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anonymous

  • one year ago

check my work

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  1. anonymous
    • one year ago
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    Consider a rod kept on the ground, it is at rest with respect to a frame of reference S of an observer standing at a distance of x1 from the rod(since both r moving along with the earth, they are at rest wrt the earth). He thus measure it's length (L) to be x2-x1, it is the rod's length in a frame of reference respect to which is at rest, so it is the rod's proper length |dw:1444042534166:dw| \[L=x_{2}-x_{1}\] Now consider the observer to be standing in the same place, but the rod is now kept in a vehicle that is moving with a velocity v in the positive x direction w.r.t. the observer The rod is now at rest with respect to the frame of reference of the bus, and it's length now measured by an observer in the bus is given as \[L'=x_{2}'-x_{1}'\]|dw:1444043022456:dw| But for the observer on ground, the coordinates are given by using inverse lorentz transformation equation(we use inverse as we want to find coordinates in S frame) \[x_{1}^{new}=\frac{x_{1}'+vt'}{\sqrt{1-\frac{v^2}{c^2}}} \space \space , \space \space x_{2}^{new}=\frac{x_{2}'+vt'}{\sqrt{1-\frac{v^2}{c^2}}}\] \[L^{new}=x_{2}^{new}-x_{1}^{new}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(x_{2}'+vt'-x_{1}'-vt')=\gamma(x_{2}'-x_{1}')=\gamma L'\] But the new length observed by the observer on the ground should be same as before, as he has not moved from his original position so we have \[L=L^{new} \space \space \space ; \space \space \space L=\gamma L' \implies L'=\sqrt{1-\frac{v^2}{c^2}}L\] @IrishBoy123

  2. anonymous
    • one year ago
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    Another assumption that is made is that the coordinates of both the end points of the rod are measured simultaneously in both frames

  3. IrishBoy123
    • one year ago
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    SR, excellent! will look later... @Michele_Laino

  4. anonymous
    • one year ago
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    ok thanks:)

  5. Michele_Laino
    • one year ago
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    let's suppose the rod lies on reference \(S'\)

  6. Michele_Laino
    • one year ago
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    so its proper length is \(L'\)

  7. Michele_Laino
    • one year ago
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    namely I mean its real length is \(L'\)

  8. Michele_Laino
    • one year ago
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    now, I have to measure its length in reference \(S\), and I have to make the measure of both ends of the rod at the same time \(t\)

  9. Michele_Laino
    • one year ago
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    of course, reference \(S'\) is moving, with respect to \(S\), with speed \(V\)

  10. Michele_Laino
    • one year ago
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    so, using \(Lorentz\; transformation\), I can write this: \[\begin{gathered} x{'_2} = \gamma \left( {{x_2} - Vt} \right) \hfill \\ x{'_1} = \gamma \left( {{x_1} - Vt} \right) \hfill \\ \end{gathered} \] and subtracring the second equation from the first one, I can write this: \[\begin{gathered} x{'_2} - x{'_1} = \gamma \left( {{x_2} - {x_1}} \right) \hfill \\ {L_O} = \gamma {L_{meas}} \hfill \\ \end{gathered} \]

  11. Michele_Laino
    • one year ago
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    where \(L_O\) is the real length and \(L_{meas}\) is the measured length. So, finally, we get: \[{L_{meas}} = \frac{{{L_O}}}{\gamma }\] namely the so called \(Length\; contraction\), being \(\gamma>1\)

  12. anonymous
    • one year ago
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    Suppose 2 events are recorded in the same position in their respective frames by 2 observers, one frame at rest w.r.t. the earth(S) and the other frame moving at a velocity v in the positive x direction w.r.t. the earth(S') The time interval between the events occurring at time t1 and t2 for the S frame and time t1' and t2' for frame S' are thus given as \[\Delta t=t_{2}-t_{1}\]\[\Delta t'=t_{2}'-t_{1}'\] But for an observer in frame S he observes the times t1 and t2 for the events occurring in frame S' according to inverse lorentz transformation as \[t_{2}=\gamma(t_{2}'+\frac{vx'}{c^2}) \space \space ; \space \space t_{1}=\gamma(t_{1}'+\frac{vx'}{c^2})\] \[\therefore \Delta t=\gamma(t_{2}'+\frac{vx}{c^2}-t_{1}'-\frac{vx}{c^2})=\gamma(t_{2}'-t_{1}')=\gamma \Delta t'\] Thus \[\Delta t'=\frac{1}{\gamma} \Delta t\]

  13. anonymous
    • one year ago
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    Why have you used lorentz instead of inverse lorentz when u need to find length in frame S?

