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Why does Einstein say that a clock slows when it is moving?

by Grimble
Tags: clock, einstein, moving, slows
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Grimble
#1
Dec27-09, 04:30 AM
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In Chapter XII Einstein says the following:
Let us now consider a seconds-clock which is permanently situated at the origin (x' = 0) of K'. t' = 0 and t' = 1 are two successive ticks of this clock. The first and fourth equations of the Lorentz transformation give for these two ticks:
t = 0
and [tex]t = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]
As judged from K, the clock is moving with the velocity v; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but [tex] \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]
seconds, i.e. a somewhat larger time. As a consequence of its motion the clock goes more slowly than when at rest.
But what is he saying here?

That 1 proper second in the reference frame K is equal to gamma proper seconds in the moving frame K0, leading to the conclusion that the seconds in K are larger than the seconds in K0 and that therefore the moving clock slows?

No it isn't that, for the proper seconds in K0 have been transformed by the application of the Lorentz Transformation equations into coordinate time (as we term it), so he is saying that 1 proper second in the reference frame K is equal to gamma coordinate seconds in the moving frame K0.

So Einstein is actually describing a conversion from one set of units to another; that conversion being performed using the Lorentz Factor.

And does the moving clock slow? Well, no, for what we see is that 1 unit of time in the reference frame K is equal to gamma coordinate units, so if anything one would say that the moving clock goes faster, albeit in smaller seconds.

Grimble
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HallsofIvy
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Dec27-09, 04:42 AM
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What he is saying is that a person in frame K sees t'= 1 second tick off a clock in frame K0 while he sees [tex]t = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex] tick off his own. t and t' do NOT measure "how long a second is", they measure what time is shown on a clock. Since t is larger than t', a clock in K0 shows time changing more slowly than in K.
Grimble
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Dec27-09, 05:04 AM
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Quote Quote by HallsofIvy View Post
What he is saying is that a person in frame K sees t'= 1 second tick off a clock in frame K0 while he sees [tex]t = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex] tick off his own. t and t' do NOT measure "how long a second is", they measure what time is shown on a clock. Since t is larger than t', a clock in K0 shows time changing more slowly than in K.
Yes but since the formula he is using is [tex]t = \frac{t^'}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex] with [itex]t^'[/itex] set to 1, we have an equation where the dimensions are not the same on each side and, as I understand it, this is just plain wrong.

Pengwuino
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Dec27-09, 05:08 AM
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Why does Einstein say that a clock slows when it is moving?

t' = 1 second, the units remain
HallsofIvy
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Dec27-09, 07:53 AM
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Quote Quote by Grimble View Post
Yes but since the formula he is using is [tex]t = \frac{t^'}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex] with [itex]t^'[/itex] set to 1, we have an equation where the dimensions are not the same on each side and, as I understand it, this is just plain wrong.
Why would you say the dimensions are wrong? v and c are both in "distance per time" units so [itex]v^2/c^2[/itex] is dimensionless and so is
[tex]\frac{1}{\sqrt{1- \frac{v^2}{c^2}}}[/tex]
Both sides have dimensions of time.

(Surely you are not thinking that "setting t= 1" removes the time units on the right? t is a time measure. Setting t= 1 means setting t equal to "1 second" or "1 minute" or "1 hour".)
Grimble
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Dec27-09, 06:53 PM
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Quote Quote by Pengwuino View Post
t' = 1 second, the units remain
What sort of second is that? A proper second, or a coordinate second, for they are surely separate units, are they not? After all the Lorentz factor is that which is used to convert between the two.


Quote Quote by HallsofIvy View Post
Why would you say the dimensions are wrong? v and c are both in "distance per time" units so [itex]v^2/c^2[/itex] is dimensionless and so is
[tex]\frac{1}{\sqrt{1- \frac{v^2}{c^2}}}[/tex]
Both sides have dimensions of time.

(Surely you are not thinking that "setting t= 1" removes the time units on the right? t is a time measure. Setting t= 1 means setting t equal to "1 second" or "1 minute" or "1 hour".)

