What happens biologically during time dilation?

[danielatha4]

If someone is traveling close to the speed of light, relative to your reference frame, then the time observed on their clocks slows. But what does that mean when it comes to aging? Do their cells that make up their body replicate more slowly? Is there any type of disagreement in the amount of cells replicated in their body between your reference frame and theirs?

 

[DaveC426913]

If someone is traveling close to the speed of light, relative to your reference frame, then the time observed on their clocks slows. But what does that mean when it comes to aging? Do their cells that make up their body replicate more slowly? Is there any type of disagreement in the amount of cells replicated in their body between your reference frame and theirs?

You are thinking about it wrong.

In the frame of reference of the dude moving near c, there is no time distortion effect. There's no experiment he can perform upon himself or anywhere inside his spaceship that will show he's moving near c.

In fact, there is no way you can show that it is not you who is moving. There is no objective frame wherein one of you is not moving. It's all relative (that's why it's called relativity). The dude's frame of reference is exactly as valid as yours - he is stationary, and it is you who are traveling at a near c.

In conclusion, to ask what is happening to his cells is to ask what is happening to your cells.

 

[danielatha4]

Well on average it takes a cell to completely replicate anywhere between 12-24 hours. So we’ll use 24 hours for this example, or 86400 seconds.

So let’s use an example where someone is traveling .8c relative to your reference frame, which we’ll assume is an inertial reference frame.

Event 1 we will call the beginning of replication for a cell, and event 2 will be the completion of replication. So in the observing reference frame

t2 – t1 = 86400 seconds.

But the time elapsed observed from your frame for the person traveling .8c is given by

t’2 – t’1 = (t2 – (vx2/c^2))(gamma) – (t1 – (vx1/c^2)(gamma)

We’ll say at t1 the x coordinate will equal 0 and with simple calculations knowing the person is traveling .8c we know that the x coordinate at t2 is 2.07x10^13m

(86400 – ((.8*2.07E13)/2.99E8)(1/.6)) – 0 = 51936.31s

Which only allows about 3/5 of the replication of the same type of cell in the reference frame of the observer. Therefore, for every 5 cells replicated on your body, you only observe 3 replicated on the person traveling .8c with respect to you.
Am I right about this?

 

[DaveC426913]

Well on average it takes a cell to completely replicate anywhere between 12-24 hours. So we’ll use 24 hours for this example, or 86400 seconds.

So let’s use an example where someone is traveling .8c relative to your reference frame, which we’ll assume is an inertial reference frame.

Event 1 we will call the beginning of replication for a cell, and event 2 will be the completion of replication. So in the observing reference frame

t2 – t1 = 86400 seconds.

But the time elapsed observed from your frame for the person traveling .8c is given by

t’2 – t’1 = (t2 – (vx2/c^2))(gamma) – (t1 – (vx1/c^2)(gamma)

We’ll say at t1 the x coordinate will equal 0 and with simple calculations knowing the person is traveling .8c we know that the x coordinate at t2 is 2.07x10^13m

(86400 – ((.8*2.07E13)/2.99E8)(1/.6)) – 0 = 51936.31s

Which only allows about 3/5 of the replication of the same type of cell in the reference frame of the observer. Therefore, for every 5 cells replicated on your body, you only observe 3 replicated on the person traveling .8c with respect to you.
Am I right about this?

You're still on the wrong track.

Let me ask you this:

If you are watching an old film (you know the ones where the camera is cranked by hand, so everyone is running around faster than normal?) would you check to see whether the cars are fast by a factor of X and the people are walking fast by a factor of Y and the clocks are fast by a factor of Z and then check to see if there's a discrepancy between X and Y and Z?

No, if the film is moving fast, then everything, everything is moving fast exactly the same. They're not moving fast individually; you can say that the entire-world-that-is-the-the-film as a single unit is moving fast.

Furthermore, if the people in the film discussed it amongst themselves (if people-on-cellulose could somehow be alive), they would not see any problem with time. The fact that you are observing them at the wrong speed does not meant here's anything wrong with their world.


So:

It is not the individual components in the dude's spaceship that are slowed down. It is all of time in his frame of reference that is slowed down. And even that is only with respect to you.

There is no absolute time against which both you and the dude can have events measured. The passage of time is a property of the frame of reference you or he are in.

 

[danielatha4]

I understand that in the moving person’s reference frame (that is, moving relative to another frame) everything remains the same, as their frame in an inertial reference frame. Just like what is observed in the other person’s inertial reference frame. I’m talking about their observations relative to each other.

There is less of a difference in time between events one and two for someone traveling an appreciable fraction of the speed of light relative to an inertial frame and IN the inertial reference frame that is observing. Is this not right?

I actually think I got my own logic wrong in the last post. If event 1 indicated the beginning of a cell replicating and time 2 indicated the ending of it. Then the cells would replicate faster for the person traveling with respect to the observer, as it took 3/5 the time to finish replicating as in the observer’s frame, and this is of course IN the observer’s reference frame. NOT in the frame of the one traveling. For the one traveling the effects should be reciprocal. However, I know your still arguing against that claim in general.

