On 8 fév, 09:39, "harry" wrote:
wrote in message

... On Thu, 7 Feb 2008 wrote:

[...]

f(x)=c [sqrt(4ax + x^2)) / (2a - x)]

For a=0 that gives:

f(x)=c [sqrt(x^2)) / (- x)] = -c

What does your negative speed mean?!

Not applicable.

In context since either energy is present
and then velocity is in the direction of motion
or energy is absent and then velocity is zero.

There is no such thing as less energy than zero
energy.


As I wrote, it's important to be clear as to what you mean by
"relativistic". I've now asked twice for clarification. Let me ask a
third
time: exactly what do you mean by "relativistic"?
I mean relativistic mass increase with velocity.

OK, that's straightforward - Newtonian mechanics has been adapted
to SRT in just that way.

Really?

Then, show me how the kinetic energy has equation has
been transformed.

[..]

What does Galilei-relativistic means ?

Newtonian mechanics uses Galilean relativity between inertial frames.

So it refers to relative localization. Not applicable in context.
Talking about absolute velocity related to local quantum of
energy.

and
this was in the context of pointing out the importance of defining terms
such as "relativistic". Most people would not say that Newtonian
mechanics is "relativistic",
Then "most people" don't know much. It is in no way relativistic.

Read that again: you answered like "most people" who "don't know much". :-)

Tried to see the shade of meaning, but didn't see it. I really meant
that those who think that Newtonian mechanics is relativistic don't
know much.

André Michaud

wrote in message
...
On 8 fév, 09:39, "harry" wrote:
wrote in message

...
On Thu, 7 Feb 2008 wrote:

[...]

f(x)=c [sqrt(4ax + x^2)) / (2a - x)]

For a=0 that gives:

f(x)=c [sqrt(x^2)) / (- x)] = -c

What does your negative speed mean?!

: Not applicable.

: In context since either energy is present
and then velocity is in the direction of motion
or energy is absent and then velocity is zero.

: There is no such thing as less energy than zero
energy.

I agree! I'm afraid that your equation isn't applicable...


As I wrote, it's important to be clear as to what you mean by
"relativistic". I've now asked twice for clarification. Let me ask a
third
time: exactly what do you mean by "relativistic"?
I mean relativistic mass increase with velocity.

OK, that's straightforward - Newtonian mechanics has been adapted
to SRT in just that way.

: Really?

: Then, show me how the kinetic energy has equation has
been transformed.

All standard textbooks of the 80s show that, such as Feynman's lectures on
physics.
See for example:
http://www.bartleby.com/173/15.html
and
http://en.wikipedia.org/wiki/Mass_in_special_relativity

[..]

What does Galilei-relativistic means ?

Newtonian mechanics uses Galilean relativity between inertial frames.

: So it refers to relative localization.

NO, it refers to laws of nature independent of used reference frame.
Classical relativity is at the basis of Newtonian mechanics.

: Not applicable in context.
: Talking about absolute velocity related to local quantum of energy.

Doesn't sound as if you talk the same language, sorry. :-)

and
this was in the context of pointing out the importance of defining
terms
such as "relativistic". Most people would not say that Newtonian
mechanics is "relativistic",

Then "most people" don't know much. It is in no way relativistic.

Read that again: you answered like "most people" who "don't know much".
:-)

: Tried to see the shade of meaning, but didn't see it. I really meant
: that those who think that Newtonian mechanics is relativistic don't
: know much.

I know that you meant that: it implies that you belong to "most people" who
don't know about classical relativity. :-O)

Cheers,
Harald

On Feb 8, 8:36 am, wrote:
On 7 fév, 22:48, Timo Nieminen wrote:



On Thu, 7 Feb 2008 wrote:

It certainly does, since the final equation could not have been
developed if I hadn't done it.

or even whether I like what you did or did not do, it doesn't make
"relativistic mass increase" exponential.

Then explain to us how this equation can be not exponential:

f(x)=c [sqrt(4ax + x^2)) / (2a - x)]

Because x does not appear in an exponent.

