# Help to clarify special relativity confusion

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I guess these questions will seem rather trivial to experts in relativity. However i think the answers will be informative to those new to the subject.

'a' is an observer at a fixed point in a frame of reference 'A ref'.

'b' is an observer at a fixed point in a frame of reference 'B ref'.

'A ref' is deemed to be the stationary reference frame and 'B ref' the moving frame.

The velocity of reference frame 'B ref' relative to reference frame 'A ref' is 0.8c.

'a' see's 'b' moving away from 'a'.

At what velocity will 'a' determine 'b' to be travelling at by use of:-

1. triangulation.

2. red/blue shift

3. please state use of other.

Those mega-minds who never tire, may care to consider the same scenario with the following difference.... 'a' see's 'b' moving towards 'a'.
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Well.... different methods for determining velocity should not disagree. If b is moving away from a at 0.8c the velocity would be measured as being 0.8c. As far as I know anyway, we'll see once someone more educated comes in.

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You can't use triangulation here can you? Because you only have two frames of reference and one of these is the moving object. I think you would deal with this problem using a linear method.

However if 'b' is moving relative to 'a' at 0.8c then that is how observer 'a' will see it. If observer 'b' looks at the system he will see 'a' moving away at 0.8c.

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You can't use triangulation here can you? Because you only have two frames of reference and one of these is the moving object. I think you would deal with this problem using a linear method.

Correct. You could use a measuring stick/ruler in optional question..

3) please state use of other.

However if 'b' is moving relative to 'a' at 0.8c then that is how observer 'a' will see it. If observer 'b' looks at the system he will see 'a' moving away at 0.8c.

The velocity of reference frame 'B ref' relative to reference frame 'A ref' is 0.8c.

The question is' date='... at what velocity will 'a' determine 'b' to be travelling at?

I put it to you that if an object was traveling at 0.8c away from you. The velocity it would be seen to be traveling at would be somewhat slower. I base this proposition on the fact that the light image of the object takes an increasing longer time to reach you.[/indent']
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I take it b is moving relative to the frame B.

*cheers for lorentz transformations*

I'm quite tired so appologies for just posting a link, but here you go

http://hyperphysics.phy-astr.gsu.edu/Hbase/relativ/veltran.html

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to apeofam's remark

yes but the rate of change in that image stays the same.

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I take it b is moving relative to the frame B.

*cheers for lorentz transformations*

I'm quite tired so appologies for just posting a link' date=' but here you go

I'm not sure why?

The question states... 'b' is an observer at a fixed point in a frame of reference 'B ref'. As 'b' moves in the direction x so does frame B.

The question is,... at what velocity will 'a' determine 'b' to be travelling at?

Inertial frame 'B ref' is parallel and adjacent to the stationary reference frame 'A ref', whereas 'a' is at an ever increasing distance from 'b'.

Had a look at ?your? link:-) , you might like this one http://magen.co.uk/calculator.html The lorentz transformations describe one of Hesheit's magic tricks. Real magic.

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Ah, I interperated the question wrong, my appologies

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Imagine that I am 'a' or in reference frame a... I will see 'b' (and it's reference frame) moving away from me at 0.8c.

Relative to me (or a) he is moving at 0.8c so that is what I will see.

Now imagine that I am 'b'. I will look at 'a' and will see it moving away from me at 0.8c.

Both of them think that they are at rest and the other observer/frame is moving away at 0.8c. Unless one of them undergoes acceleration, at which point they can feel a force acting on them and can so tell they are moving, it is impossible for the two observers to know which of them is actually moving. Both will think that they are at rest, both will think the other is moving at 0.8c and so on for other SR effects, for example both will see the other ones clock's running slow.

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to apeofam's remark yes

Thank you Luke. Nice to know someone else agrees with that point. An example may be helpful to others. I'll be posting one when i've finished checking it out for flaws that i might find.

but the rate of change in that image stays the same.

Do you care to elucidate?

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Ah, I interperated the question wrong, my appologies

Thank you for letting me know. I thought maybe i had gotten something wrong already. The effects of "Special Relativity" are hard enough to comprehend. In the example 'simple questions', SR effects are inter-twined with non-relativistic effects. It is an area which begs confusion. Hopefully this thread may clarify things a bit. No doubt i will say something completely stupid before that goal is achieved. So what.

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Imagine that I am 'a' or in reference frame a... I will see 'b' (and it's reference frame) moving away from me at 0.8c.

Relative to me (or a) he is moving at 0.8c so that is what I will see.

Now imagine that I am 'b'. I will look at 'a' and will see it moving away from me at 0.8c.

Both of them think that they are at rest and the other observer/frame is moving away at 0.8c. Unless one of them undergoes acceleration' date=' at which point they can feel a force acting on them and can so tell they are moving, it is impossible for the two observers to know which of them is actually moving. Both will think that they are at rest, both will think the other is moving at 0.8c and so on for other SR effects, for example both will see the other ones clock's running slow.[/quote']

Your explanation is fine as far as it goes, but does not address the question fully. If you are saying that the answer to my question...
at what velocity will 'a' determine 'b' to be travelling at?
is
0.8c, then i will need know how the time it takes for light to travel has been removed to obtain such a sight.

