On Feb 3, 10:09*am, "Jeckyl" wrote:
"Peri of Pera" wrote in ...





On Feb 3, 9:36 am, "Jeckyl" wrote:
"Peri of Pera" wrote in
...

On Feb 2, 1:49 pm, "Jeckyl" wrote:
"Peri of Pera" wrote in
messagenews:bb01e949-a208-4aa8-a317-e490b88df__BEGIN_MASK_n#9g02mG7!__...__END_MASK_i ?a63jfAD$__BEGIN_MASK_n#9g02mG7!__...__END_MASK_i? ...

Time Dilation achieves isotropic Speed

When Lorentz invented time dilation as part of his contraction
hypothesis he did so to allow the speed of light to remain constant.

"Peri of Pera" wrote in message
...
Jecko,
Can you answer a simple question in a simple way? How does time
dilation act to preserve
the law of physics d=v*t.

I already did .. read my replies.

On Feb 2, 12:04*am, Peri of Pera wrote:
Time Dilation achieves isotropic Speed

When Lorentz invented time dilation as part of his contraction
hypothesis he did so to allow the speed of light to remain constant.
He thought that if *the length *of a moving object contracted, its
time *had to slow down or the speed of light would not be constant.
However, Lorentz achieved the opposite effect with his thought
process.

Example:
An object of 100m length traveling with a speed of 200000km/sec *would
according to the Lorentz transformation

gamma = 1/sqrt(1-200000km/sec^2/300000km/sec^2) *= 1.3416408

shrink to 74.535599m (100/gamma). At rest, light will cover 100m in
100m/300000000m/sec = 0.000000333333sec. Time dilation will expand
this fraction of *time to 0.00000044721360 seconds
(0.000000333333secs*1.3416408) for an object with the speed of
200000km/sec.

In the dilated time of 0.00000044721360 seconds, light at 300000000m/
sec will transit a distance of 134.16408m (300000000m/
sec*0.000000044721360sec) but if light had slowed down to 223607021m/
sec, light would exactly cover the original 100m in the dilated time
of *0.000000044721360sec (223607021m/sec*0.00000044721360sec=100m).
Clearly, if the speed of light had not been reduced, the law of
physics d=v*t would have been violated.

Peter Riedt

Peter I have light speed creating a foreshortning of space at the
direction photons are traveling.At 186,242 mps spacetime is infinitly
short.Just another way of viewing SR Bert

Peri of Pera skrev:
The questions a If time slows down and there is more time to do
things will light go a longer distance or is the speed of light
reduced? Either must occur to preserve the law of physics d=v/t.
Peter Riedt

According to SR:
1. Time is what clocks show.
2. Clocks always run at their proper rate,
they never slow down or speed up.

Since time never slows down you haven't asked a question.

--
Paul

http://home.c2i.net/pb_andersen/

On Feb 1, 10:49*pm, "Jeckyl" wrote:
"Peri of Pera" wrote in ...

Time Dilation achieves isotropic Speed

When Lorentz invented time dilation as part of his contraction
hypothesis he did so to allow the speed of light to remain constant.
He thought that if *the length *of a moving object contracted, its
time *had to slow down or the speed of light would not be constant.

And there also has to be a change in simultaneity as well

However, Lorentz achieved the opposite effect with his thought
process.

This should be good for a laugh again...

Example:
An object of 100m length traveling with a speed of 200000km/sec *would
according to the Lorentz transformation

gamma = 1/sqrt(1-200000km/sec^2/300000km/sec^2) *= 1.3416408

shrink to 74.535599m (100/gamma).

If you mean that a stationary obserer would measure as the distance between
the endpoint of the moving rod at a given instance of time in the stationary
system .. then yes

At rest, light will cover 100m in
100m/300000000m/sec = 0.000000333333sec.

OK

Time dilation will expand
this fraction of *time to 0.00000044721360 seconds
(0.000000333333secs*1.3416408) for an object with the speed of
200000km/sec.

If you mean that at the location of the stationary observer, an interval
that is 0.000000333333 on the moving clock will take 0.00000044721360 on the
observers clock .. then yes.

