# A dumb question from a layman

#### briv

Forgive me if this is an ignorant question. I have only a very superficial grasp of GR and SR, but this has been really bothering me ever for a while now.

I understand that according to SR, nothing can excede the speed of light. I also understand that constant-velocity motion is relative to the observer, and that there is no absolute notion of motion. So, here's my question:

Two observers are travelling through a vast, empty region of space from opposite directions, but on parallel, nearby paths. They are both travelling at a constant speed greater than half the speed of light--say .75 the speed of light. When they passed one another, wouldnt each observe that the other was travelling at a speed greater than light-- 1.5x SOL? How can this be?

Again, I apoligise if it's a stupid question, but I got to know.

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#### marcus

Gold Member
Dearly Missed
Hello briv that is a fine question but it does not help me any for you to act humble about it. It is not a stupid question but instead it is exactly the right kind. Also this is the kind of question that PF mentors would love you for asking---you should take it to "General Physics" forum, where they will immediately hear you

Originally posted by briv

Two observers are travelling through a vast, empty region of space from opposite directions, but on parallel, nearby paths. They are both travelling at a constant speed greater than half the speed of light--say .75 the speed of light. When they passed one another, wouldnt each observe that the other was travelling at a speed greater than light-- 1.5x SOL? How can this be?
I see you have put in a "third" observer who sees the two others each coming from opposite directions at speed 3/4.
Maybe that third observer is you and you are serving as a stationary reference and watching.

There is a formula for the combined speed

(a + b)/(1 + ab)

The numerator is what you think it OUGHT to be and then just to disappoint you and prevent it from coming out to be 1.5
Einstein sneakily divides by the (1+ab) you see in the denominator

(3/4 + 3/4)/(1 + 9/16)

which equals 0.96

#### briv

Wow! that was fast reply. Thank you Marcus.

Yeah, that humility bit may have been a bit much. It was attempt by me make a preemptive defense against any impatient rolling eyes that I thought may have greeted what I'm sure was just an elementary physics question. Not very well thought out on my part.

About that formula: It's something I wasnt familiar with. Does this mean that when I'm driving down a straight stretch of highway at steady 55 mph and observe another car as it passes me on the opposite side also travelling at 55 mph but in the opposite direction, then that car does not appear to me to be moving 110 mph, as I have always thought?

#### Janus

Staff Emeritus
Gold Member
Originally posted by briv
Wow! that was fast reply. Thank you Marcus.

Yeah, that humility bit may have been a bit much. It was attempt by me make a preemptive defense against any impatient rolling eyes that I thought may have greeted what I'm sure was just an elementary physics question. Not very well thought out on my part.

About that formula: It's something I wasnt familiar with. Does this mean that when I'm driving down a straight stretch of highway at steady 55 mph and observe another car as it passes me on the opposite side also travelling at 55 mph but in the opposite direction, then that car does not appear to me to be moving 110 mph, as I have always thought?
One thing that marcus didn't quite explain is that in the forumula he gave all the velocities must be measured as fractions of the SoL.

If you want to measure the velocities some other way (meters/sec, mph, etc.) the formula would be :

(a+b)/(1+ab/c²)

where c is the SoL in a vacuum as measured in the same way.

so in your 55 mph example, you would get

(55+55)/(1+55(55)/669600000²)

(669600000 is c in mph.)

this gives 109.999999999999257857719939083099 mph

Such a small difference from 110 mph that it isn't really worth worrying about. (It amounts to a difference of about 1/21266580 of an inch per hr)

So while this formula is the technically correct one to use, when a or b are small when compared to c we can get away with just using a+b and get an answer that is close enough to the correct one for most practical uses.

#### briv

Thanks Janus. I think I understand now. I am really beginning to appreciate Einstein for the monumental genius that he was-- I say only "beginning" not because I ever doubted him, but because I was never exposed to his discoveries.

There is still something I dont understand though.

To illustrate, more questions: (I dont know the exact speed of light in mph, but here lets just call it 670 million mph.)

