[Guest]

In Einstein's paper "on the electrodynamics of moving bodies" he asserted that in all frames of reference the speed of light(3*10^8 m/s aprox.) is constant. I have also obtained one of einstiens equations for determining relative velocity.

[Rv=(V1+V2)/(1+(sqrt((V1)(V2)/(c^2))))]

read relative velocity equals the quotient of the sum of two velocities over one plus the square root of the quotient of the product of the velocites over the speed of light squared.

Now my question comes in here. What if one of the velocities was of a particle travelling at the speed of light©. say the velocity of particle 2 was the speed of light (V2=C).

then the math would simplify to [Rv=(V1+c)/(1+sqrt(V1/c))].

I thought that since the speed of light was constant in all frames of reference the relative velocity between the particle travelling at less than C and the particle travelling at C would always be C. but When I graphed it on a calculator at school the graph didn't support my thinking. Instead of it showing a constant line at y=c It showed a curve that decreased then increased. this means that the particle, that we measure as, travelling slower than C would measure the velocity of the particle travelling at, what we measure to be C, to be less than C. How is this explained?(since the previous statement contradicts what einstein said becuase if i measure a particle to be travelling at C so should the other moving particle.)

I want my question to be clear so for sake of clarity I pose this illustration. I am in space at point A (pA) and I measure Particle 1 (p1) to be travelling at a velocity (any velocity) less than the speed of light (p1 V<C) and I also measure particle 2 (p2) to be travelling at the speed of light (p2 V=C)). Einstein said that the speed of light is constant for all frames of reference, even p1's. My graph shows that the relative velocity(Rv) between P1 and P2 is <C. How can this be? Am I misunderstanding something? Am I missing something?

[Yoruichi-san]

Err...where did you get the formula for 'relative velocity'? I'm not sure what they're defining 'relative velocity' as...but...

Say you're in a frame of reference, S, and you see an object moving with velocity u, and I'm in a frame of reference, S', which is moving at a velocity v with respect to S. Then I will see the object moving with velocity u' = (u-v)/(1-uv/c²). If u=c, i.e. if the object is moving at the speed of light in frame S, then the equation becomes u'= (c-v)/(1-v/c)=c, i.e. it is also moving at the speed of light in frame S'.

Edit: Okay, I think I kind of see where you're confused...'relative velocity' is NOT the velocity of particle 2 in any frame of reference...so there's no reason for it to be equal to the speed of light...I'm too sleepy right now to figure out what it is actually is though...

Edited March 22, 2009 by Yoruichi-san

[Yoruichi-san]

Okay, being slightly more awake...there's definitely something wrong with your equation...the velocities should subtract instead of add, since if the two particles are moving in the same direction (i.e. have the same sign), then their velocity relative to each other should be less, not more, than if they are moving in opposite directions. Also, the square root term bugs me, since if the particles are moving in opposite directions, (V1)*(V2) will be negative...and the square root will be complex, which doesn't make sense. I'll try deriving an equation for relative velocity using the regular equation later when I have time...

[Guest]

Err...where did you get the formula for 'relative velocity'? I'm not sure what they're defining 'relative velocity' as...but...

Say you're in a frame of reference, S, and you see an object moving with velocity u, and I'm in a frame of reference, S', which is moving at a velocity v with respect to S. Then I will see the object moving with velocity u' = (u-v)/(1-uv/c²). If u=c, i.e. if the object is moving at the speed of light in frame S, then the equation becomes u'= (c-v)/(1-v/c)=c, i.e. it is also moving at the speed of light in frame S'.

I got my equation from the fifteenth edition of Einstien's papers on the theory of special and general relativity. going back to the book and flipping through the pages see now one of the lorentz transformation equations looks similar to my equation. so maybe my earlier attempt was just to try to explain the lorentz equation, which explains how a particle was moving along the x axis at velocity V. (this question had been on my mind for a while and was just sitting on the back burner until I found a audience to ask the question to.)

where did you get your equation from?

I found an online graphing calculator and plugged your formula in substituting V for x and 300000000 for C and it turned out that a line did form at y=3*10^8.

Edit: Okay, I think I kind of see where you're confused...'relative velocity' is NOT the velocity of particle 2 in any frame of reference...so there's no reason for it to be equal to the speed of light...I'm too sleepy right now to figure out what it is actually is though...

As for the second part... My equation was supposed to find out the relative velocity of P2 to P1 and if it was anything other than C something was arrey since that meant that in P1's frame of refference P2 wasn't traveling at the speed of light thus creating a contradiction to our view of P2. So there was a reason for P1's measures to match ours.

One thing bothers me though...when I try to use your formula to find the relative velocity between particles that are both traveling at less than light speeds it always gives me the velocity of what ever speed I put in as the velocity of u as the relative speed no matter what v is.

[Guest]

In Einstein's paper "on the electrodynamics of moving bodies" he asserted that in all frames of reference the speed of light(3*10^8 m/s aprox.) is constant. I have also obtained one of einstiens equations for determining relative velocity.

[Rv=(V1+V2)/(1+(sqrt((V1)(V2)/(c^2))))]

read relative velocity equals the quotient of the sum of two velocities over one plus the square root of the quotient of the product of the velocites over the speed of light squared.

