Ageing and travel near the speed of light

[billyharris72]

Hi all:

This post really comes from two things:

 

1) My abortive attempt at reading Einstein's Relativity (I got about halfway).

2) I remember a long time ago a documentary that discussed travelling to Alpha Centauri at 1/10 the speed of light and back (I was a small child at the time - may have been Cosmos but I can't remember). In that documentary it suggested that, for the traveller, when they returned they would find that everyone had aged more than they had.

 

I can't seem to reconcile the first (insofar as I understand it at all) and the second. My problem is as follows:

 

Going back to Einstein's famous railway platform example, I can (sort of) see why, for an observer on the platform, a clock on a speeding train (moving at relativistic speeds) would run slower. (I am assuming here that the direction of the train is not important in this, but could be wrong).

 

However, surely the platform is only "at rest" for the observer on the platform. Relative to an observer on the train the platform is also moving at relativistic speeds. The Lorentz transformation should then apply in the same way, such that if I were on the train, the platform clock would be slower than the clock on the train.

 

So, if you were on the platform and I was on the train, would we both see the other's clock move more slowly than our own?

 

If so, what happens when the train slows down and stops, and the two observers compare clocks? They can't now both be slower than one another, so what happens? Do they now show the same time, or is one slower and if so why?

 

Can anyone set me straight?

 

Billy.

[vlaiv]

Yes, at a first glance it indeed looks like both clocks should be running slower then the other one.

But things are not as straight forward as that. First, that example with someone traveling to neighboring star is oversimplified in that it does not include acceleration and deceleration phase of the journey - which is in domain of general relativity rather than special.

You might also say, ok lets imagine following scenario: no acceleration phase, just clocks. One going off to distant star, and one that you hold in your hand. Upon arriving at the distant star first clock flies by another (third) clock that is heading back to you and they synchronize time in flight, so effectively "correct time is carried back by third clock". What should clocks show (the one you hold and third one) when they meet up again and why?

Right answer is rather complicated, because in order to give right answer you must realize following few points and take them into account:

1. In special relativity simultaneity of events is not guaranteed - this means that that you might perceive distant clocks exchanging / syncing time at different moment than they do. They might see it at the time when they are passing each other by, thus being in vicinity / "same spot", while you might "see it" happen before or after that

2. In your frame of reference distance to the star first clock is traveling to is not the same distance first clock will see you travel in opposite direction. Because length contraction if you say star is 4 Ly away from you, then for flying clock that distance might be much much less (depends on speed that clock is traveling relative to you), so although you are "heading in opposite direction" to that clock at the same speed, you are going to travel much less time to cover that contracted distance.

I'm not sure if anyone worked out what will each value on clock be when they meet up, I did look on internet if above described scenario has been worked out and example values calculated with no luck. I did plan to actually at some point get to it to work thru all the math and special relativity and derive all values in framework of special relativity to show that there indeed is no contradiction.

[andrew s]

The simple answer is the time are different as they follow different world lines. 

There is an old thread "These twins are hurting my head" where I did a diagram to show what happens. I am away at the moment so can't repost the diagram. Search for "These twins"

Regards Andrew

[George Jones]

 

On 10/22/2017 at 09:23, billyharris72 said:

2) I remember a long time ago a documentary that discussed travelling to Alpha Centauri at 1/10 the speed of light and back (I was a small child at the time - may have been Cosmos but I can't remember). In that documentary it suggested that, for the traveller, when they returned they would find that everyone had aged more than they had.

 

As Andrew said, the worldlines, and the experiences on the worldlines, are very different. In order to compare clocks, one observer has to "turn around" and come back. You might argue that each observer sees the other observer turn around and come back, so the situation is symmetrical. Only one observer, however, experiences acceleration during the turn around. To see this asymmetry, consider the following.

