Attention Paid To Accelerating Reference Frames Overthrows SR

[StarThrower]

1. Let clock A and clock B be of identical construction, and let them both not be subjected to any force. Let them be at rest with respect to each other. Thus, the relative velocity v is equal to zero.

Therefore, using newtonian mechanics or SR, the clocks tick at the same rate. For the sake of simplicity, let the two clocks be synchroninzed. Thus, if X is the reading on one clock, and Y is the simultaneous reading on the other clock, then x-y=0. As long as the clocks remain at rest with respect to each other, the clocks will remain in-sync.

Suppose that clock B is inside a uniformly accelerating ship. Thus, clock B is being subjected to an OUTSIDE force of a constant value. As this happens, the relative speed v between the two clocks will continuously increase. While clock B is accelerating relative to clock A, clock B is no longer in an inertial reference frame. Let us stipulate that clock A is always in an inertial reference frame. Therefore, whilst clock B is accelerating, an observer stationed at clock A can use the time dilation formula of SR, to draw conclusions about whether or not clock B remains in-sync with clock A.

The time dilation formula is:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}

\Delta t is an amount of time measured by clock A, and \Delta t^\prime is the corresponding amount of time measured by clock B. In newtonian mechanics, these two quantities remain equal as clock B accelerates, in SR the quantities are not equal, and instead related by the time dilation formula given above.

As you can see, the conclusion arrived at using the time dilation formula, is that the clocks are no longer in-sync once clock B begins to accelerate. In fact, clock B begins to tick SLOWER. That means that if I am located at clock A, and my current reading is 5525, then I can be certain that clock B simultaneously reads something less than 5525 (even if I don't know exactly what it reads). In other words, I can be certain that X>Y, where X is the current reading on clock A, and Y is the simultaneous reading on clock B.

Now, let some amount of time have passed, and let it be the case that we have:

X = 1000 and Y = 400

Thus, the difference in the readings of the two clocks is now

X-Y= 1000-400 = 600

At the moment that clock B reads 400, let it be the case that the engines shut off, so that clock B is now in an inertial reference frame. Let V denote the final relative speed of these two clocks.


Is it the case that the clocks now tick at the same rate? If they now tick at the same rate then the difference in the readings "X-Y" is again constant, and its value is 600 ticks. But what does SR say about this?

We are still assuming that the time dilation formula is a true statement. Thus, an observer at rest with respect to clock B, can formulate statements about the rate of clock A using the time dilation formula. He will use the following formula:

\Delta t^\prime = \frac{\Delta t}{\sqrt{1-V^2/c^2}}

where delta t` is an amount of time measured by clock B, and delta t is the corresponding amount of time measured by clock A. Only if SR is wrong, are these two quantities equal. Since we are assuming SR is correct, delta t` is not equal to delta t, and instead the two quantities are related through the above formula.

Hence, by SR, clock A is now ticking slower than clock B by a factor of gamma. Thus, if an observer at rest with respect to clock B waits a sufficient length of time in his frame, the two clocks should again eventually fall into sync. In other words, an observer stationed at clock B should KNOW that X>Y but that clock A which reads X is now ticking slower than clock B which reads Y, so that at some point in the future, we will have X=Y.

The problem is now mentally visible. An observer stationed at clock A, will always use the following formula to understand how the rate of clock B relates to the rate of clock A:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}

Where delta t is an amount of time measured by clock A, and delta t` is an amount of time measured by clock B. Thus, by SR, clock B always ticks slower than clock A. Hence, once the reading of clock B is less than the reading of clock A, it will never again be equal to the reading of clock A. Hence, there is no moment in the future, at which X=Y. Thus, we have reached the following contradiction:

After the moment at which clock A reads 1000, and clock B reads 400, there will come a moment in time at which X=Y, and after the moment in time at which clock A reads 1000, and clock B reads 400, there won't come a moment in time at which X=Y.

Logical analysis ends here.
SR is overthrown

I welcome any challengers to this perfect line of reasoning.

[Severian596]

This is the second of such threads and I have to ask you if you've considered that SR does not account for gravity or acceleration. So when I hear the phrase "SR is overthrown," it's kinda meaningless. SR applies only to very particular circumstances. The more you push the boundaries of those circumstances the farther away from SR you get. You should realize that SR has limits to its applications...

do you agree?

[Chen]

You already got your anwer in the last thread you posted:
http://physicsforums.com/showthread.php?s=&threadid=16943
And exactly what made you feel the need to post this three times?

[StarThrower]

Originally posted by Severian596
This is the second of such threads and I have to ask you if you've considered that SR does not account for gravity or acceleration. So when I hear the phrase "SR is overthrown," it's kinda meaningless. SR applies only to very particular circumstances. The more you push the boundaries of those circumstances the farther away from SR you get. You should realize that SR has limits to its applications...

do you agree?

