Yes, You Can Go Faster Than Speed Of Light

[sluggo]

Moronium #209

I never got a coherent explanation for this confusing situation.  Does anyone have one?

We have two objects, A and C.  Let's look at this from A's perspective.  He sees C moving away from him at .9c.  And, of course, C will also see A receding from him at .9c, because that's their relative velocity, right?

Now then, let's say that A notices object B, which is between him and C.  He sees B as going in the same direction as C, except that B is receding from him at the rate of only .45c.

What will B see?   He will see  A receding from him at .45c, right?  He has to, because that's how A sees him. That's their relative velocity. He will also see C receding from him at the rate of .45c, but in the opposite direction, right?

 

 

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The composition of velocities can be restated as 'addiition of velocities calculated from different references is not allowed'.

A will calculate velocities for B and C using (a-v)/(1-av), with lower case corresponding to speed for selected observers. Since B is in relative motion to A, his calculations will differ for C, but the same form applies. There is no need in this case for the composition formula of SR.

Given relative to A, a=0, b=.45, c=.90, B calculates c as .45/.59=.76.

If A used B's result, C would exceed c at 1.21.

With A and C moving in opposite directions from B, the same form still works, if you remember to change the signs for direction.

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Going back to A's perspective, because he noticed B, A no longer sees C receding from him at .9c.  A now sees C receding from him at the lesser speed of (approx) .75c, right?  

 

 

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B's presence does not change A's measurements.

 

[Moronium]

So, this is your question?

 

 

That is easily obtained by using the velocity addition formula.

 

The relevant formula is:

 

....So you have V_AC = 0.9c, V_AB = 0.45c and V_BC = 0.756c

 

Do you see any problems with that answer?

 

 

 

I have no problem with your math, Popeye, and thanks for the input.  A-wal said it would be .8, but you're a little more precise with .756c.  But, no, it doesn't really answer the question I'm asking, which I have stated above multiple times.  I won't ask it again here, but you can scroll up.

 

Just curious....would you then conclude, as A-wal did, that you should add .756 to .45 to get 120.6c as B's view of the relative velocity between A and C?

Edited June 28, 2018 by Moronium

[Moronium]

B's presence does not change A's measurements.

 

No, I wouldn't think so.  Your calculations are close to Popeye's, so there's a second opinion on how the formula is to be applied.  But, as I've said, my question was not about the math, per se.

Edited June 28, 2018 by Moronium

[OceanBreeze]

I have no problem with your math, Popeye, and thanks for the input.  A-wal said it would be .8, but your a little more precise with .756c.  But, no, it doesn't really answer the question I'm asking, which I have stated above multiple times.  I won't ask it again here, but you can scroll up.

 

 

 

That is the only question I found. Is there another one?

 

Just curious....would you then conclude, as A-wal did, that you should add .756 to .45 to get 120.6c as B's view of the relative velocity between A and C?

 

 

No, I would not! The relative velocity is 0.9c as stated. But you will need to take that one up with A wall.

[Moronium]

 

Moronium #209

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If A used B's result, C would exceed c at 1.21.

 

 

But A doesn't use B's result, only B does.  So what does B get?

Edited June 28, 2018 by Moronium

[Moronium]

No, I would not! The relative velocity is 0.9c as stated. But you will need to take that one up with A wall.

 

The relative velocity, as seen by A, B, and C, you mean, or just as seen by A and C?

Edited June 28, 2018 by Moronium

[Super Polymath]

[OceanBreeze]

The relative velocity, as seen by A, B, an C, you mean, or just as seen by A and C?

 

 

The relative velocity between A and C is 0.9c

 

The relative velocity between A and B is 0.45c

 

The relative velocity between B and C is 0.756c

 

B must use the velocity addition formula to calculate the relative velocity between A and C as he is not in the frame of reference to see this velocity, he can only calculate it and he MUST use the velocity addition formula.

When he does that, he will get 0.9c.

 

If he adds the velocities naively, he will get 1.206c but that is not a velocity! A velocity must have both magnitude and direction. What direction can B assign to the number 1.206c? He cannot assign a direction, so that number is not a relative velocity. It is properly called a "closing speed" or "opening speed" and is the rate of change of the distance between A and C as seen from B. Closing speeds are not velocities.

Edited June 28, 2018 by OceanBreeze

[Moronium]

B must use the velocity addition formula to calculate the relative velocity between A and C as he is not in the frame of reference to see this velocity, he can only calculate it and he MUST use the velocity addition formula.

When he does that, he will get 0.9c.

 

 

He will?  This doesn't sound quite right to me.  What's the math associated with that?

[Moronium]

If he adds the velocities naively, he will get 1.206c but that is not a velocity! A velocity must have both magnitude and direction. What direction can B assign to the number 1.206c? He cannot assign a direction, so that number is not a relative velocity. It is properly called a "closing speed" or "opening speed" and is the rate of change of the distance between A and C as seen from B. Closing speeds are not velocities.

 

I'm not concerned with velocity here, only speed.  So you're saying that what B sees as the "opening speed" will be 1.206?

 

You're trying to make a distinction without a difference here.   When you're dealing in just one dimension, as we are here, the velocity is a scalar, not a vector.  In other words, speed and velocity are the same things, in these circumstances. Put yet another way, yes, it is a "relative velocity."

Edited June 28, 2018 by Moronium

[Moronium]

The relative velocity between B and C is 0.756c

  

Is that the speed B will actually "see," using the doppler effect to gauge his speed relative to C, or is that just what he (or I guess I should say you) will calculate?

