Does time dilation work this way?

[Prem1998]

Suppose, a ball is thrown upwards. The proper time interval between the throw and the instant at which it attains max height is t. And, the dilated time interval between the same events for a different moving observer is 2t.
Now, my question is: Both observers see the same instants happening, then what accounts for the increased time for the later observer? Do the same instant remains paused for the second observer for twice the time length than the first observer?
Suppose for the first observer the height of the ball above the ground at time T is 2m and after time dt, it is 2+dx m. By an increase of dt, just the next instant happens and nothing happens in between. Now, for the second observer, the ball remains paused at the height of 2m for a time interval 2dt. So, the universe just makes time 'more chunky' to account for the increased time. Just like when we slow down a movie, the time between two scenes gets dilated, but we still don't see what happens between the individual frames. We still see the same frames, but each frame remains paused for a longer time on the screen.
So, does the second observer see what happens to the ball between the time T and T+dt to add up to the original no. of instants to make more instants ( hence more time ) or does the same instant at time T remain paused for a longer time 2dt to account for the increased time?

 

[Ibix]

You are trying to use special relativity to analyse a situation where gravity is important. That won't work. There are an awful lot of complexities that you can't just gloss over as you are doing.

The reason for time dilation in SR is quite straightforward. A different frame is a different choice of coordinates on spacetime. Changing frames is closely analogous to rotating your axes in normal Euclidean space. Draw a line straight up a page. It has length L and height L and width 0. Now rotate the page a few degrees. The length is still L, but the height is L cosθ and the width is L sinθ.

In the same way, when you change frames, the length of a path (the proper time) is unchanged. But the "height" (the coordinate time) and the "width" (the distance traveled through space) of the path do change. You just need to get used to the notion that "now" is as flexible a term as "here" is.

 

[PeroK]

If you leave out gravity and consider the ball moving with constant velocity between points A and B, then you actually have three times involved here:

The proper time of the ball between these events
The measured time between these events for one observer
The measured time between these events for a second observer (moving relative to the first)

When you begin to study relativistic kinematics, this set-up will be important.

But, you do not need to analyse this situation in order to understand time dilation in the first place.

 

[Prem1998]

Okay, forget the bit about a ball being thrown up. That was my mistake. Let's just say gravity is negligible here.
My question was: Both observers see the same instants, then what additional instants account for the increased time interval between the events for the second observer?
For the first observer, the instant T+ dt happens after the one at T and nothing happens in between ( as dt has no finite value and is as small as you can get). Let's take the ball example without involving gravity. Now by max height I mean the height attained by the ball when it touches the ceiling of the spacecraft of the first observer. Is the second observer able to see instants between T and T+dt when ball is between 2m and 2+dx m? I don't think there are instants between T and T+dt( unless the universe creates more in-between instants to account for more time between the same events). So, the only possible explanation could be that all the instants remain paused for twice the interval dt, to account for an increased time.

 

[PeroK]

Okay, forget the bit about a ball being thrown up. That was my mistake. Let's just say gravity is negligible here.
My question was: Both observers see the same instants, then what additional instants account for the increased time interval between the events for the second observer?
For the first observer, the instant T+ dt happens after the one at T and nothing happens in between ( as dt has no finite value and is as small as you can get). Let's take the ball example without involving gravity. Now by max height I mean the height attained by the ball when it touches the ceiling of the spacecraft of the first observer. Is the second observer able to see instants between T and T+dt when ball is between 2m and 2+dx m? I don't think there are instants between T and T+dt( unless the universe creates more in-between instants to account for more time between the same events). So, the only possible explanation could be that all the instants remain paused for twice the interval dt, to account for an increased time.

There is no such thing as an "instant", in the sense of time interval "as small as you can get". It's perfectly valid to study motion by considering small increments, but generally you use the notation $\Delta t$ and $\Delta t'$ where these are small intervals of finite length (but not "infinitesimals").

$dt$ should be used to denote the derivative, which comes from taking the limit as $\Delta t \rightarrow 0$. For example:

$\frac{dv}{dt} = \lim_{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}$

Note that here $dt$ has no independent meaning as a quantity. It's just part of the notation for a derivative.

$dt$ is also used for a thing called a "differential". You can learn about those here, for example:

http://tutorial.math.lamar.edu/Classes/CalcI/Differentials.aspx

 

[Prem1998]

There is no such thing as an "instant", in the sense of time interval "as small as you can get". It's perfectly valid to study motion by considering small increments, but generally you use the notation $\Delta t$ and $\Delta t'$ where these are small intervals of finite length (but not "infinitesimals").

