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Lorentz invariant integral measure

  1. Sep 23, 2015 #1
    Hi

    I'm studying electron-muon scattering

    and now considering the Lorentz invariant integration measure.

    The textbook introduced it, which use dirac delta function to show that d3p/E is a Lorentz scalar.

    I understood it but I wanted to find other way and tried like this: 20150923_154737-1.jpg

    I need a hint on the underlined equation. The primed one is lorentz transformed one.

    How can I show that d(p^2) is Lorentz invariant?

    Just calculate it hard ?
     
    Last edited: Sep 23, 2015
  2. jcsd
  3. Sep 23, 2015 #2

    Demystifier

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    Your attempt has a mistake because ##E## is not the same as ##p^0##. The four variables ##p^0,p^1,p^2,p^3\equiv (p^0,{\bf p})## are 4 independent variables, but ##E=\sqrt{{\bf p}^2+m^2}## is not independent of ##{\bf p}##. Hence, your conclusion that ##d{\bf p}^2## should be Lorentz invariant is wrong.

    The measures ##d^4p=dp^0 d^3p## and ##\frac{d^3p}{E}## are Lorentz invariant, but ##dE d^3p## and ##\frac{d^3p}{p^0}## are not.

    If you are still confused, try first to solve the following problem. Consider 3-dimensional Euclidean space with Cartesian coordinates ##x,y,z##. In this space, consider the sphere defined by ##x^2+y^2+z^2=R^2##, which is a 2-dimensional object invariant under 3-dimensional rotations. Find the measure of area ##dA## on the upper half of the sphere, in the form ##dA=f(x,y)dxdy##. Argue that this measure is rotation-invariant simply because the sphere is rotation-invariant. Find ##f(x,y)## explicitly in two ways: (i) geometrically and (ii) by starting from volume element ##dV=dxdydz## and using a ##\delta##-function trick. Can you see the relation with the problem of Lorentz-invariant measure above? In particular, can you see the analogy between ##R## and ##m##?

    Partial solution: You should get $$f(x,y)=\frac{R}{\sqrt{R^2-x^2-y^2}}.$$Is it correct to write the right-hand side as ##R/z##?
     
    Last edited: Sep 23, 2015
  4. Sep 23, 2015 #3

    Demystifier

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    Since I have nothing more important to do at the moment, let me do this explicitly. In spherical coordinates we have
    $$dV=dA dr$$
    where ##dA## is the area element of the sphere. Since ##dr\delta(r-R)=1##, we can use this to write
    $$dA=dV\, \delta(r-R). \;\;\; (1)$$
    We also have
    $$\delta(R^2-r^2)=\delta((R-r)(R+r))=\frac{\delta(R-r)}{2R}$$
    so (1) can be written as
    $$dA=dV \, 2R\delta(R^2-r^2) \;\;\; (2)$$
    Clearly, (2) is rotation invariant. In Cartesian coordinates this can be written as
    $$dA=dxdydz \, 2R\delta(R^2-x^2-y^2-z^2) \;\;\; (3)$$
    Introducing the function
    $$w(x,y)=\sqrt{R^2-x^2-y^2}$$
    which is non-negative on the sphere of radius ##R##, (3) can be written as
    $$dA=dxdydz \, 2R\delta(w^2-z^2)=dxdydz \, 2R\delta((w-z)(w+z)) \;\;\; (4)$$
    For the upper half of the sphere we have ##z>0##, so for the upper half (4) can be written as
    $$dA=dxdydz \, 2R \frac{\delta(w-z)}{2w}=dxdy \frac{R}{w}. \;\;\; (5)$$

    Since this is rotation invariant and since ##R## is a constant, it follows that the measure
    $$\frac{dxdy}{w}$$
    is rotation invariant. This is analogous to the fact that the measure
    $$\frac{d^3p}{\omega}$$
    with ##\omega=\sqrt{m^2+{\bf p}^2}## is Lorentz invariant. The differences are that (i) for Lorentz invariance we have one dimension more, and (ii) some signs are different because Minkowski metric is different from the Euclid metric.
     
    Last edited: Sep 24, 2015
  5. Sep 23, 2015 #4

    vanhees71

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    Isn't this a bit too complicated? Usually the derivation starts from the fact that ##\mathrm{d}^4 p## is invariant under proper Lorentz transformations. Then the on-shell ##\delta##-distribution, ##\delta(p^2-m^2)=\delta[(p^0)^2-E^2]##, where ##E^2=\vec{p}^2+m^2##. Finally ##\Theta(p^0)## is a Lorentz invariant under proper orthochronous Lorentz transformations and thus finally
    $$\mathrm{d}^4 p \Theta(p^0) \delta(p^2-m^2)=\mathrm{d} p^0 \mathrm{d}^3 \vec{p} \frac{1}{2E} \delta(p^0-E)$$
    is Lorentz invariant. Thus if you have a function ##f(p)## with ##p^0=E## you can write
    $$\int \mathrm{d}^4 p \Theta(p^0) \delta(p^2-m^2)=\int \mathrm{d}^3 \vec{p} \frac{1}{2E} f(p^0=E,\vec{p}).$$
    If ##f(p)=f(p^0,\vec{p})## is a scalar field, then for the on-shell function ##f(p^0=E,\vec{p})## the 3D integral with the measure ##\mathrm{d}^3 \vec{p}/E## is a scalar.
     
  6. Sep 23, 2015 #5

    Demystifier

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    Yes, but he already indicated that he understands the standard proof of Lorentz invariance with the use of a delta function. He wanted a more direct proof without a delta function, and in his attempt he failed. I wanted to explain him what exactly was wrong with his attempt. In addition, I also thought that it would be easier to undertstand this in a case which can be more easily visualized, i.e. where 4-dimensional Lorentz invariance is replaced with 3-dimensional rotational invariance. While the calculation turned out to be longer, I thought that the resulting geometric idea would be easier to understand intuitively.
     
  7. Sep 25, 2015 #6

    ShayanJ

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    I don't understand this part.
    1) The 4-momentum is always defined to be ## p^\mu=(E,\bf p) ## in a particular frame. Now you say ## p^0 ## is not the same as E. So what is it?
    2) You say that different components of the 4-momentum are independent but what about the restriction ## p^2=-m^2 ## on any 4-momentum?
     
  8. Sep 25, 2015 #7

    Demystifier

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    Since ##E## is defined as ##\sqrt{{\bf p}^2+m^2}##, these two questions are really the same. To answer them, it should be enough to give one example where ##p^{\mu}## is not equal to ##(E,\bf p)##. So here is one example: the 4-momentum integration in a Feynman diagram with a loop.

    If you are still confused, think about doing the exercise proposed in #2.
     
  9. Sep 25, 2015 #8

    ShayanJ

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    I think I understand it now. It seems that the right way of thinking about 4-momentums is that there is a 4-dimensional momentum space and particles move on ## p^2=-m^2 ## hypersurfaces in this space. I was missing this viewpoint. thanks.
     
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