SR makes mathematical predictions about how the rates of clocks moving relative to one another are related. The time dilation formula is used to conclude that clocks which tick at the same rate and start out synchronized, are no longer synchronized if one of the clocks begins to accelerate.

Therefore, the time dilation formula can be used exactly as it was used by an observer stationed at clock A, since clock A remains in an inertial reference frame throughout the entire time that clock B is accelerating. Additionally, once clock B enters an inertial reference frame again, an observer there can use the time dilation formula. Accelerating reference frames don't present a problem for SR, as long as you understand how to use the time dilation formula. Nevertheless, you eventually reach a contradiction, hence the time dilation formula is incorrect. And this answers your other question. Simply replace the phrase, "SR is overthrown" by the phrase "the time dilation formula is incorrect". Hopefully, that has meaning for you.

As for comments about gravity, we can imagine these two clocks in a region of the universe far away from any other objects, so that the gravitational fields where the clocks are located is essentially zero. As far as the gravitational force of one clock on the other clock goes, if the clocks are sufficiently far apart, the gravitational force on them is approximately zero, and can be neglected.

You also said that, "SR applies only to very particular circumstances."

To that I only have one thing to say:

Since assuming the time dilation formula is true leads to contradiction, there is no circumstance in which SR applies.

So I don't agree.

[StarThrower]

Originally posted by Chen
You already got your anwer in the last thread you posted:
http://physicsforums.com/showthread.php?s=&threadid=16943
And exactly what made you feel the need to post this three times?

Chen you are incorrect. This was not a thermodynamic argument, it was an argument based upon accelerating frames of reference. The same stipulations were used in my thermodynamic argument, but this is an entirely different argument. The conclusion is always the same:

The time dilation formula isn't true.

Regards,

The Star

[russ_watters]

Actually, chen, this time you're wrong. This thread contains a different misunderstanding of SR: 'when' does time dilation occur.

Starthrower, SR time dilation occurs as a result of motion, not acceleration. So once the acceleration has stopped, the times shown by the two clocks will continue to diverge.

I only skimmed your argument, but it looks like after the acceleration has stopped, you flipped the time dilation equation as if either clock could be considered stationary. You can't do that: you still know which clock is moving and which isn't because of the acceleration that just stopped.

Here is a real-world engineering example of SR/GR in action: prior to launch, GPS satellites have their clock rates adjusted to comply with the predictions of SR/GR. So while on the ground, GPS clocks don't run in sync with other clocks. Once launched, the clocks maintain their syncronization with ground-based clocks.
I welcome any challengers to this perfect line of reasoning. Wow, that's some arrogance you have there. Even if you ever stop being wrong, don't expect anyone to give you any recognition unless you drop the attitude.

[Severian596]

Originally posted by StarThrower
In other words, I can be certain that X>Y, where X is the current reading on clock A, and Y is the simultaneous reading on clock B.

...

Is it the case that the clocks now tick at the same rate? If they now tick at the same rate then the difference in the readings "X-Y" is again constant, and its value is 600 ticks.
I believe these are the weak points of your logic. This is just my opinion but my ego isn't as big as yours seems to be so you may ignore it.

You must be careful with comparing simultaneity in two different reference frames. And you must not assume that just because one clock stops accelerating the clocks begin to tick at the same rate, as russ_waters already said.

Read this PDF (http://physics.nyu.edu/hogg/sr/sr.pdf) by David W. Hogg, Chapter 2 for some good examples and a good explanation. Read it more than once if necessary; you seem to be lost in your thought experiment.

[Severian596]

One thing that hasn't been mentioned so far in this thread is experimental proof of the existence (and correctness) of the time dilation equation. Here's a link that deals with just such a question:
http://www.physlink.com/Education/AskExperts/ae433.cfm

And here's another link that discusses Einstein's Equivalence Principle - the principle that states that accelerations are equivalent to gravitational fields. <
http://www.upscale.utoronto.ca/GeneralInterest/Harrison/GenRel/TimeDilation.html> This topic is further than my current place in the study of the topics of relativity, but as long as the source of this page is credible and the page is correct, I believe it is very appropriate for your thought experiement that deals with acceleration.

Notice that when discussing time dilation most text books cite situations where objects are already in motion...they completely avoid acceleration. You chose to ignore the acceleration in your thought experiment and perhaps this was in error. If in fact the last line of this page is correct (as I said I'm not up to current research in the topic of General Relativity): this completes the derivation of the fact that clocks in gravitational fields run slowly, it sounds like during an acceleration a clock is hit by twofold time dilations; one for motion and another for acceleration.

