Relativity?

[swansont]

It's a definition.

E.g. a substitution.

Why then wrong?

If it's wrong can you equate my equation

to show the error

so I can see?

!

Moderator Note

We will not be playing that game. The burden of proof is yours. It's incumbent on you to show that it's valid.

 

If you aren't prepared to do this then we're done here. It's up to you.

[Capiert]

!

Moderator Note

We will not be playing that game. The burden of proof is yours. It's incumbent on you to show that it's valid.

If you aren't prepared to do this then we're done here. It's up to you.

You've said it's wrong

but don't say where;

I've defined it as a substitution

& given the math (algebra);

& then Strange "guesses" it's wrong.

What else do you want?

He's playing games

not me.

 

I.e.

Is substituting wrong? y/n

Is that substitution wrong? y/n

What is wrong? e.g. in the algebra.

 

If

c=1 dollar, =$1

v=1 cent,

then what is the unknown

v'=?.

Answer:

99 cents.

 

c=v+v'.

 

I could just as easily say

c=v+?

where the symbol "?"

is a value we are searching for.

 

Are you trying to tell me

that substitution is wrong?

Especially, for (checking) speed.

Edited May 15, 2017 by Capiert

[pzkpfw]

Speed just doesn't work that way, as was figured out from the consequence of light speed being the same for all observers.

 

The correct way to add speeds is:

 

v3 = ( v1 + v2 ) / ( 1 + ( ( v1 x v2 ) / ( c x c ) ) )

 

e.g. A sees B moving at v1, B sees C moving at v2, A sees C moving at v3.

 

This may not match your idea of "common sense", but that's because you live in a World where speeds are low.

 

Where v1 and v2 are small compared to c, we get:

 

v3 = ( v1 + v2 ) / ( 1 + ( ( small x small ) / ( c x c ) ) )

v3 = ( v1 + v2 ) / ( 1 + ( very small / very big ) )

v3 = ( v1 + v2 ) / ( 1 + (almost zero ) )

v3 = ( v1 + v2 ) / ( pretty much just 1 )

v3 = pretty much just ( v1 + v2 )

 

... i.e. v1 + v2, as you'd expect from "common sense". Someone in a car moving at 100 km/h throws a ball forward at 100 km/h and everyone thinks that ball is moving at pretty much 200 km/h. (It's actually 199.999999 km/h or so, but nobody notices.)

 

On the other hand, where v1 and v2 are closer to c, we get this:

 

v3 = ( v1 + v2 ) / ( 1 + ( ( almost c x almost c ) / ( c x c ) ) )

v3 = ( v1 + v2 ) / ( 1 + ( almost c squared / c squared ) )

v3 = ( v1 + v2 ) / ( 1 + ( almost one ) )

v3 = ( v1 + v2 ) / ( almost 2 )

v3 = pretty much half of ( v1 + v2 )

 

So say a spaceship going almost c away from Earth has it's own shuttle leave it at almost c. Does earth see the shuttle going nearly twice c? No. Only nearly c.

 

(The spaceship would consider the distance between Earth and the shuttle to be increasing at almost twice c, because from the spaceship point of view Earth is going one way at almost c and the shuttle is going the other way at almost c: but nobody here will see anybody else actually moving faster than c.)

[Strange]

although they leave each other at light speed c, ((also) meaning wrt each other).

I can't tell you how much sense that (now) makes to me.

 

 

Do you mean it makes sense now? Or are you being sarcastic?

 

Which space ship has the slower clock (wrt the other)?

 

(Answer: the other (space ship).

Which space ship is the other?

Answer: both!

What does that tell me?

Answer: nothing!)

 

What do you mean, it tells you nothing? It tells you there relative time dilation (and length contraction). What else do you expect it to tell you?

 

Yes I did but I doubt that you recognize the task('s significance) correctly.

Einstein's "math" does not allow (adding speeds to be) a speed larger than c,

but as you might notice

it does NOT agree

with the task (observation).

