Time Dilation: Stationary vs. Moving Atomic Clocks

[spaghed87]

Hello, I'm Eddie and this is my first post here. I'm almost a Junior in college going for a B.S. In chemistry but I must say that time dilation is one of the most interesting things in science. I was wondering if anyone has actual figures with regards to the difference in time between a stationary and moving atomic clock at certain speeds? Please don't yell at me if this is already posted because I'm new here and since I'm in a rigorous field I don't have time to study much other than chemistry. If so just redirect me to that post please. Thanks for reading/responding to my question.

 

[pmb_phy]

Hello, I'm Eddie and this is my first post here. I'm almost a Junior in college going for a B.S. In chemistry but I must say that time dilation is one of the most interesting things in science. I was wondering if anyone has actual figures with regards to the difference in time between a stationary and moving atomic clock at certain speeds? Please don't yell at me if this is already posted because I'm new here and since I'm in a rigorous field I don't have time to study much other than chemistry. If so just redirect me to that post please. Thanks for reading/responding to my question.

Welcome to the forum spaghed87.

I'm sorry but off hand I don't have that data. If you're interested in the theory behind it then you can take a look at a web page I created for that puyrpose. The physics is simple and it only require simple algebra to understand. Meanwhile let's hope someone can get you that hard data.

See - http://www.geocities.com/physics_world/sr/light_clock.htm

Best wishes

Pete

 

[Jonathan Scott]

According to Special Relativity, the rate at which time elapses for a clock moving at speed v compared with one at rest is given by sqrt(1-v^2/c^2) where c is the speed of light in vacuum. When v much smaller than c, the fractional difference is about 1/2 (v^2/c^2). That is, for every 1 second of elapsed time, the moving clock loses 1/2 (v^2/c^2) seconds. For example, if the moving clock is going at about 300 m/s (as for a fast subsonic aircraft) and c is assumed to be 300,000,000 m/s, then the speed is about a millionth of the speed of light and the moving clock is slower by 1 part in 2,000,000,000,000 (around 43 nanoseconds per day).

 

[Jonathan Scott]

Sorry, I may have misunderstood your question. If you want the actual experimental figures, look up the "Hafele-Keating" experiment. There is for example an entry in Wikipedia which includes the results. This also needed to take into account General Relativity for differences in gravitational potential.

 

[jtbell]

For references to lots of experiments that test special relativity in various ways, including time dilation:

http://www.physics.adelaide.edu.au/~dkoks/Faq/Relativity/SR/experiments.html

 

[longshinewoole]

Difficult language

[I'm sorry but off hand I don't have that data. If you're interested in the theory behind it then you can take a look at a web page I created for that puyrpose.
See - http://www.geocities.com/physics_world/sr/light_clock.htm

Best wishes

Pete[/QUOTE]

Pete, it was difficult to understand your language. In the web page you created, you used the words "clock" and "measure". Such language must mean that the time intervals will be read from the clocks, not calculated by means of the classic equation t = d/v. For instance, using a clock to measure a time interval, it means a departure time and an arrival time will be read from the clock, and the arrival time minus the departure time will give us the time interval. Since you did not use the tool, the clock, how can you tell us the tool worked slower or faster.

By the word "measure", you meant you would take such a measuring action, reading a departure time and an arrival time. But you did not.

Would you explain why?

 

[pmb_phy]

Pete, it was difficult to understand your language. In the web page you created, you used the words "clock" and "measure". Such language must mean that the time intervals will be read from the clocks, not calculated by means of the classic equation t = d/v.

That is correct. The measurement of the time as read in the rest frame of the clock is what you'd intuitively think of doing. Record the reading on the clock as it passes through the first event (i.e. flash of light is emitted). Then record the reading as it passes through the second event (i.e. flash of light is detected). All this is accomplished in the rest frame, S, of the clock. Then a similar procedure is done in the rest frame of the observer, a frame in which the clock is moving. Call this frame S'. In S' (as in all frames) there are many clocks which are all synchronized (Note: While the clocks are synchronized in one frame they need not be synchronized in another). Observers in S record the reading on the S' clock which is ideally co-located with the first event. Then the recording on another clock in frame S' is taken when the clock at rest in S passes through the second event. The difference of these time readings is called the time difference.

By the word "measure", you meant you would take such a measuring action, reading a departure time and an arrival time. But you did not.

Would you explain why?

It is self explanitory. In fact you're the first person who has ever made such a comment amoung many people who have read that page over the last 5 years or so.

Pete

 

[Garth]

Hello, I'm Eddie and this is my first post here. I'm almost a Junior in college going for a B.S. In chemistry but I must say that time dilation is one of the most interesting things in science. I was wondering if anyone has actual figures with regards to the difference in time between a stationary and moving atomic clock at certain speeds? Please don't yell at me if this is already posted because I'm new here and since I'm in a rigorous field I don't have time to study much other than chemistry. If so just redirect me to that post please. Thanks for reading/responding to my question.

Hi Eddie, and also welcome from me to these Forums!

In General Relativity gravitational fields also cause time dilation between two clocks at different altitudes (potentials) in the field.

See Gravitation Research Using Atomic Clocks in Space and Space Probe to Test Einstein's "Space-Time Warp" Theory from the Gravity Probe B website. (Just click on the underlined links)

At an altitude of 10,000 km Gravity Probe A measured a time dilation of 7 parts per 10¹⁰.

Garth

 

[spaghed87]

Thanks for the replies. I now have a better understanding. Since General Relativity gravitational fields also cause time dilation are they considered for mathmatical formulas that also include the effect Special Relativity has on time dilation?

 

[Garth]

Thanks for the replies. I now have a better understanding. Since General Relativity gravitational fields also cause time dilation are they considered for mathmatical formulas that also include the effect Special Relativity has on time dilation?

Yes the field equations of GR are fully covariant which means they naturally subsume SR effects, the GR metric tends to the Minkowski SR metric when the gravitational field tends to zero.

Garth

 

[longshinewoole]

difficult in language

That is correct. The measurement of the time as read in the rest frame of the clock is what you'd intuitively think of doing. Record the reading on the clock as it passes through the first event (i.e. flash of light is emitted). Then record the reading as it passes through the second event (i.e. flash of light is detected). All this is accomplished in the rest frame, S, of the clock. Then a similar procedure is done in the rest frame of the observer, a frame in which the clock is moving. Call this frame S'. In S' (as in all frames) there are many clocks which are all synchronized (Note: While the clocks are synchronized in one frame they need not be synchronized in another). Observers in S record the reading on the S' clock which is ideally co-located with the first event. Then the recording on another clock in frame S' is taken when the clock at rest in S passes through the second event. The difference of these time readings is called the time difference.

It is self explanitory. In fact you're the first person who has ever made such a comment amoung many people who have read that page over the last 5 years or so.

Pete

Pete,
If you were truly reading time intervals from the clocks, then you were unable to write those 4 equations of yours. You were only able to write: departure time - arrival time = time interval. For instance, the train departed at 1 o'clock and arrived at 2 o'clock. 2 - 1 =1 hour.

In your web page, you used shorter and longer distances to obtain shorter and longer time intervals by means of t = d/v. You did not read from the clocks. If you would use distances to obtain time intervals, you should tell us so from the beginning.

If you told us so from the beginning, I believe you would not be able to tell us time was running slower or faster in such and such frames; you were only able to tell us this: a shorter distance resulted in a shorter time interval, while a longer distance resulted in a longer time interval.