If someone else more knowledgeable feels the desire to participate in the all-to-common theme of the defense of modern theory, maybe they could confirm or deny my statements. [;)]

[Severian596]

Originally posted by StarThrower
SR makes mathematical predictions about how the rates of clocks moving relative to one another are related. The time dilation formula is used to conclude that clocks which tick at the same rate and start out synchronized, are no longer synchronized if one of the clocks begins to accelerate.
Finally, and I hope I'm not beating a dead horse here, you need to keep in mind that the application of SR is VERY limited. You have to jump through hoops to set up an inertial reference frame that satisfies the conditions of Special Relativity instead of having to use General Relativity equations. THAT's what I meant by my first comment...the farther you push the boundaries the less SR applies and the more you need to use more complex and realistic GR equations and theories.

[StarThrower]

Actually, chen, this time you're wrong. This thread contains a different misunderstanding of SR: 'when' does time dilation occur.

Starthrower, SR time dilation occurs as a result of motion, not acceleration. So once the acceleration has stopped, the times shown by the two clocks will continue to diverge.

Russ, you need to exert an infinite amount of effort in order to prove there is a misunderstanding on my part.

We have a simple formula here, namely the time dilation formula. Take a real good look at it:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}

Firstly: Understanding of algebra, knowledge of the definition of speed, knowledge of what a clock does, and knowledge of what a coordinate system is, are necessary and sufficient knowledge order to understand the formula. All of which you have russ.

Next: Suppose that clock B is subjected to an instantaneous impulse. Let the clocks start out ticking at the same rate, and be synchronous, and read zero. At the moment in time at which both clocks read zero, let the relative speed be zero, and let the impulse be applied to clock B. Thus, at the very next moment in time, clock B is now moving relative to clock A, at a nonzero speed V. At the following moment in time, the relative speed of clock B to clock A is still V, because the impulse is over. Thus, clock B is now moving at a constant speed relative to clock A.

So russ, you have several moments to consider. There is the moment in time at which the impulse hits (namely the moment in time at which both clocks read zero).

At the moment in time at which both clocks read zero, the relative speed of the clocks is still zero, hence the time dilation formula gives:

\Delta t = \Delta t^\prime

At this moment in time.

At the very next moment in time (the important one), the clocks are now in relative motion at speed V, and so the time dilation formula gives:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}

delta t is to be obtained by subtracting two successive readings of clock A from one another, and delta t` is to be obtained by subtracting two successive readings of clock B from one another.

So we have two consecutive moments in time...

Initial reading on clock A = 0
Initial reading on clock B = 0

Final reading on clock A = R1
Final reading on clock B = R2

Thus, the elapsed time according to clock A is R1, and the elapsed time according to clock 2 is R2.

After the impulse has been applied, the time dilation formula is in full effect in either a frame at rest with respect to clock A, or a frame at rest with respect to clock B, and the conclusion must be that clock B now ticks slower than clock A, so that it is unambiguously true that R1>R2.

The readings in the clocks will continue to have a difference that is increasing, simply because clock B is now ticking slower than clock A, BUT the rate at which this is happening is constant, because V is now constant. I am not sure you understand this.

The rate at which the readings are separating is constant. But the fact that this rate is non-zero is a result of assuming the time dilation formula is true.

At any rate, using an impulse, instead of a long drawn out constant force, shows that the time dilation formula can be used at all moments in time through the impulse by an observer at rest with respect to clock A.

Kind regards,

The Star

[DrMatrix]

Let us stipulate that clock A is always in an inertial reference frame. Therefore, whilst clock B is accelerating, an observer stationed at clock A can use the time dilation formula of SR, to draw conclusions about whether or not clock B remains in-sync with clock A.

The time dilation formula is:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}

\Delta t is an amount of time measured by clock A, and \Delta t^\prime is the corresponding amount of time measured by clock B. In newtonian mechanics, these two quantities remain equal as clock B accelerates, in SR the quantities are not equal, and instead related by the time dilation formula given above.SR deals with inertial reference frames. While clock B accelerates, it is not in an inertial frame and you cannot apply SR. A quick look at the time dialation formula should show you why. v is the velocity. If B is accelerating, you don't have a constant v and cannot apply the formula

Once B stops accelerating, you can apply the time dialation formula. According to A's frame, B is ticking slower. According to B's frame, A is ticking slower. SR shows no preference.
I only skimmed your argument, but it looks like after the acceleration has stopped, you flipped the time dilation equation as if either clock could be considered stationary. You can't do that: you still know which clock is moving and which isn't because of the acceleration that just stopped.Is sounds like you'rs saying that A's frame is prefered because B underwent acceleration. I know that cannot be what you mean, but that's what it sounds like. After the acceleration stops, both are inertial frames and you can flip frames. From A's POV, B is slowed, and from B's frame A is slowed. Since velocity is relative, SR shows no preference. Either clock can be considered stationary.

