Jump to content

New look at Time Dilation


Bart

Recommended Posts

Bart - just quickly, I didnt have time to read your paper in any detail; but two things immediately jumped out at me. If the rocket is accelerating at a rate such that it completely mimics the acceleration due to gravity (ie the astronauts and the clocks always feel 9.8ms-2) at all times it will never leave the ground; to get off the ground you need to be above that amount. Secondly and more importantly - there is no time dilation within the frame, you seem to be saying that the astronaut will measure differences in the clocks - he won't

 

edit

just read one of the clock is a signal from earth - that will change in the ship

Edited by imatfaal
Link to comment
Share on other sites

The "experiment" is nonsense.

 

It talks about "measurements", but no measurements were made because the experiment was never actually performed.

 

He ignores relativistic effects for some of the clocks in the calculations of their time rate and claims because these results don't match up with the Relativistic prediction, time dilation does not exist.

 

In other words, he misapplies Relativity to a thought experiment and then thinks that the since the results he thinks he would get don't agree with Relativity that he's proved SR wrong, when all he's proven is that he doesn't understand Relativity.

Link to comment
Share on other sites

Thanks for the comments, but what about the following question:

 

Orbital speed of the ISS space station is 27 700km / h.

The accuracy of atomic clocks is 10 * 10E-10 seconds per day.

So if time dilation was indeed true, then the relativistic delay of the atomic clocks on the ISS should be highly visible and be 30 microseconds per day, this is very, very much because it is 30 000 times greater than the accuracy of these clocks.

To our knowledge, the occurrence of such delay of the clocks on the ISS has not been confirmed.

 

Edited by Bart
Link to comment
Share on other sites

Thanks for the comments, but what about the following question:

 

Orbital speed of the ISS space station is 27 700km / h.

The accuracy of atomic clocks is 10 * 10E-10 seconds per day.

So if time dilation was indeed true, then the relativistic delay of the atomic clocks on the ISS should be highly visible and be 30 microseconds per day, this is very, very much because it is 30 000 times greater than the accuracy of these clocks.

To our knowledge, the occurrence of such delay of the clocks on the ISS has not been confirmed.

 

 

 

Whether or not such an test has actually been done on the ISS is moot, since the same experiment is essentially done every day with GPS satellites. The clocks on these satellite have to be adjusted to compensate for time dilation in order for the system to remain accurate.

Link to comment
Share on other sites

Just some more detailed points in regards to the flaws in the linked paper.

 

1. In the formulas for clocks 3 and 4, the author simply replaces c with Cs (assumably the speed of light in the glass). This is incorrect as the formulas using c are based on the idea the c is invariant, and Cs is not. The proper formulas would have to reflect this and incorporate the velocity addition formula.

 

2. Simply applying the Doppler shift formula and plugging in the velocity of the rocket does not give you the time rate difference between the clocks. Instead, what you need to do is to calculate the total accumulated time for both clocks over the course of a round trip of the rocket. (and if you do, you will find it in complete agreement with time dilation. )

 

3, The formula given is fine for the time as measured in the rocket, but fails to take into account any relativistic effects. A proper analysis of the period of the pendulum would have to take into account the effect of the velocity addition theorem on the pendulum bob. In addition, when the author "corrects" for length contraction, all they did was shorten the length of the pendulum arm and reapply the period equation. Since the arm of the pendulum would not always be parallel to the motion of the rocket, length contraction on it would not constantly act along its length. (in other words as the pendulum swings from side to side, the length of the arm would vary. The author oversimplifies a complex problem and as a result come to the wrong conclusion.

Link to comment
Share on other sites

  • 4 weeks later...

Just some more detailed points in regards to the flaws in the linked paper.

 

1. In the formulas for clocks 3 and 4, the author simply replaces c with Cs (assumably the speed of light in the glass). This is incorrect as the formulas using c are based on the idea the c is invariant, and Cs is not. The proper formulas would have to reflect this and incorporate the velocity addition formula.

 

 

2. Simply applying the Doppler shift formula and plugging in the velocity of the rocket does not give you the time rate difference between the clocks. Instead, what you need to do is to calculate the total accumulated time for both clocks over the course of a round trip of the rocket. (and if you do, you will find it in complete agreement with time dilation. )

 

 

3, The formula given is fine for the time as measured in the rocket, but fails to take into account any relativistic effects. A proper analysis of the period of the pendulum would have to take into account the effect of the velocity addition theorem on the pendulum bob. In addition, when the author "corrects" for length contraction, all they did was shorten the length of the pendulum arm and reapply the period equation. Since the arm of the pendulum would not always be parallel to the motion of the rocket, length contraction on it would not constantly act along its length. (in other words as the pendulum swings from side to side, the length of the arm would vary. The author oversimplifies a complex problem and as a result come to the wrong conclusion.

