# I Equations for observed distance/velocity in SR

1. Mar 31, 2017

### Arkalius

Hello everyone. I've just recently found this forum and it has been a lot of fun browsing around. I've recently taken a stronger interest in the topics of relativity in physics and have recently developed a much better understanding of SR (and somewhat of GR) than I'd had previously and its been fun exploring that understanding in various scenarios.

Of greater interest to me recently is how things appear to observers in relativistic situations when you consider the travel time of light. Many thought experiments and scenarios are described from the "measured" viewpoint, or as if each observer could witness all instantaneous events (from their frame) at the same moment. This has its uses, and it simplifies an already complicated subject, but I find it useful to explore what observers would actually see in reality.

To that end I'd come up with some simple equations that give a ratio of observed length and speed to the actual length and speed of an object moving toward or away from you. These are \begin{align} \frac 1 {1-\beta} \\ \frac 1 {1+\beta} \end{align} with (1) being for objects moving toward you, and (2) for objects moving away, and with $\beta = \frac v c$. These were great and all, but I wanted a more general equation that worked in 3d and 4d spacetime. In those, an object doesn't always move directly toward or away from you.

So, I came up with this more generic equation for this ratio: $$\gamma^2 \left( \beta \cos \alpha + \sqrt{1-\beta^2 \sin^2 \alpha} \right)$$ Here, $\alpha$ is the (actual) angle between the relative velocity vector and position vector. You will see that when that angle is 0 or $\pi$, the equation reduces to the two I have above. You simply multiply actual length or velocity by this value to get the observed length or velocity.

Another useful equation is for the observed angle of deflection and that is given by $$\alpha_{obs} = \alpha - \arcsin \left( \beta \sin \alpha \right)$$

I'd not seen equations like these listed anywhere that I'd looked in the past, and I found them useful for examining scenarios for how they would appear to the observers. It certainly brings a different perspective to the situation. Things moving toward us won't appear contracted, but rather stretched out. Also, things can appear to move toward us at many times the speed of light because of this. It also means nothing can appear to move directly away from us at more than half the speed of light either.

Anyway, I look forward to participating in more discussions on this forum and learning more interesting things about relativity and other topics.

2. Mar 31, 2017

### BvU

Hello Arkalius,

I advise you to first and foremost become familiar with the simpler cases of special relativity. Expressions are already complicated enough there. Learn about hte Lorentz transform and its accompanying phenomena. If you are fluent with those (conceptually and with the formalism), then it's still early enough to move on to the issues you are now messing with. I don't believe a single one of your expressions -- but you may ascribe that to my ignorance.

In the mean time play with the MIT game and wonder about this strange world. Have fun !

3. Mar 31, 2017

### FactChecker

Your equations look like they are just some sort of Dopler effect. That is not correct. You are missing the main point of SR. There are simple explanations of SR already in terms of the relativity of "simultaneity". You should pay more attention to them before trying to make your own equations.

4. Mar 31, 2017

### Arkalius

I'm already quite familiar with those. I feel like I've developed a more intuitive understanding of how Minkowski spacetime works, and understand Lorentz transformations almost instinctually now. There are scenarios I run into where I can't quite visualize it right and have to rely on a Minkowski diagram (my favorite tool for that currently is http://ibises.org.uk/Minkowski.html ) but the more scenarios I encounter the better at it I've gotten.

Well.. points for being direct I guess. I assure you, they should be accurate. I'm looking at all of the fun triangle drawings I have on my desk from the work I did on it right now, and they seem to fit with the examples of relativistic Doppler effects that I've seen.

5. Mar 31, 2017

### Arkalius

You're right that it is a kind of Doppler effect transformation. But, I think you might be missing the main point of my post, which is this: I understand the main point of SR, and wanted to move beyond, into understanding how to move from what the measured view of reality is in SR to what the observed view would be, factoring travel time of light. This view of things is in some ways less bizarre, and in others, moreso. For example, in a measured view, when accelerating away from distant objects, their clocks can appear to go backwards as our plane of simultaneity shifts. But, when looking at the observed view, this effect disappears, and clocks always tick in the forward direction. On the other hand, sufficient acceleration toward something will actually cause it to seem to stretch away further into the distance, and if you settle at a high enough velocity, make it appear as if you approach it faster than light

6. Mar 31, 2017

### A.T.

This might interest you:
http://www.spacetimetravel.org/

7. Mar 31, 2017

### Staff: Mentor

8. Mar 31, 2017

### pixel

9. Mar 31, 2017

### FactChecker

10. Mar 31, 2017

### PAllen

If you want feedback on your equations, you will need to define your observational model more. For example, it is not clear at all to me how it makes sense to talk about the apparent length of a ruler moving towards you parallel to its length. Is your formula possibly for a ruler turned perpendicular to this? Then, for defining what you 'see', you need to specify e.g. whether your idealization is a tiny spherical detector, versus a pinhole camera with a flat detector read in simultaneous captures of the movie camera frame. The latter introduces additional distortions compared to the former, and makes the orientation of the camera a necessary element of the specification.

