How would the solar system appear if you approached at near c?

[Chris Miller]

It's appearance wouldn't change much until you get close. I believe when you take into account the different amounts of time for light to reach you from different points, and that you move significant distance in the meantime, although the solar system will appear rotated, its shape shouldn't change much. It won't appear squashed in the direction of motion, for example, if my reasoning is correct.

Thanks, David. Given the velocity's so near c as to length contract 100 light years into 1 Planck length, you're already pretty close. Others here have explained that you wouldn't literally "see" anything what with relativistic Doppler pushing even the CMB up into the gamma end of the spectrum. My question was more what SR would suggest it "looked" like, and which, I think, is flat (length contracted) in your direction of travel and virtually frozen in time. I'd be interested in what your reasoning is here.

 

[Nugatory]

Seems like any constant would work for c, insofar as spacetime intervals would still be the same.

$c$ is only needed to correct because we're measuring distance along the time axis in seconds and distances along the space axes in meters. It serves the same role as the factor of 36 that would appear in the Pythagorean theorem if we perversely measured the length of one side of a right triangle in feet and the other side in fathoms. And any constant will not do - you must use the value of the speed of light in whatever units you've chosen or $\Delta{s}^2$ won't come out the same in both frames. You can verify this for yourself by picking two events at random, calculating the interval between them, using the Lorentz transforms to find the coordinates of these two points in some other frame, and calculating the interval using those new coordinaes.

In just about every serious modern treatment of relativity we get rid of all the factors of $c$ by choosing units in which $c$ is equal to one: distances in light-seconds and times in seconds, for example. That unclutters the equations dramatically without losing any of the fundamental physics.

Also seems like c² * Delta t² must be < Delta x² + Delta y² + Delta z² or you'd be looking at imaginary numbers?

No, although we do work with $\Delta{s}^2$ instead of $\Delta{s}$ to avoid any complex values.

When $\Delta{s}^2$ is negative, the separation between the two events is "timelike". There exists a frame in which they happened at the same place, but no frame in which they happened at the same time, and all frames agree about which happened first. A clock following an inertial path between the two events (meaning the clock is at rest in the "both happened at the same place" frame) will register $\sqrt{-\Delta{s}^2}$ seconds passing, and this is called the "proper time".

When $\Delta{s}^2$ is positive, the separation is "spacelike", and there exists a frame in which the two events happened at the same time, but no frame in which they happened at the same place. Different frames will disagree about which happened first (that is, which one has the smaller $t$ coordinate), and all will agree that no observer can be present both events without exceeding the speed of light.

(You should be aware that there are two different sign conventions out there; some sources put the negative sign on the spatial $\Delta$ values instead of the time one. The physics comes out the same either way as long as you're consistent; you just say that it's the positive $\Delta{s}^2$ intervals that are timelike.)

 

[David Lewis]

Given the velocity's so near c as to length contract 100 light years into 1 Planck length, you're already pretty close.

From your frame of reference, that is true. But if you had a tape measure between you and the solar system, the distance would still read 100 light years. The numbers on the tape would be close together from your perspective, however.

My question was more what SR would suggest it "looked" like,

SR tells you what happens, not what you see. To figure out what you see, you add in the laws of classical optics.

I forgot to mention, I think the sun would appear very bright as you move toward it at high speed, and the light would be blue-shifted.

 

[Laurie K]

I use the word "present" vs. "see" because I have no clue how you'd observe any of this, but am more interested in how SR would expect Earth's orbit around the sun to present if you, somehow, could.

Øyvind Grøn's paper 'Space geometry in rotating reference frames: A historical appraisal'' might help, Figure 9 part C shows an interesting solution for the "optical appearance" of a relativistically rolling ring. "http://areeweb.polito.it/ricerca/relgrav/solciclos/gron_d.pdf

Remember, SR "optical appearance" wise, no matter how fast an observer is traveling that observer can only capture those photons that would have been present at that observers discrete observation location and time anyway, regardless of whether the observer was actually there or not.

 

[Chris Miller]

. But if you had a tape measure between you and the solar system, the distance would still read 100 light years.

I don't think so. Else you'd travel 100 light years in 1 Plank interval of time, and probably get a speeding ticket.