The only absolute way to compare the two is to bring them back together.

StarThrower is wrong, but not because A's frame is prefered, but because neither is preferred.

[DrMatrix]

Initial reading on clock A = 0
Initial reading on clock B = 0This is allowed because the two clocks are next to each otherFinal reading on clock A = R1
Final reading on clock B = R2
Here you have a problem. The clocks are not at the same place and are not in the same frame. Remember, simultanaity is relative. Clock A reads R1 and B reads R2 according to A or according to B? The answer cannot be both. And there is no ablsolute time.

[Hurkyl]

At any rate, using an impulse, instead of a long drawn out constant force, shows that the time dilation formula can be used at all moments in time through the impulse by an observer at rest with respect to clock A.

How, precisely? The only points in time that you've considered are the times before and after the application of the impulse.


With a proper argument involving calculus, you can derive the differential formula for proper time where t, x, y, z are the coordinates of the worldline of a particle in an inertial reference frame:


(c d\tau)^2 = (c dt)^2 - dx^2 - dy^2 - dz^2

[StarThrower]

SR deals with inertial reference frames. While clock B accelerates, it is not in an inertial frame and you cannot apply SR...

StarThrower is wrong, but not because A's frame is prefered, but because neither is preferred.


Clock A is in an inertial reference frame, and so the time dilation formula is valid there. An observer at rest with respect to clock B cannot use the time dilation formula in that frame, because that frame is non-inertial.

I have made absolutely no error, and you made a huge one.

Kind regards,

The Star

[StarThrower]

This is allowed because the two clocks are next to each otherHere you have a problem. The clocks are not at the same place and are not in the same frame. Remember, simultanaity is relative. Clock A reads R1 and B reads R2 according to A or according to B? The answer cannot be both. And there is no ablsolute time.

The answer is both, so it doesn't matter that the clocks aren't at the same place. In fact, that is the whole point of the time dilation formula, to tell you what that other clock now reads, even though it has moved away from you.

This post has arisen because of your confusion about what conditions on a frame are necessary, in order to use the time dilation formula in that frame.

Kind regards,

The Star


P.S.: This post is an example of what happens when a reasonging agent operates under a false assumption, and hasn't yet arrived at a contradiction which will allow them to negate their false assumption, and experience an increase in their knowledge.

One cannot know that which is false.
Aristotle Of Stagira

[StarThrower]

How, precisely? The only points in time that you've considered are the times before and after the application of the impulse.


With a proper argument involving calculus, you can derive the differential formula for proper time where t, x, y, z are the coordinates of the worldline of a particle in an inertial reference frame:


(c d\tau)^2 = (c dt)^2 - dx^2 - dy^2 - dz^2


Do you want me to show you how Hurkyl?

[DrMatrix]

The answer is both, so it doesn't matter that the clocks aren't at the same place. In fact, that is the whole point of the time dilation formula, to tell you what that other clock now reads, even though it has moved away from you.The whole point of the time dialation formula is to tell you what the how relative velocity affects time according another reference frame. Now depends on the frame. Now in A's frame is different from now in B's frame. Now is not absolute.

Your fundamental problem seems to be that you are assuming time is absolute while applying SR. Since SR denies absolute time, It is a given that you will find a contradiction if you assume absolute time and SR.

[StarThrower]

The whole point of the time dialation formula is to tell you what the how relative velocity affects time according another reference frame. Now depends on the frame. Now in A's frame is different from now in B's frame. Now is not absolute.

Your fundamental problem seems to be that you are assuming time is absolute while applying SR. Since SR denies absolute time, It is a given that you will find a contradiction if you assume absolute time and SR.

That isn't what you are witnessing. I am assuming the fundamental postulate of SR is true, and arriving logically at a contradiction.

Kind regards,

The Star

[StarThrower]

How, precisely? The only points in time that you've considered are the times before and after the application of the impulse.


With a proper argument involving calculus, you can derive the differential formula for proper time where t, x, y, z are the coordinates of the worldline of a particle in an inertial reference frame:


(c d\tau)^2 = (c dt)^2 - dx^2 - dy^2 - dz^2


Hurkyl seems to be in love with the following formula:

(c d\tau)^2 = (c dt)^2 - dx^2 - dy^2 - dz^2

c is greater than zero, therefore you can divide by c to get:

(d\tau)^2 = (dt)^2 - (d[x/c])^2 - (d[y/c])^2 - (d[z/c])^2

Lets consider motion in a straight line along the x axis only. Therefore

(d[y/c])^2 = 0 and (d[z/c])^2 = 0

and

(d\tau)^2 = (dt)^2 - (d[x/c])^2

Which, provided not (dt=0), can be written as:

