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

[Dale]

And length can only be measured if the ends are looked at simultaneously (in your frame), correct?

Yes, exactly. If you don't measure the distance at the same time then you can get things like "my car is 100 km long because my front bumper is 100 km away from where my rear bumper was an hour ago"

 

[Chris Miller]

In the stationary frame, where ship was initially 100 lights away from Earth? No, by definition of this frame. If we place an asteroid, or whatever, at this origin point, then we will have an "asteroid-Earth" object, that is 100 light years long and is at rest in the stationary frame. This object is 1 Planck length long and is moving towards the ship in the near-c frame. Likewise, the ship has normal rest length in the near-c frame and almost zero length in the stationary frame. Seems reciprocal enough.You can just keep hitting the wall with your head, hoping that understanding will somehow just fall onto your head, or you can try and work through a textbook. Special relativity is not too complicated a theory, but it has its subtleties, and if you want to get it, you just have to put in some serious, systematic effort. You have to learn to use and understand space-time diagrams and Lorentz transformation formulas - these are the essential concepts, that you're missing here. All the special effects (time dilation, length contraction, relative simultaneity, etc..) are derived from them.An inertial near-c frame wouldn't be inertial if it didn't exist forever, so to speak. There should be enough past in it to contain all the Earth's revolutions that have ever happened.

Again, your patience, advice and clarifications are very much appreciated. I (think I) get now that the Earth and 100 light years of interceding space represent a stationary frame, and the ~c ship another. So it makes SR sense that each would "see" the other as length contracted.

I have enough of a math background that the simpler Lorentz transformations (those not involving arbitrary direction and rotation) are easy to understand and use (mathematically). It's plugging the results into a narrative that makes sense that gives me trouble. In other words, as a computer programmer, they're easy to work with. As a science fiction writer, less so. Especially when it comes to temporal conjunctions like while, when, before, after , during, and even (and especially) now (slightly off topic, but doesn't quantum entanglement sort of imply a universal "now"?).

 

[Chris Miller]

Suppose the first of those revolutions occurs during the year 1901 and the 100^th during the year 2000. What do you find/imagine/describe happened during the year, say, 1865?

Among other things, Earth revolved about the sun one time.

Yes, of course. And so during my ~c ship's t = -n nanosecond, the Earth orbited the sun 100 times?

 

[Chris Miller]

If the bold is your main question, I think someone pointed out that an orbit can be considered a clock, and if you're at rest in the spaceship then it appears to me that the Earth's orbit would constitute a "moving clock" according to your reference frame. Yes, no? Maybe so?

Apparently, from my ~c spaceship the earth/clock would appear almost frozen in its orbit, just as my on board clocks would to Earth observers.

 

[Chris Miller]

At relativistic speeds the universe is not spatially homogenous or isotropic, so it doesn't have a single uniform temperature. It is hot in front of you and cold behind you. Similarly with density and many other properties.

Interesting. Is this due to Doppler effect? If one could measure the cosmic microwave background radiation's wavelength accurately enough, one could always point (e.g., on Earth N,S,E,W) in the direction and compute the velocity one was traveling with respect to... to what?

 

[jbriggs444]

Interesting. Is this due to Doppler effect? If one could measure the cosmic microwave background radiation's wavelength accurately enough, one could always point (e.g., on Earth N,S,E,W) in the direction and compute the velocity one was traveling with respect to... to what?

With respect to the frame in which the CMBR is isotropic.

 

[jbriggs444]

but doesn't quantum entanglement sort of imply a universal "now"?).

No. It does not.

 

[Nugatory]

but doesn't quantum entanglement sort of imply a universal "now"?).

Only because the introductory quantum mechanics that you first encounter as an undergraduate is the simpler non-relativistic form of the theory, applicable only when the speeds are small compared with the speed of light and the energies of the particles involved are small compared with the $E=mc^2$ energy implied by their rest masses. (This is why you don't find photons in your introductory QM classes - neither assumption works for them).

Quantum Field Theory, which is the full relativistic version of the quantum mechanics, does not have these limitations and does not imply any sort of universal "now". You're only likely to encounter a serious QFT class after you've completed a bachelor's in physics and are starting in on an advanced degree.

