Understanding Relativity: Chat with a Physics Professor Online for Clarity

[phillip1882]

I've tried to understand the theory of relativity several times and could never wrap my head around it. so i would like to make an offer here. i would like to chat with a physics professor online for about an hour over Skype. my discussion probably does not belong a serious physics forum which is why i am not posting my questions here. if you would prefer that, I don't mind posting my questions here instead.
if I'm breaking any forum rules i would like to apologize ahead of time, and just say I'm not quite sure where else to turn.

 

[phyzguy]

Why not just post your questions here? I'm sure people would be happy to try and answer them. You know what they say, the only stupid question is the one that is not asked!

 

[phillip1882]

if i have two light beams, traveling the same direction, both going the same speed (because the speed of light is constant) what speed would one light beam be traveling relative to the other light beam? the only logical conclusion i personally can come to is: zero.
again let's say i have two light beams, this time i fire them at each other. both light beams again are traveling at the speed of light, so the speed of light of one light beam relative to the other must therefore be 2*the speed of light. if I'm wrong I don't quite see how. let's start of with this.

 

[Mentz114]

Velocity composition in relativity uses a rule which ensures that the 'added' velocities cannot exceed c. Actually, one cannot use light to measure a velocity against - it will always be traveling at c regardless of the matter frame. If two beams are traveling parallel, one could say ( pace above ) that they had zero relative velocity.

See here http://en.wikipedia.org/wiki/Velocity-addition_formula

 

[VantagePoint72]

so let's start off by proving you can add and subtract velocity even when you're dealing with something that's moving near the speed of light.
if i have two light beams, traveling the same direction, both going the same speed (because the speed of light is constant) what speed would one light beam be traveling relative to the other light beam? the only logical conclusion i personally can come to is: zero.
again let's say i have two light beams, this time i fire them at each other. both light beams again are traveling at the speed of light, so the speed of light of one light beam relative to the other must therefore be 2*the speed of light. if I'm wrong I don't quite see how. let's start of with this.

Yes, in your frame of reference the light beams are traveling apart at 2c. That's fine, and doesn't contradict relativity: relativity says that the speed of light is c in all inertial frames. Both beams of light are traveling at c in your frame. The subtleties of SR come in when you try to convert relative velocities between frames. Now, good inertial frames of observers in SR must travel strictly less than the speed of light, so to simplify things let me change your example slightly. A high energy particle is traveling at 0.8c to my left, and a second one travels at 0.8c to my right. In my frame, I would say the relative velocity between the two is 1.6c. Again, this is fine—both velocities relative to me are less than c.

Now, what about the frame of reference of the left-going particle ("particle 1")? How fast is the right-going particle ("particle 2") moving away according to particle 1? This is where our pre-relativistic intuition doesn't serve us very well. The answer is not 1.6c again because relative velocities add differently when transforming between frames. The formula is this, and it gives about 0.98c in this case. So, according to particle 1, particle 2 is moving less than c (as it must!). The reason for this is that velocity is the ratio of distance traveled to time, but time dilation and length contraction mean that this ratio can depend on your frame of reference. If you play around with the velocity addition formula, you'll see that if you plug in any two speeds less than the speed of light, you get a relative speed in one of the moving frames that is also less than the speed of light.

 

[phillip1882]

okay i'll try again. let's take the example of two particles, both going 0.8*the speed of light form the position of an inertial observer.
if they are going in the same direction, what speed would one particle be going relative to the other?
again, i would say 0. if they are going opposite directions, they must be going 1.6 relative to one another.
now the previous poster has said this is wrong, that is, if one of the 0.8 particles had eyes, he would see the other 0.8 particle going 0.9756.
okay I'm very curious as to how exactly this would occur.
so let's do some math.
let's say we have a room that's 8 light minutes long. both particles start in the middle.
for a particle traveling at 0.8*the speed of light, to go from the half way point to the edge of the room would take 5 minutes.
the time dilation equation is: sqrt(1-v^2/c^2).
the space dilation equation is sqrt(1-v^2/c^2).
so, from the perspective of the 0.8 particle, he would see himself approaching the wall at:
new time: 5/sqrt(1-0.8^2) = 8.33 minutes.
new distance: 4/sqrt(1-0.8^2) = 6.67 light minute length.
final velocity = 0.8*the speed of light.
so he doesn't see any change in his own velocity relative to the stationary observer; even factoring in space and time dilation. so, can you show me how he would see the particle moving away from him as anything other than 1.6*the speed of light?