In the dilated time of 0.00000044721360 seconds, light at 300000000m/
sec will transit a distance of 134.16408m (300000000m/
sec*0.000000044721360sec) but if light had slowed down to 223607021m/
sec, light would exactly cover the original 100m in the dilated time
of *0.000000044721360sec (223607021m/sec*0.00000044721360sec=100m).
Clearly, if the speed of light had not been reduced, the law of
physics d=v*t would have been violated.

You are clearly confused about relativity.

Let's look at this more clearly...

Let us assume our coordinates are such that at x=0,t=0 we also have x'=0,
t'=0
Let us assume that we have a moving rod of length L travelling at speed v
Let us assume the light is shining along the rod in the direction of travel
Let us assume the light is emitted at t=t'=0, when the rear end of the rod
is at x=x'=0

In the stationary frame of reference, the rod is moving while the light is
travelling at c
So the light gets to the other end of the rod, as seen by the stationary
observer, at x, t where
t = L / (c - v)
x = c . t ... because x / t = c

That corresponds in the rods frame of reference to a point (x',t')

x' = gamma . ( x - v.t ) ... Lorentz
x' = gamma . ( c.t - v.t ) ... subs for x
x' = gamma . ( c - v ) . t ... factorise
x' = gamma . ( c - v ) . L / (c - v) ... subs for t
x' = gamma . L ... cancel

t' = gamma . ( t - v.x / c^2 ) ... Lorentz
t' = gamma . ( t - v.t / c ) ... subs for x
t' = gamma . (1 - v/c ) . t ... factorise
t' = gamma . (c - v) / c . L / (c - v) ... subs for t
t' = gamma . L / c ... cancel

so the speed of light as seen in the moving rods frame is

c' = x' / t'
c' = gamma . L / ( gamma . L / c )
c' = c

So you can see that the speed of light as measuring in the frame of the
moving frame is c as well

Jeckyl, not bad. Within such a sort time you can come up with these
calculation to explain c' = c. But unfortunately, there are some
minor problem that makes the proof not correct.

L = 100m
c = 300000000 m/s
v = 200000000 m/s
Gamma = 1/sqrt(1-v^2/c^2) = 1.3416

t = L / (c - v)
x = c . t ... because x / t = c
t = 100 / (3E8 - 2E8) = 0.000001 sec
x = 3E8 * 0.000001 = 300 m

x' = gamma . ( x - v.t ) ... Lorentz
t' = gamma . ( t - v.x / c^2 ) ... Lorentz
x' = 1.3416 * (300 - 2E8 * 0.000001) = 134.164 m == Wrong
t' = 1.3416 * (0.000001 - 2E8 * 300 / (3E8)^2) = 0.00000044721 sec

1. x' has to be 100 m.
In the moving frame the far end of the rod is always at x' = 100m.
2. The ratio of t/t' = 0.000001/0.00000044721 = 2.24 (not = gamma)

Please try and come up something better. I believe you can because
you are smarter than the other guy.

Since the rod is moving (stationary w.r.t. the moving frame) so the
conditions we know for sure a
x' = 100 m and t' = x'/c = 0.0000003333 sec
You should come up with something to calculate x and t such that x / t
= c.

"snapdragon31" wrote in message
...
On Feb 1, 10:49 pm, "Jeckyl" wrote:
Let's look at this more clearly...

Let us assume our coordinates are such that at x=0,t=0 we also have x'=0,
t'=0
Let us assume that we have a moving rod of length L travelling at speed v
Let us assume the light is shining along the rod in the direction of
travel
Let us assume the light is emitted at t=t'=0, when the rear end of the
rod
is at x=x'=0

In the stationary frame of reference, the rod is moving while the light
is
travelling at c
So the light gets to the other end of the rod, as seen by the stationary
observer, at x, t where
t = L / (c - v)
x = c . t ... because x / t = c

That corresponds in the rods frame of reference to a point (x',t')

x' = gamma . ( x - v.t ) ... Lorentz
x' = gamma . ( c.t - v.t ) ... subs for x
x' = gamma . ( c - v ) . t ... factorise
x' = gamma . ( c - v ) . L / (c - v) ... subs for t
x' = gamma . L ... cancel

t' = gamma . ( t - v.x / c^2 ) ... Lorentz
t' = gamma . ( t - v.t / c ) ... subs for x
t' = gamma . (1 - v/c ) . t ... factorise
t' = gamma . (c - v) / c . L / (c - v) ... subs for t
t' = gamma . L / c ... cancel
Please try and come up something better. I believe you can because
you are smarter than the other guy.

