Time dilation and the photon clock

[neopolitan]

Seems to me that you've dodged or abandoned the light clock issue and have decided to divert this thread into yet another discussion of your "no twin paradox" stuff. Which seems, despite your "calculations", to be "nothing more than opinion".

Yes, I must apologise to Stellar1 for that.

The "no twin paradox" is, I believe, resolved. This discussion between JesseM and myself has been more about the validity of the equation for time dilation, as calculated using the light clock scenario. You have to go through all the responses in this chain to see how we got here but, in short, I said there is a problem with the light clock derivations. The discussion between JesseM and I has since been about the equations derived from the light clock and the application of relativistic equations in the "real world".

I was going to say that if JesseM is amenable, then we can take this question to another thread and leave this thread for further discussion of the light clock. But have just seen his latest response, so I doubt that he wants to discuss this question further.

I will leave it here. If anyone wants to take up the issue further, they can find my contact details via links in the "There is no twin paradox - mathematical proof" chain to which Doc Al refers or you send an email to me via my public profile entry.

thanks all and Merry Christmas,

neopolitan

 

[JesseM]

I was going to say that if JesseM is amenable, then we can take this question to another thread and leave this thread for further discussion of the light clock. But have just seen his latest response, so I doubt that he wants to discuss this question further.

I am happy to discuss it further--the fact that I objected to your accusations of dogma does not mean I want to end the discussion, I'm just asking that you keep an open mind that you might be making some errors instead of making assumptions that your arguments are irrefutable and that anyone who disagrees must be doing so for irrational reasons.

 

[Dale]

1. Time dilation: t' = t / \sqrt{1 - v^2/c^2}

2. Length contraction: x' = x * \sqrt{1 - v^2/c^2}

3. Time elapsed for K', according to K', since K' carries a watch and measures how much time has elapsed: t' = t * \sqrt{1 - v^2/c^2}

4. Distance traveled by K', as calculated by K, given that he knows he has a speed of v: x' = x * \sqrt{1 - v^2/c^2}

None of these equations are complete.

1. Should be t'=γ(t-vx/c^2)
2. Should be x'=γ(x-vt)

Solving 1 and 2 for t and x we get

3. t=γ(t'+vx'/c^2)
4. x=γ(x'+vt')

Which are of the same form as 1 and 2. The difference in the sign is because if the primed frame is moving at v in the positive x direction then the unprimed frame is moving at v in the negative x' direction.

 

[JesseM]

None of these equations are complete.

1. Should be t'=γ(t-vx/c^2)
2. Should be x'=γ(x-vt)

No, those are not the equations for time dilation and length contraction; rather, they are the equations for transforming from the coordinates of an event (x,t) to the coordinates of the same event (x',t') in another coordinate system moving at v relative to the first. See this thread for a discussion of the difference between these two pairs of equations.

 

[Dale]

No, those are not the equations for time dilation and length contraction; rather, they are the equations for transforming from the coordinates of an event (x,t) to the coordinates of the same event (x',t') in another coordinate system moving at v relative to the first. See this thread for a discussion of the difference between these two pairs of equations.

My expressions are correct. It doesn't matter what you call the equation, the fact is that t' ≠ γt in general. If you want to write an expression for time dilation you should differentiate the Lorentz transform wrt t and write dt'/dt = γ which is not at all the same as t' = γ t. I really dislike the equation t' = γ t because it is not true in general, it is very prone to error in useage, and it confuses people like neopolitan.

 

[yuiop]

Hello,
I just baught my next set of textbooks and started reading about relativity. In one of the books it uses the example of a two clocks who "tick" every time a photon it emitted hits the mirror and returns to the sensor. It demonstrated that, if the box containing this clock is moving, it will tick slower than one that is stationary. I understand this and why, but I don't understand how this is supposed to show time dilation? If I perform the same experiment but with a clock that shoots a tennis ball, while fixing the tennis ball's speed at a constant value, the moving clock, even at speeds far below the speed of light, will still tick slower than the stationary one, yet there would not really be time dilation.

Let's analyse the tennis ball clock at slow speed.