 

[Nugatory]

so, from the perspective of the 0.8 particle, he would see himself approaching the wall at:
new time: 5/sqrt(1-0.8^2) = 8.33 minutes.
new distance: 4/sqrt(1-0.8^2) = 6.67 light minute length.
final velocity = 0.8*the speed of light.

No, he sees the wall approaching him at .8c while he's at rest.

The other particle also sees itself at rest, while the wall is moving away at .8c.

And the particles both see themselves at rest while the other particle is moving away at a speed that is not .8c+.8c, but rather is $\frac{(.8+.8)c}{1+(.8)^2}$
The left-moving particle sees the left-hand wall approaching at .8c, but the right-moving particle does not see the distance between the wall and the left-moving particle (both moving away from it) closing at that rate.

 

[ghwellsjr]

well my previous message was deleted, as was the response to it. this is why i didn't want to post a message because you are fairly strict about your rules and what constitutes fair behavior (which i admire by the way.)
okay i'll try again. let's take the example of two particles, both going 0.8*the speed of light form the position of an inertial observer.
if they are going in the same direction, what speed would one particle be going relative to the other?
again, i would say 0. if they are going opposite directions, they must be going 1.6 relative to one another.
now the previous poster has said this is wrong, that is, if one of the 0.8 particles had eyes, he would see the other 0.8 particle going 0.9756.
okay I'm very curious as to how exactly this would occur.
so let's do some math.
let's say we have a room that's 8 light minutes long. both particles start in the middle.
for a particle traveling at 0.8*the speed of light, to go from the half way point to the edge of the room would take 5 minutes.
the time dilation equation is: sqrt(1-v^2/c^2).
the space dilation equation is sqrt(1-v^2/c^2).
so, from the perspective of the 0.8 particle, he would see himself approaching the wall at:
new time: 5/sqrt(1-0.8^2) = 8.33 minutes.
new distance: 4/sqrt(1-0.8^2) = 6.67 light minute length.
final velocity = 0.8*the speed of light.
so he doesn't see any change in his own velocity relative to the stationary observer; even factoring in space and time dilation. so, can you show me how he would see the particle moving away from him as anything other than 1.6*the speed of light?

You got the right numbers for the time and distance the wall has traveled in the particle's rest frame but instead of calculating the factors in isolation and trying to interpret them just use the Lorentz Transformation on the coordinates of the rest frame of the walls and draw a diagram of the rest frame of one of the particles. Can you do that?

To make it a little easier for you, I'm providing a diagram for the rest frame of the two walls (shown in black). The two particles are shown in red and blue. The dots mark off one-minute increments of Proper Time for each object:

Now to use the Lorentz Transformation with a speed of 0.8c to get the coordinates for the rest frame of the blue particle. You only have to transform the coordinates of the five events at the extremities of each worldline. Then mark off the events on this template and draw in the worldlines (don't worrry about the dots):

Then I think all your questions will be answered.

 

[Fredrik]

How much math do you know, philip1882? Are you familiar with matrices and matrix multiplication for example?

I suggest that you start by familiarizing yourself with spacetime diagrams, as explained by Schutz here. Unfortunately not all the pages of the explanation are viewable online. But maybe you can get the book from a library or something.

 

[phillip1882]

http://postimg.org/image/d5ug05njf/
wouldn't this be it?
i guess i don't quite understand why the equation for adding velocity is what it is, given that velocity doesn't change relative to the stationary observer.

 

[Nugatory]

i guess i don't quite understand why the equation for adding velocity is what it is, given that velocity doesn't change relative to the stationary observer.

It follows directly from the Lorentz transforms, which in turn follow directly from Einstein's two postulates. You'll find derivations all over the internet, for example, here: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/veltran.html

 

[D H]

i guess i don't quite understand why the equation for adding velocity is what it is, given that velocity doesn't change relative to the stationary observer.

The simple answer is that it doesn't. Multiple experiments demonstrate that the simple vector addition of velocities just doesn't apply to objects moving at very high velocities.

One key thing that physicists have learned over the last 100+ years is that the nice and simple world of Newtonian mechanics does not extrapolate to the world of the very small, very large, or very fast. Extrapolation is always at best a dubious thing to do. It turns out to be a very bad thing to do in physics.

