# Twice the speed of light?

1. Oct 26, 2005

### Raziel McCloud

Einstein said nothing can go faster than the speed of light.If it did among other things time would slow you down to bring you back to 186,000 m/sec. Alright if you were going the speed of light and there is a beam of light right beside you,it would appear as if it wasn't moving right?Well if you had a flashlight with you in your ship,and you shined it would it leave the flashlight at 2X the speed of light or would it simply never leave the flashlight?

2. Oct 26, 2005

### Pengwuino

The light beam would travel away from you at... bum bum bummmm... the speed of light! There's Einstein's Special Theory of Relativity. Someone else will be able to give you an explanation as to why that happens. I'm way too tired to try, took me about 30 seconds to spell 'relativity' correctly.

3. Oct 26, 2005

### Raziel McCloud

but compared to something outside the ship not moving it would be 2Xthe speed of light right(1shipspeed+1flashlightspeed=2)Maybe?

4. Oct 26, 2005

### Galileo

The speed of light is 300.000 km/s. Or did you mean miles?

No. Light always travels at the speed of light, relative to the observer. If you are going very very close to the speed of light, you have to specify from which point of view (reference frame) that is. Say, as seen from a person on earth you are flying very near the speed of light. A light beam would pass you at 300.000 km/s, but as seen from the earth the light beam would only surpass you very slowly, since you are moving with such great speed.

There is no contradiction here, since observers moving relative to one another experience time and space differently.
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5. Oct 26, 2005

### Raziel McCloud

yes,miles,I believe thats correct any way?but say we are away from any planet or other veiwers if both you and the beam left facing the same direction(like to cars drag racing)and you both left at the speed of light, it wouldn't be right beside you like two cars going 90kmh right beside each other, i thought it would seem to you the driver like them beam right beside you wasn't moving at all,but if not then you say it would still go the speed of light from you the driver? Then would it not be going 2X the speed of light and what about the flashlight?

Alright if you were going the speed of light and there is a beam of light right beside you,it would appear as if it wasn't moving right?Well if you had a flashlight with you in your ship,and you shined it would it leave the flashlight at 2X the speed of light or would it simply never leave the flashlight?

6. Oct 26, 2005

### JesseM

It's not possible for a massive object like a ship to travel at the speed of light according to relativity (it would take an infinite amount of energy to accelerate it to this speed). But we can take a variation on your question--let's pretend for the sake of the argument that light moves at exactly 186,000 miles/second, and suppose on earth I observe a light beam flying by and then you in your ship chasing after it at 185,000 miles/second. Does this mean that from your own point of view, the light will only be moving at 1,000 miles/second away from you? The answer is no, because you use different rulers and clocks to measure speed, and from the earth's point of view your rulers appear shrunk and your clocks appear slowed-down and out-of-sync. In fact, using your own rulers and clocks, you will still measure this same light beam to be moving at 186,000 miles/second away from you. In general, velocities don't translate between different observer's reference frames in relativity in the same way they do in Newtonian physics--if you observe an object moving at velocity v relative to yourself, and I observe your ship moving at velocity u in the same direction relative to me, then instead of me observing the object moving at velocity w = u+v relative to me as you'd expect in Newtonian physics, I would instead observe it moving at $$w = (u+v)/(1 + uv/c^2)$$. You can see with a little algebra that if v=c (you observe a light beam moving at c relative to yourself), then w=c as well (I will also observe that same light beam to be moving at c relative to myself).

Last edited: Oct 26, 2005
7. Oct 26, 2005

### Severian596

Hi Raziel. Your questions are good questions, and they're the first questions we all ask at the threshold of studying the Special Theory of Relativity, or Special Relativity (SR). At this juncture in your career as a student you must keep some things in mind:

1) We (humans) have the luxury of using the earth as a "rest frame" for much of what we see. The earth doesn't move relative to us. Keep in mind that the earth is the "neutral observer" in your race car example. Or maybe the neutral observer is a bug who's standing still on the track, or the start/finish line. Neither of your race cars are "neutral observers" of each other because they are both in motion. Now ask yourself this question...who's the neutral observer for the universe? Whose point of view to you use when you ask the question, "Would the flashlight beam of light not appear to be moving 2X the speed of light?"

2) Newtonian physics are a subset of physics as defined by General Relativity. Newtonian physics explain many phenomena, General Relativity explains everything that Newtonian physics explain and more[b/]. To be more specific, Newtonian physics are okay as long as you are moving nowhere near the speed of light, like your race cars. As your subjects of study start to move faster and faster, your Newtonian calculations will start showing signs of inconsistency with your observed results. As you approach the speed of light Newtonian physics don't hold up any more. You need GR math to explain your observations.

