## Special Relativity Clocks

 Quote by DaleSpam If the clocks are stationary in the moving frame then their worldline is x=0.8ct, not x=0.5ct.
Dale-- The clocks are moving at 0.8cT, but X is not an indicator of the position of the clocks, its the independent space variable of the stationary frame. There is no dedicated symbol indicating the position of the moving frame.
JM

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 Quote by JM ... some values of X lead to t ≥ T,
This is correct if we consider time intervals measured by spatially separated clocks rather than time intervals measured by a single clock.

The statement 'moving clocks always run slow' applies to single clocks and not to time intervals calculated from multiple clocks far apart from each other.

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 Quote by JM Dale-- The clocks are moving at 0.8cT, but X is not an indicator of the position of the clocks, its the independent space variable of the stationary frame. There is no dedicated symbol indicating the position of the moving frame. JM
That is not helpful or meaningful. Please clarify what you were trying to say in statement (4) "For example let x=0.5ct (v/c = 0.8) in the equation to find τ =t."

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[QUOTE=JM;3866481]
 Quote by DaleSpam No, I don't accept the idea. If you have a clock which is moving in an arbitrary fashion (including, but not limited to, x=vt) you use the following formula to calculate the time displayed on the clock: $$\tau = \int \sqrt{1-v(t)^2/c^2} dt$$ QUOTE] Dale--I see where this formula comes from, and I think it is a valid result. The question is whether it is a 'universal' result, ie 'moving clocks always run slow, meaning t (moving)
I think you are confusing some concept here. Proper time as defined in the integral is not a coordinate at all. It gives time elapsed on a single clock following some spacetime path between two specific events. Two different frames may give different labels to all the events on the clocks path, but the computed proper time will come out the same (as will the time elapsed on an actual single clock between two physically defined events).

Within any one inertial frame in SR, this integral demonstrates that any moving clock, no matter its path, will seem to run slower than any stationary clock of that frame. The same will be true in every other inertial frame. The seeming discrepancy gets resolved by the frames differing on clock synchronization. Each thinks the other's clocks at different positions are out of synch.

 Quote by ghwellsjr It is justified because he takes a clock from the stationary frame located at the origin (t=0 and x=0) and puts it at the origin of the moving frame (τ=0 and ζ=0) and the origin of the moving frame is defined as x=vt according to the stationary frame.
I sense a confusion between x-the independent variable in the Lorentz Transforms, and x-indicating the position of the moving clock. Do you agree that x in the LT can assume any of a large range of values? Even some values that result in t-moving > t-stationary?
If x now indicates the position of the moving origin, isnt that different from the LT? Arent we entitled to an acknowledgement of this change, and an explanation of how the new meaning relates to the old, since the same transform is used in both?
JM

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 Quote by JM Lets use x,y,z,t = moving coordinates and X,Y,Z,T =stationary coordinates.
Better still, use $\Delta x,\Delta y, \Delta z, \Delta t$ = moving frame measurements and $\Delta X,\Delta Y, \Delta Z, \Delta T$ = stationary frame coordinates, then if $\Delta x = \Delta y = \Delta z = 0$ then t will always be less than T and represents a single clock that is at rest in the moving frame and this is called the proper time and uses the symbol tau.

Quote by ghwellsjr
So this tells us that it is not all the clocks in the moving frame that the time co-ordinate applies to but only the one at the spatial origin of the moving frame which is where the moving clock that we are considering is located.
 Again the confusion between x-variable and x-indicator of origin. Hasn't Einstein specified that all the clocks of a given frame are synch-ed using exchange of light signals? If this applies here then all the moving clocks read the same value. This doesn't add much for this case, but is important for other cases, eg Point 4
Concerning 4, I can't tell what you are doing, can you provide more detailed steps?
.
The point here was to show an example of a calculation using x-variable that resulted in t-moving not less than t-stationary. The procedure is to sub the values given into the LT for time.
JM

 Quote by PAllen As shown in Einstein's 1905 paper, every observer perceives every clock (moving or not), to go either the same rate as theirs (if not moving relative to said observer), or slower than said observer's clock.
Thats what I'm questioning, particularly the universality of his result. My questions and the replies are noted. Your input is appreciated.
JM