Say we have a ball cannon that fires a tennis ball vertically at 10 m/s at a reflector and a digital clock that times the round trip of the ball back to the cannon. If the reflector is 5m away then the round trip time is one second (ignoring gravity). Now let's mount our ball clock on a train that is also going at 10 m/s (36 kph) but horizontally. The diagonal path length of the ball from the point of view of an observer at the side of the track using pythagorus theorum is sqrt(10^2+10^2) = 14.14 meters. The velocity of the ball is total of its vertical velocity component and its horizontal component (acquired from the motion of the train). Using normal velocity vector addition the velocity of the ball 14.14 m/s. The round trip time is then one second from the point of view of the observer on the side of the track AND from the point of view of an observer on the train. So the ball clock is not ticking slower from anyone's point of view at these slow velocities.


Now, let's increase the scale and velocities of the experiment and include a light clock.

The height of the clock is half a light second. The round trip time for a photon is 1 second from the point of view of an observer on the train. The ball clock has been upgraded to fire balls at 0.5c and the ball should return in 2 seconds from the point of view of an observer on the train. 2 photon clock ticks = 1 ball clock tick.

Assume the train is going at 0.5c from the point of view of an observer beside the track.
The track side observer measures the path length taken by the photon as 1.1547 light seconds for one tick of the light clock and 2*1.1547 = 2.309 light seconds for the total distance traveled by the photon in two ticks. The path length traveled by the ball in one tick is 1.527 light seconds as measured by the track side observer. The velocity of the ball from his POV is therefore 1.527/2.309 = 0.66c if the 2nd tick of the photon clock is going to coincide with the first tick of the ball clock.

Now if we use normal velocity vector addition to calculate the speed of the ball with respect to the track side observer, we get sqrt(0.5^2+0.5^2) = 0.7071c

The discrepancy is because normal velocity addition does not work at relativistic speeds. Everything except light slows down in a reference frame that is moving with respect to the observer. The ball slows down and any clock constructed of balls or anything else slows down similarly.

The correct formulas for adding relativistic velocities can be found at this link.

http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

Using the formula given by Baez, the vertical velocity of the ball from the POV of the trackside observer is

wy = uy / [(1 + ux vx / c2) gamma(vx)]

which is 0.5/((1+0.0*0.5)/sqrt(1-0.5^2) = 0.433c (using c=1)


Now that we have the correct vertical component of the ball's velocity we can use normal vector addition to obtain the velocity of the ball = sqrt(0.433^2+0.5^2) = 0.66c which agrees with the figure I gave earlier.

I have not shown all my working but if you are puzzled how I obtained my figures, just ask ;)

 

[JesseM]

My expressions are correct. It doesn't matter what you call the equation, the fact is that t' ≠ γt in general.

Only if t and t' are taken to represent time coordinates rather than time intervals; in the time dilation equation they are time intervals, and t' does equal γt whenever t is time between two events on an inertial clock's worldline as measured by that clock, and t' is the time between those same two events in an inertial frame where the clock has speed v.

I really dislike the equation t' = γ t because it is not true in general, it is very prone to error in useage, and it confuses people like neopolitan.

It is true in general, as long as you are clear on what the symbols t and t' are supposed to represent in that equation.

 

[Dale]

Only if t and t' are taken to represent time coordinates rather than time intervals

I know this, but IMO it is a big problem pedagogically and I recommend strongly against it. Students expect t and x to refer to coordinates, and students expect Δt and Δx to refer to intervals. This is a convention that has been explained and used since the first day of introductory physics and to change the convention for no reason is a terrible idea. It is much clearer to say dt/dt' = γ or to say Δt = γ Δt'. That way you are clear that you are talking about intervals rather than coordinates.

as long as you are clear on what the symbols t and t' are supposed to represent

This is the key problem. It is inherently unclear to use the symbol t to refer to an interval. It is a constant source of confusion, as adequately evidenced by this thread.

 

[pervect]

There's no particular problem with longitudinal light clocks, except that the derivation is a bit lengthly. See for instance http://arxiv.org/abs/physics/0505134. They tick at the same rate as the non-longitudinal kind.

