Time dilation and the photon clock

[neopolitan]

I didn't stray from the challenge, I rejected it as bogus!

Okay, I understand.

 

[neopolitan]

You make it sound like you've uncovered some flaw in the way the light clock is typically used. Well, where's the flaw?

I am going to be Socratean here. I know Socrates was annoying and eventually the Atheans persuaded him to kill himself, but if I give you the answer, you won't own it. If you work it out yourself, you might just decide to give me a hand rather than fight me every step of the way.

I am going to ask you to provide for me four equations. The first two are simple: the equation for time dilation (in terms of t, c and v) and the equation for length contraction (in terms of x, c and v).

For the third and fourth equations, consider this scenario. K' knowingly travels for an extremely long time away from K at a rather slow speed of v, relative to K. Then, in an extremely short period of time, K' changes direction 180 degress and then travels back to K at a speed of v, relative to K. According to K, K' has been away for a period of t and therefore traveled (approximately) a total distance of v.t = x

Equation three: How long has K' been away, according to K', in terms of t, c and v?

Equation four: How far did K' travel, according to K', in terms of x, c and v?

I will continue the discussion once you have written up the equations. You may care to discuss them in your answer. But until you, or someone else, presents those four equations, there really is nothing to discuss. Until then, it's nothing more than opinions.

cheers,

neopolitan

 

[JesseM]

Equation three: How long has K' been away, according to K', in terms of t, c and v?

This is a well-defined question, since K' can carry a watch and measure how much time has elapsed between first passing K and then later passing K again after the turnaround (according to K' the time would be $$t * \sqrt{ 1 - v^2/c^2}$$, if t is the time measured by K between their two meetings, and K is an inertial observer).

Equation four: How far did K' travel, according to K', in terms of x, c and v?

This is not a well-defined question, since K' does not have a single inertial rest frame. You could ask how far K' traveled in the inertial frame where K' was at rest during the outbound leg, or how far K' traveled in the inertial frame where K' was at rest during the inbound leg (in both cases K' would travel zero distance during the phase where K' was at rest, and during the other phase of the trip, K' would be moving at velocity 2v/(1 + v^2/c^2) in this same frame according to the velocity addition formula, while K would be moving away at v in this frame, so the time for K' to catch up with K should be $$\frac{x * \sqrt{1 - v^2/c^2}}{(2v/[1 + v^2/c^2]) - v}$$, so to get the distance K' moved you'd multiply by the velocity of K', which was 2v/[1 + v^2/c^2], giving an answer of $$\frac{x * (2v/[1 + v^2/c^2]) * \sqrt{1 - v^2/c^2}}{(2v/[1 + v^2/c^2]) - v}$$ =
$$\frac{x * (2v/[1 + v^2/c^2]) * \sqrt{1 - v^2/c^2}}{(v*[1 - v^2/c^2]/[1 + v^2/c^2])}$$ = $$\frac{2x}{\sqrt{1 - v^2/c^2}}$$), or how far K' traveled in a non-inertial frame where K' was at rest during the entire trip (of course in this last case, since K' is always at rest, the distance traveled by K' is zero!).

 

[rbj]

The second postulate of relativity is that the speed of light (i.e. the speed of a photon) is the same in all inertial (non-acclerated) frames of reference.

doesn't the second postulate regarding the constancy of the speed of light really mean the wavespeed of propagation of E&M, what comes out as $1/\sqrt{\epsilon_0 \mu_0}$ in the solution of Maxwell's Eqs. (with no mention of the concept of photons)? That E&M has particle-like properties (besides the wave-like properties) and that the particle speed is the same as the wavespeed, is another issue. There have been some propositions that the rest mass of photons are not precisely zero and might have something like 10^-56 kg of mass which means that they do not move at precisely the same as the wavespeed c.

 

[jcsd]

doesn't the second postulate regarding the constancy of the speed of light really mean the wavespeed of propagation of E&M, what comes out as $1/\sqrt{\epsilon_0 \mu_0}$ in the solution of Maxwell's Eqs. (with no mention of the concept of photons)? That E&M has particle-like properties (besides the wave-like properties) and that the particle speed is the same as the wavespeed, is another issue. There have been some propositions that the rest mass of photons are not precisely zero and might have something like 10^-56 kg of mass which means that they do not move at precisely the same as the wavespeed c.

Strictly it's the propagation speed and that assumes that photons have zero rest mass.

