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

[Herman Trivilino]

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.

 

[Herman Trivilino]

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. :)

 

[Herman Trivilino]

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.

 

[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.

 

[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.

 

[Herman Trivilino]

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?