You must have a non-standard meaning of "exponential".

(With modern definitions of mass [essentially, "mass"="proper mass"],
there is no relativistic mass increase, but I don't think that's an issue
here.)

Do I perceive here that you don't understand relativistic mass
increase with velocity ?

Do I perceive you've never seen the expression

p = gamma*m*v

in which m is an invariant, independent of velocity? Or
this one:

E^2 = (pc)^2 + (mc^2)^2

in which similarly m is also an invariant?

The convention these days is to use the term "mass"
to refer to this invariant quantity, formerly called the
"rest mass". While I do see the idea of "relativistic mass"
persisting in some websites written for the general public,
unfortunately, I believe you will find that the practice among
working scientists in their own exchanges to reserve
the term "mass" for the invariant quantity.

- Randy

"harry" wrote in message
...
|
| wrote in message
| ...
| On 8 fév, 09:39, "harry" wrote:
| wrote in message
|
|
...
| On Thu, 7 Feb 2008 wrote:
|
| [...]
|
| f(x)=c [sqrt(4ax + x^2)) / (2a - x)]
|
| For a=0 that gives:
|
| f(x)=c [sqrt(x^2)) / (- x)] = -c
|
| What does your negative speed mean?!
|
| : Not applicable.
|
| : In context since either energy is present
| and then velocity is in the direction of motion
| or energy is absent and then velocity is zero.
|
| : There is no such thing as less energy than zero
| energy.
|
| I agree! I'm afraid that your equation isn't applicable...

If I fire a bullet away from you giving it negative energy,
I must absorb that then give it some positive energy in order
to sent it toward you. Sending it away from you is what -v
means. What else would a negative velocity mean?
-1 + 2 = 1 when I went to school.

On 8 fév, 10:43, "harry" wrote:
wrote in message

...
On 8 fév, 09:39, "harry" wrote:

wrote in message

...
On Thu, 7 Feb 2008 wrote:

[...]

f(x)=c [sqrt(4ax + x^2)) / (2a - x)]

For a=0 that gives:

f(x)=c [sqrt(x^2)) / (- x)] = -c

What does your negative speed mean?!

: Not applicable.

: In context since either energy is present
and then velocity is in the direction of motion
or energy is absent and then velocity is zero.

: There is no such thing as less energy than zero
energy.

I agree! I'm afraid that your equation isn't applicable...

You are wrong. It totally applies.

As I wrote, it's important to be clear as to what you mean by
"relativistic". I've now asked twice for clarification. Let me ask a
third
time: exactly what do you mean by "relativistic"?
I mean relativistic mass increase with velocity.

OK, that's straightforward - Newtonian mechanics has been adapted
to SRT in just that way.

: Really?

: Then, show me how the kinetic energy has equation has
been transformed.

All standard textbooks of the 80s show that, such as Feynman's lectures on
physics.
See for example:http://www.bartleby.com/173/15.html

Invalid. The gamma factor is assumed, not derived.

andhttp://en.wikipedia.org/wiki/Mass_in_special_relativity

Invalid demonstration, for the same reason.

Show me real physics.

[..]

What does Galilei-relativistic means ?

Newtonian mechanics uses Galilean relativity between inertial frames.

: So it refers to relative localization.

NO, it refers to laws of nature independent of used reference frame.
Classical relativity is at the basis of Newtonian mechanics.

Whst has inertial classical relativity got to do with
relativistic mass increase ?

: Not applicable in context.
: Talking about absolute velocity related to local quantum of energy.

Doesn't sound as if you talk the same language, sorry. :-)

Can't be helped.

and
this was in the context of pointing out the importance of defining
terms
such as "relativistic". Most people would not say that Newtonian
mechanics is "relativistic",

Then "most people" don't know much. It is in no way relativistic.

Read that again: you answered like "most people" who "don't know much".
:-)

: Tried to see the shade of meaning, but didn't see it. I really meant
: that those who think that Newtonian mechanics is relativistic don't
: know much.