Note there is a very cute trick that can be used for these problems. As long as you assume all objects are luminous at a standard frequency and intensity in their rest frame ( thus avoiding specifying lighting source position and motion), aberration plus Doppler can be used to get the exact appearance, including surface markings, without doing any form of ray tracing. This includes getting brightness and color correct.

Last edited: Mar 31, 2017
11. Mar 31, 2017

### Arkalius

Well, sure there are more complicated things to resolve if you're trying to model the precise appearance of a moving object, especially in 3d space. The distortions can be fairly strange as I understand, such as objects moving past you appearing somewhat rotated. I'm really only concerned with the more general aspects, the effects on apparent distance, speed, and general size in the direction of motion. That is what these equations are intended to help with. They are, in effect, a polar coordinate transformation.

But, I appreciate your input. Certainly the observed effects of relativistic motion can be quite bizarre. It's too bad we don't have access to macroscopic demonstrations of such things.

12. Mar 31, 2017

### Arkalius

Ah that does look pretty neat. It makes me think of Velocity Raptor, which is a web-based 2d puzzle game that uses special relativity and does kind of the same thing. The villain slows down the speed of light to 3mph so all of your motion is notably relativistic. The game then requires you to take advantages of the strange trappings of relativistic motion to overcome obstacles that would be impossible to surmount under normal circumstances. In fact, it is kind of what spurred my interest in the concepts of "observed" reality in SR. Later in the game, it moves from showing you a measured view of your environment to the observed view based on travel time of light, and the distortions there were quite wild, and I was fascinated by them and wanted to understand them better.

13. Mar 31, 2017

### PAllen

Right, but you have not specified enough for anyone but you to understand what your equations are supposed to mean. Having worked through both aberration and ray tracing computations of moving objects (thus experienced in the field), I have not a clue what your equations are supposed to describe.

14. Mar 31, 2017

### Arkalius

I see. The larger equation is a ratio between observed distance to measured distance of a moving object, as well as observed velocity to actual velocity. Take the measured distance between something moving relative to you with some velocity, and multiply by the value of the equation to get the distance the object appears to be from you at that instant. Similarly, multiply that by the object's velocity to see what velocity the object will appear to be moving. The 4th equation gives you the angle between the relative velocity vector and observed position vector of the moving object.

15. Mar 31, 2017

### PAllen

Define apparent distance and speed versus measured. You may think there are universal definitions of these, but that is not so. Give us yours, or a link to what you are using. Otherwise, no one can say if your equations are correct or not.

[edit: Let me be clear about how definition is so crucial. Suppose I define measured distance at a given time as the coordinates in a standard Minkowski inertial frame, with 'me' being the t axis throught the origin. Then, if an object is at some position, at some time, I define its apparent distance in terms of the angle subtended by the light from the object emitted at that position and time, when such light reaches me. If this angle is less than expected per rest dimensions for that distance, I am defining it as apparently further away. Using this definition (transverse angle subtended compared to expectation), then an object approaching me never has an apparent distance different from its measured distance. Rather than claim your equation is wrong, I simply want to know what definitions you are using.]

Last edited: Mar 31, 2017
16. Mar 31, 2017

### m4r35n357

17. Mar 31, 2017

### Arkalius

Measured distance is the $\Delta x$ in the spacetime interval equation $s^2 = \Delta x^2 - c^2\Delta t^2$, from our chosen frame. Apparent distance is how far the moving thing in question appears to be from that observer, based on the light information reaching him at that moment.

Actual velocity is just that. It is the velocity of the object of interest relative to the observer. Apparent velocity is how fast the object will appear to be traveling based on the light being received by the observer.

The angle $\alpha$ represents the angle between actual velocity vector and the actual position vector (from whence the aforementioned $\Delta x$ is derived). The calculated $\alpha_{obs}$ is the angle between actual velocity vector and the apparent position vector based on light being received at that moment.

Together these equations can tell you where a moving thing appears to be as told by light signals based on where they actually are in that instant for that observer, as well as telling you how fast the thing appears to be moving based on its actual velocity.

