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Why I think that SR might be wrong?


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#1 dkv

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Posted 03 August 2008 - 12:38 AM

I have issues with relativity... remember SR ?
It says that space and time transfrom based on following equation.
x'=Y(x-v.t) where x is the coordinate of a point in S frame of reference ,v is relative velocity between S and S' frame of references.. and t is the time passed since the synchronization of clocks.Y is the transformation factor.
similarly
t'=Y(t-v.x/c^2)
Note that t' depends on x...
1.If the two frame of reference are following a coordinate at an arbitary large distance then t' can become negative...
becuase t-v.x/c^2 becomes negative .. In other words the S perceives that clock in S' in running into past...(for any arbitary small value of v)

2.We also find that t' can become positive if the x coordinate is small...
In other words S percieves that S' time is not isotropic...
Which is a contradiction therefore SR is wrong.

3.One more point.In the equation
t'=Y(t-v.x/c^2)
if we put t=0 then we get
t' = - Y(v.x/c^2)
which is a negative value .. in other words at the time of synchronization the time in S' appears to be moving backward.
Which is absolutely wrong.


If SR is wrong then GR is also wrong because it based on SR.
Good amount skepticism is required to understand the idea... SR has been ingrained deeply into our psyche.

#2 Jay-qu

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Posted 03 August 2008 - 02:15 AM

1 and 3 are one in the same, what the equations are showing you is correct, your reasoning is not. The time co-ordinate is negative, that does not mean time is going backwards. Try incrementing the time in the unprimed frame you will see that the time will increment forwards in the primed frame also.

2. I dont understand your reasoning.. but consider that even direction is relative in SR

#3 dkv

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Posted 03 August 2008 - 02:41 AM

1 and 3 are one in the same, what the equations are showing you is correct, your reasoning is not. The time co-ordinate is negative, that does not mean time is going backwards. Try incrementing the time in the unprimed frame you will see that the time will increment forwards in the primed frame also.

2. I dont understand your reasoning.. but consider that even direction is relative in SR


2.In the reference frame which is at rest the time is istropic..It doesnt depend on the space coordinates of observation. There is one and only one clock is SR.
However when the observer looks at moving frame of reference he finds that there is not one clock but many depending upon the coordinates in discussion.
t'=Y(t-x.v/c^2) and for two different locations of observations the time passes anisotropically.
Which violates the principle of one frame one clock.

1,3.I say time start to move backward because observer at rest can observe a series of points such that the tranformed time can start decreasing...
There is no difference between 1,3 and 2... only the consequence of anisotropy are different. In this case it allows a backward movement of time.
x= 0 to x=infinity gives t'= t(almost) to minus (infinity)
(the time moves from positive value to negative value )

#4 modest

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Posted 03 August 2008 - 10:48 AM

t'=Y(t-v.x/c^2)
Note that t' depends on x...

It also depends on t, which is important

1.If the two frame of reference are following a coordinate at an arbitary large distance then t' can become negative...
becuase t-v.x/c^2 becomes negative


As x gets arbitrarily large, so does t. So, it's not immediately apparent what happens. But, first...

Consider you are in deep space in your motionless reference frame. The Enterprise goes flying by at sub-light speeds along the x axis. As it passes, you synchronize your clocks so that t=t'=0. Your clock is t and Captain Kirk's is t'.

Now, you have declared that as the Enterprise gets "arbitrarily" far away, their clock will appear to have negative numbers, so let's check. The first logical step is figuring what speed is our limiting factor. Will Enterprise's clocks appear to read less from your perspective if it's going slow or fast? Let's examine the equation:

[math]t' = \gamma(t - \frac{vx}{c^2})[/math]

x is how far away Enterprise is
v is how fast its going
t is what your clock reads
t' is Enterprise's clock (we want to know if we can make it negative)

It should be immediately obvious that we want the left side of the minus sign to be small and the right side to be large;

[math]t' = \gamma({\color{red} t} - \frac{{\color{blue} v}{\color{blue} x}}{c^2}) [/math]


To keep t small, and increase v and x, we want enterprise to go at the greatest value of v possible. So, we'll say Enterprise is traveling at warp 1, the speed of light. This will set our bound showing us the smallest (or most negative) that Enterprise's clock can read at an arbitrary x from your frame's perspective.

velocity = c, reducing the equation to:

[math]t' = \gamma(t - \frac{x}{c})[/math]

The distance Enterprise travels is the speed of light times time (x=ct);

[math]t' = \gamma(t - \frac{ct}{c})[/math]

and we get;

[math]t' = \gamma(t - t)[/math]

or:

[math]t' = 0[/math]


So, at any arbitrary distance, the smallest Enterprise's clock can read is zero (assuming it cannot travel faster than light). It will not read a negative number. In fact, we should stipulate that a real spaceship cannot reach the speed of light and the clock will always read positive. This should make sense considering at the speed of light time dilation is infinite and Enterprise's clock would not advance from your frame. At speeds less than c, t' will advance more slowly than your t, but it does advance in the positive direction.