Thanks for your patience .. you picked up my error in the above at the line:

Let us assume that we have a moving rod of length L travelling at speed v

It should have said that the L is the length of the moving rod measured in
the stationary frame (ie the contracted length)

Of course, one can do the analysis the other way .. by seeing that light
moving along the stationary rod at c transforms to light moving along the
moving rod at c

Lets assume the length of the stationary rod in the stationary frame is L
When light is emitted, we have at x=x'=0, t=t'=0
When light gets to the other end of the rod, we have x=L, t=L/c.
That point corresponds to (x',t') where

x' = gamma . ( x - v.t ) ... Lorentz
x' = gamma . ( L - v.L/c ) ... subs x=L, t=L/c
x' = gamma . (c - v) . L / c ... factorise

t' = gamma . ( t - v.x / c^2 ) ... Lorentz
t' = gamma . ( L/c - v.L / c^2 ) ... subs x=L, t=L/c
t' = gamma . ( c - v) L / c^2 ... factorise

x'/t' = ( gamma . (c - v) . L / c ) / ( gamma . ( c - v) L / c^2 ) = c

Again, the speed of light is c

On Feb 2, 2:17*am, "Jeckyl" wrote:
"snapdragon31" wrote in message

...
On Feb 1, 10:04 pm, Peri of Pera wrote:





Time Dilation achieves isotropic Speed

When Lorentz invented time dilation as part of his contraction
hypothesis he did so to allow the speed of light to remain constant.
He thought that if the length of a moving object contracted, its
time had to slow down or the speed of light would not be constant.
However, Lorentz achieved the opposite effect with his thought
process.

Example:
An object of 100m length traveling with a speed of 200000km/sec would
according to the Lorentz transformation

gamma = 1/sqrt(1-200000km/sec^2/300000km/sec^2) = 1.3416408

shrink to 74.535599m (100/gamma). At rest, light will cover 100m in
100m/300000000m/sec = 0.000000333333sec. Time dilation will expand
this fraction of time to 0.00000044721360 seconds
(0.000000333333secs*1.3416408) for an object with the speed of
200000km/sec.

In the dilated time of 0.00000044721360 seconds, light at 300000000m/
sec will transit a distance of 134.16408m (300000000m/
sec*0.000000044721360sec) but if light had slowed down to 223607021m/
sec, light would exactly cover the original 100m in the dilated time
of 0.000000044721360sec (223607021m/sec*0.00000044721360sec=100m).
Clearly, if the speed of light had not been reduced, the law of
physics d=v*t would have been violated.

Peter Riedt
This relates to the concept of the length contraction. *Length
contraction does not imply that the object in motion is shorter.
Actually, it is be longer.

No it is not .. why do you insist on this nonsense?

In your example, L = 100m, c = 300000000m/s and t = 0.000000333333sec
for v = 200000000m/s, L' = 74.535599m

What it means is that length of 100m in the stationary system is
equivalent to 74.5m in the moving system.

Sort of.

Length of the rod still 100m in the moving system. *Its equivalent
length in the stationary system is 100* (100/74.5) = 134.164m.

No .. it is 75.4 .. where did you pull than figure from?

The equivalent length of the rod is longer in the moving system!!!

No .. it is not

Explanation:

This should be good for a laugh

The same distance between 2 points measured in the moving
system is shorter because the measuring metre stick is longer.

The contracted length we see comes from the measurement of the length of the
moving object from within the stationary system using a stationary mater
stick . That stick is the same length as when it measures a non-moving
obejct .. it doesn't grow because you are measuring something that is moving
!!!

Object
elongates at the same rate as the metre stick so the measured length
is unchanged.

So .. now you have changed to talk about a moving metre stick measureing a
moving object. * In the moving frame there is no change to either the stick
of the object. *If the stationary frame measures the moving ruler they will
get a shorter length in exactly the same way that the object has a short
length.

