B Does Time Dilation Occur in a Perfectly Circular Orbit at Near-Light Speed?

[David Lewis]

A spaceship orbits Earth at 300,000 km distance at speed 0.5c...

At that altitude, orbital speed of the ship would be about 1140 m/s.

 

[jbriggs444]

At that altitude, orbital speed of the ship would be about 1140 m/s.

... for a circular orbit maintained by gravity.

For an orbit maintained by some unspecified (and large) force, the orbital speed can be 0.5 c as specified.

 

[David Lewis]

Good catch. I didn't notice OP's quote in post #4 positing a rocket engine supplying centripetal force.

 

[David Lewis]

The effect of gravity on time ?

If you have a Janus-style light clock in deep space ticking at the correct rate (neither fast nor slow) and then transport it to Earth, will the Earth’s gravitational field change the rate at which the clock ticks?

 

[jbriggs444]

If you have a Janus-style light clock in deep space ticking at the correct rate (neither fast nor slow) and then transport it to Earth, will the Earth’s gravitational field change the rate at which the clock ticks?

A properly running clock always ticks at one second per second.

 

[David Lewis]

Then if two properly functioning clocks disagree, they must have measured different amounts of time.

 

[Steeve Leaf]

Marcela will be at rest with respect to the ISS. If you connect a slender thread between two objects, and the thread neither goes slack nor breaks, then they are at rest with respect to each other.

A spaceship orbits Earth at 300,000 km distance at speed 0.5c broadcasting live images of the clock aboard, will the clocks difference will be 1 sec?

The orbit is a perfect circle.

Its moving close to the speed of light but stays at the same distance.

Is the clocks difference between the one on Earth and the images received by broadcast from the ship stay the same ?

Change the distance from 300,000 km to 129e6 km and the original answer of 0.866 time dilation raising investigation, do you see why ?

 

[David Lewis]

Good point, but if we assume that all clocks tick at the rate of one second per second then we have defined time the same way I have defined distance (whatever a tape measure says).

 

[Steeve Leaf]

The number should be more like 0.9943 . neglecting Earth's spin. (?)

 

[Steeve Leaf]

Good point, but if we assume that all clocks tick at the rate of one second per second then we have defined time the same way I have defined distance (whatever a tape measure says).

The difference is that when a ship return from near light speed journey, it returns to the same length while the twin returns younger.

 

[Nugatory]

The difference is that when a ship return from near light speed journey, it returns to the same length while the twin returns younger.

Yes, but that asymmetrical effect has nothing to do with time dilation. Length contraction and time dilation are present only when the two ships are in motion relative to one another.

(if you have not yet found, read, and understood the twin paradox FAQ, now is the time to do so).

 

[Steeve Leaf]

I 've waited 50 years to understand Newton.

A spaceship orbits Earth at 300,000 km distance at speed 0.5c broadcasting live images of the clock aboard, will the clocks difference will be 1 sec?

The orbit is a perfect circle.

Its moving close to the speed of light but stays at the same distance.

Is the clocks difference between the one on Earth and the images received by broadcast from the ship stay the same ?

I 've a diffrent opinion on the answer.
The ship circle the Earth at a distance of 300,000 km.

The orbit leangth is 1.88e6 km.

At speed of 0.5c it takes 12.5333 sec

Thats is relative speed of 3,351. km/s compare to surface observer.
Not 0.5c .

3351/300000=0.01117
Put in the formula it comes out 0.994 ( at 42000 km Earth circumference ):)

Am I close ?

 

[jbriggs444]

At speed of 0.5c it takes 12.5333 sec

Thats is relative speed of 3,351. km/s compare to surface observer.
Not 0.5c .

The speed of an orbiting spacecraft and the speed of the point on the Earth's surface which is directly below such a spacecraft are not the same thing.

 

[Steeve Leaf]

This is the speed to calculate time dilation. see Marcela question to understand my point.
There compare to the ISS 0.5c Marcela got 0 relative speed and no time dilation.
If you understand my point and think that i am wrong please explain.
It is definitely out of my league I knew it before looking back into it ☺

 

[David Lewis]

The difference is that when a ship return from near light speed journey, it returns to the same length while the twin returns younger.

But the universe is squashed for the fast moving ship, so the ship doesn't have to travel as far. This savings of time (from the stationary observer's frame) due to less distance traveled is retained.

 

[jbriggs444]

This is the speed to calculate time dilation. see Marcela question to understand my point.
There compare to the ISS 0.5c Marcela got 0 relative speed and no time dilation.
If you understand my point and think that i am wrong please explain.
It is definitely out of my league I knew it before looking back into it ☺

I am not talking to Marcela. I am talking to you. A ship moving at 0.5c relative to the Earth is moving at 0.5c relative to the earth. If you think otherwise, please explain your reasoning.

