# B A universal compass/speedometer

1. Nov 8, 2016

### Chris Miller

Given a perfect sphere with the ability to measure any of its diameters at any time, if traveling at relativistic velocities, could I use length-foreshortening to determine my speed and direction?

2. Nov 8, 2016

### Staff: Mentor

No; There is none universal speed/direction. You can only identify your speed/direction in the reference frame you have chosen.

 I may have misread: if you are proposing to measure absolute speed/direction using a sphere you carry with you, that is even a little bit worse (sorry). Time dilation and length contraction are not things that ever happen to you locally. They only happen when measured against objects/frames of references that you are moving with respect to.

Last edited: Nov 8, 2016
3. Nov 8, 2016

### Chris Miller

Thanks so much for the explanation. So, if someone on earth could (somehow) take measurements of my sphere, they could determine my direction and velocity relative to them?

4. Nov 8, 2016

### Orodruin

Staff Emeritus
If they knew it was a sphere to begin with and account properly for the travel time of light. Frankly, there would be easier ways of measuring the velocity.

5. Nov 8, 2016

### Chris Miller

Thank you, Orodurin. Of course. It's more a thought experiment than technological suggestion. This all stems from an obsession I've developed in knowing how the universe would appear to me as I near v=c. Like, if I'm approaching a 100 light year distant (by earth measurement) star at v sufficient to bring it to within a few miles, how does it present? How wide? How bright/hot. How do I experience its gravitational forces? And how fast is the earth spinning around the sun?

6. Nov 8, 2016

### A.T.

This might be of interest to you:
http://www.spacetimetravel.org/

7. Nov 8, 2016

8. Nov 8, 2016

### Janus

Staff Emeritus
There's a couple of things you need to take into account. As you are passing the Earth, you and the Earth are both seeing light that left that star at the same time.

Now let's use some numbers that are a bit easier to plug into an equation, so we'll say that the star is 1 light year away as measured by the Earth and you are passing the Earth at a speed such that, according to you, at that moment it is only 1 AU away. This means that you are traveling at 0.999999999875c relative to the Earth, Or from your perspective, the star is rushing towards you at that. speed. Now take that moment when the star, according to you, is 1 AU away and mentally rewind events backwards from it. The light you saw from the star is now traveling away from you and back to the star at c, the star meanwhile is rushing away from you at 0.999999999875c. If you work out how long it would take fro the light to catch up with the star, it turns out to be ~121,680 years.
This means that the star light that you see at the moment you passed the Earth, left that star when it was 121,680 light years away according to you. The star itself was following close on its heels. The star will actually visually appear much much smaller to you than it does to the Earth. This is also what you would expect to see according to the aberration of light, which causes light coming from in front of you to appear to come from a smaller and smaller angle as you move faster and faster with respect to the source.

On top of that, you wouldn't be able to see the star in the visible light range at all. At that relative velocity, even radio waves from the star would be shifted into gamma ray part of the spectrum.

9. Nov 8, 2016

### pervect

Staff Emeritus
I'd suggest thinking about in in the following way. We have two observers, Andy and Bella, who each have different velocities. Then Andy can create a perfect sphere, and so can Bella.

Now the question becomes, is there any way to accelerate Andy's sphere so that it's velocity matches Bella's velocity, without changing its shape? Can we move Andy's sphere so that it coincides with Bella's? (Of course, physical objects can't occupy the same space at the same time, so the question becomes more of a mathematical one). In other words, is there any way to make a perfect _rigid_ sphere?

The answer is a definite maybe - it also requires some discussion of what we mean by "rigid". If we consider the flat space-time of special relativity, as long as both Andy and Bella's spheres are not rotating, we can define what's called "Born rigid motion", which is a way of accelerating Andy's sphere to match Bela's in such a manner that all nearby points of the sphere maintain the same distance to all other nearby points during the acceleration process. Because of the way this works, if we accelerate Andy's sphere in the proper Born-rigid manner, we could in principle make Andy's sphere coincide with Bella's.

If we consider the non-flat space-time of general relativity, the answer in general is no. I rather suspect that it's impossible to even create a set of worldlines representing a sphere that remains perfect for all time, though without a detailed reference or a ready proof, I should be cautious about saying more.

Anyway, the key point that I want to make is this. There's probably an assumption in the original question that the perfect spheres are also rigid. This turns out to be the trickiest part of the question, an important unstated assumption.

10. Nov 9, 2016

### Chris Miller

Great forum! Your answers have really helped clear up some of my thinking.

pervcet: If I understand russ_watters correctly, a perfect sphere created in any frame of reference will be perfect in all frames of reference (and imperfect examined from any other)?

A.T. and Nugatory: Thanks for the links.

Janus: Your thoughtful remarks have raised further questions.

I understand that, "As you are passing the Earth [i.e., on my way toward the star], you and the Earth are both seeing light that left that star at the same time." But by my dilated time, wouldn't it be only about 8 minutes old?

