Am I missing something or is this faster than light

[madah12]

suppose we use a machine that can fire two lasers at opposite direction at the same instant
after 1 second the one laser would have traveled |C| distance to right
and the other laser would have traveled the same distance to the left
so from the lasers point of view the other one in moving -2c m/s in the opposite direction right?

 

[rcgldr]

From any sub-light frame of reference, multiple beams of light could be "observed" at the same time, and the relative speed between the beams can reach a maximum of 2c, but each beam only travels at 1c, regardless of the speed of the frame of reference.

What could be observed from a light speed frame of reference, such as photon, is debatable.

 

[madah12]

can we achieve a reference speed =c by moving two objects in opposite directions?
( I mean objects that we can use to observe the other particle )

 

[tiny-tim]

hi madah12!

let's suppose you have two spaceships going in opposite directions at 0.9c …

their relative speed comes from the velocity-addition formula, and is still less than c …

see http://en.wikipedia.org/wiki/Addition_of_Velocities_Formula#Special_theory_of_relativity"

 

[soothsayer]

so, for example, if we were able to move some sort of measuring device on way at 0.6c and some particle, say an electron, the other way at 0.6c, would our measuring device not observe an electron moving at 1.2c relative to it?

No, it wouldn't

Classically, this would make sense, but this is not how the universe works, nothing can go faster than light, in any reference frame. We need to use the Lorentz velocity transformation.

if you plug in 0.6 for u (speed of electron) and -0.6 for v (speed of measuring device, traveling opposite direction as electron, hence the negative sign, which could applied to either, depending on how you define the coord. system, of course) you'll come up with:
u'(speed of electron in measuring device frame) = 1.2c/[1-((0.6c*-0.6c)/c^2)) = 1.2c/1.36
u'=0.88c

You'll notice that any values you could plug in for u and v lower than c will yield a u' lower than c. It's weird, and counter-intuitive, but that's the way the word works!

 

[soothsayer]

As for what photons would "observe" as they traveled the speed of light relative to another beam of photons, well, that's not an easy question to answer. No one knows, photons are massless particles that travel, well, the speed of light, and our relativistic equations don't work when v=c.

try taking the limit of the time dilation and length contraction formulas as v-->c and you'll see the formulas suggest that time stops and all space contracts into a 2 dimensional plane perpendicular to the motion vector, weird stuff.

 

[madah12]

so, for example, if we were able to move some sort of measuring device on way at 0.6c and some particle, say an electron, the other way at 0.6c, would our measuring device not observe an electron moving at 1.2c relative to it?

No, it wouldn't

Classically, this would make sense, but this is not how the universe works, nothing can go faster than light, in any reference frame. We need to use the Lorentz velocity transformation.

if you plug in 0.6 for u (speed of electron) and -0.6 for v (speed of measuring device, traveling opposite direction as electron, hence the negative sign, which could applied to either, depending on how you define the coord. system, of course) you'll come up with:
u'(speed of electron in measuring device frame) = 1.2c/[1-((0.6c*-0.6c)/c^2)) = 1.2c/1.36
u'=0.88c

You'll notice that any values you could plug in for u and v lower than c will yield a u' lower than c. It's weird, and counter-intuitive, but that's the way the word works!

I am trying to understand this stuff since I didn't study relativity but does that change that at 1 second the distance between them is now 1.2c ?

 

[tiny-tim]

in our reference frame, it is 1.2c, so in our reference frame, they are moving apart faster than light

but in either spaceship's reference frame, the distance is reduced (by the famous Lorentz-Fitzgerald contraction)

 

[Phrak]

hi madah12!

let's suppose you have two spaceships going in opposite directions at 0.9c …

their relative speed comes from the velocity-addition formula, and is still less than c …

see http://en.wikipedia.org/wiki/Addition_of_Velocities_Formula#Special_theory_of_relativity"

Hi tiny. Does velocity composition work when v-->c, or would it involve dividing by zero?

 

[madah12]

in our reference frame, it is 1.2c, so in our reference frame, they are moving apart faster than light

but in either spaceship's reference frame, the distance is reduced (by the famous Lorentz-Fitzgerald contraction)

so how do we know which frame of reference to use so that our physics laws will be able to make accurate consistent predictions?

 

[Doc Al]

so how do we know which frame of reference to use so that our physics laws will be able to make accurate consistent predictions?

You can use whatever frame you like. The laws of physics work in all inertial frames.

 

[madah12]

You can use whatever frame you like. The laws of physics work in all inertial frames.

how can you reconcile that with what tiny tim said?

 

[Doc Al]

how can you reconcile that with what tiny tim said?

What's to reconcile? Distances and times are frame dependent. Everything is perfectly consistent.

Note that there is no frame where the spaceships are moving faster than light. In 'our' frame, the spaceships separate at a rate of 1.2c, but they still only move at 0.6c. (Nothing is moving at 1.2c in any frame.)

 

[madah12]

yes but the distance between them is changing with rate of 1.2c/s so the rate of the change of distance with respect to time isn't always velocity?

 

[Doc Al]

yes but the distance between them is changing with rate of 1.2c/s so the rate of the change of distance with respect to time isn't always velocity?

It's not the velocity of anything, just the rate at which the distance between them changes as seen in one particular reference frame. It is not the relative velocity of one ship with respect to the other. It's sometimes called the closing/separation speed. (Rindler calls it the mutual velocity, as opposed to relative velocity.)