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 Quote by JM I sense a confusion between x-the independent variable in the Lorentz Transforms, and x-indicating the position of the moving clock.
There are not two definitions of x. X always indicates the position of a particular event.
 Quote by JM Do you agree that x in the LT can assume any of a large range of values?
Yes.
 Quote by JM Even some values that result in t-moving > t-stationary?
Yes.
 Quote by JM If x now indicates the position of the moving origin,
x can represent the position of a clock or of the origin or of any other object or event. For example the origin is at x=0, the location of a clock might be x = -5 and the location of Fred might be x = 9 and the location where John crashed his car at t=7 is x=10. You just have to make it clear what you are measuring. For a single clock at rest in frame S' moving relative to S the value of x changes over time so that for example at t=0, x=0 and at t=1, x=0.8 and at t=2, x=1.6 and so on. In the moving frame spatial measurements are made relative to the origin of S' so if the clock is stationary in S', then at t=0, x'=0 and at t=1, x'=0 and at t=2, x'=0.

x and x' are just positions as measured in S and S' respectively. If we mean changes in position over a time interval then we should use $\Delta x$ and $\Delta x'$ respectively, or for brevity, just use x and x' and make it clear we mean spatial separations rather than spatial locations.

 Quote by DaleSpam OK, so x=0.5ct is the worldline of a clock which is moving at 0.5c in the +x direction in the stationary frame. Boosting by 0.8c gives you a clock which is moving at 0.5c in the -x direction in the moving frame. So yes, you have correctly determined that a clock which is moving at .5c in the +x direction in the stationary frame is slowed by the same amount as a clock which is moving at .5c in the -x direction in the moving frame.
I sense a change of model here, from LT where all clocks move only in the + x direction, to world lines where clocks can move in other directions. My intention is to stay within the 1905 model, and use x as an independent variable of the stationary frame. If t-moving < t-stationary, as in section 4,indicates a slow clock , then t-moving = t-stationary, as above, indicates that moving clocks are not always slow.

I appreciate the comments of all.
JM

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 Quote by JM The point here was to show an example of a calculation using x-variable that resulted in t-moving not less than t-stationary. The procedure is to sub the values given into the LT for time. JM
OK you have given that x = 0.5 and v = 0.8 and given the Lorentz transform:

$$t ' = \frac{t-vx}{\sqrt{1-v^2}}$$

we get:

$$t ' = \frac{t-0.8*0.5}{\sqrt{1-0.8^2}} = \frac{t - 0.4}{0.6}$$

then for any value of t>1 we get t'>t.

However t and t' as used here are just coordinates or labels for an event and are not a comparison of clock rates where we have to compare intervals between events. If we mean intervals we should use:

$$\Delta t' = \frac{\Delta t-v \Delta x}{\sqrt{1-v^2}}$$

Now if v=0.8c and $\Delta x$ =0.5 then $\Delta t$ must be 0.5/0.8 = 0.625

so referring to the conclusion above it is obvious in this case that t' <t.

When we talk about intervals (deltas) then $\Delta t$ and $\Delta x$ are not independent of each other, if we are talking about clocks at rest in S' because they are related by v.

Furthermore, if we use the reverse transformation:

$$\Delta t = \frac{\Delta t' + v \Delta x'}{\sqrt{1-v^2}}$$

and note that if the clock is at rest in S', then $\Delta x' = 0$ and we obtain:

$$\Delta t = \frac{\Delta t' }{\sqrt{1-v^2}}$$

then $\Delta t$ is always greater than $\Delta t'$ for all values of v<1 where c=1.

I am sure most of the confusion is because you are not clear on whether you mean coordinate labels or space and time intervals.

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 Quote by JM The clocks are moving at 0.8cT, but X is not an indicator of the position of the clocks, its the independent space variable of the stationary frame.
Then you cannot use x=0.5 ct to describe any of those clocks.

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 Quote by yuiop Still not seeing it. X is just a coordinate. If he said the clock started at the origin at (t,x) = (0,0) and specified a time duration of delta T = 1 then yes I would expect the clock to be at coordinates (1.0,0.8) but he did not specify a time duration. X depends on t.
Sure, but the equation for clocks that don't start at the origin is $x=0.8ct+x_0$. The problem is the velocity. With x=0.5ct you have a clock which is moving in both frames, not stationary in either one of them (and being stationary was specified in part 1).