$dt$ should be used to denote the derivative, which comes from taking the limit as $\Delta t \rightarrow 0$. For example:

$\frac{dv}{dt} = \lim_{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}$

Note that here $dt$ has no independent meaning as a quantity. It's just part of the notation for a derivative.

$dt$ is also used for a thing called a "differential". You can learn about those here, for example:

http://tutorial.math.lamar.edu/Classes/CalcI/Differentials.aspx

So, how exactly is there more time between the same two events for the second observer? I know there is more time, but how there is more time? What would account for the extra time? Are there more in-between instants for the second observer or do the same instants remain paused for a longer time for the second observer?

 

[PeroK]

So, how exactly is there more time between the same two events for the second observer? I know there is more time, but how there is more time? What would account for the extra time? Are there more in-between instants for the second observer or do the same instants remain paused for a longer time for the second observer?

There is no such thing as a finite instant. Time is continuous. If you take the example of a clock moving at $v$ in your reference frame, with $\tau$ being the (proper) time of the clock and $t$ being your measured time on a clock at rest in your frame. Then:

$\Delta t = \gamma \Delta \tau$

This relates finite measurements (by you) of the readings of the two clocks.

This can be taken to the limit (mathematically) by using calculus and you have:

$\frac{dt}{d \tau} = \gamma$

Where $t$ and $\tau$ are "continuous" variables.

 

[Ibix]

So, how exactly is there more time between the same two events for the second observer? I know there is more time, but how there is more time? What would account for the extra time? Are there more in-between instants for the second observer or do the same instants remain paused for a longer time for the second observer?

Why is the tip of a pencil higher when you stand it on end compared to when you tilt it a bit? Whatever answer you find acceptable to that is the answer to your question about time dilation.

You need to stop trying to "count instants". It makes no sense. An instant has zero duration so there are always infinitely many of them in any finite duration, no matter how short.

 

[Dale]

or does the same instant at time T remain paused for a longer time 2dt to account for the increased time?

OK, so the problem is that your question is fundamentally ill posed in terms of math, but insofar as it can be altered to make it right, this is closer to correct.

Let's just talk about Euclidean geometry for a moment. Consider a line segment (I will use the variable $s$) drawn on a page with two different coordinate systems (variables $x$ and $x'$ representing the "horizontal" position for each coordinate system) on the page. There are an uncountably infinite number of points $s$, $x$, and $x'$. There is a one to one mapping between any $s$ and the corresponding $x$ and $x'$. So it doesn't make sense to say that there are more points in one compared to the other. There also is no minimum increment or sense in which any of them are ever paused.

What you can do is take any two points and calculate $\Delta s=s_2-s_1$, and similarly $\Delta x$ and $\Delta x'$. You can then compare the ratio of $\Delta x/\Delta x'$ in the limit as $\Delta s$ goes to 0. You can say that ratio is greater than 1 or less than 1 and use that to say that one is faster or bigger than the other, but that doesn't imply that there are any more points (both are uncountably infinite) or that there is any pausing (both are continuous)

 

[phinds]

So, how exactly is there more time between the same two events for the second observer? I know there is more time, but how there is more time? What would account for the extra time? Are there more in-between instants for the second observer or do the same instants remain paused for a longer time for the second observer?

That's like saying that there is more length to a pencil held perpendicular to your line of sight vs the same pencil angled away from you. It's just not true. Time dilation is not something that happens to an object, it is a change in perception based on motion through space-time (gravitational time dilation is a different story). I'm just repeating here exactly what has already been said but I'm doing so to specifically target your statement that there is "more time".

Let's look at it another way. You, right now as you read this, are MASSIVELY time dilated according to a particle moving in the CERN accelerator. You are mildly time dilated according to an asteroid somewhere in the solar system that is moving fast relative to Earth, you are trivially time dilated according to a supersonic jet, and you are not time dilated at all according to the chair you are sitting in.

How could you possibly be all of these at the same time if there was any actual effect on you?

 

[Prem1998]

Why is the tip of a pencil higher when you stand it on end compared to when you tilt it a bit? Whatever answer you find acceptable to that is the answer to your question about time dilation.

You need to stop trying to "count instants". It makes no sense. An instant has zero duration so there are always infinitely many of them in any finite duration, no matter how short.