 

[Chris Miller]

Only because the introductory quantum mechanics that you first encounter as an undergraduate is the simpler non-relativistic form of the theory, applicable only when the speeds are small compared with the speed of light and the energies of the particles involved are small compared with the $E=mc^2$ energy implied by their rest masses. (This is why you don't find photons in your introductory QM classes - neither assumption works for them).

Quantum Field Theory, which is the full relativistic version of the quantum mechanics, does not have these limitations and does not imply any sort of universal "now". You're only likely to encounter a serious QFT class after you've completed a bachelor's in physics and are starting in on an advanced degree.

Thanks, will take your word. Seems we're closer to a unified field theory than I had believed.

 

[Dragon27]

Again, your patience, advice and clarifications are very much appreciated. I (think I) get now that the Earth and 100 light years of interceding space represent a stationary frame, and the ~c ship another.

I'm not sure what is meant by the word "represent" here. As has already been mentioned, length contraction is the effect of a decrease of the length of an object (the "proper" length of an object is measured in the frame where this object is at rest - the "asteroid-Earth" object is at rest in the stationary frame of reference and thus has a proper length of 100 light years) as seen from an inertial frame of reference with respect to which the object is moving.

I have enough of a math background that the simpler Lorentz transformations (those not involving arbitrary direction and rotation) are easy to understand and use (mathematically). It's plugging the results into a narrative that makes sense that gives me trouble.

Then you should work with them. Why don't you try it? Take our stationary frame of reference (the ship is at the origin, but moves with near light speed towards the Earth, which is stationary at the distance of 100, or whatever number seems more convenient to you, light years from the ship and has 0 revolutions/years on its clock) and near-c frame of reference (moves towards the Earth with the same velocity as the ship, and has this ship at the origin). Lorentz transformations in its standard "wikipedian" form could be directly applied here. Calculate directly with the use of Lorentz transformations in the near-c frame of reference: how far is the Earth from the ship?; what time is on the Earth's clock?; how long will it take for the ship to reach the Earth?; what time would be on the Earth's clock then?; what can you say about the speed of the Earth's clock (how fast do they tick) from this? Choose any convenient numbers for the distance between the Earth and the ship at the beginning (in the stationary frame) and the velocity of the ship. You should be able to independently derive length contraction and time dilation from this exercise, and, if you think about it and play with the formulas, why and when those formulas can and cannot be applied.

Yes, of course. And so during my ~c ship's t = -n nanosecond, the Earth orbited the sun 100 times?

You can calculate this one too!

 

[Mister T]

Yes, of course. And so during my ~c ship's t = -n nanosecond, the Earth orbited the sun 100 times?

If by $t=-n$ you're referring to what you previously mentioned as an event occurring at a time $t<0$, then no. It's a clock-reading $t$ not an elapsed time $\Delta t$. No time elapses during a clock-reading as clock-readings last for zero time. Things can happen at a clock-reading, and we call those things events. Events last for no time at all.

 

[Chris Miller]

Events last for no time at all.

So there's no ship's delta t associated with Earth events? Is this what my head can't get around?

 

[Mister T]

So there's no ship's delta t associated with Earth events? Is this what my head can't get around?

$\Delta t=0$ for any event in any frame. Events occur at a time, not for a time. When we say, for example, that an orbit is a year long, what we mean is that if we take a clock reading $t_1$ at the beginning (Event 1) and a clock reading $t_2$ at the end (Event 2) then the difference $\Delta t=t_2-t_1=1$ year.

 

[Dale]

Interesting. Is this due to Doppler effect? If one could measure the cosmic microwave background radiation's wavelength accurately enough, one could always point (e.g., on Earth N,S,E,W) in the direction and compute the velocity one was traveling with respect to... to what?

With respect to the local inertial frame in which the CMBR is isotropic. This frame is called the "co moving" frame for brevity.

 

[Dale]

So there's no ship's delta t associated with Earth events? Is this what my head can't get around?

Events are "points" in spacetime. They are 0 dimensional. They happen at one specific moment in time and one specific location in space. By definition they have no length, width, height, or duration.

