Time dilation for a moving electronic clockIs time dilation an illusion?SR time dilationIs time dilation a...
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Time dilation for a moving electronic clock
Is time dilation an illusion?SR time dilationIs time dilation a mechanical slowdown only?Question about time dilation and relativityHow time dilation happens by velocity?How is time dilation consistent between both observer?Is the time dilation experiment for real?Will moving observer see time dilation?Special relativity clarifiyng time dilation experimentShould we need to consider time dilation effect when travel in a very fast rocket?
$begingroup$
Is this a correct application for the time dilation theorem?
Suppose an electronic clock has a copper wire of length $d$, which allows the electrons to take $t$ seconds to complete a cycle on the $d$-long copper wire, in the electronic clock.
Then, the the entire clock is moving in a rocket at a speed $v$ in a straight line.
The electronic clock must slowdown. My question is, is the following the new slowed-down time (update: in the reference frame of someone stationary that is not moving in the rocket)?
$$
t_{new} = frac{t}{sqrt{1-frac{v^2}{c^2}}}
$$
where $c$ is light speed.
Is that correct? If not correct, why?
Note: just to be clear, I know that applies on the light clock. But does it also apply on other non-light clocks, such as an electronic clock? E.g. should I replace $c$ by the speed of the electron movement on that copper instead?
time-dilation
asked Mar 13 at 13:51
cavemancaveman
1436
$endgroup$
add a comment |
$begingroup$
Is this a correct application for the time dilation theorem?
Suppose an electronic clock has a copper wire of length $d$, which allows the electrons to take $t$ seconds to complete a cycle on the $d$-long copper wire, in the electronic clock.
Then, the the entire clock is moving in a rocket at a speed $v$ in a straight line.
The electronic clock must slowdown. My question is, is the following the new slowed-down time (update: in the reference frame of someone stationary that is not moving in the rocket)?
$$
t_{new} = frac{t}{sqrt{1-frac{v^2}{c^2}}}
$$
where $c$ is light speed.
Is that correct? If not correct, why?
Note: just to be clear, I know that applies on the light clock. But does it also apply on other non-light clocks, such as an electronic clock? E.g. should I replace $c$ by the speed of the electron movement on that copper instead?
time-dilation
asked Mar 13 at 13:51
cavemancaveman
1436
$endgroup$
5
$begingroup$
The speed of the electrons isn't what matters. An electromagnetic signal moving along a wire propagates at the speed of light. Adm. Grace Hopper had a habit of handing out wires cut to the length of a light-nanosecond, as a way to illustrate how that makes a hard limit on how fast data can be moved about a computer system.
$endgroup$
– Monty Harder
Mar 13 at 17:18
6
$begingroup$
@MontyHarder slightly pedantic correction - electromagnetic signals in wires propagate slower than the speed of light, sometimes very close to it (>95%), sometimes as slow as about half of it for certain types of coaxial cable, depending mostly on the dielectric spacer used.
$endgroup$
– llama
Mar 13 at 22:19
add a comment |
$begingroup$
Is this a correct application for the time dilation theorem?
Suppose an electronic clock has a copper wire of length $d$, which allows the electrons to take $t$ seconds to complete a cycle on the $d$-long copper wire, in the electronic clock.
Then, the the entire clock is moving in a rocket at a speed $v$ in a straight line.
The electronic clock must slowdown. My question is, is the following the new slowed-down time (update: in the reference frame of someone stationary that is not moving in the rocket)?
$$
t_{new} = frac{t}{sqrt{1-frac{v^2}{c^2}}}
$$
where $c$ is light speed.
Is that correct? If not correct, why?
Note: just to be clear, I know that applies on the light clock. But does it also apply on other non-light clocks, such as an electronic clock? E.g. should I replace $c$ by the speed of the electron movement on that copper instead?
time-dilation
asked Mar 13 at 13:51
cavemancaveman
1436
$endgroup$
Is this a correct application for the time dilation theorem?
Suppose an electronic clock has a copper wire of length $d$, which allows the electrons to take $t$ seconds to complete a cycle on the $d$-long copper wire, in the electronic clock.
Then, the the entire clock is moving in a rocket at a speed $v$ in a straight line.
The electronic clock must slowdown. My question is, is the following the new slowed-down time (update: in the reference frame of someone stationary that is not moving in the rocket)?
$$
t_{new} = frac{t}{sqrt{1-frac{v^2}{c^2}}}
$$
where $c$ is light speed.
Is that correct? If not correct, why?
Note: just to be clear, I know that applies on the light clock. But does it also apply on other non-light clocks, such as an electronic clock? E.g. should I replace $c$ by the speed of the electron movement on that copper instead?
time-dilation
time-dilation
asked Mar 13 at 13:51
cavemancaveman
1436
asked Mar 13 at 13:51
cavemancaveman
1436
edited Mar 13 at 19:08
caveman
asked Mar 13 at 13:51
cavemancaveman
1436
asked Mar 13 at 13:51
cavemancaveman
1436
asked Mar 13 at 13:51
cavemancaveman
1436
1436
5
$begingroup$
The speed of the electrons isn't what matters. An electromagnetic signal moving along a wire propagates at the speed of light. Adm. Grace Hopper had a habit of handing out wires cut to the length of a light-nanosecond, as a way to illustrate how that makes a hard limit on how fast data can be moved about a computer system.
$endgroup$
– Monty Harder
Mar 13 at 17:18
6
$begingroup$
@MontyHarder slightly pedantic correction - electromagnetic signals in wires propagate slower than the speed of light, sometimes very close to it (>95%), sometimes as slow as about half of it for certain types of coaxial cable, depending mostly on the dielectric spacer used.