I know that you meant that: it implies that you belong to "most people" who
don't know about classical relativity. :-O)

No it means that I belong to people that understand that classical
mechanics is not relativistic in any way shape or form.

André Michaud

On 8 fév, 10:49, Randy Poe wrote:
On Feb 8, 8:36 am, wrote:

On 7 fév, 22:48, Timo Nieminen wrote:

On Thu, 7 Feb 2008 wrote:
It certainly does, since the final equation could not have been
developed if I hadn't done it.

or even whether I like what you did or did not do, it doesn't make
"relativistic mass increase" exponential.

Then explain to us how this equation can be not exponential:

f(x)=c [sqrt(4ax + x^2)) / (2a - x)]

Because x does not appear in an exponent.

You must have a non-standard meaning of "exponential".

Not at all. I said the the "equation" was exponential,
not that the function was.

(With modern definitions of mass [essentially, "mass"="proper mass"],
there is no relativistic mass increase, but I don't think that's an issue
here.)

Do I perceive here that you don't understand relativistic mass
increase with velocity ?

Do I perceive you've never seen the expression

p = gamma*m*v

Not at all. It can even be derived from the equation I
gave.

in which m is an invariant, independent of velocity? Or

m is, but gamma m is not or maybe you don't really
understand the equation and the function of gamma.

this one:

E^2 = (pc)^2 + (mc^2)^2

in which similarly m is also an invariant?

Not so. in it, m simply represent the rest mass of
the particle. I suggest you analyze it a little
more deeply.

The convention these days is to use the term "mass"
to refer to this invariant quantity, formerly called the
"rest mass".

You don't say !

That's exactly what I have been telling you.

While I do see the idea of "relativistic mass"
persisting in some websites written for the general public,
unfortunately, I believe you will find that the practice among
working scientists in their own exchanges to reserve
the term "mass" for the invariant quantity.

Then the practice has to relate to some level of ignorance
in that part of the community regarding relativistic mass
increase with velocity. I suggest you open up any reference
work related to high energy accelerators and dig in.

André Michaud

On Feb 8, 11:29 am, wrote:
On 8 fév, 10:49, Randy Poe wrote:



On Feb 8, 8:36 am, wrote:

On 7 fév, 22:48, Timo Nieminen wrote:

On Thu, 7 Feb 2008 wrote:
It certainly does, since the final equation could not have been
developed if I hadn't done it.

or even whether I like what you did or did not do, it doesn't make
"relativistic mass increase" exponential.

Then explain to us how this equation can be not exponential:

f(x)=c [sqrt(4ax + x^2)) / (2a - x)]

Because x does not appear in an exponent.

You must have a non-standard meaning of "exponential".

Not at all. I said the the "equation" was exponential,
not that the function was.

a^x does not occur in the equation for any a.

What makes this equation exponential if it
has no exponential terms?

You do indeed have a non-standard meaning of "exponential".
What is it?

(With modern definitions of mass [essentially, "mass"="proper mass"],
there is no relativistic mass increase, but I don't think that's an issue
here.)

Do I perceive here that you don't understand relativistic mass
increase with velocity ?

Do I perceive you've never seen the expression

p = gamma*m*v

Not at all. It can even be derived from the equation I
gave.

in which m is an invariant, independent of velocity? Or

m is, but gamma m is not or maybe you don't really
understand the equation and the function of gamma.

The function of gamma is that the velocity dependence
is not part of the mass term. The mass term is m.
So on the one hand you acknowledge that m is
independent of velocity, then go on blithely talking
about "mass increase"? Pray tell, how does something
both increase and remain invariant?