Are you claiming that the rate of cell replication is modified to compensate for time dilation, once again relative to the observer?

 

[Doc Al]

If someone is traveling close to the speed of light, relative to your reference frame, then the time observed on their clocks slows. But what does that mean when it comes to aging? Do their cells that make up their body replicate more slowly?

Sure. Every process in the moving frame slows down according to the observing frame. (Of course, in the moving frame itself everything operates as usual. In fact, they see your clocks and bodily processes running slowly.)

Is there any type of disagreement in the amount of cells replicated in their body between your reference frame and theirs?

Not sure what you mean by "disagreement". Say that the rate of cell reproduction is 5 cells per hour. So, in one of your hours^(note), 5 cells replicate on your body and only 3 cells replicate in the moving frame body according to you. Everyone agrees on this.

What's more, the situation is completely symmetric (assuming the moving frame doesn't turn around and come back). The moving frame says the same thing about your cells. How can that be? To understand how they both can say these things requires you to understand not just time dilation, but the relativity of simultaneity. Note that the two events, the start of the first cell replication and the completion of the third in the moving frame, take place at different locations according to you, thus different 'clocks' must be used to record the time that they occur. The moving frame will see your clocks as being out of synch.

Which only allows about 3/5 of the replication of the same type of cell in the reference frame of the observer. Therefore, for every 5 cells replicated on your body, you only observe 3 replicated on the person traveling .8c with respect to you.
Am I right about this?

Sure. But see my comments above.

(Note: One of your hours as reckoned by you. The moving frame will see the time on your clocks change by a much smaller amount during the time that three cells have replicated. The relativity of simultaneity strikes again.)

 

[DaveC426913]

There is less of a difference in time between events one and two for someone traveling an appreciable fraction of the speed of light relative to an inertial frame and IN the inertial reference frame that is observing. Is this not right?

No.This is wrong.

The duration of a second in your frame of reference is one second. The duration of a second in the traveller's frame of reference is one second. Period.

You are mistakenly applying some sort of "universal" time passage to both participants.


Einstein's theory of relativity postulates that his frame of reference is exactly as valid as yours. i.e. you here on planet Earth have no special claim to any universal passage of time. From the frame of reference of the spaceship, it is stationary and you on Earth are moving away at near c.

 

[W.RonG]

check out the similar discussion in this thread: https://www.physicsforums.com/showthread.php?t=382591
Apparently some people think that time dilation is a real slowing down, whereas it is really only an effect of observation between two different inertial frames.
In Special Relativity, all inertial frames are equal; none can be considered an absolute reference for any other.
Ron

 

[TcheQ]

Apparently some people think that time dilation is a real slowing down, whereas it is really only an effect of observation between two different inertial frames.

Including Einstein, the majority of physics graduates, postgrads and PhD's, and the guys who ran this experiment

 

[Fredrik]

check out the similar discussion in this thread: https://www.physicsforums.com/showthread.php?t=382591
Apparently some people think that time dilation is a real slowing down, whereas it is really only an effect of observation between two different inertial frames.
In Special Relativity, all inertial frames are equal; none can be considered an absolute reference for any other.
Ron

Ron, you need to stop making false claims about SR. It's OK to be confused and to ask questions about the things you don't understand, but it's not OK to repeat the same mistakes over and over after they've been pointed out to you.

No one is saying that time is slowing down in an absolute sense, or that there's a preferred frame. When A's clock is slow in B's rest frame, B's clock is slow in A's rest frame as well. Both are correct to say that "the other clock is slow". That's not a statement about something absolute. What the statement really means is that the coordinate system that's naturally associated with B's world line is assigning time coordinates to the events where A's clock is present that are higher than the numbers that are displayed by A's clock at those events.

These are not controversial claims. Everyone who understands SR agrees that this is what SR is saying.

 

[Mentz114]

Apparently some people think that time dilation is a real slowing down, whereas it is really only an effect of observation between two different inertial frames.

I agree. Time dilation is an observer dependent effect, differential ageing is not.

(Fredrik, did you misread this post, perhaps ? It doesn't seem controversial.)

 

[W.RonG]

... No one is saying that time is slowing down in an absolute sense, or that there's a preferred frame. ...

actually that is what several posters are claiming.
Ron

 

[Mentz114]

No one is saying that time is slowing down in an absolute sense, or that there's a preferred frame. ...

actually that is what several posters are claiming.

Name names. Who is claiming this ?

 

[W.RonG]

Name names. Who is claiming this ?