 

[Chris Miller]

Øyvind Grøn's paper 'Space geometry in rotating reference frames: A historical appraisal'' might help, Figure 9 part C shows an interesting solution for the "optical appearance" of a relativistically rolling ring. "http://areeweb.polito.it/ricerca/relgrav/solciclos/gron_d.pdf

Remember, SR "optical appearance" wise, no matter how fast an observer is traveling that observer can only capture those photons that would have been present at that observers discrete observation location and time anyway, regardless of whether the observer was actually there or not.

Thanks, Laurie. I'll check that site out. I doubt that at near c you'd see much of anything due to relativistic Doppler effect shifting all EM radiation up into the gamma end of the spectrum. I was more curious as to how SR predicts/describes things than what you could optically observe.

 

[Chris Miller]

$c$ is only needed to correct because we're measuring distance along the time axis in seconds and distances along the space axes in meters. It serves the same role as the factor of 36 that would appear in the Pythagorean theorem if we perversely measured the length of one side of a right triangle in feet and the other side in fathoms. And any constant will not do - you must use the value of the speed of light in whatever units you've chosen or $\Delta{s}^2$ won't come out the same in both frames. You can verify this for yourself by picking two events at random, calculating the interval between them, using the Lorentz transforms to find the coordinates of these two points in some other frame, and calculating the interval using those new coordinaes.

In just about every serious modern treatment of relativity we get rid of all the factors of $c$ by choosing units in which $c$ is equal to one: distances in light-seconds and times in seconds, for example. That unclutters the equations dramatically without losing any of the fundamental physics.

No, although we do work with $\Delta{s}^2$ instead of $\Delta{s}$ to avoid any complex values.

When $\Delta{s}^2$ is negative, the separation between the two events is "timelike". There exists a frame in which they happened at the same place, but no frame in which they happened at the same time, and all frames agree about which happened first. A clock following an inertial path between the two events (meaning the clock is at rest in the "both happened at the same place" frame) will register $\sqrt{-\Delta{s}^2}$ seconds passing, and this is called the "proper time".

When $\Delta{s}^2$ is positive, the separation is "spacelike", and there exists a frame in which the two events happened at the same time, but no frame in which they happened at the same place. Different frames will disagree about which happened first (that is, which one has the smaller $t$ coordinate), and all will agree that no observer can be present both events without exceeding the speed of light.

(You should be aware that there are two different sign conventions out there; some sources put the negative sign on the spatial $\Delta$ values instead of the time one. The physics comes out the same either way as long as you're consistent; you just say that it's the positive $\Delta{s}^2$ intervals that are timelike.)

Thanks, Nugatory. It all makes sense now. Looking over this very basic quiz also helped clear things up: http://physics.bu.edu/~duffy/EssentialPhysics/chapter26/Chapter26_SampleProblems_Solutions.pdf

Working in Planck lengths and intervals instead of, as in the quiz, light years and years, would probably pose some huge integer/huge float challenges.

 

[Mister T]

From your frame of reference, that is true. But if you had a tape measure between you and the solar system, the distance would still read 100 light years. The numbers on the tape would be close together from your perspective, however.

If the tape measure is at rest relative to the solar system, what you say is true. But if I'm holding the tape measure it's at rest relative to me, and what you say is not true.

I don't think so. Else you'd travel 100 light years in 1 Plank interval of time, and probably get a speeding ticket.

In the rest frame of the rocket the distance is indeed $\frac{100}{\gamma}$ light years, so if $\gamma$ is large enough you can make it there traveling at a speed less than $c$ in any arbitrarily small amount of time. On the other hand, in the rest frame of the solar system the distance is $100$ light years, so the travel time would be larger than, but could be arbitrarily close to, $100$ years.

Some claims are valid in all inertial reference frames, such as the proper time that elapses between two events, the proper length of an object, and the speed of a light beam in a vacuum.

Some claims are valid in some reference frames and not in others.

Chris, the source of every unresolved confusion that's arisen in this thread (and in the others you started) can be traced to this issue.

 

[Mister T]

Working in Planck lengths and intervals instead of, as in the quiz, light years and years, would probably pose some huge integer/huge float challenges.

As long as you work in a system of units where $c=1$ you'll have no such problems. For example, light years and years, Planck lengths and Planck times, meters of length and meters of time, etc. N. David Mermin likes to work in nanoseconds and pheet, where one phoot is the distance light travels in a nanosecond of time. The foot has a length of $0.3048$ meters, the phoot has a length of $0.299\ 792\ 458$ meters, a difference of less than 2%.

