Time Dilation

[Tom Rexton]

"Is his tick rate, in his frame of reference, slower than my tick rate in my frame of reference?"

No. They are identical, because his clock is at rest in his frame of refence, and your clock is at rest in your frame of reference.

[Capitalism Forever]

that the "tick rates" of identical clocks are identical, independent of frame.  Indeed, the law of identity ought to suffice.

Ah, the confusion between identity and equivalence strikes again! (I have already written about this in that lengthy thread about Hume and his supposed case against inductive reasoning.) When you say "Clock A and clock B are identical," that means that clock A and clock B are ONE clock--IOW, the expression "clock A" refers to the same clock as the expression "clock B."

TWO clocks cannot be identical, only two references ("A" and "B") can resolve to AN identical clock.

In our example, we have TWO clocks, so clock A and clock B are not identical. Rather, they are built in an identical way so that they are equivalent, or interchangeable, for the purpose of measuring time. You could swap them without my noticing the swap.

But if one of the two clocks--say A--is at rest relative to me, while B is moving uniformly, it will appear to me that B is slower than A. Swap them, so that B is at rest in my frame and A is moving, and it will be A that appears slower than B.

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So, now, to put the question in very simple terms: Do we only say that clock A appears slower than clock B, or do we also say that clock A is slower than B? (Or do we pull a Clinton? )

[tommyedison]

So, now, to put the question in very simple terms: Do we only say that clock A appears slower than clock B, or do we also say that clock A is slower than B? (Or do we pull a Clinton?  )

I think that clock A is slower than B. I read in a book that muons (a type of particle) have an average lifespan of around 2.2 * 10^(-6) of a second. But when they were accelerated to speeds close to that of light, there lifespan was extended by seven times.

[stephen_speicher]

I am sure you are capable of following an example of currency conversion, so I take it you're not motivated to talk about exchange rates on a science thread...which is perfectly understandable.

You misunderstand. You stopped my quote in mid-sentence and left off the "and seeing its relevance" part about my assessment of your analogy. Look, the use of analogies can be of value, and I am generally all for putting ideas into one's own words. But at times there are technical issues in mathematics and physics that simply beg to be understood on their own terms. I have been very careful in my formulations and I think it would behoove you to struggle harder to understand what I wrote, rather than re-wording it in what I think to be imprecise forms, and using analogies that simply are not relevant.

Consider the following scenario ...

Your scenario involves another phenomenon than just time dilation, one of Doppler shift, and unlike time dilation the rates will vary depending upon whether the spaceship and Mission Control are converging or receeding.

You would do much better to understand the simpler scenario of time dilation first, at least with more precision than you seem to hold it in your mind. You might want to re-read some of my other posts, or, do yourself a favor and read one of the non-technical books I recommended.

[Capitalism Forever]

But at times there are technical issues in mathematics and physics that simply beg to be understood on their own terms.

You would do much better to understand the simpler scenario of time dilation first, at least with more precision than you seem to hold it in your mind. You might want to re-read some of my other posts, or, do yourself a favor and read one of the non-technical books I recommended.

Aha, now I see what you meant by the analogies being irrelevant. They are irrelevant because there IS no sufficiently good analogy.

Your scenario involves another phenomenon than just time dilation, one of Doppler shift, and unlike time dilation the rates will vary depending upon whether the spaceship and Mission Control are converging or receeding.

Yes, that's why I made it explicit that the spaceship is moving away from Mission Control, so there is a "redshift" in the signals. But if they know v, Mission Control can compute what the frequency would be without the Doppler shift, i.e. what you referred to as the "observed tick rate" in your example with the clocks.

Due to time dilation, this observed tick rate in Mission Control's frame is slower than the tick rate observed in the astronaut's frame. But, again, if they know v, Mission Control can compute the tick rate observed in the astronaut's frame, which happens to be called the "proper tick rate" of the astronaut's heart.

Is it correct so far, or am I making a mistake somewhere?

[stephen_speicher]

I think that clock A is slower than B. I read in a book that muons (a type of particle) have an average lifespan of around 2.2 * 10^(-6) of a second. But when they were accelerated to speeds close to that of light, there lifespan was extended by seven times.

