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I Laws of physics and inertial systems

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  1. Jul 21, 2017 #1
    It is not the laws of physics, but the forms of laws of physics which are the same in all inertial frames. Comment.


    "The forms of laws of physics are the same in all inertial frames" is a necessary condition (put by scientists ) to get satisfied by something which has to be called as a law of physics.

    To illustrate that it's not the laws of physics, but the forms of laws of physics which are the same in all inertial frames, let's consider Coulomb's electrostatic force law.

    Let's consider an inertial frame S in which two charged particles A and B with charges q and Q respectively are at rest and the distance measured between the two is r. Then, the force acting on Q is kQq / r2 ##\hat r## where ##\hat r ## is the unit vector along the line joining q and Q.

    W.r.t. another inertial frame S', the charges of the two particles remain same to that in S frame respectively.
    But, the distance between the two gets changed to r' (keeping special relativity) in mind.Then, the force acting on Q is kQq / r'2 ##\hat r'## where ##\hat r' ## is the unit vector along the line joining q and Q.

    So, it is observed that the Coulomb force between the two charged particles which is a law of physics is different for different inertial frames(due to special relativity), but the form of the force remains same in both inertial frames.

    The laws of physics differ only because of the special relativity,here.
    Under Galilean Transformation,both laws of physics and forms of laws of physics remain the same in all inertial frames.

    Is what I have written above correct?

    Galilean Transformations is valid only for inertial frames. Isn't it?
     
  2. jcsd
  3. Jul 21, 2017 #2
    Who's statement? Yours?

    Can you proof r' is not equal to r?
     
  4. Jul 21, 2017 #3
    How does it matter whose statement it is? It is a statement I have to comment on. So, tell me whether you consider it right or not with reason.
    r' may or may not be equal to r. It depends on the frame S'.
    If S' is moving with speed v along the line joining the two charges w.r.t. S, and the normal textbook assumptions for special relativity are taken into account,then due to Lorentz contraction, r' = √(1-v2/c2) r.
     
  5. Jul 21, 2017 #4

    Ibix

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    The Coulomb force isn't really a law of physics. It's an application of a specific case of a solution to Maxwell's equations to the case of stationary charges. It doesn't apply to moving charges (these have magnetic fields as well as electrostatic fields) so you wouldn't expect correct answers from naively transforming the Coulomb field.

    The things that are general statements about physics - Maxwell's equations in this case - do have the same form in different inertial frames. That doesn't mean that the solutions for a particular setup are the same in all inertial frames.
     
  6. Jul 21, 2017 #5
    Hi.

    S: all at rest
    q
    Q

    S":all moving with transverse velocity v for an example
    q##\rightarrow## v
    Q##\rightarrow## v

    In S" not Coulomb force but Lorentz force including contribution of magnetic field generated by motion of charges work on the particles. Things are more complicated than you expect. Best.
     
  7. Jul 21, 2017 #6

    vanhees71

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    Of course, your calculation in #1 is incomplete, because you have to use the correct transformation of the electromagnetic field, i.e., in this case a Lorentz boost. It's most simple to use the four-potential. So let's evaluate the electromagnetic field of the charge ##Q##. In its rest frame it's a Coulomb field. For simplicity let the particle sit in the origin. The four-potential is
    $$(A^{\mu})=\begin{pmatrix} \phi(\vec{x}) \\0 \\ 0 \\ 0 \end{pmatrix}$$
    with
    $$\phi(\vec{x})=\frac{Q}{4 \pi |\vec{x}|}.$$
    Now you can get the same field in the boosted reference frame, using
    $$\bar{A}^{\mu}(\bar{x})={\Lambda^{\mu}}_{\nu} A^{\nu}(x), \quad \bar{x}^{\mu} = {\Lambda^{\mu}}_{\nu} x^{\nu}.$$
    For a boost in ##z## direction
    $$({\Lambda^{\mu}}_{\nu}) = \begin{pmatrix} \gamma &0 &0 & -\beta \gamma \\
    0 & 1 &0 & 0 \\
    0& 0 &1 & 0\\
    -\beta \gamma & 0 & 0 & \gamma \end{pmatrix}.$$
    Now you can evaluate the potential in the boosted reference frame, then the field components
    $$\vec{\bar{E}}=-\partial_{\bar{t}} \vec{\bar{A}}-\vec{\bar{\nabla}} \bar{A}^{0}, \quad \vec{\bar{B}}=\vec{\bar{\nabla}} \times \vec{\bar{A}}$$
    and finally the force on the 2nd charge
    $$\vec{\bar{F}}=q (\vec{\bar{E}} + \vec{\beta} \times \vec{\bar{B}}).$$
    You'll see that everything is consistent.
     
  8. Jul 21, 2017 #7

    PeterDonis

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    This is a bad choice because it's not Lorentz invariant, so it is not a "law of physics" according to your definition. You should be using the Lorentz force law instead.

    This is not correct, because in the primed frame there is also a magnetic force acting between the charges, since in the primed frame the charges are moving so the current is nonzero (in the original frame the charges were static so the current was zero). In other words, the Coulomb force law is not invariant under Lorentz transformation. But the Lorentz force law is.

    Incorrect. See above.

    This is not correct either, at least not if you include electromagnetism. The laws of electromagnetism are not Galilean invariant.

    No. See above.
     
  9. Jul 21, 2017 #8

    PeterDonis

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    What's the difference?
     
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