1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Meaning of trajectory in four-electricity in Landau's book

  1. Dec 19, 2016 #1
    1. The problem statement, all variables and given/known data
    I think the in equation ##(28.2)##,##x^i## in ##\frac{dx^i}{dt}## and the ##x^i## decides ##\rho## is not the same,if they are equaivalent,##\rho## can not vary with position changing and time fixed, because ##\frac{dx^i}{dt}## indicate the ##x^i(t)## which means if position changed the time would also be changed.

    ##\frac{dx^i}{dt}## could define the velocity through a points' trajectory,##\rho## is not point charge,what's the meaning of the "velocity"?

    2. Relevant equations
    The four electricity vector is defined as ##j^i=\rho\frac{dx^i}{dt}##,##\rho## depend on ##(t,x^1,x^2,x^3)##

    3. The attempt at a solution
    I want to find some clues about the "velocity" in contents after the definition and only find LLs(Landau and Lifshitz) regarding the ##x^i## in ##\frac{dx^i}{dt}## as the same as the ##x^i## which decides ##\rho##-------In equation ##(28.5)## ##dS^i## was gotten through dividing ##dV## with ##dx^i##.
     
  2. jcsd
  3. Dec 27, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    When we define ##j^k(x^i) =\rho(x^i) \frac{ dx^k}{dt}##, I don't see any need to worry about whether the ##x^k## in ##\frac{ dx^k}{dt}## is the same as the ##x^i## in ##\rho(x^i)##.

    ##\rho(t, x^1, x^2, x^3)## is the charge density at some arbitrary point ##(x^1, x^2, x^3)## at time ##t##. At that point and time, the charge density has a velocity ##\vec{v} = \left( \frac{dx^1}{dt},\frac{dx^2}{dt},\frac{dx^3}{dt} \right)## which is well-defined. Thus, we can define ##\frac{dx^i}{dt} =\left( c, \vec{v} \right)= \left( c, \frac{dx^1}{dt},\frac{dx^2}{dt},\frac{dx^3}{dt} \right)##.

    Then ##j^k = \rho(x^i) \frac{dx^k}{dt}## is well defined. This could also be written as ##j^k(x^i) = \rho(x^i) \frac{dx^k}{dt}(x^i)## in order to show the explicit functional dependence of the factors on the choice of the spacetime point ##x^i = (x^0, x^1, x^2, x^3)##.

    In the expression ##j^k(x^i) = \rho(x^i) \frac{dx^k}{dt}(x^i)## , we can imagine varying the spatial point ##(x^1, x^2, x^3)## while keeping ##t## fixed.
     
  4. Dec 28, 2016 #3
    Thank you very much!You help me understand more.
    I am still confused with the velocity definition ##v^k=\frac{d{x'}^k}{dt}(t,x^1,x^2,x^3)\ \ \ (1)##
    {I use ##x'## replace ##x## in this equation,since ##\frac{dx^k}{dt}(t,x^1,x^2,x^3)## may not a single valued function suitable for defining the velocity.Because for the ##dt## combined with different ##dx^i## to form ##(dt,dx^i(1)),(dt,dx^i(2)),.....## may sucessfully make the charge density ##\rho(t+dt,x^i+dx^i)## exist.}
    Inserting equation (1) into the equation (28.5),we have ##\int\rho{dV}=\int\rho\frac{d{x'}^i}{dt}\frac{dt}{d{x'}^i}dV=\frac{1}{c}\int{j^i}dS_i## which indicates ##\frac{dt}{d{x'}^i}dV=\frac{1}{c}dS_i##,this means ##d{x'}^i## was treated as equivalent as ##dx^i##.But ##\frac{dx^i}{dt}(t,x^i)## is a multivalued function and not suitable for the velocity definition.
     
  5. Dec 28, 2016 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I'm having a hard time understanding what you are saying here.

    Note that, for example, the ##x^1## component of velocity of the charge density, ##\frac{dx'^1}{dt}##, could equal 0. Then, there is a problem writing its inverse ##\frac{dt}{dx'^1}##.

