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When the light is sent to clock A it will not only be seen until I reach C. I think the light or clock will be seen at every moment after it hits A (clock) such that it will be first seen when I reach the point B (as if I was at rest) and onwards in an infinite series of points pivoting from A i.e. from AB and onwards until to AC ad infinitum and each point will show a different time on the clock A... -- WaleedAddas ( talk) 07:27, 15 April 2009 (UTC)-- WaleedAddas ( talk) 07:27, 15 April 2009 (UTC)
“If the twin aboard the spaceship went to the nearest star, which is 4.45 light years away at 86 percent of the speed of light, when he returned, he would have aged 5 years. But the earthbound twin would have aged more than 10 years!” said Kak. ( http://www.physorg.com/news90697187.html)
Let the twin ant had also rocket away along with person twin, when returned have aged 5 years but the earthbound twin ant would have aged more than 10 years!
Let the inside width of the spaceship is x unit long and at rest an ant covers the back and forth distance x in (say) five years with constant speed for both moving observer and stationary observer (back elevation or front elevation of spaceship).
As soon as the spaceship starts its flight an ant also starts its journey along the straight line x inside the ship. As an ant can’t decrease or increase its constant speed/ pace inside ship and there is also no length contraction widthwise of the ship, so the only difference is the moving clock slow down as compared to the stationary clock, therefore find the following after the whole space flight.
How many number of trips (width of inside ship or back and forth distance x) covered by an ant inside the ship w.r.t to both clocks?
What will be the (difference in) age of an ant w.r.t to both the inside no. of trips and rocket away at 86 % of speed of light? 96.52.178.55 ( talk) 04:49, 3 June 2009 (UTC)khattak
Continued
Now if aforementioned ant (rocketing) that is back and forth with constant speed, inside widthwise of ship, increases its speed to say 0.8c and as the rocketing person doesn’t feel its traveling speed (0.86c) inside ship therefore becomes the stationary observer for this ant therewith
Now an ant (rocketing) has two speeds wrt stationary person observer on ground
One is the ship’s speed 0.86c inside which an ant (rocketing) is traveling in its forward direction (say for simplicity along Y-Axis) and the other is back and forth across the ship which is 0.8c (say for simplicity along X-Axis)
While w.r.t rocketing person (that became stationary observer for rocketing ant’s crosswise motion) inside ship (which is also moving with 0.86c in its forward direction), it has only one back and forth speed (0.8c) across the ship
Question:
What will be the clock reading of this ant (rocketing) that is back and forth with 0.8c widthwise inside ship (which is also moving with 0.86c in its forward direction) w.r.t the stationary observer on ground and rocketing person (that became stationary for rocketing ant’s crosswise motion) inside a ship (which is also moving with 0.86c in forward direction)?
I don’t know how to draw a diagram here but please draw a diagram before dealing the problem. Thanks 204.191.225.73 ( talk) 05:19, 27 September 2009 (UTC) Khattak
Let there are two planets A and B some light years apart in space. The light pulses cross[*} each other at the middle, if fired simultaneously from each planet towards each other.
[*]For crossing, imagine a two-way traffic on a divided single road OR each track of a railway line represents a pulse but their direction is opposite.
Now there is a transparent space ship of certain length which started its journey with (say) 0.95 C from A to B. After a very short interval, light pulse are fired from both A and B towards each other.
As the pulse’s velocity “c” (fired from A) which is just behind the ship towards B is greater than the ship’s 0.95C therefore after sometime this light pulse will
1- Enters the ship through its back
2- Travels inside the ship; longer( c+v) for outside observer while c for inside)
3- Leaves the ship through its cockpit towards B
While the pulse which is fired from B, from opposite direction will
1- Enters the ship through its cockpit
2- Travels for some time inside the ship; longer (c) for inside observer while c-v for outside)
3- Leaves the ship through its tail towards A
Both pulses cross each other in the ship such that one on the left side whiles the other on the right side of longitudinal axis of ship. A stationary observer on asteroid is also watching the whole scene; a ship and both pulses.
Since velocity of ship is just below the velocity of light therefore the travel time of light inside the ship for outside observer will be very very longer and for inside observer will be the only velocity of light as per Einstein’s postulate.