 

[rbj]

Strictly it's the propagation speed and that assumes that photons have zero rest mass.

i don't know why it has to assume anything about the rest mass of photons. you can have a concept of a "light clock" without any notion of photons bouncing up and down (or back and forth, whatever the orientation).

 

[jcsd]

i don't know why it has to assume anything about the rest mass of photons. you can have a concept of a "light clock" without any notion of photons bouncing up and down (or back and forth, whatever the orientation).

The rest mass of photons must me zero, assuming that they travel at c (defining c as being the all important 'Eisntein constant' in relatvity, rather than defiening it as the speed of light). Otherwise their momentum would be infinite, which it isn't.

 

[rbj]

The rest mass of photons must me zero, assuming that they travel at c (defining c as being the all important 'Eisntein constant' in relatvity, rather than defiening it as the speed of light). Otherwise their momentum would be infinite, which it isn't.

i know that. that's not the point.

you can have a notion of a light clock, discuss the principles of its operation, show that if the wavespeed of propagation remains the same speed c then time gets dilated, all without any notion of a photon. so whether not photons have no rest mass (i actually think that is the case) or that they must always travel at speed c (from the perspective of any inertial observer) is not needed to deal with this at all. a "light clock" need not be a "photon clock" and yet we can still arrive at the conclusions Einstein did a century ago.

 

[neopolitan]

K' ___knowingly___ travels for an extremely long time away from K at a rather slow speed of v, relative to K.

...

Equation three: How long has K' been away, according to K', in terms of t, c and v?

Equation four: How far did K' travel, according to K', in terms of x, c and v?

... (according to K' the time would be $$t * \sqrt{ 1 - v^2/c^2}$$, if t is the time measured by K between their two meetings, and K is an inertial observer).

You failed to provide equations 1, 2 and 4. (Equation 1, time dilation in terms of t, c and v. Equation 2, length contraction in terms of x, c and v.) Note:

According to K, K' has been away for a period of t and therefore traveled (approximately) a total distance of v.t = x

--------------
Provide all four equations together in the same post please, then we can continue.

cheers,

neopolitan

 

[JesseM]

Why did you highlight the word "knowingly" like that? The time measured on a clock carried along by K' will not depend on what he knows or doesn't know.

You failed to provide equations 1, 2 and 4. (Equation 1, time dilation in terms of t, c and v. Equation 2, length contraction in terms of x, c and v.) Note:

--------------
Provide all four equations together in the same post please, then we can continue.

cheers,

neopolitan

Equations 1 and 2 are just the standard time dilation and length contraction equations, which I made use of in my answers--if a clock moves at speed v for time t in some inertial frame (such as the frame of K), it will only elapse a time of $$t * \sqrt{1 - v^2/c^2}$$ in that time (so if K' is moving at the same speed v in opposite directions both before and after the turnaround in the frame of K, and the turnaround is negligibly short, then if the time between K' passing K the first and second time is t according to K, then the clock of K' will elapse the shorter time given by the time dilation equation above). And if the distance between two fixed objects at rest in a given frame (like K's own position and a space station at the position where K' turns around) is x, then the distance between them in a frame moving at speed v relative to the objects is $$x * \sqrt{1 - v^2/c^2}$$

As for equation 4, didn't you read my response? I said that your question as stated was not well-defined, since you didn't specify what sort of frame you wanted to use for K' to answer the question "how far did K' travel". I did point out that if we use an inertial frame where K' is at rest for one phase of the trip and moving at speed 2v/(1 + v^2/c^2) for the other phase of the trip (either the frame where K' is at rest before the turnaround and moving after, or the frame where K' is moving before the turnaround and at rest after), then the distance traveled by K' during the moving phase will be $$\frac{2x}{\sqrt{1 - v^2/c^2}}$$. And I also pointed out that if we use a non-inertial frame where K' is at rest the whole time, both before and after the turnaround, the distance traveled by K' is of course zero (but you can't use the usual equations of special relativity in this non-inertial frame, like the equations for time dilation and length contraction above).

 

[neopolitan]

Why did you highlight the word "knowingly" like that? The time measured on a clock carried along by K' will not depend on what he knows or doesn't know.

Because K' knows he is travelling, he knows that he has traveled for a set period (according to himself), he knows he had a speed of v. Knowing that he can calculate how far he has gone (according to himself). He will not calculate what you provided as the staff answer.

Please provide all four equations together please. You seem set on refusing to do so. Unfortunately this is a common response, many forum posters are quick to write up complex equations but leery of writing four rather simple equations together in the one place. I would love to know why.