Mentor
 Quote by JM Dale--I see where this formula comes from, and I think it is a valid result. The question is whether it is a 'universal' result, ie 'moving clocks always run slow, meaning t (moving)
It is not universal. It applies for inertial frames in flat spacetime only. The universal formula is:
$$\tau=\int \sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}$$

However the scenario you have described here uses only inertial frames in flat spacetime so the simplified version applies.

 Quote by JM I am hoping to get a clear answer from this discussion.
I hope my answer has been clear.

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 Quote by DaleSpam Sure, but the equation for clocks that don't start at the origin is $x=0.8ct+x_0$. The problem is the velocity. With x=0.5ct you have a clock which is moving in both frames, not stationary in either one of them (and being stationary was specified in part 1).
Ah OK. I concede your point now. Thanks. If x =0.5ct then the clock is moving at 0.5c in S and not stationary in S' which is moving at at 0.8c relative to S. I think I misread what JM intended, but I not convinced that JM is sure what he intended either. I think he needs to clear that up.

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 Quote by JM I sense a change of model here, from LT where all clocks move only in the + x direction
I don't know where you got that idea.

 Quote by JM My intention is to stay within the 1905 model
In the 1905 model he analyzed a clock which goes in a circle. Such a clock goes in the +x and +y and -x and -y directions at some point and every combination inbetween. A restriction to clocks moving in the +x direction is not a part of the 1905 model, and indeed is incompatible with the Lorentz transform for boosts to arbitrary speeds.

 Quote by JM If t-moving < t-stationary, as in section 4,indicates a slow clock , then t-moving = t-stationary, as above, indicates that moving clocks are not always slow.
Moving clocks are always slow. Your analysis above contradicts itself as I mentioned above.

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Quote by JM
 Quote by ghwellsjr It is justified because he takes a clock from the stationary frame located at the origin (t=0 and x=0) and puts it at the origin of the moving frame (τ=0 and ξ=0) and the origin of the moving frame is defined as x=vt according to the stationary frame.
I sense a confusion between x-the independent variable in the Lorentz Transforms, and x-indicating the position of the moving clock. Do you agree that x in the LT can assume any of a large range of values?
Yes, of course, x can be any value and the transform will work. But for every x, you also have y, z and t and you have to calculate all of them for a complete transformation.

The Lorentz Transform converts the four co-ordinates of an event defined according to one Frame of Reference into the four co-ordinates of the same event defined according to a second Frame of Reference moving at some speed v in the x direction with respect to the first FoR. Using Einstein's nomenclature from section 3 of his 1905 paper, the first FoR has co-ordinates with labels of t, x, y and z, while the second FoR has co-ordinates with labels of τ, ξ, η and ζ. You have to solve all four equations to get the co-ordinates in the second frame. You can't just solve for the time co-ordinate and ignore the spatial co-ordinates.
 Quote by JM Even some values that result in t-moving > t-stationary?
In Einstein's nomenclature, you are asking if τ can be greater than t. Of course, there are many events in the first FoR with a t co-ordinate less than the τ co-ordinate in the second FoR. But in general that has nothing to do with a clock moving in a stationary frame. The only time you can use the Lorentz Transform to calculate the time on a clock moving in the stationary frame is when a clock at the origin of the second FoR moves at the same velocity that the second FoR is moving and this will be indicated by the spatial co-ordinates remaining zero in the second FoR while the time co-ordinate is changing.
 Quote by JM If x now indicates the position of the moving origin, isnt that different from the LT? Arent we entitled to an acknowledgement of this change, and an explanation of how the new meaning relates to the old, since the same transform is used in both? JM
The origin of the second FoR is moving along the x-axis of the first FoR at a velocity of v so for any time t in the first FoR, we can calculate the x co-ordinate in the FoR for the origin of the second FoR by using x=vt. This gives us the t and x co-ordinates of an event in the first FoR (y and z are always 0). Then we can plug both the x and t values into the LT and calculate the co-ordinates of the same event in the second FoR and we will find that the time co-ordinate will always be less in the second FoR and the location co-ordinates will be 0.

In fact the time co-ordinate, τ, in the second FoR will be t√(1-v2/c2), making τ always less than t.