Now please answer the following questions while keeping in mind c+v and c-v equation.
1- What will be inside travel time or passing and crossing sight distance of both pulses for the both the moving and stationary observer or simply write the c+v and c-v equation for above scenario?
2- Since there is length contraction and difference in clocks therefore how would both the observers’ notices the pulses which leave the ship?
I hope I have explained things clearly enough that you can understand what I mean. 96.52.178.55 ( talk) 04:33, 18 July 2009 (UTC) Quoth Khattak #1
Your answer is fine with relativity but my intent was to bring one’s attention towards the following common sense
Let say for example, when both pulse and ship are moving in same direction
For inside observer, the pulse after entering the ship through its tail, will pass through the cockpit within one second or two, while
For outside observer, the pulse entering the ship through its tail will travel inside ship (which is also moving say with 0.99c) for hours and hours…
Thus there is an eye popping difference in duration of both events (one second and hours and hours). thanks 96.52.178.55 ( talk) 19:52, 30 July 2009 (UTC) Khattak
I must be misinterpreting the meaning of the variables in the basic formula (in the overview section), and I was hoping someone could clear up my confusion.
If you are on a spaceship traveling 4/5 light speed for 10 years (in your frame), you will age 10 years. During this time, your friend on Earth should have aged 50/3 years [(1-(4/5)^2)^.5= 3/5]. Isn't 50/3 years the "proper time" since it is the time recorded by the stationary observer? Basically, my question is: shouldn't the proper time always be greater than the moving observer's time? During any given interval between two events taking place on Earth, the clock on the moving spaceship will tick fewer times than the stationary clock on Earth; so doesn't that mean the proper time is greater?
I must have it backwards, since otherwise the formula in the overview section would be incorrect, right? —Preceding unsigned comment added by Jakemhall ( talk • contribs) 10:48, 30 July 2009 (UTC)
Those who argue that time 'dilation' is a fact usually interchange velocity with speed for light which is an error. For example, in this article the time 2D/c and 2L/c seem different, but the times are actually the same event, and both observers time the event the same IF they time it correctly. Notice that the vector component speed with which the light travels upward (North)while angle traveling toward the mirror to reach it in its shifted location is v(North)= c(L/D,)where L/D is the sine of the angle the ray travels from the horizontal as seen in the second figure. When this component for light is placed into the equation for time (stationary), it becomes t = 2L/[c(L/D)] = 2D/c and is the same as the other expression (moving). Time does not actually 'dilate' simply because an observer and his clock are in motion. It is just that if one uses light as a time-keeping device, you must use the proper component of velocity for the travel direction of the pulse involved. The magnitude of the velocity of light can vary from +c(East) to -c(West). Time dilation was invented in 1905 when Einstein mistakenly thought that (since light rays travel in straight lines) c and v could be always interchanged. But speed and velocity I assure you are different when an angle-traveling ray is used to compute time in one instance, and then that ray is used again in a different diagram without considering that its velocity in the direction of interest is now a component that is less than c. Velocity is only c in the propagation direction itself, which occurs here at an angle above the horizon toward the shifted mirror. The sationary observer does not compute time correctly because he ignores, or is unaware of, the lateral shift of the light and mirror ongoing at the same time. Time dilation is only an apparent phenomenon that occurs if calculations do not consider the component velocity for a directed pulse of light. Anyone can exceed the velocity of light. Simply point it North and you run West to reach the target. You beat the velocity of light toward the target. Or reflect a beam on a mirror; when the light returns to its starting position the velocity at that position was zero. You continue running,and don't turn around, you exceed the velocity of light.
Richard Sauerheber, Ph.D. —Preceding unsigned comment added by 205.153.156.222 ( talk)
Note to anonymous 205.153.156.222: I have reverted the modifications your made to your comments after I had replied to them. Please have a look at your talk page. Thank you. DVdm ( talk) 22:31, 18 September 2009 (UTC)
The Lorentz time dilation equation is derived using the concept of a light clock in which a pulse of light being continually reflected backwards and forwards between two mirrors for moving observer and zigzag for stationary observer.