 

[tiny-tim]

hi madah!

it doesn't matter that the relative speed is 1.2c, because nothing depends on it

for example, the "increase in mass" of each spaceship as measured by us depends on that spaceship's speed as measured by us,

and the "increase in mass" of each spaceship as measured by the other spaceship depends on that spaceship's speed as measured by the other spaceship,

but nothing depends on the relative speed of the spaceships as measured by us!

Hi tiny. Does velocity composition work when v-->c, or would it involve dividing by zero?

It doesn't work

 

[Doc Al]

Does velocity composition work when v-->c, or would it involve dividing by zero?

As long as only one of the velocities is c it will work, but not if both are c.

 

[soothsayer]

Madah, another consequence of relativistic effects and Lorentz transformations is that lengths also contract for an object as it reaches closer and closer to the speed of light. So, imagine two spaceships moving at opposite directions at 0.6c, you would think it's paradoxical that each spaceship would see the other moving at 0.88c while the distance increases at 1.2c per second, but according to the moving ships, this is not how the distance increases in the reference frame of an object moving that close to the speed of light. In fact, the equation I posted is derived partially from the Lorentz transformation for length contraction,

x' = ɣ(x-vt)

where x is the length according to a stationary observer, so 1.2c meters as you mentioned, and ɣ is the Lorentz factor
[PLAIN]http://http://upload.wikimedia.org/math/6/0/6/6068db761d2dfce507b37591197be25b.png

I think time dilation may also factor into it, as an object gets close to the speed of light (usually needs to be about 0.25c for relativistic effects to be noticeable) then time will also slow down for that object, so what you measure to be a second, the moving object will determine to actually be less than a second, that formula is:

t' = ɣ(t-(vx/c²))

 

[Phrak]

As long as only one of the velocities is c it will work, but not if both are c.

OK, then the direct answer to the question, assuming we are using infinitesimal calculus, is that the velocity of one photon with respect the other is 2C/epsilon, where epsilon is a positive/negative infinitesimal when v-->c is approached from below/above.

Physically, this doesn't help much. The other photon is going really, really fast, but we don't know if it's comin' or goin'.

 

[soothsayer]

Shouldn't each photon observe the other moving a speed c? That is the intuitive answer for me, anyway. Er, well, intuitive as far as SR is concerned, which isn't so intuitive itself. Actually, it doesn't make any sense to ask what speed a photon views another photon traveling, as according to SR, time would stop for a particle traveling at speed c, there is no interval of time with which to calculate dx/dt. Add the consequence of length contraction for a particle traveling speed c and you can see there is no way to give an answer to the original question with the physics we have.

 

[Phrak]

Shouldn't each photon observe the other moving a speed c?

Surprising, hu? Relativity doesn't supply the answer unless you are willing to accept that really fast to the left is the same as really fast to the right--or deny that one is physical. As you can see the velocity composition equation you gave in post #5 is not well behaved where v=w=c.

 

[soothsayer]

As you can see the velocity composition equation you gave in post #5 is not well behaved where v=w=c.

You mean when the photons are traveling in the same direction? Yeah, that would give you an indeterminate, wouldn't it? And it's quite a paradox if you think about it, because on one hand, light is supposed to travel the same speed in all reference frame, but in the other, a photon would have to either observe the other photon standing still in its reference frame or traveling with a velocity relative to it, both are strange to think about.

 

[Phrak]

You mean when the photons are traveling in the same direction? Yeah, that would give you an indeterminate, wouldn't it? And it's quite a paradox if you think about it, because on one hand, light is supposed to travel the same speed in all reference frame, but in the other, a photon would have to either observe the other photon standing still in its reference frame or traveling with a velocity relative to it, both are strange to think about.

Good catch, soothsayer! You've caught me in an error. Ouch. I should have broken out the crayons. It would be egregious to let such an error stand without correction...Given

u'=\frac{u-v}{1- \frac{uv}{c^2}}\ ,

assume u=c and v=c-\epsilon\ . Epsilon is an infinitesimal v and u are in the same direction, so make them both positive. The sign of epsilon washes-out.

u'=\frac{c-(c-\epsilon)}{1- \frac{c(c-\epsilon)}{c^2}}

\frac{c-(c-\epsilon)}{1- \frac{c(c-\epsilon)}{c^2}} = <br /> <br /> \frac {\epsilon} {1- \frac{c^2-c \epsilon}{c^2} } = <br /> <br /> \frac{\epsilon}{\frac{\epsilon}{c}} = cWhen v and u are in opposite directions, u=c and v=-c+\epsilon, then

u&#039;=\frac{2c-\epsilon}{2-\frac{\epsilon}{c}} = c

more or less.

 

[K^2]

The limit exists, but if both are exactly c, it's still undefined.

Light travels along null geodesic, and that's a very special world line. Relative to any other coordinate system, proper time along null geodesic is a constant. Since velocity relative to observer is rate of change of position with respect to proper time of the observer, if the proper time does not change, velocity cannot be defined. This isn't just a problem with division by zero in this particular formula. Relative velocity between two beams of light is fundamentally undefined.

 

[Phrak]

The limit exists, but if both are exactly c, it's still undefined.

Light travels along null geodesic, and that's a very special world line. Relative to any other coordinate system, proper time along null geodesic is a constant. Since velocity relative to observer is rate of change of position with respect to proper time of the observer, if the proper time does not change, velocity cannot be defined. This isn't just a problem with division by zero in this particular formula. Relative velocity between two beams of light is fundamentally undefined.

While this is true, the context has all along been special relativity, and the problem can still be presented at a single point of spacetime in the rigor of general relativity where comparison of velocities (speeds, actually) is local yielding the same result as in special relativity.