Since you asked, check out 'atyy' and 'sylas' in the thread I linked earlier, and apparently 'TcheQ' above. Not sure where 'stevmg' stands, seems to be leaning to the other side as well.
I think the difference between me and Fredrik is semantic, where I say clock B appears to observer A to be ticking slower, Fredrik says that clock B is ticking more slowly in A's inertial frame. as long as he doesn't believe clock B actually slows down in its own ('B') inertial frame, then we are on the same page. that's how I read his postings.
Ron

 

[W.RonG]

another one early on. 'Ich' typed:
"This is not about the aging process, and not abhout different conditions the twins have been living in. For one twin, 10 years passed, for the other 5. With all consequences."
Ron

 

[DrGreg]

There is a problem here over the use of language. When people try to answer questions such as "Is time dilation real?" or other variants, different people interpret the words in different ways. There's no unambiguous definition of "real" that everyone agrees with.

If two observers begin together, then separate, then come back together again, then there is no doubt, experimentally, whether the two observers have elapsed the same time or not. You just look at the two clocks , which are both co-located, before and after, and read off the values. In this case we are not making any instantaneous comparison of clock rates, we are comparing the two average clock rates over the whole journey.

On the other hand, if you want to compare two clocks which are separated by a distance, there is no unambiguous way of doing this. To make an instantaneous comparison you need to read both clocks simultaneously, and in relativity different observers disagree over what "simultaneous" means, with no one definition taking priority over any other. Thus whether clock A is instantaneously ticking faster, slower or at the same rate as clock B depends entirely on who is making the comparison. There is no absolute answer to that question.

In the case of the twins' paradox, everyone agrees that the traveling twin (who is not inertial for at least part of the journey) ages less than the stay-at-home inertial twin averaged over the whole journey, but observers disagree over the instantaneous rates at any point during the journey.

 

[W.RonG]

... In the case of the twins' paradox, everyone agrees that the traveling twin (who is not inertial for at least part of the journey) ages less than the stay-at-home inertial twin averaged over the whole journey ... .

there's another one.
I think y'all need to hash this over and decide which version of Special Relativity you believe in - the one where all inertial frames are equal, or the one where some inertial frames are more equal than others.
Either way it's been an interesting weekend. Toodles.
Ron

 

[Doc Al]

there's another one.
I think y'all need to hash this over and decide which version of Special Relativity you believe in - the one where all inertial frames are equal, or the one where some inertial frames are more equal than others.
Either way it's been an interesting weekend. Toodles.
Ron

You seemed to have missed the point that one of the twins accelerates and thus does not remain in a single inertial frame. Basic stuff.

 

[Mentz114]

W.ron.G,

I think you might be conflating two phenomena

1. Time dilation : where two IRFs see one and others clocks running slower. This depends on the relative speed and seems paradoxical ( but isn't ).

2. Differential ageing : this is not dependent on relative velocity at one instant, but the proper length of the journeys before the clocks are brought together and compared. All IRFs agree on this.

This is what DrGreg and earlier posts are saying.

 

[JesseM]

there's another one.
I think y'all need to hash this over and decide which version of Special Relativity you believe in - the one where all inertial frames are equal, or the one where some inertial frames are more equal than others.
Either way it's been an interesting weekend. Toodles.
Ron

Ron, did you look at the numerical example I gave back in post 63 of the other thread? I showed that in an example where one twin (Stella) travels away from another twin (Terence) and then turns around and returns, you can analyze it from the perspective of two different frame, both of which are considered "equal" and which disagree about the rates the two twins were aging during each phase of the trip, yet both frames end up agreeing on the total amount that each twin has aged at the point when they reunite, with the twin who turned around having aged less in total.

As Mentz114 said, you have to differentiate between the rate that each twin is aging (or that their clock is ticking) at a particular moment or during a particular phase of the trip, which different inertial frames can disagree on, and the total time elapsed for each twin during the course of the entire trip from start to finish, which all frames actually agree on despite disagreeing about the rates at different moments. It turns out that in SR all frames always agree in their predictions about local events that happen at a single point in space and time, like two twins meeting and comparing their ages at a single location and a single moment, even though they can disagree about other things like which of two twins is aging faster at a particular moment, or even things like which of two twins that are far apart celebrates their 40th birthday first (this is the relativity of simultaneity, which says that when it comes to events at different locations in space, different frames can sometimes disagree on which happened earlier and which happened later...so different frames can disagree about which twin is older when they're far apart, just not when they actually meet at a single location).

 

[Dale]

I think y'all need to hash this over and decide which version of Special Relativity you believe in - the one where all inertial frames are equal, or the one where some inertial frames are more equal than others.

All inertial reference frames are equal and they are related to one another via the Lorentz transform. The proper time elapsed by a clock is a Lorentz invariant and all reference frames agree on its value. In particular, all reference frames agree that the non-inertial twin is the one that ages less. I strongly recommend looking into JesseM's example and asking specific questions about any points that you don't understand.

