# 1G Relativistic Rocket Ride, Wheeee

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We are going to travel 113,242 light years across the Milky Way starting at rest in our galaxy's inertial frame near an outer edge. We will use a continuous 1G of acceleration the entire distance. The table below is from

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

T= time in rockets inertial frame t= time in earths inertial frame

d= distance in light years v= light speed Y= time dilation

T --------- t ----------- d ----------- v -------------- γ

1 year --- 1.19 yrs ----- 0.56 lyrs --- 0.77c ----------- 1.58

2 ------- 3.75 -------- 2.90 -------- 0.97 ----------- 3.99

5 ------- 83.7 -------- 82.7 -------- 0.99993 --------- 86.2

8 ------- 1,840 ------ 1,839 -------- 0.9999998 ------ 1,895

12 ------ 113,243 ---- 113,242 ------ 0.99999999996 -- 116,641

Relative to the clocks in the space ship and as observed by the crew, they will travel well over 100,000 light years in just 12 years. Their speed increases during the trip, as the distance covered according to their galaxy map shows.

True or False:

1. Due to time dilation, the ship's inertia frame as observed by the crew and shown by the ships clocks well exceeds the speed of light relative to the Milky Way. Especially during the last half of the trip.

2. 99%+ of the the time dilation is caused by SR with only a small part of the time dilation caused by GR because of the 1G from acceleration.

Edited by NowThatWeKnow
Fix typos

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i'll go!

interesting point they make about C.B.R. getting blue-shifted enough to melt the ship.

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i'll go!

interesting point they make about C.B.R. getting blue-shifted enough to melt the ship.

We leave tomorrow if the two questions are answered. You don't have any thought on that?

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True or False:

1. Due to time dilation, the ship's inertia frame as observed by the crew and shown by the ships clocks well exceeds the speed of light relative to the Milky Way. Especially during the last half of the trip.

2. 99%+ of the the time dilation is caused by SR with only a small part of the time dilation caused by GR because of the 1G from acceleration.

1. False. The ship will observe length contraction of the path of the trip.

2. False. Two sides of the same coin. You can approach the problem using either solution, and if you do it right you'll get the same answer.

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1. False. The ship will observe length contraction of the path of the trip.

2. False. Two sides of the same coin. You can approach the problem using either solution, and if you do it right you'll get the same answer.

Let me rephrase #1:

Speed as observed by the crew and shown by the ships clocks well exceeds the speed of light relative to the Milky Way when star map time and distance measurements are taken. This would be an observation from math measurements and not a visual view distorted by length contraction. We know that star A and star B on our map are x ly apart when at rest in the Milky Way frame.

2. I was afraid you were going to say that. Help me out here. I view the 1G force as a byproduct of acceleration. My calculations show 1 G vs way out in space as .22 seconds per year. Nothing to get excited about. Using the calculators on

http://www.cthreepo.com/cp_html/math1.htm

shows a non accelerating frame speed of .97c having a time dilation of 4.11

Consider my chart.

After 2 years accelerating you reach a speed of 0.97c with a time dilation of 3.99

Is the small difference because of 1G gravity? By looking at this it seems that speed, and not gravity, is responsible for most of the time dilation. Also as the speed increases and gravity remains constant, the time dilation and length contraction increases.

Edited by NowThatWeKnow
typos
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Let me rephrase #1:

Speed as observed by the crew and shown by the ships clocks well exceeds the speed of light relative to the Milky Way when star map time and distance measurements are taken. This would be an observation from math measurements and not a visual view distorted by length contraction. We know that star A and star B on our map are x ly apart when at rest in the Milky Way frame.

And they will observe the maps to be wrong when they are moving. You're taking measurements from two different frames, and that's not how you measure the speed of light.

2. I was afraid you were going to say that. Help me out here. I view the 1G force as a byproduct of acceleration. My calculations show 1 G vs way out in space as .22 seconds per year. Nothing to get excited about. Using the calculators on

http://www.cthreepo.com/cp_html/math1.htm

shows a non accelerating frame speed of .97c having a time dilation of 4.11

Consider my chart.

After 2 years accelerating you reach a speed of 0.97c with a time dilation of 3.99

Is the small difference because of 1G gravity? By looking at this it seems that speed, and not gravity, is responsible for most of the time dilation. Also as the speed increases and gravity remains constant, the time dilation and length contraction increases.

The acceleration is not the right thing to look at, it's the potential. Like I said, if you do it right, you'll get the same answer.

