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Here's a thought experiment I came up with in class today when my mind drifted (I however highly doubt I'm the first to think about this since it is pretty rudimentary) :

Let's say superman approaches a (flat) mirror at 200,000 km/s. At what speed does superman's image approach him?

This has given me a big headache. I have pondered a lot of solutions, but none of them really convinced me. What is the correct solution to this thought experiment?

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Additional comment: What mainly bothers me is whether you can consider the 'approaching image' as information being moved, in other words; whether the speed limit (of 300k) applies or not. –  OmnipresentAbsence Feb 4 '13 at 18:33
    
Did you mean 200,000km/s i.e. $2 \times 10^8$m/s so there would be significant relativistic effects? –  John Rennie Feb 4 '13 at 18:35
    
If the speed is only $2 \times 10^5$m/s then the relative speed between Superman and his image is just $4 \times 10^5$m/s i.e. twice Superman's speed. The question gets more interesting as Superman's speed approaches $c$ as you then get relativistic effects. –  John Rennie Feb 4 '13 at 18:51
    
@JohnRennie Wow, this is a blunder on my part. I indeed meant 200,000 kilometers per second. I'm sorry, I'll edit it! –  OmnipresentAbsence Feb 4 '13 at 18:53
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Just thought about it again, not sure whether it's $\gamma^2(1+\beta)(1-\beta)$ or $\gamma^2(1+\beta)^2$... Now I'm pretty sure the latter, but, I have history homework I'm supposed to be doing right now. (I think it's the latter because, with Lubos's answer having the image be equivalent to no mirror with an object coming towards you at V+V (w/ relativistic addition), my frequency doesn't make sense) –  NeuroFuzzy Feb 5 '13 at 0:07

4 Answers 4

up vote 11 down vote accepted

The question is not just about the question whether Superman's image represents a "real piece of information". It's clearly not actual matter located at the appropriate point in space. Instead, the image is a virtual place defined by certain criteria. But the fact that the image isn't quite material doesn't mean that you should switch back to Newtonian formulae for velocities. They're never right in a relativistic world. They're wrong for Supermen as well as their images.

We define the image as the fictitious object that, when stationary, would produce the same light as the mirror that reflects the light from the actual object. If you define the image in such a way that its $x$ coordinate (transverse to the plane of the mirror) is just the opposite, $-x$, of the coordinate of an appropriate point of Superman – while $y,z$ are the same at the same moment $t$ – then the world line of Superman's image is exactly the same thing as the world line of another, material object that is actually moving 200,000 km/s in the opposite direction.

Then the relative speed is, just like for two material objects, $$ V_{rel} = \frac{u+v}{1+uv/c^2} $$ for $u=v=2c/3$ or so. That's $$ V_{rel} = c\frac{2/3+2/3}{1+4/9} = \frac{12c}{13} \lt c$$ That's all fine except that I must say that there was an assumption in the identification of the image's world line. The assumption was that we calculate the location of the image a moment after a moment. That may require justification.

To be convinced that he sees exactly what he would see if the image at $−x$ were a real object, consider that the trajectories of the photons may be simply given the right sign to $x$ and one may switch between the light from Superman reflected from the mirror and light from the image - and this rule applies to the coordinate $t$ as well, assuming that $t$ is time measured in the mirror's reference frame. So the image works just like in normal physics and the relative speed $12c/13$ works without any disclaimers - he sees exactly what he would see if someone who is real were moving $12c/13$ towards him.

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Although there's clearly nothing wrong with your solution given your definition of the image, I feel as though, as you point out, your definition doesn't really get at the heart of the question (although admittedly it's somewhat vaguely stated anyway). I would be curious to work out what Superman actually sees. Have you thought more about this? I spent a few minutes on it then got back to research. –  joshphysics Feb 4 '13 at 19:21
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I've changed my mind. He sees exactly what he would see if the image at $-x$ were a real object. After all, the trajectories of the photons may be simply given the right sign to $x$ and one may switch between the light from Superman reflected from the mirror and light from the image - and this rule applies to the coordinate $t$ as well, assuming that $t$ is time measured in the mirror's reference frame. So the image works just like in normal physics and the relative speed $4c/13$ works without any disclaimers - he sees exactly what he wuold see if someone who is real were moving $4c/13$ agains –  Luboš Motl Feb 4 '13 at 21:29
    
@joshphysics: the apparent velocity of an object approaching at $v$ is $\frac{cv}{c-v}$, in this case $12c$. –  Retarded Potential Feb 4 '13 at 22:45
    
@joshphysics, If he was a point particle I guess he'd just see his image as it was some time $t$ beforehand. With some length/width you would get odd nonlinear effects such as diagram 2 in en.wikipedia.org/wiki/… [edit:] also, as I wrote in a comment on the question, I think the frequency of light would be unchanged. –  NeuroFuzzy Feb 4 '13 at 23:24
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@Alpha: the addition formula of velocities, $(u+v)/(1+(uv/c^2))$, can be derived from the Lorentz transform. What he sees in the mirror is different from what is "in his reference frame", due to delay of light. He sees the image oncoming at $12c$, and blueshifted by a factor of 5, as discussed in the comments. (The image is also elongated by this same factor.) –  Retarded Potential Feb 14 '13 at 16:59

Rereading the OP I found that it is the speed of approach of light that is your question, more that the actual appearance of the image.