 

[DrGreg]

... In the case of the twins' paradox, everyone agrees that the traveling twin (who is not inertial for at least part of the journey) ages less than the stay-at-home inertial twin averaged over the whole journey...

there's another one.
I think y'all need to hash this over and decide which version of Special Relativity you believe in - the one where all inertial frames are equal, or the one where some inertial frames are more equal than others.
Either way it's been an interesting weekend. Toodles.
Ron

You've just ignored the words highlighted in bold above.

I don't think there's anything for us to "hash over", we all seem to be agreeing with each other, but somehow you're not understanding that. I'm not sure why.

 

[danielatha4]

Can anyone get back to my main point?

Did I do my calculations right? and can I clearly assume that for 86400 seconds in my frame, about 53000 seconds has passed in their frame given the initial conditions I stated?

and if so, how would this pertain to aging? particularly if we were to call event 1 the beginning of the replication of a cell and event 2 the ending of the replication.

 

[Mentz114]

Can anyone get back to my main point?

Did I do my calculations right? and can I clearly assume that for 86400 seconds in my frame, about 53000 seconds has passed in their frame given the initial conditions I stated?

and if so, how would this pertain to aging? particularly if we were to call event 1 the beginning of the replication of a cell and event 2 the ending of the replication.

I have not checked your calculation but I'm assuming it's correct. I also assume you're comparing clocks that are in the same place. The difference in elapsed time you calculated applies to all processes. Biology is governed by the laws of physics, so ageing, or cell replication or any process biological or other, would agree with the clocks.

 

[JesseM]

Can anyone get back to my main point?

Did I do my calculations right?

If the cell is moving at 0.8c in your frame, and the time between the events on the cell's worldline is 86400 seconds in your frame, then just going by the time dilation equation, the time between these events in the cell's frame would actually be 86400*0.6 = 51840 seconds (the equations in your calculation looked right so it's probably just roundoff error...as a tip, it's easiest to use units where c=1, so if you measure time in seconds, measure distance in light-seconds rather than meters).

and can I clearly assume that for 86400 seconds in my frame, about 53000 seconds has passed in their frame given the initial conditions I stated?

The language here is a little ambiguous--when 86400 seconds pass on your clock, it's true that only 51840 seconds pass on their clock (including a biological clock) as seen in your frame. But in their frame, when 51840 seconds pass on their clock only 51840*0.6 = 31104 seconds have passed on your clock. There is no objective sense in which one of your clocks is running slower than the other's.

You can think of this in terms of the relativity of simultaneity, which says that for events which happen at different locations, different frames can disagree about whether they are simultaneous (happen at the same time coordinate) or non-simultaneous (happen at different time coordinates)...in this case, in your frame the event of their clock reading 51840 seconds is simultaneous with the event of your clock reading 86400 seconds, but in their frame the event of their clock reading 51840 seconds is simultaneous with the event of your clock reading 31104 seconds (assuming here that you departed from one another when both your clocks read 0 seconds).

and if so, how would this pertain to aging? particularly if we were to call event 1 the beginning of the replication of a cell and event 2 the ending of the replication.

If both of you are moving apart inertially, then each of you will say that the other one is aging slower in your own rest frame. But if one of you stops moving inertially and accelerates to turn around so you can reunite at a single location, then all frames will agree that the one who turned around will have aged less when you reunite.

 

[DaveC426913]

Did I do my calculations right? and can I clearly assume that for 86400 seconds in my frame, about 53000 seconds has passed in their frame given the initial conditions I stated?

and if so, how would this pertain to aging? particularly if we were to call event 1 the beginning of the replication of a cell and event 2 the ending of the replication.

I urge to address the points made to you, before blindly following your calculations.

Are you aware that time itself is a property of the frame you are in, not some universal property against which all observers are measured? Are you aware that it makes little sense to discuss how long cells take to divide in different frames? If a call takes 3 minutes to divide on Earth then it takes 3 minutes to divide anywhere else in the universe at any speed?

Please - before posting more calculations - acknowledge whether you accept this or whether you reject it.

 

[JesseM]

If a call takes 3 minutes to divide on Earth then it takes 3 minutes to divide anywhere else in the universe at any speed?

The phrasing here might be confusing...a cell always takes 3 minutes to divide from the perspective of the reference frame where the cell is at rest, not from the perspective of other frames. I suppose you could also say that a cell always takes 3 minutes of proper time to divide, and proper time (the time between two events on an object's worldline as measured by a clock moving along that worldline) is frame-invariant.

 

[DaveC426913]

The phrasing here might be confusing...a cell always takes 3 minutes to divide from the perspective of the reference frame where the cell is at rest, not from the perspective of other frames.

Yes. I want the OP to acknowledge this before moving on.

The OP seems to be under the impression that time passes in some absolute fashion such that cell division takes fewer or more seconds depending on how fast you're moving. I wish to disabuse him of this notion.

 

[Fredrik]

(Fredrik, did you misread this post, perhaps ? It doesn't seem controversial.)