 

 

1. In the calculation using the glass clock has to be applied to the speed of light in glass (Cs), and not in a vacuum and in my understanding the formula used by the author is correct

 

2. I do not understand your objections, may I know on what formulas is obtained the same time of the clocks? My calculations show that for the trip there (T1) and back (T2), total time measured by the clock based on the Doppler effect by the formula: T1 + T2 = 1 / (for * ((1-v / c) / (1 + v / c)) ^ 0.5) + 1 / (for * ((1 + v / c) / (1-v / c)) ^ 0.5), will always be greater than for the vacuum clock based on the relativistic formula: T1 + T2 = 2 * (2D1 / c) / (1 - (v / c) ^ 2) ^ 0.5, and it is for any speed of the rocket.

 

3. It is true that the author simplifies the calculations for the pendulum, but it has little effect on that the clock is fast, but it does not affect the correctness of conclusion.

 

4. "Whether or not such an test has actually been done on the ISS is moot, since the sameexperiment is essentially done every day with GPS satellites. The clocks onthese satellite have to be adjusted to compensate for time dilation in orderfor the system to remain accurate."

 

Correction of time of GPS clocks, is that the clocks are fast and not late, and this may be the result of the following two circumstances:

 

- To take account the orbital speed of Earth. See calculations by Professor Joe Nahhas http://www.scribd.co...-is-an-illusion

 

- The need for constant adjustment of time of the satellite clocks due to the slow degradation (shortening) of satellite's orbits.

 

Thus, the GPS does not need to be a reliable evidence of time dilation.

Edited by Bart
Link to comment
Share on other sites

4. "Whether or not such an test has actually been done on the ISS is moot, since the sameexperiment is essentially done every day with GPS satellites. The clocks onthese satellite have to be adjusted to compensate for time dilation in orderfor the system to remain accurate."[/font][/size]

 

Correction of time of GPS clocks, is that the clocks are fast and not late, and this may be the result of the following two circumstances:- To take account the orbital speed of Earth. See calculations by Professor Joe Nahhas http://www.scribd.co...-is-an-illusion

 

- The need for constant adjustment of time of the satellite clocks due to the slow degradation (shortening) of satellite's orbits.

 

Thus, the GPS does not need to be a reliable evidence of time dilation.

 

 

I don't know what crap he's trying to spout; there's little explanation of where he's getting his formulas, but … the GPS correction is 38 microseconds/day: 45 from gravitational and 7 from kinematic, with opposite signs. Secondly, the orbit is not equatorial, so that means the vector addition of the velocities is wrong.

Link to comment
Share on other sites

 

1. In the calculation using the glass clock has to be applied to the speed of light in glass (Cs), and not in a vacuum and in my understanding the formula used by the author is correct

Your understanding is wrong. Relativity is based on the fact that there is an invariant speed ©. Now. it happens that light travels at c in a vacuum, which it is convenient to use light in examples, but only if we use light traveling in a vacuum. For example, we can always assume that light in a vacuum travels at c relative to any frame. If we use the light clock example, we came use the addition rule for orthogonal velocities:

[math]w = \sqrt { v^2 + u^2 - \frac{v^2u^2}{c^2}} [/math]

For the light clock in a vacuum, u=c, and w always equals c no matter what v is. If u is equal to .567c, then W varies according to v (Cs is not invariant). This does not mean that you cannot calculate the time dilation, just that you can't use the formulation given in the link which treats Cs as invariant. If you do do the calculation, you will get the same answer as standard time dilation gives.

 

2. I do not understand your objections, may I know on what formulas is obtained the same time of the clocks? My calculations show that for the trip there (T1) and back (T2), total time measured by the clock based on the Doppler effect by the formula: T1 + T2 = 1 / (for * ((1-v / c) / (1 + v / c)) ^ 0.5) + 1 / (for * ((1 + v / c) / (1-v / c)) ^ 0.5), will always be greater than for the vacuum clock based on the relativistic formula: T1 + T2 = 2 * (2D1 / c) / (1 - (v / c) ^ 2) ^ 0.5, and it is for any speed of the rocket.

