# The Ultimate Grand Race: Will the photon cross the alpha particle?

Question:

There are $2$ points A and B, separated by a distance of $10$ light years. Suppose we have a light source at point A. An alpha particle is moving away from A at a speed of $0.99c$. As soon as it crosses B, a photon is emitted from the light source.

So:

1. Will the photon ever catch up with the alpha particle? If it does so, when will it?
2. Say an observer is situated at A, on the light source. What will he/she see?

Classically, we have that the relative velocity between the $2$ particles is $c-0.99c=0.01c$. So time taken $=\frac{10c \times 1 \text{year}}{0.01c}=1000$ years. But, this is out-and-out wrong. From the laws of special relativity , we find that the relative velocity (using the formula $\frac{u+v}{1+uv/c^2}$) of the photon with respect to the alpha particle is $c$ and the relative velocity of the alpha particle with respect to the photon is also $c$. So the time taken, with respect to an observer on any of the $2$ particles is $\frac{10c \times 1 \text{year}}{c}=10$ years.

However, one thing that bugs me is that, from the point of view of an observer standing on the alpha particle, it appears that he himself is at rest; and a photon is moving towards him at a speed $c$. Again, this can be contradicted by saying that the observer will never see the photon since the photon is coming at $c$ and the information about the position of the photon is also travelling at $c$. So the observer will not realise where the photon exactly is until it reaches him.

Is the above a flawed reasoning?

But again this is difficult to physically visualise. Also, what will the observer at A see? Will there be any relativistic effects like time dilation, length contraction in this case? Can anyone provide me with any sort of insight regarding the problem and the answer?

• In the frame of the observer sitting on the light source, your first calculation (the one you were too quick to abandon) is correct. The photon hits the alpha particle in 1000 years. This is a matter of elementary arithmetic (as you demonstrated). – WillO Feb 14 '17 at 18:51
• Observer always knows position of a photon. Observer does not have certain "localization", as a physical person. Observer is the whole reference frame with many Einstein - synchronized clocks along path of travel of the photon. en.wikipedia.org/wiki/Observer_(special_relativity) – Albert Feb 14 '17 at 21:18

The classical formula is in fact correct for an observer who is at rest relative to both "points" A and B. ("Points" here means parallel timelike world-lines, hence the scare-quotes.) Such an observer sees this process take 1,000 years. To try to make this discussion clearer, let me give these "points" some personality and say that there's a person who is at A named Alice, and a person at B named Bob, and they agree that they are both remaining stationary relative to each other. We will also assume that we're far enough in the future that everyone is immortal and has time to spare.

Okay, so this third observer Carol comes by, following the alpha-particle close by and she epically high-fives Bob on her way away from Alice at speed 0.99c, right as the alpha-particle crosses B. This high-five (or crossing of a timelike worldline or what have you) is a point in spacetime that we could call an "event."

In a separate event, Alice fires a light pulse towards Bob, just as you've described. It just happens to be fired, as Alice and Bob both see things, when this Bob-Carol high five happens. Alice of course won't know about this lucky coincidence until later when the light from the high-five reaches her, and she calculates back and determines "oh, those happened at the same time": but nevertheless, she happened to fire the pulse at this time. Now from Alice and Bob's perspective the rest of the argument does indeed proceed classically: they have to wait for 1000 years from when Alice fired the light pulse, before they see the light pulse finally hit Carol. (Bob won't know about the light pulse for 100 years, of course, but he will then calculate back and determine that Alice fired it 100 years before. Then 1800 years later he will see the reflected light pulse from the illumination of Carol's spaceship and working backward he'll reason that the light took 900 years to hit her, and therefore happened at t=1000 years. Similarly 100 years after that, Alice discovers the same timeline.)

Now, what about Carol? Carol does not actually agree that the distance between Alice's and Bob's two world-lines is $L = \text{10 light-years}$. She thinks it is somewhat shorter. This is not actually so surprising: imagine two lines that are going "upwards" through your desk and are parallel, so they have some constant distance $D$ between them. Now rotate the two lines so they come out of your desk at some angle, but all-the-while maintaining the same distance D between them: the distance between the two points on your desk will increase from this minimum of D as you rotate the lines, which corresponds to choosing a plane which goes through the two lines and measuring the distance on the plane.

The story is very similar for worldlines, but instead of preserving the metric $(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$ like rotations do in 3D space, the Lorentz transforms preserve the metric $(\Delta w)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2.$ They are therefore very similar to rotations and in fact contain rotations of the 3D part of the subspace as a special case! However when we choose this "plane" which goes between these two "lines" the $-$ sign in the metric means that the distance between the lines decreases rather than increases, as we're more familiar with from rotations. The distance between the two world-lines decreases from the minimum as you have a new "spacelike hyperplane" that characterizes the moment you're viewing as "the present," until some moment where you are travelling at the speed of light (and the two lines fuse into one and the space must become strangely one-dimensional or perhaps even zero-dimensional to you).