I took his post to be an extension of what he said in the thread he linked to. In this post, he said that the twins will be the same age when they meet again.

 

[Frame Dragger]

Yes. I want the OP to acknowledge this before moving on.

The OP seems to be under the impression that time passes in some absolute fashion such that cell division takes fewer or more seconds depending on how fast you're moving. I wish to disabuse him of this notion.

Agreed. Not to mention that as anything within its own inertial frame will have a normal tick rate "to itself at least", the idea of biological effect makes no sense unless you're in MANY different IRFs at the same time (in which case cell division is the least of your worries).

 

[sylas]

... the idea of biological effect makes no sense unless you're in MANY different IRFs at the same time (in which case cell division is the least of your worries).

I just noticed this. Spot on.. with the bonus of hilarious mental images.

 

[Frame Dragger]

I just noticed this. Spot on.. with the bonus of hilarious mental images.

Thank you very much! :) Just remember my name; when the IRFs start-ah-draggin', start runnin'!

 

[Dale]

the idea of biological effect makes no sense unless you're in MANY different IRFs at the same time (in which case cell division is the least of your worries).

But in fact you are always in ALL possible inertial reference frames and even in ALL possible non-inertial reference frames (with no resulting biological worries) whose coordinate chart includes your worldline.

 

[Frame Dragger]

But in fact you are always in ALL possible inertial reference frames and even in ALL possible non-inertial reference frames (with no resulting biological worries) whose coordinate chart includes your worldline.

Ok, yes... but to be fair that last bit "...whose coordinate chart includes your worldline" is the critical bit in this context. I was of course, thinking of a far more extreme situation, as Sylas realized; say, being in the ergoregion of a Kerr BH. You're right though, and I shouldn't forget the basics.

 

[danielatha4]

Are you aware that time itself is a property of the frame you are in, not some universal property against which all observers are measured?

Yes. In an IRF, a cell that takes, for example, 24 hours to divide (measured by a clock that is at rest to them) will divide in 24 hours in EVERY IRF no matter what. Two people on different sides of the universe will see this cell in their body divide in 24 hours, according to their clock, regardless of their velocity relative to anybody. As long as they have an inertial reference frame. I'm asking the scenario of someone observing someone else's reference frame relatively, in which their clock is no longer local to the cell division taking place.

Are you aware that it makes little sense to discuss how long cells take to divide in different frames?

Little sense, maybe. Isn't this esentially what the twin paradox comes down to? Everyone interested in physics was probably introduced to the idea of time dilation using the example that someone will age less when moving at significant velocity. I understand that it is a very elementary, and unsophisticated approach, as they don't discuss anything about how the measurements are relative. I'm trying to understand EXACTLY how someone would be younger. I find it logical to believe that for someone to be younger relative to someone then the relativistic observation had to have altered something biologically. At least, relatively, not in anybody's own inertial reference frame.

If a call takes 3 minutes to divide on Earth then it takes 3 minutes to divide anywhere else in the universe at any speed?

Yes, completely understood. As long as the cell is at rest relative to the observer watching it divide and as long as their reference frame is inertial, it will take 3 minutes. I honestly don't know if I'm COMPLETELY missing something you're trying to tell me.

 

[JesseM]

Yes. In an IRF, a cell that takes, for example, 24 hours to divide (measured by a clock that is at rest to them) will divide in 24 hours in EVERY IRF no matter what.

Every frame will agree that 24 hours elapses on the clock at rest relative to the cell, but in terms of the clocks at rest in other frames, the cell may take quite a lot less than 24 hours to divide.

I'm asking the scenario of someone observing someone else's reference frame relatively, in which their clock is no longer local to the cell division taking place.

The language is a little awkward here, a reference frame is just a coordinate system, so you can't really observe "someone else's reference frame", you can only observe physical processes like clock ticks and cell division. And inertial reference frames are defined in terms of a hypothetical network of clocks and rulers at rest relative to each other and spread throughout space, with the time of any event (including an event on the worldline of an object moving at relativistic speed relative to this network) defined in terms of local readings on this network. For example, if I want to know the coordinates in my frame of the event of an alarm going by on a clock moving at 0.8c relative to me, I might see that when the alarm went off the clock was next to the 1.3 light-years mark on my x-axis ruler, and the clock in my network that was located at the 1.3 light-year mark read 0.7 years as the speeding clock went past it and the alarm went off, so I'd assign the event of the alarm going off coordinates x=1.3 light years, t=0.7 years in my frame. Likewise, if I was measuring cell division of a cell moving in my frame, I would still (ideally) use clocks that were "local to the cell division taking place" to measure events that happen to the cell, it's just that these clocks wouldn't be at rest relative to the cell.

Anyway, sorry if I'm being overly nitpicky about your choice of words, just thought this discussion might be helpful...but if I understand your meaning, you want to know what happens to the time for cell division when it's measured in a frame where the cell is in motion. In this case the answer is that the time for the cell to divide will be lengthened in that frame (time dilation).

