Relative speeds...

[Externet]

Two light sources S simoultaneously turned on; what is the relative speed (collision speed) of both light beams when propagation wavefronts meet midway ?

 

S---------><----------S

c-(-c)=?

 

Miguel

[ParticleOne]

Beauty of relativity: Speed = c

[timo]

Two light sources S simoultaneously turned on; what is the relative speed (collision speed) of both light beams when propagation wavefronts meet midway ?

 

S---------><----------S

c-(-c)=?

? = 2c.

There is no problem having differences of speeds being >c. Velocities and therefore also differences of velocities depend on the coordinate system used (=frame of reference). The stringent from relativity is that every velocity must be <=c. So the difference of two velocities must be <= 2c.

Two massive bodies travelling away from a common midpoint at 0.8c each will have a relative velocity of 1.6c. BUT: If you go to a frame of reference where one of these two bodies is at rest the transformation laws for velocities will guarantee you that in this system the relative velocity (which simply equals the velocity of the other body there) will be <c.

[Rocket Man]

i was under the impression that speed was relative to a medium (space) and that relative velocities can be up to 2c but the relative momentum is always <C. in newtonian physics momentum p = mv

is relativity, p = gamma (mv)

so when the two momentums are added together and converted back, some how the momentum tells us that 0.5c + 0.5c = 0.8c

this has been discussed better else where.

[JaKiri]

i was under the impression that speed was relative to a medium (space)

 

No, SR states that all velocities are relative to observers, each of which are equally valid, and there is no medium to be relative against (no aether).

 

and that relative velocities can be up to 2c but the relative momentum is always <C. in newtonian physics momentum p = mv

is relativity' date=' p = gamma (mv)

so when the two momentums are added together and converted back, some how the momentum tells us that 0.5c + 0.5c = 0.8c

this has been discussed better else where.[/quote']

 

If you have two things going towards eachother, at speeds a and b, the speed at which they approach eachother (in a reference frame moving along with one of them) is given by (a+b)/(1+(ab/c^2)). "Momentum is always <C" is meaningless, I'm afraid.

 

Interactions can take place faster than c, for example if you had two stiff rods in an X shape, moving <- and ->. If they had a recession speed of c, the place where they cross would move faster than c.

[Apeofman]

No' date=' SR states that all velocities are relative to observers, each of which are equally valid, and there is no medium to be relative against (no aether).

 

 

 

If you have two things going towards eachother, at speeds a and b, the speed at which they approach eachother (in a reference frame moving along with one of them) is given by (a+b)/(1+(ab/c^2)). "Momentum is always <C" is meaningless, I'm afraid.

 

Interactions can take place faster than c, for example if you had two stiff rods in an X shape, moving <- and ->. If they had a recession speed of c, the place where they cross would move faster than c.[/quote']

Excuse me, but i am a little confused as to whom the observer is in this thread.

 

If the observers are sitting on the beams they would see nothing of each other until they meet in the middle.

 

Also, if you are going to use either a or b as a reference frame then it's velocity is 0c in the equation you cite.

[Klaynos]

Excuse me' date=' but i am a little confused as to whom the observer is in this thread.

 

If the observers are sitting on the beams they would see nothing of each other until they meet in the middle.

 

Also, if you are going to use either a or b as a reference frame then it's velocity is 0c in the equation you cite.[/quote']

 

An observer cannot be sitting on a beam, a reference frame that has a velocity of c is not valid...

[JaKiri]

If the observers are sitting on the beams they would see nothing of each other until they meet in the middle.

 

"Observers" don't actually have to be little people sitting there with stop watches and space suits. It's a thought experiment.

 

 

We're sitting in the frame of the image while this is going on. A is moving at c left. B is moving at c right.

 

C has horizontal velocity equal to B, and therefore must be moving horizontally (from A's point of view) at c as well.

 

C also has a vertical component of motion, however, and therefore will have a total relative velocity of greater than c.

 

This is not an actual object going greater than c, of course, so that's ok.

 

Also, if you are going to use either a or b as a reference frame then it's velocity is 0c in the equation you cite.

 

Yes, because each rest frame is equally valid.

[JaKiri]

An observer cannot be sitting on a beam, a reference frame that has a velocity of c is not valid...