You have to consider some timing issues when using these equations. If a ship is stationary relative to you 9 light-days away, and then begins a journey to you at 0.9c, you won't see him traveling to you at 0.9c for 10 days. From your point of view he'll appear stationary for 9 days, and then seemingly approach at 9c for 1 day. If you just naïvely apply my equation at the moment he begins moving, it would suggest he appears to be 90 light days away at that point. So you have to be careful how you apply it.

18. Mar 31, 2017

### PAllen

You keep using terms (apparent distance) with no definition whatsoever. (also, see the edit to my post asking for this, for an example of the issue of no definition).

19. Mar 31, 2017

### Arkalius

I said "Apparent distance is how far the moving thing in question appears to be from that observer, based on the light information reaching him at that moment." Another way of putting it is how far the moving object actually was from you when it emitted the photons you are now observing. I suppose I can understand some amount of confusion, in that one cannot know the distance of a source of light merely by observing that light in an instant. You have to do other things like generate parallax etc. But it doesn't change the fact that the light you see "now" from a moving object doesn't reflect where that object is "now" (from your reference frame), but rather where it was when it emitted it. This all emerged from my interest in how one's actual view of everything around us distorts as a result of relativistic motion, spurred on from seeing a simulation of this distortion in a game. The observed distortion is significantly different than the actual spacetime distortion given by the Lorentz transformation, and this was fascinating to me.

20. Mar 31, 2017

### PAllen

This definition has precisely zero content for me. Sorry. By what model? Brightness? subtended angle? I have no idea.

21. Mar 31, 2017

### PAllen

So trying to understand this, are you defining apparent distance as follows:

If at some time t, using some standard inertial SR frame/coordinates, I receive light from an object, its distance at time of emission is apparent distance, while its current (not yet observed) distance is measured distance? I find these definitions peculiar, but at least it is a definition.

But then, for a uniformly moving object, I see no corresponding definition of apparent speed.

22. Mar 31, 2017

### Arkalius

Let's look at it by example. We'll use a simple twins paradox setup. Twin leaves Earth at 0.8c and travels for 10 years, moving 8ly away from Earth (from Earth's perspective), turns around and comes home at 0.8c. Total trip time from Earth's perspective is 20 years. From the traveler's perspective, he traveled 4.8ly from Earth and back for a total time of 12 years.

What the Earth twin actually sees is the traveler flying away from earth at about 0.44c for 18 years, and then returning to Earth at 4c for 2 years. Basically, for each year after the traveler leaves, he is (from Earth's perspective) actually 0.8ly further away, but only appears to be 0.44ly further. So at year 2, he's 1.6ly away, but only seems to be .89ly away. Thus both his distance and speed seem smaller than they really are at any given instant.

Now, after he turns around, the equation won't work quite right. You have to consider the time delay between when he turns around and when you're aware of it. The equations will work if you just pretend he continues to move away for this period, but that's a little odd.

At the 18 year mark, when you're finally aware of him now heading back, he is actually only 1.6ly away, but he appears to be 8ly away for you. The next year he will be only 0.8ly away, but appear to be 4ly away to you, thus it seems he is moving at 4x the speed of light.

It also affects the apparent passage of time. If you continually observe a time signal coming at you from the ship, while it's moving away it seems time is moving at only 33% of your own (instead of the 60% the Lorentz factor would suggest). While moving toward you, it seems it is moving 3x faster than your own. This still works out correctly, since you'll observe 6 years pass on his clock during the 18 year apparent outbound journey, and 6 years pass on the 2 year inbound one.

From the traveler's side, things are quite different. Initially it is the same; Earth seems to move away at about .44c, but for only 6 years for the traveler, at which point it only seems to be 2.67ly away (despite actually being 4.8ly). But, when he turns around, his view of reality distorts into a new configuration, and Earth seems to stretch away and seem to be 24ly away, and be approaching at 4x the speed of light for 6 years. This illustrates the asymmetry of the twins paradox pretty well. While Earth sees the traveler fly away for 18 years slowly and return fast for 2, the traveler sees Earth recede slowly for 6 years and approach quickly (but from apparently much further away) for another 6. Like for Earth, for the traveler, Earth's clocks appear to only tick at 33% the rate during the outbound journey, passing only apparently 2 years during that leg. On the way back, it ticks 3x faster, which means it passes 18 years during the inbound leg.

