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4-force, 4-momentum, energy and mass relations.

  1. May 22, 2014 #1
    This is an exercise of Special Relativity the professor asked last week.
    Sorry for the long post, I hope you dont get bored reading it, also, this is my first post here :shy:

    1. The problem statement, all variables and given/known data
    Defining the 4-force that acts on a particle as the proper-time variation of the 4-momentum [itex]F^\mu := \frac{d P^\mu}{d \tau}[/itex].
    1. Justify Einsteins relation between mass and energy: [itex] E = mc^2 [/itex]
    2. Show, using the [itex]\eta _{\mu\nu}P^\mu P^\nu [/itex], that [itex] E^2 = (mc^2)^2 + (\vec{p}^2c^2) [/itex], where [itex] \vec{p} [/itex] is the 3-momentum.
    3. Show that in SR the 3-force and 3-acceleration is not always parallel to each other.

    2. Relevant equations
    I'm using the convention: [itex] x^0 = ct [/itex] and [itex] (- + + +) [/itex] for the metric tensor.

    4-velocity: [itex] u^\mu = \gamma_{(v)} (c, \vec{v}) [/itex], with [itex] \vec{v} [/itex] the 3-velocity in lab frame.

    And some results I got, they are somewhere in Wikipedia and some books also.

    gamma: [tex] \frac{d}{d \tau} = \frac{d t}{d \tau} \frac{d}{d t} = \gamma_{(v)} \frac{d}{d t} [/tex]

    I'll omit [itex](v)[/itex] in gamma for brevity.

    derivative of gamma: [tex] \dot{\gamma} = \frac{d \gamma}{dt} = \gamma ^3 \frac{\vec{v} \cdot \vec{a}}{c^2} [/tex]

    acceleration: [tex] a^\mu = \left ( \gamma ^4 \frac{\vec{v} \cdot \vec{a}}{c}, \gamma^4 \vec{a}_\parallel + \gamma ^2 \vec{a}_\perp \right ) [/tex]
    where [itex]\vec{a}_\parallel = (\vec{a} \cdot \vec{v}) \vec{v}/v^2[/itex] is the parallel component of the 3-acceleration to the 3-velocity and [itex] \vec{a}_\perp = \vec{a} - \vec{a}_\parallel[/itex] is the perpendicular one.​


    3. The attempt at a solution for question 1 and 2

    I think I got the results, but i have some doubts.

    [tex]F^\mu = \frac{d P^\mu}{d \tau} = \left(\frac{d}{d \tau}(m \gamma c), \frac{d}{d \tau}(m \gamma \vec{v}) \right )[/tex]
    I'll call this 'result' as (EQ1).

    Using the gamma relation in the relevant equations:
    [tex]F^\mu = \gamma \left ( m c \dot{\gamma}, \frac{d\vec{p}}{dt} \right ) = \left ( m c \gamma ^4 \frac{\vec{v} \cdot \vec{a}}{c^2}, \gamma \vec{f}_R \right ) [/tex]
    I'll call this 'result' as (EQ2). Where [itex] \vec{f}_R [/itex] is the 3-force. We know that [itex] F^\mu [/itex] is mass times acceleration:

    [tex]F^\mu = m a^\mu = m \left ( \gamma ^4 \frac{\vec{v} \cdot \vec{a}}{c}, \gamma^4 \vec{a}_\parallel + \gamma ^2 \vec{a}_\perp \right ) [/tex]
    This is (EQ3).

    Comparing (EQ2) and (EQ3) we get:
    [tex] \vec{f}_R = m \left ( \gamma^3 \vec{a}_\parallel + \gamma \vec{a}_\perp \right ) [/tex]

    scalar product with \vec{v} gives a variation in energy:
    [tex] \vec{f}_R \cdot \vec{v} = m \gamma^3 \vec{a} \cdot \vec{v} = \frac{dE}{d \tau} [/tex]
    This is (EQ4).
    Here is the derivative with respect to the proper-time, right? I'm not sure about this...​

    Putting (EQ4) inside (EQ3) in the first component and re-using the second component of (EQ2):
    [tex] F^\mu = \gamma \left (\frac{\vec{f}_R \cdot \vec{v}}{c}, \vec{f}_R \right ) = \left ( \frac{1}{c} \frac{d E}{d\tau}, \frac{d\vec{p}}{d\tau} \right )[/tex]
    This is (EQ5).

    Comparing (EQ5) and (EQ1) yelds:
    [tex]\frac{dE}{d\tau} = \frac{d}{d\tau} (m \gamma c^2)[/tex]

    So:

    [tex]E = m \gamma c^2 +\ constant[/tex]
    This is (EQ6).

    This is almost what the question 1 asks, but what is this constant? What is the meaning of [itex]E[/itex]? Is it kinect + 'rest energy'?​

    Question 2 I got:

    [tex]\eta _{\mu\nu}P^\mu P^\nu = m^2 \gamma ^2 (-c^2 + \vec{v}^2) = -m^2 c^2 [/tex]
    (EQ7)

    From (EQ6) we can write the 4-momentum as:
    [tex] P^\mu = \left ( \frac{E}{c}, \vec{p} \right ) [/tex]
    (EQ8)

    so:
    [tex]\eta _{\mu\nu}P^\mu P^\nu = - \frac{E^2}{c^2} + \vec{p}^2 = -m^2 c^2[/tex]

    Wich is the answer to question 2 if the constant of (EQ6) is zero...



    4. The attempt at a solution for question 3

    If we do the same trick used to split the acceleration in parallel and perpendicular parts:

    [tex] \vec{f}_R := \vec{f}_{R, \parallel} + \vec{f}_{R, \perp} [/tex]
    We build the parallel one:

    [tex] \vec{f}_{R, \parallel} = \frac{(\vec{f}_R \cdot \vec{a}) \vec{a}}{\vec{a}^2} [/tex]
    and the perpendicular one:

    [tex] \vec{f}_{R, \perp} = \vec{f}_R - \vec{f}_{R, \parallel} = \vec{f}_R - \frac{(\vec{f}_R \cdot \vec{a}) \vec{a}}{\vec{a}^2} [/tex]

    But I was unable to prove [itex] \vec{f}_{R, \perp} \neq \vec{0} [/itex].

    Do you have any hint? All that I got was some previous equations.


    If I was not clear in some statement, please tell me.

    (is there a way to put a 'name' in some equations to be displayed on the right side of it? Like latex documents?)
    (Sorry for my bad English...)

    Thanks in advance,
    Heitor.
     
  2. jcsd
  3. May 23, 2014 #2
    From the expression [tex] \vec{f}_R = m \left ( \gamma^3 \vec{a}_\parallel + \gamma \vec{a}_\perp \right ) [/tex] we see that the force is not parallel to the acceleration because [itex] \gamma^3 [/itex] is different from [itex] \gamma [/itex] so the parallel and perpendicular components scale differently.
     
  4. May 23, 2014 #3
    By the way, I think that instead of taking equations from books and wikipedia, you ought to prove them.
     
  5. May 23, 2014 #4
    Is there any interpretation for this? A 3-force in Special Relativity does not have the same meaning as in Newtonian Mechanics?


    I did proved them, it was the previous question in the list.
     
  6. May 23, 2014 #5
    The 3-force is derived from force laws just like in Newtonian mechanics but f=ma doesn't apply. instead we have f = m (γ3aparallel + γaperpendicular)
     
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