Recent content by kron

  1. K

    Stupid question about invariance of maxwell equations

    Ok, thanks Fredrik. I still could say one sums over the index 'a' so I don't have to transform that as a Tensor with three indices but as a "normal" 4-vector. \partial'_a F'^{ab}=j'^b But if I would take an arbitray linear transformation for that I wouldn't get the desired result, because the...
  2. K

    Stupid question about invariance of maxwell equations

    Yes I mean linear transformations, sorry. My problem is, what can I say about the \Lambda^b{}_e-matrices ? I can put anything I want in the lambda's, because the equations (not the 4-vectors) will remain invariant. I'm not sure if the second equality just holds for LT. Isn't this just a general...
  3. K

    Stupid question about invariance of maxwell equations

    Hi, as you all know one can write the Maxwell-equations in covariant form, namely: \partial_a F^{ab} = \frac{4\pi }{c} j^{b} and \partial_a G^{ab}=0 where \textbf{G} is the dual Tensor to \textbf{F}. Now the two simple questions. I can see that they are invariant, because I...
  4. K

    A few questions from introduction to sr by rindler.

    Hi, I know this is a very old post, but the answer x'(2)-x'(1)=Lsqrt[(1-b)/(1+b)] is obviously wrong, as indicated by the correct solution: [..]prove that in S' the distance between them is L*(c+v)^0.5/(c-v)^0.5[..] To obtain that, you have to pick a rest frame at first. If we take S' as a...
  5. K

    Solving for Current in Resistors | Voltage 40.7 V | A and B Points

    Hi, you have to calculate for the 12 and 6 Ohm the following: 1/12 + 1/6 = 1/r_1 so r_1 = 6*12/(6 + 12) = 4 = 1/0.25. So 0.25 is wrong. Do the same for 1/4 + 1/8 = 1/r_3 -> r_3. r_2 = 5 Ohm. So R = r_1 + r_2 + r_3. Now you have I = U/R (between a and b) with U = 40.7 V. Try to figure out the...
  6. K

    Solving the Problem of an Infinite Chain Slipping Down a Table

    Yes, but we need only one variable x_1 = y for example. Now you have to find an expression for the kinetic energy T which is simple and for the potential energy U which is simple as well. U depends of course at your point of reference. Make a sketch.
  7. K

    Solving the Problem of an Infinite Chain Slipping Down a Table

    Hi, do you know the Euler–Lagrange equations ? You have to find an expression for the potential and kinetic energy, the difference is the Lagrange-function. Put this function in the Langrange equations and you get a second order diff.-equation. kind regards
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