In our mundane, ordinary world, velocities apparently do add vectorially and passage of time apparently is the same to all observers. These apparently obvious facts are observably false in the world of the very, very fast. The failures of these cherished concept is itself a fact. Your choices are to

• Accept these new facts as facts and learn from them, in which case we'll be quite happy to try to help you along that road, or
• Reject these facts as apparent nonsense, in which case we'll be quite happy to tell you to peddle your nonsense elsewhere.

The choice is up to you.

 

[atyy]

http://postimg.org/image/d5ug05njf/
wouldn't this be it?
i guess i don't quite understand why the equation for adding velocity is what it is, given that velocity doesn't change relative to the stationary observer.

In Newtonian physics, we usually use inertial coordinates, which are the "common sense" coodinates with spatial axes at right angles and time such that Newton's second law holds. But although they may seem common sense, they are actually coordinates that respect an underlying symmetry of Newtonian physics.

In special relativity, the concept of inertial coordinates also exists, but the symmetry is slightly different. If the "stationary" observer sets up an inertial coordinate system, then the "moving" observer is also able to set up an inertial coordinate system - both of them use the same procedures to set them up. However, the relationship between the "stationary" and "moving" inertial coordinate systems is different from Newtonian physics, because of the different underlying symmetry of special relativity. This is why the velocity addition formula is different in special relativity.

In a sense, setting up inertial coordinates in special relativity defines the speed of light to be the same constant in all inertial frames. So it is a definition, but it ultimately makes things simpler because it respects the underlying symmetry of the theory - which also happens to be a much better description of the symmetry that is present in nature than the symmetry of Newtonian inertial coordinates.

 

[phyzguy]

Phillip1882,

Let's try and complete the exercise GHWellsJr started and maybe this will help you understand. In the frame where the two walls are at rest, and the two particles are moving left and right at 0.8c, the space-time diagram looks like the attached (Rest_Frame.png). The two walls are in black, and the two particles are in red and blue. The LH particle is watching the RH particle, and when the RH particle reaches the wall (at X = 4.0 lightminutes, T = 5.0 minutes in this frame), a light ray (green line) travels back to the LH particle, arriving at the LH particle at X = -36 light-minutes, T = 45 minutes.

Now let's look at it from the frame of the LH particle, which is shown in the attached LH_Particle_Frame.png. Because of Lorentz contraction and time dilation, the LH particle sees the two walls (in black) each 2.4 light-minutes from the origin (the rest-frame 4.0 light-minutes times the Lorentz contraction factor of 0.6), and each moving to the right at 0.8c. The light ray emitted from the RH particle when it reaches the RH wall arrives at the LH particle at T' = 27 minutes (this is T=45 minutes in the wall rest frame times the time dilation factor of 0.6). Extrapolating the light ray backwards along the green line, the LH particle concludes that the RH particle arrived at the wall at T' = 13.67 minutes, and that the RH particle traveled 13.33 light minutes in this time (because it had to travel the original 2.4 light-minutes + 13.67 minutes * 0.8c to reach the RH wall). So, he concludes that the RH particle traveled 13.33 lightminutes in 13.67 minutes, giving a velocity of 0.975c, which is just what is given by the velocity addition formula of (0.8+0.8)/(1+0.8*0.8) = 0.975.

Does this help?

 

[Naty1]

i guess i don't quite understand why the equation for adding velocity is what it is,

yeah, well join the club...

Before an 'Einstein', neither did the finest physicists in the world understand! "experts' in the early 1900's were trying to understand Maxwell's equations and were making some progress via Lorentz and Fitzgerald, but only Einstein could put the theory together.

atyy posts:

However, the relationship between the "stationary" and "moving" inertial coordinate systems is different from Newtonian physics, because of the different underlying symmetry of special relativity.

In retrospect, even Einstein did not fully understand that spacetime is a four dimensional continum...his college math professor [Herman Minkowski] had THAT insight...so Einstein put the algebra together for special relativity based in part on the work of Maxwell, Lorentz and Fitzgerald...but Minkowski provided the geometric view of a four dimensional spacetime...Einstein adopted that view and used it to develop general relativty...curved spacetime! Utterly extraordinary...

Minkowski's famous comment:

Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality."

In other words, with a fixed speed of light, not some ficticious 'ether' which was suspected in that era, space and time conspire jointly, changing in flat spacetime between the two so as to maintain the constant speed of light. When time dilates, space contracts...and 'c' stays constant.

Nobody quite understands THAT either...it's not like Einstein started from some fundamental first principles...he took available mathematics and experimental observations...and somehow found out what matched. All we know for sure is that, so far, such a relationship matches observations.