 

[Stellar1]

So here's another thought:

In the perspective of the photon, it is the box itsself that is vibrating back and fourth. From the photons perspective, gamma=sqrt(0) so therefore the height of the box is 0, no? So that would mean that the clock is constantly ticking, correct?

 

[JesseM]

So here's another thought:

In the perspective of the photon, it is the box itsself that is vibrating back and fourth. From the photons perspective, gamma=sqrt(0) so therefore the height of the box is 0, no? So that would mean that the clock is constantly ticking, correct?

Actually, since gamma = 1/sqrt(1 - v^2/c^2), if you plug in v=c you get gamma=1/sqrt(0), division by zero. In SR a photon does not have its own rest frame, so it isn't really meaningful to talk about the "photon's perspective".

 

[Stellar1]

ahh, yes, sorry, it is 1/0 in that case. So in that case the length would be infinite and it would never tick...?

What is SR? Special relativity I take it? Why does it matter that a photon does not have its own rest frame?

 

[JesseM]

ahh, yes, sorry, it is 1/0 in that case. So in that case the length would be infinite and it would never tick...?

Again, the equations for length contraction and time dilation only make sense when you are talking about length and time as measured in a given inertial frame, it is meaningless to use them outside this context, and a photon does not have an inertial rest frame.

What is SR? Special relativity I take it? Why does it matter that a photon does not have its own rest frame?

Yes, SR = special relativity. As I said, the time dilation equation and length contraction equation are specifically derived based on the assumption that you are talking about time and length as measured by a system of rulers and clocks that are moving inertially, you can't have a ruler/clock system that's moving at the speed of light, so there is no physical meaning to plugging v=c into those equations.

 

[Stellar1]

Why does a photon not have an inertial rest frame?

 

[JesseM]

Why does a photon not have an inertial rest frame?

Because a given observer's inertial frame is based on rulers and clocks at rest relative to that observer...it's impossible to accelerate rulers and clocks up to light speed.

Another way of seeing that a photon cannot have an inertial frame is that one of the two postulates of SR is that the laws of physics must be the same in every inertial frame, but in all slower-than-light frames photons always move at c, so if there was a frame where a photon could be at rest this postulate would be violated.

 

[yuiop]

So here's another thought:

In the perspective of the photon, it is the box itsself that is vibrating back and fourth. From the photons perspective, gamma=sqrt(0) so therefore the height of the box is 0, no? So that would mean that the clock is constantly ticking, correct?

It is hard to make any sense of measurements from the point of view of a photon as the answers usually involve 0/0. (Indeterminate). However, you could ponder the point of view of a ball in the Stellar ball clock ;)

 

[peter0302]

Of course, the speed of a photon isn't constant :P It's just constant in a vacuum, and that is given as c

Sorry, this always irks me. :) The speed of a photon IS always constant. The reason light appears to slow down in a given medium is a consequence of the time it takes an electron to absorb a photon and emit a new one. The photon going in is not the same photon as the one going out.

 

[rbj]

Sorry, this always irks me. :) The speed of a photon IS always constant. The reason light appears to slow down in a given medium is a consequence of the time it takes an electron to absorb a photon and emit a new one. The photon going in is not the same photon as the one going out.

well, guess what might irk some other folks (but not me, i don't really care)? it's this "photon is absorbed and re-emitted" explanation for why the apparent SOL is slower in transparent materials. i think this: http://en.wikipedia.org/wiki/Refractive_index#The_speed_of_light outlines what i mean. it says

At the microscale, an electromagnetic wave's phase velocity is slowed in a material because the electric field creates a disturbance in the charges of each atom (primarily the electrons) proportional to the permittivity of the medium. The charges will, in general, oscillate slightly out of phase with respect to the driving electric field. The charges thus radiate their own electromagnetic wave that is at the same frequency but with a phase delay. The macroscopic sum of all such contributions in the material is a wave with the same frequency but shorter wavelength than the original, leading to a slowing of the wave's phase velocity. Most of the radiation from oscillating material charges will modify the incoming wave, changing its velocity. However, some net energy will be radiated in other directions (see scattering).

 

[pervect]

One might also want to take a look at ZapperZ's physics forum FAQ on this topic at https://www.physicsforums.com/showpost.php?p=899393&postcount=4 .