(\frac{d\tau}{dt})^2 = 1 - (\frac{d[x/c]}{dt})^2

Which implies that:

(\frac{d\tau}{dt})^2 = 1 - (\frac{dx}{cdt})^2

Let v = speed = dx/dt; hence

(\frac{d\tau}{dt})^2 = 1 - (\frac{v}{c})^2

And taking the square root of both sides we obtain:


\frac{d\tau}{dt} = \sqrt{ 1 - (\frac{v}{c})^2 }

Undoing the constraint that not (dt=0) we have:

d\tau = dt\sqrt{ 1 - v^2/c^2}


Definition:

\gamma = \frac{1}{\sqrt{1-v^2/c^2}}

Thus:

\frac{1}{\gamma} = \sqrt{1-v^2/c^2}

Thus:

d\tau = \frac{dt}{\gamma}

Where d\tau is the proper time, which is time measured by a clock at rest, and dt the corresponding amount of time measured by the 'moving' clock.

Or equivalently:

dt = \gamma d\tau = \frac{d\tau}{\sqrt{1-v^2/c^2}}


At the top of this thread I introduced the time dilation formula as follows:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}

delta t was an arbitrary amount of time measured by a clock at rest, and delta t` was the corresponding amount of time measured by a clock moving relative to the clock at rest, relative speed being v.

Letting the amount of time be arbitarily small would have given:

dt = \frac{dt^\prime}{\sqrt{1-v^2/c^2}}

in the above formula, dt is an infinitessimal amount of time measured by a clock at rest, and dt` is the correspoing amount of time measured by a clock in relative motion at speed v.

To transform from my original equation to Hurkyl's requires that:

my dt is equal to his d\tau and my dt` is equal to his dt.

Making the necessary substitution of d tau in my original formula leads to:

d\tau = \frac{dt^\prime}{\sqrt{1-v^2/c^2}}

And making the necessary substitution of his dt for my dt` leads to:

d\tau = \frac{dt}{\sqrt{1-v^2/c^2}}

And writing the previous equation in terms of gamma leads to:

d\tau = \gamma dt

Hence we have:

d\tau = \gamma dt
AND
d\tau = \frac{dt}{\gamma}

From which it follows that:

\gamma dt = \frac{dt}{\gamma}

Now, we can divide both sides of the above equation by gamma, provided that gamma isn't equal to zero.

Assume that gamma isn't equal to zero. Therefore:

dt = \frac{dt}{(\gamma)^2}

Thus:

dt = dt (1-v^2/c^2)

c is a nonzero constant, hence for the above equation to be true, it MUST be the case that v=0. We can now close the scope of our current assumption:

If not (gamma=0) then v=0.

We now ask, whether or not gamma can ever be equal to zero. Suppose that gamma can equal zero, therefore we would have:

0 = \frac{1}{\sqrt{1-v^2/c^2}}

For the equation above to be true, we cannot have the division by zero error of algebra, therefore the quantity (1-v^2/c^2)^1/2 cannot equal zero. Multiplying both sides of the above equation by this quantity yeilds:

0=1

Which is false.

Therefore, it is impossible for gamma to ever equal zero. Therefore,

not (gamma = 0).

Therefore,

not (gamma = 0 ) AND if not (gamma=0) then v=0.

Therefore, v=0.

Thus, Hurkyl's formula isn't true for an arbitrary speed v, it is only true for speed 0. And Hurkyl assumed it was true for an arbitrary speed v.

Kind regards,

The Star

[Hurkyl]

Do you want me to show you how Hurkyl?

I was trying to point out the right way (in a given reference frame) to compute the time on an accelerated clock, by using the differential formula for proper time; rather than arguing heuristically with the time dilation formula, you can argue rigorously with the differential version.

Of course, I'm not particularly sure why you're trying to study the times on the clocks during the period of acceleration.

[Hurkyl]

And allow me to remind you that &tau; is the time measured by the clock under analysis and t is the time given by the current choice of coordinates.

[StarThrower]

I was trying to point out the right way (in a given reference frame) to compute the time on an accelerated clock, by using the differential formula for proper time; rather than arguing heuristically with the time dilation formula, you can argue rigorously with the differential version.

Of course, I'm not particularly sure why you're trying to study the times on the clocks during the period of acceleration.

Then do it for me.

Kind regards,

The Star

P.S. The reason I am concentrating on the times on the clocks during the acceleration, is because I am the greatest physicst on earth, and not particularly modest.

[Hurkyl]

Oh, and incidentally, Mr greatest physicist on earth, you still haven't shown how to derive the differential formula for proper time from the time dilation formula... you went the other way around. :biggrin:

[StarThrower]

And allow me to remind you that τ is the time measured by the clock under analysis and t is the time given by the current choice of coordinates.