If this is not an article of faith for you, please write all four equations together in one post. If this is an article of faith for you, just acknowlege it and I will accept that.

cheers,

neopolitan

 

[JesseM]

Because K' knows he is travelling, he knows that he has traveled for a set period (according to himself), he knows he had a speed of v.

But speed is relative, he doesn't know he has a speed of v in any absolute sense, only a speed of v in the rest frame of K. K' can certainly calculate what the time will be in the rest frame of K, but this will not be the same as the time according to his own clocks. Are you in fact asking what time K' calculates has elapsed in the frame of K? If so then of course the answer is just t.

Knowing that he can calculate how far he has gone (according to himself).

What frame do you mean when you say "according to himself"? The notion of "distance travelled" has no frame-independent meaning in relativity. If you don't explicitly say what frame you're asking for a given answer in, your questions are meaningless.

Please provide all four equations together please. You seem set on refusing to do so.

I'm not "set on refusing to do so", I figured you would have the reading comprehension necessary to put together the answers I gave in different posts, and I also requested that you clarify what frame you were asking for the calculations to be in. The answers I have already given so far (although I am not sure if they are in the frames you're thinking of, since you haven't answered my questions about this) were:

1. "if a clock moves at speed v for time t in some inertial frame (such as the frame of K), it will only elapse a time of $$t * \sqrt{1 - v^2/c^2}$$ in that time"

2. "And if the distance between two fixed objects at rest in a given frame (like K's own position and a space station at the position where K' turns around) is x, then the distance between them in a frame moving at speed v relative to the objects is $$x * \sqrt{1 - v^2/c^2}$$"

3. "K' can carry a watch and measure how much time has elapsed between first passing K and then later passing K again after the turnaround (according to K' the time would be $$t * \sqrt{1 - v^2/c^2}$$, if t is the time measured by K between their two meetings, and K is an inertial observer)."

4. "if we use an inertial frame where K' is at rest for one phase of the trip and moving at speed 2v/(1 + v^2/c^2) for the other phase of the trip (either the frame where K' is at rest before the turnaround and moving after, or the frame where K' is moving before the turnaround and at rest after), then the distance traveled by K' during the moving phase will be $$\frac{2x}{\sqrt{1 - v^2/c^2}}$$."

If this is not an article of faith for you, please write all four equations together in one post. If this is an article of faith for you, just acknowlege it and I will accept that.

Please drop the accusations of dogma, they are insulting and nothing I have said justifies this sort of accusation, I am perfectly willing to answer any well-defined question you have, and I had no way of knowing that you wouldn't consider your question answered unless I put the equations in an easy-to-read list as opposed to putting each one in different paragraphs of two different posts. And since I have answered your questions as best I could given the ambiguity in the way you stated them, I would appreciate it if you would address my requests for clarification about which frame you're asking about in each one, I may have to modify the equations depending on your answer (note that I clearly stated what frame or clock I was talking about in each of my answers).

 

[neopolitan]

JesseM,

I am willing to accept that you are not deliberately muddying the waters. So I will present the answers. You can argue them to your heart's content.

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

Note that 2 and 4 are the same. Note that 1 and 3 are not the same.

This is the problem.

Let's save some time. You will argue that this is not a problem. I will argue that not only is it a problem, but you can actually derive the last two equations as the correct equations for relativistic effects in at least four different ways, even if you use the light clock (correctly). You will argue that I don't know what I am talking about and that I must go to four years of physics studies to understand these things properly. I will argue that four years of engineering studies plus many years of application have give me more than enough mathematics to work these things out and that the only difference I can possibly see that physics studies might make involves indoctrination, rather than better mathematics skills to be applied to what mathematicians consider "trivial". You will consider that offensive but you will, in the same breath, refuse to take an open-minded look at what I have to say, which will (in my frame of reference) prove what I had just said.

Can you prove me wrong on the last step in this process?

cheers,

neopolitan

By the way, I was not accusing you of being dogmatic. I just didn't assume it wasn't the case and gave you the opportunity to clarify one way or the other. You have to admit that it did work as an incentive to write all four equations together. Being polite sure wasn't working.

 

[Doc Al]

I am going to be Socratean here. I know Socrates was annoying and eventually the Atheans persuaded him to kill himself, but if I give you the answer, you won't own it. If you work it out yourself, you might just decide to give me a hand rather than fight me every step of the way.

I am going to ask you to provide for me four equations. The first two are simple: the equation for time dilation (in terms of t, c and v) and the equation for length contraction (in terms of x, c and v).

