# Time dilation and aging

1. May 11, 2015

### Threebrane

Hi all,
First post here.
I have been debating with a collegue today about time dialation. I am certain that an astornaut in space will experience more time or age faster than people on earth due to time dialation of a massive object.
He is saying that he will come back younger. Whos right? and why?
Second. Speed also dictates time.
So if i was in a 2 ton vehicle in space how fast would that vehicle need to travel in order to experience the same time lapse as on earth?

2. May 11, 2015

### Staff: Mentor

Where in space? In low earth orbit, the effect from the high speed dominates, and the astronaut ages more slowly. In high earth orbits, the gravitational effect dominates, and the astronaut ages faster. Outside the influence of earth, it depends on what the astronaut is doing.
That question does not make sense.

3. May 11, 2015

### rede96

An astronaut who is in a geosynchronous orbit (orbiting at the same speed as the earth turns) will in fact age less than someone who stays on earth. He doesn't get 'younger' it is just that time will pass slower for him relative to someone on earth. For him time will seam to pass as normal.

Ignoring the weight of the vehicle for now, time does pass at a different rate for someone who is moving relative to you. So if an astronaut took a journey to Mars and back and you stayed on the earth, then he will have aged less than you when he returned.

Without going into the calculations, time dilation due to the mass of the earth is quite small compared to someone travelling away from us at a few thousand miles per hour for a year or so and then returning.

4. May 11, 2015

### Staff: Mentor

No, he won't; he will age more. The altitude of a geosynchronous orbit is high enough that the speeding up of clock rates due to altitude is more than the slowing down of clock rates due to motion.

5. May 12, 2015

### yoron

Time dilations are comparisons between clocks. Your proper time is always the same, as in the one you measure by your wrist watch. When you find a time dilation it will be your wrist watch you trust to define that 'time dilation' by (naively expressed), and standing by that definition your proper time becomes a constant, just as 'c' is a local constant, no matter uniform (relative) motion. You can if you like exchange your wristwatch for the 'far away' observers wrist watch, but doing so you fool yourself as the 'slower or faster' clock you now go by still is defined relative your local. Unless you 'forget' to compare, in which case it doesn't matter for the rate of 'time', just as it doesn't matter whether I measure it in minutes or in seconds.
=

Another way to express it is to call it comparisons between frames of reference. But you always need a base for your comparison, and that one is locally invariant in the same manner that you will find 'c' to be invariant. It makes very little sense to me, using any other definition than the local as the base for your measurement, or experiment. A repeatable experiment is the result of several local measurements, at different SpaceTime positions that taken together, prove a concept as valid for a 'whole SpaceTime'.

I better point out that the definition of a 'repeatable experiment' to me indicate the need of 'constants' as 'c' and a invariant local clock. Otherwise they can't be equivalent, and that should make physics a quagmire. Unless you can redefine Relativity to not being relative at all, defining some 'Globally held gold standard' of for example 'time', or for that case 'c'. And this is also a example of why I've started to wonder about 'container models', as I call them. The unspoken presumptions we all make of the, common for us all, universe existing. What I see defining it is causality, locally definable constants, rules and principles, and the scale you use.

Last edited: May 12, 2015
6. May 12, 2015

### Threebrane

The reason behind that question was probably due to an incomplete understanding of the subject...
my thought was; speed and mass affect time.
The understanding i have is that the speed of the earth combined with its mass (Speed x Mass=TD?) affects time here on earth.
Now if an object was in space that had a very small mass (such as a 2 ton space vehicle, i understand that is weight, but an estimate of its density and or mass would somehow be represented by its weight here on earth if it is made of common metal components) In order to affect time equally to the time dialation of earth, how fast would this space craft need to travel? Or possibly i have just shown my ignorance hand already on my second post... Or maybe on the first one :)

Umm... Like i mentioned earlier i am not an expert on this by any means... In fact most of my information came from Morgan Freeman, Carl Sagan and Neil Degrasse Tyson... And the book the elegant universe... I know only the explained Time dialation as expressed in Time dialation for dummies... The math escapes me...BUT...
If you were in orbit with the GPS sattelites that are orbiting at 14,000km per hour. Your clock would tick 38 microseconds faster per day. So wouldnt an astronaut experience more time while in orbit?

Yoron:
I think i get what you are saying. I am always comparing the atomic clocks on earth to our hypothetical time travelers "atomic wrist watch".
But i think i need to read your comment a couple more times to capture exactly what you are saying.

7. May 12, 2015

### yoron

Well, it's my view of it, and it's been taking quite some time for me too to see what I mean :) I don't see my view of it invalidating a 'twin experiment', although to me it place the time dilation as a result of frames of reference interacting. Which actually is one reason more why I wonder about 'container models' validity. It's not that a 'common container' doesn't exist, we communicate and we have four dimensions measurably, it's just how it come to be that keeps getting me confused.

8. May 12, 2015

### rede96

Doh... it was late :)

9. May 12, 2015

### Staff: Mentor

This is not a good way to look at it. First of all, "time" is ambiguous; you need to specify whether you mean coordinate time (the number assigned as the "time coordinate" to various events) or proper time (the time actually recorded on clocks traveling on various worldlines through spacetime). More importantly, you need to understand that we are talking about spacetime, so "time" is not something separate from space; they are connected. Also, if you are looking at "time" being different for different objects, you have to be specific about which objects and what worldlines they are traveling on.

The general rule is that an object's path through spacetime determines the proper time elapsed on its clock. See below for some examples.

Time here on earth as compared to what? Again, the general rule, as above, is that an object's path through spacetime determines the proper time elapsed on its clock. In the vicinity of Earth, the two main things that affect proper time are the object's altitude above the Earth, and its speed relative to the Earth. So, for example, if two objects are both at rest relative to the Earth, the one that is lower down will have its clock tick slower than the one higher up. And if two objects are at the same altitude, but one is moving and the other is at rest relative to the Earth, the one that is moving will have its clock tick slower than the one that is at rest. But those rules of thumb are specific to the vicinity of the Earth; they might not work in other scenarios.

There is no "time dilation of Earth"; you need to specify what object, where in relation to the Earth, and what you are comparing it to.

10. May 12, 2015

### Staff: Mentor

So the question is: how fast do you have to travel in space far away from earth for a clock on earth and in the spaceship to stay synchronized.

You would calculate this by calculating the gravitational time dilation factor on earth vs in empty space, then setting that value equal to the SR time dilation factor and solving for V.

11. May 12, 2015

### Janus

Staff Emeritus
Which turns out to be equal to what the escape velocity from the surface of the Earth is.

If you take

$$\sqrt{1-\frac{2GM}{rc^2}}$$

from the Gravitational time dilation formula:

And compare it to

$$\sqrt{1-\frac{v^2}{c^2}}$$

from the SR time dilation equation,

you will notice that first is the same as the second with the escape velocity

$$\sqrt{\frac{2GM}{r}}$$

Replacing v.

Another interesting situation is where an orbiting clock runs as the same rate as a clock sitting on the surface of the body it is orbiting (assuming the body itself does not rotate). This works out to be at a distance of 3/2 the radius of the body.

12. May 12, 2015

### Staff: Mentor

It's probably cheating if I used an online time dilation calculator to find the answer isn't it? Even worse after I suggested a more rigorous approach and then didn't follow it myself? It wasn't precise enough, so I didn't recognize escape velocity when I saw the number, though. Interesting connection.

13. May 12, 2015

### Janus

Staff Emeritus
When you think about it, it makes sense, since both are related to gravitational potential.