An instant can't have a zero duration, it's infinitesimal. That's why an infinite number of infinitesimal length instants add up to give a finite time duration. If an instant were zero, an infinite number of zero duration would still add up to zero.
Since I have to stop counting instants, I should probably say it like this: Every instant that the first observer sees can be mapped to an instant that the second observer sees. The first observer sees the ball at a height of 2m at some time, then it can be mapped to the instant at which the second observer sees it at the same height. There is a one-to-one correspondence. But the fact that the same instants take more time to complete for the second observer reflects that the instants are dilated or they have more length or they remain paused for a longer time. But this does not mean that time becomes discontinuous by this. 2*dt=2dt which is still an infinitesimal, so time does still advance in infinitesimals and there's still an infinite number of them. Even if the instant length becomes more but still an infinitesimal, then it does not mean that time is advancing in finite jumps.
So, basically, if, for the first observer, an instant represented a single point on the graph, i.e. the instant remained in front of her eyes for a time equivalent to a single point (not zero), then for the second observer the instant represents maybe two points. Each instant corresponding to each position of the ball can still be mapped one-to-one to the corresponding instant for the second observer. (But for the second observer, the same instants give more length in total).

 

[phinds]

An instant can't have a zero duration, it's infinitesimal.

So, after having been told specifically that this is wrong, you are going to insist on it anyway? See, the problem with this is that physics has certain definitions on which other things are based. When you start making up your own definitions, you have left the realm of physics and wandered off into la la land.

 

[Dale]

An instant can't have a zero duration

Uhh, an instant does have zero duration. That is the definition.

But the fact that the same instants take more time to complete for the second observer reflects that the instants are dilated or they have more length or they remain paused for a longer time.

You can say that the coordinate difference between two points is larger in one coordinates than in another without trying to ascribe a length to a point.

It would really help if you would try to learn and use the standard definitions of terms rather than trying to get the community to use your personal definitions.

 

[Prem1998]

Uhh, an instant does have zero duration. That is the definition.

You can say that the coordinate difference between two points is larger in one coordinates than in another without trying to ascribe a length to a point.

It would really help if you would try to learn and use the standard definitions of terms rather than trying to get the community to use your personal definitions.

Any time event is composed of instants. So, an infinite number of zeros add up to give a finite duration, right? I thought 0+0+0+......=0
AND, integration(dt)= t. I thought integration was more like an infinite sum. We should have used zeros instead of dt's

 

[Nugatory]

An instant can't have a zero duration, it's infinitesimal. That's why an infinite number of infinitesimal length instants add up to give a finite time duration. If an instant were zero, an infinite number of zero duration would still add up to zero.

A instant is a single point in spacetime. Just like a point on the number lin, it has size zero. And just like points on the number line, you can add an infinite number of zero-sized points to get something that doesn't add up to zero: There are an infinite number of zero-sized points between one and two on the number line, and the distance between one and two on the number line is clearly not zero

Infinite numbers of zeros are mathematically a bit tricky; much of this stuff is will be covered in a mathematically rigorous college-level calculus class. However, you might be able to form some intuition for how A can find two seconds between two events while B only finds one second between them, yet every instant on A's longer timeline can be paired with a single instant on B's shorter timeline if you google for "Hilbert's Hotel"

 

[PeterDonis]

Both observers see the same instants

Actually, they don't. The two different times refer to two different spacetime intervals, i.e., the spacetime "lengths" of two different curves. They do not refer to two different "lengths" applied to the same curve.

This doesn't in any way affect the fact that you don't have a correct understanding of how "instants", derivatives, etc. work, as other posters have pointed out. But all of that is actually irrelevant to your original question.

 

[Dale]

Any time event is composed of instants.

In standard relativistic terminology an event is a single point in spacetime, I.e. it is determined by four coordinates: t, x, y, z and has zero spatial size and zero duration.

I think that we are getting lost in mathematical details and terminology rather than physics. It would be better to just use the standard terms so that you can focus on understanding the physics instead of arguing about definitions.