This was too vague.

[StarThrower]

Oh, and incidentally, Mr greatest physicist on earth, you still haven't shown how to derive the differential formula for proper time from the time dilation formula... you went the other way around. :biggrin:

Instead of me deriving the differential formula for proper time from the time dilation formula, why don't you derive the time dilation formula from the fundamental postulate of relativity, and then after that is done, use its differential version to derive the formula for proper time.

The derivation I am thinking of uses the pythagorean theorem, and the speed of light in one frame c, cancels out with the speed of light in the other frame c`, and the time dilation formula emerges.

If you can't do it, or don't want to do it, then I can.

Kind regards,

The Star

[Hurkyl]

Let &tau; be the reading on a clock. Let t, x, y, and z be the temporal and spatial coordinates of the clock (according to a given inertial reference frame). Then, everywhere where the worldline of the clock is differentiable, (c d&tau;)2 = (c dt)2 - dx2 - dy2 - dz2

More precisely,

Let the parametrization (t(s), x(s), y(s), z(s)) (a <= s <= b) of the worldline of a clock through an inertial reference frame, and let &tau;(s) be the reading on the clock at (t(s), x(s), y(s), z(s)). Then, we have:


\left(c \frac{d\tau}{ds} \right)^2 = \left(c \frac{dt}{ds}\right)^2 -
\left(\frac{dx}{ds} \right)^2 - \left(\frac{dy}{ds} \right)^2
- \left(\frac{dz}{ds} \right)^2


Or (assuming the clock's readout isn't going backwards as t(s) increases)


\begin{equation*}\begin{split}
\tau(s) &= \tau(a) + \int_a^s \frac{d\tau(u)}{du} \, du \\
&= \tau(a) + \int_a^s \frac{1}{c} \sqrt{
( c t'(u) )^2
- x'(u)^2
- y'(u)^2
- z'(u)^2
} \, du
\end{split}\end{equation*}

[StarThrower]

Let τ be the reading on a clock. Let t, x, y, and z be the temporal and spatial coordinates of the clock (according to a given inertial reference frame). Then, everywhere where the worldline of the clock is differentiable, (c dτ)2 = (c dt)2 - dx2 - dy2 - dz2

More precisely,

Let the parametrization (t(s), x(s), y(s), z(s)) (a <= s <= b) of the worldline of a clock through an inertial reference frame, and let τ(s) be the reading on the clock at (t(s), x(s), y(s), z(s)). Then, we have:


\left(c \frac{d\tau}{ds} \right)^2 = \left(c \frac{dt}{ds}\right)^2 -
\left(\frac{dx}{ds} \right)^2 - \left(\frac{dy}{ds} \right)^2
- \left(\frac{dz}{ds} \right)^2


Or (assuming the clock's readout isn't going backwards as t(s) increases)


\begin{equation*}\begin{split}
\tau(s) &= \tau(a) + \int_a^s \frac{d\tau(u)}{du} \, du \\
&= \tau(a) + \int_a^s \frac{1}{c} \sqrt{
( c t'(u) )^2
- x'(u)^2
- y'(u)^2
- z'(u)^2
} \, du
\end{split}\end{equation*}



Hurkyl, are you trying to give me a headache? Just curious.

Regards,

StarThrower

[Hurkyl]

If you can't do it, or don't want to do it, then I can.

I'm not the one going around asserting I'm the greatest physicist of all time. :biggrin: I thought you'd jump at the chance to convince us all! After all, you even offered to do the proof. :rolleyes:

And quite frankly, I would like to see you do it, mainly to see if you really are capable of treating relativistic reference frames properly.


Hurkyl, are you trying to give me a headache?

You asked for precision. :rolleyes:

[StarThrower]

I'm not the one going around asserting I'm the greatest physicist of all time. :biggrin: I thought you'd jump at the chance to convince us all! After all, you even offered to do the proof. :rolleyes:

And quite frankly, I would like to see you do it, mainly to see if you really are capable of treating relativistic reference frames properly.




You asked for precision. :rolleyes:

I intend to prove exactly that Hurkyl, but why on earth I'm trying to do so on a message board is beyond me.

Regards,

The Star


Who knows, maybe this internet conversation will be the most famous conversation of all time. :smile:

[Severian596]

P.S. The reason I am concentrating on the times on the clocks during the acceleration, is because I am the greatest physicst on earth, and not particularly modest.

"The only true wisdom is in knowing you know nothing."
~Socrates

"Wisest is he who knows what he does not know."
~Plato (428/7-348/7 B.C.E.)

[Severian596]

...but why on earth I'm trying to do so on a message board is beyond me.

