I am trying to understand how the width of a supernova light curve depends on the redshift of its component frequencies.

Let us make the simple assumption that the light curve is Gaussian. The inverse Fourier transform of a Gaussian is given by:

$$\large e^{-\alpha t^2}=\int_{-\infty}^{\infty}\sqrt{\frac{\pi}{\alpha}}e^{-\frac{(\pi f)^2}{\alpha}}e^{2\pi ift}\ df$$

Now if all the components of the light curve are redshifted by a factor $k$ then I think the right-hand side of the above equation becomes:

$$\large \int_{-\infty}^{\infty}\sqrt{\frac{\pi}{\alpha}}e^{-\frac{(\pi f)^2}{\alpha}}e^{2\pi ikft}\ df$$

I now change variables in the integral using:

$$f'=kf$$ $$df'=k\ df$$

The above integral becomes the inverse Fourier transform of a modified Gaussian curve:

$$\large \int_{-\infty}^{\infty}\sqrt{\frac{\pi}{\alpha k^2}}e^{-\frac{(\pi f')^2}{\alpha k^2}}e^{2\pi if't}\ df'$$

Thus it seems that if the components are redshifted by a factor $k$ the light curve transforms in the following way:

$$\large e^{-\alpha t^2} \rightarrow e^{-\alpha k^2t^2}$$

Is this correct?

PS I now accept that the above calculation is correct and shows that redshift has the same effect as time dilation.

  • $\begingroup$ Why would you assume a Gaussian shape? Explosions are rather asymmetric in time. $\endgroup$ – Carl Witthoft Mar 26 '15 at 14:57
  • $\begingroup$ I'm just using a Gaussian as an example to understand how the width of a light pulse changes due to redshift. $\endgroup$ – John Eastmond Mar 26 '15 at 15:44

The width of a SN light curve is changed due to a time dilation between the source and the observer. If the source emits light with wavelength $\lambda_\text{em}$, it will be observed with wavelength $\lambda_\text{ob}$, so that its redshift is $$ 1 + z = \frac{\lambda_\text{ob}}{\lambda_\text{em}}. $$ We can also write everything in terms of frequencies $\nu_\text{em}=c/\lambda_\text{em}$ and $\nu_\text{ob}=c/\lambda_\text{ob}$:

$$ 1 + z = \frac{\nu_\text{em}}{\nu_\text{ob}}. $$ Now, suppose that the source emits $N = \nu_\text{em}\,\delta t_\text{em}$ oscillations in a time interval $\delta t_\text{em}$, then these same oscillations will be observed in a time interval $\delta t_\text{ob}$, such that $N = \nu_\text{em}\,\delta t_\text{em} = \nu_\text{ob}\,\delta t_\text{ob}$. Therefore $$ 1 + z = \frac{\delta t_\text{ob}}{\delta t_\text{em}}. $$ In other words, any time interval $\delta t_\text{em}$ is dilated into $\delta t_\text{ob}$.

  • $\begingroup$ So is it true to say that cosmological redshift is completely equivalent to time dilation? Could one consistently assert that an atomic clock ticking millions of years ago is actually slower than the equivalent clock today? $\endgroup$ – John Eastmond Mar 26 '15 at 22:43
  • $\begingroup$ @JohnEastmond If we could observe the second hand from that distance it would appear to moving slower. Not the same thing. Pulsar's answer is good and this is a major piece of evidence that redshift is due to cosmological expansion. Tired-light type hypotheses cannot explain time-dilated supernova light curves. physics.stackexchange.com/q/156618 $\endgroup$ – Rob Jeffries Mar 27 '15 at 0:25

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