The twins struggle with Minkowski

[uncool]

I understand the animation.

 

Look.

Someone said, if muons go at 0,98c in Earth's FOR, then the remaining c-0,98c=0,02c in Earth's FOR, because it is all measured in the same FOR.

 

If a child is coming to you and ask you the result of

 

0,98c+0,02c

 

What is your answer?

If both are measured in Earth's FOR, c. If the 0.02c is measured in the frame moving at 0.98c (as is assumed in the relativistic velocity addition argument), then slightly less than c.

=Uncool-

[swansont]

I understand the animation.

 

Look.

Someone said, if muons go at 0,98c in Earth's FOR, then the remaining c-0,98c=0,02c in Earth's FOR, because it is all measured in the same FOR.

 

If a child is coming to you and ask you the result of

 

0,98c+0,02c

 

What is your answer?

 

"What is the result of .98c+.02c" is an ill-formed question. What do you mean by it?

 

In our context, it means that if a source one light-second away emitted a simultaneous burst of both photons and muons, the photons would arrive after a time of one second (d/c), while the muons would arrive at time of 1.02 seconds (d/.98c). i.e. the muons would "chase" the light, and the speed difference is 0.02c

 

(ignoring the decay of the muons)

[michel123456]

0,98c+0,02c

 

0,5c+0,6c

 

0,45c+0,67c

 

0,95c+0,95c

 

are all answerable without any problem, without "if" this or "if" that. These are not ill-formed questions.

 

Relativity states very clearly that simple summation of velocities is wrong. In any case.

[Iggy]

Relativity states very clearly that simple summation of velocities is wrong. In any case.

 

No, relativity insists on the following:

 

This is due to a misunderstanding of the question which can easily be confused with the following one: Suppose the object B above is an experimenter who has set up a reference frame consisting of a marked ruler with clocks positioned at measured intervals along it. He has synchronised the clocks carefully by sending light signals along the line taking into account the time taken for the signals to travel the measured distances. He now observes the objects A and C which he sees coming towards him from opposite directions. By watching the times they pass the clocks at measured distances he can calculate the speeds they are moving towards him. Sure enough he finds that A is moving at a speed v and C is moving at speed u. What will B observe as the speed at which the two objects are coming together? It is not difficult to see that the answer must be u+v whether or not the problem is treated relativistically. In this sense velocities add according to ordinary vector addition.

 

http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/velocity.html

 

If u, v, and w are all measured in the same frame then w=u+v is the correct relationship. Rather than just repeating that everyone else is wrong and you are right, maybe you could try to explain why you think you are right or give some text that you think agrees with you.

[swansont]

Relativity states very clearly that simple summation of velocities is wrong. In any case.

 

No, it doesn't. Summation of velocities is wrong in the case where you doing a coordinate transformation, because the transformation is Lorentzian and not Galilean. You are not doing a coordinate transformation when you are looking at relative velocities of moving objects in your rest frame. There is no change of reference frame.

[michel123456]

If u, v, and w are all measured in the same frame then w=u+v is the correct relationship. Rather than just repeating that everyone else is wrong and you are right, maybe you could try to explain why you think you are right or give some text that you think agrees with you.

 

Maths agree with me.

 

I am at point B

An object A approaches me from the left at 0,9c

Another object C approaches from the right at 0,9c

 

Then the 2 objects approaches at ...1,8c?

 

But:

Me, at point B don't observe anything moving at speed 1,8c

Object A don't observe anything moving at speed 1,8c

Object C don't observe anything moving at speed 1,8c

 

No object, no wave, nothing propagates at 1,8c. Nothing from nowhere moves at 1,8c. Nothing is observed at 1,8c.

The result 1,8c is a physical nonsense.

The result 1,8c is wrong.

 

If you want to add 2 velocities, you must use the relativistic formula. Only then the result will have a physical meaning.

 

That's it.

I quit.

[Sisyphus]

Maths agree with me.

 

I am at point B

An object A approaches me from the left at 0,9c

Another object C approaches from the right at 0,9c

 

Then the 2 objects approaches at ...1,8c?

 

True. In your rest frame, the relative velocity between them is 1.8C. In their rest frames, it is less than C.

 

But:

Me, at point B don't observe anything moving at speed 1,8c

Object A don't observe anything moving at speed 1,8c

Object C don't observe anything moving at speed 1,8c

 

True, true, and true.

