<p>Yes, we always look into the past, when looking somewhere. There is for instance a mirror on the moon. When sending a laser beam to that mirror, we can detect the reflected light about 2.5 seconds later. This could be interpreted as looking 2.5 seconds into the past, when the laser has been fired. Details <a href="http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment">here</a>.</p>

CC BY-SA 3.0
2014-01-14T20:59:53.093Z
Some quantum tunneling effects (http://en.wikipedia.org/wiki/Quantum_tunnelling) have been discussed to travel faster than light, although weakening the signal; I can't rule out, that this is possible.
Gerald
CC BY-SA 3.0
2014-01-14T21:08:02.970Z
More about that discussion here: http://en.wikipedia.org/wiki/G%C3%BCnter_Nimtz
Gerald
CC BY-SA 3.0
2014-09-21T15:21:25.403Z
0.3 meters per nanosecond.
Wayfaring Stranger
CC BY-SA 3.0
2014-01-14T20:31:24.893Z
2014-01-15T00:49:52.813Z
2014-01-15T00:49:52.813Z
Gerald
<p>You always look into the past. If you look into mirror, you see yourself as you looked moment before. Specifically this is how long before, in seconds where $d$ is the distance to mirror in meters:</p>

<p>$$\frac{d*2}{300000}$$</p>

<p>If such mirror faced earth and was far enough, we would be able to see past indeed. Actually <a href="https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment" rel="noreferrer">there's a tiny mirror facing Earth on Moon</a>.</p>

<p>Also I can't believe nobody posted this one yet:</p>

<p><img src="https://imgs.xkcd.com/comics/geese.png" title="Anyway, that's a common misconception. Geese live for a long time; all the ones we can see will probably keep flying around for billions of years before they explode."></p>

CC BY-SA 3.0
2016-02-09T06:47:17.390Z
2016-02-09T06:47:17.390Z
Tomáš Zato
<p>Here are some thoughts adapted to an answer I placed on Phyiscs SE to a similar question some time ago.  In order to observe the past we need to detect light from the Earth, reflected back to us from somewhere distant in space.</p>

<p>The average albedo of the Earth is about 0.3 (i.e. it reflects 30 percent of the light incident upon it). The amount of incident radiation from the Sun at any moment is the solar constant ($F \sim 1.3 \times 10^3$ Wm$^{-2}$) integrated over a hemisphere. Thus the total reflected light from the Earth is about $L=5\times 10^{16}$ W.</p>

<p>If this light from the Earth has the same spectrum as sunlight and it gets reflected from something which is positioned optimally - i.e. it sees the full illuminated hemisphere. then, roughly speaking, the incident flux on a reflecting body will be $L/2\pi d^2$ (because it is scattered roughly into a hemisphere of the sky).</p>

<p>Now we have to explore some divergent scenarios.</p>

<ol>
<li>There just happens to be a large object at a distance that is highly reflective. I'll use 1000 light years away as an example, which would allow us to see 2000 years into the Earth's past.</li>
</ol>

<p>Let's be generous and say it is a perfect reflector, but we can't assume specular reflection. Instead let's assume the reflected light is also scattered isotropically into a $2\pi$ solid angle. Thus the radiation we get back will be
$$ f = \frac{L}{2\pi d^2} \frac{\pi r^2}{2\pi d^2} = \frac{L r^2}{4\pi d^4},$$
where $r$ is the radius of the thing doing the reflecting.</p>

<p>To turn a flux into an astronomical magnitude we note that the Sun has a visual magnitude of $-26.74$. The apparent magnitude of the reflected light will be given by
$$ m  = 2.5\log_{10} \left(\frac{F}{f}\right) -26.74 = 2.5 \log_{10} \left(\frac{4F \pi d^4}{L r^2}\right) -26.74 $$</p>

<p>So let's put in some numbers. Assume $r=R_{\odot}$ (i.e. a reflector as big as the Sun) and let $d$ be 1000 light years. From this I calculate $m=85$.</p>

<p>To put this in context, the Hubble space telescope ultra deep field has a magnitude limit of around $m=30$ (<a href="http://arxiv.org/abs/1305.1931" rel="nofollow">http://arxiv.org/abs/1305.1931</a> ) and each 5 magnitudes on top of that corresponds to a factor of 100 decrease in brightness. So $m=85$ is about 22 orders of magnitude fainter than detectable by HST. What's worse, the reflector also scatters all the light from the rest of the universe, so picking out the signal from the earth will be utterly futile.</p>

<ol start="2">
<li>A big, flat mirror 1000 light years away.</li>
</ol>

<p>How did it get there? Let's leave that aside. In this case we would just be looking at an image of the Earth as if it were 2000 light years away (assuming everything gets reflected). The flux received back at Earth in this case:
$$ f = \frac{L}{2\pi [2d]^2} $$
with $d=1000$ light years, which will result in an apparent magnitude at the earth of
$m=37$.</p>

