# I What you see vs what actually happens

Tags:
1. Jul 20, 2016

### Afterthought

My impression always was that when you describe a problem in special relativity, you are already implicitly taking into account the light that would need to travel for some person to theoretically "see" a special relativistic phenomenon. I was confronted recently in another thread that this impression was wrong.

My question is, is this correct? If so, then after you do your problem using the Lorentz Transformations, or geometrically using the space-time interval, to determine what a person actually sees, you need to do additional computations! Yet, I have never seen this step done personally (granted I learned SR on my own). Is there a reason for this? Is it perhaps that usually this additional computation doesn't change the end result much - or at all - and so we don't usually do it? Are there any guidelines then when to and not to take traveling light into account?

Or is it the case rather that the initial conditions we give already take light into effect? In either case, are there some equations to help transform between what you "see" and what actually happens? Or is it something that you always have to do geometrically? Unless, since it's possible I misunderstood, that what you see and what actually happens are one and the same.

Thanks.

2. Jul 20, 2016

### Orodruin

Staff Emeritus
Yes.

What a physical observer actually sees is not that relevant to the physical description unless this is actually your experiment.

I remember some animations on the aberration and Doppler shift of light on Wikipedia that do show this, but I am on my mobile and cannot check at the moment.

3. Jul 20, 2016

### Vitro

In my view SR is about the geometry of spacetime, events and relationships between them, and how events map to frames of reference / coordinate systems. Descriptions of SR scenarios would typically include wording like "measure", "observe" (to also mean "measure" rather that "see"), and unfortunately the sloppy "see" but also generally synonymous to the previous two. "See" rarely means what one actually sees, since that's largely irrelevant.

Also, "observer" mostly means an inertial fame of reference (IFR) rather than a physical entity (object, person) at a particular location in space. So the observer is everywhere and everywhen events happen, no need to account for light travel.

SR becomes quite straightforward once you train your mind to think in terms of events and frames of reference. The first step in analyzing a scenario should be to identify all the relevant events and IFRs and then just use those going forward.

4. Jul 20, 2016

### A.T.

http://www.spacetimetravel.org/

5. Jul 20, 2016

### pixel

Check out (at least) the first two paragraphs in The Visual Appearance of Rapidly Moving Objects by V.F. Weisskopf: http://www.phy.pmf.unizg.hr/~npoljak/files/clanci/weisskopf.pdf

Even in classical physics there are apparent, i.e. visual, changes in the shape of an object that is moving very fast. The object itself is not changing, it just appears that way to an observer or on a photograph. Likewise in relativity, what is calculated is what would be measured in a reference frame with distributed clocks and measuring rods, not what a person would see.

6. Jul 20, 2016

### m4r35n357

Not that I know of. Here's an outline of the process in 1+1D, it's not at all rigorous but it shows the extremely simple algebraic steps required:

The Lorentz Transform in 2 dimensions is:
$$t' = \gamma (t - vx)$$
$$x' = \gamma (x - vt)$$
If we apply the light travel "boundary conditions" $$t = \pm x$$ then we can easily calculate what we can see in 1+1 dimensions. Let's do time first; there are two cases, corresponding to plus and minus (left or right) respectively:
$$t' = \gamma (t - vt) = \gamma (1 - v) t = \frac { (1 - v) } { \sqrt{1 - v^2} } t = \frac { (1 - v) } { \sqrt{(1 - v) (1 + v)} } t = \sqrt {\frac { 1 - v } { {1 + v} } }t$$
$$t' = \gamma (t + vt) = \gamma (1 + v) t = \frac { (1 + v) } { \sqrt{1 - v^2} } t = \frac { (1 + v) } { \sqrt{(1 - v) (1 + v)} } t = \sqrt {\frac { 1 + v } { {1 - v} } }t$$
Similarly for space:
$$x' = \sqrt {\frac { 1 - v } { {1 + v} } }x$$
$$x' = \sqrt {\frac { 1 + v } { {1 - v} } }x$$

and thus we end up with the Doppler relations.

7. Jul 21, 2016

### Afterthought

Thank you for the links, checking them out right now.

The derivation of the Doppler relations also seem helpful, I will try seeing (pun intended) if I can apply them to a simple problem.