 

[Steeve Leaf]

The farther the ship is the longer it takes for it to complete an orbit doing the same speed 0.5c.
In one case it is 12 sec and in the other 90 min.
In the 90 min case the relative speed compare to ISS was 0 according to an answer above, if this is true than iss relative speed is not ( even close ) 0.5c.
The same principle is used on the 12 sec ship.
The distance of the 0.5c circulating ships affect the time it takes them to do so and the realistic speed accordingly.

Unless the answer above is wrong (Mercela) I don't see the flow of my logic but I'm sure you can, so explain.

 

[_PJ_]

Anytime you have a light (or other massless entity) transmitted, it must be at c speed and this speed is agreed upon by ALL reference frames.
So in any situation when you want to understand the dilation of time or contraction of space, just consider a photon in your system as the basis for time.
Distances are then all calculated as result from time and lightspeed.

 

[Steeve Leaf]

In Length Contraction , does it matter if you coming or going ?
What about a little tilt of 3 dimensional objects ?
At near relative light speed of course.☺

 

[_PJ_]

In Length Contraction , does it matter if you coming or going ?
What about a little tilt of 3 dimensional objects ?
At near relative light speed of course.☺

Length contraction applies in the direction of motion. (The same can be said for time, time metric is dilated in direction at which we travel through time)

Actual speed doesn't matter, there is always contraction and dilation with motion, only that the effects are so small (to our perception scale) that they are negligible and only readily apparent when approaching speeds comparable to c

 

[jbriggs444]

The farther the ship is the longer it takes for it to complete an orbit doing the same speed 0.5c.
In one case it is 12 sec and in the other 90 min.

Right. So you can arrange for the orbital period of a satellite moving at 0.5c to be anything you want by simply selecting the right altitude. And arranging for the requisite centripetal acceleration.

It would help if, instead of mentioning "Marcela" you had referenced the question you were trying to answer: One common technique in these forums is to mention the post number where the referenced passage exists. In this case, that would be #27:

A spaceship named "Marcela" circle kilometers (maybe about 129e6 km) away at 0.5c.
It has synchronized its time of orbit with IS that it sees it all the time at the same position and distance.
What is the speed of Marcela relative to IS and how the optic clocks explain this time dilation ?

Here you have postulated a spacecraft which is still moving at 0.5c relative to earth. You have varied its orbital radius so that it remains directly above the ISS at all times (this seems to assume that the ISS is in an equatorial orbit -- which it is not). You proceed to ask about its speed relative to the ISS.

The answer is obviously 0.5c. The velocity of the ISS is negligible compared to 0.5c. So the velocity of the 0.5c orbiter is still 0.5c in any inertial frame where the ISS is momentarily at rest, just as it is in an Earth-centered inertial frame.

However, there is a potential ambiguity. You have said that "it sees it all the time at the same position and distance". This suggests that you are adopting a frame of reference that rotates with the orbit of the ISS. In this reference frame, you have stipulated that Marcela is motionless.

The answer now is obviously 0.

But changing the reference frame does not change anything about the situation. The time dilation formula based on an object moving in an inertial frame still applies. The fast orbiter is still moving at 0.5c and its time dilation is the same as before.

 

[Janus]

This is the speed to calculate time dilation. see Marcela question to understand my point.
There compare to the ISS 0.5c Marcela got 0 relative speed and no time dilation.
If you understand my point and think that i am wrong please explain.
It is definitely out of my league I knew it before looking back into it ☺

Just because Marcela and the ISS do not have a relative motion as measured from their rotating frame, does not mean that there is not going to be no time dilation between them. This is due to the fact that they are viewing each other from a non-inertial rotating frame.

Earlier I advised you to forgo these types of non-inertial scenarios until you had come to grips with inertial motion scenarios. The above is an example of what I was talking about. You came to an incorrect conclusion by trying to analyze a situation you weren't ready to deal with.

The easiest way to analyze the Marcela-ISS scenario is to first look at it from the inertial frame which does not rotate with the pair, and then transform that into what the pair will see. First we will consider a signal transmitted from ISS to to Marcela. For ease of explanation, we will assume a tight beam transmission.
The following image will help in illustrating the scenario.

