If a star is at a distance of one lightyear, how old are its photons when they reach earth (from the photons’ perspective)? If time dilation is near zero at light speed, can we assume that the light that we see today from a distant star has the same age as when it was emitted?

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    $\begingroup$ I've just asked Have there been studies of "old photons" to see just how constant Planck constant has been? $\endgroup$
    – uhoh
    Commented Feb 25, 2020 at 10:49
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    $\begingroup$ How would you define "old" ? A photon will live as long as it can maintain its inner structural integrity. Why would one need to estimate its age ? $\endgroup$
    – Overmind
    Commented Feb 25, 2020 at 13:52
  • $\begingroup$ @Overmind more importantly, photon cannot experience any time thanks to time dilation. $\endgroup$ Commented Feb 26, 2020 at 13:18
  • $\begingroup$ This is another topic of which current view I do not agree with. A photon's speed can be altered (increased). This is a long talk not for this topic. $\endgroup$
    – Overmind
    Commented Feb 26, 2020 at 13:41
  • $\begingroup$ Scusi, comment under wrong question. $\endgroup$ Commented Feb 26, 2020 at 16:48

5 Answers 5


Photons can't have a perspective.

If we have a particle with mass, we can imagine taking a frame of reference in which that particle is at rest. We can then see things "from the particle's perspective". But there is no frame of reference in which a photon is at rest. Photons always move at the speed of light in every frame of reference.

If I try to set up a frame of reference which is moving at the speed of light there is a singularity. The universe has no time, and the whole of space is squashed into two dimensions. So in a very real way, a photon doesn't have a perspective. We can only consider time in a frame of reference that includes it. It makes perfect sense to say that the photons are one year old in our frame of reference. And that is the best we can say.

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    $\begingroup$ But photon's get red shifted and lose energy passing through dark energy expansion, so . . . that's kind of a frame of reference. (just tossing a curve-ball. I like your answer). $\endgroup$
    – userLTK
    Commented Feb 23, 2020 at 23:28
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    $\begingroup$ @userLTK: Sorry, but what would the frame of reference be ? The photon is red-shifted for us, but that's in our reference frame. $\endgroup$
    – MSalters
    Commented Feb 24, 2020 at 13:30
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    $\begingroup$ Two dimensions? $\endgroup$
    – chepner
    Commented Feb 24, 2020 at 18:49
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    $\begingroup$ @chepner at the speed of light, length contraction becomes infinite. It's like space is squeezed into 2 dimensions, or, another way to look at it, no time passes for the photon between when the photon leaves one object and hits another, even if they're light-years apart. $\endgroup$
    – userLTK
    Commented Feb 24, 2020 at 19:04
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    $\begingroup$ The universe being 2-dimensional from a photon's "perspective" implies that it is "everywhere" on its path "at once" (well, there is no "time", is there), which is quite interesting in terms of the "connectedness" of the universe. $\endgroup$ Commented Feb 25, 2020 at 8:36

There is a quantity in relativity of $s^2$ which is defined as $t^2-x^2$, where $t$ is the difference in time between two events, and $x$ is the difference in position (measured in units such that $c=1$). If $s^2$ is positive, then the square root of it is the proper time. If it's negative, then the square root of $-s^2$ is the proper distance. (BTW, proper times/distances are invariant under Lorentz transformations, that is, they are the same in every inertial frame of reference).

Proper time is how much time is experienced by an object between the two events. We can write $x = vt$, where $v$ is the velocity of the object, and then we have $s = \sqrt {t^2-(vt)^2}$, and we can factor out the $t$ to get $s = t \sqrt {1-v^2}$. This is the time dilation: the larger $v$ is, the less proper time we have for a fixed $t$. If $v=c$, then we have $s = t \sqrt {1-1}$ (remember, we have units such that $c = 1$), so the proper time is zero. So there is no proper time between the emission and absorption of a photon (apart from some complications, such as a photon traveling through a transparent medium).

So if you accept proper time as "age", then a photon has no age. If you have a different definition, then it could have an age. In this sense, photon does not "experience" time, and there is no "from the photon's perspective". A example of the implications of this is neutrino oscillation. Neutrino have three different flavors, and they will oscillate between them. It was once unknown whether neutrinos have mass, but the fact that neutrinos oscillate necessitate that they have mass. If they didn't have mass, they would travel at $c$ and have zero proper time. But oscillations take place over time, so a particle travelling at $c$ would not be able to oscillate.