I will quote the most fundamental paper on this issue, the famous paper by Einstein, June 30, 1905, ON THE ELECTRODYNAMICS OF MOVING BODIES

§ 2. On the Relativity of Lengths and Times

Let a ray of light depart from A at the time $t_A$ , let it be reflected at B at the time $t_B$, and reach A again at the time $t'_A$. Taking into consideration the principle of the constancy of the velocity of light we find that $t_B - t_A = \frac{r_{AB}}{c - v}$ and $t'_A - t_B = \frac{r_{AB}}{c + v}$ where $r_{AB}$ denotes the length of the moving rod

Clearly $c + v= \frac{r_{AB}}{t'_A - t_B }$ is the speed of approach of the light ray to the retina of Superman. Then, being $c$ the mean spead of light in a closed path (same paper §1) = 300000 Km/s then the

correct answer is $c$ + 200000 km/s = 500000 km/s.

-- original answer:
find it for yourself is the best way. useful references:

Seeing Relativity and corresponding paper

A few years ago a package light++ was available from here: Werner Benger Visualization of special relativity
Lubos RF

About image and information:
Information is any stimulus that is 'acknowledged' by any object. Stimulus is a variation over a possible undetected continuum. To be perceived the receptor has to have a non null physical dimension to act as an antenna. Image is the collection of stimulus (usually light rays) that are presented in a given moment at the detectors (your retina f.i.) irrespective of the total length of space traversed by the light, and the moment of the emission. Look at the night sky and the image has light from billions of l.y. and the one reflected by the Moon a few moments ago.

The image seen is contracted in relation to the size of the image (reversed) when the speed is near zero (in one dimension). In 3d the image is more complicated.

I hope that the picture is clear. Two light rays emitted in the past,a mirror and a retina.
enter image description here

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Rereading the OP I realize that I will have to expand on 'speed' –  Helder Velez Feb 7 '13 at 8:39
    
This answers a different question, and is also the wrong answer to that question. –  Retarded Potential Feb 7 '13 at 17:40
    
@Retarded : it's Einstein answer, and anyone can check the transcribed words that fully apply to the question. You can clarify your opinion and post your own answer. –  Helder Velez Feb 7 '13 at 18:32
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The entire point of relativity is that c + v = relative velocity is completely wrong. Superman will always see light coming towards him at c. –  P O'Conbhui Feb 9 '13 at 0:45
    
I understand that this answer is not on target because it only says about the speed of light in relation to the retina and in that particular point it is accurate and in full agree with Einstein, in spite of the downvotes. I posted another answer, this time explaining the evolution of the image. –  Helder Velez Feb 13 '13 at 14:18

enter image description here

The image describes the geometry of the problem.

I'm 10 steps away of a mirror and see my image at 20 steps.
For each step I make my image approach 2 steps. So, at low speed, non-relativistic, the image approaches me at double of my own speed.
If I'm approaching the mirror at c speed I am not able to see my image because there are no reflected rays that were emitted by me because they cannot travel faster them myself. In any other case the double of my own speed is the minimum value of the speed of the image in relation to me.

The distance to the mirror was measured by timing the reflected light when Superman (SM) was still in relation to the medium and the path was labeled at equal distances. Travelling at 2/3c as SM crosses each mark he releases a different colored light ray and he will see it at x position (blue shifted image at -x0).

To measure the speed I will not use the clock of SM because he is subjected to time dilation and length contraction. Instead and because I am 'above' the scenario and know all the distances and speeds I will use my own uniform clock because I am measuring from the 'medium' viewpoint, i.e. as observer I am not in motion I just sit at each point of observation waiting for the event: light hit. This way this is only a classical exercise, not a relativistic one.

From A) $\frac{x_0+x}{c}=\frac{x_0-x}{c*2/3}$ we get $x=x_0/5$ (and $x0=5x$)
and this is the equation that relates the position of the retina of SM and the position of the emission of light, as he sees it.
For example when SM is at x=4 the emission was at 20 and the distance to the image is 20+4, at x=3 the numbers are 15 and 18, at x=0.2 the numbers are 1 and 1.2 .