Give each observer an accelerometer. Each accelerometer consists of two main parts - a hollow sphere like a basketball inside of which is a slightly smaller sphere. Initially, the centres of the spheres coincide, so that there is a small, uniform gap between the spheres. If an observer is accelerating, the gap will be closed, and contact between the spheres will be made. An alarm that indicates accelerated motion will sound. If the observer is not accelerating, no alarm will sound, and constant speed, straight line motion is indicated.

Only one observer hears the alarm sound.

[rockystar]

George has the answer there - it's to do with turning around. Even if the acceleration were minimised as much as possible, they are still changing direction and actually entering a different frame of reference, whereas the person on earth maintains their frame of reference throughout. 

With the clocks on the train and the platform, you are correct in that both observers would experience each others clocks running slowly. In the case of the traveler to another star, when they change direction they actually "see" the clocks back on earth speeding up and over taking their own until they get back up to their relativistic, constant speed - but there is General Relativity involved her, and that's complicated

This is an excellent course to help understand it: https://www.coursera.org/learn/einstein-relativity/home/welcome it's a bit on an investment in time, but well worth it if you have a genuine interest.

 

[andrew s]

On 10/22/2017 at 17:19, vlaiv said:

First, that example with someone traveling to neighboring star is oversimplified in that it does not include acceleration and deceleration phase of the journey - which is in domain of general relativity rather than special.

This is a mistake I have made in the (recent) past. Special Relativity can handle acceleration provided the gravitational field generated by the energy momentum tensor can be neglected in just the same way as you can use Newtonian physics if the relative velocity is small compared to the speed of light.

SR remains valid for most practical purposes, such as experiments in particle accelerators e.g. LHC.

Here is a reference http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html

Regards Andrew

[George Jones]

The statement "A moving runs slow." needs to be unpacked carefully, as time dilation is not something that is seen visually, at least not directly.

 

On 10/22/2017 at 09:23, billyharris72 said:

Going back to Einstein's famous railway platform example, I can (sort of) see why, for an observer on the platform, a clock on a speeding train (moving at relativistic speeds) would run slower. (I am assuming here that the direction of the train is not important in this, but could be wrong).

 

This depends on what is meant. Suppose that the train is moving towards the platform, that Ted is on the platform, and that Alice is on the train. If Ted keeps one eye on his watch and his other eye on Alice's watch, then Ted sees Alice's second hand spinning faster than his second hand. Alice, however, sees Ted's second hand spinning faster.

Now suppose hat the train is moving away from the platform. If Ted keeps one eye on his watch and his other eye on Alice's watch, then Ted sees Alice's second spinning slower than his second hand. Alice, however, sees Ted's second hand spinning slower.

So the direction of the train does matter for visual effects. Neither of the above scenarios, however, involves the standard mathematical expression for time dilation.

 

I am going to attempt to explain the difference between the above visual effects and time dilation. Unfortunately, my explanation is going to be somewhat complicated, as I will use four clocks, the watches of Ted and Alice, and clocks A and B that are stationary with respect to Ted, and that are directly between Ted and Alice.

Suppose that

1) Alice initially is more than 20 light-years away from Ted (according to Ted)

2) Alice moves directly towards Ted at 80% of the speed of light

3) clock A is 4 light-years from Ted

4) clock B is 20 light-years from Ted.

 

As Alice moves towards Ted, she first encounters clock B and then clock A. Ted watches all this with his telescope. I want to consider three time intervals.

1) T1 = 20 years is the difference in readings on clock A when Alice encounters A and clock B when Alice encounters B.

2) T2 = 12 years is the difference in readings on Alice's watch between when she encounters A and B.

3) T3 = 4 years  is the time that elapses on Ted's watch between when Ted, with his telescope, sees Alice encounter clocks A and B.

 

T1 in Ted's frame is greater than T2 in Alice's frame. This is what is meant by "Moving clocks run slow.", and is given by the time dilation expression. Notice that two different clocks in Ted's fame are used to measure T1.

