near light-speed travel

[chrisnhowell]

Scenario: man in spaceship leaves earth at near light-speed and returns a year later.

1. Would he need to travel constantly away from earth or could he just orbit for the relative time difference to occur?

2. What would would we hear/see from earth if we were in radio/video contact with him?

Thank you for any replies.

 

[MaximT]

3 minutes ago, chrisnhowell said:

1. Would he need to travel constantly away from earth or could he just orbit for the relative time difference to occur?

Orbiting earth at such speed, can't, but if you are constantly pushing inside, why not. To me, it will suffice to achieve a time differential base on relativity.

5 minutes ago, chrisnhowell said:

2. What would would we hear/see from earth if we were in radio/video contact with him?

Time dilation should be constant, the signal won't reach him normally, it will have to be corrected every second... After a year of continuous correction, it will be the same, at correct time, the differential don't add over time.

[Strange]

You would also need to consider the Doppler effect which would increase the frequency of signals when they are approaching you and reduce them when travelling away.

[chrisnhowell]

Thanks for the reply. I apologise if I'm simply being ignorant but what do you mean by 'pushing inside'?

For radio/video I meant one-way from him to earth.

Thanks.

[studiot]

37 minutes ago, chrisnhowell said:

Scenario: man in spaceship leaves earth at near light-speed and returns a year later.

1. Would he need to travel constantly away from earth or could he just orbit for the relative time difference to occur?

2. What would would we hear/see from earth if we were in radio/video contact with him?

Thank you for any replies.

 

Kind of depends what you mean by ' the relative time difference'.

 

In all the formula the time dilation (given by the gamma factor) depends upon the relative velocity between the two spacehip and the Earth.

If the spaceship is not getting further away the radial relative velocity is zero - so there is no time dilation to be had there.

The relative velcoity due to the tangential velocity would be variable and much lower. this would impact upon your question part 2.

 

The radial velocity represents the maximum possible relative velocity and thus the maximum possible time dilation.

 

Does this help?

 

[chrisnhowell]

I appreciate the response but my knowledge in the subject is too basic. I don't really understand the answers! 

Thank you

[Strange]

2 minutes ago, chrisnhowell said:

I appreciate the response but my knowledge in the subject is too basic. I don't really understand the answers! 

Perhaps you need to ask a more specific question. Your initial questions were very general and so the answers, necessarily, are very general as well.

 

[chrisnhowell]

good point. I'll try but I'm not sure it will help me! Let me think and I shall post shortly.

Edited April 13, 2019 by chrisnhowell

[Strange]

For a satellite orbiting the Earth, there is an effect due to their velocity and also an effect due to the difference in gravity between the Earth and the height of the orbit.

For example, GPS satellites travel at about 14,000 k/h. This means their clocks run slower by about 7 microseconds per day.

However, because the satellite is on average about 19,000 km above the Earth, their clocks run faster by about 46 microseconds per day. 

So, overall the clocks run faster by about 39 microseconds per day.

[studiot]

Your question has more depth than I first thought.

Thank you for a good question +1

Here is a sketch showing two spaceships, one moving directly away from Earth the other orbiting, but both travelling at velocity v.

 

Spaceship 1 is travelling directly away from Earth at constant speed v so its relative velocity is also constant at v.

Spaceship 2 is orbiting at constant  speed but its velocity is continually changing as its direction is continually changing.
So it is accelerating.

Technically special relativity does not apply directly to this situation although it can be extended to it.

 

[Strange]

As time dilation only depends on speed, not acceleration then (if you ignore gravity) you can just take the orbital speed and use special relativity.

Wikipedia has a neat graph showing the relative contributions of gravity and orbital speed for different altitudes:

So at low altitudes, the effect of relative speed dominates (clocks on ISS run slower than those on Earth) while at higher altitudes, gravitational effects dominate and so clocks run faster than on Earth.

https://en.wikipedia.org/wiki/Time_dilation#Combined_effect_of_velocity_and_gravitational_time_dilation

1 hour ago, chrisnhowell said:

Thanks for the reply. I apologise if I'm simply being ignorant but what do you mean by 'pushing inside'?

You cannot orbit the Earth at "near light speed" (it is greater than escape velocity). So you would need rockets continuously pushing you towards the Earth.

[md65536]

5 hours ago, studiot said:

If the spaceship is not getting further away the radial relative velocity is zero - so there is no time dilation to be had there.

This is incorrect. Unlike length contraction, time dilation doesn't depend on the direction, only the relative speed.

There's no Doppler effect in this case, so you can "see" the true time dilation effect.

6 hours ago, chrisnhowell said:

Scenario: man in spaceship leaves earth at near light-speed and returns a year later.

[...]

2. What would would we hear/see from earth if we were in radio/video contact with him?

You don't say if it's a spaceship year or Earth year, but the effects are the same regardless.