You guys are missing the point. Here you have 1/2 of the facts right, and a wrong conclusion. The seminal muon experiment in accelerators was performed by Bailey et al. (Nature, Vol. 268, pp. 301-304, July 28, 1977; Nuclear Physics B, Vol. 150, p. 1, 1979) in which they accelerated the muons to .9994 c, and the 2.2 microsecond decay lifetime was measured to increase by a factor of more than 29 times.

But this is not time dilation, but rather elapsed proper time. Proper time refers to the accumulation of time in the observer's own reference frame, whereas time dilation and length contraction are a relationship established as a consequence of observation or measurement among different frames. Time dilation does not lead to physical differences between clocks when they are reunited and compared, but elapsed proper time does reflect physical differences for differing paths.

[stephen_speicher]

Aha, now I see what you meant by the analogies being irrelevant. They are irrelevant because there IS no sufficiently good analogy.

No. That is not what I said. I said that your analogy was not relevant and strongly suggested you just deal with the issue directly since it just begs to be understood in its own terms. I have noticed many people over the years unnecessarily confuse themselves about relativity, manufacturing too many scenarios of their own that do not directly relate to the subject.

As to your example: I do not know why you would want to complicate things, but adding Doppler shifts or other relativistic effects only make a simple issue more complicated. If you cannot accept and/or understand time dilation all by itself, then it is pointless to add other things to it. The simplest case of all is two observers in relative uniform motion, each measuring the other's clock to have a slower tick rate than own. That is where we started.

[Capitalism Forever]

If you cannot accept and/or understand time dilation all by itself, then it is pointless to add other things to it.

Did I give you the impression that I don't accept time dilation? That's certainly not the case! If I didn't accept time dilation, I would very plainly state that I don't accept it. I am not disputing any of the statements you have made in response to my usage of the word "really." I am glad you have made those statements because I have found them instructive; my aim with my subsequent posts was not to oppose your statements but to elaborate on them in my own words in order to see if I understood them correctly.

The observed tick rate of a clock in relative motion is slower than that of a clock at rest--that's a fairly simple idea that I have no difficulty grasping. I would like to move on and integrate this idea with the rest of my knowledge: I would like to know why it is so and what the implications of this fact are.

The Theory of Elementary Waves gives a very lucid explanation for the invariancy of the speed of light, and time dilation can be mathematically derived from that fact, so I am relatively satisfied as to the "why" ; that's why I was focusing on the implications of time dilation. Such as the implication, or lack of implication, regarding the biological processes in a human body.

Your reply to tommyedison's previous post is actually a great example of the kind of answers I am looking for:

Time dilation does not lead to physical differences between clocks when they are reunited and compared, but elapsed proper time does reflect physical differences for differing paths.

Applying this to a human body, this means that time dilation does not lead to age differences between astronauts when they are reunited, so if they have traveled symmetrical paths, their amounts of elapsed proper time will be the same and thus they will have aged the same amount. But if they take asymmetrical paths, they will have differing amounts of elapsed proper time, so their ages will be different--although not due to time dilation but due to something else.

[stephen_speicher]

Applying this to a human body, this means that time dilation does not lead to age differences between astronauts when they are reunited, so if they have traveled symmetrical paths, their amounts of elapsed proper time will be the same and thus they will have aged the same amount. But if they take asymmetrical paths, they will have differing amounts of elapsed proper time, so their ages will be different--although not due to time dilation but due to something else.

Yes, by definition you integrate the metric tensor along the path through spacetime to arrive at the elapsed proper time. So, path dependency and gravitation are essential, and for perfectly symmetrical motion, under the same conditions, there will be no difference in accumulated time for two standard synchronized clocks that are reunited. But granted asymmetrical paths there will indeed be differences in elapsed time when the clocks are reunited, and presumably the same for aging through bodily functions.

[dsandin]

Ah, the confusion between identity and equivalence strikes again! (I have already written about this in that lengthy thread about Hume and his supposed case against inductive reasoning.) When you say "Clock A and clock B are identical," that means that clock A and clock B are ONE clock--IOW, the expression "clock A" refers to the same clock as the expression "clock B."

TWO clocks cannot be identical, only two references ("A" and "B") can resolve to AN identical clock.

In our example, we have TWO clocks, so clock A and clock B are not identical. Rather, they are built in an identical way so that they are equivalent, or interchangeable, for the purpose of measuring time. You could swap them without my noticing the swap.

