# Closing door paradox?

1. Dec 4, 2007

### Jeroen L

Hi all,
I just recently introduced myself to SR theory, and now I'm a bit stuck already (probably like most beginners)
Here's my problem:
Suppose that a photon is coming straight towards my font door, which is open.
But I close the door just a bit quicker than the photon can reach it.
However, in the frame of the photon, I seem to be having a time dilation, not to mention a distance contraction, according to an observer sitting on the photon. So I will close the door slower (actually not at all) and the distance to me is shorter (actually 0).
As I understand SR now, this would mean that the photon would go through the door if I close it just in time in my viewpoint.

The only solution seems to me that the photon is already having a personal time in which I have closed the door, when it is coming towards me. Is this incorrect?
And what if the photon is released by somebody with a flashlight, standing still with respect to me, before I close the door, but the photon reaches my front door when it is closed? Is the photon accelerated into the future, when it is emitted by the flashlight? Otherwise, according to SR, there is no way that I can stop the light from going through my front door, even if I close it before the light reaches my house.

2. Dec 4, 2007

### dst

I thought photons themselves experienced no time (i.e. from one end to the other of the universe would still be no time taken)? Hence there would be no "observer on the photon"?

Don't quote me on that.

3. Dec 4, 2007

### Ich

Yes, this is incorrect.
Never try to evaluate things from a photon's reference frame. There is no such thing in SR, that's why you get all those zeros and infinities.
Take a fast particle instead. It will experience proper time, and the door will be closed during that timespan.

4. Dec 4, 2007

### Jeroen L

Oh right, the limits to infinity do not convert well into mathematics. But shouldn't their be an intuitive interpretation possible?

Then suppose the photon travels through air, having a speed below c, say 0.95c. What would happen to the space-time starting point of the photon emitted from the flashlight in the reference frame of my door and in that of the photon, after it has instantly accelerated to 0.95c? Would it still be at the location of the flashlight in both frames after the distances have contracted? What would the photon experience during this acceleration? Would the age of the door suddenly increase, just like with the acceleration in the twin paradox? There, the age increase after the turn-around depended on the distance (gravity potential), so objects far away in the direction of acceleration age more than objects nearby, while objects behind will get younger. That would mean that the flashlight instantly shoots the photon into the future of the door, but not the future of the flashlight, since that is at the same time-space location.
If the door-time that the photon skips during emission, compensates the distance contraction and the slower clock of the door during the travel towards it, the light will indeed consciously crash into my closed door. But it will be a younger photon than expected.

Last edited: Dec 4, 2007
5. Dec 9, 2007

### yuiop

Try this,

Replace the photons with small clocks. (We can't meaningfully talk about the elapsed time measured by a photon.) Imagine we have two clocks moving to the right at speed of 0.8c (time dilation gamma factor = 1.67). We let the first clock pass through the door and start shutting the door. The door takes one second to close and the second following clock collides with the door just as it shuts. If the two flying clocks are synchronized in their own frame and first showed zero as it passed the door the second would record 1.67 seconds as it hit the door. [Note 1] We also work out that it was 0.8 light seconds behind the first because that it the distance it travels at 0.8c in one second of our time. Now look at it from the point of view of an observer co-moving with the flying clocks. From his point of view the clocks are stationary and the distance between them is 1.33 light seconds. (We saw the clock separation distance as 0.8 due to length contraction) [Note 2] From his point of view the door is moving to the left at 0.8c so it takes the door 1.67 seconds to travel from the first clock to the second clock ( t= d/v = 1.33/0.8) He also sees that the clock attached to the door shows an elapsed time of one second which is what he would expect due to time dilation of the clock moving with the door. From our point of view and his point of view the second flying clock impacts the door just as it closes so everything is consistent.

[Note 1] You might be surprised that the second clock shows 1.67 seconds. If our clock records 1.0 second shouldn’t the second clock show 0.6 seconds? (i.e. less due time dilation) The answer is that 0.6 seconds elapse on the second clock but it initially started with 1.07 seconds on its dial giving a total of 1.67 seconds at the collision. This is due to the way clocks are synchronised. Clocks synchronised in a moving frame moving relative to us look out of sync by a factor of -L*v/c^2 to us. It is this (lack of ) synchronisation of moving clocks separated by a distance that is the root cause of the confusion behind many apparent relativity paradoxes.

[Note 2] Our observer on the ground who saw the second clock at the time the door started closing (Time zero on his clock) was 0.8 light seconds ground distance from the door. The observer moving with the flying clock does indeed see the ground observer only 0.48 light seconds distance from the door but not at time zero by his clock. At time zero he was further away from the door and by the time the ground observer gets to him, 1.07 seconds has elapsed on his clock. It takes another 0.6 seconds by his clock for the door to arrive.

6. Dec 9, 2007

### pervect

Staff Emeritus