Relativity?

[Capiert]

2 space ships leave

each other at light speed c.

 

What are their length contractions

& time dilations?

 

(E.g. wrt to earth,

each is travelling c/2.)

Edited May 14, 2017 by Capiert

[Sensei]

2 space ships leave

each other at light speed c.

 

What are their length contractions

& time dilations?

 

You should know already from equations that they cannot fly at speed of light (in Special Relativity)..

 

Speed of light is reserved exclusively for photons = light. That's why it's called "speed of light", not "speed of matter"..

 

If you want to redefine relativity, go ahead and propose your equations which are matching experimental data..

Edited May 14, 2017 by Sensei

[koti]

2 space ships leave

each other at light speed c.

 

What are their length contractions

& time dilations?

Their lenghts are 0 and their time stands still.

[Strange]

Their lenghts are 0 and their time stands still.

 

 

Which is why the speed of light is not a valid frame of reference.

[Capiert]

You should know already from equations that they cannot fly at speed of light (in Special Relativity)

 

Wrt earth each spaceship travels at c/2.

 

What is the answer?

[Strange]

2 space ships leave

each other at light speed c.

 

What are their length contractions

& time dilations?

 

(E.g. wrt to earth,

each is travelling c/2.)

 

The 2 space ships have a speed of 0.8c, relative to one another (http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel2.html)

* Note: the calculator on that page assumes the situation you have described - rockets going in opposite directions, so you need to enter 0.5 and -0.5 as the two speeds.

 

The time dilation and length contraction factor is 0.6 (http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html)

Edited May 14, 2017 by Strange

[koti]

 

 

Which is why the speed of light is not a valid frame of reference.

I was hoping that the OP will come to this conclusion.

[Capiert]

Which is why the speed of light is not a valid frame of reference.

 

Einstein said,

there is no preferred reference frame;

so why do you reject that 1

in (preferred, biased) favour of the others?

Einstein knew SR did NOT work for everything,

that is why he invented GR.

I was hoping that the OP will come to this conclusion.

What is OP=?

[Strange]

Einstein said,

there is no preferred reference frame;

so why do you reject that 1

in (preferred, biased) favour of the others?

 

 

This has nothing to do with preferred frames. It is basically just arithmetic. If you try and treat the speed of light as a reference frame then you end up dividing by zero.

 

The Lorentz factor is: [latex]\frac 1 { \sqrt {1 - \frac {v^2} {c^2} } }[/latex]

If you set v = c, you end up with [latex]\frac 1 { \sqrt {1 - \frac {c^2} {c^2} } } = \frac 1 { \sqrt {1 - 1 } } = \frac 1 0[/latex]

 

So it is meaningless.

 

 

 

What is OP=?

 

Original Poster (you).

 

Or sometimes Original Post (the first post in the thread)

[Sensei]

That's why I said he needs to propose alternative math equations (which won't lead to division by zero like in SR)..

SR Kinetic energy equation (indistinguishably) matches Newtonian's kinetic energy at low velocities,

so "alternative physics" should extend it,

and match either Newtonian's physics at low speed,

and SR at higher speed (which is experimentally confirmed in experiments decays of unstable particles and high energy physics at CERN),

and then introduce something at extreme speeds.... or nothing to discuss.. ?

Edited May 14, 2017 by Sensei

[Strange]

2 space ships leave

each other at light speed c.

 

What are their length contractions

& time dilations?

 

(E.g. wrt to earth,

each is travelling c/2.)

 

 

As these are just questions, why is this in Speculations? Are you planning to present an alternative?

[Capiert]

As these are just questions, why is this in Speculations? Are you planning to present an alternative?

I wanted to know how you would solve the questions

& you gave answers,

So before presenting anything (socalled new)

I would like to consider them.

 

But I have more questions

which will help me orientate.

E.g. We know we can derive rest mass m'

from energy E=m'*c^2).

Can we do the same using momentum?

(E.g. mom=2*m'*c. ?)

I suspect, due to the incompatibility,

(between mom & E)

you will say no,

& modify either the mass

&/or momentum

(with transforms).

 

My next question would then be,

why does (simple) energy receive priorty

over (simple) momentum (mom=m*v)

in relativity equations?