Measurement is alway in the same system. We cannot use a moving meter
stick to measure the length of the object in the stationary frame or
vise versa.

*So as to compare to the stationary system, the moving
object is longer.

No .. it is shorter. *GEes you are both stubborn and incredibly stupid ...
contraction means 'shorter'. *It doesn't mean longer.

Due to time dilation, the equivalent time in the stationary system is
= 0.00000044721360sec which is also longer

You are making the same mistakes as the OP regarding what time dilation
means.

d = 134.164 m (equivalent length of 100m of the moving system)

The 100m rod is 100m long in the moving system. *A 134.164 rod in the moving
system would appear to be 100m in the stationary system. *You are completely
confused about Sr and LT here.

This is the concept most relativists get wrong. A 134.164 m rod in
the moving system does not appear to be 100 m in the stationary
system. Instead, a 100 m rod in the moving system appears to be
134.164 m long in the stationary system.
------------------------------------
Let v = 0.995c, Gamma = 10
x = 10c (a star is 10c from Earth measured in the stationary system)

Due to length contraction
x' = x/Gamma = c (the star is c from Earth measured in the moving
system)

A length of c in the moving system appears to be 10c long in the
stationary system. If we divide both sides by c, we have:
A length of 1 m in the moving system appears to be 10 m long in the
stationary system.

One explanation was that the distance between Earth and the star
shrinks 10 time when an observer is moving at 0.995c. We know that it
is not possible. The only logically explanation is that the measuring
metre stick is stretched 10 times at 0.995c so that the measurement of
the same distance seems to be 10 times shorter.
Since, the moving metre stick is 10 times longer so are all the moving
objects.

v = 300000000 m/s
t = 0.00000044721360 sec (equivalent time of 0.000000333333s of the
moving system)
v * t = 3 * 10^8 * 44.72136 * 10^(-8) = 134.16 = d
Please note that both of d and t are longer in the moving system.

No .. they clearly are not .. not to anyone who has a correct understanding
of the physics involved.

You *know* that you do not (yet) understand SR .. why are you inflicting
your misunderstanding onto someone else?- Hide quoted text -

- Show quoted text -

"snapdragon31" wrote in message
...
Measurement is alway in the same system.

No .. it is not

We cannot use a moving meter
stick to measure the length of the object in the stationary frame or
vise versa.

Yes .. we can.

So as to compare to the stationary system, the moving
object is longer.

No .. it is shorter. GEes you are both stubborn and incredibly stupid ..
contraction means 'shorter'. It doesn't mean longer.

Due to time dilation, the equivalent time in the stationary system is
= 0.00000044721360sec which is also longer

You are making the same mistakes as the OP regarding what time dilation
means.

d = 134.164 m (equivalent length of 100m of the moving system)

The 100m rod is 100m long in the moving system. A 134.164 rod in the
moving
system would appear to be 100m in the stationary system. You are
completely
confused about Sr and LT here.

This is the concept most relativists get wrong.

Here we go again

A 134.164 m rod in
the moving system does not appear to be 100 m in the stationary
system.

Yes .. it does

Instead, a 100 m rod in the moving system appears to be
134.164 m long in the stationary system.

Wrong .. just plain wrong. Do you not get it .. length contraction means
length contraction !!!???

Lets watch you stuff it up again...

------------------------------------
Let v = 0.995c, Gamma = 10
x = 10c (a star is 10c from Earth measured in the stationary system)
Due to length contraction
x' = x/Gamma = c (the star is c from Earth measured in the moving
system)

Where x is a length of an object at rest in the rest frame, and x' is how
long it is measured by an observer in the moving frame

And c is less than 10c .. so the length is conrtacted

The same but opposite formula applies going in the opposite rtansform

x = x'/Gamma

Where x' is a length of an object at rest in the moving frame, and x is how
long it is measured by an observer in the moving frame. Note that the
meaning of x' and x in this equation are NOT the same as in the first
equation.

So a length of an object that is 10c at rest in the moving frame is measured
as c by an observer in the rest frame.

Both ways the length gets shorter.

That is what most people who are ignorant of SR and LT fail to understand

It might help you to understand it, if it is written as

L_moving = L_rest / Gamma

where L_rest is the length of the object at rest, and L_moving is the length
of the object as measured from a relatively moving frame of reference,.