<p>OK, this is more promising, but still 7 magnitudes below detection with the HST and perhaps 5 magnitudes fainter than might be detected with the James Webb Space Telescope if and when it does an ultra-deep field. It is unclear whether the sky will be actually full of optical sources at this level of faintness and so even higher spatial resolution than HST/JWST  might be required to pick it out even if we had the sensitivity.</p>

<ol start="3">
<li>Just send a telescope to 1000 light years, observe the Earth, analyse the data and send the signal back to Earth.</li>
</ol>

<p>Of course this doesn't help you see into the past because we would have to send the telescope there. But it could help those in the future see into <em>their</em> past.</p>

<p>Assuming this is technically feasible, the Earth will have a maximum brightness corresponding to $m \sim 35$ so something a lot better than JWST would be required and that ignores the problem of the brightness contrast with the Sun, which would be separated by only 0.03 arcseconds from the Earth at that distance.</p>

<p>Note also that these calculations are merely to detect the light from the <em>whole Earth</em>. To extract anything meaningful would mean collecting a spectrum at the very least! And all this is for only 2000 years into the past.</p>

CC BY-SA 3.0
2015-01-30T23:12:47.063Z
2015-01-30T23:12:47.063Z
ProfRob
<p>In fact, something like such a mirror do exist in the universe. Dust arround progenitor of SN 1572 still reflects light od the outburst. Spectral analysis of the light confirm that the supernova was of Ia type (the fact established long before from light curve of the supernova).</p>

<p><a href="https://arxiv.org/abs/0810.5106" rel="nofollow noreferrer">Tycho Brahe's 1572 supernova as a standard type Ia explosion revealed from its light echo spectrum</a>  </p>

CC BY-SA 4.0
2019-12-30T05:44:28.430Z
`+1` here's another "look back" in time, or "blast from the past": [1](https://astronomy.stackexchange.com/q/14781/7982), [2](https://astronomy.stackexchange.com/q/14660/7982), [3](https://astronomy.stackexchange.com/q/28239/7982)
uhoh
CC BY-SA 4.0
2019-12-25T14:42:27.027Z
2019-12-25T14:42:27.027Z
Leos Ondra
<p>I think the question is referring to situating a very large mirror in space facing earth.  If we were to put it several light minutes away, then events occurring opposite the mirror could be reviewed <em>de novo</em> with more preparation upon the warning we received upon the first light of the event arriving at earth.</p>

<p>For example, a supernova going off in M31 might not be under observation at the moment its light first arrives, and so the initial observations might be lost.  However, with a mirror facing M31, we would be able to observe that mirror as the event unfolded, having been warned that there was something worth watching, in advance.</p>

<p>Nice idea!  But it would likely be much less expensive to simply have multiple telescopes always watching "prime" starscape for unexpected events.</p>

CC BY-SA 3.0
2014-01-15T01:07:46.317Z
That's an interesting interpretation of the question.
Stan Liou
CC BY-SA 3.0
2014-01-15T04:17:53.177Z
My question exactly. So it would work?
ilarsona
CC BY-SA 3.0
2014-01-15T17:37:18.037Z
Of course it would work, @ilarsona.  It would be prohibitively expensive, though.  And a mirror for one direction would block the view to the other.
Cyberherbalist
CC BY-SA 3.0
2014-01-15T18:43:21.813Z
@ilarsona Suppose an advanced extraterrestrial race living 4.5 billion years ago constructed a perfectly flawless mirror of sufficient size 4.5 billion light years from Earth, pointed precisely at Earth. We would today be able to watch the entire history of the Earth play out by observing the Earth in that mirror through our telescopes. Whether such a mirror *could* be constructed is another matter.
called2voyage
CC BY-SA 3.0
2014-01-15T20:03:32.900Z
So what about just a couple cameras then? Sure... hubble and all that, but those work with in the field of light as well... can we record the past?
ilarsona
CC BY-SA 3.0
2014-01-15T22:15:53.263Z
@called2voyage that's a nice idea, but we'd have to wait 4.5 billion years for the first images to reach us.  To be receiving images *now* from our beginning the mirror would need to be half the age of the earth in light-years.  In other words, 2.25 billion light-years away.  It's a two-way trip, after all.
Cyberherbalist
CC BY-SA 3.0
2014-01-16T14:42:07.283Z
@Cyberherbalist Good point, the actual calculations are slightly more complex than I let on, but the general idea is--yes--you could look into the past of Earth in this theoretical scenario.
called2voyage
CC BY-SA 3.0
2014-01-16T14:43:51.323Z
@ilarsona No, the camera would have to already be pointed at us. To send a camera out takes time, and it takes time (the speed of light) for the signal to get back to us so by the time the data reaches us we don't have any information more useful than what we could record here.
called2voyage
CC BY-SA 3.0
2014-01-14T22:09:21.853Z
2014-01-14T22:09:21.853Z
Cyberherbalist
<p>I only ask this because of how fast light travels. The question remains in the title. Why, or why not, would this work? </p>