 

[bernhard.rothenstein]

Hello, I'm Eddie and this is my first post here. I'm almost a Junior in college going for a B.S. In chemistry but I must say that time dilation is one of the most interesting things in science. I was wondering if anyone has actual figures with regards to the difference in time between a stationary and moving atomic clock at certain speeds? Please don't yell at me if this is already posted because I'm new here and since I'm in a rigorous field I don't have time to study much other than chemistry. If so just redirect me to that post please. Thanks for reading/responding to my question.

Time dilation is a fundamental relativistic effect proved by experiment.
I recommend to start with
t=(t'+Vx'/cc)/sqrt(1-VV/cc) (1)
and to mention that in (1) t represents the reading of a clock C of the I frame t' representing the reading of a clock C' of the I' frame when they are instantly located at the same point in space. Clocks C and C' are synchronized with the other clocks of theirs reference frames and all the clocks read a zero time when the origins of the two frames are located at the same point in space.
Presenting (1) as
dt=(dt'+Vdx'/cc)/sqrt(1-VV/cc) (2)
dt and dt' represent non-proper (coordinate time intervals) measured as a difference between two clocks of the involved reference frames respectively located at the points where distant events take place.
For dx'=0 (2) becomes
dt=dt(0)/sqrt(1-VV/cc) (3)
dt representing a coordinate time interval whereas dt(0) represents a proper time interval measured as a difference between the readings of a single clock of I' present at the point where the involved events take place. I think that without stating all that the discussion will never end.
soft words and hard arguments

 

[MeJennifer]

A comment about "clocks slowing down"

I like to make a comment here, I think that this notion of "clocks slowing down" that is frequently presented here on PF is an incorrect description which only leads to more confusion about relativity.

Suppose there are two beacons pulsating a green light each second. Each beacon can also measure the pulse frequency and the light frequency of the other beacon. What is called "time dilation" is when each beacon measures, after discounting the Doppler effect, a reduced pulse frequency and a lower light frequency from the other beacon.

But clearly the light that each beacon sends out is green, and the clock triggers one pulse each second. Obviously something does not slow down or changes frequency just because it is in relative motion with something else.

I think it is correct to state that the rate and frequency of the light signals is somehow observed differently at the destination than at the source, in other words the light signals appear to be "time dilated". But I think it is incorrect to state that the rate and frequency of the light signals at the source are changed.

The other case where "clocks slowing down" is incorrect is when two observers travel between two events on different spacetime paths. If one observer records less time it does not mean that his clock runs slower it simply means that he took a shorter path in spacetime.

A clock that really slows down is a clock that is obviously defect!

 

[pmb_phy]

longshinewoole

Each web page I create was created to help someone with a specific problem that either came up, kept comming up, or I thought was interesting enough to discuss. Each page does not describe relativity in its entirety as this page doesn't either. The reader is assumed to know what it means to take a time measurement and what a time interval is etc. Everything in that page is as it should be. The "t" in that expression is an actual time interval and it is that time interval that is calculated from readings on clocks. In fact that is how I defined it, i.e. Let "t" equal the time time between these two events as measured in S'.. Two readings are taken from clocks. One reading is what the clocks in S' read at event #1 and what the clocks in S' read at event #2. The difference of those time measurements will then be the time interval as measured in S'. The calculations are then expressed in terms of recorded measurements such as speed, that time interval, the proper time interval etc. The "t" is not deduced from a calculation t = d/v. That is just a relationship between physical quantities which are required in the calculations. I could have written "delta t" but I really didn't like that delta in this page. It was merely a matter of aestetics for me.

If you'd like I could create another web page to describe all of this to you and others who are having a difficult time with it. The light clock web page would then have a link to that page in an appropriate place and will be required reading. As of now I don't see a need to change anything.

If the people reading this are unable to understand the calculations and how the physical quantities are obtained then I'll consider creating another page on the meaning of recording events etc. But neither of you seem to have a problem following the derivation, or have you?

Pete

 

[longshinewoole]

Pete,

I have additional difficulty in your light clock.

In your light clock setup, you showed that light departed from the clock at its second position, light emitted, event 1, while the clock was moving to the right at v. Your setup also showed light arrived back at the clock at its 4th position, light detected, event 2. It did not arrive back at the first postion. Such a departure time and arrival time must give a time interval equal to the time interval that the light took to cover the two hypotenuses, that should be t, not r.

Hence I think your r was flawed because the light did not arrive back at the first position. Instead, the light arrived back at the clock when it has reached the 4th position.

 

[pmb_phy]

Pete,

I have additional difficulty in your light clock.

In your light clock setup, ...

You do realize that the light clock is of several a standard ways to derive the time-dilation relation, right? It can be found in many textbooks.

..you showed that light departed from the clock at its second position, ...

That is its first position. You're looking at two diagrams right next to each other.

..light emitted, event 1, while the clock was moving to the right at v. Your setup also showed light arrived back at the clock at its 4th position,..

Actually its the third position.

.. light detected, event 2. It did not arrive back at the first postion.

It wasn't supposed to. If you measure the time interval between two events and the two events occur at the same location in one frame then they won't occur at the same location at the other. This fact is very important in understanding time dilation.

I highly recommend that you surf the internet for a "Light Clock" since I, of course, did not originate this idea. This is a standard derivation for the time-dilation formula as I mentioned above. If you come across a website which does a much better job at it than I do then please post the URL so that I can see a better way to explain it. Okay? Thanks.

Pete

 

[bernhard.rothenstein]

time dilation, light clocks and wrist watches

[I'm sorry but off hand I don't have that data. If you're interested in the theory behind it then you can take a look at a web page I created for that puyrpose.
See - http://www.geocities.com/physics_world/sr/light_clock.htm

Best wishes

Pete

Pete, it was difficult to understand your language. In the web page you created, you used the words "clock" and "measure". Such language must mean that the time intervals will be read from the clocks, not calculated by means of the classic equation t = d/v. For instance, using a clock to measure a time interval, it means a departure time and an arrival time will be read from the clock, and the arrival time minus the departure time will give us the time interval. Since you did not use the tool, the clock, how can you tell us the tool worked slower or faster.

By the word "measure", you meant you would take such a measuring action, reading a departure time and an arrival time. But you did not.

Would you explain why?[/QUOTE]
The light clock is involved with some "wristwatches" located located on each mirror in each of the involved inertial reference frames I and I' respectively. Let C'(0) and C(0) be the clocks on the mirrors located on the overlapped OX(O'X') axes in I and I' respectively. When they read a zero time they are located at the same point in space and the light ray is emitted. When the light rays (emitted and reflected) return to the OX axis clock C'(0) arrives in front of a clock C of the I frame syncronized with C(0) using Einstein's clock synchronization procedure. Clock C'(0) reads now t' whereas clock C reads t. Clock C'(0) measures a proper time interval t'-0 whereas in I observers measure a nonproper time interval t-0 related by
(t-0)=(t'-0)/sqrt(1-VV/cc)
a relationship which can be derived without using the light clock as shown in Asher Peres, "Relativistic telemetry," Am.J.Phys. 55 943-946 (1987).
soft words and hard arguments

 

[robphy]

The light clock is involved with some "wristwatches" located located on each mirror in each of the involved inertial reference frames I and I' respectively.

I disagree.
Given an inertial worldline, a light clock [constructed with a light source and receiver and an inertial distant mirror traveling alongside] functions as a wristwatch for that worldline. There is no need to consider a "wristwatch" on the distant mirror for this light clock.

 

[bernhard.rothenstein]

light clock and wrist watch

I disagree.
Given an inertial worldline, a light clock [constructed with a light source and receiver and an inertial distant mirror traveling alongside] functions as a wristwatch for that worldline. There is no need to consider a "wristwatch" on the distant mirror for this light clock.