But if one of the two clocks--say A--is at rest relative to me, while B is moving uniformly, it will appear to me that B is slower than A. Swap them, so that B is at rest in my frame and A is moving, and it will be A that appears slower than B.

Confusion strikes to be sure!

First of all, raising this identity/equivalance issue is entirely out of place. There is exactly one meaning of "identical clocks" here. And it isn't that all those clocks are metaphysically the same entity(!). Even a single inertial frame does not have just one clock, it has an indefinite multitude of identical clocks.

Second, the topic here is your question, "Is his tick rate, in his frame of reference, slower than my tick rate in my frame of reference?". And it seems to be a result of more confusion, as I have already pointed out. To answer this question doesn't require computation based on a relative velocity of inertial frames. And it doesn't require a fine grasp of time dilation. We just cut the Gordian knot with the simple realization that identical clocks are identical clocks. Because the clocks are identical (i.e., they physically indicate the same amount of change via the same number of the same physical units) -- and because this is the straightforward case of the uniform translation of frames -- they necessarily accumulate the identical amount of proper time per tick.

[Capitalism Forever]

There is exactly one meaning of "identical clocks" here.  And it isn't that all those clocks are metaphysically the same entity(!).

I don't approve of the use of "identical" in the sense of "equivalent." It's a very misleading way to use the word. It misled you too:

Indeed, the law of identity ought to suffice.

The law of identity says: "A is A." In our example, clock A is NOT clock B. It's another clock that is located elsewhere and even moves at a different velocity.

Indeed, if the two clocks were both set to 12:00 while at rest, then moved away from each other and subsequently reunited, one of them may end up reading 3:00 and the other 3:30 due to differences in elapsed proper time. If one applied the law of identity the way you did, they would have to show the same time.

[dsandin]

Re: Capitalism Forever, Jul 25 2004, 02:39 PM

This is getting somewhat ridiculous. You persist in picking at a perfectly justified word and concept, the use of which I have fully defended. You continue to disregard its context. And in the process, you express an off-the-wall objection to my minor point that the law of identity demands like results from like entities in like circumstances. Meanwhile my substantial point -- the one and only object of my several recent postings here -- is no longer acknowledged by you. Maybe you now see its validity and cogency and have put it behind you. Well, okay.

Except I'm not sure I fully believe it, since your last paragraph shows that you are still confused. Again: the situation is the symmetrical case of uniform translatory motion, not asymmetric space-time paths that cause differing proper times in identical clocks.

[Capitalism Forever]

You persist in picking at a perfectly justified word and concept, the use of which I have fully defended.  You continue to disregard its context.

I do not disregard its context. I know the sense in which you used the word and interpreted your statements in that sense of the word. I just remarked that I don't LIKE the word being used in that sense because I think it's misleading.

And in the process, you express an off-the-wall objection to my minor point that the law of identity demands like results from like entities in like circumstances.

The law of identity demands like results from like non-volitional entities in like circumstances. Two people with an identical genetic makeup may face the same circumstances and react differently. It is only when you know that the clocks are non-volitional that you know they must act the same way.

Meanwhile my substantial point -- the one and only object of my several recent postings here -- is no longer acknowledged by you.  Maybe you now see its validity and cogency and have put it behind you.  Well, okay.

I never questioned its validity! My question that you initially responded to was in quotation marks:

"Is his tick rate, in his frame of reference, slower than my tick rate in my frame of reference?"

I didn't post that question because I was asking it. I posted it as an answer to Tom's question "from what frame of reference?" It was a part of our discussion about the meaning (or meaninglessness) or "really." I knew the answer to the question I "asked" !

[stephen_speicher]

I do not disregard its context. I know the sense in which you used the word and interpreted your statements in that sense of the word. I just remarked that I don't LIKE the word being used in that sense because I think it's misleading.

You claimed more. In reference to dsandin you claimed that "It misled you too." Personally, I find your argument completely specious and, quite frankly, it just deflects from the substantive issues that dsandin identified and discussed. Instead of defending the indefensible you would do much better to actually process the excellent ideas that dsandin has offerred. He is right on the money and you are somewhere out in left field.

[Capitalism Forever]

You claimed more. In reference to dsandin you claimed that "It misled you too."

And I stand by that claim. I explained in my previous post why:

The law of identity demands like results from like non-volitional entities in like circumstances. Two people with an identical genetic makeup may face the same circumstances and react differently. It is only when you know that the clocks are non-volitional that you know they must act the same way.