CC BY-SA 3.0
2014-01-15T00:47:52.430Z
Mirror or not, we are always seeing into the past. When you read a book you are seeing the book as it was a very small fraction of a second ago.
astromax
CC BY-SA 3.0
2014-01-15T16:55:30.717Z
You couldn't use this method to see times before the mirror was put in place. For example, a mirror 1 light-year away would (in principle) let us see Earth as it was 2 years ago -- but it would take more than a year to get it there.
Keith Thompson
CC BY-SA 3.0
2014-11-24T18:09:43.330Z
I can't believe somebody else asked this, I almost finished asking this question when I saw a link to this one.
SSH This
<light><space-time>
By putting a mirror in space, would we be able to see into the past?
ilarsona
# By putting a mirror in space, would we be able to see into the past?

(Asked by: username_0 on 2014-01-14)

I only ask this because of how fast light travels. The question remains in the title. Why, or why not, would this work?

**Tags:** light, space-time

> **Comments:**
> **username_1**: Mirror or not, we are always seeing into the past. When you read a book you are seeing the book as it was a very small fraction of a second ago.  
> **username_2**: You couldn't use this method to see times before the mirror was put in place. For example, a mirror 1 light-year away would (in principle) let us see Earth as it was 2 years ago -- but it would take more than a year to get it there.  
> **username_3**: I can't believe somebody else asked this, I almost finished asking this question when I saw a link to this one.  

---

## 5 answers

### Answer by username_10 on 2014-01-14. Score: 15 | Accepted answer

I think the question is referring to situating a very large mirror in space facing earth. If we were to put it several light minutes away, then events occurring opposite the mirror could be reviewed *de novo* with more preparation upon the warning we received upon the first light of the event arriving at earth.

For example, a supernova going off in M31 might not be under observation at the moment its light first arrives, and so the initial observations might be lost. However, with a mirror facing M31, we would be able to observe that mirror as the event unfolded, having been warned that there was something worth watching, in advance.

Nice idea! But it would likely be much less expensive to simply have multiple telescopes always watching "prime" starscape for unexpected events.

> **Comments:**
> **username_11**: That's an interesting interpretation of the question.  
> **username_0**: My question exactly. So it would work?  
> **username_10**: Of course it would work, @username_0.  It would be prohibitively expensive, though.  And a mirror for one direction would block the view to the other.  
> **username_12**: @username_0 Suppose an advanced extraterrestrial race living 4.5 billion years ago constructed a perfectly flawless mirror of sufficient size 4.5 billion light years from Earth, pointed precisely at Earth. We would today be able to watch the entire history of the Earth play out by observing the Earth in that mirror through our telescopes. Whether such a mirror *could* be constructed is another matter.  
> **username_0**: So what about just a couple cameras then? Sure... hubble and all that, but those work with in the field of light as well... can we record the past?  
> **username_10**: @username_12 that's a nice idea, but we'd have to wait 4.5 billion years for the first images to reach us.  To be receiving images *now* from our beginning the mirror would need to be half the age of the earth in light-years.  In other words, 2.25 billion light-years away.  It's a two-way trip, after all.  
> **username_12**: @username_10 Good point, the actual calculations are slightly more complex than I let on, but the general idea is--yes--you could look into the past of Earth in this theoretical scenario.  
> **username_12**: @username_0 No, the camera would have to already be pointed at us. To send a camera out takes time, and it takes time (the speed of light) for the signal to get back to us so by the time the data reaches us we don't have any information more useful than what we could record here.  

---

### Answer by username_4 on 2014-01-14. Score: 10

Yes, we always look into the past, when looking somewhere. There is for instance a mirror on the moon. When sending a laser beam to that mirror, we can detect the reflected light about 2.5 seconds later. This could be interpreted as looking 2.5 seconds into the past, when the laser has been fired. Details [here](http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment).

> **Comments:**
> **username_4**: Some quantum tunneling effects (http://en.wikipedia.org/wiki/Quantum_tunnelling) have been discussed to travel faster than light, although weakening the signal; I can't rule out, that this is possible.  
> **username_4**: More about that discussion here: http://en.wikipedia.org/wiki/G%C3%BCnter_Nimtz  
> **username_5**: 0.3 meters per nanosecond.  

---

### Answer by username_6 on 2016-02-09. Score: 5

You always look into the past. If you look into mirror, you see yourself as you looked moment before. Specifically this is how long before, in seconds where $d$ is the distance to mirror in meters:

$$\frac{d\*2}{300000}$$

If such mirror faced earth and was far enough, we would be able to see past indeed. Actually [there's a tiny mirror facing Earth on Moon](https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment).