Wait a minute there, you cannot be suggesting that an equation that has units of days on one side and milliseconds on the other has the same units, just because they both are units of time, can you? For that would be the equivalent of saying we could have feet on one side of an equation and centimetres on the other and that would be valid, as they are both dimensions of distance.

No, what the formula is saying is that 1 unit of proper time is equal to gamma seconds of coordinate time, that is coordinate time has units that are smaller than proper units by a factor of gamma.

Grimble
meopemuk
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Dec27-09, 07:30 PM
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Quote Quote by Grimble View Post
What sort of second is that? A proper second, or a coordinate second, for they are surely separate units, are they not? After all the Lorentz factor is that which is used to convert between the two.
I don't know what you mean by "proper second" and "coordinate second", and what is the difference between them. As far as physics is concerned, there is only one type of "second" - the second shown by a well-designed clock. Einstein predicted that the "second" shown by a moving clock lasts longer in comparison to the "second" shown by the clock at rest. This is the relativistic time dilation.

Eugene.
edpell
#8
Dec27-09, 07:49 PM
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Quote Quote by HallsofIvy View Post
What he is saying is that a person in frame K sees t'= 1 second tick off a clock in frame K0 while he sees [tex]t = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex] tick off his own. t and t' do NOT measure "how long a second is", they measure what time is shown on a clock. Since t is larger than t', a clock in K0 shows time changing more slowly than in K.
likewise by symmetry a person in frame K0 sees t=1 seconds tick off a clock in frame K while he sees [tex]t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex] tick off his own. Since t' is larger than t, a clock in K shows time changing more slowly than in K0

Have I stated this correctly? Velocity is relative yes?
meopemuk
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Dec27-09, 07:56 PM
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Quote Quote by edpell View Post
likewise by symmetry a person in frame K0 sees t=1 seconds tick off a clock in frame K while he sees [tex]t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex] tick off his own. Since t' is larger than t, a clock in K shows time changing more slowly than in K0

Have I stated this correctly? Velocity is relative yes?
If you and I are moving with respect to each other, then I say that your second (the second shown by your co-moving clock) is longer than mine. On the other hand, you say that my second is longer than your second. Both of us are correct. That's relativity.
matheinste
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Dec27-09, 08:03 PM
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Hello Grimble

All this has already been addressed at length in a thread started by you entitled, as far as I can remember, Time Dilation Formula.
Matheinste
edpell
#11
Dec27-09, 08:31 PM
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Observation versus clock count. I see how both parties see the frequency of light flashes emitted by the other as slower due to their relative velocity away from each other. But suppose both observers agree to fire retro rockets after 1 million flashes (one flash per second) locally in their own frame and come to a mutual relative velocity of zero. And at that point they both agree to turn off the flasher. How many total flashes will each see before they see the flashing stop? I say both will count 1 million flashes. So what do you mean when you say a different amount of time passes in one frame versus the other?
Grimble
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Dec31-09, 05:18 AM
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Let us say we have two identical clocks, one in system A, and one in system B.

Clock A records 10 seconds while clock B records only 8 seconds.

So for clock B more time is passing between successive 'ticks' (as Einstein put it): clock B is running slow.

But is it? Or is time running faster? If more time is passing between two ticks of an identical clock does that not imply that it is time passing at a faster rate?
HallsofIvy
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Dec31-09, 05:40 AM
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Quote Quote by Grimble View Post
What sort of second is that? A proper second, or a coordinate second, for they are surely separate units, are they not? After all the Lorentz factor is that which is used to convert between the two.
In this formula, a "second" is "a second as measured on the observer's clock". No, that does not convert between two different "kinds" of seconds.





Wait a minute there, you cannot be suggesting that an equation that has units of days on one side and milliseconds on the other has the same units, just because they both are units of time, can you? For that would be the equivalent of saying we could have feet on one side of an equation and centimetres on the other and that would be valid, as they are both dimensions of distance.
No, I am not suggesting nor did I say such a thing. There is only one "kind" of second here, one second as measured on the observers clock.