While we're at it, in addition to Carol measuring a shorter distance between Alice and Bob than they measure between themselves, Carol also does not agree that Alice emitted this light when she high-fived Bob. This is actually the fundamental prediction of special relativity, in that all other predictions come by simply integrating this one. The prediction is that if I move relative to you by speed $v$ in some direction $\hat z$, and you have two clocks which you think are synchronized separated by some position difference $(\Delta x, \Delta y, \Delta z)$, then I will think that they are out-of-sync. For small speeds $v$ this de-synchronization has the magnitude $\Delta t = \Delta z~v/c^2$ and it takes the form that the one which is further away in the $+z$ direction that I'm moving is "ahead" by $\Delta t$ of the one which is "behind" it in that direction. So if $\Delta t$ is 5 minutes and I'm traveling in the "upward" direction then the clock which is "higher" is going to show 10:05 when the one which is "lower" shows 10:00. Equivalently you can transform the fixed time your-10:00 into my frame of reference and you can find that at the "higher" clock this happens at my-9:55, when we agree for reference that the "lower" clock is showing our-10:00. (For this to happen at low velocities of course that means $\Delta z$ must be large, so that's why you are not very familiar with this effect in day-to-day life. If you are travelling in a car at 60mph relative to me, the distance is something like 100 light years away to get this 5-minute desynchronization.)

We physicists would use matrices to say that writing $w = c t$ the "boost" transform, (in the z direction) is for small velocities, $$\begin{bmatrix}w'\\z'\end{bmatrix} = \begin{bmatrix}1 & -v/c\\-v/c&1\end{bmatrix} \begin{bmatrix}w\\z\end{bmatrix}.$$If you know how to exponentiate a matrix, and you know the limit $\lim_{N\to\infty} [1 \pm (\phi/N)]^N = e^{\pm\phi}$ you can use this to figure out what the relation is for larger velocities; you will naturally discover the $\sinh$-and-$\cosh$ version of the Lorentz boost matrix and this leads straight into a discussion of rapidities and your velocity-addition formula. You will have to cautiously re-derive that then the primed frame is at rest with respect to the worldline $z=w~\tanh \phi$ and hence the relative speed between the two frames is $v = c \tanh \phi,$ but that's the only subtlety.

Anyway: so Alice and Bob both think that Alice fired this pulse at the same time $t$ as when Carol and Bob high-fived; but Carol sees things fundamentally differently, where she thinks that Alice fired this pulse much, much later than that.

If you fully correct for these two effects, integrating properly as we get to higher and higher velocities, then yes, you can also treat the light in Carol's frame of reference at moving at speed $c$ and you will find that it hits her after a time which is a fraction $1/\sqrt{1 - (v/c)^2}$ of the time that Alice and Bob calculate, so in this case that factor is roughly 7 and it will only take $1000/7$ of her years between when she high-fives Bob and when she sees Alice's light-pulse. Of course Alice and Bob must be able to record her in her spaceship between these two events and they will see her only age 1000/7 years as they both age 1000 years; this phenomenon is known as "time dilation" and is reciprocated by Carol: she also sees them move in "slow motion." (Note that this is after correcting for the Doppler shift that also naturally exists between them; when Carol was moving towards Bob pre-high-five, they actually saw each others' clocks moving fast due to the Doppler shift overpowering this effect, but then if they divided out the Doppler shift they would have seen this slowdown.)

This leads to the thought experiment, "why don't they just call each other up and see who's talking in slow motion?" and the answer is, "as long as the phone they use works by exchanging information slower than the speed of light, the propagation delay will perfectly hide the discrepancy and they will never be able to answer that question."

• +1! Thanks for the answer. However I want to ask you a few questions to clarify doubts regarding your answer. Firstly, what do you mean by " the observer who is at rest relative to both points "? And who is Carol i.e. which frame does she belong to? – SchrodingersCat Feb 14 '17 at 19:23
• "The observer who is at rest" is either Alice or Bob. Carol is moving relative to Alice and Bob at a speed of 0.99c; in other words she is in a reference frame wherein the alpha-particle is at rest. When the alpha-particle crosses A, she epically high-fives Bob. (I originally said Alice but I will change it to match your A vs. B.) – CR Drost Feb 14 '17 at 21:00
• Hope that's a bit clearer! – CR Drost Feb 14 '17 at 21:56