 

The formula

[math] t_1 \left( \sqrt{\frac{1-\frac{v}{v}}{\frac{1+\frac{v}{v}}}+t_2 \left( \sqrt{\frac{1+\frac{v}{v}}{\frac{1-\frac{v}{v}}}[/math]

 

Gives the time that passes for the transmitter according the the rocket as the rocket travels away and then returns where t1 and t2 is the time for each leg as measured by the rocket. Assuming that t1=t2, this reduces to

 

[math] \frac {2t_2}{\sqrt{1-\frac{v^2}{c^2}}}[/math], which is exactly what the standard relativistic formula predicts. If you you want to go the other way where the receiver is on Earth and the transmitter on the rocket, then you cannot make t1=t2 (where t1 is the period in which the Earth sees the Rocket as red shifted and t2 is the period in which it sees it as blue shifted.) This is due to the signal delay between the time when the rocket turns around and when the Earth sees the rocket turn around. (If the rocket travels at .5c to distance of 1 light year, and returns, the Earth will see it red-shifted for 3 years (2 year for the Rocket reach 1 light year away and 1 year for the light from the turn around to reach Earth. ) and then see it red shifted for 1 yr. (the rocket will give chasing after its own light signal. Once you take this into account, you also find that the the Doppler shift the Earth sees agrees with the standard time dilation prediction. Far from disagreeing with Relativity, the Doppler shift agrees with it.

 

 

3. It is true that the author simplifies the calculations for the pendulum, but it has little effect on that the clock is fast, but it does not affect the correctness of conclusion.

He does not just "simplify" the calculations, he uses completely the wrong ones.

 

His whole line of reasoning is one straw man argument after another.

Edited by Janus
Link to comment
Share on other sites

Your understanding is wrong. Relativity is based on the fact that there is an invariant speed ©. Now. it happens that light travels at c in a vacuum, which it is convenient to use light in examples, but only if we use light traveling in a vacuum. For example, we can always assume that light in a vacuum travels at c relative to any frame. If we use the light clock example, we came use the addition rule for orthogonal velocities:

[math]w = \sqrt { v^2 + u^2 - \frac{v^2u^2}{c^2}} [/math]

For the light clock in a vacuum, u=c, and w always equals c no matter what v is. If u is equal to .567c, then W varies according to v (Cs is not invariant). This does not mean that you cannot calculate the time dilation, just that you can't use the formulation given in the link which treats Cs as invariant. If you do do the calculation, you will get the same answer as standard time dilation gives.

 

 

 

 

Iam very sorry, but I still do not understand this question. Please, explain how time dilation will be seen by an observer on the clocks in the following description, taken from Wikipedia, and assuming that there are two light clocks side by side, a vacuum clock and the other glass clock.

Simpleinference of time dilation due to relative velocity http://en.wikipedia.org/wiki/Time_dilation

 

 

Observer at restsees time 2L/c.

 

 

 

Observer movingparallel relative to setup, sees longer path, time > 2L/c, same speedc.

 

Time dilation can be inferredfrom the observed fact of the constancy of the speed of light in all referenceframes.[2][3][4][5]

 

This constancy of the speed oflight means, counter to intuition, that speeds of material objects and lightare not additive. It is not possible to make the speed of light appear fasterby approaching at speed towards the material source that is emitting light. It isnot possible to make the speed of light appear slower by receding from thesource at speed. From one point of view, it is the implications of thisunexpected constancy that take away from constancies expected elsewhere.

 

Consider a simple clock consistingof two mirrors A and B, between which a light pulse is bouncing. The separationof the mirrors is L and the clock ticks once each time it hits a givenmirror.

 

In the frame where the clock isat rest (diagram at right), the light pulse traces out a path of length 2Land the period of the clock is 2L divided by the speed of light:

 

 

Fromthe frame of reference of a moving observer traveling at the speed v(diagram at lower right), the light pulse traces out a longer, angledpath. The second postulate of special relativity states that the speed of lightis constant in all frames, which implies a lengthening of the period of thisclock from the moving observer's perspective. That is to say, in a frame movingrelative to the clock, the clock appears to be running more slowly.Straightforward application of the Pythagorean theorem leads to the well-knownprediction of special relativity:

 

Thetotal time for the light pulse to trace its path is given by

 

 

Thelength of the half path can be calculated as a function of known quantities as

 

 

SubstitutingD from this equation into the previous and solving for Δt' gives:

 

 

andthus, with the definition of Δt:

 

 

whichexpresses the fact that for the moving observer the period of the clock islonger than in the frame of the clock itself.

 

 

Link to comment
Share on other sites

Iam very sorry, but I still do not understand this question. Please, explain how time dilation will be seen by an observer on the clocks in the following description, taken from Wikipedia, and assuming that there are two light clocks side by side, a vacuum clock and the other glass clock.

Simpleinference of time dilation due to relative velocity http://en.wikipedia.org/wiki/Time_dilation

 

 

Observer at restsees time 2L/c.