No, what the formula is saying is that 1 unit of proper time is equal to gamma seconds of coordinate time, that is coordinate time has units that are smaller than proper units by a factor of gamma.
No, that is NOT what the formula is saying.

Grimble
HallsofIvy
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Dec31-09, 05:43 AM
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Quote Quote by edpell View Post
Observation versus clock count. I see how both parties see the frequency of light flashes emitted by the other as slower due to their relative velocity away from each other. But suppose both observers agree to fire retro rockets after 1 million flashes (one flash per second) locally in their own frame and come to a mutual relative velocity of zero. And at that point they both agree to turn off the flasher. How many total flashes will each see before they see the flashing stop? I say both will count 1 million flashes. So what do you mean when you say a different amount of time passes in one frame versus the other?
Yes, of course, both will count 1 million flashes. But each will see the others flashes as coming slower than their own and will believe that he started his retro rockets before the other.
HallsofIvy
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Dec31-09, 05:47 AM
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Quote Quote by Grimble View Post
Let us say we have two identical clocks, one in system A, and one in system B.

Clock A records 10 seconds while clock B records only 8 seconds.

So for clock B more time is passing between successive 'ticks' (as Einstein put it): clock B is running slow.

But is it? Or is time running faster? If more time is passing between two ticks of an identical clock does that not imply that it is time passing at a faster rate?
No, there is NOT "more time passing between two ticks" because it is those ticks that are measuring time. An observer in system A sees B's clock record 10 seconds only 2 seconds after his. He concludes that B's clock is running slower than his and that therefore time is running more slowly in B's system.
matheinste
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Dec31-09, 06:37 AM
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Quote Quote by Grimble View Post
Let us say we have two identical clocks, one in system A, and one in system B.

Clock A records 10 seconds while clock B records only 8 seconds.

So for clock B more time is passing between successive 'ticks' (as Einstein put it): clock B is running slow.

But is it? Or is time running faster? If more time is passing between two ticks of an identical clock does that not imply that it is time passing at a faster rate?
The rate of ticking of any clock IS the rate at which time is passing. There is no "real" rate of time passing that a clock's rate of ticking can be compared with. The rate of ticking is THE time for any clock. Time does not pass at some absolute rate, some clocks recording more of it and some less depending on their relative motion. All clocks are on an equal footing.

Matheinste.
Grimble
#17
Dec31-09, 10:31 AM
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Quote Quote by HallsofIvy View Post
There is only one "kind" of second here, one second as measured on the observers clock.
OK, so if two clocks measure the same second, and it is longer for one clock than the other, but it is still the same second?

A measure, of any kind, is surely based upon some standard, and that standard is what measurements are relative to?

Then if the length of a second is different for two observers, yet is still equal to one second in each case, then the standard second for each observer has to be different?

And if the standard is different for each observer, then, surely, they must be measuring time on separate scales that are appliccable to each observer?

Or are you saying that there is, somehow a single "universal" second, or somekind of "Absolute" second that is measured differently by observer's moving at different relative velocities?

If a second has different durations, depending on the conditions under which it is measured the does this not imply defferent scales depending on those conditions?

And if all seconds are the same then where do the concepts of proper time and coordinate time come from?

Grimble
edpell
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Jan1-10, 02:00 PM
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Quote Quote by Grimble View Post
Or are you saying that there is, somehow a single "universal" second, or somekind of "Absolute" second that is measured differently by observer's moving at different relative velocities?
I would say there is only the local time in the observers inertia frame.

Let's say we made ten clocks all the same and tested them while they were all in the same inertia frame. We find they all run at the same rate. We then send all ten into unique inertia frames. In each frame we do various low speed physics experiments we will find that physics is working as expected when we use our local copy of the clock. Each local clock, each local time is as expected in terms of physical/chemical/biological processes.

If we want to translate between frames we can. The amount of local clock time in frame A (call it Ta) will be different than the amount of local clock time in frame B (Tb).

All our experience is in a slow Newtonian world we have no intuitive understanding of the relativistic mix of time and place. We have to use the math. We can always translate time and place in frame X into time and place in frame Y. It just feels wrong.


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