Where you've run into the integer float issues is when you try to use a system where $c \neq 1$, specifically when $c << 1$ or $c >> 1$.

 

[Dale]

Working in Planck lengths and intervals instead of, as in the quiz, light years and years, would probably pose some huge integer/huge float challenges.

I have repeatedly recommended how you can fix that.

 

[David Lewis]

If the tape measure is at rest relative to the solar system, what you say is true [the tape will read distance correctly]. But if I'm holding the tape measure it's at rest relative to me, and what you say is not true.

If the tape travels with you, you won’t be able to measure how far you’ve traveled, or how far away you are from the solar system at each point in time.

If you fix one end of the tape 100 light years from the solar system, and the other end at the solar system, the tape will measure distances correctly both from the Earth, and from the moving ship.

If the tick marks on the tape are one light-minute apart then, looking out the ship’s window, you will see a tick mark go by every (just slightly over a) minute. At high speeds, the numbers will appear close together from the ship's frame, but the distance measurements will still be correct.

 

[Mister T]

If the tape travels with you, you won’t be able to measure how far you’ve traveled, or how far away you are from the solar system at each point in time.

Sure you will. Suppose you're at the origin of your tape measure. Just observe the solar system's location on your tape measure, and that tells you how far away it is. Observe another one later. Subtract the two readings and that will tell you how far you've traveled between the readings. Relative to the solar system, of course. Which goes without saying because otherwise there's no meaning to the phrase "how far you've traveled".

 

[Dale]

If you fix one end of the tape 100 light years from the solar system, and the other end at the solar system, the tape will measure distances correctly both from the Earth, and from the moving ship.

This is not correct. Indeed, it is not possible to make a device which can do that since the Earth and the ship disagree.

 

[Chris Miller]

Right, Dale. I was agreeing with you (re units of measure). I'm more interested in the equations than plugging arbitrary numbers into them, though I can see where this would be a good exercise. This little quiz from Boston U made it all pretty clear to me: http://physics.bu.edu/~duffy/EssentialPhysics/chapter26/Chapter26_SampleProblems_Solutions.pdf

 

[Mister T]

If the tick marks on the tape are one light-minute apart then, looking out the ship’s window, you will see a tick mark go by every (just slightly over a) minute. At high speeds, the numbers will appear close together from the ship's frame, but the distance measurements will still be correct.

This would be a strange way to use the word "correct". Using that contracted tape measure you'd conclude that your ship's length is larger than it's proper length.

If you looked out the window on the other side and saw a tape measure moving at a different speed from the other one you'd again get a different length. Maybe this is what people mean by the term length dilation.

 

[David Lewis]

If the tape measure is subdivided into light-years then, after (just over) one year of Earth time has passed, you should be able to look out the window of your spaceship and see the number "1" go by.

An Earth telescope watching that event will also (when the light reaches the astronomer 99 years later) show you pass by the number "1" on the tape.

 

[Ibix]

If the tape measure is subdivided into light-years then, after (just over) one year of Earth time has passed, you should be able to look out the window of your spaceship and see the number "1" go by.

An Earth telescope watching that event will also (when the light reaches the astronomer 99 years later) show you pass by the number "1" on the tape.

Of course. But that's measuring distances in the Earth frame. The 1ly divisions won't be a light year apart in the ship frame - so this isn't measuring distance traveled in the ship's frame. Indeed, the ship isn't moving in its own rest frame, so there's no distance traveled to measure.

 

[Dale]

If the tape measure is subdivided into light-years then, after (just over) one year of Earth time has passed, you should be able to look out the window of your spaceship and see the number "1" go by.

This is not how distances in the ship frame are measured. The tape measure cannot "measure distances correctly both from the Earth, and from the moving ship" as you claimed above.

 

[Mister T]

If the tape measure is subdivided into light-years then, after (just over) one year of Earth time has passed,

That's an event occurring on Earth ...

you should be able to look out the window of your spaceship and see the number "1" go by.

... and that's an event occurring on the ship.

If it's your intention that these two events are simultaneous in either the rest frame of Earth or the rest frame of the ship, then they won't be simultaneous in the other. Moreover, if they are indeed simultaneous in any frame then they have a spacelike separation, and therefore they could occur in different temporal order in different frames of reference.