You failed to mention the horrible manner in which you present your arguments, and the circles in which you lead your "proofs."

[StarThrower]

"The only true wisdom is in knowing you know nothing."
~Socrates

"Wisest is he who knows what he does not know."
~Plato (428/7-348/7 B.C.E.)

"Wisdom is a poor substitute for knowledge."
~Author Unknown

[StarThrower]

You failed to mention the horrible manner in which you present your arguments, and the circles in which you lead your "proofs."

My arguments may be presented haphazardly, but I assure you there is nothing circular about my reasoning.

Kind regards,

The Star

[Severian596]

My arguments may be presented haphazardly, but I assure you there is nothing circular about my reasoning.

Kind regards,

The Star

I guess it's too bad that I can't base my beliefs on your own assurance of your methods.

[Michael F. Dmitriyev]

StarThrower,
This mathematics is incorrect.
You are ignoring the fact, that achievement of zero value is infinite process demanding infinite energy (time). Any value can come nearer to zero only (the speed too).
Do you have any own explanation of a constancy of the speed of light?

[StarThrower]

StarThrower,
This mathematics is incorrect.
You are ignoring the fact, that achievement of zero value is infinite process demanding infinite energy (time). Any value can come nearer to zero only (the speed too).
Do you have any own explanation of a constancy of the speed of light?

Micheal you have misunderstood something; my guess is that you think the speed of light is c in any inertial frame.

Kind regards,

The Star

[DrMatrix]

Are you saying that the speed of light depends upon the inertial frame? Because if you are, that is contrary to one of the fundamental assumptions of SR.

You are assuming the speed of light depends upon the reference frame, and (by trying to apply SR) assuming the speed of light is independent of the reference frame. It is no surprise that you can find a contradiction.

[Michael F. Dmitriyev]

Micheal you have misunderstood something; my guess is that you think the speed of light is c in any inertial frame.

Kind regards,

The Star
May be you have missed my reply on the previous pages. My point is:
all conclusions of SR are fair for all objects except of light itself.

Speed of light is unique and absolute speed.
It does not subjected to relativity at all.

[StarThrower]

Are you saying that the speed of light depends upon the inertial frame? Because if you are, that is contrary to one of the fundamental assumptions of SR.

You are assuming the speed of light depends upon the reference frame, and (by trying to apply SR) assuming the speed of light is independent of the reference frame. It is no surprise that you can find a contradiction.

Fundamental Postulate SR: The speed of a photon is c=299792458 meters per second in any inertial reference frame.

Here is what I am saying. I am saying that I can find at least one statement X such that:

If the fundamental postulate of SR is true then (X and not X).
At which point it follows that the fundamental postulate of SR is false.
At which point it follows that the following statement is true:

There is at least one inertial reference frame in which the speed of a photon isn't c=299792458.

So yes, the speed of a photon does depend upon what inertial frame you are in. In some inertial reference frames the speed of a photon will be 299792458 m/s, but in other inertial reference frames, the speed of the exact same photon is something other than 299792458 m/s.

I did not "assume" the speed of light depends upon inertial frame, in fact, I assumed the exact opposite (that it doesn't), and subsequently arrived at a contradiction.

Kind regards,

The Star

[StarThrower]

Let τ be the reading on a clock. Let t, x, y, and z be the temporal and spatial coordinates of the clock (according to a given inertial reference frame). Then, everywhere where the worldline of the clock is differentiable, (c dτ)2 = (c dt)2 - dx2 - dy2 - dz2

More precisely,

Let the parametrization (t(s), x(s), y(s), z(s)) (a <= s <= b) of the worldline of a clock through an inertial reference frame, and let τ(s) be the reading on the clock at (t(s), x(s), y(s), z(s)). Then, we have:


\left(c \frac{d\tau}{ds} \right)^2 = \left(c \frac{dt}{ds}\right)^2 -
\left(\frac{dx}{ds} \right)^2 - \left(\frac{dy}{ds} \right)^2
- \left(\frac{dz}{ds} \right)^2


Or (assuming the clock's readout isn't going backwards as t(s) increases)


\begin{equation*}\begin{split}
\tau(s) &= \tau(a) + \int_a^s \frac{d\tau(u)}{du} \, du \\
&= \tau(a) + \int_a^s \frac{1}{c} \sqrt{
( c t'(u) )^2
- x'(u)^2
- y'(u)^2
- z'(u)^2
} \, du
\end{split}\end{equation*}



Hurkyl, you have made an error in all of this, and it is too hard to find with the symbolism you are using. Really, you should do what I asked, which is to derive the time dilation formula from the assumption that the speed of light is c in any inertial reference frame.

Kind regards,

The Star

[Hurkyl]

So you're not gonna try and do it yourself, then? I think I'm gonna have a heart attack and die from that surprise! What's wrong with the symbolism? Looks like ordinary calculus to me...