T3 in Ted's frame is less than T2 in Alice's frame, because, visually, Ted sees Alice's second hand spinning faster than his second hand. This is given by the relativistic Doppler effect, because the Doppler effect applies to all periods, including the periods of second hands.

 

I hope I haven't screwed up my calculations.

[billyharris72]

Thanks George, that's really useful. I'll need to take some time on the calculations, but your response and a few others have made it clear (in a way Einstein's initial presentation did not, at least to me) that the speed of the clock depends not just on the speed of the train relative to light but on its velocity (so whether it is approching or receeding matters). I had suspected that might be a factor but was not sure. That at least provides some sort of intuitive framework for how the problem can be reconciled, as to compare clocks one person would need to turn round and come back to the other one.

Billy.

[vlaiv]

7 minutes ago, billyharris72 said:

Thanks George, that's really useful. I'll need to take some time on the calculations, but your response and a few others have made it clear (in a way Einstein's initial presentation did not, at least to me) that the speed of the clock depends not just on the speed of the train relative to light but on its velocity (so whether it is approching or receeding matters). I had suspected that might be a factor but was not sure. That at least provides some sort of intuitive framework for how the problem can be reconciled, as to compare clocks one person would need to turn round and come back to the other one.

Billy.

Not really true, please have a look at excellent link Andrew posted, there is comprehensive FAQ where you can find all sorts of answers (it is a bit old, so some things have changed in the mean time, like stuff related to gravitational waves and such, but SR is "up to date"). Especially pay attention to following links:

http://math.ucr.edu/home/baez/physics/Relativity/SR/movingClocks.html

http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html

there is something called "Clock Postulate". Measured clock rate is always related exclusively to its speed (not velocity) at any given point in time / instant. For constantly moving clocks it will always run slower. For accelerated clocks there is nice explanation in "moving clocks" link. What it really tells is that when clock is accelerating (decelerating) at any given instant clock rate will depend on speed, but since speed is changing from moment to moment, that change in speed will affect clock rate that we measure in rather unexpected ways. So we can have clock rates going faster and slower (and even in reverse) as seen from our reference frame if our reference frame is not inertial.

[George Jones]

2 hours ago, billyharris72 said:

Thanks George, that's really useful. I'll need to take some time on the calculations, but your response and a few others have made it clear (in a way Einstein's initial presentation did not, at least to me) that the speed of the clock depends not just on the speed of the train relative to light but on its velocity (so whether it is approching or receeding matters).

Yes, it is absolutely true that the visual speed of a watched moving clock depends on velocity, not just speed.

 

2 hours ago, vlaiv said:

Not really true,

 

What is not really true?

 

I use a mish-mash of the Doppler effect, Lorentz contaction, and time dilation in

https://www.physicsforums.com/threads/basic-problem.177939/#post-1384776

to do the calculation.

[vlaiv]

1 hour ago, George Jones said:

Yes, it is absolutely true that the visual speed of a watched moving clock depends on velocity, not just speed.

 

 

What is not really true?

 

I use a mish-mash of the Doppler effect, Lorentz contaction, and time dilation in

https://www.physicsforums.com/threads/basic-problem.177939/#post-1384776

to do the calculation.

I was referring to the statement that time depends on velocity (being a vector) rather than just speed of observed frame of reference, or more precisely that whether it is approaching or receding matters.

It always depends only on observed speed of frame of reference at particular time. Even for accelerated frames of reference, rate of passage of time measured depends on relative speed in that particular instance (granted when there is acceleration speed will not be the same in "next instance" so measured rate of time will also differ). There is additional twist when observer is in accelerated frame of reference: it turns out that measured rate of time in another reference frame depends also on distance to it. This is consequence of fact that information / light propagates at finite constant speed, and in time information reaches distant observer due to acceleration of one of the frames speed difference will increase. Another interesting fact is that if we have single accelerated frame of reference and two clocks in it, spatially separated in direction of acceleration - they will tick at different rates (first clock will see second as ticking slower, while second will see first as ticking faster).