# Looking backwards when traveling at c

1. Apr 29, 2014

### johann1301

If i am moving at the speed of light, facing the direction that i am going in, everything in front of me appears flat.(and every event i see in front of me is happening very fast)

What will i see if i look backwards?

I guess i will see a still image, and all events i see are "frozen" but what happens to space? Will it look contracted in this case also?

(Lets say that every object i see is at rest in my referance system. And i am moving at c relative to the system)

2. Apr 29, 2014

### .Scott

The system?

3. Apr 29, 2014

### johann1301

Sorry; frame, referance frame. In my language (norwegian) we say system...

4. Apr 29, 2014

### Staff: Mentor

You can't, according to relativity theory. Therefore it doesn't make sense to ask what relativity theory says you would see if it were actually possible. We would need a new theory that replaces relativity, in order to answer this question.

What you can ask and get an answer to, according to relativity theory, is something like "if I am moving very very close to the speed of light in some reference frame in which other objects are at rest, then what would I see if I look backwards?"

You should also make clear whether you are asking about what you actually see visually, keeping in mind that you can see objects only via light emitted or reflected from them; or about how those objects "actually behave" in your reference frame, after "subtracting" light-propagation effects from what you see.

[aha, Nugatory's post arrived while I was editing this one to add the last paragraph!]

Last edited: Apr 29, 2014
5. Apr 29, 2014

### Staff: Mentor

First, a small problem that can be fixed by tweaking the wording a bit:
You cannot move at the speed of light if the laws of physics are correct, so your question as worded comes down to "If I'm in a situation where the laws of physics don't apply, what do they tell me?". The answer to that question is of course "nothing - they don't apply here". The infinities that you see in the time dilation and length contraction formulas do not mean that you get infinite dilation and contraction at the speed of light, they mean that those formulas do not apply at all when $v=c$.

However, we can fix this by rewording the question to be "if I am moving at very close to the speed of light, say 99.9999%, ....". I'll assume that that's the question you really meant to ask.

It's good that you say:
as you've successfully avoided the enormous huge pitfall of specifying a speed without saying what it's relative to. Physically, this situation (you moving relative to everything else) is equivalent to you being at rest and everything else moving relative to you.
And when you look backwards, it's a lot like looking forward except that the light that reaches you from the objects behind you is redshifted because they're moving away from you. In contrast, he light from the objects in front of you is blueshfted because they're moving towards you. But the length contraction and time dilation effects are the same in both directions.

You do have to be careful with that word "see". Taken literally, you're asking what images will form on the retina of your eye, and that's influenced by some things that you probably don't care about: If the light is sufficiently red-shifted, you won't see it at all because it will be redshifted out of the visible spectrum. More subtly, light from different parts of the same object will have to travel different distances to reach your fast-receding eyeballs, so you won't be able to directly see some length contraction effects.

But if by "see" you mean what people usually mean in this context, which is to say that we have all the fancy sophisticated detectors we could ask for, capable of picking up and interpreting all sorts of signals from the objects around us, and compuers that can correct for light travel time so that we know when light signals that reach our ship at different times were emitted at the same time (using our ship's time, of course).... Then yes, what we see looking behind is pretty much what we see looking ahead, except that everything is redshifted and moving away instead of blueshifted and moving towards us.

6. Apr 29, 2014

### Staff: Mentor

That might be a good reason to use Norwegian instead of English in these discussions

Seriously, kidding aside, I find that the word "frame" confuses many beginners, who would have been better served by the phrase "coordinate system in which I am at rest relative to the origin".

7. Apr 29, 2014

### ghwellsjr

How do these fancy sophisticated detectors of signals and these computers correct for light travel time so that they can determine which remote events are simultaneous?

8. Apr 29, 2014

### ghwellsjr

I'm a beginner and I'm confused. Just about everybody admits that they use these terms interchangeably. Why is it important for beginners to know that there is a difference? And what is that difference?

I know that for every frame there are an infinite number of coordinate systems, for example, you could use Cartesian or polar or cylindrical coordinates and for each of those, you can put the origin in different places and at different times, and for each of those you can have the axes pointing in different directions, and for each of those you can use different units for distance and time, but beside all that, is there some other difference that beginners should know about?

You see, I only get confused when someone says there is some important difference that beginners should know about but then won't say what that difference is.