I agree with you. But for a more sophisticated reader, the clock on the distant mirror in the moving light clock is synchronized with the clock on the ground mirror by the outgoing light signal in the moving inertial reference frame I'. Detecting that synchronization from the stationary reference frame I, we arrive at a tringle to which we can apply Pithagoras' theorem in the reference frame of the stationary light clock

cctt=vvtt+dd (1)
d invariant distance between the mirrors, ct distance traveled by the light signal that performs the synchronization in the stationary reference frame, vt distance traveled by the wrist watch on the distant mirror of the moving light clock detected from I.
From (1)
ct=d/sqrt(1-VV/cc)=ct'/sqrt(1-VV/cc). (2)
Somebody on the Forum told me that when it is about special relativity we are all childish. Am I?
Thanks and regards

 

[bernhard.rothenstein]

light clock and wrist watches again

I disagree.
Given an inertial worldline, a light clock [constructed with a light source and receiver and an inertial distant mirror traveling alongside] functions as a wristwatch for that worldline. There is no need to consider a "wristwatch" on the distant mirror for this light clock.

Thanks. As you can see the "wristwatch" on the distant mirror is not taken into account in my message.

 

[robphy]

I disagree.
Given an inertial worldline, a light clock [constructed with a light source and receiver and an inertial distant mirror traveling alongside] functions as a wristwatch for that worldline. There is no need to consider a "wristwatch" on the distant mirror for this light clock.

Thanks. As you can see the "wristwatch" on the distant mirror is not taken into account in my message.

Right... that's why we don't need to add unnecessary complications.

This reminds me of a short conversation with a math professor of mine.
I said "A rectangle is a parallelogram with four right angles."
He stopped me and said "A rectangle is a parallelogram with one right angle."
While what I said is correct, it has unnecessary complication... and follows from his statement [which is easier to prove].

 

[longshinewoole]

language problem

I highly recommend that you surf the internet for a "Light Clock" since I, of course, did not originate this idea. This is a standard derivation for the time-dilation formula as I mentioned above. If you come across a website which does a much better job at it than I do then please post the URL so that I can see a better way to explain it. Okay? Thanks.

Pete

Pete, it is not necessary to look at any other light clocks because all are the same: language problem. Saying one thing and doing quite the opposite. Namely saying to use clocks to measure time intervals, but instead using t=d/v to produce them.

Let us look at yours again. Light departed from the clock. There must be a departure time, say det. Light arrived back at the clock and there must be an arrival time, say art. To derive the time interval is to deduct det from art, namely t = art - det. Since there is only one departure, only one arrival, there can be only one time interval.

But yours or anybody else showed us two different time intervals. These were produced from a shorter distance and a longer distance by means of t=d/v. You turned around and said these were read from the clocks. I am sorry I think there must be something wrong with your doing that.

 

[JesseM]

Let us look at yours again. Light departed from the clock. There must be a departure time, say det. Light arrived back at the clock and there must be an arrival time, say art. To derive the time interval is to deduct det from art, namely t = art - det. Since there is only one departure, only one arrival, there can be only one time interval.

But this is the whole point, in relativity there is no single way to measure the time between any two events, if you use clocks in different states of motion you'll get different answers, and you have no basis for saying one clock is measuring the "real" time while the others are not.

In the frame where the light-clock is in motion, clocks at rest in that frame will measure t = art - det to be greater than the time between these same two events as measured in the frame where the light-clock is at rest.

 

[bernhard.rothenstein]

wrist watch kight clock

Right... that's why we don't need to add unnecessary complications.

This reminds me of a short conversation with a math professor of mine.
I said "A rectangle is a parallelogram with four right angles."
He stopped me and said "A rectangle is a parallelogram with one right angle."
While what I said is correct, it has unnecessary complication... and follows from his statement [which is easier to prove].

A single question. We use clocks in order to assingn a time coordinate to an event. Please define the point where takes place the event to which we assign the time displayed by a light clock?
Regards

 

[robphy]

A single question. We use clocks in order to assingn a time coordinate to an event. Please define the point where takes place the event to which we assign the time displayed by a light clock?
Regards

We assume a suitable distant mirror of the light clock [as described earlier].
Call the emission event on the inertial worldline of the source "tick 0".
Call the first reception event (after reflection) on the inertial worldline of the source "tick 1".
Call the second reception event (after reflection) on the inertial worldline of the source "tick 2".
etc...

Note: the units of my clock are in "ticks" [of this light clock].

 

[pmb_phy]

A single question. We use clocks in order to assingn a time coordinate to an event. Please define the point where takes place the event to which we assign the time displayed by a light clock?
Regards

The events that I used were the emission and detection of a flash of light. The emission event was event 1 while the detection event was event 2.

Pete

 

[bernhard.rothenstein]

light clock and wrist watches

The events that I used were the emission and detection of a flash of light. The emission event was event 1 while the detection event was event 2.

Pete

Event 1 is E(1)(0,0,0) in I and E'(1)(0,0,0) in I'. In order to define them we should involve the wrist watches on the close mirrors C(0) and C'(0) respectively. Involving the light clock event 2 is E(2)(x,0,t) in I and
E'(2)(0,0,t'=2d/c), d representing the invariant distance between the mirrors. In order to define them we should involve the wrist watch C'(0,0) and C(x,0) of I, synchronized with C(0,0). Pythagoras' theorem, Asher Peres and the Lorentz-Einstein transformations as well, tell us that
t=t'/sqrt(1-VV/cc). Introducing the wrist watches it is obvious that t'-0 is a proper time interval and that t-0 is a non-proper one.
I think that the introduction of the wrist watches adds some physics to the light clock. Involving the clocks on the distant mirrors we can solve some interesting problems concerning length and time in two space dimensions. Of course cooking and teaching are questions of taste and we know that de gustibus nihil disputandum. You helped me to clarify for myself some problems concerning the largely uses pedagogical device the "light clock".
Thanks and regards
Bernhard

 

[longshinewoole]

difficult language

In the frame where the light-clock is in motion, clocks at rest in that frame will measure t = art - det to be greater than the time between these same two events as measured in the frame where the light-clock is at rest.

I am sorry, JesseM. I think there was language problem in your statement: clock at rest vs clock at rest.

Even though you were right, but my difficult with the light clock had nothing to do with motion or non-motion. The difficult I raised had something to do with this: whether the clock was read or not?

My finding was: it was not read. My finding cannot be false because all such thought experiments showed d/v as the time interval. None showed art - det.

 

[JesseM]

In the frame where the light-clock is in motion, clocks at rest in that frame will measure t = art - det to be greater than the time between these same two events as measured in the frame where the light-clock is at rest.

I am sorry, JesseM. I think there was language problem in your statement: clock at rest vs clock at rest.

I was comparing the time as measured by clocks at rest in a frame where the light-clock is moving with the time as measured by clocks at rest in the frame where the light-clock is at rest.

Even though you were right, but my difficult with the light clock had nothing to do with motion or non-motion. The difficult I raised had something to do with this: whether the clock was read or not?

My finding was: it was not read. My finding cannot be false because all such thought experiments showed d/v as the time interval. None showed art - det.

In every frame, art - det = d/v, if you use that frame's measure of the distance d the signal traveled and the speed v of the signal, along with the times art and det measured according to that frame's clocks.

 

[rewebster]

Hickory, dickory, doc

Einstein ran up the clock

and when he came down

and what was there found

by the guy on the ground

was that Einstein was off by a 'toc'

 

[longshinewoole]

language problem

In every frame, art - det = d/v, if you use that frame's measure of the distance d the signal traveled and the speed v of the signal, along with the times art and det measured according to that frame's clocks.

