1. Not finding help here? Sign up for a free 30min 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!

Time translation invariant equation

  1. Aug 15, 2013 #1

    fluidistic

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    I must show that the equation ##\frac{\partial E}{\partial t}=t\nabla \times E## is invariant under time translations and I must also find its symmetries if it has any. Where E is a function of time and position.


    2. Relevant equations

    Already given.

    3. The attempt at a solution
    I assume that ##\vec E (\vec x ,t)## is a solution of the equation ##\frac{\partial \vec E (\vec x ,t)}{\partial t}=t\vec \nabla \times \vec E (\vec x , t)##.
    I want to show that ##\vec E _T (\vec x , t)=\vec E (\vec x , t-T)=\vec E(\vec x , u)## is not a solution of the equation ##\frac{\partial \vec E _T (\vec x ,t)}{\partial t}=t\vec \nabla \times \vec E_T (\vec x , t)##
    Let's go:

    ##\frac{\partial \vec E_T(t,\vec x )}{\partial t}=\frac{\partial \vec E_T(u, \vec x )}{\partial t}=\frac{\partial u}{\partial t} \cdot \frac{\partial \vec E ( u , \vec x)}{\partial t}=\frac{\partial \vec E ( u , \vec x)}{\partial t}=u \vec \nabla \times \vec E (u , \vec x )=u \vec \nabla \times \vec E_T (\vec x , t)=(t-T) \vec \nabla \times \vec E_T (\vec x , t) \neq t\vec \nabla \times \vec E_T (\vec x , t)##.
    Is what I've done okay/valid?

    For the symmetries, I guess I must check out if it has a translational invariance and/or rotation invariance.

    Edit: I checked out and if I didn't make any error the equation is invariant under space translation. I think this means that if the equation represents a physical system then the linear momentum is conserved, but since it isn't time invariant, the energy isn't conserved. I'd have to check out if it's invariant under rotations...
     
    Last edited: Aug 15, 2013
  2. jcsd
  3. Aug 19, 2013 #2

    BruceW

    User Avatar
    Homework Helper

    I think you probably understand what you are meant to do. But the way you have written the answer is really confusing as to which step you are doing and why. And yeah, I'd guess you are meant to check for spatial translation and rotation. Maybe rotations around the origin is enough.
     
  4. Aug 20, 2013 #3

    fluidistic

    User Avatar
    Gold Member

    Thanks for the feedback. Could you tell me please which steps (all?) are confusing and/or how would you do it? Maybe it will make me understand something that I don't currently.
     
  5. Aug 20, 2013 #4

    BruceW

    User Avatar
    Homework Helper

    well, you do the entire thing in one line. But the order of logic seems weird to me. Mathematically it is right. But instead you could use several lines (one for each equality) and you could write why you can say that the two things are equal. This would make it easier to follow.
     
  6. Aug 22, 2013 #5

    fluidistic

    User Avatar
    Gold Member

    I see, thanks.
    Let ##u=t-T##. I define ##\vec E _T (t, \vec x )=\vec E (t-T, \vec x)##. So that ##\vec E_T (t, \vec x)=\vec E(u, \vec x)##.
    I want to see whether ##\vec E _T(t, \vec x )## is a solution to the equation ##\frac{\partial \vec E _T(\vec x ,t)}{\partial t}=t\vec \nabla \times \vec E _T (\vec x , t)##. If it is, then the equation is invariant under time translations. If it is not a solution, then it's not invariant under time translations.

    I start with the left hand side of the equation: ##\frac{\partial \vec E _T(\vec x ,t)}{\partial t}=\frac{\partial \vec E (u, \vec x) }{\partial t}##.
    Using the chain rule (I believe?), I can rewrite the last expression as ##\frac{\partial u}{\partial t} \frac{\partial \vec E (u, \vec x )}{\partial u}##. But looking at the definition of u, it is easy to see that ##\frac{\partial u}{\partial t}=1##. Thus, one reaches that ##\frac{\partial \vec E _T(\vec x ,t)}{\partial t}=\frac{\partial \vec E (u,\vec x)}{\partial u}##.

    Now I use the equation given in the problem statement (considering t as a dummy variable), so I can write the right hand side of the last equation as ##u \vec \nabla \times \vec E (u, \vec x )## (I can do this because ##\vec E (\vec x , t)## is a solution to the equation given in the problem statement).
    Replacing back u by t-T, I reach that ##\frac{\partial \vec E _T (t, \vec x) }{\partial t}=(t-T) \vec \nabla \times \vec E (t-T, \vec x) ## This is true because the curl acts on the spatial coordinates, not on time.
    And so ##\frac{\partial \vec E _T (t, \vec x) }{\partial t}=(t-T) \vec \nabla \times \vec E _T (t, \vec x)##.

    Therefore, if ##\vec E (\vec x ,t)## is a solution to the equation ##\frac{\partial \vec E (\vec x ,t)}{\partial t}=t\vec \nabla \times \vec E (\vec x , t)##, then ##\vec E _T (t , \vec x )## is a solution to the equation ##\vec E _T (t , \vec x )=(t-T) \vec \nabla \times \vec E_T (\vec x , t)##.
    Thus ##\vec E _T (t , \vec x )## cannot be a solution to the equation ##\frac{\partial \vec E _T(\vec x ,t)}{\partial t}=t\vec \nabla \times \vec E _T (\vec x , t)## unless that ##T=0##.
    So that the equation is not invariant under time translations.
    What do you think now?
     
  7. Aug 22, 2013 #6

    BruceW

    User Avatar
    Homework Helper

    excellent :) nice and clear to read. also, I agree with the method and answer.
     
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: Time translation invariant equation
  1. Time invariance (Replies: 2)

Loading...