 

[Prem1998]

A instant is a single point in spacetime. Just like a point on the number lin, it has size zero. And just like points on the number line, you can add an infinite number of zero-sized points to get something that doesn't add up to zero: There are an infinite number of zero-sized points between one and two on the number line, and the distance between one and two on the number line is clearly not zero

Infinite numbers of zeros are mathematically a bit tricky; much of this stuff is will be covered in a mathematically rigorous college-level calculus class. However, you might be able to form some intuition for how A can find two seconds between two events while B only finds one second between them, yet every instant on A's longer timeline can be paired with a single instant on B's shorter timeline if you google for "Hilbert's Hotel"

I googled it and I don't think it had anything to do with what I'm asking here. The basic problem in it was when it considered that an infinite number of rooms were already occupied. That's like treating infinite like finite. Well, first think of the process of filling those rooms with people. I don't think you can have them fully occupied even with an infinite number of people.
And, coming back to the topic, we're taught that a point has zero length because it is convenient with most problems. AND, we usually work with the real number system. I don't think infinitesimals are allowed here. We can't talk about the number next to zero. The best thing we can have close to infinitesimals is zero.
You don't have to go to college to learn about infinitesimals. In high school calculus, an increment by a single point in a variable x is represented by dx. An infinite number of dx's add up to x. That's where you can't treat a point as zero.
You should google about whether a point has zero or infinitesimal length.

 

[Prem1998]

In standard relativistic terminology an event is a single point in spacetime, I.e. it is determined by four coordinates: t, x, y, z and has zero spatial size and zero duration.

I think that we are getting lost in mathematical details and terminology rather than physics. It would be better to just use the standard terms so that you can focus on understanding the physics instead of arguing about definitions.

I'm sorry about that. I meant a time interval, not an event. SORRY

 

[Dale]

Any time interval is composed of instants. {Quote modified per above}

Ok, you can say something like a given time interval is the set of all instants such that the time is within some specified range. The operation that you use to "compose" a time interval from instants would be a union rather than an addition. Addition of points is not a defined operation.

Again, I think that we are getting lost in terminology and math, I think we have lost sight of your physics question.

 

[Prem1998]

Actually, they don't. The two different times refer to two different spacetime intervals, i.e., the spacetime "lengths" of two different curves. They do not refer to two different "lengths" applied to the same curve.

This doesn't in any way affect the fact that you don't have a correct understanding of how "instants", derivatives, etc. work, as other posters have pointed out. But all of that is actually irrelevant to your original question.

I don't know what that's supposed to mean. Both observers don't see the same instants? Suppose each instant in the ball experiment is defined by the height of the ball at that instant. Suppose the initial height is 1m and final one is 4m. Each of the observers see the ball traveling from 1m to 4m and they see each and every instant that comes in between. They see the instant when the ball is at 2m, at 2.1m, at 2.01m etc. There is a one-to-one correspondence between what they both see. Then, what does it mean that they don't see the same instants?

 

[PeterDonis]

I don't know what that's supposed to mean.

Consider two observers, A and B. We use coordinates in which A is at rest and B is moving at speed $v$ in the $x$ direction. At the spacetime origin, $(x, t) = (0, 0)$, the two observers pass each other. 1 unit of time later (the unit can be seconds, minutes, hours, years, whatever), i.e., when 1 unit has elapsed on A's clock, A is at the event (point in spacetime) $(x, t) = (0, 1)$. The event on B's worldline which is simultaneous with this event, in this frame (A's rest frame), is $(x, t) = (v, 1)$ (I'm using units in which $c = 1$ for simplicity). But on B's clock, only $\sqrt{1 - v^2}$ units of time elapse between $(0, 0)$ and $(v, 1)$.

The shorter elapsed time for B is called "time dilation". But the two times--A's 1 unit and B's $\sqrt{1 - v^2}$ units--refer to the spacetime "lengths" along two different curves: the curve A follows, from $(0, 0)$ to $(1, 1)$, and the curve B follows, from $(0, 0)$ to $(v, 1)$. So the two times do not refer to "the same instants"; they refer to two different sets of "instants", belonging to two different curves.

 

[Prem1998]

Ok, you can say something like a given time interval is the set of all instants such that the time is within some specified range. The operation that you use to "compose" a time interval from instants would be a union rather than an addition. Addition of points is not a defined operation.

No. The whole observation is the union of the individual instantaneous observations. But, a time interval is a number associated with the observation, more like it's magnitude. You have to add up the individual ones to get the total.
And, I don't get why you're still saying a point is zero. We're taught that in sixth standard. In calculus, an increment by a point in a variable is represented by an infinitesimal increment. Why can't we use zeros there? And, again I'm saying the integration (or infinite sum) of dx is x, just like the how an infinite number of points add up to give a line. What about the integration of zero. I don't think that would give you anything finite.
I've not created the definition that a point has an infinitesimal length. It is acceptable in science and mathematics wherever it is required. It is according to the problem whether we think of a point as zero or of infinitesimal length. You should google about it.