If you were right, then the language should never mention the tool clock. It should be saying: d/v will be used to produce time intervals. As such, the theorist cannot blame the clock for producing the different time intervals?

But I think you were wrong in that: (art - det) can produce only one time interval. Or, show me how to produce two different time intervals by means of art-det.

 

[JesseM]

If you were right, then the language should never mention the tool clock. It should be saying: d/v will be used to produce time intervals.

A basic principle of relativity is that the laws of physics are such that there is absolutely no way to distinguish between different inertial frames--if you are in a windowless box in space and you perform different experiments inside, the results will be exactly the same regardless of whether you are at rest relative to the galaxy or traveling at 0.99c relative to it. A consequence of this is that if two experimenters traveling at high speed have identical light clocks traveling alongside them, each experimenter should measure their own light clock to tick at the same rate relative to any other clocks they happen to have along (a wristwatch, an atomic clock, etc.) So, if we can show that each experimenter must measure the other experimenter's light clock to be running slower than their own, this means that each experimenter must measure all of the other experimenter's clocks to be running slow by the same rate. It's convenient to focus on the light clock because it's fairly easy to derive the amount that it must slow down, given that each experimenter measures the speed of light as c (which is another basic principle of relativity).

But I think you were wrong in that: (art - det) can produce only one time interval. Or, show me how to produce two different time intervals by means of art-det.

Because, as I said, any clock used by the experimenter#1 who's moving alongside the light clock will run at the same rate relative to the light clock. So, if experimenter#2 sees the light clock running slower, he'll also measure all of experimenter#1's other clocks running slower than his own. So if each experimenter uses his own clocks to measure the time-interval (art - det), they'll disagree about the length of the time interval.

For example, suppose the two mirrors of the light clock are 1 light-second apart, so that in the frame of experimenter #1 who's at rest relative to the light-clock, any other clock he uses--a stopwatch, say--will measure 2 seconds between the time the light leaves the base and the time it returns. Now, experimenter #2 sees the light-clock moving at 0.6c, which means he'll measure the light clock slowed down by a factor of 1.25, so if he has a stopwatch at the position of the base when the signal left it and another synchronized stopwatch at the position of the base when the signal returns to it, he'll measure a time of 2.5 seconds between these events. But experimenter#2 also measures experimenter #1's stopwatch to be slowed down by a factor of 1.25, so in 2.5 seconds he'll find that experimenter#1's stopwatch only ticked forward by 2.5/1.25 = 2 seconds.

 

[YellowTaxi]

For example, suppose the two mirrors of the light clock are 1 light-second apart, so that in the frame of experimenter #1 who's at rest relative to the light-clock, any other clock he uses--a stopwatch, say--will measure 2 seconds between the time the light leaves the base and the time it returns. Now, experimenter #2 sees the light-clock moving at 0.6c, which means he'll measure the light clock slowed down by a factor of 1.25, so if he has a stopwatch at the position of the base when the signal left it and another synchronized stopwatch at the position of the base when the signal returns to it, he'll measure a time of 2.5 seconds between these events. But experimenter#2 also measures experimenter #1's stopwatch to be slowed down by a factor of 1.25, so in 2.5 seconds he'll find that experimenter#1's stopwatch only ticked forward by 2.5/1.25 = 2 seconds.

Surely that's totally wrong again. In your last sentence you suggest that #2 observes time dilation AND time contraction (!) just depending on which moving clock he looks at... You're just trying to confuse everybody ?

It should actually say this:
In 2.5 seconds #2 would see #1's stopwatch ticked forward by 2.5*1.25=3.125 seconds.
Or more simply,
In 2 seconds #2 would see #1'swatch/clock/atom ticked froward by 2*1.25=2.5 seconds

 

[JesseM]

Surely that's totally wrong again. In your last sentence you suggest that #2 observes time dilation AND time contraction (!) just depending on which moving clock he looks at... You're just trying to confuse everybody ?

I'm not sure what you mean by "time contraction". Time dilation means that if a clock is moving in your frame, you will see the time between ticks be larger in your frame, which is equivalent to saying that the clock will tick less time than your own clocks in a given time-interval as measured in your frame. For example, if a moving clock takes twice as long to advance forward by one second, that's the same as saying that every 2 seconds on my clock, the moving clock advances forward only 1 second.

It should actually say this:
In 2.5 seconds #2 would see #1's stopwatch ticked forward by 2.5*1.25=3.125 seconds.

No, that would mean #2 was seeing #1's clock ticking faster than his own. That's the opposite of time dilation!

 

[YellowTaxi]

Ooops, sorry you're right
the way you worded it did seem very contradictory the first time I read it.

 

[longshinewoole]

language problem

For example, suppose the two mirrors of the light clock are 1 light-second apart, so that in the frame of experimenter #1 who's at rest relative to the light-clock, any other clock he uses--a stopwatch, say--will measure 2 seconds between the time the light leaves the base and the time it returns. Now, experimenter #2 sees the light-clock moving at 0.6c, which means he'll measure the light clock slowed down by a factor of 1.25, so if he has a stopwatch at the position of the base when the signal left it and another synchronized stopwatch at the position of the base when the signal returns to it, he'll measure a time of 2.5 seconds between these events. But experimenter#2 also measures experimenter #1's stopwatch to be slowed down by a factor of 1.25, so in 2.5 seconds he'll find that experimenter#1's stopwatch only ticked forward by 2.5/1.25 = 2 seconds.

Let me draw two figures to represent the experiment you described above.
Figure 1:

#1>>>>v
A_______________________________________B

#2

The light signal departs at point A. Both #1 and #2 have a clock to record the departure. Here I believe both will record the same departure time. Let us called the departure time det.

#1 moves at v or 0.6c toward B.

Figure 2:
When #1 arrives at B, the light signal arrives at B too.

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>#1
A_______________________________________B
...............#2's clock

#2 also has a clock at B to record the arrival of the light signal. Here I believe both will record the same arrival time. Let us call it art.

art - det = t, one time interval. So, I could not get two different time intervals 2 seconds and 2.5 seconds even if I follow your description.

I wish to repeat that, the light signal departed at A, and arrived at B, constituting the two events. One departure and one arrival can only produce one identical time interval. That was my understanding.

It was a fact that you were still using distances to produce your time interval, the vertical distance between the two mirrors for the 2 seconds. Such a fact will mean light departed at point A and arrived back at point A. But your words included the motion of #1, which meant that light did not arrive until #1 arrived at point B. Hence it means that there was contradiction (A or B) in your own words.

My difficulty was, when you use clocks to measure, you cannot obtain two different time intervals. You did not solve my difficulty. Instead, you arbitrarily used two different time intervals to show that there were two different time intervals. Don't you think it was illogical to do so?

From the beginning when I took part in this thread of time dilation, I said it had language problem. It said observers would use clocks, but clocks were never used. If a teacher gives an assignment to some student to use light clocks (or radar) to measure the time interval light would take to travel between two mountains. The students found out from the map that the distance between the two mountains was n miles. They did not go to the mountains and do the measurement; they just used n/c = t and reported to their teacher that they have measured light would take t seconds to travel between the mountains. It you were the teacher, what would you think of the students?

