Looking to understand time dilation

[ghwellsjr]

...
So I resolved to look at one of the most common starting points: the moving light clock with which I am sure you are all familiar.
...
Also I was concerned about length contraction (getting smaller) and time dilation (getting bigger) for a moving object when speed = distance / time.
...

It's easy to misunderstand how length contraction combined with time dilation would result in the same speed (for light) being measured for a moving clock/observer as compared to a stationary one. Let me explain:

You seem to understand how a light clock works. In a rest frame where the speed of light is the same in all directions, it is easy to understand that if we place two mirrors a fixed distance apart in any orientation, the light will take the same time to make a round trip back and forth between the mirrors, no matter what the orientation of the two mirrors is.

Now we consider a light clock in motion. First we position the two mirrors so the light reflects back and forth at right angles to the direction of motion. You probably understand that the light takes longer to traverse the path between the two mirrors because it has to take diagonal paths and you can probably figure out that the total round-trip path is gamma multiplied by the previous path length (when the light clock was at rest) multiplied by 2 and that this demonstrates time dilation.

But what happens if we rotate the light clock 90 degrees so that the light is going back and forth along the direction of the motion of the mirrors? Well now if you follow the details (which I'm sure you can), you will need to move the mirrors closer together in order for the light to make the round trip in the same time as it did before we rotated the light clock. This demonstrates length contraction which is the original distance divided by gamma.

Now if we want to calculate the speed of light as demonstrated by this light clock, we might naively say that the speed is equal to the distance divided by the time and since the distance was the orginal distance divided by gamma and the time was the original time multiplied by gamma we should get the speed as distance divided by time divided by the square of gamma. This is what you calculated in a previous post, but can you see how this is wrong?

There are two ways to demonstrate that it is wrong. If we look at the first orientation of the light clock where length contraction is not involved, we use the actual distance that the light traveled along the diagonals, not the distance between the mirrors, correct? In other words, we make mental note of where the mirrors were when the light struck them and we use the actual distance traversed by the light when we calculate the total distance. This distance is gamma multiplied by the original distance multiplied by 2 (for the round trip).

And if we look at the second orientation of the light clock where length contraction is involved, we have to do the same thing as before: make mental note of where the mirrors were when the light struck them and again use the actual distance traversed by the light to calculate the total round trip distance. This distance is again gamma multiplied (not divided) by the original distance multiplied by 2.

If you actually do this exercise you will also see that in the last orientation, the two halves of the round trip are not equal in length--when the light is going in the same direction as the mirrors are moving it is longer, when the light is going in the opposite direction as the mirrors are moving it is shorter, and both these distances are different than the two halves of the distances in the first orientation. And in the same way we can say that the time intervals involved for light to traverse each path between the mirrors is different. It is this difference that is the reason for the relativity of simultaneity.

So you can see that even though we talk about length contraction for a moving object, for the light path, it is actually a length stretching because the light is striking the mirrors at different times between which the mirrors are moving to new locations. The Lorentz Transform takes care of the correct calculations.

 

[Grimble]

To be clear, are you saying the clock is at rest in the "stationary frame" or the "travelling frame"? Einstein said the clock is permanently situated at the origin (x' = 0) of K' so I guess since you use a primed t' to denote the "travelling frame" you're saying the clock is at rest in the traveling frame, correct? In that case, if t' is the time between two ticks of the clock in the traveling frame, while t is the time between the same two ticks in the stationary frame, then t should be larger than t', meaning you're correct to say t = γt'. Nowadays the more common convention with primed vs. unprimed time intervals in the time dilation equation is to say unprimed is the time in the clock's rest frame while primed is the time in the frame where the clock is moving at speed v, the reverse of your (and Einstein's) notation, so probably that's why ghwellsjr thought you got it wrong.

Yes absolutely! Thank you Jesse.

Let me refer back to my post 263 where I wrote

This took me into another little puzzle that had been niggling away at me. If in the rest frame c = x/t, then we should also find that c = x'/t' …
But x = x'/γ, while t = γt'
which gives us c = x/t = x'γ²/t'

Also I was concerned about length contraction (getting smaller) and time dilation (getting bigger) for a moving object when speed = distance / time.

In order to address these concerns I turned to Chapter XII: the Behaviour of Measuring-Rods and Clocks in Motion.

Examining this all becomes clear.

Einstein places a metre rod in the moving frame and asks what size will it be as observed from the stationary frame. He is comparing the unit size between observers.
Whereas for the clock he takes the the time between two ticks (one second) in the resting frame and then asks how many seconds it takes in the moving frame. So here he is comparing, not the unit size but the number of units between observers.

And if the ratio of lengths is x/x' = 1/γ then that is also the ratio of unit sizes
while the ratio of times is t/t' = γ then that is the ratio of the number of units.
Which gives us the unit size * the number of units = γ/γ = 1.

In that post I was trying to explain where I was coming from and the logic behind some confusing and apparently conflicting aspects of SR, as I see it

And yes, if we change the time measurement to unit size instead of number of units, then the time formula becomes t = t'/γ and the speed of light become x'/t'.

As you can see I am not trying to be difficult.

Grimble

 

[Grimble]

Jesse, this is what Grimble is claiming (I've corrected his typo):

And if you read that passage in full you will see that I raise that as one of the puzzles that were niggling me and how it occurs and how to resolve it.

It is an APPARENT discrepancy that disappears upon investigation.