1- The above concept was taken from a moving train (falling of object from ceiling of train under the influence of earth's gravity) but since in space there is no GRAVITY therefore an emerging pulse should follow the straight downward path from where it is emerged, and shouldn’t move along the direction of ship for both stationary and moving observers. (In other words the free fall of pulse shouldn’t be influenced by the ship’s velocity)
2- As light carry momentum and can push/ veer objects easily in SPACE therefore, should a ship swerve with a new resultant velocity when a pulse touches the second mirror? Further, any flat object or flanks of ship could reflect a pulse in thought experiment, so is there any special reason in using a mirror?
The same is applied to the Einstein’s elevator
Your comment please! Thanks
68.147.38.24 ( talk) 01:57, 4 October 2009 (UTC) Khattak
Sorry about the aforementioned mistakes
The description of pulse bounces between top and bottom mirror, shown in article is very fine with question.
Moving train is not stated in the article but my intent was basic relativity 68.147.38.24 ( talk) 16:34, 4 October 2009 (UTC) Khattak #1
A recent edit has been reverted because it added errors and seems to have been made without reading the article: the new material is reproduced here, some reasons are given below why the edit is erroneous and misleading, pls discuss before replacing the material.
Sample of reasons against the material in the edit:
Terry0051 ( talk) 21:28, 19 October 2009 (UTC)
Thank you for the revert; peer review is never bad, and when it rejects your edits, that's positive evidence that it has served its purpose. In the interest of developing the article, I'd like to respond, though. Let's see if we can synthesize something better as a result some discussion.
Decoy ( talk) 21:59, 20 October 2009 (UTC)
[From Terry0051] Thanks for your thoughtful comments. Can I suggest that one hopefully fruitful route might be, to identify/seek consensus on which core facts should be communicated in the intro or opening sections (checking that there's consensus on the facts themselves!) and then working out how to put them briefly & clearly in suitable encyclopedic style.
I have to apologise for having only a short time right now, I'll get back to it in a couple of days, but points I'd suggest are:
With best wishes in the meantime Terry0051 ( talk) 13:36, 21 October 2009 (UTC)
(Returning to offer supplementary suggestions answering in part the question posed about 'what points are unsourced & would need [specific] RS support?') The relevant points for specific support seem to include the suggested facts about:
With good wishes, Terry0051 ( talk) 19:31, 28 October 2009 (UTC)
Anonymous user 213.100.87.94 has repeatedly introduced edits stating that the earth's orbital motion and the galactic motion of the solar system should be taken into account for terrestrial time scales.
Standard time scales used on the surface of the Earth include Terrestrial Time, which is closely related (by an offset) to international atomic time and (by a scale factor) to Geocentric Terrestrial Time. But these are coordinate time scales, and the coordinate system to which they refer is a geocentric one, moving with the Earth. Accordingly, the Earth's orbital motion around the Sun and the galactic motion of the solar system are not relevant to these time scales. (There are indeed other coordinate time scales referred to the solar-system barycenter, including TCB and TDB, for which the earth's orbital motion and its (varying) position in the relevant gravitational potential wells are taken into account.)
As already stated in DVdm's edit summary, the effect of the Earth's rotational motion has indeed been taken into account for the synchronization of Earth-based atomic clocks. The concept 'stationary clock' was not present in the article before the edits by 213.100.87.94, and really does not play a part. Accordingly this edit (*) and others like it by 213.100.87.94 do appear to be in error: (*: "In reality however, there are no stationary clocks, given that the earth rotates on it's axis with a surface velocity of 465 m/s, and orbits the sun at a mean velocity of 14,893 m/s, within a solar system which itself orbits the outer edge of the Milky Way Galaxy at a velocity of 2,092,990 m/s, these velocities must also be included in the equation, but in fact they never are.") It seems appropriate to remove this statement.