Let's say we have a stretch of track 1.005 billion mileslong--the distance that light will travel in 1.5 hrs. At the exact same moment, two racers start at opposite ends at 3/4 the speed of light. After one hour, shouldn't they pass on another at the middle of the track? If so, both racers covered 1.005 billion miles of track in one hour making their combined speed 1.5 the SOL, or would it? According to Einstein's formula, combined they only travelled 643.2 million miles, which would leave 361.8 million miles still untraversed after 1 hour. How can this be if each racer travels over 502.5 miles of track during this time?

I really do appreciate the help you guys are giving. Thank You.

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#### marcus

Gold Member
Dearly Missed

" When they passed one another, wouldnt each observe that the other was travelling at a speed greater than light-- 1.5x SOL?"

so you were asking about the approach speed of the other guy as observed by ONE OF THE RACERS.

each of the two sees the other approach at 0.96 sol
or so we predict using that formula (a+b)/(1+ab)

but a third observer stuck to the road, standing at the midpoint as you suggest, still sees the distance between the two racers decrease at a simple 1.5 sol rate.

the rate of decrease of distance is not technically a velocity since it is not the speed of any particular thing moving relative to something else----each racer is approaching the man at center at speed 3/4, so the man at center observes the distance between the two as decreasing at 3/2

BUT WHEN YOU LOOK AT IT FROM THE PERSPECTIVE OF ONE OF THE RACERS as you introduced the topic in your first post, then
that racer guy has a different clock and he sees the road as shortened

You, standing beside the road at midpoint, have to reckon how it will seem to one of the guys who is moving-----figure his clock, figure his yardstick----the way he sees it with his sense of time and distance the opposite racer is approaching him at only 0.96 sol.

"SOL" for speed-0f-light seems like a good name for that unit of speed, dont remember other people at PF using it before now but probably some do

Originally posted by briv

Let's say we have a stretch of track 1.005 billion mileslong--the distance that light will travel in 1.5 hrs. At the exact same moment, two racers start at opposite ends at 3/4 the speed of light. After one hour, shouldn't they pass on another at the middle of the track? If so, both racers covered 1.005 miles of track in one hour making their combined speed 1.5 the SOL, or would it? According to Einstein's formula, combined they only travelled 643.2 million miles, which would leave 361.8 million miles still untraversed after 1 hour. How can this be if each racer travels over 502.5 miles of track during this time?

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#### briv

Marcus and Janus, thanks for helping me to understand. I now accept the fact that there's no way of getting around Einstein's speed limit. He has essentially rigged the universe to do his bidding---bloody genius!

Im off to try to learn more about these theories which I find utterly, utterly fascinating. When I'm done you can count on me to be back to pester you kind folks with more questions.

Thanks guys!

#### Mr.E Man

Howdy, I loved the question and the groping answers. The original query was one of perception and based, in part, on the age old problem of "If you're travelling at exactly the speed of light and you shine a flashlight ahead of yourself, won't the beam be traveling at twice the speed of light?"
The glib answer is, that since nothing can travel faster than the speed of light, given the currently understood constraints of 3D space/time, the flashlight will not be able to shine a beam directly in front of you, but will shine the light at an ever increasing pseudoperceptual velocity, relative to your position, as you swing the beam around until it is pointing directly behind you.
However, in your question you requested the perceptual response of each nearlight traveler (not an outside observer) to the approach of the other. It should be noted at this time that the inverse temporal dialation effect is in force, whereby the closer you come to the speed of light, the slower you move in your own little time bubble, relative to surrounding space. Also keep in mind that if measurement is done by visual observation, that under this temporal effect the human brain is also processing the visual cues at a diminished capacity. Also keep in mind that your awareness of the oncoming traveler is visually light based, and is determined by the compressed visual event horizon precedeing the other traveler.
So, long story short, by the time that you've realized that you are appraching the oncoming traveler, you will have already passed them long ago, your relative time depending upon how fast your individual brain processes information. So how fast, is based upon the amount of incoming light information where you will see them at the beginning, but they will suddenly wink out of existance as you seem to approach.

Light speed travel under normal space/time is NOT tourist friendly.

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