The distance to the image is: $i(x) = x+x0 = x + 5x = 6x$
The rate of change of the distance i(x) in relation to the observer position at x is $\frac{d(i(x))}{dx} = 6$.
In terms of time evolution of the distance to the image, in relation to the medium, is $\frac{d(i(x))}{dt}=\frac{d(i(x))}{dx}\frac{dx}{dt}= 6*2/3 c = 4c$.

It is hard to accept an answer that is above c but we have to realize that light and the SM, the only ‘objects’ in scene do not travel above c .


-- measuring from the mobile viewpoint is different because his own clock rate changes when he is in motion. To do that one must enter the relativistic approach that I avoided in my answer. In spite of that my answer is correct, I believe.

To show the length contraction you can think in this way: when he was at x0=20 he wanted to measure the distance to the mirror and received the reflection at x=4, 24 seconds after the emission. Then he divided by 2 and got a distance of only 12 if he uses the same clock rate as obtained at rest, but we know that he cannot do this because of time dilation.

I understand that my previous answer is not on target because it only says about the speed of light in relation to the retina and in that particular point it is accurate and in full agree with Einstein. To perceive the motion of the reflected image we have to take more than one snapshot and because of that I colored the emitted light.

Everyone knows that the speed of propagation of the light is c but it should be emphasized that this value is only in relation to the medium. Exactly because of this I quote what Einstein wrote in his paper of June 30, 1905, ON THE ELECTRODYNAMICS OF MOVING BODIES

§ 2. On the Relativity of Lengths and Times

Let a ray of light depart from A at the time $t_A$ , let it be reflected at B at the time $t_B$, and reach A again at the time $t'_A$. Taking into consideration the principle of the constancy of the velocity of light we find that $t_B - t_A = \frac{r_{AB}}{c - v}$ and $t'_A - t_B = \frac{r_{AB}}{c + v}$ where $r_{AB}$ denotes the length of the moving rod

Clearly $c+v$ and $c-v$ are the relative speeds of light in relation to the mobile.

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I have trouble understanding what you mean. It is true that, for an observer at the mirror, Superman appears to be oncoming at $2c$. More simply we can say that generally, for oncoming objects, $v_{apparent}=cv/(c-v)$, and here $v=2c/3$). And his image appears to be oncoming at the same speed from the opposite direction. I think that's what you're saying in the first bit. Superman also sees his image oncoming at $12c$. That is also correct, and consistent with Luboš Motl's answer of $v=12c/13$ giving $v_{apparent}=12c$. I think that's what the next bit means. No idea about the rest. –  Retarded Potential Feb 14 '13 at 16:51
    
@Retarded do you agree with the statements A) B) C) in the image? If yes then clocks and photo-detectors still in the medium will measure the way I wrote. There is no need to use relativity as long as the observer (taking measures) is still i.r.t. to the medium, as I did. It is a simple classical problem. I can not relate your comment with my answer and in fact I do not understand it. –  Helder Velez Feb 14 '13 at 19:28
    
My comment was basically that I do not understand your answer :-) Okay, an observer at the mirror sees Superman approaching at $2c$, and sees the image approaching at $2c$ from the other side, and you can get this classically. But you do need relativity if you want to say anything about Superman's frame of reference, or what Superman sees. –  Retarded Potential Feb 16 '13 at 19:43
    
I may be stupid but I'm not wrong :-) The number $2c$ comes from the formula I posted, $v_{apparent} = cv/(c-v)$, and in this case $v=2c/3$. If you want to know where the formula comes from I'll make a separate question and answer. I won't post a separate answer to this question because the question has already been correctly answered by Luboš Motl. –  Retarded Potential Feb 17 '13 at 19:29
    
@RetardedPotential I didn't mean to be disrespectful and I deleted my previous comment. Sorry. –  Helder Velez Feb 18 '13 at 12:35

Superman's image approaches him at c, it will be blue shifted a bit towards xrays. Luckily Superman has xray vision so he should be able to see himself.

For the side question of what you would see at c, try this link.

http://www.space.com/19268-star-wars-hyperspace-physics-reality.html

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How did you get $c$? And there's no hyperspace or faster-than-light travel here, so how is the link relevant? –  Retarded Potential Feb 6 '13 at 1:55
    
I got c because light travels at c. What speed do you think it travels at? –  Jitter Feb 6 '13 at 5:56
    
I just looked at my mirror image, and it wasn't travelling at $c$. It sort of mirrored my motion, actually... –  Retarded Potential Feb 6 '13 at 6:00
    
The link is there becauce it mentions approaching c as well as ftl. –  Jitter Feb 6 '13 at 6:02
    
The light is still traveling at c to Superman's eye. That's what the question asked. What do you think approach means? –  Jitter Feb 6 '13 at 6:11

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