It is my belief that in SR, the relative velocity between frames, is measured by observers spaced out along the axis of travel ( i.e. the x axis of the lorenz transform ) using synchronised clocks. Under such circumstance there is almost no delay between observation and event. The circumstance of observers 'a' and 'b' is different, it takes time for the light to travel between them. How can 'a' see 'b' move away a distance of 0.8 light-second in one second, when it will take 0.8 second for the image of 'b' being there to come back to where 'a' is at, to be observed.

I find relativity confusing on a good day.
If there is something wrong with my reasoning let me know.
Same goes if you agree with the points made;)

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there are no other observers spaced out, the only observation of any variable occurs when "a" sees "b" move away

you can derive the lorentz transformations in this example by saying that observer "a" uses some sort of radar gun to determine the position of b.

anyway, as I was saying before when b is say 1 lightyear away from "a" when he starts moving, it will take 1 year before "a" knows that "b" is moving however "a" can tell by observing the light that "b" emitted what velocity he was traveling at, there is no transformation in this case.

however if you add a third observer "C" who sits at rest at the origin, and he observes "B" moving at .5c in the positive x direction, and he observes "A" moving with .8c in the negative x direction, then the question of what velocity does "A" have relative to "B" is far more complicated, and you must use something called relativistic addition of velocities, as "a" can't move away from "B" at a velocity greater than c

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Thank you for the reply Luke.

there are no other observers spaced out' date=' the only observation of any variable occurs when "a" sees "b" move away

you can derive the lorentz transformations in this example by saying that observer "a" uses some sort of radar gun to determine the position of b.[/quote']

I agree.
The values will be the same as those obtained by measurements made along the x axis in the stationary frame. Effectively the light is the ruler and the common clock of the radar gun provides the synchronous clock time measurements.

Given that the velocity of light is a constant c, the distance that a light/electromagnetic-wave has travelled can be derived from it's transit/journey time by multiplying the velocity of light c, by the elapsed time of the journey.. i.e. d=ct. In the case of a radar gun the wave is emitted, and travels away from the gun until it is reflected off the moving object, when it catches up to the object's current position. The distance between the radar gun and the object at the time of reflection is half the round trip..i.e. d/2.

By the way. I don't consider this method as being synonymous with 'a' seeing the velocity of 'b',.. the context of my original question. However i did think it worthy of further comment, as it demonstrates that different ways of determining velocity can yield different results. That difference's are found, does not mean that the velocities found by one method is right and another method wrong. They are as valid as one another.

anyway, as I was saying before when b is say 1 lightyear away from "a" when he starts moving, it will take 1 year before "a" knows that "b" is moving however "a" can tell by observing the light that "b" emitted what velocity he was traveling at, there is no transformation in this case.

I agree with you.

however if you add a third observer "C" who sits at rest at the origin, and he observes "B" moving at .5c in the positive x direction, and he observes "A" moving with .8c in the negative x direction, then the question of what velocity does "A" have relative to "B" is far more complicated, and you must use something called relativistic addition of velocities,

I agree so far.
However you say..

as "a" can't move away from "B" at a velocity greater than c

observe/measure
"A" to be moving away from "B" at a velocity greater than the velocity of light c.

Otherwise yes. In fact i posted a link in this thread to a calculator that uses geometry (my avatar is the result of one such calculation) to solve the addition of relativistic velocities problem and others. The results are in accord with using the Lorentz transforms.

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shouldn't that read,.. "B" can't observe/measure[/u'] "A" to be moving away from "B" at a velocity greater than the velocity of light c.
Yeah, that's what he means.

Observer "C" sees "A" moving at 0.5c, lets say to the left and sees "B" moving at 0.8c to the right. What he was saying is that the relative velocity between "A" and "B", that is, when one measures the other's speed it cannot simply be 0.5c + 0.8c so you have to do some maths.

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There is another side to special relativity. It has less to do with relative reference as it has to do with velocity differences due to energy input. For example, we can not make even a small mass reach C because the mass would become infinite. It would take infinite energy input to make even a small amount of mass reach C. From a stationary reference, watching this impossible feat, even though we see ourselves as the moving reference, no energy is being expended within our reference. Our reference mass is still where it was when the accelerated mass began it journey even though it is beginning to gain relativistic mass.

Why do we only see time and distance special relativity from our relative reference but not relativistic mass increase also?

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Yeah' date=' that's what he means.

Observer "C" sees "A" moving at 0.5c, lets say to the left and sees "B" moving at 0.8c to the right. What he was saying is that the relative velocity between "A" and "B", that is, when one measures the other's speed it cannot simply be 0.5c + 0.8c so you have to do some maths.[/quote']

In that case I agree with that too.
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Why do we only see time and distance special relativity from our relative reference but not relativistic mass increase also?
We do. We also observe effects such as Doppler Shift, aberration, Terrell rotation, angular compression as well as the more well known length contraction and time dilation.

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