Also I can't believe nobody posted this one yet:

![](https://imgs.xkcd.com/comics/geese.png "Anyway, that's a common misconception. Geese live for a long time; all the ones we can see will probably keep flying around for billions of years before they explode.")


---

### Answer by username_7 on 2015-01-30. Score: 4

Here are some thoughts adapted to an answer I placed on Phyiscs SE to a similar question some time ago. In order to observe the past we need to detect light from the Earth, reflected back to us from somewhere distant in space.

The average albedo of the Earth is about 0.3 (i.e. it reflects 30 percent of the light incident upon it). The amount of incident radiation from the Sun at any moment is the solar constant ($F \sim 1.3 \times 10^3$ Wm$^{-2}$) integrated over a hemisphere. Thus the total reflected light from the Earth is about $L=5\times 10^{16}$ W.

If this light from the Earth has the same spectrum as sunlight and it gets reflected from something which is positioned optimally - i.e. it sees the full illuminated hemisphere. then, roughly speaking, the incident flux on a reflecting body will be $L/2\pi d^2$ (because it is scattered roughly into a hemisphere of the sky).

Now we have to explore some divergent scenarios.

1. There just happens to be a large object at a distance that is highly reflective. I'll use 1000 light years away as an example, which would allow us to see 2000 years into the Earth's past.

Let's be generous and say it is a perfect reflector, but we can't assume specular reflection. Instead let's assume the reflected light is also scattered isotropically into a $2\pi$ solid angle. Thus the radiation we get back will be
$$ f = \frac{L}{2\pi d^2} \frac{\pi r^2}{2\pi d^2} = \frac{L r^2}{4\pi d^4},$$
where $r$ is the radius of the thing doing the reflecting.

To turn a flux into an astronomical magnitude we note that the Sun has a visual magnitude of $-26.74$. The apparent magnitude of the reflected light will be given by
$$ m = 2.5\log\_{10} \left(\frac{F}{f}\right) -26.74 = 2.5 \log\_{10} \left(\frac{4F \pi d^4}{L r^2}\right) -26.74 $$

So let's put in some numbers. Assume $r=R\_{\odot}$ (i.e. a reflector as big as the Sun) and let $d$ be 1000 light years. From this I calculate $m=85$.

To put this in context, the Hubble space telescope ultra deep field has a magnitude limit of around $m=30$ (<http://arxiv.org/abs/1305.1931> ) and each 5 magnitudes on top of that corresponds to a factor of 100 decrease in brightness. So $m=85$ is about 22 orders of magnitude fainter than detectable by HST. What's worse, the reflector also scatters all the light from the rest of the universe, so picking out the signal from the earth will be utterly futile.

2. A big, flat mirror 1000 light years away.

How did it get there? Let's leave that aside. In this case we would just be looking at an image of the Earth as if it were 2000 light years away (assuming everything gets reflected). The flux received back at Earth in this case:
$$ f = \frac{L}{2\pi [2d]^2} $$
with $d=1000$ light years, which will result in an apparent magnitude at the earth of
$m=37$.

OK, this is more promising, but still 7 magnitudes below detection with the HST and perhaps 5 magnitudes fainter than might be detected with the James Webb Space Telescope if and when it does an ultra-deep field. It is unclear whether the sky will be actually full of optical sources at this level of faintness and so even higher spatial resolution than HST/JWST might be required to pick it out even if we had the sensitivity.

3. Just send a telescope to 1000 light years, observe the Earth, analyse the data and send the signal back to Earth.

Of course this doesn't help you see into the past because we would have to send the telescope there. But it could help those in the future see into *their* past.

Assuming this is technically feasible, the Earth will have a maximum brightness corresponding to $m \sim 35$ so something a lot better than JWST would be required and that ignores the problem of the brightness contrast with the Sun, which would be separated by only 0.03 arcseconds from the Earth at that distance.

Note also that these calculations are merely to detect the light from the *whole Earth*. To extract anything meaningful would mean collecting a spectrum at the very least! And all this is for only 2000 years into the past.


---

### Answer by username_8 on 2019-12-25. Score: 2

In fact, something like such a mirror do exist in the universe. Dust arround progenitor of SN 1572 still reflects light od the outburst. Spectral analysis of the light confirm that the supernova was of Ia type (the fact established long before from light curve of the supernova).

[Tycho Brahe's 1572 supernova as a standard type Ia explosion revealed from its light echo spectrum](https://arxiv.org/abs/0810.5106)

> **Comments:**
> **username_9**: `+1` here's another "look back" in time, or "blast from the past": [1](https://astronomy.stackexchange.com/q/14781/7982), [2](https://astronomy.stackexchange.com/q/14660/7982), [3](https://astronomy.stackexchange.com/q/28239/7982)  

---

