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I Zeroth component of 4-momentum & energy-momentum relation

  1. May 2, 2016 #1
    As I understand it one is forced to use 4-vectors since we require objects that transform as vectors under application of Lorentz transformations and 3-vectors do not (technically they do under rotations, but not under boosts). Equivalenty, if one starts off with Minkowski spacetime from the outset, then 4-vectors are the natural quantities to consider since we are already dealing with a 4-dimensional space.

    Given this, one then introduces the notion of 4-momentum, defining it as $$p^{\mu}=m\frac{dx^{\mu}}{d\tau}=\gamma m\frac{dx^{\mu}}{dt}$$ where ##x^{\mu}## is the position 4-vector, ##\tau## is the proper time, ##t## is the coordinate time and ##\gamma =\frac{1}{\sqrt{1-v^{2}/c^{2}}}## is the Lorentz factor (with ##\mathbf{v}## the relative velocity between two inertial frames).

    I'm trying to reason why the zeroth component of 4-momentum of a particle, i.e. ##p^{0}## is proportional to its energy ##E##, $$p^{0}=\frac{E}{c}$$ where ##c## is the speed of light.

    Can one do this as follows.
    We first note that, by definition, ##p^{0}=\gamma mc=\frac{mc}{\sqrt{1-v^{2}/c^{2}}}##, and so for ##v<<c## we have that $$ p^{0}=\gamma mc=\frac{mc}{\sqrt{1-v^{2}/c^{2}}}\approx mc(1+\frac{v^{2}}{2c^{2}})=\frac{1}{c}(mc^{2}+\frac{mv^{2}}{2})$$ We recognise ##\frac{mv^{2}}{2}## as the kinetic energy of a non-relativistic particle and hence interpret ##mc^{2}+\frac{mv^{2}}{2}## as the total energy of the particle in the non-relativistic limit. Thus we are motivated to interpret ##p^{0}## as being proportional to the energy of the particle, i.e. ##p^{0}=\frac{E}{c}##.

    Given this we can then proceed to derive the relativistic energy-momentum relation: $$ (p^{0})^{2}=\frac{E^{2}}{c^{2}}=\gamma^{2}m^{2}c^{2}= \frac{m^{2}c^{2}}{1-v^{2}/c^{2}}= \frac{m^{2}c^{2}+m^{2}v^{2}-m^{2}v^{2}}{1-v^{2}/c^{2}}\\ =\frac{m^{2}v^{2}}{1-v^{2}/c^{2}}+m^{2}\frac{c^{2}-v^{2}}{1-v^{2}/c^{2}}=p^{2}+m^{2}c^{2}$$ and hence $$E^{2}=m^{2}c^{4}+p^{2}c^{2}.$$

    Have I understood this concept correctly? Is the derivation correct? I'd really appreciate any feedback.
     
  2. jcsd
  3. May 2, 2016 #2

    PeroK

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    Looks good to me.
     
  4. May 2, 2016 #3
    OK great, thanks for taking a look.

    One other thing that I'm slightly unsure about. Is it possible to motivate the usage of four-vectors (as opposed to three-vectors) in special relativity without first introducing Minkowski spacetime? I mean, if one introduces the notion of spacetime first, then it is natural to consider four-vectors as opposed to three-vectors since the spacetime is four-dimensional from the outset (time being considered as a fourth dimension as it is no longer absolute and as such frame-dependent an hence each coordinate frame has its own temporal coordinate, to distinguish the time in that frame. Thus each frame is labeledby a set of four coordinates, would this be correct?), but are there any other intuitive heuristic arguments as to why four-vectors are the correct objects to consider? (I've heard that it's partly because three-vectors do not transform as vectors under Lorentz transformations, but I'm not sure if this is correct or not?)
     
    Last edited: May 2, 2016
  5. May 2, 2016 #4

    PeroK

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    I would say that four-vectors arise naturally both geometrically and algebraically. Once you have the Lorentz Transformation involving four variables, it's natural to consider four-vectors, purely in terms of 4-tuples: ##(t, x, y, z)##. The algebra and the geometry go hand-in-hand, each being equivalent to the other.

    If you wanted, you could do SR algebraically without considering the geometric spacetime view. It would be a bit like doing complex numbers purely algebraically, and never considering the geometry of the complex plane. But, of course, it's much more powerful to have the algebra and geometry at your disposal.
     
  6. May 2, 2016 #5
    So would it be correct to say that it is natural to consider 4-vectors since Lorentz transformations transform both temporal and spatial coordinates, so in order to construct Lorentz covariant and invariant objects we must consider 4-tuples, i.e. four component vectors?!

    Also, would what I said about why time is considered a fourth dimension in special relativity correct? i.e. "time being considered as a fourth dimension as it is no longer absolute and as such frame-dependent an hence each coordinate frame has its own temporal coordinate, to distinguish the time in that frame. Thus each frame is labeledby a set of four coordinates, would this be correct?"

    What about the statement that 3-vectors don't transform correctly, i.e. as vectors, under Lorentz transformations? I think I read this somewhere but can't for the life of me remember where.
     
  7. May 2, 2016 #6

    PeroK

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    I'm not sure what you are worried about. I think you know all this stuff. Time to move on!
     
  8. May 2, 2016 #7
    You're right, I should stop dwelling on things and trying to "trip" myself up.
     
  9. May 2, 2016 #8
    Yes, you can derive the invariant interval equation, the energy-momentum relation, etc., without recourse to four-vectors. They're just "magnitudes" without direction at that point, each in the form of ##(ds \text{ something})^2 = (c \, dt \text{ something})^2 - (d \vec r \text{ something})^2##. The left side is always Lorentz-invariant (because ##ds## is), and the four terms on the right side always obey the Lorentz transformation (because ##c \, dt##, ##dx##, ##dy##, and ##dz## do). Add spacetime direction to the mix, and you've got the four-vector formalism.
     
  10. May 2, 2016 #9
    I also think that, geometry aside, four-vectors are convenient conceptually and algebraically. It's nice to have all that information wrapped up in a single symbol.
     
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