Even if you substitute the definition of proper time by "distance traveled", you cannot guarantee you got that distance right. Suppose photon was absorbed by hydrogen interstellar gas, then re-emitted with different wavelength, was it a new photon or an old one?

Time has no change for system associated with photon. So technically speaking, the photons you have, which were not collided with anything, would be of same age as the star itself.

But, when scientists introduce the concept of "photon age", they mostly consider it the function of its wavelength. So photons emitted near red zone of black hole will be losing their energy $E=h \omega$, have increased wavelenght $\lambda=c\cdot 2\pi/\omega$ up to infinite wavelenght (when you start at $g_{00}=0$). In other words, such photons will have "infinite age".

Because the photon was emitted by distant star, it will expirience red shift according to Hubble's law $v=H\cdot L$. This red shift could be instead associated with "photon aging". Which will have exact same meaning that entire Universe have expanding tendency with curvature according to General Relativity. So "photon aging" in that sense will be the same as space-time curvature already considered in GR, which will be the duplicate.


This depends on who exactly is doing the measuring.

There is a very real sense, despite the answers here perhaps either downplaying or not quite getting directly at it enough, in which you can reasonably say that the photons that have been emitted are "frozen" and ageless, and that is that neither photons, nor any other kind of massless particles, undergo any kind of internal change or evolution.

The other answers are right in suggesting there isn't such a thing as a "perspective" from a photon - but in fact, the reasons for it lend weight to the idea of "ageless photons": you cannot have a "point of view" of a photon is because a photon cannot undergo any internal change, and the reason it does that is that it does not age (in a very broad physical sense). For you to "experience", your material organism needs to undergo some sort of internal dynamism - such as the neuronal firings that form the information processing patterns within your brain which, at the very least, allows you to mark time and mark the receipt of information from outside sources. But a photon cannot do either.

Mathematically, the concept of photons "not aging" can be given in terms of the proper time metric of their worldline: the "length", or metric, of a curve $\gamma$ between any two points in space-time, under the usual Minkowski coordinates $(t, x, y, z)$, is

$$\tau = \int_\gamma ds = \int_\gamma \sqrt{dt^2 - \frac{1}{c^2}\left(dx^2 + dy^2 + dz^2\right)}$$

. This in, in fact, the closest thing to what I think you're asking about: while it's not a "perspective", it is an "absolute" measure of "how much something has aged" during its transit between two points in space-time. And for photons, $\tau = 0$ always - they are ageless.

But, of course, from the viewpoint of someone on Earth, then this depends on our simultaneity standard. In Minkowski spacetime there is a natural - but not exclusive - one to use, which is that of the "Minkowski coordinates" I just mentioned with a set value of $t$, and that's the one you usually hear about, and by that standard, the photons are 1 year old. From the viewpoint of general-relativistic thinking though, which, by the way, is actually the more "complete" theory of relativistic mechanics and doesn't just fail in its lessons if we consider the flat spacetime case, there actually isn't anything essential about the Minkowski coordinates. It's the spacetime metric that I mentioned above, that counts, and hence there isn't a fully honestly non-arbitrary sense at which "simultaneity" even makes sense at all - it's better to talk of "causally connected" and "causally disconnected.": whether two space-time points can send a message from one to the other.

If we choose this route, to take a disciplined general-relativity consciously aware view of the situation, the statements we can make are: the emission of photons is "just barely" causally connected to their reception at Earth, the interval crossed is zero so the photons are ageless "from their own [non-]point of view", and the round-trip time for sending a photon out and then back is two years, and there is no point in talking "now" unless you define for me which one out of $\beth_1$ possibilities you want for me as the one you'd like to use.

So the answer to your question "Are photons aged?" is:

With a little adjustment to the definitions to get at what I bet you're really after, no. But if we choose to become a bit more explorative, yes, aged one year. And then even moreso, on the last hand, it doesn't make sense.


No, at the speed of light time is stationary I think Einstein says and it will only degrade when interacting with something else.And as we know a photon will continue forever in a vacuum if unhindered and if so there can be no ageing


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