Edited June 28, 2018 by Moronium

[OceanBreeze]

I'm not concerned with velocity here, only speed.  So you're saying that what B sees as the "opening speed" will be 1.206?

 

You're trying to make a distinction without a difference here.   When you're dealing in just one dimension, as we are here, the velocity is a scalar, not a vector.  In other words, speed and velocity are the same things, in these circumstances. Put yet another way, yes, it is a "relative velocity."

 

 

Velocity is always a vector while speed is a scalar. That is not a distinction that I am "trying to make" it is a distinction that exists in the correct precise usage of language! Being  sloppy and imprecise is what leads to confusion and needless arguments.

 

These three relative velocities are what are actually seen by the respective observers:

 

The relative velocity between A and C is 0.9c, as seen by A and C with respect to each other

 

The relative velocity between A and B is 0.45c, as seen by A and B with respect to each other

 

The relative velocity between B and C is 0.756c, as seen by B and C with respect to each other

 

Now, B sees A moving away at a velocity of 0.45c to the East and B also sees C moving away with a velocity of 0.756c to the West, so in opposite directions. He cannot assign one direction to the simple sum of this motion. Can you see that? Since he cannot assign one direction, the simple sum of 1.206 c is not a velocity. It is an opening speed (usually referred to as a closing speed)

 

If B wants to know the relative velocity between A and C he uses the velocity addition formula as follows:

 

[math]{ V }_{ AC }\quad =\quad \frac { { V }_{ AB }\quad +\quad { V }_{ BC } }{ 1\quad +\quad \frac { ({ V }_{ AB })\quad ({ V }_{ BC }) }{ { c }^{ 2 } }  }[/math]

 

 

Punching in the numbers:

 

[math]{ V }_{ AC }\quad =\quad \frac { { 0.45c }\quad +\quad { 0.756c } }{ 1\quad +\quad \frac { ({ 0.45c })\quad ({ 0.756c }) }{ { c }^{ 2 } }  }[/math]

 

You get V_AC = 0.9c  calculated by B using the velocity addition formula.

 

Does that answer all your questions?

Edited June 28, 2018 by OceanBreeze

[Vmedvil2]

The point of special relativity is that you cannot move faster than C away from a rest frame with velocity you could argue that two objects are moving opposite directions at greater than C are moving faster than the speed of light but from the rest frame's view they are not, you can never move a object with mass at the speed of light from a rest frame's view. Final Verdict unfortunately. Note how in the equations that the REST frame is always the variable to take in account, it is all about moving from a rest frame's view in special relativity. If two objects are moving relative from each other faster than C in different directions that is irrelevant it only matters what a rest frame view of the situation which is both are moving less than C away from the rest frame.

Edited June 28, 2018 by VictorMedvil

[OceanBreeze]

The point of special relativity is that you cannot move faster than C away from a rest frame's with velocity you could argue that two objects are moving opposite directions at greater than C are moving faster than the speed of light but from the rest frame's view they are  not, you can never move a object with mass at the speed of light from a rest frame's view. Final Verdict unfortunately.

 

Yes, exactly! Two objects moving in opposite directions as seen from an observer in the middle may have closing speed greater than c but a closing speed is not a velocity therefore it does not violate SR. To get the relative velocity between the two objects the observer at B must use the velocity addition formula and the result will always be less than c.

 

Thank you for your input!

[Moronium]

Velocity is always a vector while speed is a scalar. That is not a distinction that I am "trying to make" it is a distinction that exists in the correct precise usage of language! Being  sloppy and imprecise is what leads to confusion and needless arguments.

 

  

Well, this wiki author seems to disagree with you:

 

In the one-dimensional case,[2] the velocities are scalars and the equation is either:

{\displaystyle \,v_{rel}=v-(-w)} \,v_{rel}=v-(-w), if the two objects are moving in opposite directions, or:

{\displaystyle \,v_{rel}=v-(+w)} \,v_{rel}=v-(+w), if the two objects are moving in the same direction.

 

https://en.wikipedia.org/wiki/Velocity

 

He cannot assign one direction to the simple sum of this motion. Can you see that? Since he cannot assign one direction, the simple sum of 1.206 c is not a velocity. It is an opening speed (usually referred to as a closing speed

 

 

 

No, I honestly can't see that.  If I put $10 in the bank, then withdrew five, I wouldn't find it impossible to conclude that there was $5 left in the bank just because the two transactions were in "opposite directions."  $5 would still be the "simple sum."

Edited June 28, 2018 by Moronium

[Vmedvil2]

Well, this wiki author seems to disagree with you:

 

 

 

No, I honestly can't see that.  If I put $10 in the bank, then withdrew five, I wouldn't find it impossible to conclude that there was $5 left in the bank just because the two transactions were in "opposite directions."  $5 would still be the "simple sum."

 

That is false velocity is a vector and not a scalar so you must use the Pythagorean theorem which is A² + B² = C²  , it is not a linear relationship between the vectors, then that wiki Author is wrong and I will correct it being a actual scientist. It is part of Vector Calculus and Linear Algebra.

 

 

Edited June 28, 2018 by VictorMedvil

[Moronium]

Does that answer all your questions?

 

 

No, actually I still have others about what you've said here, but, for now, scroll down (from where you were), you'll see another question.

Edited June 28, 2018 by Moronium