8. Nov 2, 2005

### LeBrok

Hi folks, I’ve got a good one.

Let’s clean up the space, no stars, light other matter, so we have no reference points but two bodies A and B. A and B fly away from each other with speeds close to c each, so they recede at close to 2c.
Now, looking from reference frame of body A; I could say that A is stationary and B is flying away faster than light. Only I’m not sure that I could see B (flashlight from B), because it is flying faster than light. :surprised
No light from B, no information, no violation of laws of physics? :grumpy:

Also from your discussion here it looks like the light is “escaping” relativity. No mater what reference frame c is always going constant speed. If I understood right, the clocks are ticking differently in different frames so that’s why this is happening. It’s like a grand illusion, lol.

Or maybe there is an universal reference frame based in quantum of space-time, fabric of space. The speed of light could be limited by a maximum speed that information can travel from one quantum to the other.

I appreciate any help in understanding this, thanks.

9. Nov 2, 2005

### pervect

Staff Emeritus
10. Nov 3, 2005

### Severian596

See pervect's post and my earlier note. You mistakenly apply Newtonian physics's velocity addition to a situation that requires General Relativity (because you deal with speeds near c).

11. Nov 3, 2005

### jackle

The slightly tricky bit to understand is that if I am in the middle of A and B, space ships travelling in opposite directions at nearly c, appear to me, after 10 seconds, to be seperated by nearly 20 light seconds (10s x 2 x c)

A<-----10 light seconds----->ME<-----10 light seconds----->B

However, if you sit on one such space ship, you might think you would see a seperation of nearly 20 light seconds, increasing at a speed of nearly 2c. You won't! Now is the right time to use your relativistic formula for adding velocities.

You can only get a sub-light velocity by adding two sub-light velocities. Two objects moving at nearly c relative to you in opposite directions can appear to be seperating at nearly 2c (but they won't see this from their own perspectives). It is a common misunderstanding that this would lead to relative speeds of up to 2c, and a lot of the general public seem to get stuck here. Use the relativistic formula to find out what you would see from the spaceships.

Last edited: Nov 3, 2005
12. Nov 4, 2005

### LeBrok

Thanx guys, it sounds very intriguing and a new challenge in understanding the world, it should be a great reeding. As soon as I find some quiet moment or two, as it will require analyzing of those equations, and retraining my brain for relativistic thinking.

From the first glance at relativistic velocity equation it looks like the traveler on body A is seeing the flashlight from B, but in different frequencies like microwaves, caused by red shift due to B speed.
On the other hand the time on A traveling at speed almost c is slowed down. In this case the man on body A might see same frequencies light as is made by flashlight if they are traveling with same speed but in opposite directions. The slowed time could compensate fully for the red shift and allow to see flashlight with naked eye……hmmm, very interesting. Wrong or right it is a very good brain exercise, hehe!

Talk to you later.

13. Nov 8, 2005

### Severian596

Remember that this is only true for observers who do not travel with A. B will report that A's time is slowed by a certain factor (often referred to as $\gamma\hspace{1}'$), and A will report that B's time is slowed by that same factor $\gamma\:'$.

But neither A nor B says to themselves, "Hey, things are happening more slowly on my ship." They see time for themselves pass normally. Only other observers (who reside in a different rest frame or frame of motion) report that time passes more slowly for A and B.

Last edited: Nov 8, 2005
14. Nov 8, 2005

### eon_rider

So well put. Agreed.

EDIT: Forget what we see for a moment. Let's presume the 10 second .99c journey is over. And now A,B and the ME in the middle communicate with each other. (time dialation and length contraction are now pretty much zero. We've stopped)

Two further questions.

1) Is it correct that the reason ship A can’t see ship B moving faster than light because positional information on the location of the other ship can't be communicated at “faster than light" speed?

2) Regardless, will both ships A and B after stopping, then resting a moment, then sending communications to each other and the guy in the middle find out that they (A and B) are positioned just under 20 light seconds apart even though they both only saw each other travel for 10 seconds at sub light speed? (meaning they would think they were only 10 light seconds apart.)

The guy in the middle could confirm that they are indeed 20 light seconds apart. Is that right?

Eon.