 

[Mister T]

Quick question regarding this: since the Lorentz transform is Δx = γ(Δx'+vΔt') and Length contraction is L₀/γ = L, is it technically speaking correct that for a transformation where Δt'=0 that you have the length contraction formula? [i.e. Δx = γ(Δx'+vΔt') at Δt' = 0 is Δx = γΔx', which is Δx'/γ = Δx]

Note that if $\Delta x=\gamma \Delta x'$ then $\frac{\Delta x}{\gamma}=\Delta x'$. You somehow got your last equation wrong.

Perhaps this explains why the responses confuse me.

The way I see it $\Delta x$ is the proper length. If it's the length $L_o$ of a rod that's at rest in the unprimed frame, and $\Delta x=x_2-x_1$ where $x_1$ and $x_2$ are the coordinates of the rod's endpoints, then the value of $\Delta t$ is not relevant. Since the rod's not moving in the unprimed frame, it doesn't matter when you measure the location of its endpoints.

Thus it correctly describes length contraction, as you intended.

 

[Chris Miller]

Thanks Dale and Miser T. Clearing up my terminology (the math is dead simple) might be freeing some conceptual blocks for me. So, if my ship's t₁ = Earth begins its first of 100 solar revolutions and t₂ = Earth finishes its 100th revolution, then my t₁ and t₂ describe the end points of a line through 4D spacetime, and a completely different line than described by Earth's t'₁ and t'₂ for these same events? Of course this is still a lot easier to envision geometrically than existentially.

 

[jbriggs444]

Thanks Dale and Miser T. Clearing up my terminology (the math is dead simple) might be freeing some conceptual blocks for me. So, if my ship's t₁ = Earth begins its first of 100 solar revolutions

That terminology needs work. If you want to understand it right, you have to say it right. [In my opinion]

"The event where and when your ship's clock reads t₁ is simultaneous according to the ships rest frame with the event where and when the Earth begins the first of 100 solar revolutions".

That would be a meaningful statement.

[...]then my t₁ and t₂ describe the end points of a line through 4D spacetime

"The event where and when your ships clock reads t₁ and the event where and when your ship's clock reads t₂ are the end points of a line through 4D spacetime".

That would be a meaningful statement

and a completely different line than described by Earth's t'₁ and t'₂ for these same events?

It is not clear what you are saying there. If the two events being specified are the same then the line between those two events is the same. Regardless of whether their primed and unprimed coordinates are the same.

But you have not specified the events corresponding to t'₁ and t'₂ yet. You've just said when and not carefully specified where.

 

[Mister T]

So, if my ship's t₁ = Earth begins its first of 100 solar revolutions and t₂ = Earth finishes its 100th revolution, then my t₁ and t₂ describe the end points of a line through 4D spacetime,

They do not because you have specified only the time coordinates. You need to also include the position coordinates.

and a completely different line than described by Earth's t'₁ and t'₂ for these same events?

Again, you'd need to specify the position coordinates in the primed frame.

Note that without this information there's really no way to make sense of what you're saying. I tried, but I couldn't do it. For example, I need to know how you, in your space ship, are measuring the position of Earth's sun. You could be doing it using a position axis in which you're at rest, the sun is at rest, or neither are at rest. Without this framework there's no way to sketch up a spacetime diagram, and without coordinates, no way to do a Lorentz transformation. This is essentially the source of the confusions that have arisen in these threads.

 

[Battlemage!]

Apparently, from my ~c spaceship the earth/clock would appear almost frozen in its orbit, just as my on board clocks would to Earth observers.

To me this is one of the least confusing aspects of special relativity. The underlying principle is that all inertial frames are equally valid, and combined with the light clock example we all learn in week one of a modern physics class, it really makes perfect sense to me. But then, everyone is different. Some of the subtleties are definitely confusing.

If it applies to a man made clock, it should apply to any periodic motion. Because after all, clocks are made out of matter. They are only special because we assign meaning to them.

Oddly I've had people argue with me that just because this happens with clocks doesn't mean it "really" happens with time, but what is time if not a rate at which we measure how matter changes? As I see it, if the motion of the atoms that make an object run slower, and all systems that interact with those atoms also run slower, what precisely is the difference between time moving slower and "just the physical stuff" moving slower?

