Time dilation for a moving electronic clock
$begingroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
1
$begingroup$
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
4 hours ago
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
New contributor
$endgroup$
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
New contributor
New contributor
edited 2 hours ago
caveman
New contributor
asked 7 hours ago
cavemancaveman
1285
1285
New contributor
New contributor
1
$begingroup$
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
4 hours ago
add a comment |
1
$begingroup$
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
4 hours ago
1
1
$begingroup$
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
4 hours ago
$begingroup$
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
4 hours ago
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.
$endgroup$
add a 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).
$endgroup$
15
$begingroup$
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
6 hours ago
9
$begingroup$
hats off to, "give them a boost".
$endgroup$
– JEB
5 hours ago
1
$begingroup$
@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
5 hours ago
2
$begingroup$
Well I was shooting for the double entendre.
$endgroup$
– Aaron Stevens
5 hours ago
1
$begingroup$
@AaronStevens Double entendres are a degenerate form of humor.
$endgroup$
– Acccumulation
2 hours ago
|
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.
$endgroup$
$begingroup$
So, my equation is correct to calculate the dilation of the electronic clock?
$endgroup$
– caveman
2 hours ago
$begingroup$
@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
2 hours ago
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.
$endgroup$
add a comment |
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4 Answers
4
active
oldest
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4 Answers
4
active
oldest
votes
active
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$begingroup$
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.
$endgroup$
add a 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.
$endgroup$
add a 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.
$endgroup$
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 6 hours ago
Žarko TomičićŽarko Tomičić
1,003511
1,003511
add a comment |
add a 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).
$endgroup$
15
$begingroup$
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
6 hours ago
9
$begingroup$
hats off to, "give them a boost".
$endgroup$
– JEB
5 hours ago
1
$begingroup$
@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
5 hours ago
2
$begingroup$
Well I was shooting for the double entendre.
$endgroup$
– Aaron Stevens
5 hours ago
1
$begingroup$
@AaronStevens Double entendres are a degenerate form of humor.
$endgroup$
– Acccumulation
2 hours ago
|
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).
$endgroup$
15
$begingroup$
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
6 hours ago
9
$begingroup$
hats off to, "give them a boost".
$endgroup$
– JEB
5 hours ago
1
$begingroup$
@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
5 hours ago
2
$begingroup$
Well I was shooting for the double entendre.
$endgroup$
– Aaron Stevens
5 hours ago
1
$begingroup$
@AaronStevens Double entendres are a degenerate form of humor.
$endgroup$
– Acccumulation
2 hours ago
|
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).
$endgroup$
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 6 hours ago
JEBJEB
6,1681718
6,1681718
15
$begingroup$
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
6 hours ago
9
$begingroup$
hats off to, "give them a boost".
$endgroup$
– JEB
5 hours ago
1
$begingroup$
@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
5 hours ago
2
$begingroup$
Well I was shooting for the double entendre.
$endgroup$
– Aaron Stevens
5 hours ago
1
$begingroup$
@AaronStevens Double entendres are a degenerate form of humor.
$endgroup$
– Acccumulation
2 hours ago
|
show 3 more comments
15
$begingroup$
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
6 hours ago
9
$begingroup$
hats off to, "give them a boost".
$endgroup$
– JEB
5 hours ago
1
$begingroup$
@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
5 hours ago
2
$begingroup$
Well I was shooting for the double entendre.
$endgroup$
– Aaron Stevens
5 hours ago
1
$begingroup$
@AaronStevens Double entendres are a degenerate form of humor.
$endgroup$
– Acccumulation
2 hours ago
15
15
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
9
9
$begingroup$
hats off to, "give them a boost".
$endgroup$
– JEB
5 hours ago
$begingroup$
hats off to, "give them a boost".
$endgroup$
– JEB
5 hours ago
1
1
$begingroup$
@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
5 hours ago
$begingroup$
@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
5 hours ago
2
2
$begingroup$
Well I was shooting for the double entendre.
$endgroup$
– Aaron Stevens
5 hours ago
$begingroup$
Well I was shooting for the double entendre.
$endgroup$
– Aaron Stevens
5 hours ago
1
1
$begingroup$
@AaronStevens Double entendres are a degenerate form of humor.
$endgroup$
– Acccumulation
2 hours ago
$begingroup$
@AaronStevens Double entendres are a degenerate form of humor.
$endgroup$
– Acccumulation
2 hours ago
|
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.
$endgroup$
$begingroup$
So, my equation is correct to calculate the dilation of the electronic clock?
$endgroup$
– caveman
2 hours ago
$begingroup$
@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
2 hours ago
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.
$endgroup$
$begingroup$
So, my equation is correct to calculate the dilation of the electronic clock?
$endgroup$
– caveman
2 hours ago
$begingroup$
@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
2 hours ago
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.
$endgroup$
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 6 hours ago
PM 2RingPM 2Ring
2,9612922
2,9612922
$begingroup$
So, my equation is correct to calculate the dilation of the electronic clock?
$endgroup$
– caveman
2 hours ago
$begingroup$
@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
2 hours ago
add a comment |
$begingroup$
So, my equation is correct to calculate the dilation of the electronic clock?
$endgroup$
– caveman
2 hours ago
$begingroup$
@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
2 hours ago
$begingroup$
So, my equation is correct to calculate the dilation of the electronic clock?
$endgroup$
– caveman
2 hours ago
$begingroup$
So, my equation is correct to calculate the dilation of the electronic clock?
$endgroup$
– caveman
2 hours ago
$begingroup$
@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
2 hours ago
$begingroup$
@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
2 hours ago
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 2 hours ago
Andrew SteaneAndrew Steane
5,289735
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caveman is a new contributor. Be nice, and check out our Code of Conduct.
caveman is a new contributor. Be nice, and check out our Code of Conduct.
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caveman is a new contributor. Be nice, and check out our Code of Conduct.
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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
4 hours ago