$endgroup$
– llama
Mar 13 at 22:19
add a comment |
5
$begingroup$
The speed of the electrons isn't what matters. An electromagnetic signal moving along a wire propagates at the speed of light. Adm. Grace Hopper had a habit of handing out wires cut to the length of a light-nanosecond, as a way to illustrate how that makes a hard limit on how fast data can be moved about a computer system.
$endgroup$
– Monty Harder
Mar 13 at 17:18
6
$begingroup$
@MontyHarder slightly pedantic correction - electromagnetic signals in wires propagate slower than the speed of light, sometimes very close to it (>95%), sometimes as slow as about half of it for certain types of coaxial cable, depending mostly on the dielectric spacer used.
$endgroup$
– llama
Mar 13 at 22:19
5
5
$begingroup$
The speed of the electrons isn't what matters. An electromagnetic signal moving along a wire propagates at the speed of light. Adm. Grace Hopper had a habit of handing out wires cut to the length of a light-nanosecond, as a way to illustrate how that makes a hard limit on how fast data can be moved about a computer system.
$endgroup$
– Monty Harder
Mar 13 at 17:18
$begingroup$
The speed of the electrons isn't what matters. An electromagnetic signal moving along a wire propagates at the speed of light. Adm. Grace Hopper had a habit of handing out wires cut to the length of a light-nanosecond, as a way to illustrate how that makes a hard limit on how fast data can be moved about a computer system.
$endgroup$
– Monty Harder
Mar 13 at 17:18
6
6
$begingroup$
@MontyHarder slightly pedantic correction - electromagnetic signals in wires propagate slower than the speed of light, sometimes very close to it (>95%), sometimes as slow as about half of it for certain types of coaxial cable, depending mostly on the dielectric spacer used.
$endgroup$
– llama
Mar 13 at 22:19
$begingroup$
@MontyHarder slightly pedantic correction - electromagnetic signals in wires propagate slower than the speed of light, sometimes very close to it (>95%), sometimes as slow as about half of it for certain types of coaxial cable, depending mostly on the dielectric spacer used.
$endgroup$
– llama
Mar 13 at 22:19
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
To answer your first question, yes, the formula is correct, and it shows changed time interval for a moving observer. And, yes, the formula can be applied to all clocks and to all events.Time interval that takes one event to happen changes depending on the motion of the observer. In order to see this for all events, look for some derivation of Lorentz boosts that is, Lorentz transformations regarding observers in constant speed motion in regard to each other. One way to derive these is to observe what happens to a spherical light pulse for two different observers in relative motion. When you get Lorentz transformations for the time and space then you just check out how does a time interval behave, that is, how does difference betwen two time points look in one and how in some other frame. So you can have t1 and t2. Time interval is t1-t2...transform t1 and t2 and check how does transformed time interval, T1 - T2, behave.
answered Mar 13 at 15:20
Žarko TomičićŽarko Tomičić
1,013511
$endgroup$
$begingroup$
Thanks. But to be honest, I am a bit surprised why the light speed, which is derived from the light-clock, still applies to non-light clock ones. Is there any proof for this applicability to non-light clocks? I mean, I can re-prove the time dilation by using an electronic clock, then instead of $c^2$ I'll have $v_e^2$ denoting electron's speed on that copper. The resultant time dilation would be different than the light-clock ones. So why should I believe the light-based one, and not the electron-based one for the dilation?
$endgroup$
– caveman
Mar 13 at 22:55
4
$begingroup$
@caveman: The starting point of relativity is the postulate that all fundamental laws of nature work equally well in a moving inertial frame. That this postulate is true about our universe is ultimately an empirical fact -- it's not a matter of proof, but of (abundant) empirical verification.
$endgroup$
– Henning Makholm
Mar 13 at 23:20
2
$begingroup$
So therefore if a light clock ticks slower by a certain factor in a moving frame, that factor has to hold for all other clocks too, because the laws that we have just postulated are invariant say they must tick at the same rate as a light clock.
$endgroup$
– Henning Makholm
Mar 13 at 23:21
$begingroup$
@caveman: So why should I believe the light-based one, and not the electron-based one for the dilation? Because the speed of light is frame-invariant, and the speed of electrons is not. And the derivation depends on that frame-invariance.
$endgroup$
– WillO
Mar 14 at 4:17
1
$begingroup$
@caveman: light speed is not just about light. It is fundamental in many ways, it is invariant when you change your coordinates from one frame to another and it is also limiting speed for every object. But, rather than thinkikng about light speed as the fundamental property we could think this way: everything in the universe moves with the light speed if we look at it from the 4 dim space-time perspective. So, as you sit in your chair, we could say that you are moving through time with the light speed in your rest frame, of course...but, when you start to gain speed you move less in time....
$endgroup$
– Žarko Tomičić
Mar 14 at 13:36
|
show 1 more comment
$begingroup$
Note that the clock slows down as viewed from someone else's reference frame.
In the rocket: clocks, melting ice, human biology, time, and whatever else you can think of all proceed normally. (Same with length contraction: right now we are both pancake flat in the reference frame of a cosmic ray, but we don't really notice it--at all).
Of course life on Earth proceeds normally too, even though the rocket's occupants see Earth's time ticking slowly (not to mention an, ahem, flat Earth).
answered Mar 13 at 15:21
JEBJEB
6,3331718
$endgroup$
23
$begingroup$
Ah, so flat Earthers aren't necessarily wrong, they are just in the wrong reference frame! :) Someone should really give them a boost.