  14. Michele_Laino
    • one year ago
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    because I have to measure the length in \(S\) and I have to make the measure of position of the ends of the rod at the same tame \(t\)

  15. Michele_Laino
    • one year ago
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    same \(t\), not same \(t'\)

  16. anonymous
    • one year ago
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    Ohk, since we are measuring in frame S, we require the same time in frame S. This point was always bothering me, that why do we use the lorentz instead of inverse, if we are calculating for frame S

  17. Michele_Laino
    • one year ago
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    since \(x'_2\) is a function of \(x_2,t\), similarly for \(x'_1\), which is a function of \(x_1,t\)

  18. anonymous
    • one year ago
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    Then my derivation for time dilation wrong....If I correct my error I should arrive at the result \[\Delta t'=\gamma \Delta t\]

  19. anonymous
    • one year ago
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    is that right?

  20. Michele_Laino
    • one year ago
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    let's see: if the real interval time is \(t'_2-t'_1=\Delta t'\), then I have to measure \(Delta t\) in reference \(S\) at the same position in \(S'\), namely \(x'_2=x'_1\)

  21. Michele_Laino
    • one year ago
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    \(\Delta t\)*

  22. Michele_Laino
    • one year ago
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    so we can write this: \[\begin{gathered} {t_2} = \gamma \left( {t{'_2} - \frac{V}{{{c^2}}}x'} \right) \hfill \\ {t_1} = \gamma \left( {t{'_1} - \frac{V}{{{c^2}}}x'} \right) \hfill \\ \end{gathered} \] since the two events occurred at the same position \(x'_2=x'_1=x'\)

  23. Michele_Laino
    • one year ago
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    again, subtracting the first equation, from the second one, we get: \[\Delta t = \gamma \Delta t'\] namely the so called \(time\;dilation\), being \(\gamma >1\)

  24. anonymous
    • one year ago
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    So my derivations r correct !!

  25. Michele_Laino
    • one year ago
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    the second one is correct!

  26. anonymous
    • one year ago
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    what's wrong with the first one ??

  27. Michele_Laino
    • one year ago
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    sorry, also the first derivation is correct!

  28. anonymous
    • one year ago
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    can u help me with another question

  29. Michele_Laino
    • one year ago
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    ok!

  30. anonymous
    • one year ago
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    Prove that the 4-dimensional volume element \[dxdydzdt\] Is invariant under Lorentz transformations From our previous derivations we have \[L=\gamma L'\] \[\implies \Delta x=\gamma \Delta x'\] Taking the limits \[\Delta x \rightarrow 0 \space \space \space ; \space \space \space \Delta x' \rightarrow 0\] \[dx=\gamma dx'\]\[\implies dx'=\frac{1}{\gamma }dx\] Obviously \[dy'=dy \space \space \space ; \space \space \space dz'=dz\] and from our time dilation we get \[\Delta t'=\frac{1}{\gamma}\Delta t\] Again taking the limits \[\Delta t \rightarrow 0 \space \space \space ; \space \space \space \Delta t' \rightarrow 0\]\[dt'=\frac{1}{\gamma}dt\] thus we get \[dx'dy'dz'dt'=\frac{1}{\gamma^2}dxdydzdt\] Which is not the same

  31. Michele_Laino
    • one year ago
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    the natural way to express the invariance of volume, is to consider the fundamental tensor \(g_{ik}\), namely we have to show the truth of a more general statement: \[\sqrt {g'} dV' = \sqrt g dV\] where \(\sqrt g\)is the square root of the determinant of the tensor \(g_{ik}\) Since in special relativity, we have \(g=g'=-1\), then we have showed your statement, namely: \(dV'=dV\)

  32. Michele_Laino
    • one year ago
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    oops..\(g=g'=1\)

  33. anonymous
    • one year ago
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    ooh no idea what you talking about, I haven't read about tensors yet, is there an easier way to represent?

  34. Michele_Laino
    • one year ago
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    I'm sorry, it is the only way, which I know, in order to solve your exercise. Are you familiar with tensor calculus?

