# Gedanken experiment - perpendicularly fired photon

(The premise of this question is has been shown to be false in the accepted answer. Read only if you have similar query, but not clear if it has an answer.)

Imagine a railway station and trains which are equipped with single photon sources (one each in the platform and inside the trains) and a narrow strip of detector which is positioned right above the photon sources. The photons are fired vertically upwards, perpendicular to the train’s direction of travel. In addition, let us assume that apart from the photo detectors on the train and the platform, the roof of the train is open, and that the ceiling of the station above the train also has photo detectors, corresponding to the position of the photo detectors on the platform. Now, the photon sources in the station and the trains are set to trigger at the same moment – i.e. when the train’s photon source passes in front of Alice. Now, Bob is sitting in front of the photon source in the train which is travelling at approximately half the speed of light, towards the station.

When the train in which Bob travels crosses Alice, both photon sources emit a photon (on platform and on the train). Hence, one photon is emitted towards the roof of the railway station by the photon source stationed in the platform; another is emitted inside the train above which is the photo detector on the train (moving w.r.t. Alice) and the photo detector on the station’s roof (stationary w.r.t. Alice) above the train.

Which means, Alice will (? might) record two photons hitting the photo detectors on the ceiling of the station (the photon from the platform and the photon from the train). The photon emanating from inside the train should hit the station’s ceiling since the train’s detector would have moved away by the time the photon travels upwards. Now from Bob’s view point, the photon emitted inside the train will be detected by the sensor IN the train; the photon emitted by the emitter on the platform will miss the detector on top of it (since the station is moving relative to Bob’s train). Either way, SR would not work for both observers at the same time.

In case you are thinking that there is some trickery with which this question can be answered in a manner similar to the barn door paradox, let Charlie travel in the track parallel to Bob, in the opposite direction (both Bob and Charlie are equidistant from Alice’s view point and are approaching with the same speed). Now we will have three different observers with different predictions; but due to law of conservation of energy, the photons can be detected by only one of the sensors, which should settle the confusion (or if they do separate, we would be able to tell because of the lesser energy liberated by the photon). In general, the experiment suggests that depending on the relative speed of the train to the station, there are an infinite number of positions in which the photon can be found, which could either support the many worlds theory, or is at least at odds with special relativity.

can some body tell me if I'm right or wrong - I REALLY need an answer

Your premise that Alice and Bob observe both photon trajectories as vertical is false.

The locus of events that form the worldline of either photon are absolute but the coordinates Alice and Bob assign to these events are not.

If Alice observes the station emitted photon to have a constant $x$ coordinate $x = 0$ then, assuming the photon is emitted at $t = t' = 0$, the Lorentz transformation to Bob's coordinate is

$$x' = -\gamma vt$$

In other words, Bob does not observe the photon to have a vertical trajectory but, rather, to have a horizontal component too.

Similarly, Alice observes the train emitted photon to have x coordinate

$$x = \gamma vt$$

Are you not familiar with the well-known light clock thought experiment? If the Galilean transformation were true, Alice would observe Bob's photon to have a vertical speed of $v_z = c$ and a horizontal speed of $v_x = v$ and, thus, the photon would have a total speed of $u = \sqrt{c^2 + v^2}$

According to the Lorentz transformation, Alice observes Bob's photon to have a total speed $u = c$ and a horizontal speed of $v_x = v$ thus, the vertical speed of the photon is reduced, according to Alice, to $v_z = \sqrt{c^2 - v^2}$.

• I think I'm missing elementary things - I will read throug the answers thoroughly (I think all answers point in the same direction, that one of my assumptions is wrong - will correct myself). Thanks. – Vignesh Mar 14 '14 at 9:48
• Many thanks for your detailed answer - this clears it! – Vignesh Mar 17 '14 at 18:09

No issue with SR and no many worlds - when Bob is firing his photon upwards towards the train's detector, Alice as well sees the photon of Bob moving to the train's detector, but for her the photon's direction is not upwards but diagonal. For her the photon is taking more time than it takes for Bob because it is going diagonally - this phenomenon is called time dilation.

• Hmmm.. That doesn't answer my question - Why would stationary Alice see a diagonal path for the photon? – Vignesh Mar 12 '14 at 21:06
• Why would stationary Alice see a diagonal path for the photon? For the exact same reason Bob observes a diagonal path for the station emitted photon. This is elementary. The events that constitute the worldline of the photon are absolute - the coordinates Alice and Bob assign to those events are not. If Alice observes that the station emitted photon has constant x coordinate of zero, then by the Lorentz transformation, Bob observes the photon to have non-constant x' coordinate: $x' = -\gamma vt$ – Alfred Centauri Mar 12 '14 at 22:35
• Sorry, did not get it the first time around. I stand corected. – Vignesh Mar 17 '14 at 18:03

The point you are confused about is the very fundamental point that the laws of physics are the same in all inertial frames. Let's look at what Bob sees. When the photon is emitted, bob sees the photon go straight up and hit the detector on the train, becuase photons move in straight lines in inertial frames, and the train is an intertial frame. Since the photon is going straight up in Bob's frame, and Bob's frame is moving with respect to Alice's frame, Alice sees the photon going up diagonally. Since the photon was emitted directly underneath the detector on the station, and the photon travels diagonally, Alice will not observe the photon from the train hitting the detector in the station. Thus both people will observe the station photon hitting the station detector, and the train photon hitting the train detector.