$endgroup$
– Aaron Stevens
Mar 13 at 15:23
11
$begingroup$
hats off to, "give them a boost".
$endgroup$
– JEB
Mar 13 at 15:36
2
$begingroup$
@AaronStevens I read it as "that should really give them a boost [ideologically]", which is great double entendre. No offense, but I am putting my hat back on.
$endgroup$
– JEB
Mar 13 at 15:45
2
$begingroup$
Well I was shooting for the double entendre.
$endgroup$
– Aaron Stevens
Mar 13 at 15:46
2
$begingroup$
@AaronStevens "(of an encoding or function) Having multiple domain elements correspond to one element of the range." en.wiktionary.org/wiki/degenerate
$endgroup$
– Acccumulation
Mar 13 at 19:00
|
show 3 more comments
$begingroup$
All (properly functioning) clocks must experience the same time dilation, no matter what mechanism they use to tell time. Otherwise, you could use 2 different clocks to determine your absolute speed, but relativity says that absolute speed isn't a thing.
answered Mar 13 at 15:06
PM 2RingPM 2Ring
3,0072922
$endgroup$
$begingroup$
So, my equation is correct to calculate the dilation of the electronic clock?
$endgroup$
– caveman
Mar 13 at 19:11
1
$begingroup$
@caveman Yes, it is. If it's the right equation for a lightclock, it's the right equation for an electronic clock, or a caesium atomic clock, or a mechanical pendulum clock, or an hourglass. ;)
$endgroup$
– PM 2Ring
Mar 13 at 19:23
add a comment |
$begingroup$
I recommend Feynmann's chapter on this. It is included in Six easy pieces I think. He points out that once you've found the effect for one type of clock (e.g. light pulse clock) then you have found it for all types, because otherwise the different types of clock would get out of step with one another when observed by one observer but stay in step when observed by another. That would be a direct impossibility. For example if one observer says that two clocks go "bong" at the same moment at the same place, then all observers must agree that the bongs happened at the same moment at the same place.
You can put your electronic clock next to a light pulse clock, both in the same state of overall motion, and then whenever they tick at the same moment in their joint rest frame, that is one joint 'bong'.
This is also discussed in other introductory texts of course.
answered Mar 13 at 19:24
Andrew SteaneAndrew Steane
5,478735
$endgroup$
add a comment |
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4 Answers
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oldest
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4 Answers
4
active
oldest
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oldest
votes
$begingroup$
To answer your first question, yes, the formula is correct, and it shows changed time interval for a moving observer. And, yes, the formula can be applied to all clocks and to all events.Time interval that takes one event to happen changes depending on the motion of the observer. In order to see this for all events, look for some derivation of Lorentz boosts that is, Lorentz transformations regarding observers in constant speed motion in regard to each other. One way to derive these is to observe what happens to a spherical light pulse for two different observers in relative motion. When you get Lorentz transformations for the time and space then you just check out how does a time interval behave, that is, how does difference betwen two time points look in one and how in some other frame. So you can have t1 and t2. Time interval is t1-t2...transform t1 and t2 and check how does transformed time interval, T1 - T2, behave.
answered Mar 13 at 15:20
Žarko TomičićŽarko Tomičić
1,013511
$endgroup$
$begingroup$
Thanks. But to be honest, I am a bit surprised why the light speed, which is derived from the light-clock, still applies to non-light clock ones. Is there any proof for this applicability to non-light clocks? I mean, I can re-prove the time dilation by using an electronic clock, then instead of $c^2$ I'll have $v_e^2$ denoting electron's speed on that copper. The resultant time dilation would be different than the light-clock ones. So why should I believe the light-based one, and not the electron-based one for the dilation?
$endgroup$
– caveman
Mar 13 at 22:55
4
$begingroup$
@caveman: The starting point of relativity is the postulate that all fundamental laws of nature work equally well in a moving inertial frame. That this postulate is true about our universe is ultimately an empirical fact -- it's not a matter of proof, but of (abundant) empirical verification.
$endgroup$
– Henning Makholm
Mar 13 at 23:20
2
$begingroup$
So therefore if a light clock ticks slower by a certain factor in a moving frame, that factor has to hold for all other clocks too, because the laws that we have just postulated are invariant say they must tick at the same rate as a light clock.
$endgroup$
– Henning Makholm
Mar 13 at 23:21
$begingroup$
@caveman: So why should I believe the light-based one, and not the electron-based one for the dilation? Because the speed of light is frame-invariant, and the speed of electrons is not. And the derivation depends on that frame-invariance.
$endgroup$
– WillO
Mar 14 at 4:17
1
$begingroup$
@caveman: light speed is not just about light. It is fundamental in many ways, it is invariant when you change your coordinates from one frame to another and it is also limiting speed for every object. But, rather than thinkikng about light speed as the fundamental property we could think this way: everything in the universe moves with the light speed if we look at it from the 4 dim space-time perspective. So, as you sit in your chair, we could say that you are moving through time with the light speed in your rest frame, of course...but, when you start to gain speed you move less in time....