Terry0051 ( talk) 20:35, 10 December 2009 (UTC)
I created the following diagram which show's time dilation Δt' (Y axis) from the frame of reference of a moving observer traveling at the speed V (X axis in natural unit c),i think this is helpful diagram because it clearly show that for small friction of the speed of light, time dilation is approximately one (almost flat line for x from 0 to 0.1). what do you think? should i add the diagram to the article? —Preceding unsigned comment added by Zayani ( talk • contribs) 18:17, 19 December 2009 (UTC)
This page is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
When the light is sent to clock A it will not only be seen until I reach C. I think the light or clock will be seen at every moment after it hits A (clock) such that it will be first seen when I reach the point B (as if I was at rest) and onwards in an infinite series of points pivoting from A i.e. from AB and onwards until to AC ad infinitum and each point will show a different time on the clock A... -- WaleedAddas ( talk) 07:27, 15 April 2009 (UTC)-- WaleedAddas ( talk) 07:27, 15 April 2009 (UTC)
“If the twin aboard the spaceship went to the nearest star, which is 4.45 light years away at 86 percent of the speed of light, when he returned, he would have aged 5 years. But the earthbound twin would have aged more than 10 years!” said Kak. ( http://www.physorg.com/news90697187.html)
Let the twin ant had also rocket away along with person twin, when returned have aged 5 years but the earthbound twin ant would have aged more than 10 years!
Let the inside width of the spaceship is x unit long and at rest an ant covers the back and forth distance x in (say) five years with constant speed for both moving observer and stationary observer (back elevation or front elevation of spaceship).
As soon as the spaceship starts its flight an ant also starts its journey along the straight line x inside the ship. As an ant can’t decrease or increase its constant speed/ pace inside ship and there is also no length contraction widthwise of the ship, so the only difference is the moving clock slow down as compared to the stationary clock, therefore find the following after the whole space flight.
How many number of trips (width of inside ship or back and forth distance x) covered by an ant inside the ship w.r.t to both clocks?
What will be the (difference in) age of an ant w.r.t to both the inside no. of trips and rocket away at 86 % of speed of light? 96.52.178.55 ( talk) 04:49, 3 June 2009 (UTC)khattak
Continued
Now if aforementioned ant (rocketing) that is back and forth with constant speed, inside widthwise of ship, increases its speed to say 0.8c and as the rocketing person doesn’t feel its traveling speed (0.86c) inside ship therefore becomes the stationary observer for this ant therewith
Now an ant (rocketing) has two speeds wrt stationary person observer on ground
One is the ship’s speed 0.86c inside which an ant (rocketing) is traveling in its forward direction (say for simplicity along Y-Axis) and the other is back and forth across the ship which is 0.8c (say for simplicity along X-Axis)
While w.r.t rocketing person (that became stationary observer for rocketing ant’s crosswise motion) inside ship (which is also moving with 0.86c in its forward direction), it has only one back and forth speed (0.8c) across the ship
Question:
What will be the clock reading of this ant (rocketing) that is back and forth with 0.8c widthwise inside ship (which is also moving with 0.86c in its forward direction) w.r.t the stationary observer on ground and rocketing person (that became stationary for rocketing ant’s crosswise motion) inside a ship (which is also moving with 0.86c in forward direction)?
I don’t know how to draw a diagram here but please draw a diagram before dealing the problem. Thanks 204.191.225.73 ( talk) 05:19, 27 September 2009 (UTC) Khattak
Let there are two planets A and B some light years apart in space. The light pulses cross[*} each other at the middle, if fired simultaneously from each planet towards each other.
[*]For crossing, imagine a two-way traffic on a divided single road OR each track of a railway line represents a pulse but their direction is opposite.
Now there is a transparent space ship of certain length which started its journey with (say) 0.95 C from A to B. After a very short interval, light pulse are fired from both A and B towards each other.
As the pulse’s velocity “c” (fired from A) which is just behind the ship towards B is greater than the ship’s 0.95C therefore after sometime this light pulse will
1- Enters the ship through its back
2- Travels inside the ship; longer( c+v) for outside observer while c for inside)
3- Leaves the ship through its cockpit towards B
While the pulse which is fired from B, from opposite direction will
1- Enters the ship through its cockpit
2- Travels for some time inside the ship; longer (c) for inside observer while c-v for outside)
3- Leaves the ship through its tail towards A
Both pulses cross each other in the ship such that one on the left side whiles the other on the right side of longitudinal axis of ship. A stationary observer on asteroid is also watching the whole scene; a ship and both pulses.
Since velocity of ship is just below the velocity of light therefore the travel time of light inside the ship for outside observer will be very very longer and for inside observer will be the only velocity of light as per Einstein’s postulate.
