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

[jbriggs444]

Right, thanks. But its end point coordinates (and so length, etc.) will vary by observer, won't it?

Its endpoint coordinates will vary by observer, yes.
Its length will not vary by observer, no. What is the formula for length in 4D space time?

 

[Chris Miller]

Its endpoint coordinates will vary by observer, yes.
Its length will not vary by observer, no. What is the formula for length in 4D space time?

Interesting (though guessing rhetorical) question. Would have to incorporate both distance and duration. How "long" is the line from x_b,y_b,z_b,t_b to x_e,y_e,z_e,t_e? Hmm... I'll commence to cipher on it.

EDIT:
But I'm sure it won't be the same as for x'_b,y'_b,z'_b,t'_b to x'_e,y'_e,z'_e,t'_e

 

[Dragon27]

Right, thanks. But its end point coordinates (and so length, etc.) will vary by observer, won't it?

The coordinates of the event vary by observer. And they did so in classical, non-relativistic mechanics too. The formulas for the coordinates transformation were different, of course. Galilean.

But I'm sure it won't be the same as for x'_b,y'_b,z'_b,t'_b to x'_e,y'_e,z'_e,t'_e

Will it be, or will it not? But more importantly - how would you know the answer to this question? Do you hope to just intuitively guess the answer, or would it be better to read up on it in a textbook? Because that could happen to be one of the most basic pieces of knowledge about Special Relativity, essential to understanding of anything about 4 dimensional Minkowski space.

 

[Nugatory]

Right, thanks. But its end point coordinates (and so length, etc.) will vary by observer, won't it?

The coordinates of the end points will be different with different frames - not surprising when you consider that a "frame" is just a convention used to assign these coordinates. However, all the geometric facts about the line, anything that you can say about it without reference to coordinates (it does or does not pass through a given event, it does or does not intersect another line or a surface, ...) will of course hold in all frames. Less obviously, but at least as important, the length (properly termed a "spacetime interval") of the line will be the same in all frames, even though the coordinates of the endpoints are different.

The spacetime interval $\Delta{s}$ between two events $(t_1,x_1,y_1,z_1)$ and $(t_2,x_2,y_2,z_2)$ is given by $\Delta{s}^2={-c^2\Delta{t}^2+\Delta{x}^2+\Delta{y}^2+\Delta{z}^2}$ where $\Delta{x}$ is defined as $x_2-x_1$ and likewise for the other three coordinates, and it comes out the same in all frames even though the individual coordinate values are different.

 

[Herman Trivilino]

But its end point coordinates (and so length, etc.) will vary by observer, won't it?

The "length" of an observer's worldline is the proper time that elapses for that observer. It's a relativistic invariant, meaning it will have the same value in all coordinate systems. The coordinates of the endpoints will vary, though.

 

[Herman Trivilino]

Interesting (though guessing rhetorical) question. Would have to incorporate both distance and duration. How "long" is the line from x_b,y_b,z_b,t_b to x_e,y_e,z_e,t_e? Hmm... I'll commence to cipher on it.

Only someone of the caliber of a would-be Nobel prize winner would be able to figure that out on his or her own.

EDIT:
But I'm sure it won't be the same as for x'_b,y'_b,z'_b,t'_b to x'_e,y'_e,z'_e,t'_e

It will be the same.

I recommend that you state your original query in terms of these coordinates. For example, you could let all the y's and z's be zero, and let the x-axis and the x'-axis lie upon the same line, the line that passes through both the spaceship and Earth's sun.

 

[Chris Miller]

The coordinates of the end points will be different with different frames - not surprising when you consider that a "frame" is just a convention used to assign these coordinates. However, all the geometric facts about the line, anything that you can say about it without reference to coordinates (it does or does not pass through a given event, it does or does not intersect another line or a surface, ...) will of course hold in all frames. Less obviously, but at least as important, the length (properly termed a "spacetime interval") of the line will be the same in all frames, even though the coordinates of the endpoints are different.

The spacetime interval $\Delta{s}$ between two events $(t_1,x_1,y_1,z_1)$ and $(t_2,x_2,y_2,z_2)$ is given by $\Delta{s}^2={-c^2\Delta{t}^2+\Delta{x}^2+\Delta{y}^2+\Delta{z}^2}$ where $\Delta{x}$ is defined as $x_2-x_1$ and likewise for the other three coordinates, and it comes out the same in all frames even though the individual coordinate values are different.

Thanks very much, Nugatory. Helpful and interesting.

EDIT
Seems like any constant would work for c, insofar as spacetime intervals would still be the same. Also seems their 3D (x,y,z only) lengths could be different and that t is what then equalizes, and which I see as length contraction being compensated/corrected by time dilation (or vice versa).

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

 

[Chris Miller]

Only someone of the caliber of a would-be Nobel prize winner would be able to figure that out on his or her own.

I've always disliked school. Always felt like I was being fed solutions faster than I could understand (or at least develop an interest in) the problems. I'd rather spend time trying to solve a problem than have the answer handed to me. After I've failed, the solution is a lot more relevant and interesting to me. That would have been a fun geometric problem to try to tackle. I feel like I've finally got a handle on some of the basic SR concepts, and this site has been a huge help.

 

[Chris Miller]

The coordinates of the event vary by observer. And they did so in classical, non-relativistic mechanics too. The formulas for the coordinates transformation were different, of course. Galilean.
Will it be, or will it not? But more importantly - how would you know the answer to this question? Do you hope to just intuitively guess the answer, or would it be better to read up on it in a textbook? Because that could happen to be one of the most basic pieces of knowledge about Special Relativity, essential to understanding of anything about 4 dimensional Minkowski space.

Seems I guessed wrong about the "spacetime interval." This, unfortunately, is how I learn (I've learned a lot today!) I imagine it's all online now. May look around, now that I know better where to look. You guys have been very helpful and generous with your responses.

 

[David Lewis]

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

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.

 

[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.

 

[Herman Trivilino]

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.

 

[Herman Trivilino]

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.

 

[Herman Trivilino]

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

 

[Herman Trivilino]

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.

 

[Herman Trivilino]

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.