# Redshift of supernova light curve

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.

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