 

No object, no wave, nothing propagates at 1,8c. Nothing from nowhere moves at 1,8c. Nothing is observed at 1,8c.

 

True, true, and true.

 

The result 1,8c is a physical nonsense.

The result 1,8c is wrong.

 

False. It would be nonsense to have anything moving at 1.8C, but nothing is.

 

If you want to add 2 velocities, you must use the relativistic formula. Only then the result will have a physical meaning.

 

False.

[Iggy]

Rather than just repeating that everyone else is wrong and you are right, maybe you could try to explain why you think you are right or give some text that you think agrees with you.

...The result 1,8c is a physical nonsense ...If you want to add 2 velocities, you must use the relativistic formula. Only then the result will have a physical meaning.

 

The physical meaning of velocity is change in distance divided by change in time. If both ships are approaching you at .9c from opposite directions then the distance between the ships in your frame changes at a rate of 1.8c. Do a thought experiment to prove this to yourself.

 

The ships are .9 light-minutes away from you in each direction. The distance between the ships in your frame is 1.8 light-minutes. One minute later the distance between the ships is zero. Change in distance is 1.8 light-minutes and change in time is 1 minute. V=1.8c.

 

The result has the physical meaning it should.

 

Is there any other reason you think this is wrong?

[Janus]

Maths agree with me.

 

I am at point B

An object A approaches me from the left at 0,9c

Another object C approaches from the right at 0,9c

 

Then the 2 objects approaches at ...1,8c?

 

But:

Me, at point B don't observe anything moving at speed 1,8c

Object A don't observe anything moving at speed 1,8c

Object C don't observe anything moving at speed 1,8c

 

No object, no wave, nothing propagates at 1,8c. Nothing from nowhere moves at 1,8c. Nothing is observed at 1,8c.

The result 1,8c is a physical nonsense.

The result 1,8c is wrong.

 

If you want to add 2 velocities, you must use the relativistic formula. Only then the result will have a physical meaning.

 

That's it.

I quit.

 

1.8c is the "closing speed" between A and C in the rest FoR of B. It is the result of taking the starting distance between A and C according to someone at rest with respect to B and dividing it by the time it takes for A and C to meet according to the same FoR. There is nothing nonsensical about it. It is just as real physically as the fact that A measures its velocity with respect to C as being 0.994475c and measures the respective speed between B and C as 0.094475c ( or conversely that C measures the respective speed between A and B as being 0.094475c.) These are all perfectly valid and physically real measurements in each FoR.

Edited March 8, 2011 by Janus

[swansont]

No object, no wave, nothing propagates at 1,8c. Nothing from nowhere moves at 1,8c. Nothing is observed at 1,8c.

 

You're right. Nothing is observed moving at 1.8c. What one thing do you think is claimed to be moving at 1.8c?

[michel123456]

I quit.

 

I changed my mind.

 

It seems you give an example where relativistic summation does not apply, Iggy. At first glance I wonder with 3 or 4 objects, what the summation of distances between them would physically mean, but with 2 objects you made a point. I have to think about it.

 

 

from mathpages.com

 

2.4 Doppler Shift for Sound and Light

For historical reasons, some older text books present two different versions of the Doppler shift equations, one for acoustic phenomena based on traditional Newtonian kinematics, and another for optical and electromagnetic phenomena based on relativistic kinematics. This sometimes gives the impression that relativity requires us to apply a different set of kinematical rules to the propagation of sound than to the propagation of light, but of course that is not the case. The kinematics of relativity apply uniformly to the propagation of all kinds of signals, provided we give the exact formulae. The traditional acoustic formulas are inexact, tacitly based on Newtonian approximations, but when they are expressed exactly we find that they are perfectly consistent with the relativistic formulas.

 

And the whole chapter after.

[Spyman]

I have to think about it.

Here is a thread where LightHeavyW8 tries to argue Superluminality due to closing speed: If Two Spaceships Close at 1.4c...?

 

Maybe you will find it interesting while thinking...

(Nice animation by Janus in post #34.)

[michel123456]

Well that was not very clever from LightHeavyW8 to think that an object traveling at 0,7c will go faster than c. I hope I am not that bad... I am not arguing in favor of superluminal velocity, quite the contrary.

Nice work from Janus, indeed.*

 

Still thinking, that Iggy showed the speed of the reduction of a distance which was part of the definition of the speed.