Stop taking it out of context. PLEASE:tongue:

Grimble

 

[Grimble]

Hi Grimble, you haven't responded to my repeated suggestions to avoid these formulas (they are never necessary and often problematic) and stick to the Lorentz transform only. The reason you are getting a bad answer is because you are using formulas that do not apply. The time dilation formula applies when the clock is at rest in one of the frames, which is never the case for light. The length contraction formula only applies when you have a pair of events which are simultaneous in each frame and associated with the ends of a single object, which is also never the case for a single pulse of light.

Now, with that motivation, please try to re-do your line of reasoning using the Lorentz transform.

Once again, please read what I wrote in post 263. Not one part of it but all of it. I was explaining how I had arrived at my understanding.

I was saying that there was potential confusion with those formulae and why and how to understand it.

 

[Grimble]

Correct so far.

In the rest frame (S) of the light clock, the time taken is t = 1 s and the distance traveled by the photon is d = 1 ls.

In the frame in which the light clock is moving (S') the distance traveled by the photon is d' = γ ls and the time taken according to a clock moving relative to the light clock is t' = γ s.

This is where the confusion sets in. You say "both take the same time (relativity)" without specify according to what clocks or observers. If you are not always careful to specify the observer that makes the measurement in relativity you are doomed to be perpetually confused.

As I was discussing the longer path and it is only longer to the moving observer I think it reasonable to assume that that is the observer concerned.

Relativity states that the speed of light is always c in any inertial reference frame so you can rule out the first option. Since you have already stated that the photon takes one second in the rest frame of the light clock it can be be reasonably assumed you are talking about frame S' when you say "while on the contrary if it travels at 'c' it must take longer than 1 second to hit the mirror and return". This is correct. It DOES take longer than 1 second in frame S' for the photon to complete its round trip. In frame S' it takes t' = γ s > (1 s).

Which is what I am saying ...

x = x'/γ implies x' = xγ which is correct.
t' = γt' is obviously not correct. I assume you meant t' = tγ.
(Remember you said earlier " if it travels at 'c' it must take longer than 1 second to hit the mirror and return") .. so t' must be greater than t.

Again please read this passage in context...
[/QUOTE]

 

[Grimble]

OK, the problem here is that x = x'/γ and t = γt' don't work for any arbitrary pair of events, the time dilation formula t = γt' is only valid when you are talking about the time between a pair of events that both happened at the same position in the primed frame (like ticks of a clock at rest in the primed frame), while the length contraction formula is x = x'/γ only valid when you are talking about the length in both frames of an object at rest in the primed frame, or equivalently the distance in both frames between a pair of events simultaneous in the unprimed frame (if the events were on either end of an object at rest in the primed frame, then since they are simultaneous in the unprimed frame the distance between these events is the object's length in the unprimed frame, whereas even though they are non-simultaneous in the primed frame, since the object is at rest in the primed frame the distance between them still counts as the object's 'length' in the primed frame).

Yes exactly, that is what I am saying - one is measuring the size of the units and the other is counting the number of units.

If you are talking about two events on the worldline of a light beam such that c=x/t and c=x'/t', neither of these conditions would be satisfied so you can't use the length contraction and time dilation formulas. As DaleSpam says, you should really use the general Lorentz contraction formulas to avoid this sort of confusion:

x' = γ(x - vt)
t' = γ(t - vx/c^2)

Let me just point out here that in appendix 1, where Einstein shows a simple derivation of the Lorentz Transformations, in step 6 he writes viz

For the origin of k' we have permanently x' = 0, and hence according to the first of the equations (5)

If we call v the velocity with which the origin of k' is moving relative to K, we then have

i.e. substituting v with x/t

If we follow that substitution by replacing x with vt this gives us t' = γ(t - v²t/c²) or t' = γt(1 - v²/c²) = t/γ for t' = γ(t - vx/c^2)

And

x = γ(x' + vt')
t = γ(t' + vx'/c^2)

You can see that in the special case where two events happened at the same position in the primed frame so x'=0, the equation t = γ(t' + vx'/c^2) reduces to the time dilation equation t = γt'. Likewise in the special case where the two events were simultaneous in the unprimed frame so t=0, the equation x' = γ(x - vt) reduces to x = x'/γ. But again, it's not valid to use the time dilation and length contraction equations in cases where the events you're considering don't satisfy the required conditions, whereas it's always valid to use the more general Lorentz transformation equations.

 

[Dale]

And yes, if we change the time measurement to unit size instead of number of units, then the time formula becomes t = t'/³ and the speed of light become x'/t'.

As you can see I am not trying to be difficult.

No. The formulas DO NOT APPLY here. It is not simply a matter of switching unit size and number of units.

If you want to use those formulas (which I strongly recommend against) then you need to learn their domain of applicability.

 

[nismaratwork]

Grimble, you are of course free to argue this point until the end of time (har har), but if you actually TRY to take DaleSpam's advice and listen to JesseM and Yuiop, you'll learn something. I know I usually do when I that route.

 

[ghwellsjr]

Let me refer back to my post 263 where I wrote

This took me into another little puzzle that had been niggling away at me. If in the rest frame c = x/t, then we should also find that c = x'/t' …
But x = x'/γ, while t = γt'
which gives us c = x/t = x'γ²/t'

Also I was concerned about length contraction (getting smaller) and time dilation (getting bigger) for a moving object when speed = distance / time.

In order to address these concerns I turned to Chapter XII: the Behaviour of Measuring-Rods and Clocks in Motion.

Examining this all becomes clear.

Einstein places a metre rod in the moving frame and asks what size will it be as observed from the stationary frame. He is comparing the unit size between observers.
Whereas for the clock he takes the the time between two ticks (one second) in the resting frame and then asks how many seconds it takes in the moving frame. So here he is comparing, not the unit size but the number of units between observers.