If you find you still can't get these transformations to work, then give us a specific example of speed, distance and time and we'll work it out.

~modest

#5 dkv

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Posted 04 August 2008 - 06:04 AM

I think you are mistaking the formula.
v is the velocity of the frame S' with respect to S
x is the point under observation which need not be measured because it is the x-coordinate in the S frame.. it is sufficient to know that such a point exists.
Point at x is not moving
t is the time since the synchronization of clocks between S and S'
(i.e when the origins of S and S' were coincidental or when the x was equal to 0..once the clocks were synchronized the forumla works well for any arbitary point x..clock synchronization is necessary to have meaningful relative time)

S frame has following values : x,t , v
The observer in S wishes to find out the coordinates x' ,t' in S' relatively.

we know time is related by equation
t'=Y(t-x.v/c^2)
if the point x is 4 light years away (c^2 meters distance away) then the equation becomes
t'=Y(t-v)
if v>t then t' <0

Which is wrong.

Another main point is that the t' varies if the point under discussion changes..
You can imagine that multiple persons sitting in the S frame and each observing {x1,x2,x3...} points then every observer in S will find a different time in S'.
t'1=Y(t-x1.v/c^2)
t'2=Y(t-x2.v/c^2)
...
WHich means that time in S' is not isotropic.
Where as in S the time is isotropic...

The formula intrinsically handles time delays.
And there is no need to worry about actual measurement because in SR coordinates can be extended till infinity ... I used a point which was reachable...(a star 4 light years away)

#6 Erasmus00

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Posted 04 August 2008 - 10:35 AM

2.In the reference frame which is at rest the time is istropic..It doesnt depend on the space coordinates of observation. There is one and only one clock is SR.
However when the observer looks at moving frame of reference he finds that there is not one clock but many depending upon the coordinates in discussion.


All of the clocks that are stationary in the observers rest frame tick at the same rate, this is what we mean when we say time is isotropic in an inertial frame.

However, if a stationary observer watches moving clocks, the isotropy is ALREADY broken- there is a preferred direction (that of the motion of the observed). Hence, a grid of clocks moving together will not tell time isotropically.

1,3.I say time start to move backward because observer at rest can observe a series of points such that the tranformed time can start decreasing...


But no one clock can occupy those points. You cannot have any single clock that ever has time tick backwards.
-Will

#7 dkv

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Posted 05 August 2008 - 07:43 AM

Time backwards doesnt mean that if change in t>0 then the change in t' <0
What it means is that certain points might appear in past with respect to present...
All changes are in future.. i.e del t > 0 and del t'>0 but orgins in time can vary...
When observing x=0 at t=0 the t'=0 but at x>0 the t'<0 (for t=0) which means at x= 0 I see today... but at x>0 I might see yesterday.. however as I progress in time the point at x>0 also moves into future...
S finds that S' is living with multiple realities...which is nonsense...

However, if a stationary observer watches moving clocks, the isotropy is ALREADY broken- there is a preferred direction (that of the motion of the observed). Hence, a grid of clocks moving together will not tell time isotropically.

There is a preferred direction which results in anisotropic time... but it is physically unacceptable because we know that moving frame carries only ONE measuring rod and only one clock... It binds the relative observer's time with the observables in space.
Imagine S and S' observing an atom (it has a finite size) The S frame finds that S' is observing that different points on the atoms are living at different times.
Such a conclusion violates the integrity of Atom.. If we replace atom with human then SR predicts that S' perceives the points on human body as aging anisotropically. Why should the relative velocity be allowed to violate the integrity of Unity? I am one.. I am 30 years old. Why should anyone find that my body is living at different times ?If I catch a disease now then in some reference frame half of my body will appear diseased and half will catch the disease later.
I had never read such a nonsense called SR in my life before.

