Clocks and rulers

[swansont]

 

But there is NO frame in which there is no time dilation AND no length contraction- in that frame practically none of the muons should make it to the earth's surface - but they do! This magic preferred frame is both a physical impossibility and worthless in that it would not give the correct predictions.

 

 

Right. That's true as soon as there is relative motion.

[michel123456]

In SR, no, but in GR, yes. If you go higher in a potential well, clocks speed up.

So then, when clocks speed up, what about length? Does that suppose length increase (instead of length contraction)?

[Mordred]

you can honestly and accurately answer this question yourself. Just look at the two Lorentz formulas.

 

[latex]L=L_o\sqrt{\frac{v^2}{c^2}}=\frac {L_o}{\gamma}[/latex]

 

[latex]T=\frac{T_o}{1-\sqrt {\frac{v^2}{c^2}}}=T_o\gamma [/latex]

 

the longest length measured will be the rest frame. the time interval will be shortest in the rest frame. Which equates to fastest time.

Edited August 4, 2016 by Mordred

[swansont]

So then, when clocks speed up, what about length? Does that suppose length increase (instead of length contraction)?

 

 

Length is a bit less straightforward, since you are dealing with curved space. Someone better versed in GR should answer this.

[Mordred]

 

 

Length is a bit less straightforward, since you are dealing with curved space. Someone better versed in GR should answer this.

for now Im just answering in the SR limit without adding extra complexities. In order to give Michel a chance to understand the basic relationships

 

the formulas above assume a rest frame and no other influence. Which gets extremely complex in GR as you can have numerous geodesics.

 

Though clarity of which length is being examined will help. (length of object parallel to direction of motion.) or seperation length between two events.?

Edited August 4, 2016 by Mordred

[Markus Hanke]

So then, when clocks speed up, what about length? Does that suppose length increase (instead of length contraction)?

 

 

I think I should point out that we need to distinguish between kinematic effects in SR ( due to relative motion ), and gravitational effects in GR ( due to the metric not being constant ). They are best treated as distinct effects, and do not follow the same laws.

In SR, relative motion always leads to time dilation and length contraction, never to the opposite, because inertial frames are always symmetric.

In GR, there is generally no symmetry between frames, so while one observer may see clocks dilated and lengths contracted, another observer may disagree and say that clocks are sped up and lengths are stretched out. What's more, if you have both gravity and relative motion simultaneously, then these effects will overlap and combine ( but not linearly ), so it is not always easy or straightforward to figure out who measures what; oftentimes brute force is required, in the sense that one needs to put pen to paper and actually do the maths. Once again, this is best avoided from the outset by using covariant quantities.

[michel123456]

you can honestly and accurately answer this question yourself. Just look at the two Lorentz formulas.

 

[latex]L=L_o\sqrt{\frac{v^2}{c^2}}=\frac {L_o}{\gamma}[/latex]

 

[latex]T=\frac{T_o}{1-\sqrt {\frac{v^2}{c^2}}}=T_o\gamma [/latex]

 

the longest length measured will be the rest frame. the time interval will be shortest in the rest frame. Which equates to fastest time.

Oh. That is the answer in SR.

 

 

I think I should point out that we need to distinguish between kinematic effects in SR ( due to relative motion ), and gravitational effects in GR ( due to the metric not being constant ). They are best treated as distinct effects, and do not follow the same laws.

In SR, relative motion always leads to time dilation and length contraction, never to the opposite, because inertial frames are always symmetric.

The same answer. Fine, very clear, thank you.

 

 

In GR, there is generally no symmetry between frames, so while one observer may see clocks dilated and lengths contracted, another observer may disagree and say that clocks are sped up and lengths are stretched out. What's more, if you have both gravity and relative motion simultaneously, then these effects will overlap and combine ( but not linearly ), so it is not always easy or straightforward to figure out who measures what; oftentimes brute force is required, in the sense that one needs to put pen to paper and actually do the maths. Once again, this is best avoided from the outset by using covariant quantities.

Does that mean that all the interminable discussions about time dilated orbiting clocks* and other twins should be calculated before getting an answer?

 

*Because it should be obvious that when a clock orbiting the Earth is observed time dilated from us at rest, from the frame of the clock the Earth is time "contracted". It is not a symmetric situation. In a symmetric situation the Earth should be time dilated too, as observed from the clock.

Edited August 5, 2016 by michel123456

[Strange]

*Because it should be obvious that when a clock orbiting the Earth is observed time dilated from us at rest, from the frame of the clock the Earth is time "contracted".