(If momentum is derived from energy

& not in reverse.)

Why is the priority not reversed (or reversable)

so that momentum dominates

(the simplicity)

& energy needs the transforms (instead)?

(Einstein's relativity

seems (to me)

to prefer energy (evaluation)

(being simple),

over momentum.)

 

&

 

What is wrong

with the syntax

of letting

c=v+v'

where

v' symbolizes

the compliment

(or missing speed (difference))

needed

to add with v

to give c.

(I.e. Having nothing to do with (Lorentz) transforms.)

Nothing complicated.

Edited May 14, 2017 by Capiert

[studiot]

I wanted to know how you would solve the questions

& you gave answers,

So before presenting anything (socalled new)

I would like to consider them.

 

But I have more questions

which will help me orientate.

E.g. We know we can derive rest mass

from energy.

Can we do the same using momentum?

I suspect, due to the incompatibility,

(between mom & E)

you will say no,

& modify either the mass

&/or momentum

(with transforms).

 

My next question would then be,

why does (simple) energy receive priorty

over (simple) momentum (mom=m*v)

in relativity equations?

(If momentum is derived from energy

& not in reverse.)

Why is the priority not reversed (or reversable)

so that momentum dominates

(the simplicity)

& energy needs the transforms (instead)?

(Einstein's relativity

seems (to me)

to prefer energy (evaluation)

(being simple),

over momentum.)

 

The way to teat momentum, velocity, etc relativistically is to use the 4D versions

 

That is use 4-momentum, 4-velocity and so on.

 

These physical quantities are much better behaved, just as you would be with two arms and two legs, rather than one chopped off.

 

 

&

 

What is wrong

with the syntax

of letting

c=v+v'

where

v' symbolizes

the compliment

(or missing speed (difference))

needed

to add with v

to give c.

(I.e. Having nothing to do with (Lorentz) transforms.)

Nothing complicated.

 

If you must use 3-velocity then you have been told several times now that there are no two velocities v and v' that add to c.

 

The process of addition (or subtraction) is not one of simple adding two plain numbers together.

Edited May 14, 2017 by studiot

[Capiert]

& If

(v)^2=((vx)^2) + ((vy)^2) + ((vz)^2)

&

(v')^2=((vx')^2) + ((vy')^2) + ((vz')^2),

then

why is c not v+v'?

[Strange]

why is c not v+v'?

 

 

Because v1 + v2 < c

[Capiert]

Because v1 + v2 < c

Please explain.

(c^2)=((vx+vx')^2)+((vy+vy')^2)+((vz+vz')^2).

[Strange]

Please explain.

 

http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel2.html

 

 

(c^2)=((vx+vx')^2)+((vy+vy')^2)+((vz+vz')^2).

 

Where does that equation come from?

[Capiert]

http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel2.html

 

Where does that equation come from?

Common sense.

[Strange]

Common sense.

 

 

Well it appears to be wrong.

[Capiert]

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?

Edited May 14, 2017 by Capiert

[Strange]

Velocities don't add like that.

 

Did you actually read this: http://hyperphysics....iv/einvel2.html

[Capiert]

The 2 space ships have a speed of 0.8c, relative to one another

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.

 

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!)

Velocities don't add like that.

 

Did you actually read this: http://hyperphysics....iv/einvel2.html

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?

 

(If you don't like using departure speed c,

(to keep the math simple); then

something like 0.999998*c,

(e.g. 2 millionths less than c)

(for dramatic results)

could be used instead,

to avoid divide by zero.

 

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.)

Edited May 15, 2017 by Capiert

[koti]

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

 

Start with the fact that there is no such thing as velocity v=c for massive objects.

[KipIngram]

In the OP he noted that each had speed wrt Earth of 0.5c; it would just be that an observer on earth would measure their speed wrt one another as c. But that's not the same as what was just said.

Edited May 15, 2017 by KipIngram

[koti]

In the OP he noted that each had speed wrt Earth of 0.5c; it would just be that an observer on earth would measure their speed wrt one another as c. But that's not the same as what was just said.

We both know that this is where it has to start as literally nothing is being absorbed. Its going to stay that way untill basics are going to be understood.