Notice that the only thing used here was the laws of physics being the same in all reference frames. Thus we can do the same thought experiment in newtonian dynamics with a ball being thrown into the air (ignore gravity). Then you would have said that the ball thrown upward from inside the train "should hit the station’s ceiling since the train’s detector would have moved away by the time the ball travels upwards." But this is not true because we know that the train is an inertial frame and so in the frame of the train the ball must move straight up and so Alice will see the ball moving diagonally, so the ball will hit the train's detector and miss the station's detector.

• Thanks for the answer, I think I got it this time. I have one more question regarding this - 'Do photons have a preferential frame?' - I thinks that is another question, so I will frame a new query... – Vignesh Mar 17 '14 at 18:06

When the photons are emmited, Alice will see that Bob's photon has a finite horizontal momentum component. That is, it's not travelling vertically towards the roof, but diagonally towards the place where Bob's sensor will be some time later (how much time depends on Bob's speed relative to Alice, which determines the horizontal component of Bob's photon momentum). Both Alice and Bob will agree that Bob detects his own photon. Think of it as if Bob threw a ball upwards inside the train (an ultrarelativistic, massless ball traveling at exactly the speed of light).

By the way, you may want to reformulate your question. Saying that the photon sources trigger "at the same moment" does not make sense: the same moment respect to which reference frame? Simultaneity notions change when you go from one frame to another in SR.

• Got it! But instead of reformulating the silly question, I will ask another question which might be more relevant! Thanks for your answer. – Vignesh Mar 17 '14 at 18:08

The problem is that geometry is being confused with real world momentum. The photon on the stationary train and the one on the moving train are both being emitted vertically with the same momentum vector as per Newton's first law. When they return after bouncing off the mirrors they are still vertical and this is why they are both seen by the stationary witness to arrive at the same moment. (the rider also sees the same thing.) But how can this be if the mirror has moved down the track while the photon was in transit thus creating a longer light path? The answer is that at the instant it was emitted the photon on the moving train picked up a second momentum vector from the energy of the train itself. So during its transit it moved along its own vertical momentum vector while moving simultaneously horizontally along the transport vector given to it by the train. At no time did the photon turn and move diagonally. The photon need not travel at light speed along the transport vector. To make sense of this imagine that the vertical vector is subdivided into a million sub units as is the horizontal vector. The photon travels the first vertical increment, then the first horizontal, then the second vertical and so on like climbing a stair casing. The increments are so small that our perception is tricked into thinking that we are seeing a smooth continuous geometric diagonal line. If that's how we see things then we will have to patch up our perception with an ad-hoc time dilation and Lorentz transformation. A minimum of two vectors are required for a state of momentum to exist between two reference frames. When the train is stationary, the transport vector is set to zero. The photon vector is always set to "C". The transport vector velocity varies between zero and "C". There are as many transport vectors as there are reference frames in motion relative to the photons own reference frame.

• If the photon on the train was fired horizontally does it mean it has its own momentum plus the second momentum due to the moving train that you talk about above? Because its velocity cannot increase as it is a photon, does it mean the extra momentum is accounted for by increased frequency? – Sandip Chitale Mar 5 at 7:31

As the speed of light is supposed to be independent of the motion of the source, the light clock can't work.The beam emitted from the source will fall short of the mirror, which will have moved on. Otherwise the Einstinian postulate would have been violated, as the beam would have to exceed the speed of light, being c plus the movement along the x axis. Impossible.

• The light clock can work. Please see en.wikipedia.org/wiki/Aberration_of_light which not only gives the explanation from special relativity, it also gives all significant historical theories for the aberration of starlight. – PM 2Ring Mar 6 '19 at 9:42
• The aberration of light makes perfect sense. The "light clock" in the Wikipedia article on Time Dilation is rather a Langevin clock. Einstein's clocks were positioned on laterally moving rods. Sound speed is governed by its "ether", usually air. But light, SR tells us, needs no such medium. So the pulse of light emitted in the Langevin clock will fall short of hitting the perpendicularly mounted mirror, as that mirror will have moved on. So therefore the Langevin clock is impossible, unless Einstein (and others) are wrong about the speed of light being independent of the motion of its source. – James West Mar 29 '19 at 2:35