The photon is emitted and absorbed, but the absorption can be regarded as an absorption by the lattice to create a phonon.

I've seen explanations similar to the Wiki one posted above in Feynman's popularizations (i.e. QED).

 

[peter0302]

well, guess what might irk some other folks (but not me, i don't really care)? it's this "photon is absorbed and re-emitted" explanation for why the apparent SOL is slower in transparent materials. i think this: http://en.wikipedia.org/wiki/Refractive_index#The_speed_of_light outlines what i mean. it says

I think that's saying the same thing. The bottom line is that the photon going in is not the same photon as the one going out. The only reason for the apparent delay is due to the interaction with the atoms in the media. But inbetween atoms the speed is always c. And as soon as the light leaves the medium, its speed is likewise c. No?

 

[NtroPunDfeetd]

Does the photon actually travel at light speed? When your on a flatbed truck traveling down the highway at 70mph with a photon clock in hand, does the photon for you on the truck still travel at the speed of light and count your seconds accurately? To an observer on the ground does the photon look as if it's going slower? Has this ever been tested and proven?

Please answer my questions shortly and staying on track completely with what I'm asking; do not diverge from my question into other explanations.

 

[Doc Al]

Does the photon actually travel at light speed?

Yes. According to everyone.

When your on a flatbed truck traveling down the highway at 70mph with a photon clock in hand, does the photon for you on the truck still travel at the speed of light and count your seconds accurately?

Sure.

To an observer on the ground does the photon look as if it's going slower?

No. The photon travels at light speed. The clock ticks slower, since for the ground observer the photon travels a greater distance.

Has this ever been tested and proven?

No one has ever done experiments with photon clocks, but the effect it illustrates--time dilation--has been tested.

 

[Austin0]

Doc Al is Offline:
Posts: 23,346
Blog Entries: 1 Re: Time dilation and the photon clock

--------------------------------------------------------------------------------

Originally Posted by NtroPunDfeetd
Does the photon actually travel at light speed?

Yes. According to everyone.

.

To an observer on the ground does the photon look as if it's going slower?

No. The photon travels at light speed. The clock ticks slower, since for the ground observer the photon travels a greater distance.

What is the interpretation in the case of a photon clock at a location of low gravitational potential?
Is it traveling slower or traveling a greater distance due to the geometry at a small Scwarzschild radius?
Thanks

 

[yuiop]

What is the interpretation in the case of a photon clock at a location of low gravitational potential?
Is it traveling slower or traveling a greater distance due to the geometry at a small Scwarzschild radius?
Thanks

To keep it simple take the case of a short horizontal light clock. (Horizontal rulers are not length contracted by gravity.) To an observer higher up, the clock lower down is running slow and the speed of light lower down is slower than the speed of light higher up. These two effects conspire in such a way, that to a local observer co-located with the light clock at a small Schwarzschild radius, the speed of light appears normal. This slowing down of clocks and light, deep in a gravitational well is not an optical illusion, because by lowering a clock down and bringing it back up again, it can be demonstrated that the lowered clock really does run slower than a clock that remained higher up.

It follows that if the clock lower down is really running slower than the clock higher up, then the speed of light lower down must really be slower than the speed of light higher up, in order for a local observer lower down to measure the local speed of light as c.

 

[Austin0]

To keep it simple take the case of a short horizontal light clock. (Horizontal rulers are not length contracted by gravity.) To an observer higher up, the clock lower down is running slow and the speed of light lower down is slower than the speed of light higher up. These two effects conspire in such a way, that to a local observer co-located with the light clock at a small Schwarzschild radius, the speed of light appears normal. This slowing down of clocks and light, deep in a gravitational well is not an optical illusion, because by lowering a clock down and bringing it back up again, it can be demonstrated that the lowered clock really does run slower than a clock that remained higher up.

It follows that if the clock lower down is really running slower than the clock higher up, then the speed of light lower down must really be slower than the speed of light higher up, in order for a local observer lower down to measure the local speed of light as c.

Thanks for your very clear explanation. It was pretty much what my own logic arrived at but I no longer have the implicit faith in "logic" I once had.