[snip more nonsense from completely backward misintpretation of SR and LT]

On Feb 3, 11:20*pm, "Jeckyl" wrote:
"snapdragon31" wrote in message

...





On Feb 1, 10:49 pm, "Jeckyl" wrote:
Let's look at this more clearly...

Let us assume our coordinates are such that at x=0,t=0 we also have x'=0,
t'=0
Let us assume that we have a moving rod of length L travelling at speed v
Let us assume the light is shining along the rod in the direction of
travel
Let us assume the light is emitted at t=t'=0, when the rear end of the
rod
is at x=x'=0

In the stationary frame of reference, the rod is moving while the light
is
travelling at c
So the light gets to the other end of the rod, as seen by the stationary
observer, at x, t where
t = L / (c - v)
x = c . t ... because x / t = c

That corresponds in the rods frame of reference to a point (x',t')

x' = gamma . ( x - v.t ) ... Lorentz
x' = gamma . ( c.t - v.t ) ... subs for x
x' = gamma . ( c - v ) . t ... factorise
x' = gamma . ( c - v ) . L / (c - v) ... subs for t
x' = gamma . L ... cancel

t' = gamma . ( t - v.x / c^2 ) ... Lorentz
t' = gamma . ( t - v.t / c ) ... subs for x
t' = gamma . (1 - v/c ) . t ... factorise
t' = gamma . (c - v) / c . L / (c - v) ... subs for t
t' = gamma . L / c ... cancel
Please try and come up something better. *I believe you can because
you are smarter than the other guy.

Thanks for your patience .. you picked up my error in the above at the line:

Let us assume that we have a moving rod of length L travelling at speed v

It should have said that the L is the length of the moving rod measured in
the stationary frame (ie the contracted length)
Your speed is incredible. I admit that I cannot think that fast.
The equations look good. But I still need to plug in some numbers to
have a better feeling of what they are.

L' = 100m, Gamma = 1.34164
== L = 100/1.34164 = 74.54m


Of course, one can do the analysis the other way .. by seeing that light
moving along the stationary rod at c transforms to light moving along the
moving rod at c

Lets assume the length of the stationary rod in the stationary frame is L
When light is emitted, we have at x=x'=0, t=t'=0

When light gets to the other end of the rod, we have x=L, t=L/c.
That point corresponds to (x',t') where
x = L = 74.54 m
t = L/c = 2.48E-07 = 0.000000248 sec
Should t be L/(c-v)?

x' = gamma . ( x - v.t ) ... Lorentz
x' = 1.34 * (74.54 - 2E8 * 2.48E-07) = 33.33 m
(Sorry, it is not equal to 100m)

x' = gamma . ( L - v.L/c ) ... subs x=L, t=L/c
x' = gamma . (c - v) . L / c ... factorise

t' = gamma . ( t - v.x / c^2 ) ... Lorentz
t' = gamma . ( L/c - v.L / c^2 ) *... subs x=L, t=L/c
t' = gamma . ( c - v) L / c^2 *... factorise

x'/t' = ( gamma . (c - v) . L / c ) / ( gamma . ( c - v) L / c^2 ) = c

Again, the speed of light is c- Hide quoted text -

- Show quoted text -

Another way to prove x' / t' = c, given x = c * t is
x' = gamma . ( x - v.t ) ... Lorentz
t' = gamma . ( t - v.x / c^2 ) ... Lorentz

x'/t' = [gamma . (x - v.t)]/[gamma . (t - v.x / c^2)]
= (x - v.t)/(t - v.x / c^2)
= (c.t - v.t)/(t - v.c.t / c^2) substitute x = c.t
= [t.(c-v)]/[t.(1-v/c)]
= (c-v)/[(c-v)/c]
= c

In this case, there is no way to check what x, t, x' and t' are. But
one thing is sure that
if t t' due to time dilation then
x must be greater than x', right?

Talk to you next week.