And if the ratio of lengths is x/x' = 1/γ then that is also the ratio of unit sizes
while the ratio of times is t/t' = γ then that is the ratio of the number of units.
Which gives us the unit size * the number of units = γ/γ = 1.

In that post I was trying to explain where I was coming from and the logic behind some confusing and apparently conflicting aspects of SR, as I see it

And yes, if we change the time measurement to unit size instead of number of units, then the time formula becomes t = t'/γ and the speed of light become x'/t'.

As you can see I am not trying to be difficult.

Grimble

So you started with t = γt' and you ended up with t = t'/γ? They're both true? Something to do with the first one uses number of units and the second one uses unit size? Can you show us where exactly in your referenced text Einstein explains this?

 

[JesseM]

Yes exactly, that is what I am saying - one is measuring the size of the units and the other is counting the number of units.

I have no idea what that means, both can be understood as the difference in coordinate position or coordinate time between some specific pair of events. Do you agree that the time dilation formula you wrote down is only applicable when you want to compare the time-interval in both frames between two events that happened at the same position in the primed frame, and that the length contraction formula you wrote down is only applicable when we're talking about the length in both frames of an object at rest in the primed frame, or equivalently when we're talking about the distance-interval in both frames between two events which are simultaneous in the unprimed frame? If you agree with that, then do you agree that we can not use these formulas when we want to know the separation in both frames between a pair of events which lie on the worldline of a light beam, i.e. a pair of events such that x/t = c (where x and t stand for the coordinate separation between the two events)?

Let me just point out here that in appendix 1, where Einstein shows a simple derivation of the Lorentz Transformations, in step 6 he writes viz

For the origin of k' we have permanently x' = 0, and hence according to the first of the equations (5)

If we call v the velocity with which the origin of k' is moving relative to K, we then have

i.e. substituting v with x/t

If we follow that substitution by replacing x with vt this gives us t' = γ(t - v²t/c²) or t' = γt(1 - v²/c²) = t/γ for t' = γ(t - vx/c^2)

No, that conclusion isn't valid in general, although it may be valid for some specific pairs of events. When reading Einstein's derivation in Appendix 1, you have to keep track of which equations are supposed to apply generally to the coordinates of arbitrary events like the Lorentz transformation, and which equations are only supposed to hold for events on a specific worldline, like the worldline of the light-signal which he introduces at the start of the derivation, or the worldline of the origin of k' for which that first equation x=(bc/a)*t above is supposed to hold. Note that at the start of the derivation he introduces the equation x=ct, hopefully it's obvious that this holds for events on the worldline of the light-signal sent from the origin at t=0, but that it is not a general equation that applies to the x,t coordinates of any arbitrary event in spacetime! You may want to take a look at this thread about Einstein's derivation in Appendix 1, where I wrote:

The confusing part may be that he wants a general transformation which can translate any event with coordinates x and t to corresponding coordinates x' and t', but he starts by considering the special case of a light beam emitted from the origin at t=0, so x=ct for any point on the light beam's path (this would obviously not be true for arbitrary events that don't lie on this path). He's pointing out that as long as the general transformation has the property (x'-ct') = λ(x-ct) (regardless of the value of x and t), that will guarantee that the light beam has the same speed of c in both coordinate systems , because if x=ct, that equation implies x'=ct' too. I don't know if it'd be possible to come up with a coordinate transformation where it was true that any (x,t) satisfying x=ct would also satisfy x'=ct', but it wasn't true that any arbitrary x,t would satisfy (x'-ct') = λ(x-ct). I guess you could come up with a coordinate transformation that did satisfy that equation but where λ was a function of x and t rather than being a constant, but then it wouldn't be a linear coordinate transformation...

Anyway, the specific step you mention, going from x=(bc/a)*t to v=(bc/a), only works because we are talking about the position as a function of time for a specific entity, namely the origin of k'. If you are talking about the position and time intervals between an arbitrary pair of events it doesn't work, because that "v" symbol specifically refers to the velocity of the k' frame relative to the K frame (note the line "If we call v the velocity with which the origin of k' is moving relative to K"), so for a pair of events on the worldline of an object not at rest in k' it wouldn't be true that x/t = v (of course if you have an arbitrary pair of events on the worldline of some object not at rest in k', you could use a different symbol like v₂ to refer to the object's velocity in the K frame, and then it would be true that x/t = v₂ for the pair of events on its worldline, but you can't use the same symbol v to denote both the velocity of the object and the velocity of the k' frame relative to K). If you are specifically comparing a pair of events on the worldline of an object at rest in k' (i.e. two events that have the same position-coordinate in k', so x'=0), then it would be true that x=vt and for those specific events your derivation t' = γ(t - v2t/c2) or t' = γt(1 - v2/c2) = t/γ would be valid, but of course I already specified that the time dilation equation t = γt' (which is just a rearrangement of t' = t/γ which you got there) only holds in the case where you're looking at two events which occurred at the same position in the t' frame.

 

[JesseM]

Einstein places a metre rod in the moving frame and asks what size will it be as observed from the stationary frame. He is comparing the unit size between observers.
Whereas for the clock he takes the the time between two ticks (one second) in the resting frame and then asks how many seconds it takes in the moving frame. So here he is comparing, not the unit size but the number of units between observers.