 

[longshinewoole]

For example, suppose the two mirrors of the light clock are 1 light-second apart, so that in the frame of experimenter #1 who's at rest relative to the light-clock, any other clock he uses--a stopwatch, say--will measure 2 seconds between the time the light leaves the base and the time it returns. Now, experimenter #2 sees the light-clock moving at 0.6c, which means he'll measure the light clock slowed down by a factor of 1.25, so if he has a stopwatch at the position of the base when the signal left it and another synchronized stopwatch at the position of the base when the signal returns to it, he'll measure a time of 2.5 seconds between these events. But experimenter#2 also measures experimenter #1's stopwatch to be slowed down by a factor of 1.25, so in 2.5 seconds he'll find that experimenter#1's stopwatch only ticked forward by 2.5/1.25 = 2 seconds.

JesseM, I had a new discovery of your experiment. I think you have predetermined the outcome of your experiment.

As a rule, whether an experiment will give a certain outcome or not is determined at the end of an experiment. In your case, whether the clocks would measure two different time intervals or not should be determined at the end of the experiment--comparing the clocks. But in the experiment described by you, the outcome had been predetermined before the experiment was underway.

What do you think?

 

[JesseM]

Let me draw two figures to represent the experiment you described above.
Figure 1:

#1>>>>v
A_______________________________________B

#2

The light signal departs at point A. Both #1 and #2 have a clock to record the departure. Here I believe both will record the same departure time. Let us called the departure time det.

#1 moves at v or 0.6c toward B.

Figure 2:
When #1 arrives at B, the light signal arrives at B too.

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>#1
A_______________________________________B
...............#2's clock

#2 also has a clock at B to record the arrival of the light signal. Here I believe both will record the same arrival time. Let us call it art.

art - det = t, one time interval. So, I could not get two different time intervals 2 seconds and 2.5 seconds even if I follow your description.

First of all, your diagram does not resemble a light clock! With a light clock you have two mirrors which should be arranged vertically in your diagram, and in experimenter #2's frame, the light clock is moving perpendicular to the vertical axis between the mirrors, with the light bouncing diagonally from the bottom mirror to the top mirror and back down to the top mirror. Look at the diagrams on this page--I can't post images here, but this diagram shows how the light clock and bouncing light beam look in the frame of experimenter #1 who is at rest relative to the clock, and this diagram shows how the clock and beam look in the frame of experimenter #2 who sees the light clock to be moving to the right. What I was saying that if experimenter #1 had a wristwatch affixed to the bottom mirror and measured the time the light departed from the bottom and the time it returned, and then experimenter #2 had two synchronized wristwatches, one at the position of the bottom mirror when the light departed and another at the position of the bottom mirror when the light returned (in his frame these are different positions, since the light clock has moved sideways in between these events, as shown in the second diagram), and he used these wristwatches to measure the time between departure and return, then he will measure a different time than experimenter #1. But to understand why this is, you have to remember that in experimenter #2's frame, experimenter #1's wristwatch is slowed down relative to his own wristwatches! If you assume both experimenter #1's wristwatch and experimenter #2's wristwatch tick at the same rate, you will indeed conclude they both get the same value for art - det, but this assumption is false according to relativity.

I wish to repeat that, the light signal departed at A, and arrived at B, constituting the two events. One departure and one arrival can only produce one identical time interval. That was my understanding.

Again, not if their wristwatches tick at different rates! If I measure your wristwatch to be ticking at half the rate of my own, then naturally when you measure the time between two events next to you, you'll find only half the time that I measure between those same two events.

It was a fact that you were still using distances to produce your time interval, the vertical distance between the two mirrors for the 2 seconds. Such a fact will mean light departed at point A and arrived back at point A. But your words included the motion of #1, which meant that light did not arrive until #1 arrived at point B. Hence it means that there was contradiction (A or B) in your own words.

Since your diagrams don't show two mirrors I don't know how to relate them to the light clock experiment. Each experimenter is measuring the time between event of the light departing the bottom mirror and the event of the light returning to the bottom mirror; in experimenter #1's frame both these events happen at the same point in space, so he can use a single wristwatch at rest next to the bottom mirror, while in experimenter #2's frame these events happen at different points in space, so if he wants to use wristwatches which are at rest in his frame, he has to use two synchronized watches at different positions, one at the position of the bottom mirror when the light departed and one at its position when the light returned. Again, just look at the diagrams I linked to and this should be fairly clear.

My difficulty was, when you use clocks to measure, you cannot obtain two different time intervals. You did not solve my difficulty. Instead, you arbitrarily used two different time intervals to show that there were two different time intervals. Don't you think it was illogical to do so?

I don't know what you mean. Each experimenter uses watches at rest in their frame to measure the time between two definite events, the event of the light leaving the bottom mirror and the event of the light returning to the bottom mirror (after being reflected by the top mirror). Naturally, if each experimenter's wristwatches tick at the same rate, they will find the same time interval; but if they tick at different rates, they will not. Relativity predicts they will not. This is an empirical question, to be settled by experiment--and huge numbers of experiments have shown that relativity's predictions about clock rates are correct, and the prediction that all clocks tick at the same rate is incorrect! What part of this is unclear?

From the beginning when I took part in this thread of time dilation, I said it had language problem. It said observers would use clocks, but clocks were never used. If a teacher gives an assignment to some student to use light clocks (or radar) to measure the time interval light would take to travel between two mountains. The students found out from the map that the distance between the two mountains was n miles. They did not go to the mountains and do the measurement; they just used n/c = t and reported to their teacher that they have measured light would take t seconds to travel between the mountains. It you were the teacher, what would you think of the students?

The point of the light clock thought-experiment is just to show that if the fundamental assumptions of relativity are correct--namely, the assumption that the laws of physics work exactly the same in all inertial frames, and the assumption that light travels at c in all frames--then it must be true that clocks which are moving relative to a given observer will be measured to run slower than the same clocks measured by observers at rest relative to them. The point of the thought-experiment is not to try to prove these assumptions are correct in the first place! That's a matter to be settled by experiment, and many many experiments have shown that the predictions of relativity are correct.

 

[longshinewoole]

First of all, your diagram does not resemble a light clock!

JesseM, though my diagrams did not show the vertical distance and the slanted distances, yet my language showed I was talking in terms of these. The light-clock setup was a familiar thing everywhere. In yours, the red slanted line shows the light departs at one point, but you did not name the departure point. In my diagram, I named it A. Your setup shows that light did not return to this point A. Instead your setup shows light returns to a new point on the right. Yours did not name this arrival point either. In mine, I named it B. Hence my diagrams included the meaning identical to yours. If we truly used clock to do the measurement, we would have a departure time at A, as shown by the red line in your setup. And we would have an arrival time at B, your red line on the right. These two events can only produce one identical time interval; that is, if we used clocks to do the measurement.

I would say, the purpose for setting up the light-clock in your way was to use distances to produce time intervals. Measurements by clocks play no part. Don't you agree? So why you included wristwatches etc. in your language?

 

[JesseM]

JesseM, though my diagrams did not show the vertical distance and the slanted distances, yet my language showed I was talking in terms of these. The light-clock setup was a familiar thing everywhere. In yours, the red slanted line shows the light departs at one point, but you did not name the departure point. In my diagram, I named it A. Your setup shows that light did not return to this point A. Instead your setup shows light returns to a new point on the right. Yours did not name this arrival point either. In mine, I named it B. Hence my diagrams included the meaning identical to yours. If we truly used clock to do the measurement, we would have a departure time at A, as shown by the red line in your setup. And we would have an arrival time at B, your red line on the right. These two events can only produce one identical time interval; that is, if we used clocks to do the measurement.