I'm trying to understand EXACTLY how someone would be younger. I find it logical to believe that for someone to be younger relative to someone then the relativistic observation had to have altered something biologically. At least, relatively, not in anybody's own inertial reference frame.

The problem with seeing it in terms of any process being objectively slowed down is that although all frames agree on the total aging of each twin in the twin paradox, they disagree about which twin was aging faster during particular phases of the journey, and there is no objective basis for judging one frame more correct than another. I gave an example of this in post 63 of this thread:

you can analyze the problem from any inertial frame and all will have the same answer about the age of the inertial twin and the age of the non-inertial twin when they reunite. Let's call the inertial (Earth-bound) twin "Terence" and the traveling twin "Stella", following the Twin Paradox FAQ. First let's look at the numbers in Terence's rest frame. Suppose that in this frame, Stella travels away from Terence inertially at 0.6c for 10 years, at which point she is at a distance of 0.6*10 = 6 light-years from Earth in this frame, then she turns around (i.e. she accelerates, a non-inertial motion which will cause her to experience G-forces that show objectively that she wasn't moving inertially) and heads back towards Terence at 0.6c, finally reuniting with Terence after 20 years have passed since her departure in this frame. Since Terence is at rest in this frame, he has aged 20 years. But since Stella was moving at 0.6c in this frame, the time dilation formula tells us her aging was slowed down by a factor of \sqrt{1 - 0.6^2} = 0.8, so she only aged 0.8*10 = 8 years during the outbound leg of her trip, and another 0.8*10 = during the inbound leg, so she has only aged 16 years between leaving Earth and returning.

Now let's analyze the same situation in a different inertial frame--namely, the frame where Stella was at rest during the outbound leg of her trip (she can't also be at rest during the inbound leg in this frame, since this is an inertial frame while Stella accelerated between the two legs of the trip). In this frame, Terence on Earth is initially moving away from Stella at 0.6c while she remains at rest. In Terence's frame, remember that Stella accelerated when she was 6 light-years away from Earth, so we can imagine she turns around when she reaches the far end of a measuring-rod at rest in Terence's frame and 6 light-years long in that frame, with Terence sitting on the near end; in the frame we're dealing with now, the measuring-rod will therefore be moving along with Terence at 0.6c, so it'll be shrunk via length contraction to a length of only 0.8*6 = 4.8 light-years. So, Stella accelerates when the distance between her and Terence is 4.8 light-years in this frame, and since Terence as moving away from her at 0.6c in this frame, they will be 4.8 light-years apart after 4.8/0.6 = 8 years have passed. During these 8 years, it is Terence's aging that is slowed down by a factor of 0.8, so while Stella ages 8 years during this leg, Terence only ages 0.8*8 = 6.4 years. Then Stella accelerates to catch up with Terence, while Terence continues to move inertially at 0.6c. Using the relativistic velocity addition formula, if Stella was moving at 0.6c in Terence's frame and Terence is moving at 0.6c in the same direction in this frame, then in this frame Stella must be moving at (0.6c + 0.6c)/(1 + 0.6*0.6) = 0.88235c during the inbound leg. And since Terence is still moving at 0.6c in the same direction, the distance between Stella and Terence will be closing at a "closing speed" of 0.88235c - 0.6c = 0.28235c. Since the distance was initially 4.8 light years at the moment Stella accelerated, in this frame it will take 4.8/0.28235 = 17 years for Stella to catch up with Terence on Earth. During this time Terence has aged another 0.8*17 = 13.6 years, so if you add that to the 6.4 years he had aged during the outbound leg, this frame predicts he has aged 20 years between Stella leaving and Stella returning, same as in Terence's frame. And since Stella is traveling at 0.88235c her aging is slowed by a factor of \sqrt{1 - 0.88235^2} = 0.4706, so during those 17 years in this frame she only ages 0.4706*17 = 8 years during the inbound leg. If you add that to the 8 years she aged during the outbound leg, you find that this frame predicts she has aged 16 years between departing and returning, which again is the same as what was predicted in Terence's frame.

So, to sum up:

Aging between event of Stella departing Earth and event of Stella turning around to return to Earth (i.e. total aging during the outbound leg of the journey):

--in Terence's rest frame, Terence (who was at rest) aged 10 years and Stella (who was moving at 0.6c) aged 8 years between these events (so Stella was aging slower during the outbound leg in this frame).
--in second frame where Stella was at rest during outbound leg, Terence (who was moving at 0.6c) aged 6.4 years and Stella (who was at rest) aged 8 years between these events (so Terence was aging slower during the outbound leg in this frame).
--the coordinate time between these events was 10 years in Terence's frame, 8 years in the second frame.