They are both "number of units". Suppose we have little lights on both the front and back end of an object which is moving relative to the K frame, and both lights flash simultaneously in the K frame. In this case, the distance between the positions of each flash qualifies as the "length" of the object in the K frame (since 'length' is always the distance between the positions of the front and back of an object at a single moment), so you're counting the number of distance-units between the flashes using a ruler at rest in the K frame and calling that the length x of the moving object in the K frame. Meanwhile the object is at rest in the k' frame, so although the flashes at either end of the object happen at different times in this frame (relativity of simultaneity), the distance between them in the k' frame is exactly the same as the distance you'd get if you set off two flashes on either end of the object which were simultaneous in the k' frame. Either way, you're counting the number of distance-units between the flashes using a ruler at rest in the k' frame and calling that the length x' of the object in the k' frame. Then you will find that x = x'/γ, so you can see that the length contraction equation is still talking about the "number of units" between a pair of events (the two flashes) as measured in each frame.

 

[Grimble]

So you started with t = γt' and you ended up with t = t'/γ? They're both true? Something to do with the first one uses number of units and the second one uses unit size? Can you show us where exactly in your referenced text Einstein explains this?

In chapter XII as I said back in post 263.

It therefore follows that the length of a rigid metre-rod moving in the direction of its length with a velocity v is

of a metre.

and

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

seconds, i.e. a somewhat larger time.

 

[Grimble]

I have no idea what that means, both can be understood as the difference in coordinate position or coordinate time between some specific pair of events. Do you agree that the time dilation formula you wrote down is only applicable when you want to compare the time-interval in both frames between two events that happened at the same position in the primed frame, and that the length contraction formula you wrote down is only applicable when we're talking about the length in both frames of an object at rest in the primed frame, or equivalently when we're talking about the distance-interval in both frames between two events which are simultaneous in the unprimed frame? If you agree with that, then do you agree that we can not use these formulas when we want to know the separation in both frames between a pair of events which lie on the worldline of a light beam, i.e. a pair of events such that x/t = c (where x and t stand for the coordinate separation between the two events)?

As I said (in that very quote) yes exactly.
All I am saying there in addition is that when Einstein says that the length contracts and the time gets larger it is because he is considering different aspects of the measurements. For the time he is measuring how the length is less and the number of seconds is greater. One may see this as when we calculate dime dilation we again use the size of the unit and t = t'/γ

No, that conclusion isn't valid in general, although it may be valid for some specific pairs of events. When reading Einstein's derivation in Appendix 1, you have to keep track of which equations are supposed to apply generally to the coordinates of arbitrary events like the Lorentz transformation, and which equations are only supposed to hold for events on a specific worldline, like the worldline of the light-signal which he introduces at the start of the derivation, or the worldline of the origin of k' for which that first equation x=(bc/a)*t above is supposed to hold. Note that at the start of the derivation he introduces the equation x=ct, hopefully it's obvious that this holds for events on the worldline of the light-signal sent from the origin at t=0, but that it is not a general equation that applies to the x,t coordinates of any arbitrary event in spacetime! You may want to take a look at this thread about Einstein's derivation in Appendix 1, where I wrote:

Anyway, the specific step you mention, going from x=(bc/a)*t to v=(bc/a), only works because we are talking about the position as a function of time for a specific entity, namely the origin of k'. If you are talking about the position and time intervals between an arbitrary pair of events it doesn't work, because that "v" symbol specifically refers to the velocity of the k' frame relative to the K frame (note the line "If we call v the velocity with which the origin of k' is moving relative to K"), so for a pair of events on the worldline of an object not at rest in k' it wouldn't be true that x/t = v (of course if you have an arbitrary pair of events on the worldline of some object not at rest in k', you could use a different symbol like v₂ to refer to the object's velocity in the K frame, and then it would be true that x/t = v₂ for the pair of events on its worldline, but you can't use the same symbol v to denote both the velocity of the object and the velocity of the k' frame relative to K). If you are specifically comparing a pair of events on the worldline of an object at rest in k' (i.e. two events that have the same position-coordinate in k', so x'=0), then it would be true that x=vt and for those specific events your derivation t' = γ(t - v2t/c2) or t' = γt(1 - v2/c2) = t/γ would be valid, but of course I already specified that the time dilation equation t = γt' (which is just a rearrangement of t' = t/γ which you got there) only holds in the case where you're looking at two events which occurred at the same position in the t' frame.

Thank you yes that does explain another little puzzle

 

[Grimble]

They are both "number of units". Suppose we have little lights on both the front and back end of an object which is moving relative to the K frame, and both lights flash simultaneously in the K frame. In this case, the distance between the positions of each flash qualifies as the "length" of the object in the K frame (since 'length' is always the distance between the positions of the front and back of an object at a single moment), so you're counting the number of distance-units between the flashes using a ruler at rest in the K frame and calling that the length x of the moving object in the K frame. Meanwhile the object is at rest in the k' frame, so although the flashes at either end of the object happen at different times in this frame (relativity of simultaneity), the distance between them in the k' frame is exactly the same as the distance you'd get if you set off two flashes on either end of the object which were simultaneous in the k' frame. Either way, you're counting the number of distance-units between the flashes using a ruler at rest in the k' frame and calling that the length x' of the object in the k' frame. Then you will find that x = x'/γ, so you can see that the length contraction equation is still talking about the "number of units" between a pair of events (the two flashes) as measured in each frame.

But are you not then assuming that the length unit is the same in both frames even though one is length contracted?

Consider too the moving clock, and imagine a clock with a display showing the time. It cannot show a different time when it is moving can it - it is still the same clock face displaying the same time wherever it is seen from, whoever it is observed by.
Surely the difference in the time displayed is the length of the time units the ratio of t/t' is the ratio of the lengths of the time units.
A clock cannot tell different times to different observers, yet it may display different durations.