Radio waves are light. You can describe it based on how it looks. The video below has some effects from the spaceship's perspective, including orbit at the end.

Let's say we're talking one Earth year and one spaceship day. The spaceship would orbit about seven times a second in lowest orbits. It would appear slowed to 1/365 our time, so the one day of spaceship time would be seen over our year. Since the spaceship sends out a day's worth of signal and light, the light that it emits (not reflects) would appear very dark... 1/365th as bright. It would also be red-shifted by that factor.

The spaceship would see blue-shifting and a year's worth of Earth's light/signals (and the sun's too) in a day, so it would be intensely lit, which is why it wouldn't necessarily appear darkly reflective.

 

By the way, the relativistic Doppler ratio is \( \sqrt{\frac{1 + \beta}{1 - \beta}} \), where beta is the speed as a fraction of c. If you negate the speed, you get the inverse Doppler ratio. If you take the average of the Doppler ratio and its inverse, you get the Lorentz factor.

So, if you have a spaceship traveling at constant speed through flat spacetime on a path that you can divide up into equal-length segments in one direction and the opposite, and have it arriving at its starting point in some inertial frame (eg. some point on an Earth orbit), you can see that the average rate of time seen passing in inertial frame is equal to the Lorentz factor.

(Actually that's true for any constant-speed closed loop through flat space-time, but I can't think of how to see that intuitively.)

Edited April 13, 2019 by md65536

[Janus]

5 hours ago, chrisnhowell said:

Scenario: man in spaceship leaves earth at near light-speed and returns a year later.

1. Would he need to travel constantly away from earth or could he just orbit for the relative time difference to occur?

2. What would would we hear/see from earth if we were in radio/video contact with him?

Thank you for any replies.

 

Okay let's start with the scenario where he leaves in a straight-line away from the Earth, turns around and returns along a straight-line.

We'll pick 0.99c as his relative speed to the Earth.  This gives a time dilation of ~ 1/7.   So if he is gone for 1 year Earth time, he returns having aged  1/7 of a year.

What would he measure in terms of radio/video from the Earth?  While going away he will receive signal that are Doppler shifted by a ratio of

sqrt((1-v/c)/(1+v/c))  where v is positive when he is receding from the Earth.  

Since by his clock it takes 1/14 of a year for him to reach the turn around, he gets ~0.005 y worth of Earth time transmissions (a bit under two days). 

It takes another 1/14 of a year to make the return trip. But now that he is approaching the Earth, we need to make v negative in our equation.  This means he gets Earth transmissions at a much faster rate, and gets just a under a year's worth of transmission over the return leg ( if you plug the numbers I used above in you actually get a bit over a year, But this is due to my using rounded out numbers rather than more accurate ones.)

The end result is that he will see a total of 1 yr worth of signals from the Earth during the 1/7 of a year he measures by his clock.

From the Earth, you will also measure a Doppler shift in the signals coming from the ship while it is receding and approaching, and we use the same formula as above. The difference is that while the Ship sees a shift in the signals received immediately upon his turn-around, this is not the case for the Earth.  For the Earth the ship is nearly 1/2 a light year away when it turns around.  So while by the Earth clock, 6 mo pass before the ship turns around, nearly another 6 months must pass before the signal carrying that info can get to Earth. The upshot is that the Earth receives "slowed down" transmissions from the Ship for just a bit under a full year ( about 2 days short), and gets ~1/14 yr ship time worth of transmissions.  Then for the rest of the year( ~2 days), measures sped up transmissions, where he gets 1/14 of a years worth of ship time transmissions), the end result is that he gets 1/7 of a years worth of ship transmissions upon its return.

Now imagine the ship circling the Earth at some fixed distance (while ignoring gravitation) . In this case, the distance never changes.  For the Earth, All we have to worry about is the relative speed of the Ship.  This generates a time dilation of 1/7.  This is also the rate at which the Earth would receive signals from the Ship.  This is the "transverse Doppler effect" The Doppler effect you would measure from an object passing you, but at the moment it is neither approaching or receding. 

For the Ship, things are a bit different.  For one, unlike the Earth, observations made from it are not being made from an inertial frame.  Since it is circling the Earth, it is constantly changing its velocity. ( velocity combines both speed and direction) , which means it is constantly accelerating, and that acceleration is always towards the Earth.  Observations made from within a non-inertial frame of reference follow different rules than those made from inertial ones.  

  In an non-inertial frames, clocks placed in different positions relative to the acceleration vector run at different rates even if they are not moving relative to each other.  So while the ship can consider the Earth as hovering directly "overhead" with no relative motion, it is also displaced with respect with the acceleration, and the clocks there would be expected to run fast.