 

[phillip1882]

Now let's look at it from the frame of the LH particle, which is shown in the attached LH_Particle_Frame.png. Because of Lorentz contraction and time dilation, the LH particle sees the two walls (in black) each 2.4 light-minutes from the origin (the rest-frame 4.0 light-minutes times the Lorentz contraction factor of 0.6), and each moving to the right at 0.8c. The light ray emitted from the RH particle when it reaches the RH wall arrives at the LH particle at T' = 27 minutes (this is T=45 minutes in the wall rest frame times the time dilation factor of 0.6). Extrapolating the light ray backwards along the green line, the LH particle concludes that the RH particle arrived at the wall at T' = 13.67 minutes, and that the RH particle traveled 13.33 light minutes in this time (because it had to travel the original 2.4 light-minutes + 13.67 minutes * 0.8c to reach the RH wall). So, he concludes that the RH particle traveled 13.33 lightminutes in 13.67 minutes, giving a velocity of 0.975c, which is just what is given by the velocity addition formula of (0.8+0.8)/(1+0.8*0.8) = 0.975.

hmm. This does help me understand how you're deriving the equation, but i must confess i have a problem with this derivation. you're using light itself to determine the speed of two objects that could potentially be moving faster than the speed of light away from each other.
i mean, imagine i did the following. let's say i used two baseballs moving apart at the rate of 100 miles an hour (combined, so 50 miles apiece), and then i tried to measure the speed between them by using a 60 mile an hour ball. you understand i may get a vastly different velocity than what's actually occurring?
so, let me try asking the question i have again. if, after factoring in time and space dilation, the velocity of an object remains the same, how can the velocity of two objects not simply add and subtract?

 

[Nugatory]

so, let me try asking the question i have again. if, after factoring in time and space dilation, the velocity of an object remains the same, how can the velocity of two objects not simply add and subtract?

I'm not sure what exactly what you mean by "the velocity of an object remains the same" (remains the same as what?), but if you actually calculate the effects of time dilation and length contraction you get the relativistic formula. I pointed you to one such derivation in post #11, but could probably find another derivation out there that uses no calculus if you would prefer.

But sometimes... There really is no substitute for doing the math.

 

[jartsa]

okay i'll try again. let's take the example of two particles, both going 0.8*the speed of light form the position of an inertial observer.
if they are going in the same direction, what speed would one particle be going relative to the other?
again, i would say 0. if they are going opposite directions, they must be going 1.6 relative to one another.
now the previous poster has said this is wrong, that is, if one of the 0.8 particles had eyes, he would see the other 0.8 particle going 0.9756.
okay I'm very curious as to how exactly this would occur.
so let's do some math.
let's say we have a room that's 8 light minutes long. both particles start in the middle.
for a particle traveling at 0.8*the speed of light, to go from the half way point to the edge of the room would take 5 minutes.
the time dilation equation is: sqrt(1-v^2/c^2).
the space dilation equation is sqrt(1-v^2/c^2).
so, from the perspective of the 0.8 particle, he would see himself approaching the wall at:
new time: 5/sqrt(1-0.8^2) = 8.33 minutes.
new distance: 4/sqrt(1-0.8^2) = 6.67 light minute length.
final velocity = 0.8*the speed of light.
so he doesn't see any change in his own velocity relative to the stationary observer; even factoring in space and time dilation. so, can you show me how he would see the particle moving away from him as anything other than 1.6*the speed of light?

The correct calculation of final velocity:

new distance: 4*sqrt(1-0.8^2) = 2.4 light minute length.

new time: 5*sqrt(1-0.8^2) = 3.0 minutes.

final velocity = new distance / new time = 2.4 / 3.0 = 0.8 *speed of light

 

[Dale]

I mean, imagine i did the following. let's say i used two baseballs moving apart at the rate of 100 miles an hour (combined, so 50 miles apiece), and then i tried to measure the speed between them by using a 60 mile an hour ball. you understand i may get a vastly different velocity than what's actually occurring?

You should work out the math. You would get the right answer. In fact, you would still get the right answer if the measurement balls were going 10 mph.

The one difference is that the speed of baseballs is not frame invariant, so you could only do this procedure in the frame where your measurement balls speed was known.

 

[phyzguy]

hmm. This does help me understand how you're deriving the equation, but i must confess i have a problem with this derivation. you're using light itself to determine the speed of two objects that could potentially be moving faster than the speed of light away from each other.