In the same way a ruler alongside an object will always shew the same length wherever it is observed from, it is the size of the units that will change not the number of them.

 

[JesseM]

But are you not then assuming that the length unit is the same in both frames even though one is length contracted?

What do you mean "the same in both frames"? They are both using markings on their own rulers to measure the distance between the two events, and each one would say that the markings on the other guy's ruler are shrunk down to a smaller distance apart than the markings on their own ruler. Of course their rulers are identically constructed so each one looks the same in its own rest frame, but the same is true of the clocks they use to measure the time between two events.

Part of the confusion here may be that the time dilation equation t = γt' and the length contraction equation x = x'/γ are usually seen as "going together" in a pair because the first compares the time in both frames between ticks of a clock at rest in the primed frame, while the second compares the length in both frames of an object at rest in the primed frame. But if you translate them into the language of events, then the first is dealing with the time between two events which occurred at the same spatial position in the primed frame, but the second is dealing with the distance between two events which occurred at the same time in the unprimed frame (I explained why this is equivalent to looking at the length in both frames of an object at rest in the primed frame in post #291). If instead you wanted to pair the time dilation equation with an equation giving the distance between two events which occurred at the same time in the primed frame (so in both cases, the pair of events have a 'special' property in the primed frame), then the pair would look like this:

t = γt'
x = γx'

This sort of confusing aspect of the meaning of the two equations (and why they are normally given as a matched pair) is why it may be a good idea to just take Dalespam's advice and use the more general Lorentz transformation equations, which always work for any pair of events...

Consider too the moving clock, and imagine a clock with a display showing the time. It cannot show a different time when it is moving can it - it is still the same clock face displaying the same time wherever it is seen from, whoever it is observed by.

If you are looking at some specific event on the worldline of the clock, then all frames agree on what reading it shows at that event. But if you are asking a question like "what reading does the clock show 20 seconds after it passed by observer A", then different frames can disagree about this, because they disagree which event on the clock's worldline happened "20 seconds after it passed by observer A". For example, if the clock is moving at 0.6c relative to observer A, and the clock showed a reading of t'=0 at the time it passed observer A, then in the clock's own rest frame it shows a reading of t'=20 after 20 seconds of coordinate time in this frame, but in the rest frame of observer A it shows a reading of t'=16 after 20 seconds of coordinate time. It may help to realize that each frame is imagined to define the time-coordinates of events using clocks at rest in that frame (and synchronized in that frame according to the frame's definition of simultaneity). So in the clock's own rest frame, the clock itself is a valid way of measuring coordinate time, so 20 seconds passing on the clock means 20 seconds of coordinate time have passed. But in the rest frame of observer A, you'd need two clocks at rest and synchronized in this frame to measure the elapsed time on the moving clock. If a clock right next to observer A read t=0 at the moment the moving clock passed him reading t'=0, and then later the moving clock passed another clock at rest in A's frame (and synchronized with the one next to A) when that clock read t=20 and the moving clock read t'=16, then that's the physical meaning of the claim that the moving clock takes 20 seconds of coordinate time to tick forward by 16 seconds of its own time.

Surely the difference in the time displayed is the length of the time units the ratio of t/t' is the ratio of the lengths of the time units.

No, it's meaningless to talk about "the ratio of the lengths of the time units" because different frames disagree on the ratio--in my frame your time units are longer than mine (because your clock is ticking slower and thus takes longer to tick forward by a given amount), while in your frame your time units are shorter than mine, while in a third frame where we are both moving at the same speed in opposite directions, the ratio of my time units to your time units would be 1:1 (both our clocks are ticking at the same rate in this frame because they both have the same speed).

Again, the time dilation equation t = γt' is looking at some specific pair of events which occurred at the same position in the primed frame (like two readings on a clock at rest in the primed frame), and t is the amount of coordinate time between these specific events in the unprimed frame (as measured by clocks at rest and synchronized in the unprimed frame), while t' is the amount of coordinate time between these specific events in the primed frame (as measured by a clock at rest in the primed frame which is at the same position as each event when they occur).

A clock cannot tell different times to different observers

As I said I think that statement is overly vague. If you specify a specific event on the clock's worldline there can't be disagreement about what reading it shows at that event, but if you use some other type of specification like "the time on the clock 20 seconds of coordinate time after some event" then there can be disagreement on the reading.

 

[Grimble]

If you are looking at some specific event on the worldline of the clock, then all frames agree on what reading it shows at that event. But if you are asking a question like "what reading does the clock show 20 seconds after it passed by observer A", then different frames can disagree about this, because they disagree which event on the clock's worldline happened "20 seconds after it passed by observer A". For example, if the clock is moving at 0.6c relative to observer A, and the clock showed a reading of t'=0 at the time it passed observer A, then in the clock's own rest frame it shows a reading of t'=20 after 20 seconds of coordinate time in this frame, but in the rest frame of observer A it shows a reading of t'=16 after 20 seconds of coordinate time. It may help to realize that each frame is imagined to define the time-coordinates of events using clocks at rest in that frame (and synchronized in that frame according to the frame's definition of simultaneity). So in the clock's own rest frame, the clock itself is a valid way of measuring coordinate time, so 20 seconds passing on the clock means 20 seconds of coordinate time have passed. But in the rest frame of observer A, you'd need two clocks at rest and synchronized in this frame to measure the elapsed time on the moving clock. If a clock right next to observer A read t=0 at the moment the moving clock passed him reading t'=0, and then later the moving clock passed another clock at rest in A's frame (and synchronized with the one next to A) when that clock read t=20 and the moving clock read t'=16, then that's the physical meaning of the claim that the moving clock takes 20 seconds of coordinate time to tick forward by 16 seconds of its own time.