That is the rate at which a distance between 2 different objects diminishes as observed from a third. The fact that the 3 objects are aligned is just a special configuration. If there were 60 objects placed radialy around the observer, all rushing to him, what is the physical meaning of the sum of 60 distances and the rate they diminish (that can be many times the speed of light, if the 'objects' are light rays)?

 

I feel misengaged.

 

What i wanted to say previously, but Iggy's post cut my enthusiasm, is that the formula of relativity for compositions of velocities can always replace the Newtonian (Galilean) one. Relativity is more precise, it is never wrong.

 

Now I have to think about it again. Maybe Newton wins...

 

*but the animation shows only what happens following the FOR of the point where the light source originated. IMHO. Because a light ray emitted one second after departure, from the left object (his own image for example) travel much faster, as seen by the object. That is another animation.

 

editing: oops, Janus explained everything in post #42 of the same thread.

Edited March 8, 2011 by michel123456

[Sisyphus]

Well that was not very clever from LightHeavyW8 to think that an object traveling at 0,7c will go faster than c. I hope I am not that bad... I am not arguing in favor of superluminal velocity, quite the contrary.

Nice work from Janus, indeed.*

 

What you're arguing is almost identical, just inverted. He was arguing that because the closing speed was 1.4C (true), that the speed limit isn't C (false). You are arguing that since the speed limit is C (true), the closing speed isn't 1.4C (false)! You're both operating on the same flawed premise that the two things are mutually exclusive. You just picked a different statement as "true."

 

Still thinking, that Iggy showed the speed of the reduction of a distance which was part of the definition of the speed.

 

Surely that's the entire definition.

 

If there were 60 objects placed radialy around the observer, all rushing to him, what is the physical meaning of the sum of 60 distances and the rate they diminish (that can be many times the speed of light, if the 'objects' are light rays)?

 

You wouldn't sum 60 distances. You would sum at most 2, depending on what you're looking for. And since no velocity can be greater than C, the sum is at most 2C.

 

What i wanted to say previously, but Iggy's post cut my enthusiasm, is that the formula of relativity for compositions of velocities can always replace the Newtonian (Galilean) one. Relativity is more precise, it is never wrong.

 

Now I have to think about it again. Maybe Newton wins...

 

It's not a matter of winning, because there is no conflict. The relativistic velocity additions are for translating between reference frames.

[michel123456]

You are arguing that since the speed limit is C (true), the closing speed isn't 1.4C (false)!

 

Did I say that?

 

You're both operating on the same flawed premise that the two things are mutually exclusive. You just picked a different statement as "true."

 

I suppose.

 

I still cannot swallow:

 

_that c-0,98c=0,02c is correct, even in a single FOR.

_that 0,98c + 0,02c= c is correct, even in a single FOR (that must be my error)

_that the result 0,02c has a physical meaning.

_that the meaning is that the muon is "chasing its own image", as observed from our FOR.

 

Since in its own frame, the muon is not "chasing its own image"

 

_that it is necessary to make different diagrams (animations) in order to explain one single phenomenon, each one from a specific FOR

_that those diagrams do not present the same physical result, as I understand.

 

What I could swallow is that all observers are not measuring the same "thing".

Edited March 9, 2011 by michel123456

[swansont]

I still cannot swallow:

 

_that it is necessary to make different diagrams (animations) in order to explain one single phenomenon, each one from a specific FOR

 

 

That's the idea behind relativity. The measurements from each FOR are different.

[swansont]

Perhaps this will help.

 

The velocity addition formula takes into account the differences in length contraction and time dilation when you switch frames. These are incorporated into the velocity addition formula.

 

Closing speed does not involve a frame switch. You are always in the rest frame of the observer, so there is no time dilation or length contraction for which you need to adjust. All of the measurements take place in the frame of the observer.

[michel123456]

I have wood in my head.

Staying in my own FOR, without frame switch, observing an object that moves at close to c speed, don't I observe length contraction & time dilation of this object?

[swansont]

I have wood in my head.

Staying in my own FOR, without frame switch, observing an object that moves at close to c speed, don't I observe length contraction & time dilation of this object?

 

Of the object itself, but not of the path it travels, i.e. a meter stick that you use to measure distance is unchanged. And that's what you used to find speed. Your meter stick and your clock. Both are unchanged.