$endgroup$
– Žarko Tomičić
Mar 14 at 13:36
|
show 1 more comment
$begingroup$
To answer your first question, yes, the formula is correct, and it shows changed time interval for a moving observer. And, yes, the formula can be applied to all clocks and to all events.Time interval that takes one event to happen changes depending on the motion of the observer. In order to see this for all events, look for some derivation of Lorentz boosts that is, Lorentz transformations regarding observers in constant speed motion in regard to each other. One way to derive these is to observe what happens to a spherical light pulse for two different observers in relative motion. When you get Lorentz transformations for the time and space then you just check out how does a time interval behave, that is, how does difference betwen two time points look in one and how in some other frame. So you can have t1 and t2. Time interval is t1-t2...transform t1 and t2 and check how does transformed time interval, T1 - T2, behave.
answered Mar 13 at 15:20
Žarko TomičićŽarko Tomičić
1,013511
$endgroup$
$begingroup$
Thanks. But to be honest, I am a bit surprised why the light speed, which is derived from the light-clock, still applies to non-light clock ones. Is there any proof for this applicability to non-light clocks? I mean, I can re-prove the time dilation by using an electronic clock, then instead of $c^2$ I'll have $v_e^2$ denoting electron's speed on that copper. The resultant time dilation would be different than the light-clock ones. So why should I believe the light-based one, and not the electron-based one for the dilation?
$endgroup$
– caveman
Mar 13 at 22:55
4
$begingroup$
@caveman: The starting point of relativity is the postulate that all fundamental laws of nature work equally well in a moving inertial frame. That this postulate is true about our universe is ultimately an empirical fact -- it's not a matter of proof, but of (abundant) empirical verification.
$endgroup$
– Henning Makholm
Mar 13 at 23:20
2
$begingroup$
So therefore if a light clock ticks slower by a certain factor in a moving frame, that factor has to hold for all other clocks too, because the laws that we have just postulated are invariant say they must tick at the same rate as a light clock.
$endgroup$
– Henning Makholm
Mar 13 at 23:21
$begingroup$
@caveman: So why should I believe the light-based one, and not the electron-based one for the dilation? Because the speed of light is frame-invariant, and the speed of electrons is not. And the derivation depends on that frame-invariance.
$endgroup$
– WillO
Mar 14 at 4:17
1
$begingroup$
@caveman: light speed is not just about light. It is fundamental in many ways, it is invariant when you change your coordinates from one frame to another and it is also limiting speed for every object. But, rather than thinkikng about light speed as the fundamental property we could think this way: everything in the universe moves with the light speed if we look at it from the 4 dim space-time perspective. So, as you sit in your chair, we could say that you are moving through time with the light speed in your rest frame, of course...but, when you start to gain speed you move less in time....
$endgroup$
– Žarko Tomičić
Mar 14 at 13:36
|
show 1 more comment
$begingroup$
To answer your first question, yes, the formula is correct, and it shows changed time interval for a moving observer. And, yes, the formula can be applied to all clocks and to all events.Time interval that takes one event to happen changes depending on the motion of the observer. In order to see this for all events, look for some derivation of Lorentz boosts that is, Lorentz transformations regarding observers in constant speed motion in regard to each other. One way to derive these is to observe what happens to a spherical light pulse for two different observers in relative motion. When you get Lorentz transformations for the time and space then you just check out how does a time interval behave, that is, how does difference betwen two time points look in one and how in some other frame. So you can have t1 and t2. Time interval is t1-t2...transform t1 and t2 and check how does transformed time interval, T1 - T2, behave.
answered Mar 13 at 15:20
Žarko TomičićŽarko Tomičić
1,013511
$endgroup$
To answer your first question, yes, the formula is correct, and it shows changed time interval for a moving observer. And, yes, the formula can be applied to all clocks and to all events.Time interval that takes one event to happen changes depending on the motion of the observer. In order to see this for all events, look for some derivation of Lorentz boosts that is, Lorentz transformations regarding observers in constant speed motion in regard to each other. One way to derive these is to observe what happens to a spherical light pulse for two different observers in relative motion. When you get Lorentz transformations for the time and space then you just check out how does a time interval behave, that is, how does difference betwen two time points look in one and how in some other frame. So you can have t1 and t2. Time interval is t1-t2...transform t1 and t2 and check how does transformed time interval, T1 - T2, behave.
answered Mar 13 at 15:20
Žarko TomičićŽarko Tomičić
1,013511
answered Mar 13 at 15:20
Žarko TomičićŽarko Tomičić
1,013511
answered Mar 13 at 15:20
Žarko TomičićŽarko Tomičić
1,013511
answered Mar 13 at 15:20
Žarko TomičićŽarko Tomičić
1,013511
1,013511
$begingroup$
Thanks. But to be honest, I am a bit surprised why the light speed, which is derived from the light-clock, still applies to non-light clock ones. Is there any proof for this applicability to non-light clocks? I mean, I can re-prove the time dilation by using an electronic clock, then instead of $c^2$ I'll have $v_e^2$ denoting electron's speed on that copper. The resultant time dilation would be different than the light-clock ones. So why should I believe the light-based one, and not the electron-based one for the dilation?
$endgroup$
– caveman
Mar 13 at 22:55
4
$begingroup$
@caveman: The starting point of relativity is the postulate that all fundamental laws of nature work equally well in a moving inertial frame. That this postulate is true about our universe is ultimately an empirical fact -- it's not a matter of proof, but of (abundant) empirical verification.
$endgroup$
– Henning Makholm
Mar 13 at 23:20
2
$begingroup$
So therefore if a light clock ticks slower by a certain factor in a moving frame, that factor has to hold for all other clocks too, because the laws that we have just postulated are invariant say they must tick at the same rate as a light clock.
$endgroup$
– Henning Makholm
Mar 13 at 23:21
$begingroup$
@caveman: So why should I believe the light-based one, and not the electron-based one for the dilation? Because the speed of light is frame-invariant, and the speed of electrons is not. And the derivation depends on that frame-invariance.
