<p>Photons can't have a perspective.</p>

<p>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.  </p>

<p>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.</p>

CC BY-SA 4.0
2020-02-23T23:28:11.830Z
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).
userLTK
CC BY-SA 4.0
2020-02-24T13:30:02.743Z
@userLTK: Sorry, but what would the frame of reference be ? The photon is red-shifted **for us**, but that's in **our** reference frame.
MSalters
CC BY-SA 4.0
2020-02-24T18:49:55.050Z
Two dimensions?
chepner
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2020-02-24T19:04:59.687Z
@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.
userLTK
CC BY-SA 4.0
2020-02-24T19:09:49.043Z
Ah right. I always (probably incorrectly) picture this as the photon having a fixed position in a 4d space, with the entire history of the universe existing "around" the photon, rather than unfolding as the photon moves in any particular direction.
chepner
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2020-02-25T00:50:07.183Z
The concept of frequency doesn't make sense from the photon's perspective.
CJ Dennis
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2020-02-25T03:02:55.033Z
@chepner The Lorentz boost is how you translate between the two perspective. Both are equally valid.
Aron
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2020-02-25T08:36:53.213Z
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.
Peter - Reinstate Monica
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2020-02-25T09:52:27.707Z
@MSalters counting the upvotes, I'm sure you're right, I just have a hard time picturing it.
userLTK
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2020-02-25T12:39:19.700Z
@James K: Refer to the "twins paradox". From the photon's perspective, it does not age AT ALL.
Vincent Otieno
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2020-02-25T15:39:48.160Z
This is a great answer! A [follow-up question](https://physics.stackexchange.com/questions/533101/how-does-a-photon-perceive-the-2-d-plane-in-which-it-perceives-itself) addresses how the photon 'perceives' the 2-D space in which it perceives itself.
dotancohen
CC BY-SA 4.0
2020-02-25T19:10:03.747Z
From the reference frame of a photon it has T=0. That means that the photon experiences no passage of time from the moment it is created to the moment of measurement.
Dean Pitts
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2020-02-25T21:17:59.580Z
@VincentOtieno et. al. The point of this answer is to say that the words "From the photon's perspective" have no meaning.  You can't have a frame of reference in which the photon is at rest.
James K
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2023-10-22T10:49:02.053Z
@JamesK But is it not true that 'photons' do not travel through space. It is energy that 'moves through the medium' of space (as water does not move, only the energy _in_ the water 'moves'). Is it not true that we only see 'particle characteristics' when we cause collapse of the wave form due to our interference of measuring ? (Or have I misunderstood ?) Therefore the 'photon' does not have an 'age'. It is only the movement of the energy that took a measurable (or immeasurable) amount of time to travel.
Nigel J
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2023-10-22T11:08:08.670Z
Not just "energy" but a quantum of the electromagnetic field. The same is true of electrons.  Just as a photon is an quantum excitation of the electromagnetic field, an electron is a quantum of the Dirac field.  So would you say an electron "moves" down a wire, or out of an electron gun. Or Muons... but muons certainly age, since they decay... It's certainly the case that "everything is fields" but it doesn't follow that "nothing has an age, since everything is a field, or an interaction of fields.
James K
CC BY-SA 4.0
2020-02-23T18:34:50.860Z
2020-02-23T18:34:50.860Z
James K
<p>There is a quantity in relativity of <span class="math-container">$s^2$</span> which is defined as <span class="math-container">$t^2-x^2$</span>, where <span class="math-container">$t$</span> is the difference in time between two events, and <span class="math-container">$x$</span> is the difference in position (measured in units such that <span class="math-container">$c=1$</span>). If <span class="math-container">$s^2$</span> is positive, then the square root of it is the proper time. If it's negative, then the square root of <span class="math-container">$-s^2$</span> 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).</p>

<p>Proper time is how much time is experienced by an object between the two events. We can write <span class="math-container">$x = vt$</span>, where <span class="math-container">$v$</span> is the velocity of the object, and then we have <span class="math-container">$s = \sqrt {t^2-(vt)^2}$</span>, and we can factor out the <span class="math-container">$t$</span> to get <span class="math-container">$s = t \sqrt {1-v^2}$</span>. This is the time dilation: the larger <span class="math-container">$v$</span> is, the less proper time we have for a fixed <span class="math-container">$t$</span>. If <span class="math-container">$v=c$</span>, then we have <span class="math-container">$s = t \sqrt {1-1}$</span> (remember, we have units such that <span class="math-container">$c = 1$</span>), 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). </p>

<p>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 <span class="math-container">$c$</span> and have zero proper time. But oscillations take place over time, so a particle travelling at <span class="math-container">$c$</span> would not be able to oscillate.</p>

CC BY-SA 4.0
2020-02-25T07:10:15.960Z
2020-02-25T07:10:15.960Z
Acccumulation
<p>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?</p>

<p>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.</p>

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

<p>Because the photon was emitted by distant star, it will expirience red shift according to Hubble's law <span class="math-container">$v=H\cdot L$</span>. 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.</p>

CC BY-SA 4.0
2020-02-25T17:45:41.503Z
2020-02-25T17:45:41.503Z
sanaris
<p>This depends on who exactly is doing the measuring.</p>

<p>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 <em>ageless</em>, and that is that neither photons, nor any other kind of massless particles, undergo <em>any kind of internal change or evolution.</em></p>