So the moving clock reads 20seconds in each frame? And in one 20 seconds has passed, but only 16 seconds in the other (measured on a clock in that frame?) But read from that frame the 20 seconds shown have taken only 16 seconds to pass? so those seconds are only 80% of the duration of 1 second in either frame measured from within that same frame?

Excuse me Jesse but are you not agreeing with me here?

No, it's meaningless to talk about "the ratio of the lengths of the time units" because different frames disagree on the ratio--in my frame your time units are longer than mine (because your clock is ticking slower and thus takes longer to tick forward by a given amount), while in your frame your time units are shorter than mine, while in a third frame where we are both moving at the same speed in opposite directions, the ratio of my time units to your time units would be 1:1 (both our clocks are ticking at the same rate in this frame because they both have the same speed).

No I am saying the ratio for the two times, (proper and coordinate) measured by the same observer.

Again, the time dilation equation t = γt' is looking at some specific pair of events which occurred at the same position in the primed frame (like two readings on a clock at rest in the primed frame), and t is the amount of coordinate time between these specific events in the unprimed frame (as measured by clocks at rest and synchronized in the unprimed frame), while t' is the amount of coordinate time between these specific events in the primed frame (as measured by a clock at rest in the primed frame which is at the same position as each event when they occur).

Yes, t is coordinate time (measured in one frame from the other frame) and t' is proper time(measured in the same frame it occurs in, local time as measured by an adjacent clock, time measured on that clocks worldline)

As I said I think that statement is overly vague. If you specify a specific event on the clock's worldline there can't be disagreement about what reading it shows at that event, but if you use some other type of specification like "the time on the clock 20 seconds of coordinate time after some event" then there can be disagreement on the reading.

I'm not quite sure what you are trying to say here?
The time a clock reads is the time on that clock at anyone instant An Event if you will.
Saying that it will read differently 20 seconds after an event as measured in two different frames means nothing, for then you are comparing two different events.

 

[JesseM]

So the moving clock reads 20seconds in each frame?

At some time in frame #1 the clock reads 20 seconds, and at some time in frame #1 it reads 19 seconds and 18 seconds and so forth, and likewise the same is true in frame #2. But there's no basis for saying it reads 20 seconds at the "same instant" in both frames or anything like that, each observer agrees it goes through a sequence of readings at different time-coordinates but they don't agree on the the time-coordinate of each reading. So if you were to say that you want to know the moving clock's reading when 20 seconds of coordinate time have elapsed since it passed observer A, then in observer A's frame the answer is that the moving clock's reading is 16 seconds, while in the clock's own frame the answer is that the reading is 20 seconds. In that sense one can disagree with your statement that "It cannot show a different time when it is moving can it - it is still the same clock face displaying the same time wherever it is seen from, whoever it is observed by", or at least say that the way you worded it was too vague.

And in one 20 seconds has passed, but only 16 seconds in the other (measured on a clock in that frame?)

No, you got it backwards. 20 and 16 were different opinions about the clock's reading after 20 seconds of coordinate time, not different opinions about the amount of coordinate time when the clock's reading is 20 seconds.

No I am saying the ratio for the two times, (proper and coordinate) measured by the same observer.

So you agree both are measuring the "number of units" of time that have passed between two events on a clock's worldline, with one measuring the number of seconds of proper time that passed between the events, and one measuring the number of seconds of coordinate time?

I'm not quite sure what you are trying to say here?
The time a clock reads is the time on that clock at anyone instant An Event if you will.

Because of the relativity of simultaneity "any one instant" isn't really meaningful. An "event" refers to a specific point in spacetime (like how we can talk about a specific geometrical point on a 2D plane even though that point may be assigned different coordinates by different cartesian coordinate systems on the plane), with there being some unique fact about what physical occurrences do or don't happen at each point (for example, we can specify that we are talking about the event of one clock passing next to another one, and all frames must agree on what reading each clock shows at that event).

Saying that it will read differently 20 seconds after an event as measured in two different frames means nothing, for then you are comparing two different events.

That's exactly why I said the following:

If you are looking at some specific event on the worldline of the clock, then all frames agree on what reading it shows at that event. But if you are asking a question like "what reading does the clock show 20 seconds after it passed by observer A", then different frames can disagree about this, because they disagree which event on the clock's worldline happened "20 seconds after it passed by observer A".

The point is just that your language was ambiguous, when you said "Consider too the moving clock, and imagine a clock with a display showing the time. It cannot show a different time when it is moving can it" you didn't specify that you were asking if it could "show a different time" at the same event, you could instead have been talking about the reading the clock was showing at a specific coordinate time in each frame, in which case the different frames could disagree about the reading at that coordinate time.

 

[Grimble]

The point is just that your language was ambiguous, when you said "Consider too the moving clock, and imagine a clock with a display showing the time. It cannot show a different time when it is moving can it" you didn't specify that you were asking if it could "show a different time" at the same event, you could instead have been talking about the reading the clock was showing at a specific coordinate time in each frame, in which case the different frames could disagree about the reading at that coordinate time.

I think, Jesse, that it would be prudent for me to start with a simple definition and see if you agree with it, for if you don't nothing built upon it will make any sense in your eyes.

So, if I say that I have a clock in space and specify no relationship to anything else; let there be an observe A with the clock and as far as he knows he and the clock are alone in space.

Then, as there is nothing to relate to he is neither stationary nor moving but he is inertial.
The clock will be on its world line and will be keeping proper time.
For A and the clock proper time and coordinate time will be identical.

