Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electromagnetic tensor

  1. May 22, 2012 #1
    dear all,

    I have a question about Electromagnetic field tensor.
    As I was reading introduction to relativity by Hobson, I saw this sentence:
    "in order that the rest mass of a particle is not altered by the action of the erforce we require the latter to be a pure force, so:
    u.f=0
    1) what does it mean?

    The other question is about the rest of this text:

    in the component form this reads:

    f_μ u^μ=q F_μσ u^μ u^σ=0
    so that F must be an anti symmetric tensor. 2) why?!!!!!
     
  2. jcsd
  3. May 22, 2012 #2

    Mentz114

    User Avatar
    Gold Member

    If this holds

    Fμσuμ uσ = 0

    then Fμμ = 0, and Fμσ = -Fσμ

    It seems obvious to me if the full expansion is written out, but I'm not sure if what I've written is rigorous or helpful.
     
  4. May 22, 2012 #3
    Dear friend,

    I agree with he first one but could you please explain the second one more?
    Fμσ = -Fσμ

    I don't know why it seems so odd to me!
     
  5. May 22, 2012 #4

    Mentz114

    User Avatar
    Gold Member

    F12u1 u2 + F21 u2 u1 = 0 => F12 = -F21 ?
     
  6. May 23, 2012 #5
    U r right!
    Thank you!
     
  7. May 23, 2012 #6
    does anyone know the answer to the first question?
    why should it be like this:
    u.f=0?!!
    what does hobson mean by that?
     
  8. May 23, 2012 #7

    Bill_K

    User Avatar
    Science Advisor

    Relativistic mechanics can be expressed in terms of 4-vectors f, a, v, related to each other in the same way that we're used to seeing in three dimensions except for the use of proper time τ in place of coordinate time, f = ma = m dv/dτ.

    Consequently f·v = m v·dv/dτ = ½ m d(v·v)/dτ. But the 4-velocity has constant length c, so this requires that f·v = 0. A realistic 4-force must always be orthogonal to the 4-velocity.

    The only way to have f·v ≠ 0 is to consider a system with variable mass.
     
  9. May 23, 2012 #8
    Thank you. I understand it now.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook