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Galilean and Lorentz invariant

  1. Oct 19, 2016 #1
    1. The problem statement, all variables and given/known data
    Professor C. Rank claims that a charge at [itex](r_1, t_1) [/itex] will contribute to the air pressure
    at [itex](r_2, t_2) [/itex] by an amount [itex] B \sin[C(|r_2 − r_1|^2− c^2|t_2 − t_1|^2)] [/itex], where B and C are constants.
    (A) Is this effect Galilean invariant?
    (B) Is this effect Lorentz invariant?

    2. Relevant equations
    Galilean transformation:
    [itex]x'=x-vt, t'=t[/itex]
    Lorentz transformation:
    [itex]'x=\frac{x-vt}{\sqrt{1-(\frac{v}{c})^2}}, t'=\frac{t-\frac{tv}{c}}{\sqrt{1-(\frac{v}{c})^2}}[/itex]

    3. The attempt at a solution
    (A) I suppose that part with [itex]t_2,t_1[/itex] is not important for Galilean transformation it is same. And if I transform [itex] r_1, r_2[/itex] to [itex]r'_1=r_1-vt, r'_2=r_2-vt[/itex] the extra terms are deducated.
    So it is Galilean invariant? True?
    (B) In Lorentz case I have problem with times. Could you suggest some point of view?
     
  2. jcsd
  3. Oct 20, 2016 #2

    kuruman

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    The Galilean part is correct.
    Your problem with times in the Lorentz case may have something to do with the incorrect expression for ##t'## that you have posted.
     
  4. Oct 20, 2016 #3
    Ehm no :) Sorry it just typo :) Ok the reason why I'm dealing with such easy things is that I have results that claim opposite than me. Check attached file. Is there mistake or I'm wrong?
     

    Attached Files:

  5. Oct 20, 2016 #4

    kuruman

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    I see no mistake in the file that you attached.
    Can you post exactly what you did and what kind of problem you have with time?
     
  6. Oct 21, 2016 #5
    You didn't? I claimed that this is Galilean invariant and you agree. In this file is an answer NO. So what I'm missing? And for Lorentzian there is an answer YES. I mean it is reversely.
     
  7. Oct 21, 2016 #6
    I intuitively mean, that it is not Lorentz invariant. But if I would like to do it carefully I would I put there the transformed terms. And I'm not sure if the time which appears in r_1 and r_2 transformation is same or what should be there.
    [tex] r'_1=\frac{r_1-vT_1}{\sqrt{1-(\frac{v}{c}})^2} [/tex]
    [tex] r'_2=\frac{r_2-vT_2}{\sqrt{1-(\frac{v}{c}})^2} [/tex]
    My question is if [itex]T_1==T_2[/itex]? But if I think about that now...[itex]r_1[/itex] and [itex]r_2[/itex] are in same system so if something move with respect to one it moves same in respect to second so times should be same. Am I right? And in this file it is wrong. Do you agree?
     
  8. Oct 21, 2016 #7

    kuruman

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    Sorry for confusing you. I read the question rather hastily. No, the expression is not Galilean invariant. Did you actually calculate ##|{r'}_2^2-{r'}_1^2|## under the Galilean transformation? If so, what did you get?
    For the Lorentz part you don't even need to do that. The sine function can be thought of representing a traveling wave. What is the speed of that wave?

    On edit: In view of @vela's comment below, I meant ## \lvert \vec{r}'_2 - \vec{r}'_1 \rvert^2 ##.
     
    Last edited: Oct 21, 2016
  9. Oct 21, 2016 #8

    vela

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    The problem involves ##\lvert \vec{r}_2 - \vec{r}_1 \rvert^2##, not ##|{r'}_2^2-{r'}_1^2|##. It looks invariant under a Galilean transformation to me since it depends on the spatial distance between the two points.
     