The low-speed approximation for the frequency shift in a pseudo-gravitational potential from accelerating at g is g*d/c^2. If you recast that in terms of speed, it's v^2/2c^2, which is what you get if you assume it's all from the speed. The full equations are more complicated, and you can't use the approximations when the speed starts to get as big as 0.1c

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Let me rephrase #1:

Speed as observed by the crew and shown by the ships clocks well exceeds the speed of light relative to the Milky Way when star map time and distance measurements are taken. This would be an observation from math measurements and not a visual view distorted by length contraction. We know that star A and star B on our map are x ly apart when at rest in the Milky Way frame.

Length contraction is not a "visual distortion", according to the crew, the Milky way really shrinks in the direction of travel. You can't mix measurements from different frames (distance from the frame of the Galaxy and Time from the frame of the ship) and get a result that has a physical meaning.

2. I was afraid you were going to say that. Help me out here. I view the 1G force as a byproduct of acceleration. My calculations show 1 G vs way out in space as .22 seconds per year. Nothing to get excited about.

How did you arrive at this answer? Did you use the standard gravitational time dilation formula? If so, it is invalid for this situation. It is for use for gravitational fields that fall off by the square of the distance. Remember Gravitational time dilation is dependent on difference in gravitational potential, not differences in local gravitational strength or local acceleration due to gravity.

Using the calculators on

http://www.cthreepo.com/cp_html/math1.htm

shows a non accelerating frame speed of .97c having a time dilation of 4.11

Consider my chart.

After 2 years accelerating you reach a speed of 0.97c with a time dilation of 3.99

Is the small difference because of 1G gravity? By looking at this it seems that speed, and not gravity, is responsible for most of the time dilation. Also as the speed increases and gravity remains constant, the time dilation and length contraction increases.

In this situation you can either calculate strictly using speed and SR or using GR (ship standing still in 1g uniform gravity field while the galaxy falls past you), but you do not combine them.

This is a result of the clock postulate. IOW, acceleration does not cause any additional time dilation beyond that due to the change in relative.

This has been demonstrated by using high speed centrifuges (up to 10^18g). It has been shown that the time dilation only depends on the speed at which the object is moving and not the g's it experiences. With different radii of centrifuges you can have the same speed but experience different g forces, but the sample on the centrifuge will show the same time dilation.

As for the differences you note, they are just due to rounding errors. For instance, with the formula given in the web page quote, for a 2 year trip, v actually works out to 0.9680303033155948312177397420156c which he rounds up to 0.97c. But the difference in gamma between these two velocities is not insignificant. Using the more accurate value for v gives you a gamma of 3.986711890106634759197778382839 when using the Lorentz transformation, and calculating using the acceleration formula gives you 3.986711890106634759197778382839, the exact same answer.

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I have a couple of questions later for Janus and swansont but it is Saturday night and time to chase skirt.

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as strange as it is to imagine, time and distance are not the reliable measure of things they seem to be in everyday life.

distance/time < c is the measure. distance changes depending on the direction you're moving and time varies depending on the speed you're moving.

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And they will observe the maps to be wrong when they are moving. You're taking measurements from two different frames, and that's not how you measure the speed of light.

The acceleration is not the right thing to look at, it's the potential...

...gravitational potential from accelerating at g is g*d/c^2. If you recast that in terms of speed, it's v^2/2c^2, which is what you get if you assume it's all from the speed....

I was mixing frames trying to better visualize what is happening in both frames. In the end they traveled over 100,000 ly and and aged 12 years while the galaxy aged over 100,000 years but never exceeded the speed of light.

I understand the definition of gravitational potential but am not getting a clear picture on how it fits in to this picture.

Is the "g" in the above formula gravitational potential that should be "G" so google can work with it? Or is it # of g's acceleration.

Merged post follows:

Consecutive posts merged
Length contraction is not a "visual distortion", according to the crew, the Milky way really shrinks in the direction of travel. You can't mix measurements from different frames (distance from the frame of the Galaxy and Time from the frame of the ship) and get a result that has a physical meaning.

It was more like keeping up with both frames while being temporarily in one. The space travelers real world is at rest in the Milky way.

How did you arrive at this answer? Did you use the standard gravitational time dilation formula? If so, it is invalid for this situation. It is for use for gravitational fields that fall off by the square of the distance. Remember Gravitational time dilation is dependent on difference in gravitational potential, not differences in local gravitational strength or local acceleration due to gravity.

(G * mass of Earth) / (radius of Earth * (c^2)) = 6.95453588 × 10^-10

But as you pointed out it does not apply here.

If Gravitational time dilation is dependent on difference in gravitational potential, what influences the change in gravitational potential? Not at all clear here.

In this situation you can either calculate strictly using speed and SR or using GR (ship standing still in 1g uniform gravity field while the galaxy falls past you), but you do not combine them.

But who gets old is determined by who accelerates to a different frame?

This has been demonstrated by using high speed centrifuges (up to 10^18g). It has been shown that the time dilation only depends on the speed at which the object is moving and not the g's it experiences. With different radii of centrifuges you can have the same speed but experience different g forces, but the sample on the centrifuge will show the same time dilation.