$endgroup$
– WillO
Mar 14 at 4:17
1
$begingroup$
@caveman: light speed is not just about light. It is fundamental in many ways, it is invariant when you change your coordinates from one frame to another and it is also limiting speed for every object. But, rather than thinkikng about light speed as the fundamental property we could think this way: everything in the universe moves with the light speed if we look at it from the 4 dim space-time perspective. So, as you sit in your chair, we could say that you are moving through time with the light speed in your rest frame, of course...but, when you start to gain speed you move less in time....
$endgroup$
– Žarko Tomičić
Mar 14 at 13:36
|
show 1 more comment
$begingroup$
Thanks. But to be honest, I am a bit surprised why the light speed, which is derived from the light-clock, still applies to non-light clock ones. Is there any proof for this applicability to non-light clocks? I mean, I can re-prove the time dilation by using an electronic clock, then instead of $c^2$ I'll have $v_e^2$ denoting electron's speed on that copper. The resultant time dilation would be different than the light-clock ones. So why should I believe the light-based one, and not the electron-based one for the dilation?
$endgroup$
– caveman
Mar 13 at 22:55
4
$begingroup$
@caveman: The starting point of relativity is the postulate that all fundamental laws of nature work equally well in a moving inertial frame. That this postulate is true about our universe is ultimately an empirical fact -- it's not a matter of proof, but of (abundant) empirical verification.
$endgroup$
– Henning Makholm
Mar 13 at 23:20
2
$begingroup$
So therefore if a light clock ticks slower by a certain factor in a moving frame, that factor has to hold for all other clocks too, because the laws that we have just postulated are invariant say they must tick at the same rate as a light clock.
$endgroup$
– Henning Makholm
Mar 13 at 23:21
$begingroup$
@caveman: So why should I believe the light-based one, and not the electron-based one for the dilation? Because the speed of light is frame-invariant, and the speed of electrons is not. And the derivation depends on that frame-invariance.
$endgroup$
– WillO
Mar 14 at 4:17
1
$begingroup$
@caveman: light speed is not just about light. It is fundamental in many ways, it is invariant when you change your coordinates from one frame to another and it is also limiting speed for every object. But, rather than thinkikng about light speed as the fundamental property we could think this way: everything in the universe moves with the light speed if we look at it from the 4 dim space-time perspective. So, as you sit in your chair, we could say that you are moving through time with the light speed in your rest frame, of course...but, when you start to gain speed you move less in time....
$endgroup$
– Žarko Tomičić
Mar 14 at 13:36
$begingroup$
Thanks. But to be honest, I am a bit surprised why the light speed, which is derived from the light-clock, still applies to non-light clock ones. Is there any proof for this applicability to non-light clocks? I mean, I can re-prove the time dilation by using an electronic clock, then instead of $c^2$ I'll have $v_e^2$ denoting electron's speed on that copper. The resultant time dilation would be different than the light-clock ones. So why should I believe the light-based one, and not the electron-based one for the dilation?
$endgroup$
– caveman
Mar 13 at 22:55
$begingroup$
Thanks. But to be honest, I am a bit surprised why the light speed, which is derived from the light-clock, still applies to non-light clock ones. Is there any proof for this applicability to non-light clocks? I mean, I can re-prove the time dilation by using an electronic clock, then instead of $c^2$ I'll have $v_e^2$ denoting electron's speed on that copper. The resultant time dilation would be different than the light-clock ones. So why should I believe the light-based one, and not the electron-based one for the dilation?
$endgroup$
– caveman
Mar 13 at 22:55
4
4
$begingroup$
@caveman: The starting point of relativity is the postulate that all fundamental laws of nature work equally well in a moving inertial frame. That this postulate is true about our universe is ultimately an empirical fact -- it's not a matter of proof, but of (abundant) empirical verification.
$endgroup$
– Henning Makholm
Mar 13 at 23:20
$begingroup$
@caveman: The starting point of relativity is the postulate that all fundamental laws of nature work equally well in a moving inertial frame. That this postulate is true about our universe is ultimately an empirical fact -- it's not a matter of proof, but of (abundant) empirical verification.
$endgroup$
– Henning Makholm
Mar 13 at 23:20
2
2
$begingroup$
So therefore if a light clock ticks slower by a certain factor in a moving frame, that factor has to hold for all other clocks too, because the laws that we have just postulated are invariant say they must tick at the same rate as a light clock.
$endgroup$
– Henning Makholm
Mar 13 at 23:21
$begingroup$
So therefore if a light clock ticks slower by a certain factor in a moving frame, that factor has to hold for all other clocks too, because the laws that we have just postulated are invariant say they must tick at the same rate as a light clock.
$endgroup$
– Henning Makholm
Mar 13 at 23:21
$begingroup$
@caveman: So why should I believe the light-based one, and not the electron-based one for the dilation? Because the speed of light is frame-invariant, and the speed of electrons is not. And the derivation depends on that frame-invariance.
$endgroup$
– WillO
Mar 14 at 4:17
$begingroup$
@caveman: So why should I believe the light-based one, and not the electron-based one for the dilation? Because the speed of light is frame-invariant, and the speed of electrons is not. And the derivation depends on that frame-invariance.
$endgroup$
– WillO
Mar 14 at 4:17
1
1
$begingroup$
@caveman: light speed is not just about light. It is fundamental in many ways, it is invariant when you change your coordinates from one frame to another and it is also limiting speed for every object. But, rather than thinkikng about light speed as the fundamental property we could think this way: everything in the universe moves with the light speed if we look at it from the 4 dim space-time perspective. So, as you sit in your chair, we could say that you are moving through time with the light speed in your rest frame, of course...but, when you start to gain speed you move less in time....