 

It's perfectly valid, especially in the concept of a thought experiment where people are sitting on poles travelling at or close to the speed of light (the argument still applies if the speeds are less than c, it's just easier to demonstrate that the speed will be greater than the speed of light if you set the speeds to c because you don't have to do any maths at all).

[timo]

Excuse me, but i am a little confused as to whom the observer is in this thread.

The observer is simply someone who sits around and watches two light beams aproach each other (a third entity if you wish to see it like that). What might have confused you is that in this case "watching an object" does not mean that some light emitted from that object hits your eye but that the object has the respective coordinates in the observers frame or reference. The additional problem that to visually see something it must emit light which must reach your eye is usually not considered in relativistic toy-problems like this one.

[Apeofman]

"Observers" don't actually have to be little people sitting there with stop watches and space suits. It's a thought experiment...

 

...We're sitting in the frame of the image while this is going on. A is moving at c left. B is moving at c right.

 

C has horizontal velocity equal to B' date=' and therefore must be moving horizontally (from A's point of view) at c as well.

 

C also has a vertical component of motion, however, and therefore will have a total relative velocity of greater than c.

 

This is not an actual object going greater than c, of course, so that's ok.

 

 

 

Yes, because each rest frame is equally valid.[/quote']

 

The point i made was about JaKiri's use of the relativistic velocity addition equation.

If you have two things going towards eachother' date=' at speeds a and b, the speed at which they approach eachother (in a reference frame moving along with one of them) is given by (a+b)/(1+(ab/c^2)). "Momentum is always <C" is meaningless,..

[/quote']

a and b are stated to be the speeds of approaching objects. There is no mention of what these speeds are or relate to. I could assume that a and b are the velocities of aproaching light beams. But JaKiri was responding to ...

so when the two momentums are added together and converted back, some how the momentum tells us that 0.5c + 0.5c = 0.8c

In which case velocities a and b are relative to an unmentioned stationary refererence frame containing an observer who i will call O.

i.e Vab= (Vao+Vbo)/(1+(VaoVbo/c^2)) gives the right answer to "0.5c + 0.5c = 0.8c"

 

JaKiri however chose to ..

If you have two things going towards eachother' date=' at speeds a and b, the speed at which they approach eachother (in a reference frame moving along with one of them) is given by (a+b)/(1+(ab/c^2)).

[/quote']

... make one of the two objects the reference frame (let's say the object is a), which raises the question as to what frame of reference the velocity of b relates to, in order to justify using the equation cited.

 

Jakari now state's..

C also has a vertical component of motion, however, and therefore will have a total relative velocity of greater than c.

..What vertical component:confused: Relative to what:confused: Where: was this mentioned in your previous post. When the quantities are properly specified the equation Jakari cited..(a+b)/(1+(ab/c^2)) does not produce a sum greater than c. So why now state " and therefore will have a total relative velocity of greater than c.":confused:

 

It is my understanding that, in the context of relativity, relative velocity, is measured in the observers frame of reference along the direction of the observed objects motion. This relative measured velocity can be used in the Lorentz transforms and it's derivatives such as the relativistic velocity addition equation cited. For example..

With permission from magen.co.uk

Note. This is an old picture. The word "see":eek: is incorrect. Should be, measures/observe's (along the axis of motion!).

Look's like i need to update:embarass:

 

Time for sleep.

Maybe reality will different then:D

[Apeofman]

The observer is simply someone who sits around and watches two light beams aproach each other (a third entity if you wish to see it like that). What might have confused you is that in this case "watching an object" does not mean that some light emitted from that object hits your eye but that the object has the respective coordinates in the observers frame or reference. The additional problem that to visually see something it must emit light which must reach your eye is usually not considered in relativistic toy-problems like this one.

I understand that, but i don't yet see how Jakari's propositions give rise to a result greater than c, if he is taking that into account. Maybe things will be clearer after some sleep. Interesting topic.

[timo]

It [an observer sitting on a beam, a reference frame that has a velocity of c'] is perfectly valid.

Lets say it this way: What is the transformation matrix between the coordinates (t,x) I see and the coordinates (u,y) the guy on the beam sees ?