"The light you saw from the star is now traveling away from you and back to the star at c."
How is the light from the star now traveling (reflected?) back to the star?

"The star will actually visually appear much much smaller to you than it does to the Earth."
So, if I could instantaneously accelerate to 0.999999999875c from earth toward the sun, the sun would appear smaller than if it were 100 light years away from earth's vantage? At what v would this hold true when I was only 1 km from the sun? Blows my mind a little.

"At that relative velocity, even radio waves from the star would be shifted into gamma ray part of the spectrum."
Very interesting. Is this because of length foreshortening? If so, wouldn't they revert to their original wavelengths after entering my frame of reference?

11. Nov 9, 2016

### Staff: Mentor

There is no such thing as "entering a frame of reference" - everything is always in all frames at all times. The frame is a convention that you as an observer use to assign times and positions to events.

The blueshifting to shorter wavelengths and higher frequencies is just the Doppler effect, the same phenomenon (although with relativistic corrections the math is more complicated) that causes sounds emitted by an approaching source to reach your eardrum with a higher frequency than the same sound emitted by the same source when it is approaching. Say the star is emitting radio waves at 100 MHz; that means that a wavecrest leaves the star every ten nanoseconds, and if you were at rest relative to the star that's the frequency they would reach you at; the two crests leaving the star at time zero and time 10 nsecs would reach you at time $T$ and $10+T$ nsec where $T$ is the time it takes for the light to travel through the space between you and the star. However, if you are moving towards the star at a very large fraction of the speed of light, each successive wave crest will have to travel a shorter distance to get to you, so those two crests will reach you at time $T$ and $10+T-\Delta$ where $T-\Delta$ is the shorter time the light spends in flight because you've moved closer to the star; the frequency of crests reaching you is greater than 100 MHz.

You'll want to google for "relativistic Doppler effect" to see exactly how to calculate this effect while allowing for time dilation, length contraction, and the constant speed of light.

12. Nov 9, 2016

### jbriggs444

Frames of reference are not things that objects enter into or exit from. All objects at all times are "in" all frames of reference. Frames of reference should be thought of as different ways of assigning coordinates to events. The light does not change wavelength. It is measured to have different wavelengths depending on exactly how one does the measurement.

Edit: And @Nugatory wins again. :-)

13. Nov 9, 2016

### Chris Miller

Thanks you two. All I wanted to do was set out at 1g acceleration until I hit ~c (a bit < year, my time) and see what happened to the universe in that last second. Probably best I look out my back window?

I understand relativistic Dopler now, again thanks. So length foreshortening doesn't impact EM wavelengths? It would compression waves though, right?

14. Nov 9, 2016

### jbriggs444

In a year of proper time, you'll still be at rest in your instantaneous tangent inertial rest frame. And you won't be going at c in anyone's inertial frame. Relativistic velocity addition prevents you from ever accelerating to c.

But perhaps you are groping toward something like... http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html

Change reference frame and you change wavelength. But the light didn't change. Only the way you labelled it changed.

15. Nov 9, 2016

### Mister T

An object that's measured to be a sphere by Reference Frame A will not be a perfect sphere when measured by any other reference frame that's in motion relative to Reference Frame A.

16. Nov 9, 2016

### Janus

Staff Emeritus
As far as you are concerned, you are not time dilated, everything else is moving at close to c with respect to you and it time dilated.
You misunderstood. I'm not talking about light being reflected back to the star, I'm asking you to start at the moment you pass the Earth and mentally run time backwards from that moment. You are in essence back-tracking the light and the star to the moment the light you see at the moment you pass the Earth left the star. It's like this. You have two friends that left the same gas station at the same time on their way to you. One is driving at 60 mph and the other at 50. At the moment the first friend arrives, you get a call from the second one saying he is five miles away. How far away is the gas station that they left? The answer (30 miles), is he same whether you run events forward from when they left the gas station or if you run them backwards from the moment one friend arrives and you get the call.
It's best not to bring acceleration into a discussion of Relativity until you are really comfortable with it under uniform velocity conditions, as it opens a whole new can of worms. But you can think of it like this. Imagine sitting in a parked car during a snowfall. The snow is falling straight down, so the snow falling on the car is coming from directly over head. Now start driving, The same snow now appears to be coming from somewhere ahead of the car at an angle, the faster you drive the closer this angle becomes to being horizontal. You get the same effect with light. Light that would appear to come from a point 45 degrees from directly in front of you if you are at rest with respect to the source, will come at you from a narrower angle if you are moving towards the source, the greater your speed, the narrower the angle. It is caller the aberration of light.
It is Doppler shift. And while there are Relativistic effects that contribute to the exact degree of the shift, it can be basically explained as due to a decrease in propagation delay as the distance between you and the source decreases. Imagine a peak of an electromagnetic wave emitted by a source 10 light sec away and moving at .5c towards you. The peak will take 10 sec to reach you. The source is emitting a 1 hz wave, so 1 sec later it emits the next peak. By this time its has moved 0.5 light seconds closer to you, and takes 9.5 sec to reach you. Since it was emitted 1 sec after the first peak was, it reaches you 1+9.5 =10.5 secs after the first peak was emitted, this means you will only see a 10.5-10 = 0.5 sec time difference between the arrival of the peaks of the successive waves. The waves arrive at a higher frequency for you than emitted at the source.
Of course in this example I didn't account for Relativity. When you do, the answer comes out to 1.732 times the source frequency and not 2 times.