  35. anonymous
    • one year ago
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    Nope not at all, I'm a freshman

  36. anonymous
    • one year ago
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    I don't believe it is even in our 3 year course

  37. Michele_Laino
    • one year ago
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    since, what is invariant is the volume of the 4-dimensional space, namely what is invariant is the subsequent quantity: \[d\tau = \int {dx{\kern 1pt} dy{\kern 1pt} dz{\kern 1pt} cdt} \]

  38. Michele_Laino
    • one year ago
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    oops... \[\sqrt g d\tau = \sqrt g \int {dx{\kern 1pt} dy{\kern 1pt} dz{\kern 1pt} cdt} \]

  39. Michele_Laino
    • one year ago
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    it is a proposition from General Relativity

  40. anonymous
    • one year ago
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    If you don't know an easier way, that's alright, I can leave this exercise for later, I doubt it will even come in the examination, mostly they want us to do the derivations and a numerical related with it,however I was curious but in the end this problem was only confusing me more to be honest. In the book they have taken \[dx'=\gamma dx\]\[dy'=dy\]\[dz'=dz\]\[dt'=\frac{dt}{\gamma}\] if you multiply that out the gamma cancels

  41. Michele_Laino
    • one year ago
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    I think that it is correct!

  42. anonymous
    • one year ago
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    The part I don't understand is why \[dx'=\gamma dx\]

  43. Michele_Laino
    • one year ago
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    here we have to specify the role of reference system \(S\) and \(S'\) first

  44. anonymous
    • one year ago
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    Physically I can realize it as the contraction of the length exactly cancelling out the dilation of time, but mathematically I can't interpret the equation of dx'

  45. Michele_Laino
    • one year ago
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    let's suppose that \(S'\) is moving, with respect to \(S\) with a speed \(V\)

  46. anonymous
    • one year ago
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    yep!

  47. Michele_Laino
    • one year ago
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    along a \(x-\)direction, for example

  48. anonymous
    • one year ago
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    ok

  49. Michele_Laino
    • one year ago
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    then we can write this: \[dx = \frac{{dx'}}{\gamma }\] which is the length contraction

  50. Michele_Laino
    • one year ago
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    furthermore, we can write this: \[dt = \gamma dt'\] which is the time dilation

  51. Michele_Laino
    • one year ago
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    as you can see the gamma's cancel

  52. Michele_Laino
    • one year ago
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    oops.. I meant gamma factors

  53. anonymous
    • one year ago
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    shouldn't it be \[dx'=\frac{dx}{\gamma}\] Since a length in frame S' will be smaller than S, we observer the length in S' to be a fraction of the length in S

  54. Michele_Laino
    • one year ago
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    I suppose that I'm in reference \(S\)

  55. Michele_Laino
    • one year ago
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    so: \(dx'\) is the real length, and \(dx\) is the measured length

  56. Michele_Laino
    • one year ago
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    similarly for \(dt',dt\)

  57. anonymous
    • one year ago
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    Then why is the expression for length contraction has the primes interchanged?Is it because I've measured in S and observed in S'? where as if for time im measuring in t' and observing for t, I should respectively measure length in S' and then observe in S?

  58. anonymous
    • one year ago
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    that would make sense then

  59. Michele_Laino
    • one year ago
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    I have used terms "observed" and "measured" with the same meaning

  60. anonymous
    • one year ago
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    Actually I've measured the proper length in frame S in my derivation, that is why my length element is coming different, I should measure the proper length in frame S' because im measuring proper time in frame S'

  61. Michele_Laino
    • one year ago
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    I have supposed to be in reference \(S\) and then to measure the quantities \(dx',dt'\) in reference \(S\)If you are in reference \S\), then you have to measure length and time in \(S\), namely you have to write what you measure when you are in \(S\)

  62. Michele_Laino
    • one year ago
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    more precisely: \(dx',dt'\) are the given quantities, and you have to write what are \(dx,dt\) when you are in \(S\)

  63. anonymous
    • one year ago
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    Yeah, makes sense, i was doing opposite in length contraction, starting with dx and going to dx'

  64. Michele_Laino
    • one year ago
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    yes! I think so

  65. anonymous
    • one year ago
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    good thread !! thx

  66. Michele_Laino
    • one year ago
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    :)

  67. IrishBoy123
    • one year ago
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    awesome thread!!! thank you both. @Abhisar @Astrophysics @arindameducationusc @iambatman @Radar @Robert136 @rvc @UnkleRhaukus

  68. Abhisar
    • one year ago
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    Wow!, that was some nice stuff!

  69. Astrophysics
    • one year ago
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    Thanks for the tag @IrishBoy123 This is a great thread, thank you! Special relativity at work :)

  70. rvc
    • one year ago
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    Thanks @IrishBoy123 :)

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