$endgroup$
– Žarko Tomičić
Mar 14 at 13:36
$begingroup$
@caveman: light speed is not just about light. It is fundamental in many ways, it is invariant when you change your coordinates from one frame to another and it is also limiting speed for every object. But, rather than thinkikng about light speed as the fundamental property we could think this way: everything in the universe moves with the light speed if we look at it from the 4 dim space-time perspective. So, as you sit in your chair, we could say that you are moving through time with the light speed in your rest frame, of course...but, when you start to gain speed you move less in time....
$endgroup$
– Žarko Tomičić
Mar 14 at 13:36
|
show 1 more comment
$begingroup$
Note that the clock slows down as viewed from someone else's reference frame.
In the rocket: clocks, melting ice, human biology, time, and whatever else you can think of all proceed normally. (Same with length contraction: right now we are both pancake flat in the reference frame of a cosmic ray, but we don't really notice it--at all).
Of course life on Earth proceeds normally too, even though the rocket's occupants see Earth's time ticking slowly (not to mention an, ahem, flat Earth).
answered Mar 13 at 15:21
JEBJEB
6,3331718
$endgroup$
23
$begingroup$
Ah, so flat Earthers aren't necessarily wrong, they are just in the wrong reference frame! :) Someone should really give them a boost.
$endgroup$
– Aaron Stevens
Mar 13 at 15:23
11
$begingroup$
hats off to, "give them a boost".
$endgroup$
– JEB
Mar 13 at 15:36
2
$begingroup$
@AaronStevens I read it as "that should really give them a boost [ideologically]", which is great double entendre. No offense, but I am putting my hat back on.
$endgroup$
– JEB
Mar 13 at 15:45
2
$begingroup$
Well I was shooting for the double entendre.
$endgroup$
– Aaron Stevens
Mar 13 at 15:46
2
$begingroup$
@AaronStevens "(of an encoding or function) Having multiple domain elements correspond to one element of the range." en.wiktionary.org/wiki/degenerate
$endgroup$
– Acccumulation
Mar 13 at 19:00
|
show 3 more comments
$begingroup$
Note that the clock slows down as viewed from someone else's reference frame.
In the rocket: clocks, melting ice, human biology, time, and whatever else you can think of all proceed normally. (Same with length contraction: right now we are both pancake flat in the reference frame of a cosmic ray, but we don't really notice it--at all).
Of course life on Earth proceeds normally too, even though the rocket's occupants see Earth's time ticking slowly (not to mention an, ahem, flat Earth).
answered Mar 13 at 15:21
JEBJEB
6,3331718
$endgroup$
23
$begingroup$
Ah, so flat Earthers aren't necessarily wrong, they are just in the wrong reference frame! :) Someone should really give them a boost.
$endgroup$
– Aaron Stevens
Mar 13 at 15:23
11
$begingroup$
hats off to, "give them a boost".
$endgroup$
– JEB
Mar 13 at 15:36
2
$begingroup$
@AaronStevens I read it as "that should really give them a boost [ideologically]", which is great double entendre. No offense, but I am putting my hat back on.
$endgroup$
– JEB
Mar 13 at 15:45
2
$begingroup$
Well I was shooting for the double entendre.
$endgroup$
– Aaron Stevens
Mar 13 at 15:46
2
$begingroup$
@AaronStevens "(of an encoding or function) Having multiple domain elements correspond to one element of the range." en.wiktionary.org/wiki/degenerate
$endgroup$
– Acccumulation
Mar 13 at 19:00
|
show 3 more comments
$begingroup$
Note that the clock slows down as viewed from someone else's reference frame.
In the rocket: clocks, melting ice, human biology, time, and whatever else you can think of all proceed normally. (Same with length contraction: right now we are both pancake flat in the reference frame of a cosmic ray, but we don't really notice it--at all).
Of course life on Earth proceeds normally too, even though the rocket's occupants see Earth's time ticking slowly (not to mention an, ahem, flat Earth).
answered Mar 13 at 15:21
JEBJEB
6,3331718
$endgroup$
Note that the clock slows down as viewed from someone else's reference frame.
In the rocket: clocks, melting ice, human biology, time, and whatever else you can think of all proceed normally. (Same with length contraction: right now we are both pancake flat in the reference frame of a cosmic ray, but we don't really notice it--at all).
Of course life on Earth proceeds normally too, even though the rocket's occupants see Earth's time ticking slowly (not to mention an, ahem, flat Earth).
answered Mar 13 at 15:21
JEBJEB
6,3331718
answered Mar 13 at 15:21
JEBJEB
6,3331718
answered Mar 13 at 15:21
JEBJEB
6,3331718
answered Mar 13 at 15:21
JEBJEB
6,3331718
6,3331718
23
$begingroup$
Ah, so flat Earthers aren't necessarily wrong, they are just in the wrong reference frame! :) Someone should really give them a boost.
$endgroup$
– Aaron Stevens
Mar 13 at 15:23
11
$begingroup$
hats off to, "give them a boost".
$endgroup$
– JEB
Mar 13 at 15:36
2
$begingroup$
@AaronStevens I read it as "that should really give them a boost [ideologically]", which is great double entendre. No offense, but I am putting my hat back on.
$endgroup$
– JEB
Mar 13 at 15:45
2
$begingroup$
Well I was shooting for the double entendre.