 

 

As Markus says, it is not that simple. The combination of velocity and gravitational potential might lead to that result. Or might not. It depends.

[swansont]

 

 

As Markus says, it is not that simple. The combination of velocity and gravitational potential might lead to that result. Or might not. It depends.

 

 

Right. There is a circular orbit with no net time dilation. There is a class of orbits where the orbiting clocks run fast, and a class where they run slow, relative to earth.

[Mordred]

Oh. That is the answer in SR.

Correct

you can honestly and accurately answer this question yourself. Just look at the two Lorentz formulas.

 

[latex]L=L_o\sqrt{\frac{v^2}{c^2}}=\frac {L_o}{\gamma}[/latex]

 

[latex]T=\frac{T_o}{1-\sqrt {\frac{v^2}{c^2}}}=T_o\gamma [/latex]

 

the longest length measured will be the rest frame. the time interval will be shortest in the rest frame. Which equates to fastest time.

These two formulas only hold true when the metric is approximately Euclidean.

 

the line element in this case is

 

[latex]ds^2=dx^2+dy^2+dz^2 [/latex]

 

when the spacetime geometry is best approximated by that line element the first two equations work to good approximation.

 

Those equations won't be accurate in curved spacetime which in spherical polars the line element becomes.

 

[latex]ds^2=dr^2+r^2d\theta^2+r^2sin^2\theta d\phi[/latex]

 

(without expansion or rotation)

 

Which is why I find most claims we see in Speculations of "I solved blah blah in GR but they use nothing more than SR formulas Humorous.

Edited August 5, 2016 by Mordred

[Markus Hanke]

Does that mean that all the interminable discussions about time dilated orbiting clocks* and other twins should be calculated before getting an answer?

*Because it should be obvious that when a clock orbiting the Earth is observed time dilated from us at rest, from the frame of the clock the Earth is time "contracted". It is not a symmetric situation. In a symmetric situation the Earth should be time dilated too, as observed from the clock.

 

 

The case of a satellite orbiting a central body that is approximately spherical and has only a small amount of angular momentum, is very straightforward - even without explicit calculation you can easily guess what the outcome would be in general terms. Only if you need actual numbers ( e.g. the amount of clock adjustment necessary for GPS satellites ) would you have to get down and dirty with the maths. That being said, not all scenarios are this straightforward - in more complicated setups such as multi-body systems involving magnetic fields for example, the situation requires some serious calculations involving a lot of computing power to arrive at a conclusion.

 

 

 

*Because it should be obvious that when a clock orbiting the Earth is observed time dilated from us at rest, from the frame of the clock the Earth is time "contracted". It is not a symmetric situation. In a symmetric situation the Earth should be time dilated too, as observed from the clock.

 

Correct, but if you mean gravitational time dilation, then it is the other way around - the clock closer to the central body is dilated with respect to a more distant clock. But yes, the situation is not symmetric - if you swap frames, then the more distant clock is "sped up" relative to the close-by clock. For simple scenarios such as stationary spherically symmetric bodies, gravitational time dilation can be taken as a function of gravitational potential ( but remember that this does not generalise to more complex scenarios ).

[Mordred]

To give an example of the need to get dirty with the math as Markus and I both stress.

 

Lets take an example that relates. Most ppl know that the Schwartzchild metric has an event horizon. However in the Kerr metric you have 4 event horizons. The rotation of the BH has particular observer influences that do not occur in the static BH solution.

 

Simple metrics and verbal descriptions rarely describe every scenario.

Edited August 5, 2016 by Mordred

[michel123456]

 

 

The case of a satellite orbiting a central body that is approximately spherical and has only a small amount of angular momentum, is very straightforward - even without explicit calculation you can easily guess what the outcome would be in general terms. Only if you need actual numbers ( e.g. the amount of clock adjustment necessary for GPS satellites ) would you have to get down and dirty with the maths. That being said, not all scenarios are this straightforward - in more complicated setups such as multi-body systems involving magnetic fields for example, the situation requires some serious calculations involving a lot of computing power to arrive at a conclusion.

 

 

Correct, but if you mean gravitational time dilation, then it is the other way around - the clock closer to the central body is dilated with respect to a more distant clock. But yes, the situation is not symmetric - if you swap frames, then the more distant clock is "sped up" relative to the close-by clock. For simple scenarios such as stationary spherically symmetric bodies, gravitational time dilation can be taken as a function of gravitational potential ( but remember that this does not generalise to more complex scenarios ).

The 2 effects do not cancel each other exactly?

[Markus Hanke]

The 2 effects do not cancel each other exactly?

 

 

Which two effects do you mean ?