Now my question comes in here. What if one of the velocities was of a particle travelling at the speed of light©. say the velocity of particle 2 was the speed of light (V2=C).

then the math would simplify to [Rv=(V1+c)/(1+sqrt(V1/c))].

The terms in your equation are not quite correct. First of all, the sqrt does not belong. And, the way you have it written, V1 is the velocity of object 1 relative to an outside observer, V2 would be the velocity of object 2 relative to object 1, and Rv would be the velocity of object 2 relative to the observer. The equation provided by Y-san gives the velocity of object 2 relative to object 1 (u') in terms of the velocities of (1) and (2) relative to the observer, which is what I think you were looking for.

I thought that since the speed of light was constant in all frames of reference the relative velocity between the particle travelling at less than C and the particle travelling at C would always be C. [...] My graph shows that the relative velocity(Rv) between P1 and P2 is <C. How can this be? Am I misunderstanding something? Am I missing something?

I didn't quite catch all of that, but just because a particle is travelling at c in one frame of reference does not mean it does so in every frame of reference. Its velocity is zero, e.g. in its own frame. Anyway, I think if you use Y-san's equation, you will find that object 2 is always travelling at c relative to both (1) and the observer.

I hope this helps...

D.

[Guest]

The origins of the equation are unknown but I did get it from some reliable source. maybe one of my parents old physical science textbooks. It's in a storage trailer somewhere. I'll try to find it. Lets just assume that the two particles are travelling in the same direction for the sake of simplicity. and Of course in everyone's own frame of reference they are at rest and everything else is moving. in the case of particle 2; it is at rest and everything else is moving at the speed of light.

[Yoruichi-san]

As for the second part... My equation was supposed to find out the relative velocity of P2 to P1 and if it was anything other than C something was arrey since that meant that in P1's frame of refference P2 wasn't traveling at the speed of light thus creating a contradiction to our view of P2. So there was a reason for P1's measures to match ours.

One thing bothers me though...when I try to use your formula to find the relative velocity between particles that are both traveling at less than light speeds it always gives me the velocity of what ever speed I put in as the velocity of u as the relative speed no matter what v is.

Then you're misinterpreting/misreading the equation...using your example, v would be equivalent to your V1 (velocity of particle 1, which is also the velocity of particle 1's frame of reference relative to the initial inertial frame), u is V2 (velocity of particle 2 in the initial inertial frame), and u' would be the velocity of particle 2 in the frame of reference of particle 1.

The origins of the equation are unknown but I did get it from some reliable source. maybe one of my parents old physical science textbooks. It's in a storage trailer somewhere. I'll try to find it. Lets just assume that the two particles are travelling in the same direction for the sake of simplicity. and Of course in everyone's own frame of reference they are at rest and everything else is moving. in the case of particle 2; it is at rest and everything else is moving at the speed of light.

Honestly...I think you're mis-defining the variables...d3k3's interpretation of your equation makes more sense...and keep in mind that what's constant is the speed of light , i.e. the speed of particle 2 is constant in all frames and is equal to c only if particle 2 is a photon...

[Guest]

The origins of the equation are unknown but I did get it from some reliable source. maybe one of my parents old physical science textbooks.

But... you still have to get the correct answer at non-relativistic speeds. Based on your interpretation of the variables, when V1, V2 << c, I should still get Rv = V2 - V1...

Also, I'm positive the sqrt does not belong. Start with the Galilean/Newtonian definition and apply the Lorentz transformation...

[Guest]

In Einstein's paper "on the electrodynamics of moving bodies" he asserted that in all frames of reference the speed of light(3*10^8 m/s aprox.) is constant. I have also obtained one of einstiens equations for determining relative velocity.

[Rv=(V1+V2)/(1+(sqrt((V1)(V2)/(c^2))))]

Right, I don't know where you got that equation from. It is much like this one:

Vr = (V1 + V2) / ( 1 + ((V1*V2)/(c²)))

Which is his equation for relative velocity. (also see here; same equation, different variable names.)

Except for your inclusion of the square root.

This calculates the relative velocities of two objects, showing that they never exceed the speed of light. It is best known in the train paradox; when one runs or fires a bullet or whatever on a moving train, and calculating the actual speed of the bullet. The another common example is the idea of a vehicle moving at the speed of light and turning on it's headlights (thus we have V1 and V2 both equalling the speed of light.) The calculation shows that the combined speed (that is the speed of the light from the head lights relative to a stationary object) will never exceed the speed of light.

In that example the calculation would be simplicity itself:

Vr = (c + c) / ( 1 + ((c*c)/(c²)))

Vr = (c + c) / ( 1 + ((c²)/(c²)))

Vr = (c + c) / ( 1 + (1))

Vr = 2c / 2

Vr = c

So basically this was the resulting equation to show that the speed of light is indeed a maximum speed possible. Actually what it is, is a constant in spacetime; everything moves at "the speed of light" (299,792,458 metres per second) in spacetime.

[Guest]

Okay.... I gotta say ONE thing.

YOU GUYS AMAZE ME WITH YOUR SMARTNESS!

I can't even read the original topic with out making my brain hurt!