 

 

 

Observer movingparallel relative to setup, sees longer path, time > 2L/c, same speedc.

 

Time dilation can be inferredfrom the observed fact of the constancy of the speed of light in all referenceframes.[2][3][4][5]

 

This constancy of the speed oflight means, counter to intuition, that speeds of material objects and lightare not additive. It is not possible to make the speed of light appear fasterby approaching at speed towards the material source that is emitting light. It isnot possible to make the speed of light appear slower by receding from thesource at speed. From one point of view, it is the implications of thisunexpected constancy that take away from constancies expected elsewhere.

 

Consider a simple clock consistingof two mirrors A and B, between which a light pulse is bouncing. The separationof the mirrors is L and the clock ticks once each time it hits a givenmirror.

 

In the frame where the clock isat rest (diagram at right), the light pulse traces out a path of length 2Land the period of the clock is 2L divided by the speed of light:

 

 

Fromthe frame of reference of a moving observer traveling at the speed v(diagram at lower right), the light pulse traces out a longer, angledpath. The second postulate of special relativity states that the speed of lightis constant in all frames, which implies a lengthening of the period of thisclock from the moving observer's perspective. That is to say, in a frame movingrelative to the clock, the clock appears to be running more slowly.Straightforward application of the Pythagorean theorem leads to the well-knownprediction of special relativity:

 

Thetotal time for the light pulse to trace its path is given by

 

 

Thelength of the half path can be calculated as a function of known quantities as

 

 

SubstitutingD from this equation into the previous and solving for Δt' gives:

 

 

andthus, with the definition of Δt:

 

 

whichexpresses the fact that for the moving observer the period of the clock islonger than in the frame of the clock itself.

 

 

 

 

Again, whenever we talk about the speed of light being the same for all observers, we are only talking about the speed of light in a vacuum. We call this speed "c". c is always equal to 299,792,458 m/s. There is nothing special about light itself, only c, which happens to be the speed light travels in a vacuum. So if you ever have an equation, such as the time dilation formula or the addition of velocities formula, which has "c" in it you can only use c in that formula. You cannot, as the author of the link did, simply replace c with Cs, the speed of light through a medium other than a vacuum.

 

If you want to determine the time dilation factor for a clock in which the light passed through glass light clock, you have to use the velocity addition equation that I gave in the last post to find the speed of the light on the diagonal according to the "rest" Observer. So if the speed of light through the glass is 170,000 kps and the clock is traveling at 169,000 kps then the speed of the light on the diagonal would be ~219749 kps. You would then use this speed And the Pythagorean theorem to determine the time dilation for the light clock. If you did so, you would arrive at the same relative time dilation as you would using a standard light clock.

Link to comment
Share on other sites

Again, whenever we talk about the speed of light being the same for all observers, we are only talking about the speed of light in a vacuum. We call this speed "c". c is always equal to 299,792,458 m/s. There is nothing special about light itself, only c, which happens to be the speed light travels in a vacuum. So if you ever have an equation, such as the time dilation formula or the addition of velocities formula, which has "c" in it you can only use c in that formula. You cannot, as the author of the link did, simply replace c with Cs, the speed of light through a medium other than a vacuum.

 

If you want to determine the time dilation factor for a clock in which the light passed through glass light clock, you have to use the velocity addition equation that I gave in the last post to find the speed of the light on the diagonal according to the "rest" Observer. So if the speed of light through the glass is 170,000 kps and the clock is traveling at 169,000 kps then the speed of the light on the diagonal would be ~219749 kps. You would then use this speed And the Pythagorean theorem to determine the time dilation for the light clock. If you did so, you would arrive at the same relative time dilation as you would using a standard light clock.

 

Thank you Janus for clarifying. But even with the adoption of the calculated speed of light for the glass clock as 219749 kps, it still does not agree with the lengthwise clock.

Link to comment
Share on other sites

Thank you Janus for clarifying. But even with the adoption of the calculated speed of light for the glass clock as 219749 kps, it still does not agree with the lengthwise clock.

 

219749 kps is the speed of the light traveling on the diagonal relative to the the rest observer, which resulted from using the orthogonal velocity addition formula. For the lengthwise clock you have to use the standard velocity addition formula:

 

[math]\frac{u+v}{1+\frac{uv}{c^2}}[/math]

 

Again notice that if u=c (as for the light clock in vacuum) then the answer is always c.

 

However if u = 170,000kps,(glass light clock) then for the light traveling in the direction of v (fro v =169,000 kps) you get:

 

~256970 kps

 

and for the light traveling in the opposite direction (after being reflected), you get.

 

~-1470 kps

(the minus indicates that the velocity of the light is in the opposite direction.)