#8 modest

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Posted 05 August 2008 - 01:46 PM

t'=Y(t-v)
if v>t then t' <0


Why don't you show a situation with a distance, velocity, time elapsed since t=t'=0 for each clock. Give us an example where t' is less than zero or trust my proof above showing that it never is with the speed of light constraint.

Fact is, if two clocks are at x=x'=t=t'=0 they can then go anywhere and do anything they want (under the speed of light) and never observe the other clock at less than t=0. NEVER.

~modest

#9 dkv

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Posted 05 August 2008 - 11:52 PM

at x=x'=t=t'=0 means that the clocks at the position x=x'=0 were synchronized.
there is no doubt that formula t'=Y(t-x.v/c^2) uses the time 't' fixed by the synchronization at t=t'=0 and x=x'=0.
let us put t=0 in the equation. (I take a more simple case...probably you think that x should be measured when actually it is nothing but a coordinate point...)
we get t'= -Y(x.v/c^2) i.e at any distance x<>0 the t' is not zero which means that synchronization is not simultaneous spatially.
And this violates the integrity of references ... every reference frame can have only one clock... no matter from where we see it.
As I said the problem doesnt look obvious unless we choose something which
carries a sense of unity in time.. like Self or Atom or Insects etc...
If we replace the point at x with a collection of points such that they represent an insect or atom or yourself.. then will it make sense to know that person in S frame thinks that S' is observing a collective unity as a distributed one? I am made up of several points in space.. and no matter which frame of reference you take all points on body have exactly lived the same amount... it doesnt vary with respect to any event measurement..like synchronization.
t'=-Y(x.v/c^2) then at the time of synchronization i.e t=0 some points on my body are in past with respect to the event at t=0 and some points on my body are in future with for x<0 with respect to the event at t=0....
as t increases to 1 all the points on my body move into the future but still they appear to be living at different times. Which is wrong.
Einstein has committed a horrrendous mistake. I am surprised that he was chosen Times personality of the century!!

#10 modest

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Posted 06 August 2008 - 04:35 AM

there is no doubt that formula t'=Y(t-x.v/c^2) uses the time 't' fixed by the synchronization at t=t'=0 and x=x'=0.


Good, that is the standard configuration and I'm glad you use it. I will depict the standard configuration for the sake of our conversation:

Posted Image

let us put t=0 in the equation.


Ok.

(I take a more simple case...probably you think that x should be measured when actually it is nothing but a coordinate point...)


I've made no statements regarding what should be or should not be measured.

we get t'= -Y(x.v/c^2) i.e at any distance x<>0 the t' is not zero which means that synchronization is not simultaneous spatially.


Ok. I see where we’ve been talking past each other. When you’ve been referring to a change in x, I figured you meant a change in x due to v (over t) -- rather than *different values in x’ along the x’ axis (at the same t)* That was a misunderstanding on my part. Ok, I think I can answer your question.

Consider two spaceships that are each one lightyear long in their own frame. The captain of each ship is in the center, half way (or half a lightyear) from both the front or back of the ship. The first ship, S, sits in a ‘motionless’ reference frame while the second ship, S’, passes (along the X axis) going half the speed of light relative to the first ship.

There are three clocks on each ship. One at in the front, one with the captain in the middle, and one in the back. Both captains synchronize the clocks on their ship. They do this with light-pulses such that the captain will see both clocks one half year slow. The captain, however, knows light travels one lightyear per year, so he knows the clocks all display the same time.

The question is, according to the Lorentz transformations, what does each captain see when they pass each other. To answer this, we’ll say both captain’s clocks read zero at the moment the two captains pass. According to the captain in S, we can say this about S’:

The front of S’ is .43 lightyears away. The back of S’ is .43 lightyears away. The clock at the front of S’ reads 0.22 years less than a synchronized clock and the one in the rear reads 0.22 years more than a synchronized clock. Moreover (and this is the important part):

If the captain of S (the ship not moving) sees at this moment the captain of S’ flash a light bulb then the light from that bulb will reach the rear of S’ before it reaches the front (according to S) [if you disagree with this, I will explain more fully]. So, NO, other frames of reference do not preserve simultaneity - nor should they.

Relativity of simultaneity - Wikipedia, the free encyclopedia

And this violates the integrity of references ... every reference frame can have only one clock... no matter from where we see it.


Do you have a source agreeing with you on that?

Every coordinate in a reference frame has the same time - from the perspective of that reference frame only. The fact that frames of relative motion do not preserve this is the whole reason we need the Lorentz transformations.