For the third and fourth equations, consider this scenario. K' knowingly travels for an extremely long time away from K at a rather slow speed of v, relative to K. Then, in an extremely short period of time, K' changes direction 180 degress and then travels back to K at a speed of v, relative to K. According to K, K' has been away for a period of t and therefore traveled (approximately) a total distance of v.t = x

Equation three: How long has K' been away, according to K', in terms of t, c and v?

Equation four: How far did K' travel, according to K', in terms of x, c and v?

I will continue the discussion once you have written up the equations. You may care to discuss them in your answer. But until you, or someone else, presents those four equations, there really is nothing to discuss. Until then, it's nothing more than opinions.

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".

And what gives with the bogus "challenges"?

 

[JesseM]

JesseM,

I am willing to accept that you are not deliberately muddying the waters. So I will present the answers. You can argue them to your heart's content.

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

This is correct if t represents the time measured on the moving clock, and t' is the time between these same two readings as measured in the frame where the clock is moving at speed v. My equation assumed t was the time in the frame where the clock was in motion, so these are equivalent.

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

Same as my equation, so presumably x is the distance between ends of the moving object in its own frame, x' is the distance between the ends of the same object in the frame where it's moving at speed v.

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

Same as my equation, so again, I presume that here you are assuming t is the time as measured in the K rest frame where K' is moving at speed v, while t' is the time elapsed on the clock of K' (this is the opposite of the convention in equation 1).

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

Meaningless unless you specify which frame you are doing the calculation in. K' does not have a speed of v in his own rest frame during either phase of the trip, obviously. K' has a velocity of v in the rest frame of K, but if you're using the rest frame of K, then there is no need to apply length contraction, since x was already supposed to be the distance in the frame of K.

From your answer here, and your unwillingness to answer my repeated requests for clarification about what frame you're using, I gather you are fairly confused about frame-dependent vs. frame-independent quantities in relativity, and the fact that claims about distance (unlike time) are always specific to a particular frame.

Note that 2 and 4 are the same. Note that 1 and 3 are not the same.

1 and 3 are only "not the same" because you have switched the meaning of t and t'. In 1 you seem to be using t to represent the time elapsed on the clock of K', and t' to represent the corresponding time elapsed in the K frame; but in 3 you seem to be doing the opposite, with t as the time in the K frame, and t' as the time elapsed on the clock of K' during this time.

Let's save some time. You will argue that this is not a problem. I will argue that not only is it a problem, but you can actually derive the last two equations as the correct equations for relativistic effects in at least four different ways, even if you use the light clock (correctly). You will argue that I don't know what I am talking about and that I must go to four years of physics studies to understand these things properly.

Again with the thinly-veiled accusations of dogma. I would not answer your questions by saying something like that, since it would be little more than an appeal to authority and would show that I was not able to find any specific fault in your analysis; in fact the problem is just that you are making some rather simple conceptual errors, which I tried to explain above.

the only difference I can possibly see that physics studies might make involves indoctrination

More accusations of dogma! You seem to be supremely confident that you are right without even waiting for my response, and you seem to totally discount the possibility that you might be making some errors in your analysis. This is a terrible way to approach any intellectual subject! Unless one is open to the possibility that they may have made a mistake when they reach a conclusion that seems to differ from what the experts say, then any initial misconceptions they may have when starting to study a subject will become ossified, and they will invent grand theories of collective delusions throughout the community of experts in order to preserve the ego-gratifying certainty that they are right and everyone else is wrong.

Can you prove me wrong on the last step in this process?

Your last step is wrong because you have not specified what frame you are using, and your answer wouldn't be right in either the rest from of K' or the rest frame of K. If you have a ruler moving inertially, then whether it is at rest relative to K' (during one phase of the trip) or at rest relative to K, in neither case will the difference between the initial position and the final position of K' be equal to $$x * \sqrt{1 - v^2/c^2}$$. If you think there is some other physically meaningful way to define "distance travelled" besides difference in starting and ending position on some ruler, please specify it.

By the way, I was not accusing you of being dogmatic. I just didn't assume it wasn't the case and gave you the opportunity to clarify one way or the other. You have to admit that it did work as an incentive to write all four equations together. Being polite sure wasn't working.

Jeez, nice rationalization for rude behavior! You never even asked politely that I group them all together, you just jumped directly into trying to provoke me. Like I said, I had already provided all four equations, how was I supposed to know that you wouldn't consider the request answered unless I put them all in one place?

 

[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?