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Special Relativity, four-momenta questions.

  1. May 18, 2013 #1
    Hi,
    I have just recently begun delving into SR, and there are a few questions I would like clarified if possible, with your assistance. NB These questions were discussed during some of the lectures and are NOT HW assignments.

    1) If a particle of rest mass m breaks down, whilst at rest, into a particle of mass m' and a photon, what would be the energies of the particle with mass m' and the photon created in the process?:
    I understand that the four-momentum of the original particle of mass m is (m,0) and those of m' and the photon are (E,p) and (ε,εn^). I also understand that since the center of mass reference frame of the original particle is also that of the products (m' and the photon), the overall four-momentum must be conserved and the following equality must hold:
    Ptot = (m,0)=(E+ε,p+n^).
    What is not clear to me is why would that entail E + p = m? I don't quite understand this equality. What does it designate? What is its meaning?

    2) In reference frame O two photons with frequencies [itex]\nu[/itex]1 and[itex]\nu[/itex]2 are moving in the positive and negative x^ directions, respectively. What would be the CM of the two photon's velocity wrt O?
    The four-momentum of the first photon may be written thus: h[itex]\nu[/itex]1(1,n^)
    Similarly for the second: h[itex]\nu[/itex]2(1,-n^).
    Now, the total four-momentum must be conserved: P = h([itex]\nu[/itex]1 + [itex]\nu[/itex]2, ([itex]\nu[/itex]1 - [itex]\nu[/itex]2)n^).
    What I don't quite understand is how applying Boost in the n^ direction at a velocity u would yield:
    h[itex]\gamma[/itex]([itex]\nu[/itex]1 - [itex]\nu[/itex]2-u([itex]\nu[/itex]1 + [itex]\nu[/itex]2)).
    Could someone please explain this delicate point to me step-by-step?

    3) Suppose Pμ = (M + h[itex]\nu[/itex],h[itex]\nu[/itex]n^). Why does that squared yield: M2 + 0 + 2h[itex]\nu[/itex]M? What does the zero denote?
    Also, what is the difference between the following notations: Pμ and Pμ?

    I'd truly appreciate any insightful remarks.
     
  2. jcsd
  3. May 18, 2013 #2
    Hi

    1) By emitting a photon, the particle loses its mass and get minus momentum.
    4-momtum of the system remains (m,0) even after the decay.
     
    Last edited: May 18, 2013
  4. May 18, 2013 #3
    Hi.

    2) In CM frame, two photons of same frequency of ν running to opposite directions are observed. You can easily find the relative velocity and value of ν so that v becomes v1 and v2 for the photons going in opposite direction.
     
  5. May 18, 2013 #4
    Hi,

    3) M2 + 2h[itex]\nu[/itex]M is the value that in any frame of reference E^2 - p^2 is.

    Pμ = -Pμ for μ=1,2,3
    Pμ = Pμ for μ=0 in usual notation for me.
     
    Last edited: May 18, 2013
  6. May 18, 2013 #5

    Mentz114

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    Your notation is not clear to me so I can't answer any but this question.

    A vector written with the index raised, ##v^\mu## is in the tangent vector space of a worldline and can be written in terms of the basis of this space ##v^\mu=a_1\partial_t + a_2\partial_x + a_3\partial_y + a_4\partial_z ## where the ##a_i## are coordinates. With the index down, ##v_\mu## this is the covector corresponding to ##v^\mu## and can be written in terms of the basis of the covector space ##v_\mu=b_1dt + b_2dx + b_3dy + b_4dz ##, where the ##dx^i## are the same differentials found in the metric, which connects the two spaces thus ##v^\mu= g^{\mu\nu}v_\nu##.
     
  7. May 18, 2013 #6

    WannabeNewton

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    Strictly speaking, just to keep the notation tractable and to drive home the important point that the vector and covector are geometric objects that exist independently of the coordinates, we write the 4-vector as ##\vec{p} = p^{\mu}\partial_{\mu}## and the 4-covector as ##\tilde{p} = p_{\mu}dx^{\mu}## point being that ##p^{\mu}## and ##p_{\mu}## aren't the vector and covector, respectively, but rather their components in the coordinate representation.
     
  8. May 18, 2013 #7

    Fredrik

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    Recall that instead of the usual dot product, we're using another function called the metric tensor, which can be defined like this:
    $$g(x,y)=x^0 y^0-x^1 y^1-x^2 y^2-x^3 y^3.$$ Note that the superscripts are indices, not exponents. The right-hand side is often written as ##x^0y^0-\mathbf x\cdot\mathbf y##.
    $$P^2=g(P,P)=(P^0)^2-\mathbf P^2=(M+h\nu)^2-(h\nu)^2\hat{\mathbf n}^2=M^2+2h\nu M +(h\nu)^2-(h\nu)^2.$$ The 0 in your result is the sum of the last two terms.