Anything wrong with that?

 

[Dale]

Are observer A and the clock co-moving?

 

[phyti]

grimble said:
"So, if I say that I have a clock in space and specify no relationship to anything else; let there be an observe A with the clock and as far as he knows he and the clock are alone in space.

Then, as there is nothing to relate to he is neither stationary nor moving but he is inertial.
The clock will be on its world line and will be keeping proper time.
For A and the clock proper time and coordinate time will be identical."

Isn't he saying that the person and clock are moving together at some indeterminant speed?

 

[Grimble]

Are observer A and the clock co-moving?

Yes the observer is permanently adjacent to the clock, he may even be holding it.

grimble said:
"So, if I say that I have a clock in space and specify no relationship to anything else; let there be an observe A with the clock and as far as he knows he and the clock are alone in space.

Then, as there is nothing to relate to he is neither stationary nor moving but he is inertial.
The clock will be on its world line and will be keeping proper time.
For A and the clock proper time and coordinate time will be identical."

Isn't he saying that the person and clock are moving together at some indeterminant speed?

One could certainly say that, but would it not be true to say that they were stationary as they cannot be moving if there is nothing to relate the movement to. i.e. movement has to be relative to something and if there is nothing to move relative to they cannot be moving?

 

[ghwellsjr]

...would it not be true to say that they were stationary as they cannot be moving if there is nothing to relate the movement to. i.e. movement has to be relative to something and if there is nothing to move relative to they cannot be moving?

In Special Relativity, things move or are stationary in a defined frame of reference. So you could say that this observer and his clock are stationary in a frame at the co-ordinates of x=1000 m, y=-2345 m, z-0 m at time 345 s and he is facing in the x direction with y to his right and z above his head OR you could define a different scenario in which he is moving in a frame with the same conditions as before but that he is traveling at vx=345 m/s, vy=0 m/s and vz=-4235 m/s, but most people just like to say he is stationary in an inertial frame or he's traveling a .6c in the x direction or not even specify a direction. We usually know what they mean.

But even if there is nothing else in the universe, you can still define him to be stationary or have any kind of motion and/or acceleration you desire (short of c) within a frame.

 

[Grimble]

In Special Relativity, things move or are stationary in a defined frame of reference. So you could say that this observer and his clock are stationary in a frame at the co-ordinates of x=1000 m, y=-2345 m, z-0 m at time 345 s and he is facing in the x direction with y to his right and z above his head OR you could define a different scenario in which he is moving in a frame with the same conditions as before but that he is traveling at vx=345 m/s, vy=0 m/s and vz=-4235 m/s, but most people just like to say he is stationary in an inertial frame or he's traveling a .6c in the x direction or not even specify a direction. We usually know what they mean.

But even if there is nothing else in the universe, you can still define him to be stationary or have any kind of motion and/or acceleration you desire (short of c) within a frame.

OK, then let us define him as stationary. Then can I say:
The FoR, that he is permanently at the origin of, is an Inertial Frame of Reference.
The clock will be on its world line and will be keeping proper time.
For A and the clock proper time and coordinate time will be identical.?

 

[Dale]

Yes the observer is permanently adjacent to the clock, he may even be holding it.

Then the proper time measured by the clock is indeed equal to the coordinate time in A's frame.

 

[phyti]

Yes the observer is permanently adjacent to the clock, he may even be holding it.



One could certainly say that, but would it not be true to say that they were stationary as they cannot be moving if there is nothing to relate the movement to. i.e. movement has to be relative to something and if there is nothing to move relative to they cannot be moving?

I don't accept Newtons definitions about motion and rest. Rest is not a state of motion, but a relation about the difference in motion of two objects. For me, when two objects have the same velocity (vector), one is at rest relative to the other, i.e., rest is a special case of motion but not a 'lack of motion' as Newton defines it.
Since as you say, there is no outside reference object (excepting dark matter), there is no motion of the frame, which by the previous definition eliminates 'rest'.
I agree with your 2nd statement in red.

But...if we look deeper, there is motion at the molecular and atomic levels, therefore the inertial condition is only an approximation, although a practical one.

Now if you could just get light-clock 101...

 

[Grimble]

Let us now add a second clock, 'held' by observer B, at rest at the origin of a second Inertial Frame of Reference.

The clock will be on its world line and will be keeping proper time.
For B and the clock proper time and coordinate time will be identical.?

As A and B are each at rest at the origin of their own Inertial Frames of Reference; we can say that their identical clocks will be keeping identical time. Identical in that their units of time will be of the same duration as judged by an independent observer.

This is necessary as they are both keeping proper time and as that is a requirement of Einstein's first postulate.

 

[Dale]

Let us now add a second clock, 'held' by observer B, at restat the origin of a second Inertial Frame of Reference.

The clock will be on its world line and will be keeping proper time.
For B and the clock proper time and coordinate time will be identical.?

Yes. Assuming that B is also inertial then the proper time measured by the clock held by B will be equal to the coordinate time in B's frame.

As A and B are each at rest at the origin of their own Inertial Frames of Reference; we can say that their identical clocks will be keeping identical time. Identical in that their units of time will be of the same duration as judged by an independent observer.

How does the independent observer judge the duration of their units of time? This is a non-standard phrase so you need to specify what you mean by this.

 

[Grimble]

Yes. Assuming that B is also inertial then the proper time measured by the clock held by B will be equal to the coordinate time in B's frame.

How does the independent observer judge the duration of their units of time? This is a non-standard phrase so you need to specify what you mean by this.

Well you are right of course.