Now please answer the following questions while keeping in mind c+v and c-v equation.
1- What will be inside travel time or passing and crossing sight distance of both pulses for the both the moving and stationary observer or simply write the c+v and c-v equation for above scenario?
2- Since there is length contraction and difference in clocks therefore how would both the observers’ notices the pulses which leave the ship?
I hope I have explained things clearly enough that you can understand what I mean. 96.52.178.55 ( talk) 04:33, 18 July 2009 (UTC) Quoth Khattak #1
Your answer is fine with relativity but my intent was to bring one’s attention towards the following common sense
Let say for example, when both pulse and ship are moving in same direction
For inside observer, the pulse after entering the ship through its tail, will pass through the cockpit within one second or two, while
For outside observer, the pulse entering the ship through its tail will travel inside ship (which is also moving say with 0.99c) for hours and hours…
Thus there is an eye popping difference in duration of both events (one second and hours and hours). thanks 96.52.178.55 ( talk) 19:52, 30 July 2009 (UTC) Khattak
I must be misinterpreting the meaning of the variables in the basic formula (in the overview section), and I was hoping someone could clear up my confusion.
If you are on a spaceship traveling 4/5 light speed for 10 years (in your frame), you will age 10 years. During this time, your friend on Earth should have aged 50/3 years [(1-(4/5)^2)^.5= 3/5]. Isn't 50/3 years the "proper time" since it is the time recorded by the stationary observer? Basically, my question is: shouldn't the proper time always be greater than the moving observer's time? During any given interval between two events taking place on Earth, the clock on the moving spaceship will tick fewer times than the stationary clock on Earth; so doesn't that mean the proper time is greater?
I must have it backwards, since otherwise the formula in the overview section would be incorrect, right? —Preceding unsigned comment added by Jakemhall ( talk • contribs) 10:48, 30 July 2009 (UTC)
Those who argue that time 'dilation' is a fact usually interchange velocity with speed for light which is an error. For example, in this article the time 2D/c and 2L/c seem different, but the times are actually the same event, and both observers time the event the same IF they time it correctly. Notice that the vector component speed with which the light travels upward (North)while angle traveling toward the mirror to reach it in its shifted location is v(North)= c(L/D,)where L/D is the sine of the angle the ray travels from the horizontal as seen in the second figure. When this component for light is placed into the equation for time (stationary), it becomes t = 2L/[c(L/D)] = 2D/c and is the same as the other expression (moving). Time does not actually 'dilate' simply because an observer and his clock are in motion. It is just that if one uses light as a time-keeping device, you must use the proper component of velocity for the travel direction of the pulse involved. The magnitude of the velocity of light can vary from +c(East) to -c(West). Time dilation was invented in 1905 when Einstein mistakenly thought that (since light rays travel in straight lines) c and v could be always interchanged. But speed and velocity I assure you are different when an angle-traveling ray is used to compute time in one instance, and then that ray is used again in a different diagram without considering that its velocity in the direction of interest is now a component that is less than c. Velocity is only c in the propagation direction itself, which occurs here at an angle above the horizon toward the shifted mirror. The sationary observer does not compute time correctly because he ignores, or is unaware of, the lateral shift of the light and mirror ongoing at the same time. Time dilation is only an apparent phenomenon that occurs if calculations do not consider the component velocity for a directed pulse of light. Anyone can exceed the velocity of light. Simply point it North and you run West to reach the target. You beat the velocity of light toward the target. Or reflect a beam on a mirror; when the light returns to its starting position the velocity at that position was zero. You continue running,and don't turn around, you exceed the velocity of light.
Richard Sauerheber, Ph.D. —Preceding unsigned comment added by 205.153.156.222 ( talk)
Note to anonymous 205.153.156.222: I have reverted the modifications your made to your comments after I had replied to them. Please have a look at your talk page. Thank you. DVdm ( talk) 22:31, 18 September 2009 (UTC)
The Lorentz time dilation equation is derived using the concept of a light clock in which a pulse of light being continually reflected backwards and forwards between two mirrors for moving observer and zigzag for stationary observer.
