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I have seen this question but I still have problem with the meaning of this symbol. From this book:

The time-dependent angle may be defined from the components of the wave vector in order to determine its sense of rotation
$$\tan\xi(t)=\frac{E_y(z_0,t)}{E_x(z_0,t)}=\frac{E_{0y}\cos(\omega t-kz_0+\delta_y)}{E_{0x}\cos (\omega t-kz_0+\delta_x)}$$ The sense of rotation may then be related to the sign of the ellipticity $\tau$, with $$\frac{\partial \xi(t)}{\partial t}\propto-\sin\delta\Rightarrow \text{sign}\left(\frac{\partial\xi(t)}{\partial t}\right)=-\text{sign}(\tau)$$

What does the symbol '$\propto$' mean in this context?

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  • $\begingroup$ Mightn't it mean "proportional to," as it often does? $\endgroup$
    – Brian Tung
    Aug 25, 2015 at 19:49
  • $\begingroup$ Your link explains it, does it not ? $\endgroup$
    – Dietrich Burde
    Aug 25, 2015 at 19:50
  • $\begingroup$ @DietrichBurde Does it mean that $\frac{\partial \xi(t)}{\partial t}$ is proportional to $-\sin\delta$? Then how do we conclude from that $\text{sign}\left(\frac{\partial\xi(t)}{\partial t}\right)=-\text{sign}(\tau)$? $\endgroup$
    – Sepideh Abadpour
    Aug 25, 2015 at 21:00
  • $\begingroup$ That will probably be hard to say without looking at the original. The likelihood that someone will scan the ~400 pages of the book for that equation when you could have just provided the page number is rather low. $\endgroup$
    – joriki
    Aug 26, 2015 at 5:30
  • $\begingroup$ ok @joriki It's on page 36 $\endgroup$
    – Sepideh Abadpour
    Aug 26, 2015 at 6:40

1 Answer 1

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This symbol means "proportional to". Generally it might be taken to include the case of a negative constant of proportionality, but here the minus sign would seem to indicate that the intended meaning is with a positive constant of proportionality. The right-hand side of the implication then follows in view of equation $(2.20)$ in the book, taking into account that $|\tau|\in[0,\frac\pi4]$ and that $E_{0x}$ and $E_{0y}$ are amplitudes and hence positive.

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