17. Nov 9, 2016

### pervect

Staff Emeritus
Yes, I'd just like to emphasize that I agree with this point for the benefit of the OP.

To go further, I'll try and explain what I see as the key underlying issue. Given a moving object, when measuring it's shape, one needs to define the concept of "now" in order to measure the shape. I.e. one takes the moving object, identifies a moment in time, and determines exactly where every point on the object is at that instant. If the concept of "now" changes, and the object is moving, the shape of the object changes, too. It seems to be hard to get across the idea that the concept of "now" isn't universally shared, that one's idea of "now" dependent on the frame of reference. But even though it's hard to get across, I'll try to mention the issue, and hope for the best. Using ordinary English to express the concept (as I've attempted to do) is very imprecise, but it seems that the more exact mathematical statements are just too long, so I'll go with the imprecise, informal description and hope for the best.

18. Nov 10, 2016

### Chris Miller

Again, what a great forum. Thank you all for taking the time to address and discuss my (I now realize) dumb question(s). I believe that forums like this should serve as a model for future education, replace (or at least augment) our current antediluvian pedagogy (especially pre U). Your generous feedback has raised more questions, which are, unfortunately, all I have to offer, and which I will put in this thread rather than cluttering up the site.

Assume my bubble's made of "google glass" that shifts/translates EM radiation from Doppler-compressed/expanded frequencies into those I evolved under.

1. My perfectly spherical (to me) space bubble zips past you on earth at ~c (very near c). So you see (i.e., by whatever means, measure) my bubble as a pancake, flat in my direction of travel. How do I see earth?

2. To you, my clock is running very slow. How does your clock appear to be running to me? E.g., How fast do you appear to be orbiting the sun?

jbriggs444: "groping" is definitely the right word here. Interesting link, though mostly way over my head. The idea that light would not catch up to me from behind or the sides seems to clash with c being a constant in all frames of reference?

"Relativistic velocity addition prevents you from ever accelerating to c." Are you saying that if I accelerate under constant thrust (say 5 kilonewtons), my acceleration (as measured by me) will ever less gradually decrease to 0 (at c)? Still curious as to what the U looks like in that last nanosecond. E.g., how is H impacted by my frame's time dilation?

Janus: Thanks for your clarifications, and explanation of light aberration and EM Doppler effect. I didn't intend to bring acceleration into the mix. Assume I'm passing earth at ~c. In your snow/car analogy, the faster I drive, the more snow hits my windshield. Wouldn't this be true of light from that encroaching star (and which would be more analogous to my driving straight up into the snow)?

"Of course in this example I didn't account for Relativity. When you do, the answer comes out to 1.732 times the source frequency and not 2 times." So relativistic effects partially counteract the Doppler effect? I'd have thought the opposite.

MisterT: "It seems to be hard to get across the idea that the concept of "now" isn't universally shared, that one's idea of "now" dependent on the frame of reference." By "now", do you mean measurement of time? Then I'd understand. Otherwise, even though I'm so over my head already that I'm nervous to ask, doesn't atomic entanglement (synchronization of potentially infinitely remote particles ) sort of belie that statement? I do understand (as an SF writer) that words make poor metaphors for math.

19. Nov 10, 2016

### jbriggs444

Constant in all inertial frames of reference using the simultaneity convention that is natural in such frames (Einstein clock synch). If you base your frame of reference on a constantly accelerating rocket, you're no longer using an inertial frame and there is no natural simultaneity convention to use.

The speed of light depends crucially on what time a pulse of light is emitted what time it is when it arrives. Those are time coordinate values. Without a natural simultaneity convention, those values are somewhat arbitrary and the "speed of light" depends on the choices you make when drawing up your coordinate system. So the speed of light as measured in a non-inertial frame is as much an attribute of the coordinate system you use as it is an attribute of the light that you consider. In general, the speeds of anything far away cease to have much meaning when using non-inertial frames.

20. Nov 14, 2016

### Janus

Staff Emeritus
The light would hit with a higher frequency. The difference between the snow and light is that as you drive faster and faster, the snow hits your car at a higher and higher speed. Light always travels at c with respect to you, so it show this as an increase in its frequency. you measure light emitted at one frequency from the star as having a higher frequency.
As far as you are concerned, it is the star that is moving towards you, and thus its clock runs slow compared to yours. So even after you account for Doppler shift, it is, according to you, emitting light at a lower frequency than what be measured by the star itself.