The end result will be that the ship, will measure sped up transmissions from the Earth, even though it is not approaching it and those signals will be ~7 times fast.

So in this case, both the Ship and Earth will agree that Earth and ship time differ by a factor of 7 and that the ship clock is the slower of the two.

 

[beecee]

9 hours ago, chrisnhowell said:

Scenario: man in spaceship leaves earth at near light-speed and returns a year later.

1. Would he need to travel constantly away from earth or could he just orbit for the relative time difference to occur?

2. What would would we hear/see from earth if we were in radio/video contact with him?

Thank you for any replies.

 

 

8 hours ago, chrisnhowell said:

I appreciate the response but my knowledge in the subject is too basic. I don't really understand the answers! 

Thank you

From one lay person to another the simple reason why time dilation occurs is due to the fact that the speed of light remains constant always in any frame of reference. The faster you travel through space, the slower you travel through time is the second important concept. The third is that time will always pass, both by mechanical clocks and biologically, at one second per second, in anyone's own frame of reference. In effect that means if you and I were twins, and I set off in a warp drive spaceship at 99.999% "c", time for me would pass as per normal, and time for you would also pass as per normal. But as we both have infinitely powerful telescopes, you would see my time as slowed down, and I would see your time as slowed down. While that may seem paradoxical, if I'm to return home in say 12 months time, I need to decelerate and accelerate again. This is best explained here, far better then I can.....https://www.askamathematician.com/2010/09/q-how-does-the-twin-paradox-work/

With the scenario I have given you, when I do return to Earth 12 months later according to my onboard clocks, I will be returning to an Earth 230 years in the future, with you long dean and buried.

Edited April 13, 2019 by beecee

[studiot]

19 hours ago, md65536 said:

This is incorrect. Unlike length contraction, time dilation doesn't depend on the direction, only the relative speed. 

There's no Doppler effect in this case, so you can "see" the true time dilation effect.

Thank you for that comment, you are right of course I thought in haste.

But I later amended my thoughts.

Thank you for the video, twas interesting.

However a small nit.

You are considering the Earth receiver as a point and that is not satisfactory for a low orbit.

Due to the Earth's rotation the equator has an tangential speed of some 1600 km/hr.

Though much smaller than the speed of the satellite Strange mentioned it is still a significant proportion, enough to create a variable relative speed depending upon where on Earth the receiver is located. This effect is doubled between receivers located on opposite sides of the globe, so potentiall a diffence of 3600 km/hr.

So there will be (a small) variation and  Doppler effect.

[swansont]

On 4/13/2019 at 7:30 AM, chrisnhowell said:

Thanks for the reply. I apologise if I'm simply being ignorant but what do you mean by 'pushing inside'?

Gravity won’t result in an orbit like that, so you would need some other force, directed toward the center, to allow such speeds. I think that’s what it means.

 

[md65536]

On 4/14/2019 at 6:23 AM, studiot said:

You are considering the Earth receiver as a point and that is not satisfactory for a low orbit.

Due to the Earth's rotation the equator has an tangential speed of some 1600 km/hr.

Though much smaller than the speed of the satellite Strange mentioned it is still a significant proportion, enough to create a variable relative speed depending upon where on Earth the receiver is located. This effect is doubled between receivers located on opposite sides of the globe, so potentiall a diffence of 3600 km/hr.

So there will be (a small) variation and  Doppler effect.

Yes, I misread your post as talking about a circular orbit centered on the observer's viewpoint, which is possible for larger orbits. It should be possible to completely negate the relativistic Doppler effect with such an orbit, leaving only negligible(?) gravitational Doppler shift including due to the observer's acceleration as the Earth rotates.

In the case you're speaking of, where the orbit is not centered on the viewpoint, the spaceship approaches then recedes as it passes nearest the observer, and the Doppler shift from that would be much much greater than the shift due to the Earth's rotation. This is true even at normal "slow" satellite speeds.

https://en.wikipedia.org/wiki/Doppler_effect#Satellite_communication --- Partial info and does not handle satellite at relativistic speeds. 

[studiot]

11 hours ago, md65536 said:

Yes, I misread your post as talking about a circular orbit centered on the observer's viewpoint, which is possible for larger orbits. It should be possible to completely negate the relativistic Doppler effect with such an orbit, leaving only negligible(?) gravitational Doppler shift including due to the observer's acceleration as the Earth rotates.

In the case you're speaking of, where the orbit is not centered on the viewpoint, the spaceship approaches then recedes as it passes nearest the observer, and the Doppler shift from that would be much much greater than the shift due to the Earth's rotation. This is true even at normal "slow" satellite speeds.

https://en.wikipedia.org/wiki/Doppler_effect#Satellite_communication --- Partial info and does not handle satellite at relativistic speeds. 

You have added more than I was thinking of. Thanks. +1