"snapdragon31" wrote in message
...
On Feb 3, 11:20 pm, "Jeckyl" wrote:
"snapdragon31" wrote in message
...
On Feb 1, 10:49 pm, "Jeckyl" wrote:
Let's look at this more clearly...
Let us assume our coordinates are such that at x=0,t=0 we also have
x'=0,
t'=0
Let us assume that we have a moving rod of length L travelling at
speed v
Let us assume the light is shining along the rod in the direction of
travel
Let us assume the light is emitted at t=t'=0, when the rear end of the
rod
is at x=x'=0

In the stationary frame of reference, the rod is moving while the
light
is
travelling at c
So the light gets to the other end of the rod, as seen by the
stationary
observer, at x, t where
t = L / (c - v)
x = c . t ... because x / t = c

That corresponds in the rods frame of reference to a point (x',t')

x' = gamma . ( x - v.t ) ... Lorentz
x' = gamma . ( c.t - v.t ) ... subs for x
x' = gamma . ( c - v ) . t ... factorise
x' = gamma . ( c - v ) . L / (c - v) ... subs for t
x' = gamma . L ... cancel

t' = gamma . ( t - v.x / c^2 ) ... Lorentz
t' = gamma . ( t - v.t / c ) ... subs for x
t' = gamma . (1 - v/c ) . t ... factorise
t' = gamma . (c - v) / c . L / (c - v) ... subs for t
t' = gamma . L / c ... cancel

Please try and come up something better. I believe you can because
you are smarter than the other guy.

Thanks for your patience .. you picked up my error in the above at the
line:

Let us assume that we have a moving rod of length L travelling at
speed v

It should have said that the L is the length of the moving rod measured
in
the stationary frame (ie the contracted length)

Your speed is incredible.

It helps when you know what you're talkin about .. even if one can stuff up
occasionaly on expressing it . Indeed .. the equations would look better if
I used L' instead of L throughout .. not that it changes anything

I admit that I cannot think that fast.
The equations look good. But I still need to plug in some numbers to
have a better feeling of what they are.

L' = 100m, Gamma = 1.34164
== L = 100/1.34164 = 74.54m

OK


Of course, one can do the analysis the other way .. by seeing that light
moving along the stationary rod at c transforms to light moving along the
moving rod at c

Lets assume the length of the stationary rod in the stationary frame is L
When light is emitted, we have at x=x'=0, t=t'=0

When light gets to the other end of the rod, we have x=L, t=L/c.
That point corresponds to (x',t') where

x = L = 74.54 m

No .. in this case, L = 100m

t = L/c = 2.48E-07 = 0.000000248 sec
Should t be L/(c-v)?

No .. but your L is wrong .. L is the length of the stationary rod in the
stationary frame

x' = gamma . ( x - v.t ) ... Lorentz
x' = 1.34 * (74.54 - 2E8 * 2.48E-07) = 33.33 m
(Sorry, it is not equal to 100m)

Because you stuffed up your L

x' = gamma . ( L - v.L/c ) ... subs x=L, t=L/c
x' = gamma . (c - v) . L / c ... factorise

t' = gamma . ( t - v.x / c^2 ) ... Lorentz
t' = gamma . ( L/c - v.L / c^2 ) ... subs x=L, t=L/c
t' = gamma . ( c - v) L / c^2 ... factorise

x'/t' = ( gamma . (c - v) . L / c ) / ( gamma . ( c - v) L / c^2 ) = c

Again, the speed of light is c

Another way to prove x' / t' = c, given x = c * t is
x' = gamma . ( x - v.t ) ... Lorentz
t' = gamma . ( t - v.x / c^2 ) ... Lorentz
x'/t' = [gamma . (x - v.t)]/[gamma . (t - v.x / c^2)]
= (x - v.t)/(t - v.x / c^2)
= (c.t - v.t)/(t - v.c.t / c^2) substitute x = c.t
= [t.(c-v)]/[t.(1-v/c)]
= (c-v)/[(c-v)/c]
= c

Indeed

In this case, there is no way to check what x, t, x' and t' are.

But if they are related by a Lorentz transformation, then the speed of light
is c.

But
one thing is sure that
if t t' due to time dilation then

You don't understand time dilation and simulatneity .. t is less that t'
because we are looking at different positions in space

x must be greater than x', right?

At least you're thinking .. but you need to do some more learning first.