The small circle is the ISS path and the larger one is for Marcela. The blue line joins the ISS and Marcela. If the ISS wants to transmit a signal to Marcela, it can't just point along the blue line. Because in the time it takes for the Signal to travel from the inner circle to the outer one, Marcela (the solid M) will have maoved to the Position shown by the Outline M, and the signal would miss the target. Instead the ISS has to "lead" Marcella and aim for where Marcella will be when the signal arrives.
The other thing we have to note that the signal. traveling outward at c meets Marcela at a right angle to Marcela's own 0.5c. This means that the Closing speed between Marcela and the signal will, in this frame be greater than c. Thus the momentum involved in their meeting will be greater than if the closing speed had been just c.
If we jump to what Marcela detects, the signal will appear to come from some where a bit ahead of it in the direction of its motion. Since it will measure the signal arriving at c, the extra momentum will be seen as an increase in frequency, or Doppler shift, of the incoming signal at a rate of ~1.155
Thus if the ISS sent a ten second message at 100 khz, this represents 1 million waves transmitted. Marcella will receive the signal at 115.5 khz and thus get those same 1 million waves in just 8.66 seconds. He will see things happening on the ISS as happening 1.155 times faster.

Next we consider a signal sent from Marcela to the ISS:

Again, if Marcela trys to send a signal to the ISS by pointing it antenna towards the ISS, the signal will actually follow the path shown by the dotted line*. This is similar to what is shown in the earlier light clock example where the light travels at a diagonal as seen by someone watching the clock pass by. In order to get the signal to travel along the solid yellow arrow, Marcela actually has to aim its antenna in the direction indicated by the orange arrow. Since this points somewhat in the opposite direction of Marcella's direction of motion, this subtracts momentum from the signal.
Again the ISS measures the signal as coming in at c, and thus measures this as a decrease in frequency of 0.866. If Marcella sends a 10 sec message at 100 khz, the ISS will receive those 1000,000 waves making up the signal over 11.55 sec. The ISS sees everything occurring on Marcella as happening 0.866 as fast as it does for itself.

Now since neither the ISS or Marcella measure the distance between themselves as changing, they can not attribute any of the Doppler shift they see as being due to a change in propagation delay and can only put it down to events actually running faster for the ISS than they do on Marcella.

Changing the distance at which Marcela circles at 0.5c will complicate the signal analysis some, but will not change the degree of time dilation between the two.

*This also happens with the ISS, but since the speed at which it travels is so small compared to the speed of light, the difference between the direction of aim and the actual light path direction will be extremely small. It does have a small effect of the measured time dilation, but it is only the difference between the time dilation factor at 0.5c and at 4.99974c.

 

[Steeve Leaf]

Thanks,
Frequency of wave doesn't change its speed.
Light speed is the same for every observer.
Think instant of shooting light at a target or tennis ball at AO (Australia Open), each transmitting a tv signal of the clock .( and you adjust the frequency on the receiver.).
How you factor in if an object traveling at 0.5c toward you (+) and away from you (-) in the formula ?
Where 4.99974c # come from ?
I still think that in Marcella case at least another point of view will be good.
You are right this subject is difficult , at least for me , so the part of acceleration effect on gravity is the issue for me now, at least its got direction symbol in it.:)

 

[Janus]

Thanks,
Frequency of wave doesn't change its speed.
Light speed is the same for every observer.
Think instant of shooting light at a target or tennis ball at AO (Australia Open), each transmitting a tv signal of the clock .( and you adjust the frequency on the receiver.).
How you factor in if an object traveling at 0.5c toward you (+) and away from you (-) in the formula ?

Relativistic Doppler shift.
f_o = f_s \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}
This incorporates both time dilation and the changing propagation delay caused by increasing or decreasing distance.
fo is the frequency observed and fs is the frequency at the source

Where 4.99974c # come from ?

it is the difference between the ISS orbital speed and Marcela's speed of 0.5c

I still think that in Marcella case at least another point of view will be good.
You are right this subject is difficult , at least for me , so the part of acceleration effect on gravity is the issue for me now, at least its got direction symbol in it.:)

 

[Steeve Leaf]

4.99974c is a little over the speed limit .
In the doppler effect the frequency comes up when the signal emitter come closer and the other way when it is traveling away , they don't cancel out I agree.

 

[Janus]

4.99974c is a little over the speed limit .
In the doppler effect the frequency comes up when the signal emitter come closer and the other way when it is traveling away , they don't cancel out I agree.

Sorry, that should be .499974c
Sometimes when I type too fast, I'll transpose characters. With words, I usually catch it. In this case since I knew the number I meant, It actually took me a while for me to even notice the misplaced decimal.

 

[Steeve Leaf]

Yes, ISS speed is 27,576 - 28080 km/h.
I admit that I don't completely understand this, but let's leave it for now.
( I can't tell how fast M is moving if it is broadcasting tv signal of the clock and you don't know the original broadcast frequency ).

Acceleration and gravity affect on visualized clocks, that need now your appreciated explanation.