Aging between event of Stella turning around and event of Stella arriving back at Earth (i.e. total aging during the inbound leg of the journey):

--in Terence's rest frame, Terence (who was at rest) aged 10 years and Stella (who was moving at 0.6c) aged 8 years between these events (so Stella was aging slower during the inbound leg in this frame).
--in second frame where Stella was at rest during the outbound leg, Terence (who was moving at 0.6c) aged 13.6 years and Stella (who was moving at 0.88235c) aged 8 years between these events (so Stella was aging slower during the inbound leg in this frame, by an even greater ratio than in Terence's frame).
--the coordinate time between these events was 10 years in Terence's frame, 17 years in the second frame.

Both frames agree that when Stella returns to Earth and meets Terence, Terence has aged 20 years while Stella has only aged 16 years. But clearly they don't agree on the details of the rates each of them were aging during each phase of the journey, and there is no basis for preferring one perspective over the other.

 

[DaveC426913]

Yes, completely understood. As long as the cell is at rest relative to the observer watching it divide and as long as their reference frame is inertial, it will take 3 minutes. I honestly don't know if I'm COMPLETELY missing something you're trying to tell me.

OK, good. Just wanted to make sure we weren't on completely different pages. Sorry to beat it to death.

 

[danielatha4]

Every frame will agree that 24 hours elapses on the clock at rest relative to the cell, but in terms of the clocks at rest in other frames, the cell may take quite a lot less than 24 hours to divide.

The language is a little awkward here, a reference frame is just a coordinate system, so you can't really observe "someone else's reference frame", you can only observe physical processes like clock ticks and cell division. And inertial reference frames are defined in terms of a hypothetical network of clocks and rulers at rest relative to each other and spread throughout space, with the time of any event (including an event on the worldline of an object moving at relativistic speed relative to this network) defined in terms of local readings on this network. For example, if I want to know the coordinates in my frame of the event of an alarm going by on a clock moving at 0.8c relative to me, I might see that when the alarm went off the clock was next to the 1.3 light-years mark on my x-axis ruler, and the clock in my network that was located at the 1.3 light-year mark read 0.7 years as the speeding clock went past it and the alarm went off, so I'd assign the event of the alarm going off coordinates x=1.3 light years, t=0.7 years in my frame. Likewise, if I was measuring cell division of a cell moving in my frame, I would still (ideally) use clocks that were "local to the cell division taking place" to measure events that happen to the cell, it's just that these clocks wouldn't be at rest relative to the cell.

Anyway, sorry if I'm being overly nitpicky about your choice of words, just thought this discussion might be helpful...but if I understand your meaning, you want to know what happens to the time for cell division when it's measured in a frame where the cell is in motion. In this case the answer is that the time for the cell to divide will be lengthened in that frame (time dilation).

The problem with seeing it in terms of any process being objectively slowed down is that although all frames agree on the total aging of each twin in the twin paradox, they disagree about which twin was aging faster during particular phases of the journey, and there is no objective basis for judging one frame more correct than another. I gave an example of this in post 63 of this thread:

So, to sum up:

Aging between event of Stella departing Earth and event of Stella turning around to return to Earth (i.e. total aging during the outbound leg of the journey):

--in Terence's rest frame, Terence (who was at rest) aged 10 years and Stella (who was moving at 0.6c) aged 8 years between these events (so Stella was aging slower during the outbound leg in this frame).
--in second frame where Stella was at rest during outbound leg, Terence (who was moving at 0.6c) aged 6.4 years and Stella (who was at rest) aged 8 years between these events (so Terence was aging slower during the outbound leg in this frame).
--the coordinate time between these events was 10 years in Terence's frame, 8 years in the second frame.

Aging between event of Stella turning around and event of Stella arriving back at Earth (i.e. total aging during the inbound leg of the journey):

--in Terence's rest frame, Terence (who was at rest) aged 10 years and Stella (who was moving at 0.6c) aged 8 years between these events (so Stella was aging slower during the inbound leg in this frame).
--in second frame where Stella was at rest during the outbound leg, Terence (who was moving at 0.6c) aged 13.6 years and Stella (who was moving at 0.88235c) aged 8 years between these events (so Stella was aging slower during the inbound leg in this frame, by an even greater ratio than in Terence's frame).
--the coordinate time between these events was 10 years in Terence's frame, 17 years in the second frame.

Both frames agree that when Stella returns to Earth and meets Terence, Terence has aged 20 years while Stella has only aged 16 years. But clearly they don't agree on the details of the rates each of them were aging during each phase of the journey, and there is no basis for preferring one perspective over the other.

