# (v) and (-v)

#### Quantum Quack

##### Life's a tease...
Valued Senior Member
Hi Guys,

The issue I want to address is :

When looking at two objects in relativity we take the postion of either one of the two objects.

The scenario that I woudl like to maintain as common through out this question:

Object A and Object B are separating at a "speed" of 0.6c. There is no other references available.

From what I have gathered when looking at these two objects relativity demands that we take one object as a frame of reference and deem that objeect as at rest thus giving us the following

A = v= 0
B = v = 0.6c

But of course we have no way of knowing whether A or B are at rest so to use this perspective we take a relativistic assumption and place A at rest and B as moving.

Now please correct me if I am wrong.

We know intuitiveley that the objects are separating equally to each other.

We also know that we can use B as our rest frame just as easilly as we can choose to use A.

If I use B as the rest frame is A's velocity 0.6c or is it considered as -0.6c.

Again intuitively A and B would both have positive velocities regardless of frame use to calculate from. Am I correct in saying that this is invalid in that relativity requires one frame to record a negative velocity?

The reference frame you choose does not matter at all - it's just a frame for calculations.

Within your calculations, you must be consistent in your reference frame - if you need to switch to another frame, you must do so carefully otherwise the results of your calculations will not correspond to reality.

It is a common mistake to use values from two different reference frames and use them together - this leads to erroneous results.

The positive-negative velocity issues is related to converting between reference frames. The standard lorentz transform requires the positive velocity direction to stay the same in the two frames.

This doesn't mean you can't change the positive velocity direction, it just means that if you do, you can't simply use the standard transform. You have to know what you're doing, otherwise the results of your calculations will be meaningless - they'll just be numbers with no correspondence to reality.

Note also that the frames in which one object is at rest are not the only frames you could choose. You could choose to work in any frame at all - as long as you were consistent.

NOw Pete you are tlking from a perspective from with in relativity, were as I am talking from a perspecting from outside of relativity.

How do we decide which object has positive velocity and which has negative velocity?

Hi QQ,
I'm talking from an analytical perspective which includes the relativity model as the most useful tool of analysis.

How do we decide which object has positive velocity and which has negative velocity?
How do you want to decide?

I think the choice is arbitrary, and it's not an either-or choice - you can happily have both objects with negative velocity, or both with positive if you prefer.

Pete can I throw another example at you just for curiocity....

we have three objects

A, B and C

C is central to A and B

A and B have velocitied directly opposite to each other with a stationary C in between them, [stationary when looking at all three frames]

A =v= 0.06c
B = v= 0.06c
using C as the rest frame.

How fast are the objects A and B separating according to relativity?
[ I can imagine this would be a text book scenario used some where]

[stationary when looking at all three frames]
That doesn't make sense - if it's stationary in more than one frame, then those frames are the same frame.
I think you mean that in C's frame, the mean velocity of A and B is zero.

A =v= 0.06c
B = v= 0.06c
using C as the rest frame.

How fast are the objects A and B separating according to relativity?

They're not - they have the same velocity.

But since you said before that they are moving in opposite directions, one of them must have a negative velocity in C's frame.

If in C's frame:
velocity of A = 0.6c, and velocity of B = -0.6c, then:

In C's frame, the separation between A and B is changing at 1.2c.

In A's frame and B's frame, the rate of separation will be different.

That doesn't make sense - if it's stationary in more than one frame, then those frames are the same frame.
I think you mean that in C's frame, the mean velocity of A and B is zero.
sorry if I have confused the issue.....

as an excersise in relative frames velocity i have placed C as central and stationary relative to both A and B.....in other words C is at rest from both A and B's perspective.

I know this is contra to relativity...ok...A ( being at rest) would consider both C and B as at velocity and so forth......

In A's frame and B's frame, the rate of separation will be different.

and this is the question I guess....why would A and B NOT agree on the rate of separation?
I undersatand that velocity is a separate issue to separation rate or speed.

What can A and B agree upon about themselves and object C?

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Everything that isn't frame-dependent. Things too numerous to mention!

I have the feeling you're leading up to something. Why not just say it?

Quantum Quack said:
and this is the question I guess....why would A and B NOT agree on the rate of separation?
I undersatand that velocity is a separate issue to separation rate or speed.
A and B agree with each other on the rate of separation (symmetrical), but they disagree with C.

This is because space and time aren't absolute, they are frame-dependant.

A & B can always agree on the Interval between events. If Event one happened at (X<sub>1</sub>, Y<sub>1</sub>, Z<sub>1</sub>, T<sub>1</sub>) and event two happened at (X<sub>2</sub>, Y<sub>2</sub>, Z<sub>2</sub>, T<sub>2</sub>) then
Interval<sup>2</sup> = DeltaX<sup>2</sup> + DeltaY<sup>2</sup> + DeltaZ<sup>2</sup> - DeltaT<sup>2</sup>

DeltaX = X<sub>2</sub> - X<sub>1</sub>
DeltaY = Y<sub>2</sub> - Y<sub>1</sub>
DeltaZ = Z<sub>2</sub> - Z<sub>1</sub>
DeltaT = T<sub>2</sub> - T<sub>1</sub>

All observers agree on the Interval between two events.

DeltaX<sup>2</sup> + DeltaY<sup>2</sup> + DeltaZ<sup>2</sup> is the square of the distance between the events and DeltaT<sup>2</sup> is the square of the time differenece between the events.

Observers usually cannot (never can?) agree on the time and distance differences.

well......hmmmmm...if you are a amatuer scientist that mixes intuition with science Like I am it is really easy to see how a liitle math skill can end up so damn confusing.

Fortunately for me I have very little math skill so I don't get into to much deep water....I guess what I am saying is that because reativity is a mathematical model that has certain rules and conditions of use and can seem to be counter intuitive it can lead an intuitivley orientated mathematician in to a dreadful mess......

IN another thread I talked of the 'g' perspective and this is a frame that can't be used in relativity.....I suppose.

If you were on object C and observing A and B together as a a group and not as separate entities, and to do so impies a constant frame of reference that being the abstract observer. [I have dubbed this the 'g' perspective]

A and B agree with each other on the rate of separation (symmetrical), but they disagree with C.

This is because space and time aren't absolute, they are frame-dependant.

according to the model of relativity........ from a abstract 'g' perspective space and time could be seen to be absolute.

IN another thread I mentioned universal mean time.....a new set of words for an old concept....

eg. if we take all dilations in the universe and average them out we have a universal mean time. [just a new idea that has sprung up in my thinking]