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Understanding Einstein's Math

  1. Oct 21, 2005 #1
    Hello all,

    My personal experience with differential equations is from a long time ago. I wold be interested in someone showing me the details of Einstein's differential equation derivation of the Lorentz transformations in "On the Electrodynamics of Moving Bodies" - 1905. i understand what he does after he gets the differential equation but not how he gets it. He starts with time T as a function of coordinates in the stationary frame based on his method of synchronizing clocks;

    1/2[T(0,0,0,t) + T(0,0-,0,t+x'/(c-v)+x'/(c+v))] = T(x',0,0,t+x'/(c-v))

    He states "Hence if x' be chosen infintesimally small,

    1/2(1/(c-v)+1/(c+v))dT/dt = dT/dX' + 1/(c-v)dT/dt

    or

    dT/dx' + v/(c^2-v^2)dT/dt = 0 "

    Where the coordinates are (x,y,z,t) and the small letter "d" in the equations is the lower case Greek delta indicating partial derivatives.

    Anybody?
     
  2. jcsd
  3. Oct 21, 2005 #2

    Tide

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    It's just a Taylor series expansion (on both x and t) retaining only first degree terms in the differentials.
     
  4. Oct 21, 2005 #3

    Integral

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    Chew on this for a bit. I sorted this out in the early days of PF. I spent about 3 weeks, taking long walks in the park thinking about how to fill in that little gap in his algebra. Be sure you understand the notation of the first equation. What I do is simple algebra with no motivation other then I know where I need to go.

    Enjoy.
     
  5. Oct 29, 2005 #4
    Einstein's calculus

    Tide,

    Thanks for the reply. I ran into and former associate who was pretty good with math and he took the same route and showed me the details.

    I've had quite a bit of math training but it was long time ago.

    (Don't ask. :-) )

    It's a personal quirk, but I guess I'm not comfortable with the expanded-series-throw-away-the-higher-order-terms approach. Not that I don't believe it works, I guess I just prefer closed form solutions. That's why I didn't recognize what Albert did.

    An interesting historical note on such things is the analysis Michelson did on his famous experiment. He too resorted to an expanded series to calculate the time difference expected in the experiment. When Fitzgerald and Lorentz looked at it years later, they used and exact equation and deduced the contraction formula from it by inspection.

    One wonders if Michelson might not have deduced it himself.

    If he had wondered about it, he might have fallen on relativity himself 20 years before Einsein. The seeds of what might have convinced him are in the appendix of his paper. He was concerned that the motion (through space) of the device might cause the perpendularly reflected beam of light to miss returning to the splitter. He works out where the beam of light returns using Huygens theory and finds that it doesn't quite get back to the same place but assume the error won't harm the experiment.

    The interesting thing is that if he had deduced the length contraction and if he had applied it to the reflection problem, it predicts a return to the exact spot on the splitter mirror. It would have been difficult to explain that as coincidence and might have prompted him and others to start thinkingin the right direction.
     
  6. Oct 29, 2005 #5

    Integral

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    CinciRob,

    no comment on my PDF? Sounds to me like it is just what you are looking for.
     
  7. Nov 29, 2005 #6

    DrGreg

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    I’ve only just noticed this thread, so this response is a month too late, but if anyone’s still interested here goes…

    I’m sure Einstein didn’t really leave out three pages of calculations. I believe he was applying the chain rule for partial derivatives.

    First, I should point out there’s scope for notational confusion, as the symbols x’ and t are used with two subtly different meanings. To avoid this confusion, I will use X’, Y, Z and T to denote the four co-ordinate variables that [itex]\tau[/itex] is a function of. As Y and Z take constant zero values throughout, the chain rule in this case is:

    [tex] \frac{\partial \tau}{\partial x ^{\prime}} = \frac{\partial \tau}{\partial X ^{\prime}} \frac{\partial X ^{\prime} }{\partial x ^{\prime}} + \frac{\partial \tau}{\partial T} \frac{\partial T}{\partial x ^{\prime}} [/tex]

    Apply this to each of the three functions of [itex]\tau[/itex] within the equation.

    In the first case X’ = 0 and T = t.

    In the second case X’ = 0 and [tex] T = t + \frac {x ^{\prime}} {c + v} + \frac {x ^{\prime}} {c - v} [/tex].

    In the third case X’ = x’ and [tex] T = t + \frac {x ^{\prime}} {c - v} [/tex].
     
    Last edited: Nov 29, 2005
  8. Jun 27, 2007 #7
    Hi,

    I was already about to post a new thread, but found this old one which addresses exactly the issue I wanted to address.
    So I thought it would be better to just continue the present thread.

    I had a look at the reference, and there is one thing which is not correct in my opinion: on the left hand side of the equation one has dt=dx'*(1/(c+v)+1/(c-v)) , but one the right hand side dt=dx'/(c-v), which is inconsistent , i.e. one should not be entitled to use the same expression dtau/dt on both sides of the equation.
     
  9. Jun 27, 2007 #8

    Basically, the same applies what I said above: your Ts are a different from each other, so they shouldn't be referenced with the same variable T.
     
  10. Jun 27, 2007 #9

    Integral

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    You need to be more specific, which reference and where in that reference are you talking about?
     
  11. Jun 27, 2007 #10
    Well, at the top of page 2 of your derivation http://home.comcast.net/~integral50/Math/Specialrel.PDF , you are making x' infinitesimal and then set dx'*(1/(c+v)+1/(c-v))=dt (this applies to the left hand side of the equation). Towards the bottom of the same page you are doing the same thing for the right hand side (not fully written out in the same detail) and set dx'/(c-v)=dt , i.e. you are assigning the same variable dt to different quantities.
     
  12. Jun 27, 2007 #11

    Integral

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    dt is not a variable it is differential element. Both expressions I develop satisfy the definition of [itex] \frac {\partial \tau} {\partial t} [/itex]. It seems to me that requiring a single fixed [itex] \Delta t [/itex] in the definition of the partial is counter to the concept of a differential element.
     
  13. Jun 28, 2007 #12
    It is not about the general concept of a differential element here but its particular value. The point here is that on the left hand side it has the value dt=dx'*(1/(c+v)+1/(c-v)) but on the right hand side the value dt=dx'/(c-v). So they are actually two different differential elements and can't therefore both be denoted dt.
     
  14. Jun 28, 2007 #13

    Integral

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    That is the same thing you said before, it is still wrong. It IS about a differential and certainly NOT about a fixed quantity. That is the same t on both sides of the equation. Both satisfy the definition of [itex] \frac {d\tau} {dt} [/itex].

    Since when are differential elements fixed quantities?
     
  15. Jun 29, 2007 #14
    They are not fixed, but their ratio is fixed:

    if you have dt1=dx'/(c-v) and dt2=dx'*(1/(c-v)+1/(c+v)), then

    dt1/dt2= [1/(c-v)] / [1/(c-v)+1/(c+v)] ,

    but you assume dt1/dt2=1 in your derivation.
     
  16. Jun 30, 2007 #15
    Hello "NeverMind".
    I assume you are Thomas Smid, the one who runs the crackpot site http://www.physicsmyths.org.uk/ , correct?

    You resurrected a post over a year and a half old. And now you resurrect it AGAIN after another year.

    People have debunked your incorrect statements many times on many forums. Please do not bring these with you to PhysicsForums.com

    Thank you.
     
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