  10. Oct 21, 2016 #9

    vela

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    No, you can't assume ##T_1 = T_2##. As measured in the S frame, event 1 occurs in space at ##\vec{r}_1## at time ##T_1##, and event 2 occurs in space at ##\vec{r}_2## at time ##T_2##. There's no reason to assume the events are simultaneous.

    By the way, you should find that the expression is Lorentz invariant.
     
  11. Oct 21, 2016 #10

    kuruman

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    I interpreted "Galilean invariant" to mean that
    ##\sin \left[ C \left( | \vec{r}'_2-{\vec{r}'}_1|^2-c^2|{t'}_2-{t'}_1|^2\right) \right]=\sin \left[ C \left( | (\vec{r}_2-\vec{r}_1|^2-c^2|{t}_2-{t}_1|^2 \right) \right] ##
    under the Galilean transformation, ##\vec{r}'=\vec{r}-\vec{v}~t;~~t'=t##. Was I wrong (again)?
     
  12. Oct 21, 2016 #11
    I understand, Even if in S system ##t_1=t_2## in S' ##t'_1=t'_2## may not be true.
     
  13. Oct 21, 2016 #12
    So for Galilean transformation:
    ##r'_1=r_1-vt, r'_2=r_2-vt, t'_1=t_1, t'_2=t_2## => ##(r_1-r_2)=(r'_1+vt -r'_2-vt)=(r'_1-r'_2)## and ##(t_1-t_2)=(t'_1-t'_2)##. Am I right? And from this it has to be Galilean invariant. Do you agree?
     
  14. Oct 21, 2016 #13
    For Lorentz transformation:
    I'm not rewriting the rules again.
    ##(r_1-r_2)=\gamma (r'_1+v t'_1-r'_2-v t'_2) ## and ##(t_1-t_2)=\gamma (t'_1+\frac{r'_1 v}{c}-t'_2 - \frac{r'_2 v}{c}) ## where ##\gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}## and ##v## is velocity between S and S'. From this I suppose it isn't Lorentz invariant.
     
  15. Oct 21, 2016 #14

    vela

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    No, that's what I took it to mean as well, but I see the mistake I made. I agree with the provided answer it's not invariant under a Galilean transformation, which makes makes me feel better now.
     
  16. Oct 21, 2016 #15

    vela

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    Think about what value of ##t## you're referring to when you say ##r \to r' = r-vt##.
     
  17. Oct 21, 2016 #16
    Ha :) That's the reason why I wrote I have problem with meaning of times. Now I probably understand...now recompute it. ##t## is in such transformation difference between the time when the systems met each other and time of measurement. True?
     
  18. Oct 21, 2016 #17
    So Galilean (again):
    ##(r_1−r_2)=(r′_1+v dt_1−r′_2−v dt_2)## , ##dt_1\neq dt_2## because ##r_1## and ##r_2## has generaly different position so met S' system in diferent time. So it is not Galilean invarinat. Do you agree with my explanation?
     
  19. Oct 21, 2016 #18

    kuruman

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    Me too.
     
  20. Oct 22, 2016 #19
    Is my ##dt## definition right? I'm struggling with Lorentz invariant. There is many terms but I can't see any connection among them.
     
  21. Oct 22, 2016 #20

    kuruman

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    That's not what you are asked to find if it is Galilean invariant. Start with
    ## \vec{r}'_1=\vec{r}_1-\vec{v}~t_1##, ## \vec{r}'_2=\vec{r}_2-\vec{v}~t_2##, ##t'_1=t_1## and ##t'_2=t_2##.
    Then note that ## | \vec{r}'_2-{\vec{r}'}_1|^2-c^2|{t'}_2-{t'}_1|^2= (\vec{r}'_2-\vec{r}'_1) \cdot (\vec{r}'_2-\vec{r}'_1)-c^2|{t'}_2-{t'}_1|^2##
    Then replace the primed quantities on the right hand side with unprimed ones as given by the Galilean transformation and see what you get.
     
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