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4-Dimensions Problem

  1. May 3, 2006 #1
    How do we derive the lagrangnian of a charged particle in 3 dimentions?

    Actually this is not the real question but it's a step to it, In three dimentions we have:

    [tex]{\cal L} = - m_ \circ c^2 \sqrt {1 - \frac{{v^2 }}{{c^2 }}} + q\overrightarrow v \overrightarrow A - q\phi [/tex]
    And As we Know in four dimentional system:
    We Have:

    Speed:U_\mu = \gamma \left( {\overrightarrow v ,ic} \right) \Rightarrow U_\mu ^2 = \gamma ^2 \left( {v^2 - c^2 } \right) = - c^2 \cr
    ,,,{\rm{Vector Potential: }}A_\mu = \left( {\overrightarrow A ,i{\phi \over c}}\right)\Rightarrow \cr { \rm{A}}_\mu U_\mu = \overrightarrow A \cdot \overrightarrow v - \phi \cr}
    Substituting This to the Lagrangian we get, and note that gamma is taken in hamiltons principle ds=L dt in 4-d becomes ds= L dt/Gamma:
    {\cal L} = m_ \circ U_\mu ^2 + qA_\mu U_\mu
    My Professor says that we should write it like this with out doing anything else:
    {\cal L} = {1 \over 2}m_ \circ U_\mu ^2 + qA_\mu U_\mu
    And that is to make it like the 3d system, 1/2 mv^2, IS THIS CORRECT?????? THIS IS MY FIRST QUESTION

    And my second question, If we used the 4d motion equation on this lagrangian:
    {d \over {dt}}{{\partial {\cal L}} \over {\partial U_\mu }} - {{\partial {\cal L}} \over {\partial x_\mu }} = 0
    do we get the lorentz force like the 3d system? i didn't get it when i tried, Any one can help?

    Can you explain the solution of this motion equation? my professor didn't know it even !!

    Thanks, Please hurry with the answer
    Last edited: May 3, 2006
  2. jcsd
  3. May 4, 2006 #2

    Physics Monkey

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    The Lagrangian [tex] L = \frac{1}{2} m \mathbf{v}^2 + q \mathbf{v}\cdot\mathbf{A} - q \phi [/tex] does give the correct physics for low speeds. The more general Lagrangian [tex] L = - m \sqrt{1-\mathbf{v}^2} + q \mathbf{v}\cdot \mathbf{A} - q \phi [/tex] is applicable even to relativistic particles. You can easily check that the more general Lagrangian reduces to the other when v is small (up to an additive constant).

    It seems to me that you may be confusing three and four dimensional notation, but I'm not really sure. Let me know if this doesn't help, and I can try again to address your question.
  4. May 4, 2006 #3
    No this doesn't help, because i want the lagrangian in four dimentions, i gave the notations for speed and vector potentional above, and mu under them can take values 1,2,3,4, and the brackets after them gives in the first parameter the three position dimention or in another word when mu take the values 1,2,3, and the second one gives the value 4 for mu, and in another word the time component for the vector,

    I've written the lagrangian in the four dimentional system, and my question is what my professor insist to put 1/2 beside m while we can't get it mathematically?

    and my second question how do we apply the four-dimentional lagrangian on the motion equation specified in the last equation

    and note somthing, the result we must get is ofcourse the lorentz force this, I'll write it, and what inside brackets are indexes for tensors or vectors:

    m B(mu) = q F(mu,nu) U(nu)

    Where B(mu) is the vector of four dimentional acceleration, F(mu,nu) us and U(nu) is the four dimentional speed vector i mentioned above, the tensor of magnetical force in four dimentional system,

    Professor says this must be the result, but how, actually i'm getting difficulties using matrices and tensors (he also not only me), plus i asked him if i can use dirac symbols and he refused, i don't know why !!

    Anyone can show me the way to get the motion equation?

    Note: the relativistic lagrangian is not correct, you've forgot c^2,

    And thanks for trying to help
  5. May 5, 2006 #4


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    a first comment: it is a bit confusing because you are using "3-dimensional" and "four-dimensional" in a different way that what is standard! The standard way of saying what you want is that you want a *covariant expression* (that is, in terms of four-vectors). You should not say that you want a "four-dimensional expression", that will confuse everybody!
    Neither is correct. The first term is actually [itex] - mc {\sqrt {U_\mu U^\mu}} [/itex].

  6. May 6, 2006 #5
    Actually Our professor is trying to simplify everything to avoid using tensors, that's why, he've written it that simple, and i've written the way to get as the professor told us, he's trying to compare everything with 3 dimentional system the way i expressed,

    I'm using 4-d vector, for example A is the vector potential has 3 position dimentions and 1 time dimention i*phi/c, and i used the 4-dim speed multiplication like when we multiply any 2 vectors, just multiply x component with the other x component, y with other y, z with other z, and the same about the time dim, time with time to get a scalar !!!! what can i explain more? if i'm wrong or the simplicity of the professor is for childs just tell me everything you can, and i would be very grateful, thanks

    If you believe it i'm now in the third year in physics college and i'm gonna finish it and till now i don't know much about tensors !! so please explain everything you can,

    Anyone can tell me how to continue with my details? or just tell me how to continue IN ANY OTHER WAY? but detailed way not in words, math + physical expression
  7. May 7, 2006 #6
    Lol, No Body Has An Answer?????????????
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