particles. The mass, m, of a particle (not to be confused with proper mass m0) is defined as the m such that the quantity m[sb]v[/sub] is conserved in all inertial frames of reference. The quantity m[sb]v[/sub] is then defined as the (three) "momentum" of the particle. This is true in both Newtonian physics and relativistic physics. It can then be shown that, if the particle is a tardyon (i.e. moves at speeds less than c) then the mass is related to the proper mass by [itex]m = \gamma m_0[/itex]. For derivation please see - http://www.geocities.com/physics_world/sr/inertial_mass.htm

    It can be shown that the mass of a tardyon transforms as

    [itex]m' = \gamma(m - v p_x /c^2)[/itex]

    For derivatioin please see - http://www.geocities.com/physics_world/sr/energy_momentum_trans.htm

    Thus m transforms in the exact same way that time does if momentum takes the role of position. Anything which transforms from one inertial frame to another in the same way as (ct, x, y, z) is called a Lorentz 4-vector. Therefore P = (mc, p) is a 4-vector called the 4-momentum (aka energy-momentum 4-vector).

    By demanding that 3-momentum be conserved in all inertial frames of reference it can then be shown that this means that m is also a conserved quantity. The derivation is a bit long and I have not been able to create a new web page to post the derivation yet.
    Yes.
    See derivation in above URL.
    If by "m" you mean "proper mass" (sometimes called "rest mass") then yes. That's true. But momentum is defined as the product of (inertial) mass times velocity. Inertial mass is defined such as to allow this.
    Sounds good to me.
    No! We do not guess it. It can be proved by deduction.
    The principle of the conservation of energy cannot be proved. It is a postulate. However certain energy conservation relations can be proved. In this case it can be proved that E is conserved.
    The time component of a 4-momentum is not kinetic energy. The time component is the sum of rest energy and kinetic energy.

    Note: Do not assume that [itex]E = mc^2[/itex] is valid in all cases and thus try to replace m with E. It won't work in general. In fact it won't work for objects which have a finite extent (like a rod). For a free particle in an inertial frame the relation [itex]m/v = E/c^2[/itex] is always valid. However if the body is an object (body with a finite size - not pointlike) rather than a particle then this relation is invalid. Missing this very important point has incorrctly led people to a paradox or two.

    Take as an example a rod, or proper mass m0 which lies parallel to the x-axis of an inertial frame S. Assume there are no forces acting on the rod and it is therefore stress/tension free. Consider the mass as observed from the inertial frame S' which is in standard configuration with S moving with speed v in the +x direction. The mass (as defined above -> m = p/v) in S' has the value [itex]m = \gamma m_0[/itex] as expected.

    Now try this again but this time exert a force on each end of the rod such that the rod remains at rest in S. Assume that the object does not compress a noticible amount. The mass as measured in S is still m0. The total force is zero but this is not a free object - there are forces acting on it. Now observe this from S'. The mass is now no longer equal to [itex]m = \gamma m_0[/itex]. The mass is now a function of the stress acting on the body and therefore does not have the value [itex]m = \gamma m_0[/itex]. The momentum of the rod is now no longer [itex]p = \gamma m_0 v[/itex].

    I have not worked this particular example out on a web page. But I have worked out a similar example. See -- http://www.geocities.com/physics_world/sr/rd_paradox.htm

    When I'm able to I will create a new page to explain why this is true.

    Pete
     
    Last edited: Oct 25, 2004
  13. Oct 25, 2004 #12

    learningphysics

    User Avatar
    Homework Helper

    Thanks to everyone that replied. It was all very helpful. I read Einstein's paper on mass and energy equivalence. He uses the situation of an elastic collision to derive formulas for momentum and kinetic energy. Then he uses the situation of an inelastic collision to derive the formula for rest energy.
     
  14. Oct 26, 2004 #13
    Okay. Last night I created a web page in order to provide this proof. I've been putting it off for some time but since there are so many people who don't understand that the conservation of P0 can be proved through conservation of 3-momentum I thought it was about time I put this up.

    Please see - http://www.geocities.com/physics_world/sr/mass_conservation.htm

    (P0 is what you and some others have refered to as "energy" but what I and others refer to as "mass")

    Pete
     
    Last edited: Oct 26, 2004
  15. Oct 26, 2004 #14

    Garth

    User Avatar
    Science Advisor
    Gold Member

    Pete - on your web page you use P0 not P0, which is it?
    Also the real problem arises when the particles are freely falling in a general external gravitational field, then energy is not conserved even though no work is being done on or by them by that field.
    Garth
     
  16. Oct 26, 2004 #15
    I made an error typing. I typed "sub" when I meant to type "sup". It has been corrected.
    This thread seems to be about SR and inertial frames in flat spacetime. That's what I was addressing. I see nothing which indicates otherwise.
    Incorrect
     
  17. Oct 26, 2004 #16

    Garth

    User Avatar
    Science Advisor
    Gold Member

    We all do that!
    Indeed in SR the problem disappears, however it is always good to generalise concepts, in this case to non flat space-times.
    You will find that most relativists disagree. The reason why I asked whether you meant P0 or P0, is that it is P0 that is conserved in the static gravitational field in the centre of mass frame in which the metric components are static, as you prove in your lecture notes. These are the same only in a Minkowski metric. I believe the reason there is confusion as to which one is to be used as energy is because in fact in general energy is not conserved in GR. (Though of course it is in SR)
    To obtain a value for energy in a frame of reference Ua, the scalar product of four-momentum with Ua has to be taken,
    E = - PaUa . [Minus sign to agree with most common sign conventions]

    Garth
     
    Last edited: Oct 26, 2004
  18. Oct 26, 2004 #17

    learningphysics

    User Avatar
    Homework Helper

    Thanks Pete! Your conservation of mass proof is exactly what I was looking for.
     