So let us specify an independent inertial observer, who can measure the time passing in A and B's own frames and that, by applying the LT transforms, he can calculate how time passes in each of those frames according to an observer in that frame.

Secondly, as each clock is keeping proper time each tick of those clocks must be an event in space-time. And as JesseM pointed out

If you are looking at some specific event on the worldline of the clock, then all frames agree on what reading it shows at that event.

So if each clock's one second ticks are events in space-time they must be ticking at identical rates.

 

[Dale]

Well you are right of course.

So let us specify an independent inertial observer, who can measure the time passing in A and B's own frames and that, by applying the LT transforms, he can calculate how time passes in each of those frames according to an observer in that frame.

So the independent observer calculates the ratio of (proper time A)/(coordinate time A) in A's frame and (proper time B)/(coordinate time B) in B's frame? If that is your intention then what is the purpose of the independent observer? Since he is transforming into each frame he doesn't add anything to the question just makes the problem more confusing. And we have already specified that both of those ratios are 1.

 

[Grimble]

So the independent observer calculates the ratio of (proper time A)/(coordinate time A) in A's frame and (proper time B)/(coordinate time B) in B's frame? If that is your intention then what is the purpose of the independent observer? Since he is transforming into each frame he doesn't add anything to the question just makes the problem more confusing. And we have already specified that both of those ratios are 1.

Yes, you are quite right of course that is all unnecessary!

 

[ghwellsjr]

Let us now add a second clock, 'held' by observer B, at rest at the origin of a second Inertial Frame of Reference.

The clock will be on its world line and will be keeping proper time.
For B and the clock proper time and coordinate time will be identical.?

As A and B are each at rest at the origin of their own Inertial Frames of Reference; we can say that their identical clocks will be keeping identical time. Identical in that their units of time will be of the same duration as judged by an independent observer.

This is necessary as they are both keeping proper time and as that is a requirement of Einstein's first postulate.

Why do you want to introduce a duplicate of your first scenario with no realtionship between them? If you want to introduce as second observer/clock, then you should define their position/motion/whatever in your first frame OR you could define your second frame in relation to your first frame and then define the second observer/clock in that second frame and then you could transform the observer/clock from the second frame to see how they appear in the first frame. Eventually you need to "put" all observers/clocks into a single frame, that is, if you want to discuss this in the context of Special Relativity.

 

[Grimble]

Why do you want to introduce a duplicate of your first scenario with no realtionship between them? If you want to introduce as second observer/clock, then you should define their position/motion/whatever in your first frame OR you could define your second frame in relation to your first frame and then define the second observer/clock in that second frame and then you could transform the observer/clock from the second frame to see how they appear in the first frame. Eventually you need to "put" all observers/clocks into a single frame, that is, if you want to discuss this in the context of Special Relativity.

Because now we have established that as two individual inertial FoR A and B are, as far as can be determined in identical situations. The one second 'ticks' of one clock keeping proper time, and therefore being Time-space Events, have to be equal in duration and frequency to the one second ticks of the the second clock also keeping proper time, and its 'ticks' therefore, also being Time-space Events.
There is nothing to say, so far concerning the simultaneity of the said cloaks' 'ticks'.

Let us therefore have these two clock's, and as they are at rest at the origin of their respective frames, the frames themselves moving at 0.6c relative to one another.

 

[ghwellsjr]

Because now we have established that as two individual inertial FoR A and B are, as far as can be determined in identical situations. The one second 'ticks' of one clock keeping proper time, and therefore being Time-space Events, have to be equal in duration and frequency to the one second ticks of the the second clock also keeping proper time, and its 'ticks' therefore, also being Time-space Events.

They are identical simply and only because you have defined them to be identical. That's why I asked why you wanted to make a duplicate of your first scenario with no relationship to your original one. I think your understanding of Special Relativity would be advanced more directly if you simply follow Einstein's example from your referenced book in which he looks at an object first at rest in one frame and then from another frame in motion relative to the first one. But if you want to take this indirect approach, then let's see where you go with it.

There is nothing to say, so far concerning the simultaneity of the said cloaks' 'ticks'.

Let us therefore have these two clock's, and as they are at rest at the origin of their respective frames, the frames themselves moving at 0.6c relative to one another.

Are you implying that now there is something to say concerning the simultaneity of the said clocks' 'ticks'?

 

[Grimble]

They are identical simply and only because you have defined them to be identical. That's why I asked why you wanted to make a duplicate of your first scenario with no relationship to your original one. I think your understanding of Special Relativity would be advanced more directly if you simply follow Einstein's example from your referenced book in which he looks at an object first at rest in one frame and then from another frame in motion relative to the first one. But if you want to take this indirect approach, then let's see where you go with it.

Are you implying that now there is something to say concerning the simultaneity of the said clocks' 'ticks'?

Not at all I'm trying to be clear and accepted that this is the case. Or someone will say I am not being specific enough.

Next point let us say that these two clocks pass one another and at the point of their passing they synchronise their clocks.
And let us say that there is another observer, C, who is at rest at the point of their intersection who observes A and B each pass him at speeds of 0.3c in opposite directions
Giving us a single space-time event involving all 3 observers.

 

[Dale]

Not at all I'm trying to be clear and accepted that this is the case. Or someone will say I am not being specific enough.

Next point let us say that these two clocks pass one another and at the point of their passing they synchronise their clocks.
And let us say that there is another observer, C, who is at rest at the point of their intersection who observes A and B each pass him at speeds of 0.3c in opposite directions
Giving us a single space-time event involving all 3 observers.

Actually, relative to observer C the other two observers will be going at 0.33333333 c.