1- The above concept was taken from a moving train (falling of object from ceiling of train under the influence of earth's gravity) but since in space there is no GRAVITY therefore an emerging pulse should follow the straight downward path from where it is emerged, and shouldn’t move along the direction of ship for both stationary and moving observers. (In other words the free fall of pulse shouldn’t be influenced by the ship’s velocity)
2- As light carry momentum and can push/ veer objects easily in SPACE therefore, should a ship swerve with a new resultant velocity when a pulse touches the second mirror? Further, any flat object or flanks of ship could reflect a pulse in thought experiment, so is there any special reason in using a mirror?
The same is applied to the Einstein’s elevator
Your comment please! Thanks
68.147.38.24 ( talk) 01:57, 4 October 2009 (UTC) Khattak
Sorry about the aforementioned mistakes
The description of pulse bounces between top and bottom mirror, shown in article is very fine with question.
Moving train is not stated in the article but my intent was basic relativity 68.147.38.24 ( talk) 16:34, 4 October 2009 (UTC) Khattak #1
A recent edit has been reverted because it added errors and seems to have been made without reading the article: the new material is reproduced here, some reasons are given below why the edit is erroneous and misleading, pls discuss before replacing the material.
Sample of reasons against the material in the edit:
Terry0051 ( talk) 21:28, 19 October 2009 (UTC)
Thank you for the revert; peer review is never bad, and when it rejects your edits, that's positive evidence that it has served its purpose. In the interest of developing the article, I'd like to respond, though. Let's see if we can synthesize something better as a result some discussion.
Decoy ( talk) 21:59, 20 October 2009 (UTC)
[From Terry0051] Thanks for your thoughtful comments. Can I suggest that one hopefully fruitful route might be, to identify/seek consensus on which core facts should be communicated in the intro or opening sections (checking that there's consensus on the facts themselves!) and then working out how to put them briefly & clearly in suitable encyclopedic style.
I have to apologise for having only a short time right now, I'll get back to it in a couple of days, but points I'd suggest are:
With best wishes in the meantime Terry0051 ( talk) 13:36, 21 October 2009 (UTC)
(Returning to offer supplementary suggestions answering in part the question posed about 'what points are unsourced & would need [specific] RS support?') The relevant points for specific support seem to include the suggested facts about:
With good wishes, Terry0051 ( talk) 19:31, 28 October 2009 (UTC)
Anonymous user 213.100.87.94 has repeatedly introduced edits stating that the earth's orbital motion and the galactic motion of the solar system should be taken into account for terrestrial time scales.
Standard time scales used on the surface of the Earth include Terrestrial Time, which is closely related (by an offset) to international atomic time and (by a scale factor) to Geocentric Terrestrial Time. But these are coordinate time scales, and the coordinate system to which they refer is a geocentric one, moving with the Earth. Accordingly, the Earth's orbital motion around the Sun and the galactic motion of the solar system are not relevant to these time scales. (There are indeed other coordinate time scales referred to the solar-system barycenter, including TCB and TDB, for which the earth's orbital motion and its (varying) position in the relevant gravitational potential wells are taken into account.)
As already stated in DVdm's edit summary, the effect of the Earth's rotational motion has indeed been taken into account for the synchronization of Earth-based atomic clocks. The concept 'stationary clock' was not present in the article before the edits by 213.100.87.94, and really does not play a part. Accordingly this edit (*) and others like it by 213.100.87.94 do appear to be in error: (*: "In reality however, there are no stationary clocks, given that the earth rotates on it's axis with a surface velocity of 465 m/s, and orbits the sun at a mean velocity of 14,893 m/s, within a solar system which itself orbits the outer edge of the Milky Way Galaxy at a velocity of 2,092,990 m/s, these velocities must also be included in the equation, but in fact they never are.") It seems appropriate to remove this statement.
Terry0051 ( talk) 20:35, 10 December 2009 (UTC)
I created the following diagram which show's time dilation Δt' (Y axis) from the frame of reference of a moving observer traveling at the speed V (X axis in natural unit c),i think this is helpful diagram because it clearly show that for small friction of the speed of light, time dilation is approximately one (almost flat line for x from 0 to 0.1). what do you think? should i add the diagram to the article? —Preceding unsigned comment added by Zayani ( talk • contribs) 18:17, 19 December 2009 (UTC)