Personally, I find your argument completely specious and, quite frankly, it just deflects from the substantive issues that dsandin identified and discussed.

He misidentified my answer to Tom as a question and he thought I didn't know the answer. He then gave the answer, and used the word "identical" in a sense I don't like. I agreed with his answer, so I didn't comment on the substance of what he said; I just made a side comment on the form in which he presented it.

Instead of defending the indefensible

When I defend something, I do it because I think it's defensible. I will stop defending it as soon as you convince me it isn't.

you would do much better to actually process the excellent ideas that dsandin has offerred. He is right on the money and you are somewhere out in left field.

If he is right on the money and I agree with him (on the substance of his answer), how can it be that I am so wrong?

[stephen_speicher]

And I stand by that claim. I explained in my previous post why:

When I defend something, I do it because I think it's defensible. I will stop defending it as soon as you convince me it isn't.

I would try to convince you but I have to leave now and check to see if my toaster is still non-volitional. I guess you never know.

[Capitalism Forever]

I would try to convince you but I have to leave now and check to see if my toaster is still non-volitional.

LOL Enjoy your meal and (just in case it's become volitional) greetings to your toaster!

[y_feldblum]

Yes, by definition you integrate the metric tensor along the path through spacetime to arrive at the elapsed proper time.

You integrate the line element, which is the value of the metric tensor on differential (infinitesimal) displacement, to arrive at proper time. (I don't mean to offend, but as long as we're keeping terms clearly defined....)

The line element is the invariant definition of distance; the metric gives the magnitude of any vector - and if that vector is displacement, it will give you the magnitude of that displacement. A vector is a geometrical construct having magnitude and direction.

[stephen_speicher]

You integrate the line element, which is the value of the metric tensor on differential (infinitesimal) displacement, to arrive at proper time. (I don't mean to offend, but as long as we're keeping terms clearly defined....)

Physicists commonly refer to the line element as a metric. This is a holdover from a classical distinction held between mathematicians and physicists as between tensors and their components. The line element provides the components of the metric, given in terms of the coordinates.

The line element is the invariant definition of distance;

Technically, the line element is the contraction of the metric tensor with some given set of tangent vectors that are directed along the coordinate axes. A Riemannian metric on a manifold is a rank-2 tensor field defined on the tangent spaces of the manifold, such that it provides, in a differentiable manner, a positive definite inner product. The manifold for general relativity, however, is semi-Riemannian, and the metric is not positive definite, and we refer to that metric as a semi-Riemannian (or pseudo-Riemannian) metric. In either case, speaking as a physicist, the metric is the definition of distance on a manifold; the component matrix of the metric tensor helps establish the invariance of the line element under transformations of the general Poincare group. Speaking as a mathematician, just writing down the line element in some chart defines the metric tensor in the same chart.

All in all, if you read the literature, you will find that the line element is usually just a shorthand way to express the metric tensor field in reference to some particular coordinate chart.

[y_feldblum]

Stephen,

The second paragraph was actually to introduce some terms to those who might not already be familiar with them. I know you know what those terms mean.

In the first, I did forget the word contract. Silly me.

By the way, can the metric be defined, more fundamentally, as a certain map from vectors to one-forms?

And on a wildly unrelated note, is four-force better defined as vector or as one-form?

[stephen_speicher]

By the way, can the metric be defined, more fundamentally, as a certain map from vectors to one-forms?

It is not defined that way, but you can think of its use that way because a 1-form is really a dual vector.

And on a wildly unrelated note, is four-force better defined as vector or as one-form?

Same concerns as in the prior post, re mathematicians and physicists. In relativity force is a 4-vector and you measure the components of the 4-vector by projecting the vector onto an apparatus. Historically the 4-force is the rate of change of the 4-momentum along the world line. For some general spacetime this rate of change should be defined by covariant differentiation. And yet usually in physics the components of a 1-form are covariant while the components of a vector are contravariant. So, you take your pick. This is part of a long-standing confusion that exists between mathematicians and physicists and between traditional tensor analysis and modern forms.

There is one old traditional book which bridges the gap in terms of differential forms, and is well worth studying for those interested in this. It is a very inexpensive republication by Dover of Harley Flanders' classic work, Differential Forms with Applications to the Physical Sciences.