No, two events can produce different time intervals, if you use clocks that run at different rates. Remember, I'm using three wristwatches--one is moving along with the bottom mirror, and the other two are at rest in the frame where the mirror is moving (and synchronized in this frame), positioned at the points you call "A" and "B". If the wristwatch moving along with the mirror is ticking slower than the other two, isn't it obvious that it'll record a smaller time-interval? For example, suppose it's ticking at half the rate of the other two, do you disagree it'll record half the time that the other measure?

I would say, the purpose for setting up the light-clock in your way was to use distances to produce time intervals. Measurements by clocks play no part. Don't you agree? So why you included wristwatches etc. in your language?

If you grant that every light clock is ticking at the "correct" rate in the frame where it is at rest, then you don't need watches. But this is the whole premise you're questioning with your argument about art-det, is it not? Obviously if the two mirrors of the clock are 0.5 light seconds apart, so that you say in the light clock's rest frame it takes 1 second for the light to go from the bottom mirror to the top mirror, in the frame where the light clock is moving the light must take longer than 1 second to go from the bottom to the top and back, because it has to go further than 0.5 light seconds each way thanks to the horizontal motion of the clock. If you like, we could place two additional light clocks at A and B, which also have their mirrors 0.5 light seconds apart but which are at rest in the observer's frame. Do you disagree that if light is shot vertically up from the bottom mirrors of these two clocks at A and B at the same moment the moving clock passes A and light is shot diagonally up from the bottom mirror, then in this case the light will return to the bottom mirror of the two light clocks at A and B earlier than it returns to the bottom mirror of the moving light clock? Do you disagree that if we measure the time for the light to return to the bottom mirror of the moving clock using these two fixed light clocks at A and B, they will measure that time to be d/c, where the distance is the length of the two diagonal paths shown in the diagram as measured in the observer's frame?

 

[longshinewoole]

difficult language

Do you disagree that if we measure the time for the light to return to the bottom mirror of the moving clock using these two fixed light clocks at A and B, they will measure that time to be d/c, where the distance is the length of the two diagonal paths shown in the diagram as measured in the observer's frame?

I had difficulty with your language. Though you had the tool clocks, but you never used them to do the measurement. Instead all your time intervals were obtained from d/v. Here again, d/c, with the d representing the two diagonal distances and c for the speed of light.

Let me follow your practice, using a distance for a time interval. Your clock had speed v and covered a distance AB. Hence AB/v represents a time interval. I believe, d/c = AB/v = t.

This equation means, the time interval taken by light to cover the two diagonal distances is equal to that taken by the clock to cover the distance AB, as can be interpreted from your light-clock setup.

If there is nothing wrong with d/c = AB/v = t, then the time dilation idea is in doubt. Or, tell me what is wrong with my equation.

I think the time dilation idea originated from twisted language. You did not read the clock but you said you did. Though in fact you obtained a shorter time interval from a shorter distance, yet you told us you read it from the clock.

 

[JesseM]

I had difficulty with your language. Though you had the tool clocks, but you never used them to do the measurement. Instead all your time intervals were obtained from d/v.

Can you quote the post you're talking about? I don't remember saying one would obtain the time intervals from d/v, I said one should use wristwatches at rest in the frame to measure the times. According to relativity's predictions (which have plenty of experimental confirmation), the times art - det measured on the watches will in fact be equal to d/c if the distance d is measured from the same frame as the watches, but you are using the watches to measure the time and confirm this. I think I was pretty clear that the watches were being used to measure time, I'm not sure where you got the idea I was saying something different--for example, in post #32 I said:

Because, as I said, any clock used by the experimenter#1 who's moving alongside the light clock will run at the same rate relative to the light clock. So, if experimenter#2 sees the light clock running slower, he'll also measure all of experimenter#1's other clocks running slower than his own. So if each experimenter uses his own clocks to measure the time-interval (art - det), they'll disagree about the length of the time interval.

And in post #38 I said:

What I was saying that if experimenter #1 had a wristwatch affixed to the bottom mirror and measured the time the light departed from the bottom and the time it returned, and then experimenter #2 had two synchronized wristwatches, one at the position of the bottom mirror when the light departed and another at the position of the bottom mirror when the light returned (in his frame these are different positions, since the light clock has moved sideways in between these events, as shown in the second diagram), and he used these wristwatches to measure the time between departure and return, then he will measure a different time than experimenter #1.

And later in that same post I said:

Each experimenter uses watches at rest in their frame to measure the time between two definite events, the event of the light leaving the bottom mirror and the event of the light returning to the bottom mirror (after being reflected by the top mirror).

And in post #40 I said:

Remember, I'm using three wristwatches--one is moving along with the bottom mirror, and the other two are at rest in the frame where the mirror is moving (and synchronized in this frame), positioned at the points you call "A" and "B". If the wristwatch moving along with the mirror is ticking slower than the other two, isn't it obvious that it'll record a smaller time-interval? For example, suppose it's ticking at half the rate of the other two, do you disagree it'll record half the time that the other measure?

So, I think I was consistently pretty clear that I was talking about using the watches to measure the time intervals, and that the watches at rest in the observer's frame (where the light clock was moving) would measure a different amount of time than the watch at rest relative to the light clock. Again, the point of the light clock thought-experiment is not to prove that this must be the case! Please make sure you have read and understood this paragraph from post #38 where I went over the point of the thought-experiment:

The point of the light clock thought-experiment is just to show that if the fundamental assumptions of relativity are correct--namely, the assumption that the laws of physics work exactly the same in all inertial frames, and the assumption that light travels at c in all frames--then it must be true that clocks which are moving relative to a given observer will be measured to run slower than the same clocks measured by observers at rest relative to them. The point of the thought-experiment is not to try to prove these assumptions are correct in the first place! That's a matter to be settled by experiment, and many many experiments have shown that the predictions of relativity are correct.

Just to be clear that you have read and understood this, please answer the following question: when you dispute the light clock thought-experiment, are you disputing the claim that if the two basic postulates of special relativity (namely, the postulate that all laws of physics work the same way in every inertial reference frame, and the postulate that light travels at the same speed in every inertial frame), then that implies that all clocks (including wristwatches, atomic clocks etc.) which are moving in a given frame will be measured to slow down by the amount predicted by the time dilation equation? Or are you agreeing that this conclusion follows from the two postulates, but disputing the claim that the two postulates are correct in the first place?

Let me follow your practice, using a distance for a time interval.

That was not my practice, although I did claim that if you measured the time interval with wristwatches, you would find it was equal to the distance/speed in that frame.

Your clock had speed v and covered a distance AB. Hence AB/v represents a time interval.

No, the distance here is the distance covered by the light, which was traveling diagonally, not by the mirrors of the clock which were traveling horizontally. Since the light had to cover the horizontal distance between A and B and the vertical distance between the bottom mirror and the top mirror, it covered a larger distance than just the distance from A to B. If H is the height of the top mirror above the bottom mirror, and X is the distance from A to B, then you can just use the pythagorean theorem to see that the light must cover a distance of \sqrt{(X/2)^2 + H^2}, so if you add the two diagonal sections, the total distance covered by the light between A and B must be twice this.

If there is nothing wrong with d/c = AB/v = t, then the time dilation idea is in doubt. Or, tell me what is wrong with my equation.

Just look at the diagrams I linked to, where the two diagonal red lines represent the path taken by the light, and the base of the triangle represents the horizontal path of the bottom mirror. You can see that the length of the red lines is greater than the horizontal line, no?

 

[longshinewoole]

language problem

Can you quote the post you're talking about? I don't remember saying one would obtain the time intervals from d/v, I said one should use wristwatches at rest in the frame to measure the times.

Just look at the diagrams I linked to, where the two diagonal red lines represent the path taken by the light, and the base of the triangle represents the horizontal path of the bottom mirror. You can see that the length of the red lines is greater than the horizontal line, no?