Right, so Stella ages 16 years and Terence has aged 20 years. Now let's suppose that each were carrying a petri dish with, two cells initially. Let's say that both start with the same types of cells and cell divides once every day. Let's give the number of cells an ideal exponential equation to explain it's growth.

n=2^365t where t is in years

So, Stella's cells have now created 2^16*365 cells which is 1.03*10¹⁷⁵⁸ cells

However, Terence's cells have created 2^20*365 cells which is 3.30*10²¹⁹⁷

Terence's 2 cells have created, in only 25% more time, 3.2*10⁴⁴¹% more cells

 

[JesseM]

Terence's 2 cells have created, in only 25% more time, 3.2*10⁴⁴¹% more cells

Yes, that's just how exponential growth works, nothing to do with relativity. If you had two petri dishes on Earth at rest relative to each other, and you allowed one to grow for 16 years before freezing it and the other to grow for 20 years before freezing it, then despite the fact that the second had only 25% more time to grow before being frozen, it grew 3.2*10⁴⁴¹% more cells too.

Anyway, did you get my point about how different frames disagree about whether Terence or Stella aged faster during the outbound leg of the trip, even though they all agree on how much they aged over the entire trip? And do you understand that in SR there can be no basis for preferring one frame's perspective over another's?

 

[danielatha4]

Yes, I understood your point. Let me see if I got this right.

In Terence’s frame:

During outbound trip: Terence aged 10 years, and Stella aged 8 years. Therefore, Stella
aged slower by 20%

During inbound trip: Terence aged 10 years, and Stella aged 8 years. Therefore, Stella aged slower by 20%

In Stella’s frame:

During outbound trip: Terence aged 6.4 years, and Stella aged 8 years. Therefore, Terence aged slower by 20%

During inbound trip: Terence aged 13.6 years, and Stella aged 8 years. Therefore, Stella aged slower by 41%

I think the point I was trying to make about the exponential growth is that according to the billions of people on Earth Stella has created significantly less cells than everyone else in 20 years.

However, if time is slowing for Stella at any point in the trip then wouldn’t that increase a rate, which inversely proportional to time passed? For example, in the exponential growth equation n=2^t the change in time does not decrease for Stella, but more cells are able to divide during that time due to a decrease in the flow of time. Therefore, effectively increasing the rate at which the cells divide.

 

[sylas]

Yes, I understood your point. Let me see if I got this right.

In Terence’s frame:

During outbound trip: Terence aged 10 years, and Stella aged 8 years. Therefore, Stella
aged slower by 20%

During inbound trip: Terence aged 10 years, and Stella aged 8 years. Therefore, Stella aged slower by 20%

Yes.

In Stella’s frame:

Oops. There is no one inertial frame associated with Stella.

During outbound trip: Terence aged 6.4 years, and Stella aged 8 years. Therefore, Terence aged slower by 20%

During inbound trip: Terence aged 13.6 years, and Stella aged 8 years. Therefore, Stella aged slower by 41%

No. "During" the two trips, both inbound and outbound, Stella is motionless in an inertial frame, and in that frame Terence ages 6.4 years, each time. The only reason this appears paradoxical is because people simply assume that the end of the first period is the same as the start of the second period.

What makes a difference is that when Stella turns around, you get a change in what is simultaneous. Hence, as Stella approaches the turn around point (or, in Stella's rest frame, as the turn around point approaches Stella) Terence is "now" 6.4 years older than at departure.

Then Stella turns around, and is in a new rest frame. In this new rest frame, Terence is "now" 13.6 years older than at the departure point. This is the key thing that lies behind pretty much every confusion on the twin thought experiment. Simultaneity is relative. When you try to identify things happening "at the same time" as Stella suddenly reversing direction, it makes a difference whether you use simultaneous in the inward rest frame or the outward rest frame.

Cheers -- sylas

 

[JesseM]

Yes, I understood your point. Let me see if I got this right.

In Terence’s frame:

During outbound trip: Terence aged 10 years, and Stella aged 8 years. Therefore, Stella
aged slower by 20%

During inbound trip: Terence aged 10 years, and Stella aged 8 years. Therefore, Stella aged slower by 20%

In Stella’s frame:

During outbound trip: Terence aged 6.4 years, and Stella aged 8 years. Therefore, Terence aged slower by 20%

During inbound trip: Terence aged 13.6 years, and Stella aged 8 years. Therefore, Stella aged slower by 41%

As sylas said, the second frame cannot be said to be "Stella's frame" throughout the trip, it is the inertial frame where Stella was at rest during the outbound trip but moving at 0.88235c during the inbound trip. One could equally well do the analysis from the perspective of an inertial frame where Stella was at rest during the inbound trip, and in this frame it'd be the reverse of the above, with Stella aging 41% slower during the outbound trip and Terence aging 20% slower during the inbound trip. This frame, too, would agree that Stella had aged 16 years and Terence had aged 20 when they reunited.

However, if time is slowing for Stella at any point in the trip then wouldn’t that increase a rate, which inversely proportional to time passed?

But do you see from the above that there can be no frame-independent sense in which "time is slowing for Stella" during any phase of the trip? No matter what phase you're talking about, there are some inertial frames that say Stella was aging slower during that phase and other inertial frames that say Terence was aging slower, but they all make the same predictions about local physical events like what age each twin is when they reunite, so there's no physical basis for preferring any of these frames.