As set up and described by Professor Shankar, I was trying to derive the Lorentz transformation with his equations...

$$\frac{t'}{t} = \frac{c-v}{c},\qquad \frac{t}{t'} = \frac{c+v}{c}$$ After adding gamma $$\frac{t'}{t} = \gamma\frac{c-v}{c},\qquad \frac{t}{t'} = \gamma\frac{c+v}{c}$$

I solved for gamma and obtained the correct result, but I just don't understand what adding gamma does? Does it correct for something? Is it a "fudge factor" as Professor Shankar says? What is the logic behind it then?

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    $\begingroup$ It's not "just a fudge factor"! What professor said that? $\endgroup$ Commented Jun 30, 2016 at 0:42
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    $\begingroup$ Ouch... OK... that's not Shankar's finest hour. He isn't completely wrong... there is a theoretical argument why the transformation has to look that way, but he hides it behind his "fudge factor". That's not the best way of teaching this. We had a beautiful discussion about that a while ego, but I can't find it. One does not start with a fudge factor, but with the insight that all transformations have to be linear to preserve homogeneity and isotropy of space. The form of the transformations follows from there. $\endgroup$
    – CuriousOne
    Commented Jun 30, 2016 at 1:01
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    $\begingroup$ Have you looked at https://en.wikipedia.org/wiki/Lorentz_factor and https://en.wikipedia.org/wiki/Lorentz_transformation? If not, the latter does a decent job of explaining what exactly a Lorentz transformation involves and from where it derives... $\endgroup$ Commented Jun 30, 2016 at 1:02
  • $\begingroup$ There are approaches to constructing the Lorentz transformation as the most general member of a class of symmetric transformation, in which $\gamma$ is introduced as a undefined function of the relative velocity to be found later. This may be a variation of that approach. $\endgroup$ Commented Jun 30, 2016 at 1:22
  • $\begingroup$ In the YouTube video that you mentioned, starting at approximately 50:00, Shankar also describes two light clocks and how from either point of view the "other" clock seems to be ticking slower. But then he talks about mechanical clocks and why the "other" mechanical clock also appears to be slower, and says "..... we don't know exactly how to explain that clock......". Well he should be able to explain it because it is quite simple to understand and/or figure it out by yourself. $\endgroup$
    – Sean
    Commented Jun 30, 2016 at 4:19

1 Answer 1


It took me three days and many pages of calculations, but I think I've solved it.

Lorentz Transformation

  • $\begingroup$ Unfortunately this looks like a case of two wrongs make a right. It doesn't make sense to multiply your "$\gamma$ correction factor equation" by the first two equations because you've already proven that they are contradictory. The contradiction pops up again when you obtain the "correction factor equation" a second time with an extra factor of $\gamma$, which explicitly contradicts the first one unless $\gamma=1$. $\endgroup$
    – adipy
    Commented Jul 4, 2016 at 16:00
  • $\begingroup$ @adipy I've gone over my calculations over and over and I still can't find the "two wrongs that make a right." $\endgroup$ Commented Jul 5, 2016 at 2:22
  • $\begingroup$ In $1 = \frac{(c^2-v^2)}{c^2}\gamma$, you've introduced a gamma factor correction but it's actually the "wrong" one, it should be squared. You then multiply this equation through by your expressions for $(t/t')$ and $(t'/t)$, which are incorrect as you demonstrate in your question (hence the need for the correction factor). These two issues cancel out to give you the correct gamma factor. $\endgroup$
    – adipy
    Commented Jul 5, 2016 at 2:48
  • $\begingroup$ If $t/t'$ is NOT equal to $(c+v)/c$, then you can't multiply an equation by $t/t'$ on the left and $(c+v)/c$ on the right and expect the equality to hold. $\endgroup$
    – adipy
    Commented Jul 5, 2016 at 2:51
  • $\begingroup$ @adipy Still, don't you find it incredibly weird that I managed to calculate the correct answer despite making two "mistakes" that somehow correct themselves in the end? I still can't find where I violated the laws of algebra? $\endgroup$ Commented Jul 5, 2016 at 7:02

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