  19. Oct 26, 2004 #18

    learningphysics

    User Avatar
    Homework Helper

    Thanks Pete! Your proof of conservation of mass was exactly what I was looking for.
     
  20. Oct 27, 2004 #19
    I seriously doubt that.

    Garth - I've explained all of this to before, several times in fact. But you're not listening to me. You have something in mind which is incorrect and you seem to be refusing to understand the correct concept. I.e. you're confusing the concept of the conservation of energy with the concpet of the constancy of the energy of a particle. These are two different things which are simply related to each other. But you constantly confuse them as being identical concepts.

    The principle of energy conservation does not mean that the energy of a single particle remains constant, i.e. it doesn't change with time.
    The principle of energy conservation refers to a sum of energies of a closed system. I.e. if the sum of the energies of a closed system is not a function of time then it is said that energy of the system is conserved.

    Recall what you said ..
    So what? Why did you say this? What was your point? Since when should work being done on a particle imply that the energy of the particle is constant? That has never been true in Newtonian physics so I can't fathom why you're always focusing all of your energy on it. Simply because work is being done on a particle is no guarantee that the total energy of the particle will remain constant. This is the same in Newtonian mechanics too. The only way that the energy of a single particle will remain constant is when the force is derivable from a potential energy function which is not an explicit function of time. When that is the case then E = K + V = constant. When the field is time dependant and V = V(t) then K + V is not a constant. This does not mean I'll win the Nobel prize for figuring out how to violate the principle of energy conservation since that principle only applies to closed systems.

    Tell me something Garth - In classical (non-relativistic) electrodynamics the total energy of a charged particle in an EM field is not a constant in general. So why haven't you focused on that too instead of focusing exclusively on GR???

    One a similar point so as to illustate what I'm saying - Consider the meaning of the phrase "momentum is conserved". Does this mean that a law of nature states that the momentum of a particle can never be changed? Of course it doesn't. Does it mean that the momentum of a charged particle in an EM field will remain constant? No. Of course not. It quite literally means that the total momentum of all particles and fields of a closed system is a constat.

    Who's confused?? And here once again your misusing the concept of energy conservation.
    That is incorrect. This is not what the notion that P0 = constant refers to. E = - PaUa is the energy of a particle as measured by an observer whose 4-velocity is U. P0 is the energy of a particle as measured by a (for lack of a better term) coordinate observer. E.g. in Schwarzschild coordinates this observer is refered to as the Schwarzschild observer aka "far away observer".

    Before you continue with this please explain what you think "conservation of energy" means as it is pertains to Newtonian mechanics. Thanks.

    Your welcome. I wanted to get this up again before I had to leave the internet again.

    I also created a new web page to explain why its incorrect to claim that m is defined as E/c2[/sub]. See

    http://www.geocities.com/physics_world/sr/mass_momentum_density.htm

    To see a worked out example of this please see
    http://www.geocities.com/physics_world/sr/rd_paradox.htm

    Notice that the mass of the struts is not proportional to the energy of the struts, i.e. In this case m does not equal E/c2 for the strut. This holds only for closed systems and thus does not hold in general. E.g. if you have a charged body in an electric field then m = E/c2 is invalid.

    Pete
     
  21. Oct 27, 2004 #20

    Garth

    User Avatar
    Science Advisor
    Gold Member

    Pete - let me take this slowly:
    A freely falling particle has no forces acting on it - according to GR - therefore it can be considered as a closed system. In which case the fact that its energy does not change means that energy is conserved and the opposite is true when its energy does change.

    This is a very interesting case to illustrate my point. In such a case the closed system consists of the total energy of the charged particle and the energy of the field, this is conserved. The energy of the charged particle varies because of the electromagnetic force acting on it from the field, work is done on the particle and its energy changes, however the system energy is conserved. However in the case of GR gravitation the energy of a freely falling and accelerating particle changes and no forces are acting (to first order ignoring tidal forces) as the Newtonian force is explained by space-time curvature. Energy is not conserved in this small closed system.

    The four velocity Ua refers to the frame of reference of the observer, we do not disagree over this, if I have been ambiguous I apologise.

    Energy is a slippery concept we have to be careful how we define it and how we measure it, however that being said the fact that GR does not in general or locally conserve energy is well accepted from Einstein onwards.

    Noether's theorem showed how the conservation of energy required a symmetry under time translations. This is only possible in GR under certain special circumstances when there is a time-like Killing vector. Such as when the metric components in the observer's frame of reference are static.

    As we know one of the basic principles of GR is that of the conservation of energy-momentum, but this is not a concatenation of the two principles of the conservation of energy and conservation of momentum. Energy and momentum are frame dependent whereas energy-momentum stands alone as a geometric object.


    Garth
     
    Last edited: Oct 27, 2004
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Four momentum: why four?
  1. Four Momentum of Photon (Replies: 13)

Loading...