It's not a derivation. It's a description of how the universe behaves, first worked out by Einstein, and verified by a huge body of experimental data until it is currently beyond doubt. The description assumes as one of the postulates that the speed of light is the same in any reference frame. This is why light plays such a central role.

i mean, imagine i did the following. let's say i used two baseballs moving apart at the rate of 100 miles an hour (combined, so 50 miles apiece), and then i tried to measure the speed between them by using a 60 mile an hour ball. you understand i may get a vastly different velocity than what's actually occurring?

For baseballs moving at 50 miles an hour, the difference from simple velocity addition is extraordinarily tiny. For two baseballs moving apart at 50 miles per hour, an observer sitting on one baseball would measure the other baseball as moving at 99.9999999999994 miles per hour, an immeasurably small difference from 100 miles per hour.

so, let me try asking the question i have again. if, after factoring in time and space dilation, the velocity of an object remains the same, how can the velocity of two objects not simply add and subtract?

It's certainly not the same. In the rest frame of the walls, the LH particle is moving at 0.8c and the walls are stationary. In the rest frame of the LH particle, the velocity of the LH particle is zero and the walls are moving at 0.8c. The number is the same, but the physical situation is very different.

 

[Naty1]

Quote by phillip1882
so, let me try asking the question i have again. if, after factoring in time and space dilation, the velocity of an object remains the same, how can the velocity of two objects not simply add and subtract?

You have it exactly backwards.

If space and time WERE fixed entities, like in Newtonian physics, velocities WOULD add and subtract as we all learned in grade school. They are not, so you can't. Newtonian physics is only a good low speed approximation.

 

[phyzguy]

hmm.
i mean, imagine i did the following. let's say i used two baseballs moving apart at the rate of 100 miles an hour (combined, so 50 miles apiece), and then i tried to measure the speed between them by using a 60 mile an hour ball. you understand i may get a vastly different velocity than what's actually occurring?

I think I see what you mean here. You're saying, "Suppose I used baseballs going at 60 mph in place of the light rays." If the universe were such a place that all 60 mph baseballs traveled at the same speed regardless of your state of motion, then you would in fact get the vastly different velocities you were referring to. But the universe doesn't work like that. If I go 60 miles per hour alongside a 60 mph baseball, I see the 60 mph baseball sitting at rest. But there is no frame of reference in which I see light rays at rest. They are always moving at the speed of light, regardless of my state of motion. This is a counter-intuitive fact which you just need to accept.

 

[VantagePoint72]

I think I see what you mean here. You're saying, "Suppose I used baseballs going at 60 mph in place of the light rays." If the universe were such a place that all 60 mph baseballs traveled at the same speed regardless of your state of motion, then you would in fact get the vastly different velocities you were referring to. But the universe doesn't work like that. If I go 60 miles per hour alongside a 60 mph baseball, I see the 60 mph baseball sitting at rest. But there is no frame of reference in which I see light rays at rest. They are always moving at the speed of light, regardless of my state of motion. This is a counter-intuitive fact which you just need to accept.

That's a bingo! Fortunately, you don't have to (and shouldn't) just accept this counter-intuitive fact because we say so. You can accept it because it's been repeatedly and to great precision confirmed by over a century of experiments.

 

[Fredrik]

Consider a train going at a velocity of 40 m/s relative to the ground, and a person on the train walking towards the front of the train at a velocity of 1 m/s.

You have correctly argued that SR says that the velocity of the ground relative to the train will be -40 m/s. (Edit: I see now that your argument in post #6 is not correct. You divided by that square root when you should have multiplied by it. But you made that mistake twice, and the two mistakes cancel each other in the calculation of the final result.) But then you're saying (the equivalent of) "since the ground has velocity -40 m/s relative to the train, and the person has velocity +1 m/s relative to the train, the person's velocity relative to the ground should be exactly 41 m/s".

This conclusion is false. Unfortunately the only thing that's obvious here is that your conclusion is not an immediate consequence of the facts that you used to come to that conclusion. It takes some work to find the actual velocity.

A straightforward way to do it is to choose three inertial coordinate systems, comoving with the ground, the train, and the person, that all assign coordinates (0,0) to a point on the person's world line. Then we can use a Lorentz transformation to calculate the coordinates of another point on the person's world line in the inertial coordinate system that's comoving with the ground, and finally calculate the person's velocity relative to the ground as "position coordinate"/"time coordinate".