<p>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 <em>to</em> the idea of "ageless photons": you cannot have a "point of view" of a photon is <em>because</em> 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.</p>

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

<p><span class="math-container">$$\tau = \int_\gamma ds = \int_\gamma \sqrt{dt^2 - \frac{1}{c^2}\left(dx^2 + dy^2 + dz^2\right)}$$</span></p>

<p>. 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,  <span class="math-container">$\tau = 0$</span> always - <strong>they are ageless.</strong></p>

<p><em>But</em>, of course, from the viewpoint of <em>someone on Earth</em>, 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 <span class="math-container">$t$</span>, and that's the one you usually hear about, and by that standard, the photons are <strong>1 year old.</strong> 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.</p>

<p>If we choose this route, to take a disciplined <em>general</em>-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 <em>round-trip</em> time for sending a photon out and then back <em>is two years</em>, and there is no point in talking "now" unless you define for me which one out of <span class="math-container">$\beth_1$</span> possibilities you want for me as the one you'd like to use.</p>

<p>So the answer to your question "<em>Are photons aged?</em>" is:</p>

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

CC BY-SA 4.0
2020-02-26T12:17:00.583Z
2020-02-28T06:52:01.963Z
2020-02-28T06:52:01.963Z
The_Sympathizer
<p>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</p>

CC BY-SA 4.0
2020-02-27T00:56:57.367Z
2020-02-27T00:56:57.367Z
John
<p>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?</p>

CC BY-SA 4.0
2020-02-25T10:49:50.953Z
I've just asked [Have there been studies of "old photons" to see just how constant Planck constant has been?](https://astronomy.stackexchange.com/q/35263/7982)
uhoh
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2020-02-25T13:52:47.053Z
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 ?
Overmind
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2020-02-26T13:18:00.297Z
@Overmind more importantly, photon cannot experience any time thanks to time dilation.
Tomáš Zato
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2020-02-26T13:41:30.827Z
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.
Overmind
CC BY-SA 4.0
2020-02-26T16:48:23.497Z
Scusi, comment under wrong question.
rackandboneman
<light><speed><photons><age>
Are photons aged?
Emir
# Are photons aged?

(Asked by: username_0 on 2020-02-23)

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?

**Tags:** light, speed, photons, age

> **Comments:**
> **username_1**: I've just asked [Have there been studies of "old photons" to see just how constant Planck constant has been?](https://astronomy.stackexchange.com/q/35263/7982)  
> **username_2**: 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 ?  
> **username_3**: @username_2 more importantly, photon cannot experience any time thanks to time dilation.  
> **username_2**: 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.  
> **username_4**: Scusi, comment under wrong question.  

---

## 5 answers

### Answer by username_5 on 2020-02-23. Score: 72

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.

> **Comments:**
> **username_6**: 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).  
> **username_7**: @username_6: Sorry, but what would the frame of reference be ? The photon is red-shifted **for us**, but that's in **our** reference frame.  
> **username_8**: Two dimensions?  
> **username_6**: @username_8 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.  
> **username_8**: Ah right. I always (probably incorrectly) picture this as the photon having a fixed position in a 4d space, with the entire history of the universe existing "around" the photon, rather than unfolding as the photon moves in any particular direction.  
> **username_9**: The concept of frequency doesn't make sense from the photon's perspective.  
> **username_10**: @username_8 The Lorentz boost is how you translate between the two perspective. Both are equally valid.  
> **username_11**: 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.  
> **username_6**: @username_7 counting the upvotes, I'm sure you're right, I just have a hard time picturing it.  
> **username_12**: @username_5: Refer to the "twins paradox". From the photon's perspective, it does not age AT ALL.  
> **username_13**: This is a great answer! A [follow-up question](https://physics.stackexchange.com/questions/533101/how-does-a-photon-perceive-the-2-d-plane-in-which-it-perceives-itself) addresses how the photon 'perceives' the 2-D space in which it perceives itself.  
> **username_14**: From the reference frame of a photon it has T=0. That means that the photon experiences no passage of time from the moment it is created to the moment of measurement.  
> **username_5**: @username_12 et. al. The point of this answer is to say that the words "From the photon's perspective" have no meaning.  You can't have a frame of reference in which the photon is at rest.  
> **username_15**: @username_5 But is it not true that 'photons' do not travel through space. It is energy that 'moves through the medium' of space (as water does not move, only the energy _in_ the water 'moves'). Is it not true that we only see 'particle characteristics' when we cause collapse of the wave form due to our interference of measuring ? (Or have I misunderstood ?) Therefore the 'photon' does not have an 'age'. It is only the movement of the energy that took a measurable (or immeasurable) amount of time to travel.  
> **username_5**: Not just "energy" but a quantum of the electromagnetic field. The same is true of electrons.  Just as a photon is an quantum excitation of the electromagnetic field, an electron is a quantum of the Dirac field.  So would you say an electron "moves" down a wire, or out of an electron gun. Or Muons... but muons certainly age, since they decay... It's certainly the case that "everything is fields" but it doesn't follow that "nothing has an age, since everything is a field, or an interaction of fields.  

---

### Answer by username_16 on 2020-02-25. Score: 8

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.


---

### Answer by username_17 on 2020-02-25. Score: 3

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.


---

### Answer by username_18 on 2020-02-26. Score: 3

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**.


---

### Answer by username_19 on 2020-02-27. Score: 0

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


---