 

Perhaps you would like to pinpoint an error?

 

The error is that you won't accept reality. You think the the spaceships will see the other receding at c. They won't. Observation would disagree with your intuition.

 

 

 

But simply (2 space ships departing at)

0.98*c (wrt each other; or -0.49*c & 0.49*c wrt earth),

gives 0.79*c.

(E.g. similar (problematic) results.)

 

This is correct. There is nothing problematic about it.

You've said it's wrong

but don't say where;

I've defined it as a substitution

& given the math (algebra);

& then Strange "guesses" it's wrong.

What else do you want?

He's playing games

not me.

 

 

As arithmetic, it isn't wrong.

 

As physics it is wrong because the world doesn't work that way.

[swansont]

I've defined it as a substitution

& given the math (algebra);

& then Strange "guesses" it's wrong.

What else do you want?

He's playing games

not me.

 

!

Moderator Note

 

You need to take a valid expression and substitute it into another valid expression for it to be properly labeled a substitution.

 

You didn't do that. You just said v + v' = c. No derivation, no justification. That's called "pulling it out of your ass". We're not going to do that.

 

Either justify your expression with valid math, or with experiment.

 

[Capiert]

Speed just doesn't work that way..

Thanks. I'm slowly getting the message. Your picture was the most persuasive; & the high speed ball throwing,

but reminding me of other limiting math phenomena.

(All like Einstein said, but I guess his train was too slow (in my fantasy) for me to accept.)

Do you mean it makes sense now?

No I meant that it had not made sense to me.

 

Or are you being sarcastic?

I'd say I was being quantative in my evaluation,

but I must admit

there does seem to be

a sarcastic note in it.

 

What do you mean, it tells you nothing?

No distinction (time dilation or contraction)

between both space ships

each travelling at high speed,

but "opposite" directions.

 

If each ship

receives the same amount of contraction

then compared to each other,

they would not observe a difference

(between themselves).

 

If both ships dilate

the same amount

then they will not

observe a time difference

between them(selves).

 

It tells you there relative time dilation (and length contraction). What else do you expect it to tell you?

Each shows no difference from the other

although going at high speed

wrt to another.

In other words

both the space ships

do not notice relativistic effects

although there is a relativistic (hi) speed difference

between them.

I personally don't expect SR to tell the truth;

but standardwise

relativistic effects would be expected

for 2 objects

with a very high speed difference between them.

I did not say which space ship exerted thrust,

& that is not important

according to Einstein's non_prefered referencing.

 

The error is that you won't accept reality. You think the the spaceships will see the other receding at c. They won't. Observation would disagree with your intuition.

I tend to agree. They won't see c departure, they would see less than c.

 

This is correct. There is nothing problematic about it.

Relative to each other

hi_speed relativitic effects (dilation & contraction)

should happen

but don't

(although moving in opposite directions).

 

As arithmetic, it isn't wrong.

That's nice to hear.

 

As physics it is wrong because the world doesn't work that way.

I still question that

because I'm only asking

for the difference

(a(n unknown) number value)

between

light speed c

& an earth speed v.

I haven't changed anything

(e.g. c is a constant,

& no complicated transforms)

& can always find the original speed v (=c-v').

Edited May 15, 2017 by Capiert

[Strange]

No distinction (time dilation or contraction)

between both space ships

each travelling at high speed,

but "opposite" directions.

 

 

If by "no distinction" you mean that they both see the same time dilation and length contraction in the other ship, then that is correct. You should not be surprised by this. It is a result of the "no preferred frame" you mentioned before. All frames are equivalent.

 

 

 

Each shows no difference from the other

although going at high speed

wrt to another.

In other words

both the space ships

do not notice relativistic effects

although there is a relativistic (hi) speed difference

between them.

 

Wrong. They both notice the same relativistic effects. As you said earlier:

 

 

Which space ship has the slower clock (wrt the other)?