There’s a good webpage here that explains in more detail:

The Lorentz Transformations

Now consider how this string of clocks appears as viewed by O from frame S. First, since they are all moving at speed v, they will be registering time more slowly by the usual time dilation factor than O’s own physically identical clocks. Second, they will not be synchronized. From the clocks on a train argument in the last lecture, if the clocks are L apart as measured by O′, successive clocks to the right (the direction of motion) will be behind by Lv/c2 as observed by O.

It should be mentioned that this lack of synchronization as viewed from another frame only occurs for clocks separated in the direction of relative motion. Consider two clocks some distance apart on the z′ axis of S′. If they are synchronized in S′ by both being started by a flash of light from a bulb half way between them, it is clear that as viewed from S the light has to go the same distance to each of the clocks, so they will still be synchronized (although they will start later by the time dilation factor).


~modest

#11 dkv

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Posted 06 August 2008 - 06:46 AM

Consider two spaceships that are each one lightyear long in their own frame. The captain of each ship is in the center, half way (or half a lightyear) from both the front or back of the ship. The first ship, S, sits in a ‘motionless’ reference frame while the second ship, S’, passes (along the X axis) going half the speed of light relative to the first ship.

There are three clocks on each ship. One at in the front, one with the captain in the middle, and one in the back. Both captains synchronize the clocks on their ship. They do this with light-pulses such that the captain will see the forward clock reading one half year ahead and the rear clock reading one half year behind. The captain, however, knows light travels one lightyear per year, so he knows the clocks all display the same time.

The question is, according to the Lorentz transformations, what does each captain see when they pass each other. To answer this, we’ll say both captain’s clocks read zero at the moment the two captains pass. According to the captain in S, we can say this about S’:

The front of S’ is .43 lightyears away. The back of S’ is .43 lightyears away. The clock at the front of S’ reads 0.22 years less than a synchronized clock and the one in the rear reads 0.22 years more than a synchronized clock.

Why should S see the S' having different clocks... The problem doesnt reduce to just 3 clocks... all the intermediate points appear to have different times which can be more of less than the clock at center.

Moreover (and this is the important part):

If the captain of S (the ship not moving) sees at this moment the captain of S’ flash a light bulb then the light from that bulb will reach the rear of S’ before it reaches the front (according to S) [if you disagree with this, I will explain more fully]. So, NO, other frames of reference do not preserve simultaneity - nor should they.


As you said from the point of view of S different parts of the ship S' is living at different times.
This leads to loss of integrity... I emphasize this because if we replace the ship with some monster man or monster snake then the problem of time becomes very clear... If the snake is poisoned at t=0 then S will find that Half the snake is dead and half is alive because some of his body parts are still living in past.(i.e it is still waging head but its tail is already dead or vice versa)
In other words unity of experience gets violated...

Suppose a Universe exists in which only moving unstoppable inertial frames are found then the Observer S who thinks that he is at rest will fail to comprehend the laws of physics in the moving frame of reference...because he will have no means to find out whether the moving frame is having any uniform experience.
We are priveldged to find inertial frames which can be stopped to find the actual experience of the observers.
The communications between them will also suffer from irresolvable communication dispute. What S thinks as not simultaneous is communicated as simultaneous....This information paradox can not resolved without stopping the moving frame because he will have no means to find out whether there is one time in the frame S' or not?
Note that one can not create a inertial frame without loosing one's own inertial state because force is required to accelerate a body which is at rest.


Do you have a source agreeing with you on that?

I already said that SR has been accepted and "validated" by experiments.
There is no one you will find who will agree with me.

Every coordinate in a reference frame has the same time - from the perspective of that reference frame only. The fact that frames of relative motion do not preserve this is the whole reason we need the Lorentz transformations.

~modest

Because speed of light must remain constant we are violating integrity of physical experience ... and we are also making the laws of Universe underivable in a Universe made up of only inertial frames.


It is important to note that point x in the formula need not lie on the frame S'. In other words x can be stationary with respect to S.
The time changes simply because the point of discussion x between S and S' changes.

murder

#12 Erasmus00

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Posted 07 August 2008 - 11:25 PM

Why should S see the S' having different clocks... The problem doesnt reduce to just 3 clocks... all the intermediate points appear to have different times which can be more of less than the clock at center.


This is the basic relativity of simultaneity problem. It lies at the heart of relativity. Points that are simultaneous in one frame (all clocks ticking 1:00 for instance) are not simultaneous in another.