 

[Laurie K]

This is not how distances in the ship frame are measured. The tape measure cannot "measure distances correctly both from the Earth, and from the moving ship" as you claimed above.

If a series of pulses was sent from a point stationary wrt the solar systems rotation (non moving frame) and each individual pulse was encoded to contain the sequential time of emission it could be used as a 'reference tape' by the observer in the space ship. The observer in the ship frame would know the time between the pulses on the ships clock and would also know the actual emission times between the pulses through the encoded data within each pulse and would therefore know the actual non moving distances between the pulses. If the observer calculated the distance to the solar system while stationary, prior to departing, they could also calculate their progress wrt the location of the emission point in the non moving frame.

 

[Dale]

would therefore know the actual non moving distances between the pulses

But those distances would not be the distances in the ship frame.

It isn't a matter of finding a clever measurement technique. The two frames fundamentally disagree about lengths, so if a given measurement is a valid measurement of length in one frame then it cannot be a valid measure of length in the other because the two frames disagree.

 

[David Lewis]

The 1ly divisions won't be a light year apart in the ship frame -

The marks on the tape are always at rest with respect to each other. Assume the ship begins at rest (with respect to the tape). Start the engine and the ship will accelerate. All the marks on the tape move together (with respect to the ship) at the same velocity. The only way for the distance between marks to change would be for one mark to move faster than another.

 

[Ibix]

The only way for the distance between marks to change would be for one mark to move faster than another.

Indeed. Accelerating frames of reference are odd. You may wish to look up Bell's spaceship paradox and its resolution for a related problem with a lot of discussion.

Alternatively, consider that if the different parts of an unaccelerated object don't move at different speeds as seen in your frame as you accelerate, length contraction could never happen.

 

[Laurie K]

The two frames fundamentally disagree about lengths, so if a given measurement is a valid measurement of length in one frame then it cannot be a valid measure of length in the other because the two frames disagree.

I'm probably thinking of more like how an airports ILS operates Dale.

The observer in the ships frame can measure the difference between any two consecutive pulses based on the ships time. They can then decode the emission times to determine the equivalent non moving frame (wrt solar system rotation) time between the two pulses. If the encoded part of the pulse also includes the positions of the major planets/sun in the solar system at the time of sending of the pulse further data would be available to the ships observer that can be compared with the locations of the major planets/sun obtained from the observations of the solar system made from the ship in the period between the two pulses.

I don't see any good reason why the encoded details contained in the two pulses should be much different to the details as observed on the ship as both were emitted from the solar system at the same time and the ship is a long way away.

 

[Dale]

Assume the ship begins at rest (with respect to the tape).

First, until now the discussion has been about an inertial ship and inertial frame. Now you are making it non-inertial. There is nothing wrong with that in principle, but i want it clear to you that this assumption that you are making here is an unnecessary and significant departure from the remainder of the thread.

Start the engine and the ship will accelerate. All the marks on the tape move together (with respect to the ship) at the same velocity.

Non inertial frames are not so trivial to specify, but in any non-inertial frame that locally matches the momentarily comoving inertial frame your statement is false. The marks on the tape do not move together at the same velocity.

One common non inertial frame is the radar coordinates frame, see https://arxiv.org/abs/gr-qc/0104077 In that paper the fact that the marks do not move together is called a shock discontinuity.

 

[Dale]

The observer in the ships frame can measure the difference between any two consecutive pulses based on the ships time. They can then decode the emission times to determine the equivalent non moving frame (wrt solar system rotation) time between the two pulses

Those two times would not be the same.

If the encoded part of the pulse also includes the positions of the major planets/sun in the solar system at the time of sending of the pulse further data would be available to the ships observer that can be compared with the locations of the major planets/sun obtained from the observations of the solar system made from the ship in the period between the two pulses.

And the observations made from would be found to disagree with the encoded data. I.e the solar system could send that information to the ship, but it would be wrong in the ship's frame and they could perform measurements to verify that it is wrong.

I don't see any good reason why the encoded details contained in the two pulses should be much different to the details as observed on the ship

Then you have not worked through the math on this or similar problems. I would strongly recommend that you do so before posting any further comments along this line.

 

[georgir]

tl;dr. don't know if someone mentioned relativistic aberration yet so let me.
I see someone said the universe is "hot in front of you and cold behind you", but that's really mostly because everything comes in front of you. Aberration of light changes apparent angles and makes everything more and more focused in your direction of motion.