If you did use clocks to measure, you should have got art - det = time interval. For instance, the train departs at 1 o'clock and arrives at 2, then 2 - 1 = 1 hour, the time interval. I said you never used the clocks because you never gave us such equations. All your equations come from d/v = t. You did say you would use wristwatches to measure the light traveling between the bottom and the top mirror, but you did not give us 1-0 = 1 second. Instead you gave us H/c = 1 second because its height was 0.5 light seconds apart. Your meth showed your language did not agree with your action.

I was looking at the diagrams. I do understand the two diagonal red lines are longer than the horizontal line. If you did use clocks to do the measurement, you would have got a departure time at point A (the first red line but not named in your diagram), and an arrival time at point B (the second red line but not named in your diagram). It the departure time was zero second and the arrival time was nth second, then you should have got n-0 =n seconds. Did you give us such an equation? No, what you gave us was d/c, d representing the two diagonals. You said you would use clocks but what you did was using the classic equation d/v = t. Your giving us the d/c was the action you took. I called it your practice. I followed this practice of yours to write AB/v, the time interval your moving clock needs to move from point A to B. I said, this time interval AB/v must be equal to your d/c. Were they equal or not?

I was not challenging any postulates, relativistic or else. I was challenging your language. I was challenging why you said one thing and did quite a different thing. This "you" represents all the people who used the thought experiment to prove the time dilation idea.

Let me repeat, to obtain time intervals with clocks, we shall have: art - det = t. To obtain time intervals with distances, we shall have d/v. If you gave us d/v, it means you did not use the clocks, though you might have said many times you would use wristwatches, atomic ones, etc.

 

[robphy]

I was not challenging any postulates, relativistic or else. I was challenging your language. I was challenging why you said one thing and did quite a different thing. This "you" represents all the people who used the thought experiment to prove the time dilation idea.

Let me repeat, to obtain time intervals with clocks, we shall have: art - det = t. To obtain time intervals with distances, we shall have d/v. If you gave us d/v, it means you did not use the clocks, though you might have said many times you would use wristwatches, atomic ones, etc.

I'm joining this conversation late (mainly because I didn't catch it from the beginning and didn't know where to jump in).

What the lightclock construction does is to establish a standard of time using mirrors and light rays. Making reference to d/v is a way to connect the general idea to time standards we might be familiar with like a wristwatch. However, one can dispense with this and just use the lightclock construction to compare time-intervals in units of the "tick of the [standard] lightclock". In particular, the time-dilation effect can be established using the lightclock alone [actually a pair of identically constructed lightclocks], without reference to another type of timepiece. In fact, one can use this standard to set up a coordinate system everywhere in Minkowski spacetime... and make measurements of all spacetime-intervals, in units of the "tick of the [standard] lightclock". d/v then is used to determine a distance in terms of the lightclock standard. (One can then build other mechanical clocks, like a wristwatch, and calibrate it with the lightclock.)

 

[MeJennifer]

The time it takes a photon to make a full roundtrip between two mirrors that are not in relative motion with each other is fixed. An observer who is in motion with respect to these mirrors will receive lights signals that will give him the impression that this time is different, but that is basically a relativistic effect. The reason for this is simple: light signals are always received at the speed of light even when they were emitted from an object that is in relative motion with the receiver.

Frames, coordinate systems, planes of simultaneity, relativity of simultaneity, tangent spaces, intervals between spacetime events, slowing down of clocks, contraction of objects, etc.
I fail to see how we help people understand relativity better by making the problem more difficult than it really is.

 

[JesseM]

If you did use clocks to measure, you should have got art - det = time interval. For instance, the train departs at 1 o'clock and arrives at 2, then 2 - 1 = 1 hour, the time interval. I said you never used the clocks because you never gave us such equations. All your equations come from d/v = t. You did say you would use wristwatches to measure the light traveling between the bottom and the top mirror, but you did not give us 1-0 = 1 second. Instead you gave us H/c = 1 second because its height was 0.5 light seconds apart. Your meth showed your language did not agree with your action.

You misunderstood, I never calculated the times on the wristwatches at all, I just claimed that if you actually performed this experiment using the wristwatches, the times they would measure would agree with the times calculated with the d/v equation. This must be true if you accept the two postulates of relativity: 1) the laws of physics work the same in every inertial frame, and 2) the speed of light is c in every frame. If postulate 2 is correct, this must mean that the time for light to travel a distance of d in any inertial frame must always be d/c. And if postulate 1 is correct, that means that the time measured by "light clocks" must always agree with the time measured by other types of clocks (such as wristwatches) which are at rest with respect to them--if they agreed in some frames but not others, this would mean the laws of physics don't work the same way in every frame.

I was looking at the diagrams. I do understand the two diagonal red lines are longer than the horizontal line. If you did use clocks to do the measurement, you would have got a departure time at point A (the first red line but not named in your diagram), and an arrival time at point B (the second red line but not named in your diagram). It the departure time was zero second and the arrival time was nth second, then you should have got n-0 =n seconds. Did you give us such an equation? No, what you gave us was d/c, d representing the two diagonals.

Yes, if the departure time was 0 and the arrival time was n, you would always find that n = d/c, assuming the second postulate of relativity is correct. Do you have any doubt that this follows automatically from the postulates of relativity? Do you think it's possible that the postulates of relativity could be correct, yet n would be something different than d/c?

You said you would use clocks but what you did was using the classic equation d/v = t.

No, I said that if you did use clocks in reality, then the actual time measured by these clocks would agree with the d/v calculation. I wasn't "calculating" what the clocks would read in some other way, I was making an empirical claim about what you would actually measure with clocks. Do you understand the difference?

Your giving us the d/c was the action you took. I called it your practice. I followed this practice of yours to write AB/v, the time interval your moving clock needs to move from point A to B. I said, this time interval AB/v must be equal to your d/c. Were they equal or not?

Ah, I didn't realize that the "v" in your equation was the velocity of the clock rather than the velocity of light. Yes, in this case I agree that AB/v = d/c, where d is the sum of the two diagonal distances and AB is the horizontal distance. But since AB, v and c are all measured in the frame where the light-clock is moving, these are only the times in that frame, which would be measured using synchronized clocks at rest in that frame--if you measure the time between the departure and arrival in another frame, like the rest frame of the light clock (using a wristwatch attached to the bottom mirror, for example), the time will be different.

I was not challenging any postulates, relativistic or else. I was challenging your language. I was challenging why you said one thing and did quite a different thing. This "you" represents all the people who used the thought experiment to prove the time dilation idea.

But it's not meant to "prove the time dilation idea" from nothing! It's just meant to prove that if the two basic postulates of relativity are true, then time dilation must occur, with moving clocks slowed down by a factor of \sqrt{1 - v^2/c^2}. Do you disagree that the light clock experiment does prove this? If you disagree, that must mean that you think it would be possible for the two basic postulates of relativity to be correct, yet for time dilation not to occur, or to obey a different equation. Do you think that's possible?

 

[longshinewoole]

Yes, in this case I agree that AB/v = d/c, where d is the sum of the two diagonal distances and AB is the horizontal distance.

From the beginning of my taking part in this time dilation thread, I pointed out I disagreed with the language you people were using. By language, I meant, if we obtained time intervals from speed and distance, our right to expain (why longer and shorter time intervals were obtained) must be restricted to using distances and speeds; we cannot use clocks, nor the action of measuring, since no measurements were ever done.

I also gave examples of how to read times from clocks, 1 o'clock, 2 o'clock, etc. We never can read distances and speeds from clocks. Another simple example.

distance D_________________E

distance F______________________________G.