This leads to loss of integrity... I emphasize this because if we replace the ship with some monster man or monster snake then the problem of time becomes very clear... If the snake is poisoned at t=0 then S will find that Half the snake is dead and half is alive because some of his body parts are still living in past.(i.e it is still waging head but its tail is already dead or vice versa)
In other words unity of experience gets violated...


Unity of experience is a phrase you've invented with no real useful application. Consider your poisoned snake- you can't poison a snake all at once, the poison circulates through its blood, and parts will die at different rates (depending on how the toxin works).

Suppose a Universe exists in which only moving unstoppable inertial frames are found then the Observer S who thinks that he is at rest will fail to comprehend the laws of physics in the moving frame of reference...because he will have no means to find out whether the moving frame is having any uniform experience.


An inertial frame is a coordinate construct, not a physical object. There cannot be a universe where only inertial frames exist, any observer can always use any coordinates- coordinate independence is a basic property of coordinates that comes with the territory- you can't have coordinates and not coordinate independence.

Note that one can not create a inertial frame without loosing one's own inertial state because force is required to accelerate a body which is at rest.


Your clearly thinking of coordinates as something more than labels, which isn't the best way of thinking.

Because speed of light must remain constant we are violating integrity of physical experience ... and we are also making the laws of Universe underivable in a Universe made up of only inertial frames.


Only by postulating relativity can you derive the physical effects of things like magnetism. Far from making the physical laws underivable, we allow them to be derived in a simple, elegant fashion.
-Will

#13 dkv

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Posted 08 August 2008 - 01:35 AM

We should have expected a relativity of a different kind in which the every reference frame should have had one and only clock... The relative tranformation should have not have violated the one clock principle..
From the S frame of reference the Snake appears to be living at different times at different points. This not due to the delay in the spread of the poison..(which is vaid in S' frame) but due to the transformation. If the head is poisoned now then the tail remains unaware of the event because it is living in the past.

In other words the tail is jumping with pride but the head is already dead.

Can awareness of an event be subjected to the limits of light? Dont we know without even knowing what we know?
Can we define Unit existence in the Special relativity ?
Can we define integrity of life in SR ? No we cant. The whole idea of nonsimultaneous relative points destroys the very basis of physics... i.e the observer. Relativity destroys the observer.

An inertial frame is a coordinate construct, not a physical object. There cannot be a universe where only inertial frames exist, any observer can always use any coordinates- coordinate independence is a basic property of coordinates that comes with the territory- you can't have coordinates and not coordinate independence.

I can imagine a universe in which there are no forces(not even gravity) but only moving particles..
There is no law which says that space and time must and must create the forces of gravity. GR uses the observation to relate Gravity and geometry of space time but it doesnt say why gravity must exist or why mass should curve the space and time.

Relativity makes the solutions complicated and complex...
Relativity uses only coordinates to derive the space and time transformations.
When actually no measurement can be made without changing the inertial state of the system. This gives me another reason why it must be wrong because SR is senstive to symmetries(twin paradox).Symmetries get broken as when observation is made and the inertial frames become non-inertial due to change in the momentum of the S and S'..
we consider only geometries(and not masses of S and S') thats why we are able to frame the equations.

#14 dkv

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Posted 12 August 2008 - 02:07 AM

Someone said I have a hidden agenda against SR..
I challenge you to provide reasons why the following should be true:
t'=Y(t-x.v/c^2)

You cant provide any reason other than what Relativity postulates.
I use the same postulates to prove that the equation is wrong. Simply wrong.

In the equation t'=Y(t-x.v/c^2) if we put v=0 we t'=t i.e the t' is same as t no matter where it exists... i.e independent of space.

However note that because speed of light is constant the observer at x=0 and t=0 is not at existing simultaneously with respect to the space...
Clock at distance x from the observer begins at a different time.
The two clocks are related by equation
t'=t-x/c
That is why when we look into the sky we say that we are look into the past.

BUt the SR equation predicts that t=t' if v=0 ... which is absolutely wrong...

v=0 nothing but one single frame at rest.

#15 modest

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Posted 12 August 2008 - 03:07 PM

Someone said I have a hidden agenda against SR..


Not hidden.

I challenge you to provide reasons why the following should be true:
t'=Y(t-x.v/c^2)


Experimental verification of Lorentz invariance and confirmation of SR:

In the equation t'=Y(t-x.v/c^2) if we put v=0 we t'=t i.e the t' is same as t no matter where it exists... i.e independent of space.