The online dictionary says:

"an accelerating dragster or space shuttle. Spinning objects such as merry-go-rounds and ferris wheels are subjected to gravitational time dilation as an effect of their angular spin."

Is there a contradiction here?

Thank you for your participation.

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I was mixing frames trying to better visualize what is happening in both frames. In the end they traveled over 100,000 ly and and aged 12 years while the galaxy aged over 100,000 years but never exceeded the speed of light.

You can't apply the equations or the concept of constant c if you mix the frames. In each frame the view is consistent: the fixed observers sees them travel 100,00 LY but observes that their clock ran slow. The spaceship observer sees the travel as being shortened. Neither observer sees that they have exceeded c.

I understand the definition of gravitational potential but am not getting a clear picture on how it fits in to this picture.

Is the "g" in the above formula gravitational potential that should be "G" so google can work with it? Or is it # of g's acceleration.

As I had said, it's an acceleration at g, or 1g of acceleration — the acceleration of gravity at the earth's surface. The distance you accelerate at g tells you the change in your (pseudo)potential.

It was more like keeping up with both frames while being temporarily in one. The space travelers real world is at rest in the Milky way.

There is no preferred frame of reference. If you couldn't observe the Milky Way there is no way to tell if you were moving or at rest with respect to it. You can tell if you are accelerating, but not if that's because of e.g. thrust of your rocket or being at rest in a gravitational field.

If Gravitational time dilation is dependent on difference in gravitational potential, what influences the change in gravitational potential? Not at all clear here.

Location, location, location. The potential is the depth of the well. The acceleration is the slope of the sides. e.g. if g is large, it says that a small change in position gives a large change in potential, but it doesn't tell you what the potential is.

But who gets old is determined by who accelerates to a different frame?

The one who doesn't accelerate. Being at rest in a gravity field is an accelerated frame, which is one of the quirky concepts of GR.

The online dictionary says:

"an accelerating dragster or space shuttle. Spinning objects such as merry-go-rounds and ferris wheels are subjected to gravitational time dilation as an effect of their angular spin."

Is there a contradiction here?

No. You can treat these as a kinetic dilation or a pseudo-gravitational dilation, but you don't do both. The observer in a gravitational field is at rest. If undergoing an indistinguishable acceleration, he can assume he is at rest. There would be no kinetic term for that observer. However, an external observer can just say that the person is moving, and solve for the kinetic dilation.

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I understand the definition of gravitational potential but am not getting a clear picture on how it fits in to this picture.

(G * mass of Earth) / (radius of Earth * (c^2)) = 6.95453588 × 10^-10

But as you pointed out it does not apply here.

If Gravitational time dilation is dependent on difference in gravitational potential, what influences the change in gravitational potential? Not at all clear here.

Thank you for your participation.

Difference in gravitational potential can be characterized as the amount of work needed to lift an object from one point to another.

To demonstrate how this applies to here, first imagine lifting an object from the surface of the Earth to the distance of the Moon. As you move further from Earth, the force of gravity declines until, at the distance of the Moon it has dropped to 1/3609 of that at the surface of the Earth. IOW, it takes less work to lift the object from the surface to 1km altitude than it does to lift it from 1 km to 2 km, etc, with the amount of energy needed to lift it each additional Km decreasing. The total energy needed to lfit the object would be the sum of all these intermediate energies.

Now imagine you are lifting the object against a 1g field for the entire distance, The froce of gravity is 1g at the surface, 1g at the distance of Moon and 1g at every point in between. It would take the same amount of work to lift the object through each successive km. Again the total energy needed would be the sum of all the smaller steps

As you can see, it would take more work to lift the object agains 1g the whole way than it would to lift it against Earth's decreasing energy.

And even though thare is no difference in g-force in the uniform field between the surface and the distance of the Moon, there is a large differnce in potential, which is what effects gravitational time dilation.

With the spaceship accelerating at 1g, you have to assume a uniform gravitational field that extends for inifinity, and that the Milky Way is falling through that field if you want to use the GR approach to the problem.

The standard Gravitational time dilation equation is for dealing with a typical gravity field that decreases in strength over distance.

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As you can see, it would take more work to lift the object agains 1g the whole way than it would to lift it against Earth's decreasing energy.

And even though thare is no difference in g-force in the uniform field between the surface and the distance of the Moon, there is a large differnce in potential, which is what effects gravitational time dilation.

With the spaceship accelerating at 1g, you have to assume a uniform gravitational field that extends for inifinity, and that the Milky Way is falling through that field if you want to use the GR approach to the problem.