- Randy

On 8 fév, 10:43, "harry" wrote:
wrote in message

...
On 8 fév, 09:39, "harry" wrote:

wrote in message

...
On Thu, 7 Feb 2008 wrote:

[...]

f(x)=c [sqrt(4ax + x^2)) / (2a - x)]

Sorry all, that's a typo

the equation is

f(x)=c [sqrt(4ax + x^2)) / (2a + x)]

André Michaud

On 8 fév, 11:48, Randy Poe wrote:
On Feb 8, 11:29 am, wrote:



On 8 fév, 10:49, Randy Poe wrote:

On Feb 8, 8:36 am, wrote:

On 7 fév, 22:48, Timo Nieminen wrote:

On Thu, 7 Feb 2008 wrote:
It certainly does, since the final equation could not have been
developed if I hadn't done it.

or even whether I like what you did or did not do, it doesn't make
"relativistic mass increase" exponential.

Then explain to us how this equation can be not exponential:

f(x)=c [sqrt(4ax + x^2)) / (2a - x)]

Because x does not appear in an exponent.

You must have a non-standard meaning of "exponential".

Not at all. I said the the "equation" was exponential,
not that the function was.

a^x does not occur in the equation for any a.

It doesn't. Never said it did.

There is no doubt however that the function
can be expressed in exponential form as you
mean.

What makes this equation exponential if it
has no exponential terms?

Because it is not linear. It is exponential.

Exponential increase is quite different from
linear increase.

You go ahead and express it in exponential
form if you wish. I describe my stuff so
it is accessible to most. I don't care if
hair-splitting "experts" disapprove.

You do indeed have a non-standard meaning of "exponential".
What is it?

Just given.

(With modern definitions of mass [essentially, "mass"="proper mass"],
there is no relativistic mass increase, but I don't think that's an issue
here.)

Do I perceive here that you don't understand relativistic mass
increase with velocity ?

Do I perceive you've never seen the expression

p = gamma*m*v

Not at all. It can even be derived from the equation I
gave.

in which m is an invariant, independent of velocity? Or

m is, but gamma m is not or maybe you don't really
understand the equation and the function of gamma.

The function of gamma is that the velocity dependence
is not part of the mass term. The mass term is m.
So on the one hand you acknowledge that m is
independent of velocity,

No. Rest mass is independent of velocity. But the
instantaneous relativistic increase in mass is not.

then go on blithely talking
about "mass increase"?

Read back. I talk of relativistic mass increase, not
mass increase.

Pray tell, how does something both increase and
remain invariant?

Simple. It it simply doesn't. Relativistic mass
increase is not the same as rest mass.

André Michaud

On Feb 8, 12:00 pm, wrote:
On 8 fév, 11:48, Randy Poe wrote:



On Feb 8, 11:29 am, wrote:

On 8 fév, 10:49, Randy Poe wrote:

On Feb 8, 8:36 am, wrote:

On 7 fév, 22:48, Timo Nieminen wrote:

On Thu, 7 Feb 2008 wrote:
It certainly does, since the final equation could not have been
developed if I hadn't done it.

or even whether I like what you did or did not do, it doesn't make
"relativistic mass increase" exponential.

Then explain to us how this equation can be not exponential:

f(x)=c [sqrt(4ax + x^2)) / (2a - x)]

Because x does not appear in an exponent.

You must have a non-standard meaning of "exponential".

Not at all. I said the the "equation" was exponential,
not that the function was.

a^x does not occur in the equation for any a.

It doesn't. Never said it did.

There is no doubt however that the function
can be expressed in exponential form as you
mean.

What makes this equation exponential if it
has no exponential terms?

Because it is not linear. It is exponential.

OK, so by "exponential" you mean "nonlinear". That
is not the standard meaning of "exponential".

Exponential increase is quite different from
linear increase.

Yes it is, but there are many other nonlinear
forms which are not exponential. Quadratic
increase for instance.

You go ahead and express it in exponential
form if you wish.

It isn't an exponential form. That's the point,
that you are using the word "exponential"
to mean "nonlinear", which is not the standard
meaning.

I describe my stuff so
it is accessible to most. I don't care if
hair-splitting "experts" disapprove.

So your "accessible" version is to use a technical
term like "exponential" when something is not
exponential?

Don't you think "not linear" would be a
more accessible way of expressing the concept
of not being linear?

- Randy