Basically, looking at the scenario this way provides a new and interesting perspective on the scenario. It may provide a somewhat more intuitive way of understanding the age difference between the twins. The "fast forward" of Earth time during the traveler's acceleration to return home (from the shift in simultaneity) can seem odd to some people when explaining the resolution to the paradox.

Now, this example only involves motion directly toward and away from you, so the first two equations in my post are sufficient to handle all of that. The other equations are only needed in more complex scenarios involving 2d or 3d space and velocities that don't move directly toward or away.

23. Mar 31, 2017

### PAllen

Already, we have some issues. The traveler didn't travel at all relative to themselves. How much the earth moved depends on which model of non-intertial coordinates you choose, specifically, your simultaneity model. Two concrete models are radar simultaneity and momentarily comoving inertial simultaneity. They give different answers, and neither is 9.6 light years total. You appear to be using an 'odometer' notion of travel distance for the traveler, which is a perfectly fine choice: pretend there is a surface at rest with respect to earth, and track its local motion relative to you, totaling it up, never worrying about measures of distance to earth.
As most people define 'see', earth twin sees no such thing. What they directly see is a certain constant redshift, with decreasing luminosity for 18 years, then blueshift with increasing luminosity for two years; and if they use the standard SR doppler formula, they get .8c away and .8c back, for speed. However, I can now guess (you still have not defined them) your terminology:

1) apparent distance is distance at emission event measured in observer centered inertial frame. 'emission time distance' would be a more natural term.
2) measured distance is current distance in the observer centered inertial frame. This irks me because it is in principle unmeasurable. Current distance would be a better term.
3) apparent velocity is rate of change of apparent distance (your definition) with observer time, and corresponds to no observable feature of the image over time. I have no suggestion of a reasonable term for this quantity.

With these definitions, of course, I agree with your numbers. I don't see why you couldn't have given the equivalent of these definitions in your OP, or when asked many times.
Thus, you extend your definitions to a non-inertial observer by the momentarily comoving inertial frame model (MCIF). The is one choice for analysis, but it leads to unnecessarily odd descriptions. Along with your definition of apparent distance which rapidly (or instantly) jumped, if you were to similarly define apparent time, you would now claim this to be before you left earth, even though there is no jump whatsoever in the image of an earth clock as the traveler turns around. The wild discrepancy between MCIF quantities versus actual observations is why a number of us on this forum dislike this approach, and why calling these quantities apparent is disturbing to me.

A more natural description of the direct observations is that for the traveler, at the mid point, the relative velocity of traveler and earth at emission changed. This changed doppler plus aberration description from seeing an enlarged, reddened image to a shrunken, bluer image.

Now that I understand your definitions, i don't really have any interest in checking your equations, because I am not very interested in your definitions.

Last edited: Mar 31, 2017
24. Mar 31, 2017

### Arkalius

Well, okay, refer to it how you will. I'm basically again using the $\Delta x$ of the spacetime interval equation as it would be calculated by the traveler's inertial frame. In both of those frames, at the point where he turns around, that value is 4.8ly when $\Delta t$ is 0.

It's unmeasurable at the moment it happens, but later analysis of observation could reconstruct it. But sure, current distance is probably better term.

It wasn't to be frustrating, I assure you. The definitions I gave made sense to me, and I was having trouble understanding what your confusion was. Obviously, you and I work within somewhat different vocabularies. I don't have a formal background in this topic; most of my understanding is self-taught. Perhaps that's why.

Why? When turning around, you're still receiving the same light from Earth. However, the shift in reference frame will serve to blue-shift the Earth and reduce it's apparent size (this reduction in apparent size the ultimate reason it appears to move further away). But, you're still seeing the same light, and thus will not see time reverse in any way.

I'm not proposing this as an approach to solving problems within special relativity. It was focused entirely on developing an understanding of what relativistic situations look like to an observer. Most thought experiments in relativity involve instantaneous observation, which is useful in understanding Lorentz transformations and the effects of relativistic velocities, but doesn't depict accurate observations (at least, when the distances involved are large).

That's fine. I wasn't presenting them to be checked (though I certainly wouldn't mind someone doing so). I was simply expressing the results of following an avenue of thought on the topic on which I hadn't seen much discussion, and which interested me greatly. I enjoy deriving interesting equations that beget a new angle of understanding and give me a chance to exercise my math and problem solving skills. They may not be extraordinarily useful, but I found them interesting and wanted to share.

25. Mar 31, 2017

### PAllen

If you look at it like that, purely as an optical effect, fine. If you look at where the 24 ly comes from in the coordinates of the new frame, it corresponds to an emission time of 18 years before you left earth.