[michel123456]

 

 

Which two effects do you mean ?

Yes. Both effects at the same time.

[Janus]

The 2 effects do not cancel each other exactly?

According to who? For example, for a stationary observer some distance from a gravity source, there will be an orbit( at 1 1/2 times the distance from the center of the gravity source than he is) where, by his measurement, the orbiting clock runs at the same rate as his own. A lower orbit would result with the clock running slower than his and a higher one results in the clock running faster. So in specific instances you can have a scenario where the two time dilations "cancel out" for a particular frame, but this is not generally the case.

[michel123456]

For an observer on Earth looking at a clock orbiting the Earth.

According to who? For example, for a stationary observer some distance from a gravity source, there will be an orbit( at 1 1/2 times the distance from the center of the gravity source than he is) where, by his measurement, the orbiting clock runs at the same rate as his own. A lower orbit would result with the clock running slower than his and a higher one results in the clock running faster. So in specific instances you can have a scenario where the two time dilations "cancel out" for a particular frame, but this is not generally the case.

So.

An observer on Earth is roughly 12700 km away from the gravity source (the center of the Earth). An orbit 1 1/2 away is 6350km above. From what you say, a clock orbiting at this distance runs at the same time as his own. A lower orbit would result with the clock running slower than his. It means when my clock says 5 minutes, the orbiting clock says 4 (exaggerated for the sake of simplicity).

Do I understand correctly?

Edited August 6, 2016 by michel123456

[robinpike]

You need to think of what is meant by synchronized and clock rate.

 

The clock will no longer be synchronized. However both length and clock rate will return to match the original conditions.

 

Albiet the clock wont be synchronized.

 

Mordred, when you say 'both length and clock will return to match the original conditions', can you elaborate a bit more please as to what, if anything, happens to the ruler?

 

Is a change to the ruler's length just an apparent change - not an actual physical change?

 

I ask, because if the ruler does get physically shorter on its journey away from our frame of reference - by what action allows it to return to 'our synchronized original length'?

[swansont]

 

Mordred, when you say 'both length and clock will return to match the original conditions', can you elaborate a bit more please as to what, if anything, happens to the ruler?

 

Is a change to the ruler's length just an apparent change - not an actual physical change?

 

I ask, because if the ruler does get physically shorter on its journey away from our frame of reference - by what action allows it to return to 'our synchronized original length'?

 

 

 

Nothing physical changes. That implies it goes from one state to another, and it doesn't. The frame changes, and the length depends on the frame.

[robinpike]

 

 

 

Nothing physical changes. That implies it goes from one state to another, and it doesn't. The frame changes, and the length depends on the frame.

 

So just to be clear, the physical length of a ruler does not change when it moves from one frame of reference to another, it is the ability to measure that length of the ruler that changes?

[swansont]

 

So just to be clear, the physical length of a ruler does not change when it moves from one frame of reference to another, it is the ability to measure that length of the ruler that changes?

 

 

The ruler can only be in its own frame. It never moves to a different frame. Observers in other frames measure it as having a different length.

[J.C.MacSwell]

 

 

The ruler can only be in its own frame. It never moves to a different frame. Observers in other frames measure it as having a different length.

I don't understand this. Does a (moving) ruler not exist in frames other than it's own rest frame? It is moving wrt the other frames. I would not think of it as "not in" them. If that is in fact the definition in physics, what advantage is there to thinking of frames of reference in this manner? It implies (to me) like something is not there.

[Delta1212]

I don't understand this. Does a (moving) ruler not exist in frames other than it's own rest frame? It is moving wrt the other frames. I would not think of it as "not in" them. If that is in fact the definition in physics, what advantage is there to thinking of frames of reference in this manner? It implies (to me) like something is not there.

I guess it depends on what the meaning of 'be' is.

[swansont]

I don't understand this. Does a (moving) ruler not exist in frames other than it's own rest frame? It is moving wrt the other frames. I would not think of it as "not in" them. If that is in fact the definition in physics, what advantage is there to thinking of frames of reference in this manner? It implies (to me) like something is not there.

Frames are defined by relative motion. Any object is "in" the frame where it is at rest. The ruler can't be moving with respect to itself.

 

Remember the context of this - questioning if the ruler physically changes.

[StringJunky]

Frames are defined by relative motion. Any object is "in" the frame where it is at rest. The ruler can't be moving with respect to itself.

 

Remember the context of this - questioning if the ruler physically changes.

As an analogy, would you say each frame represents a lens with different perspective views but all give representations that are equally valid?

Edited August 8, 2016 by StringJunky