    Also, when you write ##\int\rho\frac{d{x'}^i}{dt}\frac{dt}{d{x'}^i}dV=\frac{1}{c}\int{j^i}dS_i##, the index ##i## is summed in the integral on the right. So, you must be summing over ##i## also in the integral on the left. But, ##\frac{d{x'}^i}{dt}\frac{dt}{d{x'}^i} \neq 1## if you are summing over ##i##. So, it would not be correct to write ##\int\rho{dV}=\int\rho\frac{d{x'}^i}{dt}\frac{dt}{d{x'}^i}dV##.
     
  6. Dec 29, 2016 #5
    Thanks.
    What I mean is that:
    for a trajectory of a point charge the velocity could be defined as ##\frac{dx^i}{dt}##,because the trajectory ##x^i## could be written as ##x^i(t)##.

    however, for the movement of density of charge----##\rho(t,x^i)##,I think the velocity ##v(t,x^i)## can not be expressed as ##\frac{dx^i}{dt}##,because at the time ##t##,there will be more than one ##x^i## to make ##\rho(t,x^i)## sense(##x^i## can vary with a fixed time ##t##).
    As a result,##x^i## can not be written as ##x^i(t)##.
    So I use ##\frac{d{x'}^i}{dt}## to express ##v(t,x^i)##,I can not clarify what ##x'## exactly is.........

    In the book,the equation ##(28.5)## is ##\int\rho{dV}=\frac{1}{c}\int{j^0}dV=\frac{1}{c}\int{j^i}{dS_i}##,I think the index ##i## is not summed and equation ##(28.5)## is derived from equation ##(28.2)##----##j^i=\rho{\frac{dx^i}{dt}}##----which means inverse ##\frac{dt}{dx^i}## is used and we have problem when ##dx^i## equal ##0##.

    I feel it is hard to read this book coutinuously if confusion was not solved.Any suggestion for learning?
     
  7. Dec 29, 2016 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Your description seems OK to me. The continuous distribution of charge is like a fluid. If you arbitrarily pick a time ##t_0## and a point in space ##\vec{x}_0##, then the charge density at that time and place is ##\rho(t_0, \vec{x}_0)##. The fluid at that time and place will generally be in motion with some velocity ##\vec{v}(t_0, \vec{x}_0)##. This velocity could be written as ##\frac{\vec{dx}}{dt}## where ##dt## is an infinitesimal time interval from ##t = t_0## to ##t = t_0 + dt##. ##\vec{dx}## is the corresponding displacement of the fluid from ##\vec{x}_0## to ##\vec{x}_0 + \vec{v} dt##.

    Consider ##\int\rho{dV}=\frac{1}{c}\int{j^0}dV##. The integration is done in a particular inertial reference frame and the integral is over all of space at one fixed time ##t_0## in this frame of reference. All of space at one particular time is an example of a "hypersurface" in spacetime, as discussed in the text starting at the bottom of page 19. An element of volume ##dV## at fixed time ##t_0## can be considered as the zeroth component, ##dS^0##, of a four-vector ##dS^i##, as explained on page 20. Since ##t_0## is considered fixed, the components ##dS^1, dS^2## and ##dS^3## are all zero in this inertial frame. Thus, it is then true that we can trivially write ##j^0 dV = j^0 dS_0 = j^0 dS_0 + j^1 dS_1 + j^2 dS_2 + j^3 dS_3 = j^i dS_i##, where we sum over ##i## in the last expression. This expresses the integrand in an invariant form.

    I find the Landau Lifshitz books to be quite challenging. They are awesome but somewhat intimidating (for me). The writing is elegant, but terse. There is very little elaboration or repetition. Sometimes I need to spend quite a bit of time trying to decipher and understand a paragraph or equation. I often find that my difficulty was due to overlooking one particular phrase in one particular sentence. But the time spent is often well worth it.
     
    Last edited: Dec 30, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Meaning of trajectory in four-electricity in Landau's book
  1. Four electric currents (Replies: 3)

Loading...