Let u be the velocity of the train relative to the ground. Let v be the velocity of the person relative to the train. We know that the inertial coordinate system comoving with the train assigns coordinates (1,v) to some point on the person's world line. (I'm writing the time coordinate first, so this is (t,x) with x=vt and t=1. I'm setting t=1, because the t cancels out from the calculation anyway, so we might as well use the value that makes the calculation as simple as possible). Since the velocity of the ground relative to the train is -u (note that I'm using the result that you thought seemed to imply that the end result is going to be 41 m/s), all we have to do is to apply a Lorentz transformation with velocity -u to the coordinate pair (1,v). This is how I do it, in units such that c=1:
\begin{align}\Lambda(-u)\begin{pmatrix}1\\ v\end{pmatrix} &=\frac{1}{\sqrt{1-(-u)^2}}\begin{pmatrix}1 & -(-u)\\ -(-u) & 1\end{pmatrix}\begin{pmatrix}1\\ v\end{pmatrix} =\frac{1}{\sqrt{1-u^2}}\begin{pmatrix}1 & u\\ u & 1\end{pmatrix}\begin{pmatrix}1\\ v\end{pmatrix}\\
&=\frac{1}{\sqrt{1-u^2}}\begin{pmatrix}1+uv\\ u+v\end{pmatrix}.\end{align} That square root cancels out when we compute the "position coordinate"/"time coordinate", so all we get is
$$\frac{u+v}{1+uv}.$$ If we want to restore factors of c, i.e. if we want to use something like SI units instead of units such that c=1, it's easy to see where to put the c's. The result of the division is supposed to be a velocity, and the numerator already has dimensions of velocity, so we need the denominator to be dimensionless (i.e. "unitless"). So the velocity we seek is
$$\frac{u+v}{1+\frac{uv}{c^2}}.$$ If we type (40+1)/(1+(40*1)/299792458^2) into Wolfram Alpha, we get approximately 40.99999999999998175.

 

[phillip1882]

edit: apologies, didn't see the remaining posts.

i accept that no matter how much electrical energy you add a light beam, it won't travel any faster. i accept that if you try to push a light beam in the same direction it's traveling, it also won't go any faster. I would like to point out that sound has similar properties.

i accept the possibility that time and space dilation might be occurring. i find it hard to believe that time is infinitely dilated as you approach the speed of light, because light doesn't travel instantaneously. and what i see is that, factoring in time and space dilation, the velocity of an object relative to a "stationary" observer, remains the same. if velocity is unchanging, even factoring in space and time dilation, then velocity should be relative to your motion.
you guys seem to think I'm wrong on this point, which; okay you're free to think so, i just personally don't see how it logically adds up.

 

[phyzguy]

remains the same relative to a stationary observer.
i have a calculation of space and time dilation in an earlier post, and i got the same final velocity.
are my calculations wrong?

Now I'm really confused what you mean. A stationary observer experiences no length contraction or time dilation. It seems like you are saying that one stationary observer measures the same velocity as another stationary observer - of course this is true.

By the way, your calculations were wrong, in that you got the effects backwards. jartsa corrected this in an earlier post.

 

[epovo]

"velocity remains the same". What I think you mean by "the velocity remains the same" is this: if I measure the velocity of an object as v (with respect to me) and that object measures MY velocity with respect to it, the result will be -v. Is that what you mean?

 

[ghwellsjr]

If you would perform the exercise that I presented to you in post #8, you will be on your way to understanding why the relative speed between two objects never reaches c. A lot of people are spending a lot of time and doing a lot of work to help you. Can't you do a little bit of work yourself and do what I ask?

 

[Fredrik]

remains the same relative to a stationary observer.
i have a calculation of space and time dilation in an earlier post, and i got the same final velocity.
are my calculations wrong?

The calculation (in post #6) is wrong, but the result is correct, because you made the same mistake twice and they cancel each other out.

However, all you're doing is to argue that if the velocity of B in A's coordinates is v, then the velocity of A in B's coordinates is -v. This doesn't imply anything about addition of velocities.

 

[phillip1882]

However, all you're doing is to argue that if the velocity of B in A's coordinates is v, then the velocity of A in B's coordinates is -v. This doesn't imply anything about addition of velocities.

well sure it does.
if A is traveling at 0.8 relative to B
and -A is traveling at -0.8 relative to B,
then A relative to -A should be 1.6.

i mean; if you agree B would see the two A's moving apart at 1.6, because each individual one is traveling below the speed of light, then i fail to see why A cannot see -A as 1.6.
and again, as far as i can tell, you cannot use space and time dilation to get a different velocity.
so, what mechanism is causing the different velocity?