Excellent! Another piece of the puzzle starts to come into focus. Could you explain the difference between 1G from the rocket engines or 1G from (the apparent) centrifugal force. Inertia could keep a rocket spinning for 1G and not use fuel. Does the 1G have to be consistent with the direction of travel?

Merged post follows:

Consecutive posts merged

Location, location, location. The potential is the depth of the well. The acceleration is the slope of the sides. e.g. if g is large, it says that a small change in position gives a large change in potential, but it doesn't tell you what the potential is.

The one who doesn't accelerate. Being at rest in a gravity field is an accelerated frame, which is one of the quirky concepts of GR.

Between you and Janus I am making progress. It does seem that when things seem clear, I can view things from a slightly different angle and take them out of focus again. The first paragraph above is an example of what sinks in and should stick. The second is what makes my head hurt. I am hoping that others might benefit from my questions and your efforts.

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Excellent! Another piece of the puzzle starts to come into focus. Could you explain the difference between 1G from the rocket engines or 1G from (the apparent) centrifugal force. Inertia could keep a rocket spinning for 1G and not use fuel. Does the 1G have to be consistent with the direction of travel?

There is a difference in the potential — a rotating system is at a fixed potential, while a linear (i.e. a and v in the same direction) system's potential is changing.

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Cool questions. I was just wondering what sort of "Hubble far-field" would be observed by a "Hubble" zipping by at a significant fraction of c. Also, it is so that the younger of the twins experienced acceleration, right? There has to be such a logical way to distinguish them, at least if they start in the same frame and end up similarly. It gets worse, though, and somewhere else I described the old "flying through and punching the time-clock" idea. You can start with identical clocks but reckonings will be screwy at distant points which are conceived differently in the two frames of reference, from the start.

Edited by Norman Albers
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Also, it is so that the younger of the twins experienced acceleration, right? There has to be such a logical way to distinguish them, at least if they start in the same frame and end up similarly.

I think the younger twin is the one that accelerated but am confused by "The one who doesn't accelerate. Being at rest in a gravity field is an accelerated frame, which is one of the quirky concepts of GR."

I posted these four twin scenarios in another thread and there were no replies. I know that 1 and 2 are true so 4 should also be true. The calculators say 3 is not true unless Jane stops separating from the Earth and waits for Dick to make a similar trip speed and acceleration wise.

Since I am wrong much of the time would someone confirm that.

1. Dick takes a space trip at relativistic speeds and when he comes back to Earth he is younger the Jane.

2. Jane doesn't want to be old so she takes a similar space trip and comes back the same age as Dick.

3. Jane wants to be younger then Dick so she starts off an a space trip. Some time later Dick decides he doesn't want to be older so rather then wait for Jane to turn around and come back, he jumps in his space ship and goes to her. They are the same age?

4. They both want to be younger and take off an similar space trips but in the opposite direction. When they get back they will be the same age even though they were in different inertial frames the entire trip. Is that right? The separation and closure during the trip was equal.

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They would have been better off spending their time together, no? Sorry, I do agree with you. I think it's fun to ask how about the moving Dick flying thru nearby and we can synchronize clocks. We Janes have an extended (way) system of coordinated clocks which we think express a common time in our system, as we can listen over light-speed signal to each of their time values, and know we have corrected for the time of travel, right? Also, and here is the tweak, Dick has an extended system corrected for the delays they perceive between themselves, all moving to our observations. This is like the "police cars with looong antennas, or verrrry looong trains", whatever, babe. Anyway, this is gonna be a mess because y'all at your different points of view do not agree on simultaneity at a distance.

Edited by Norman Albers
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They would have been better off spending their time together, no?

I wouldn't want to spend that much time with my brother or sister but maybe twins would.

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Oh, sorry, I didn't figure the relation.

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Oh, sorry, I didn't figure the relation.

Twins

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Still, the fizziks IS. Now help me see the far-field question. I guess that this does not magically make the "locally speeding" observer see farther than we can in terms of the early universe. They should perceive different distances, and times???

Edited by Norman Albers
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...Dick has an extended system corrected for the delays they perceive between themselves, all moving to our observations. ....Anyway, this is gonna be a mess because y'all at your different points of view do not agree on simultaneity at a distance.

I have "simultaneity at a distance" when I run these scenarios through my head.

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You beee cool.

Edited by Norman Albers
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Still, the fizziks IS. Now help me see the far-field question. I guess that this does not magically make the "locally speeding" observer see farther than we can in terms of the early universe. They should perceive different distances, and times???

After you looking at my questions I am sure you do not want me answering your questions. My question is have we actually observed anything moving close to c that had transverse movement?

Edit - You bring up a very good point. The edge of our known universe would be very close at .99c.

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Whatever, make it 1/50 or any significant, measurable fraction. I guess this will distort our perception of a 'spherical far-field'. I do want to hear your take on my questions!

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