 

 

Notice that unlike when the light travels at c, these two speeds are different.

 

Once you factor in length contraction, you can determine how long it would take the light with these speeds to travel back and forth between the mirrors of the light clock according to the "stationary" observer. If you do the math, you will find that is is consistent with the standard time dilation prediction.

 

 

I will go further. No thought experiment of this type can ever expose a flaw or internal contradiction in SR. If you think you've found one, you have made a mistake or misapplied the theory. It is internally consistent in all ways.

 

This does not mean that SR could never be overturned or found in need of correction, but just that the only way to show this is for real life experiment or observation to give a result that is in conflict with what SR predicts. Attempts to disprove SR by thought experiment are doomed to failure from the get go.

Link to comment
Share on other sites

219749 kps is the speed of the light traveling on the diagonal relative to the the rest observer, which resulted from using the orthogonal velocity addition formula. For the lengthwise clock you have to use the standard velocity addition formula:

 

[math]\frac{u+v}{1+\frac{uv}{c^2}}[/math]

 

Again notice that if u=c (as for the light clock in vacuum) then the answer is always c.

 

However if u = 170,000kps,(glass light clock) then for the light traveling in the direction of v (fro v =169,000 kps) you get:

 

~256970 kps

 

and for the light traveling in the opposite direction (after being reflected), you get.

 

~-1470 kps

(the minus indicates that the velocity of the light is in the opposite direction.)

 

 

Notice that unlike when the light travels at c, these two speeds are different.

 

Once you factor in length contraction, you can determine how long it would take the light with these speeds to travel back and forth between the mirrors of the light clock according to the "stationary" observer. If you do the math, you will find that is is consistent with the standard time dilation prediction.

 

 

I will go further. No thought experiment of this type can ever expose a flaw or internal contradiction in SR. If you think you've found one, you have made a mistake or misapplied the theory. It is internally consistent in all ways.

 

This does not mean that SR could never be overturned or found in need of correction, but just that the only way to show this is for real life experiment or observation to give a result that is in conflict with what SR predicts. Attempts to disprove SR by thought experiment are doomed to failure from the get go.

 

 

Janus, thank you very much again for your clarification. You are a very patient guy, and you made hard work, thanks. Math is correct now, but the reality seems to be a little simpler. Thank you all for your posts.

Bart

Link to comment
Share on other sites

Janus, thank you very much again for your clarification. You are a very patient guy, and you made hard work, thanks. Math is correct now, but the reality seems to be a little simpler. Thank you all for your posts.

Bart

 

 

Math is correct? In what way, Sir ? I checked the calculation for the speeds given by Janus 256 979 kps and -1470 kps for the glass clock (No 4) and the results still do not match. There is still a large discrepancy between the glass clocks and vacuum. Could someone explain it to me, where I am wrong?

Edited by ravell
Link to comment
Share on other sites

Math is correct? In what way, Sir ? I checked the calculation for the speeds given by Janus 256 979 kps and -1470 kps for the glass clock (No 4) and the results still do not match. There is still a large discrepancy between the glass clocks and vacuum. Could someone explain it to me, where I am wrong?

 

Then you did it wrong.

 

Lengthwise clock with the distance between mirrors of .085 km as measured in the rocket frame (this makes the the round trip time 1 microsecond in the rocket frame.):

 

The speed of the rocket is 169000 kps which is 0.5633c, which means that the gamma factor is ~1.21, thus the distance between the mirrors will be length contracted to ~0.07 km.

 

If mirror 2 is in the direction of movement from mirror 1, then the distance that the light leaving mirror 1 has to travel to reach mirror 2 according to our stationary observer is 0.14km + the distance that mirror 2 moves between the time it leaves mirror 1 and catches up to mirror 2. This distance divided by 256,979 kps gives the time it takes for this leg of the trip according our stationary observer. This time can also be found by dividing the difference between the light's velocity(256,979 kps) and the rocket's velocity(169,000 kps) into the distance between the mirrors (0.07 km) this gives an answer of .798 microseconds.

 

For the return leg, the rocket is moving at 169,000 kps in one direction while the light is moving at -1470 kps in the other. The minus sign indicates the direction. Again we take the difference between the two(remembering that subtracting -1470 is the same as adding 1470 ), and divide this into the distance between the mirrors (0.07km ) which equals 0.41 microseconds. Adding this to the time for the first leg gives a total round trip time of 1.21 microseconds. Thus 1 microsecond as measured in the rocket frame is 1.21 microseconds in the stationary frame and 1.21 is the gamma factor arrived at using the vacuum clock time dilation formula. Vacuum clock and glass clock agree upon the time dilation.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.