    WannabeNewton's answer is good, but the question can also be answered without differential geometry:

    ##P^\mu## is the ##\mu##th component of ##P##

    ##P_\mu## is the ##\mu##th component of ##\eta P##, where
    $$\eta=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\end{pmatrix}.$$ So
    $$\begin{pmatrix}P_0\\ P_1\\ P_2\\ P_3\end{pmatrix}=\eta P=\begin{pmatrix}P^0\\ -P^1\\ -P^2\\ -P^3\end{pmatrix}.$$ Note by the way that for all ##x,y\in\mathbb R^4##, we have
    $$g(x,y)=x^T\eta y.$$
    If we choose the coordinate system such that ##\mathbf n## is a unit vector in the 1 direction, we have
    $$P=\begin{pmatrix}h(\nu_1+\nu_2)\\ h(\nu_1-\nu_2)\\ 0\\ 0\end{pmatrix}.$$
    $$\Lambda(u\mathbf n)P=\gamma\begin{pmatrix}1 & -u & 0 & 0\\ -u & 1 & 0 & 0\\ 0 & 0 & 1/\gamma & 0\\ 0 & 0 & 0 & 1/\gamma\end{pmatrix} \begin{pmatrix}h(\nu_1+\nu_2)\\ h(\nu_1-\nu_2)\\ 0\\ 0\end{pmatrix} =h\gamma\begin{pmatrix}(\nu_1+\nu_2)-u(\nu_1-\nu_2)\\ -u(\nu_1+\nu_2)+(\nu_1-\nu_2)\\ 0\\ 0\end{pmatrix}.$$
     
    Last edited: May 18, 2013
  9. May 18, 2013 #8
    A basic question: how did you know where and how to pluginthe values of u and 1/gamma? Why are those entries 1/gamma?
     
  10. May 18, 2013 #9

    jtbell

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    Have you studied how the Lorentz transformation is derived?
     
  11. May 18, 2013 #10

    Fredrik

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    If a Lorentz boost in 1+1 dimensions is
    $$\begin{pmatrix}a & b\\ c & d\end{pmatrix},$$ then a Lorentz boost in the 1 direction in 3+1 dimensions is
    $$\begin{pmatrix}a & b & 0 & 0\\ c & d & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}.$$ So we only have to find out what a Lorentz boost in 1+1 dimensions looks like. There are many ways to approach this problem. First of all, we can take
    $$\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}$$ as the definition of a Lorentz boost in 1+1 dimensions, and then nothing else needs to be said.

    Most books like to present some sort of argument for why a Lorentz transformation should look like this, based on Einstein's postulates. My version of that is here. Start reading at "copied from".

    Two more options (very similar to each other) are to define a Lorentz transformation as a linear transformation ##\Lambda## such that ##g(\Lambda x,\Lambda x)=g(x,x)## for all x, or as a surjective map ##\Lambda## such that ##g(\Lambda(x),\Lambda(y))=g(x,y)## for all x and y. Each of these these definitions imply that the set of all Lorentz transformations is the set of all 2×2 matrices ##\Lambda## such that ##\Lambda^T\eta\Lambda=\eta##. And then we can use this to show that an arbitrary proper and orthochronous Lorentz transformation can be written in the form above. See posts #8 and #14 in this thread.

    Note that I'm using units such that c=1, and that I define ##\eta## differently (opposite sign) in those threads.
     
    Last edited: May 18, 2013
  12. May 18, 2013 #11
    You used the term “rest” twice here. Please help me understand why. Are you assuming that the particle emitting the photon remained at rest after the photon was emitted? If so then you made an incorrect assumption.

    The mass of the particle and the energy of the photon are not uniquely determined in this problem. I.e. it can happen in a variety of ways so long as both energy and momentum are conserved. To know the final energies of both the photon and particle after the photon is emitted you need more information than that already given. The larger the energy of the emitted photon, the smaller the final energy of the particle and vice versa.

    When using non-standard notation like this it’s a good idea to define your terms. For example; what does the following symbol mean - εn^
    What exactly is this symbo? – n^

    I’ve never seen that notation used before

    From the context in which you used it appears that n^ is the momentum of a photon. Where in the world did you get that notation from?

    You made an error here somewhere. It might have been caused by not writing all of this out cleanly and defining your terms thus expecting others to understand what you meant by non-standard notation. There is nothing in your post where you should have gotten E + p = m. Perhaps you thought that was taken from (m,0)=(E+ε,p+n^)? Consider what this says

    m = E + e,
    0 = p+n^

    So somewhere you made an error in copying the notation along the way.

    I'm going to stop here until I get feed back from you.
     
  13. May 18, 2013 #12

    Fredrik

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    I assumed that n^ was the OP's way of writing ##\hat{n}## or ##\mathbf{\hat n}##, and that this was a unit vector in ##\mathbb R^3## in the direction of the momentum.
     
  14. May 19, 2013 #13
    Indeed it was, Fredrik
     
  15. May 19, 2013 #14
    Ah! Okay. Thanks. When brace notation is used (ε,εn^) the term in the second place can be either the components o of the spatial vector or the spatial vectors itself. I forgot the later. Thanks!
     
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