At v=0 a Lorentz transformation should be equivalent to a Galilean transformation which is t' = t. It is.

However note that because speed of light is constant the observer at x=0 and t=0 is not at existing simultaneously with respect to the space...


"with respect to the space" is a phrase you should avoid as Will explains above. You are defining simultaneity wrong. Two events are simultaneous if an observer half way between those two events and comoving with them sees both happen at the same time.

BUt the SR equation predicts that t=t' if v=0 ... which is absolutely wrong...

v=0 nothing but one single frame at rest.


For relative speeds much less than the speed of light, the Lorentz transformations reduce to the Galilean transformation in accordance with the correspondence principle.

-wikipedia


In other words, [math]t' = \gamma(t - \frac{vx}{c^2})[/math] reduces to t' = t. Which it does. There is no problem.

~modest

#16 dkv

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Posted 12 August 2008 - 11:35 PM

The Galilean Transformation assumes infinite speed of light.
Therefore the actual Galilean Tranformation ,if we assume finite speed of light, are:
x'=x-vt
and
t'=t-a/c where 'a' is distance between two points seperated by distance 'a'.(for simplicity we assume 'a' to lie on the x coordinate)
if point 'a' is not moving then
del t'= del t i.e the clocks tick at equal intervals.

if point 'a' is moving away from the point at origin then
del t'= del t(1-v/c) where v is velocity of reference frame at 'a' or anywhere on x axis. The clocks do not tick at equal intervals.
If v<<c then again we get
del t'=del t

Let us call those equations as modified Galilean Transformation due to constant speed of light.

(Now if we further assume that due to relativity the time and space transform
then
x' = Y'' x (x-vt)
t'=Y'' x t(1-v/c) which offcourse is a matter of speculation.. )

Since we do not discuss the tranformations of equation as above we can assume that SR tranformations are true but we should expect that SR equations to reduce to modified Galilean tranformations at low speeds.

Let us find out whether this happens or not.
We take only the time coordinate.

we know t'=Y(t-x.v/c^2)
just to repeat what those variables stand for.
x is the point under observation which may or may not be moving with respect to frame S which is at rest.
t and t' are measured from the point of origin i.e when x=0 t=0 we get t'=0
v is velocity of frame S'

At the time when t'=t=0 we can find out poisition of other clocks with respect to any point under discussion
t'(at x)=-Yx.v/c^2 (t' at x with respect to t=0 at x=0)

Note that x is point on the x coordinate if x>0 t'<0 if x<0 t'>0 (which is wrong because clocks can not start in future no matter where it is with respect to the instant t=0.. but we give benefit of doubt and use the equation as it is)

If v=0 we get t'(at any x with respect to t=0) = 0 (i.e if I start my stop watch now then all points in space start their clocks now which is a wrong conclusion.)

Compare this with modified Galilean tranformation
if v=0 we get t'= t- a/c ; a is distance on x-axis from the point of origin so we can write for convinience t'=t-x/c ie all point points with respect to origin lie in the past which is a correct conclusion!!
If I start my stopwatch now then the clocks near the stars start in the distant past because they are billions of light year away.

Hence the time equation of SR does not reduce to modified Galilean tranfromation at 0 velocity therefore the SR is inconsistent.

The same problem can found if we compare the equations at some finite velocity of S'
At x=0 ,as per the SR ,del t'= Ydel t
but the modified Galilean tranformation says that
del t'= Y'del t = del t x (1-v/c) ;Y<>Y'
in other words even at small velocities at x=0 the clocks in S' tick slowly as compared to S because Y'<1.
Whereas SR claims that at x=0 the clock of S' ticks more frequently than S because Y >1.
Which is again inconsistent.
Therefore,
the SR transformation do not reduce to Modified Galilean Transformations.

#17 Erasmus00

Erasmus00

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Posted 13 August 2008 - 12:05 AM

Can we define Unit existence in the Special relativity ?
Can we define integrity of life in SR ?


I fail to see why either of these are meaningful concepts, with or without relativity. What do you mean by unit existence? What do you mean by integrity of life? These aren't well defined anyway!

I can imagine a universe in which there are no forces(not even gravity) but only moving particles..


Thats fine, but even still there are infinitely many choices for coordinate systems, some of which will be inertial, some of which won't be. I can make any coordinate system I want, regardless of the universe I am in.

Relativity makes the solutions complicated and complex...
Relativity uses only coordinates to derive the space and time transformations.


Because relativity is a theory that is really about coordinates!
-Will