 

[jbriggs444]

No. The whole observation is the union of the individual instantaneous observations. But, a time interval is a number associated with the observation, more like it's magnitude. You have to add up the individual ones to get the total.
And, I don't get why you're still saying a point is zero. We're taught that in sixth standard. In calculus, an increment by a point in a variable is represented by an infinitesimal increment. Why can't we use zeros there? And, again I'm saying the integration (or infinite sum) of dx is x, just like the how an infinite number of points add up to give a line. What about the integration of zero. I don't think that would give you anything finite.
I've not created the definition that a point has an infinitesimal length. It is acceptable in science and mathematics wherever it is required. It is according to the problem whether we think of a point as zero or of infinitesimal length. You should google about it.

Integration is not adding up an infinite number of zeroes. It is more subtle than that. What you have written here is wrong and does not support the point you are trying to make.

 

[Prem1998]

Consider two observers, A and B. We use coordinates in which A is at rest and B is moving at speed $v$ in the $x$ direction. At the spacetime origin, $(x, t) = (0, 0)$, the two observers pass each other. 1 unit of time later (the unit can be seconds, minutes, hours, years, whatever), i.e., when 1 unit has elapsed on A's clock, A is at the event (point in spacetime) $(x, t) = (0, 1)$. The event on B's worldline which is simultaneous with this event, in this frame (A's rest frame), is $(x, t) = (v, 1)$ (I'm using units in which $c = 1$ for simplicity). But on B's clock, only $\sqrt{1 - v^2}$ units of time elapse between $(0, 0)$ and $(v, 1)$.

The shorter elapsed time for B is called "time dilation". But the two times--A's 1 unit and B's $\sqrt{1 - v^2}$ units--refer to the spacetime "lengths" along two different curves: the curve A follows, from $(0, 0)$ to $(1, 1)$, and the curve B follows, from $(0, 0)$ to $(v, 1)$. So the two times do not refer to "the same instants"; they refer to two different sets of "instants", belonging to two different curves.

Okay, they don't see same instants. They see different sets of instants. But there is still one-to-one correspondence between the elements (instants) of these different sets, right? An element belonging to the first set when the ball was at 2m can be mapped to a similar element in the second set, right? And, there is a number associated with each element and, hence, with both sets called its duration, right? Then, why do these numbers on the second set add up to give a longer duration? Do the elements of the second set get dilated? Do the instants of the ball experiment remain paused or exist for a longer time when seen by the second observer?

 

[Prem1998]

Integration is not adding up an infinite number of zeroes. It is more subtle than that. What you have written here is wrong and does not support the point you are trying to make.

We're getting a bit off-topic, but I again want to say that we consider a point to be zero whenever it is relevant to the topic. In some problems, we have to consider a point to be something infinitesimal.
Let's integrate x^2 from x=2 to x=4 as the limit of a sum:
We have to add up all the infinite number of values that x^2 takes from x=2 and x=3 at all the infinite points (that will be the length of each elementary rectangle) and we have to multiply that length by the breadth of a point (i.e. h (not zero)).
What's the value of x^2 at x=2? It's 4
Now, what's the next value of x^2 at the point next to 2? Is it (2+0)^2 (0 is a single point's contribution according to you) or should it be (2+h)^2
What about the next value? It is (2+2h)^2, NOT (2+2*0)^2
I think you get that sometimes the contribution of a point is infinitesimal ( and not zero).

 

[PeterDonis]

But there is still one-to-one correspondence between the elements (instants) of these different sets, right?

You are dealing with a continuum, so your intuitions about this, which appear to be derived from discrete sets like the integers, will not work. Yes, there is a one-to-one correspondence between the points on the two curves, but that does not mean the two curves have the same length. You need to take some time to learn real analysis, which is the mathematically rigorous formulation of apparently counterintuitive statements like the one I made in the previous sentence.

what's the next value of x^2 at the point next to 2?

There isn't one. In a continuum, there is no "next point" after a given point. That's one of the wrong intuitions from the integers that you need to unlearn. In the meantime, this thread is closed since you continue to make incorrect statements and are not accepting corrections.