$endgroup$
– Aaron Stevens
Mar 13 at 15:46
2
$begingroup$
@AaronStevens "(of an encoding or function) Having multiple domain elements correspond to one element of the range." en.wiktionary.org/wiki/degenerate
$endgroup$
– Acccumulation
Mar 13 at 19:00
|
show 3 more comments
23
$begingroup$
Ah, so flat Earthers aren't necessarily wrong, they are just in the wrong reference frame! :) Someone should really give them a boost.
$endgroup$
– Aaron Stevens
Mar 13 at 15:23
11
$begingroup$
hats off to, "give them a boost".
$endgroup$
– JEB
Mar 13 at 15:36
2
$begingroup$
@AaronStevens I read it as "that should really give them a boost [ideologically]", which is great double entendre. No offense, but I am putting my hat back on.
$endgroup$
– JEB
Mar 13 at 15:45
2
$begingroup$
Well I was shooting for the double entendre.
$endgroup$
– Aaron Stevens
Mar 13 at 15:46
2
$begingroup$
@AaronStevens "(of an encoding or function) Having multiple domain elements correspond to one element of the range." en.wiktionary.org/wiki/degenerate
$endgroup$
– Acccumulation
Mar 13 at 19:00
23
23
$begingroup$
Ah, so flat Earthers aren't necessarily wrong, they are just in the wrong reference frame! :) Someone should really give them a boost.
$endgroup$
– Aaron Stevens
Mar 13 at 15:23
$begingroup$
Ah, so flat Earthers aren't necessarily wrong, they are just in the wrong reference frame! :) Someone should really give them a boost.
$endgroup$
– Aaron Stevens
Mar 13 at 15:23
11
11
$begingroup$
hats off to, "give them a boost".
$endgroup$
– JEB
Mar 13 at 15:36
$begingroup$
hats off to, "give them a boost".
$endgroup$
– JEB
Mar 13 at 15:36
2
2
$begingroup$
@AaronStevens I read it as "that should really give them a boost [ideologically]", which is great double entendre. No offense, but I am putting my hat back on.
$endgroup$
– JEB
Mar 13 at 15:45
$begingroup$
@AaronStevens I read it as "that should really give them a boost [ideologically]", which is great double entendre. No offense, but I am putting my hat back on.
$endgroup$
– JEB
Mar 13 at 15:45
2
2
$begingroup$
Well I was shooting for the double entendre.
$endgroup$
– Aaron Stevens
Mar 13 at 15:46
$begingroup$
Well I was shooting for the double entendre.
$endgroup$
– Aaron Stevens
Mar 13 at 15:46
2
2
$begingroup$
@AaronStevens "(of an encoding or function) Having multiple domain elements correspond to one element of the range." en.wiktionary.org/wiki/degenerate
$endgroup$
– Acccumulation
Mar 13 at 19:00
$begingroup$
@AaronStevens "(of an encoding or function) Having multiple domain elements correspond to one element of the range." en.wiktionary.org/wiki/degenerate
$endgroup$
– Acccumulation
Mar 13 at 19:00
|
show 3 more comments
$begingroup$
All (properly functioning) clocks must experience the same time dilation, no matter what mechanism they use to tell time. Otherwise, you could use 2 different clocks to determine your absolute speed, but relativity says that absolute speed isn't a thing.
answered Mar 13 at 15:06
PM 2RingPM 2Ring
3,0072922
$endgroup$
$begingroup$
So, my equation is correct to calculate the dilation of the electronic clock?
$endgroup$
– caveman
Mar 13 at 19:11
1
$begingroup$
@caveman Yes, it is. If it's the right equation for a lightclock, it's the right equation for an electronic clock, or a caesium atomic clock, or a mechanical pendulum clock, or an hourglass. ;)
$endgroup$
– PM 2Ring
Mar 13 at 19:23
add a comment |
$begingroup$
All (properly functioning) clocks must experience the same time dilation, no matter what mechanism they use to tell time. Otherwise, you could use 2 different clocks to determine your absolute speed, but relativity says that absolute speed isn't a thing.
answered Mar 13 at 15:06
PM 2RingPM 2Ring
3,0072922
$endgroup$
$begingroup$
So, my equation is correct to calculate the dilation of the electronic clock?
$endgroup$
– caveman
Mar 13 at 19:11
1
$begingroup$
@caveman Yes, it is. If it's the right equation for a lightclock, it's the right equation for an electronic clock, or a caesium atomic clock, or a mechanical pendulum clock, or an hourglass. ;)
$endgroup$
– PM 2Ring
Mar 13 at 19:23
add a comment |
$begingroup$
All (properly functioning) clocks must experience the same time dilation, no matter what mechanism they use to tell time. Otherwise, you could use 2 different clocks to determine your absolute speed, but relativity says that absolute speed isn't a thing.
answered Mar 13 at 15:06
PM 2RingPM 2Ring
3,0072922
$endgroup$
All (properly functioning) clocks must experience the same time dilation, no matter what mechanism they use to tell time. Otherwise, you could use 2 different clocks to determine your absolute speed, but relativity says that absolute speed isn't a thing.
answered Mar 13 at 15:06
PM 2RingPM 2Ring
3,0072922
answered Mar 13 at 15:06
PM 2RingPM 2Ring
3,0072922
answered Mar 13 at 15:06
PM 2RingPM 2Ring
3,0072922
answered Mar 13 at 15:06
PM 2RingPM 2Ring
3,0072922
3,0072922
$begingroup$
So, my equation is correct to calculate the dilation of the electronic clock?