 

[Bartolomeo]

In fact this is very interesting question, because it shows the ambiguity of the notion of relative motion of two point objects.

I think that @Janus gives correct and very spectacular answer.

I also like Bertrand Russell’s quote in his post.

Yes, distance between the Earth and rotating spaceship doesn’t change. However, the orbiter moves not relatively to the Earth but relatively to the reference frame. Imagine grid, which covers painting with the Earth and spaceship. Spaceship moves and changes its spatial position in this frame.

Amount of time dilation depends solely on linear velocity. According to time dilation tests based on Mossbauer Effect, rotating absorber (source is at rest) measures increase of source’s frequency at gamma times, i.e. clock in the center actually runs faster. (Kholmetskii et all 2008). There were some related polemical articles in Physica Scripta recently – Zanchini, Chen, Kholmetskii. Everything is very simple though.

If absorber is placed in the center and source rotates, absorber will measure decrease of source’s frequency gamma times. I.e. rotating clock actually dilates.

Why rotating clock “sees” acceleration of clock in the center? Because it ACTUALLY dilates. If you would become much slower than usual, you would see that everything around you runs faster.

All that can be analyzed through the Transverse Doppler effect. Transverse Doppler Effect (deviation of measured frequency from proper one) is in accordance with dilation of moving clock. If observer on the Earth releases beam of light, he keeps laser pointer at right angle to direction of motion of the orbiter. Photon approaches orbiter at oblique angle from the front, thus for orbiter the source on Earth appears to be in the front (aberration of light). Photon blue shifts, this appears to be in the shorter range of wave spectrum.

Angles of emission and reception tied with relativistic aberration formula.

Motion in arbitrary direction: https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

If the orbiter holds a mirror, he has to keep this mirror at tangential to circumference “as he moves”. Photon bounces back from the mirror and travels back to the Earth. Orbiter will emit the beam at oblique angle again and photon will come back to the Earth and will enter right into laser pointer’s mouth.

Obviously, mirror was a moving source and moving clock. Moving clock appears to be slower at gamma, thus received frequency was also gamma times lower than it was at the orbiter.

If the ISS releases a photon and wants to hit an orbiter at a higher orbit, it has to tilt the laser pointer a bit backward, as @Janus correctly explains.

Animation in the internet, rotational motion episode:

Therefore it is interesting to note that if two orbiters rotate on the same orbit but wish to send a beam from certain point of circumference to opposite point of circumference, one of them (source) has to keep his laser pointer backward and another (target) has to keep his telescope forward at equal angles. Good to know, that in this case they will not see any frequency change, neither dilation nor acceleration of opposite clock. This was confirmed by Champeney and Moon time dilation test. There are two interesting articles in Nature by I. Essen (1964, 1965) in regard of this time dilation test. Also related articles are: “Ray path in a rotating system”, R.C. Jennison, Nature (No. 4898, 1963, p. 739) and “Reflection from a transversely moving mirror” , R.C. Jennison, Nature (Vol. 248, 1974, p. 660).

Thus, orbiters at different orbits, but rotating with the same angular velocity will actually accumulate different amount of time dilation and their clocks will show different time at the meeting on the Earth.

Also the last paragraph: http://www.mathpages.com/home/kmath587/kmath587.htm

If we will place a photo camera on the Earth and will make a picture of the orbiter, the orbiter will appear Lorentz – contracted in direction of its motion. If we will place a photo camera on the orbiter and will take a picture of a measuring rod on the Earth, the rod will appear not contracted, but stretched. Why? Because film in the photo – camera Lorentz contracts itself.

In these reflections we omit gravitational time dilation.

 

[Steeve Leaf]

Thanks,
ISS got a clock on it , it send a tv signal of this clock to M which broadcast it with no delay back to ISS monitor.
Now at ISS we can see the actual watch and the tv image after the round journey.
Why the difference between the actual clock and the monitor clock will change and not stay the same if
1. the distance stays the same.
2. The change in frequency don't change the time the signal travel.
3.This is ISS own clock.
4. No looking through telescope.
The difference is the time it takes to cover the unchanged distance.
Where the complication occur ?.☺

 

[jbriggs444]

Thanks,
ISS got a clock on it , it send a tv signal of this clock to M which broadcast it with no delay back to ISS monitor.
Now at ISS we can see the actual watch and the tv image after the round journey.
Why the difference between the actual clock and the monitor clock will change and not stay the same if

A hypothetical technician on M would see a high frequency carrier wave encoding a high speed video. As I understand the scenario, the high speed video is rebroadcast on a normal (according to M) frequency carrier wave.

Back on ISS, we would receive a low frequency carrier wave encoding a normal speed video. What would lead you to expect otherwise?