Note that if $\Delta x=\gamma \Delta x'$ then $\frac{\Delta x}{\gamma}=\Delta x'$. You somehow got your last equation wrong.

Perhaps this explains why the responses confuse me.

The way I see it $\Delta x$ is the proper length. If it's the length $L_o$ of a rod that's at rest in the unprimed frame, and $\Delta x=x_2-x_1$ where $x_1$ and $x_2$ are the coordinates of the rod's endpoints, then the value of $\Delta t$ is not relevant. Since the rod's not moving in the unprimed frame, it doesn't matter when you measure the location of its endpoints.

Thus it correctly describes length contraction, as you intended.

Yeah "somehow" is the right word ahaha. Don't know how I messed that up. :)

 

[Mister T]

Oddly I've had people argue with me that just because this happens with clocks doesn't mean it "really" happens with time,

And note that there is no meaningful way to dispute that claim. My response is to ask that if clocks don't measure time, then how would you propose to measure it? The point can be made that it's a nonphysical issue because there is no way to tell the difference between a universe in which clocks measure time and a universe in which they don't. Thus one is left only with the belief in some notion of a time that can't be measured. To me it's a supernatural belief that time is not an invention of the human intellect. Certainly anyone would agree that the standards we use to measure time are conventions invented by humans. I don't understand how one would propose to measure time without one of those standards, I don't understand how to give meaning to the notion of time without a way to measure it, so I therefore don't understand how time can't be a human invention.

 

[Battlemage!]

And note that there is no meaningful way to dispute that claim. My response is to ask that if clocks don't measure time, then how would you propose to measure it? The point can be made that it's a nonphysical issue because there is no way to tell the difference between a universe in which clocks measure time and a universe in which they don't. Thus one is left only with the belief in some notion of a time that can't be measured. To me it's a supernatural belief that time is not an invention of the human intellect. Certainly anyone would agree that the standards we use to measure time are conventions invented by humans. I don't understand how one would propose to measure time without one of those standards, I don't understand how to give meaning to the notion of time without a way to measure it, so I therefore don't understand how time can't be a human invention.

Well, I would say time is just that thing that keeps everything from happening at once, but regardless, any and all clocks are just objects with periodic motion, and if EVERY object with periodic motion in a region changes with the same "speed", then what other possibility is there other than "that which keeps everything from happening at once" is evolving at that rate? I know talking of rates of time would be dimensionless (time/time), but clearly time is related in some way to rates of change, and if we're talking about how clocks behave in one region with respect to another in the context of SR, we're talking about "natural" clocks as well as human made clocks. And if every "natural" clock in the region is doing the same thing, then how could anyone distinguish between "time" behaving in a certain way versus "just clocks" behaving that way?

It would be like if you woke up and everything in your region got ten times bigger, and you could not possibly tell unless you compared it to another region that didn't (i.e. as far as you can tell in your reference frame nothing changed). You can't just say the measuring rods got bigger, because every thing that could possibly be used to measure things got bigger, too. Which also means the distance between objects got bigger.

Okay sorry this is getting way into philosophy.

 

[Chris Miller]

"The event where and when your ship's clock reads t₁ is simultaneous according to the ships rest frame with the event where and when the Earth begins the first of 100 solar revolutions".

That would be a meaningful statement.

"The event where and when your ships clock reads t₁ and the event where and when your ship's clock reads t₂ are the end points of a line through 4D spacetime".

That would be a meaningful statement

Thanks Schultzy (aka jbriggs444) et al. for help with the language. (Just if you care, periods go inside closing quotes and posessive-indicating apostrophes are not optional.)

I believe I understand now how events are points in 4D spaceftime whose x,y,z,t coordinates are relative to the observer (since there's no universal coordinate system [akin say to Earth's longitude/latitude]), so that any two events are the end points of a line unique to each observer.

(I also understand better why SAE wins so many "bad writing" contests.)

 

[jbriggs444]

so that any two events are the end points of a line unique to each observer

The line is not unique to each observer. There is just one line. It is there for all observers. It is the same line.

 

[Chris Miller]

The line is not unique to each observer. There is just one line. It is there for all observers. It is the same line.

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

 

[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.

 

[Mister T]

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.

 

[Mister T]

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.