PS. When I say light speed I mean .999999999c

Last edited: Nov 8, 2005
15. Nov 8, 2005

### JesseM

No, even if A takes into account the light-speed delay she still won't conclude that B was moving faster than light. For example, A might see an image of B that appears to be 4.999 light-seconds away at t=9.999 seconds, and conclude that since the light took 4.999 seconds to reach her, B was actually 4.999 light-seconds away at t=9.999-4.999=5 seconds in her frame. Then A might later see an image of B that appears to be 9.998 light-seconds away at t=19.998 seconds, and conclude that B was actually 9.998 light-seconds away at t=19.998-9.998=10 seconds in her frame.

You could also imagine that, to avoid the issue of light delays, A constructs a large network of rulers at rest in her frame and synchronized clocks fixed at regular intervals along the rulers, so that to assign coordinates to an event, she looks at the marking on the ruler that was right next to the event when it happened and the reading on the clock that was right next to the event as it happened. This is how Einstein originally imagined that inertial coordinate systems should be physically constructed in his 1905 paper. The tricky part is that each observer must decide how to "synchronize" different clocks in their system-they can't just synchronize them at a single location and then move them apart, because moving them apart will introduce time dilation. Instead, Einstein suggested that each observer synchronize his clocks using the assumption that light travels at the same speed in all directions, so that if you set off a flash at the midpoint of two clocks in your system, you call them "synchronized" if both clocks read the same time at the moment the light from the flash hits them. But this implies that different observers must have different definitions of simultaneity (they will disagree about whether two distant events happened 'at the same time'), since if I see your clocks moving in my frame, I will see the back clock moving towards the point where the flash was emitted and the front clock moving away from it, so if I assume light travels at the same speed in all directions in my frame I must conclude that the light will catch up with the back clock before it catches up with the front one.
Yes, in his frame, although in other frames the distance would be different because of Lorentz contraction.

16. Nov 8, 2005

### Taviii

This topic is so interesting :surprised

17. Nov 8, 2005

### eon_rider

Yes this makes sense. Brilliantly put. Thank you. I believe I get it.
Now onwards below.
It's the bolded part just seems so strange. A paradox.
Because after the journey we don't have to think of frames anymore. We can communicate over these distances normally and easily establish positions. Can't we? Mars is just 180 light seconds away. So what's to disagree about when working out our true positions relative to "ME" after a .999c 10 or 20 second journey.

Why would A,B and the "ME" in the middle not be able to agree on where they actually are. (AFTER the trip's done).

I guess I'm just not getting this part.

I was thinking it was due to length contraction and time dilation that would have the "ME" in the middle measure that because time had slown down and distance had contracted for A and B as they travelled away from him/her that his/her final findings would agree with A and B's final findings.

So... I get the first part about nothing travelling faster than light , but I don't understand the second part about them all disagreeing on their final locations due to frame differences. (eg. Why can't they agree on positions AFTER the short and fast trip with normal communications gear?)

best, :)

Eon.

Last edited: Nov 8, 2005
18. Nov 8, 2005

### JesseM

A, B and "Me" would agree after the trip because they are all at rest wrt each other and are thus in the same frame. But I meant that in the frame of some other observer C who has been moving at constant velocity relative to "Me" the whole time, the final distance when A and B come to rest relative to "Me" would be different. And note that A and B themselves have not stuck to a single inertial frame, and in non-inertial coordinate systems it is quite possible that the distance between you and some other object can increase faster than c.

19. Nov 8, 2005

### Severian596

This is the point where General Relativity steps in. Special Relativity assumes a very strict set of conditions, the foremost of which is no acceleration (and therefore no gravity). If, at any point in your experiment, you break these rules and someone experiences an acceleration, you must use General Relativity to complete the experiment and break the paradox.

The most classic example of breaking this paradox is the twins paradox. The twin who underwent the acceleration (from speed zero to near-c and back to zero) is the one whose time slowed down, and the twin who experienced no acceleration is referred to as the "rest frame". If you never broke this symmetry, neither twin would serve as a reference frame and both would report that their brother was aging slower than they were! This would not end until one of them underwent an acceleration.

I can't remember exact timelines, but SR took a fraction of the time that GR took to develop. Hmmm...wonder why?

Anyway, to stress, any time you start a SR experiment and then you find yourself saying, "Then A slows down...or A turns around...or A's direction changes..." these all kick the experiment out of the realm of SR and into more complicated GR.

Last edited: Nov 8, 2005
20. Nov 8, 2005

### JesseM

Only if you want to analyze the experiment from the point of view of a coordinate system where the guy who accelerated (from the point of view of inertial systems) is at rest the whole time. It is certainly possible to analyze situations involving acceleration from the point of view of inertial observers, you don't need GR for that.