 

(Answer: the other (space ship).

Which space ship is the other?

Answer: both!

 

 

 

 

relativistic effects would be expected

for 2 objects

with a very high speed difference between them.

 

And there will be. What makes you think otherwise?

 

 

 

 

Relative to each other

hi_speed relativitic effects (dilation & contraction)

should happen

but don't

(although moving in opposite directions).

 

Of course they do. What makes you think they don't?

 

 

 

I still question that

because I'm only asking

for the difference

(a(n unknown) number value)

between

light speed c

& an earth speed v.

 

There are two answers to this depending which frame of reference you use.

 

If you see a spaceship moving at v = 0.2c then the difference between v and c is 0.8c. This tells you absolutely nothing useful.

 

However, the people on the spaceship will see their velocity as 0 and therefore the difference from the speed of light is c. This actually tells you something very profound about the way the world works.

[Capiert]

If by "no distinction" you mean that they both see the same time dilation and length contraction in the other ship, then that is correct.

 

Yes that is what I mean.

You should not be surprised by this.

I am however, because relativistic effects are missing

for a high speed example

(due to the "opposite" stipulation=condition)).

It is a result of the "no preferred frame" you mentioned before.

But it's not making sense

to have no contraction nor dilation

although hi_speed.

I doubt that you recognize that (consequence, correctly).

All frames are equivalent.

We agree there,

that's why relativistic effects (contraction & dilation)

should exist

in "any" (hi_speed case=example)

but don't there.

 

Wrong. They both notice the same relativistic effects. As you said earlier:

 

"They" don't notice anything (different, caused by the hi_speed,

because both differences are the "same").

 

And there will be (relativistic effects on 2 hi_speed ships moving apart). What makes you think otherwise?

They (=the ship's capitains)

don't notice any time dilation

of the other ship,

although they should

because they are traveling

at hi_speed (away).

(=Non_preferred ref, e.g.

either ship could be used as reference frame.

Hi speed, no dilation observed,

SR fails!)

 

(Relative to each other

both ships should experience

dilation & contraction,

but don't.)

Of course they do. What makes you think they don't?

"Relative to each other."

That's what relativity is all about:

what the other ship sees (=observes).

In that example,

they observe no difference

(as though SR does not exist)

although they are moving

at hi_speed relative

to each other.

That's a paradox, or a flaw!

 

(Math c=v+v', v'=c-v.)

There are two answers to this depending which frame of reference you use.

 

If you see a spaceship moving at v = 0.2c then the difference between v and c is 0.8c. This tells you absolutely nothing useful.

I disagree, in the sense that it will give me a (new) different (unknown) number,

which I can "use" to check calculations (for accuracy, or error).

It's only a (useful) tool (for me).

So it is not telling "you" anything about relativity;

(or is it?).

 

However, the people on the spaceship will see their velocity as 0 and therefore the difference from the speed of light is c. This actually tells you something very profound about the way the world works.

 

I'm not (really) disagreeing with you there. Edited May 15, 2017 by Capiert

[Strange]

If you are just going to insist that there are no relativistic effects when there obviously are, then this is pointless.

 

We know relativity works, so why deny it?

[swansont]

I disagree, in the sense that it will give me a (new) different (unknown) number,

which I can use to check calculations.

It's only a (useful) tool (for me).

So it is not telling "you" anything about relativity;

(or is it?).

 

 

 

!

Moderator Note

 

I said stop with the games. If it has useful information to tell us, it is your job to explain what that is. As it stands, it looks like simple algebra with no connection to any physics. Since you have failed to make your case, despite several chances, this is closed. Don't start up any new speculations threads on this. (Or abuse the PM or reported post systems to argue your case)

 

As to the rest, that's part of the mainstream theory of relativity and should be discussed in the relativity section of physics, up to the point where you insist that relativity is wrong. You have presented no evidence of that, and unsupported assertions will be a quick ticket to closure and possibly more action on the part of the staff.