$endgroup$
– caveman
Mar 13 at 19:11
1
$begingroup$
@caveman Yes, it is. If it's the right equation for a lightclock, it's the right equation for an electronic clock, or a caesium atomic clock, or a mechanical pendulum clock, or an hourglass. ;)
$endgroup$
– PM 2Ring
Mar 13 at 19:23
add a comment |
$begingroup$
So, my equation is correct to calculate the dilation of the electronic clock?
$endgroup$
– caveman
Mar 13 at 19:11
1
$begingroup$
@caveman Yes, it is. If it's the right equation for a lightclock, it's the right equation for an electronic clock, or a caesium atomic clock, or a mechanical pendulum clock, or an hourglass. ;)
$endgroup$
– PM 2Ring
Mar 13 at 19:23
$begingroup$
So, my equation is correct to calculate the dilation of the electronic clock?
$endgroup$
– caveman
Mar 13 at 19:11
$begingroup$
So, my equation is correct to calculate the dilation of the electronic clock?
$endgroup$
– caveman
Mar 13 at 19:11
1
1
$begingroup$
@caveman Yes, it is. If it's the right equation for a lightclock, it's the right equation for an electronic clock, or a caesium atomic clock, or a mechanical pendulum clock, or an hourglass. ;)
$endgroup$
– PM 2Ring
Mar 13 at 19:23
$begingroup$
@caveman Yes, it is. If it's the right equation for a lightclock, it's the right equation for an electronic clock, or a caesium atomic clock, or a mechanical pendulum clock, or an hourglass. ;)
$endgroup$
– PM 2Ring
Mar 13 at 19:23
add a comment |
$begingroup$
I recommend Feynmann's chapter on this. It is included in Six easy pieces I think. He points out that once you've found the effect for one type of clock (e.g. light pulse clock) then you have found it for all types, because otherwise the different types of clock would get out of step with one another when observed by one observer but stay in step when observed by another. That would be a direct impossibility. For example if one observer says that two clocks go "bong" at the same moment at the same place, then all observers must agree that the bongs happened at the same moment at the same place.
You can put your electronic clock next to a light pulse clock, both in the same state of overall motion, and then whenever they tick at the same moment in their joint rest frame, that is one joint 'bong'.
This is also discussed in other introductory texts of course.
answered Mar 13 at 19:24
Andrew SteaneAndrew Steane
5,478735
$endgroup$
add a comment |
$begingroup$
I recommend Feynmann's chapter on this. It is included in Six easy pieces I think. He points out that once you've found the effect for one type of clock (e.g. light pulse clock) then you have found it for all types, because otherwise the different types of clock would get out of step with one another when observed by one observer but stay in step when observed by another. That would be a direct impossibility. For example if one observer says that two clocks go "bong" at the same moment at the same place, then all observers must agree that the bongs happened at the same moment at the same place.
You can put your electronic clock next to a light pulse clock, both in the same state of overall motion, and then whenever they tick at the same moment in their joint rest frame, that is one joint 'bong'.
This is also discussed in other introductory texts of course.
answered Mar 13 at 19:24
Andrew SteaneAndrew Steane
5,478735
$endgroup$
add a comment |
$begingroup$
I recommend Feynmann's chapter on this. It is included in Six easy pieces I think. He points out that once you've found the effect for one type of clock (e.g. light pulse clock) then you have found it for all types, because otherwise the different types of clock would get out of step with one another when observed by one observer but stay in step when observed by another. That would be a direct impossibility. For example if one observer says that two clocks go "bong" at the same moment at the same place, then all observers must agree that the bongs happened at the same moment at the same place.
You can put your electronic clock next to a light pulse clock, both in the same state of overall motion, and then whenever they tick at the same moment in their joint rest frame, that is one joint 'bong'.
This is also discussed in other introductory texts of course.
answered Mar 13 at 19:24
Andrew SteaneAndrew Steane
5,478735
$endgroup$
I recommend Feynmann's chapter on this. It is included in Six easy pieces I think. He points out that once you've found the effect for one type of clock (e.g. light pulse clock) then you have found it for all types, because otherwise the different types of clock would get out of step with one another when observed by one observer but stay in step when observed by another. That would be a direct impossibility. For example if one observer says that two clocks go "bong" at the same moment at the same place, then all observers must agree that the bongs happened at the same moment at the same place.
You can put your electronic clock next to a light pulse clock, both in the same state of overall motion, and then whenever they tick at the same moment in their joint rest frame, that is one joint 'bong'.
This is also discussed in other introductory texts of course.
answered Mar 13 at 19:24
Andrew SteaneAndrew Steane
5,478735
answered Mar 13 at 19:24
Andrew SteaneAndrew Steane
5,478735
answered Mar 13 at 19:24
Andrew SteaneAndrew Steane
5,478735
answered Mar 13 at 19:24
Andrew SteaneAndrew Steane
5,478735
5,478735
add a comment |
add a comment |
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$begingroup$
The speed of the electrons isn't what matters. An electromagnetic signal moving along a wire propagates at the speed of light. Adm. Grace Hopper had a habit of handing out wires cut to the length of a light-nanosecond, as a way to illustrate how that makes a hard limit on how fast data can be moved about a computer system.
$endgroup$
– Monty Harder
Mar 13 at 17:18
6
$begingroup$
@MontyHarder slightly pedantic correction - electromagnetic signals in wires propagate slower than the speed of light, sometimes very close to it (>95%), sometimes as slow as about half of it for certain types of coaxial cable, depending mostly on the dielectric spacer used.
$endgroup$
– llama
Mar 13 at 22:19