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Redshift of supernova light curve

  1. Mar 26, 2015 #1
    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 [itex]k[/itex] 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:
    $$f'=kf$$
    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 [itex]k[/itex] the light curve transforms in the following way:
    $$\large e^{-\alpha t^2} \rightarrow e^{-\alpha k^2t^2}$$
    Is this correct?
     
    Last edited: Mar 26, 2015
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  3. Mar 26, 2015 #2

    Chalnoth

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    It's much simpler than that. It's just time dilation. The factor of redshift is also the time dilation factor. Intuitively this should make sense: if you imagine viewing an atomic clock, and we use the frequency of light observed from the clock to count time, then a redshifted clock will appear to be slowed by exactly the factor of the redshift.

    So yes, I'm pretty sure your math is correct.
     
  4. Mar 26, 2015 #3
    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?
     
  5. Mar 26, 2015 #4

    Chalnoth

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    Not quite. It's purely a perspective effect. We view the far-away object time-dilated.
     
  6. Mar 27, 2015 #5

    PeterDonis

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    "Time dilation" is a general term; it can refer to any situation in which one clock appears to be ticking slower relative to another clock. Note, though, that the "appears" is relative to a choice of coordinates; "time dilation" is never directly observed, it's always calculated. The cosmological redshift, like any redshift, is directly observed. So I'm not sure I would say the redshift is "equivalent" to time dilation. Given the observed redshift and a choice of coordinates, one can calculate the time dilation for the observed object.

    Not with the "actually" included.
     
  7. Mar 27, 2015 #6

    Chalnoth

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    Right. But in this case, the cosmological redshift is exactly the same as the time dilation of the far-away object, as observed by us. That is to say, if ##z=1##, then we will see clocks on that far-away object ticking at half the speed.

    One way to understand that this has to be the case is to consider a hypothetical scenario where we have a source emitting a continuous stream of photons at a specific wavelength. At the source, there's a clock. Every time this clock ticks, the light source's emission goes through 1000 oscillations.

    If the redshift of this object is ##z=1##, how long will it take for an observer on Earth to observe 1000 oscillations of the incoming light wave? The answer: exactly two ticks of an identical clock placed on Earth.

    This does get a bit confusing when considering the Doppler effect, because usually the redshift/blueshift from the moving object is corrected for when we estimate the "time dilation" of that object. But in the case of the cosmological redshift, there is no Doppler shift to confuse definitions. The time dilation is exactly the redshift.
     
  8. Mar 27, 2015 #7

    PeterDonis

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    No, this isn't "time dilation", because "time dilation" is supposed to correct for the change in light travel time due to the relative velocity. For example, in flat spacetime, if an object is receding at 0.87c, the observed Doppler redshift factor is 4, but the time dilation factor is only 2; the object's clock is actually observed to be running 4 times slower, but when those observations are corrected for light travel time, the object's clock is only running twice as slow--its proper time "ticks" twice as slowly as coordinate time.

    For cosmological redshift, how this calculation is done depends on your choice of coordinates. (I assumed the "natural" choice of inertial coordinates in which the observer is at rest for the flat spacetime case above.) If you use standard FRW coordinates, the time dilation is zero if both objects are "comoving"; all "comoving" objects have proper time equal to FRW coordinate time. If you choose other coordinates, you can make the "time dilation" of the faraway object be nonzero; for example, if you choose "inertial" coordinates centered on the observer, the calculation would be the same as above. But in this case, the coordinates may only cover a limited portion of spacetime; for example, these "inertial" coordinates won't cover anything beyond the Hubble radius.

    My personal preference would be to not use the term "time dilation" at all, since it's coordinate-dependent anyway. If we're talking about the redshift factor, which is what's actually involved in the supernove light curve observations, we can just say "redshift factor".
     
  9. Mar 27, 2015 #8

    Chalnoth

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    If you're talking about the actual observed image of the clock, it will always tick slower by exactly the factor of redshift.
     
  10. Mar 27, 2015 #9

    PeterDonis

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    Yes, agreed. My point is that that is not the same as "time dilation", which takes the observed redshift factor and corrects it for light travel time in order to calculate the clock's rate in some chosen coordinate chart. So, in the flat spacetime example I gave, the redshift factor is 4 but the time dilation factor is only 2. Mathematically, for this case, the redshift factor is ##\sqrt{(1 - v) / (1 + v)}##, and the time dilation factor is ##1 / \sqrt{1 - v^2}##. (Note that I should have said the Doppler factor is "about" 4 for this case, for ##v = \sqrt{3} / 2## .)
     
    Last edited: Mar 27, 2015
  11. Mar 27, 2015 #10

    Chalnoth

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    Well, I do admit that I was being a bit cavalier with my notation. I certainly wasn't talking about the time dilation in special relativity, but then special relativity does not apply in this instance. I was specifically talking about how rapidly the image of the far-away "clock" appears to be ticking to us. That is well-defined.
     
  12. Mar 27, 2015 #11

    PeterDonis

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    Yes, agreed. I would call that the "redshift factor". There is no invariant way to translate this into a "time dilation", since the latter depends on the choice of coordinates.
     
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