 

[Mister T]

The marks on the tape are always at rest with respect to each other.

No one disagrees with this, or the other basic statements you've made, such as the claims about the marks seen on the tape from the ship's window. But in your responses you don't address the objections we've made to the conclusions you keep drawing from them.

The marks on the tape are at rest relative to each other. That doesn't mean that all observers agree on how far apart they are. There is an overwhelming amount of evidence to support Einstein's relativity. The tape does not correctly measure distances for observers moving relative to it.

 

[Dale]

No one disagrees with this

I disagree with it. The statement would be true in any inertial frame, but he specifically changed the scenario to a non inertial frame. The change to a non inertial frame was key to his argument, but it made the statement false. In many non inertial frames the marks on the tape are not always at rest to each other.

 

[Raymond Potvin]

Hi everybody,

This thread made me imagine a mind experiment about time: using only classical doppler effect, would we count more or less Earth cycles around the sun than those who really happened on Earth (years), during a round trip at 100 light years away from Earth at any speed, including at close to the speed of light, accelerations and decelerations included?

 

[PeterDonis]

using only classical doppler effect,

This question is not answerable in the context of relativity, because the Doppler effect in relativity is not an isolated thing; it's connected to all the other characteristic effects of relativity, like time dilation, length contraction, relativity of simultaneity, etc. You can't change just one thing.

If you want to pose this question in the context of straight Newtonian physics, you can do that in the Classical Physics forum. But you have to pick either the full theory of SR or the full theory of Newtonian mechanics; there is no consistent theory that picks parts of one and parts of the other.

 

[Raymond Potvin]

Hi Peter,

It's a mind experiment about how we would see a clock while moving wrt it, the same kind as the light clock one, but when the clock moves directly to or from the observer. So I think it is in fact a mind experiment about the relativity of motion, not a Newtonian one. As in the light clock mind experiment, the only way to measure the Earth cycles for us would be to use light, and considering doppler effect, there would be less Earth cycles than on our own inboard clock while we would be going away from the earth, and more Earth cycles while we would be going towards the earth. Back to Earth after a while, whatever the speeds involved, I think there would be no difference between our inboard clock and the earthbound one, thus if Einstein had chosen this mind experiment in stead of the light clock one, he would not have been able to prove his point. How come the two mind experiments do not give the same result?

 

[Janus]

Hi Peter,

It's a mind experiment about how we would see a clock while moving wrt it, the same kind as the light clock one, but when the clock moves directly to or from the observer. So I think it is in fact a mind experiment about the relativity of motion, not a Newtonian one. As in the light clock mind experiment, the only way to measure the Earth cycles for us would be to use light, and considering doppler effect, there would be less Earth cycles than on our own inboard clock while we would be going away from the earth, and more Earth cycles while we would be going towards the earth. Back to Earth after a while, whatever the speeds involved, I think there would be no difference between our inboard clock and the earthbound one, thus if Einstein had chosen this mind experiment in stead of the light clock one, he would not have been able to prove his point. How come the two mind experiments do not give the same result?

Your traveler would only see the same elapsed time on his own clock and the Earth's if you assume that light travels with respect to some medium.
However, Einstein did not assume this ( if he had, he wouldn't gotten time dilation with his light clock thought experiment either.)
If we assume an invariant speed of light, like Einstein does, then the amount of time you see the Earth clock losing while traveling away will not equal the time you see it gain on the return leg and you will see it gain a greater net time than your own clock during the trip.

 

[Raymond Potvin]

You are using the conclusions of SR in a pre-SR mind experiment. With doppler effect, what you see is what you get. It happens the same way for light or for sound, it only depends on motion between source and observer. If we are moving away from the earth, we will measure more seconds per year than if we are moving towards it. With the light clock, we could assume that light took more time while traveling sideways to the motion, but not this time. This time, it is traveling directly between us and the earth, so it doesn't take more time wether we are traveling one way or another: when we will get at one light year from the earth, we will see a light that has traveled during a year. It will only be blueshifted or redshifted whether we are traveling towards the Earth or away from it at that moment.

 

[PeterDonis]

It's a mind experiment

Which, as you pose it, is not using a self-consistent set of assumptions. You can conclude anything from an inconsistent set of assumptions. So your mind experiment is not telling you anything.