Using speed c, DE will produce a shorter time interval while FG a longer time interval. If I told you I obtained the two different time intervals by means of clocks, what would you think of me?

Looking back at your thought experiment, you produced a shorter time interval from a shorter distance, the height of the mirror; and a longer time interval from a longer distance, the diagonals. Then you told us you obtained them from your clocks. Then you expalined to us that, why the shorter time interval? it was bacause one clock was running slower than the other. What would I think of your language?

If you agree you had no right to do so, then what would happen to the said thought experiment?

If you agree d/c = AB/v, then what would happen to the said thought experiment? as it means the thought experiment produced one identical time interval by using the distance traveled by the clock.

 

[JesseM]

Looking back at your thought experiment, you produced a shorter time interval from a shorter distance, the height of the mirror; and a longer time interval from a longer distance, the diagonals. Then you told us you obtained them from your clocks.

Why do you keep saying this? I've told you over and over again that I'm just saying that if you actually measure the time with clocks, you will find that this agrees with the time calculated by d/v. They are two separate methods of getting the time, which always end up giving the same result. To use the clock method, you'd have to actually check the times on the clocks, you couldn't just calculate d/v and assume that'd be the time interval.

Maybe it would help to give some numbers? Say the mirrors are 0.5 light-seconds apart, and in my frame they're moving at 0.6c. I place synchronized stopwatches at the positions A and B in your diagram, and there is also a stopwatch moving along with the mirrors, attached to the bottom one. At the moment the bottom mirror is at position A and the light is emitted, the moving stopwatch starts, reading a time of "0 seconds", and my stopwatch at A also starts, reading "0 seconds as well" (since my stopwatch at B is synchronized with the one at A, in my frame it reads '0 seconds' at the same moment as well, although since different frames disagree about simultaneity they would not agree about this). Then at the moment the bottom mirror is at position B and the light returns to it, the moving stopwatch reads a time of "1 second", and my stopwatch at B reads "1.25 seconds". So, the time interval art - det according to the moving stopwatch is 1 - 0 = 1 second, and the time interval art - det according to my two synchronized stopwatches is 1.25 - 0 = 1.25 seconds. You could predict these numbers using the time dilation equation, but you could also obtain them empirically, by actually setting things up this way and performing the measurement.

Now, we could also figure out the distance traveled by the light in both frames. In the rest frame of the light clock, since the light moves only vertically and the mirrors are 0.5 light-seconds apart, the distance is obviously 1 light second, so d/c = 1 second. In the frame where the light clock is moving, we have to figure out how far the horizontal distance is as well as the vertical distance. Let time t be the time it takes for the light to go from the bottom mirror to the top mirror; then the length of a diagonal is ct, while the length of the horizontal distance covered by the clock in that time (half the distance from A to B) is vt, with v=0.6c. Since we know the vertical distance is 0.5 light seconds, we can use the pythogorean theorem here:

(ct)^2 = (vt)^2 + (0.5 ls)^2

And then to find the value of the time t, we can use this and solve for t. Rearranging, we have:

(c^2 - v^2)*t^2 = (0.5 ls)^2
t^2 = (0.5 ls)^2 / (c^2 - v^2)
t = 0.5 ls / sqrt(c^2 - v^2) = 0.5 ls / sqrt(c^2 - (0.6c)^2)
= 0.5 ls / sqrt(c^2 - 0.36c^2) = 0.5 ls / sqrt(0.64c^2) = 0.5 ls/ 0.8c
= 0.625 seconds.

So, the time for the light to move on the diagonal from the bottom mirror to the top must be 0.625 seconds, and the time for it to return from the top to the bottom on a symmetrical diagonal must also be 0.625 seconds, for a total time of 1.25 seconds in this frame.

Then you expalined to us that, why the shorter time interval? it was bacause one clock was running slower than the other. What would I think of your language?

I don't understand your question here. If we actually measured the time using real stopwatches and obtained the times that I said you would, then if the moving stopwatch measured the departure time as 0 seconds and the arrival time as 1 second, and the other pair of synchronized stopwatches measured the departure time as 0 seconds and the arrival time as 1.25 seconds, wouldn't you say the moving stopwatch is running slower in this frame?

If you agree you had no right to do so, then what would happen to the said thought experiment?

Agree I had no right to do what?

If you agree d/c = AB/v, then what would happen to the said thought experiment? as it means the thought experiment produced one identical time interval by using the distance traveled by the clock.

There's no way to figure out the distance AB without using the assumption that light moves at c--the way you calculate the horizontal distance is by using the pythagorean theorem with the height and half the horizontal distance as two sides of a right triangle, and the diagonal as the hypotenuse, as I did in my calculation above. So whether you calculate the time using d/c or AB/v, you're basing this on the assumption that light moves at c in this frame, and that's the whole point of the thought-experiment, to derive the prediction of time dilation from the assumption that the two fundamental postulates of relativity are correct, one of which says light moves at c in every frame, the other which says all the laws of physics work the same in every frame (which necessarily implies that if a stopwatch and a light clock are both at rest in my frame and they both tick at the same rate, then it must be true that the stopwatch would continue to agree with the light clock if they were both moving at a high velocity relative to me).

I am still not sure if you actually understand the point above. I asked you some questions in my previous post which you didn't answer, please answer them (or ask questions if you're not sure what I'm asking) next time you reply to me:

But it's [the light clock thought-experiment] not meant to "prove the time dilation idea" from nothing! It's just meant to prove that if the two basic postulates of relativity are true, then time dilation must occur, with moving clocks slowed down by a factor of \sqrt{1 - v^2/c^2}. Do you disagree that the light clock experiment does prove this? If you disagree, that must mean that you think it would be possible for the two basic postulates of relativity to be correct, yet for time dilation not to occur, or to obey a different equation. Do you think that's possible?

 

[robphy]

JesseM,
I'm not sure if this will help address longshinewoole's comments...
[I'm not sure I can pinpoint his concern] but...
elaborating my previous comment...

if you express all of your measurements in units of "ticks" [for times] or "ticks/c" [for distances] (rather than "seconds" or "light-seconds"), then you have not made any reference to any other type of timepiece.

In addition, it might help to make explicit that one must use identically constructed lightclocks [produced by a factory]. And one need not specify the separation of the parallel mirrors explicitly. What is important is that they are produced identically. In this way, ALL measurements of spacetime intervals are based on the tick of this standard lightclock. (If you need a finer resolution, then choose a new standard lightclock with a smaller separation.)

 

[JesseM]

if you express all of your measurements in units of "ticks" [for times] or "ticks/c" [for distances] (rather than "seconds" or "light-seconds"), then you have not made any reference to any other type of timepiece.

Well, in the last post when I gave some specific numbers, I did give the times on the stopwatches in terms of seconds. And when I said "ticks" I was basically just thinking of the tick of the second hand on a clock.

In addition, it might help to make explicit that one must use identically constructed lightclocks [produced by a factory].

Yes, that's an important point, the light clocks in different frames must be identical.

And one need not specify the separation of the parallel mirrors explicitly. What is important is that they are produced identically. In this way, ALL measurements of spacetime intervals are based on the tick of this standard lightclock. (If you need a finer resolution, then choose a new standard lightclock with a smaller separation.)

But it's also important to understand that even if you measure time using a different type of clock, like an atomic clock based on regular oscillations of cesium atoms, then if there's one frame where a light clock and a non-light-clock agree on the length of 1 second when they're at rest next to each other, then as long as the postulates of relativity are true, they must agree with one another in all frames.