9. Apr 29, 2014

### micromass

Staff Emeritus
10. Apr 29, 2014

### HallsofIvy

Staff Emeritus
Moving at the speed of light (or, better, at 99.9% the speed of light) relative to what? A key point of "relativity" (even "Galilean" relativity) is that it makes no sense to state a speed without saying what it is relative to.

Whether you say "reference system" or "reference frame", you are always at rest yourself in your reference system. Essentially, you are saying that you are at rest observing objects moving at (near) light speed. Since you are at rest in your reference frame, the difference becomes observing objects moving toward you or moving away from you. But only the speed of objects, not the velocity (i.e. not the direction) is used in the relativity theory, there will be no difference.

11. Apr 29, 2014

### D H

Staff Emeritus
Aside: "I" referring to yourself is capitalized. Apparently that's somewhat unique to English because of our use of a single letter for the first person singular subjective pronoun.

As has already been noted, you can't travel at the speed of light. But you can get arbitrarily close.

Rather than asking about traveling at the speed of light, I'll assume you are asking about traveling at a highly speed relative to the stars in the galaxy through which you are traveling. So what will you see of you look backwards?

The inside of your spaceship will look quite normal. The stars that you saw behind you when you started your journey? That's a different question. You won't see much, if anything, behind you. There are two factors involved: aberration of light and redshift.

Consider what happens to a star that is behind you, but not directly behind you, when you start your journey. (I'm assuming you start at rest with respect to the stars). As you pick up speed, aberration of light means that that star will appear to move from behind you to beside you and then to in front of you.

The few stars that you could see behind you because they were directly behind you at the start of your journey, you can't see them thanks to redshift. Their frequency has been redshifted to below the visible range and their apparent intensity has dropped, too.

What you see in front of you changes a lot, too. As you get very close to the speed of light, almost all of the stars will have aberrated to appear to be in front of you. Their frequency has blue shifted to dangerously high values, and the intensity has increased, too. You don't want to look out the front window!

12. May 3, 2014

### ghwellsjr

I'm still waiting for an answer to my question in post #8...
...especially after this quote of yours from another thread:

13. May 3, 2014

### Staff: Mentor

Like many others, I sometimes find it tedious to say "using coordinates from one member of the family of inertial coordinate systems in which <something> is at rest..." when I can just say "using the frame of <something>....", and that's what I was doing in that thread.

However, I also consider that thread to be an example of the sort of confusion that the "frame" terminology can create in the non-expert. Look at the question to which I was replying: "Can we really say that because the math works the physical universe must use every reference frame?". The question is nonsensical, but its nonsensical nature would be more apparent if it were phrased in terms of choice of coordinate systems, something that the physical universe obviously doesn't do.

14. May 4, 2014

### ghwellsjr

Since you seem agreeable to use terms like "my frame" interchangeably with "my coordinate system" was the problem that you thought confused beginners leaving out "my" or its equivalent? Because originally you said:
But the OP was just as confused after he quoted you regarding coordinates:
And now I'm more confused. You emphasized over and over again, both in this thread and in the other thread a reference to the origin of a coordinate system which I don't understand. The origin of a coordinate system is an event which does not have velocity associated with it, not even a velocity of zero (rest), isn't that correct?

So why do you say in this thread, "I am at rest relative to the origin"? If you really meant you are at rest relative to the spacial coordinate of the origin, why single that point out? You are also at rest relative to every spacial coordinate in the system. I would think the best way to state it is that your spacial coordinates are constant, whatever they are.

And why do you say in the other thread, "The v that appears in the Lorentz transforms is the velocity of the origin of an inertial coordinate system" when in fact the origin of the primed and unprimed systems are the same event? Again, if you meant the spacial coordinates equal to zero, why single them out? It is also true for every point with constant spacial coordinates, isn't it? I would think a better way to say it is that v is the velocity of the primed inertial coordinate system relative to the unprimed one along the x axis, wouldn't you say?

15. May 4, 2014

### pervect

Staff Emeritus
Nugatory's point seems clear to me. The basic point is the one made in the FAQ https://www.physicsforums.com/showthread.php?t=511170 "Rest Frame of a Photon".

IT's difficult for me to tell what really works with beginners, if there is a way to make the FAQ more clear to them, though, I'm for it. (The logical procedure would be to propose the chages in the appropriate FAQ forum).

That said, I'm not aware of any way to word the FAQ better, and I currently don't see why what Nugatory said would be unclear. But I may not be able to really get into the mindset of someone who asks these sorts of questions.

16. May 4, 2014

### ghwellsjr

I don't see how that FAQ has anything to do with Nugatory's point.