[StarThrower]

So you're not gonna try and do it yourself, then? I think I'm gonna have a heart attack and die from that surprise! What's wrong with the symbolism? Looks like ordinary calculus to me...

Well you are still missing the error, so rather than put the blame on you, I blamed your needless parametrization of the worldline curve.

Regards,

The Star

[StarThrower]

So you're not gonna try and do it yourself, then? I think I'm gonna have a heart attack and die from that surprise! What's wrong with the symbolism? Looks like ordinary calculus to me...

You don't want to derive the time dilation formula do you.

Maybe I can find the derivation I am thinking of on the web already, and then just paste the link. It is of an experiment on measuring the speed of light using a mirror.

Kind regards,

The Star

P.S. I too can do melt your brain mathematics, but I don't find that kind of flowerly display of intelligence necessary.

[Hurkyl]

Don't complain about a statement being vague if you don't want to see it written more precisely. :rolleyes:

Shall I try again? &tau; is the time on the clock under analysis, and t is the time given by the coordinate system.


You don't want to derive the time dilation formula do you.

Are you serious that you would like to see a derivation, or are you trying to deflect attention from the fact that you offered to try and prove something first, and have yet to make that attempt?

[StarThrower]

Don't complain about a statement being vague if you don't want to see it written more precisely. :rolleyes:

Shall I try again? τ is the time on the clock under analysis, and t is the time given by the coordinate system.




Are you serious that you would like to see a derivation, or are you trying to deflect attention from the fact that you offered to try and prove something first, and have yet to make that attempt?

I really am serious. I know what has to be done to accomplish the proof. The first step is going to be to derive the time dilation formula. What is going to happen, is that you will set up two coordinate systems in relative motion at speed v. Then we can constantly refer to the way you set up the coordinate systems in order to make comments about the times in either system.

I tried to search the web for the exact derivation I am referring to, so that neither of us would have to repeat the work, here is a site that uses the derivation I am referring to, but I don't like the guy's symbolism.

Derivation Of The Time Dilation Formula (http://www.drphysics.com/syllabus/time/time.html)

Here is the specific experiment (an experiment to measure the speed of light) we will need analyzed:

You are going to make a measurement of the speed of a photon. The photon is going to be fired from a photon gun, straight to a mirror, and then be reflected back to you. So you have a clock (clock A) at rest in this lab frame F1 which will mark the precise moment the photon is emitted, and the precise moment the photon returns. Denote this time by \Delta t . Now, you have a ruler set up just sitting at rest in the lab frame. It runs from the photon gun to the mirror. Let the distance from the photon gun to the mirror be measured to be D. Thus, the total distance the photon travels is 2D. Let c denote the speed of the photon in this frame. Thus, the speed of the photon in F1 is by definition:

c = \frac{2D}{\Delta t}

Now, consider this event as viewed from a reference frame F2 which is moving at speed v relative to F1. In this frame, the path of the photon is not a straight line, but rather the path is triangular (that is how the pythagorean theorem enters the derivation). Let it be the case that if you are at rest in F2, that you see the photon gun moving from left to right. So you see the photon travel diagonally upwards until it strikes the mirror, and then diagonally downwards until it strikes clock A(which is attached to the photon gun). There is a second clock (clock B) at rest in F2, and let the time it takes the photon to move from the photon gun to the mirror and back to the photon gun in F2 be measured by clock B to be \Delta t^\prime

Thus, the total distance traveled by the photon in F2, is the sum of the two sides of an isosceles triangle. Let the side length of this triangle be denoted by S. Thus, the total distance traveled by the photon in F2 is 2S. Let c` denote the speed of the photon in F2. Thus, the speed of the photon in F2 is by the same definition as before:


c^\prime = \frac{2S}{\Delta t^\prime}

Now, the base of the isosceles triangle is the product of the relative speed v between the two frames F1,F2, and the total time it takes the photon to strike the mirror and return to the photon gun, as measured by clock B. Thus, the base of the isosceles triangle is:

v \Delta t^\prime

This is certainly enough information for you to use the pythagorean theorem to derive the time dilation formula. The height of the isosceles triangle is D, hence the following equation is true:

S^2 = D^2 + (\frac{v\Delta t^\prime}{2})^2

In the derivation of the time dilation formula, the fundamental postulate of the special theory of relativity is invoked the moment you equate c with c`. In classical mechanics they are unequal, in SR they are equal by hypothesis. The exact equation you get (by hypothesizing that c=c`) is:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}

Once that is done, all quantities in the formula are clear. Once you have carried out the derivation, more discussion will be necessary. Let me show you the logic I am after by asking you to do this for everyone:

Fundamental postulate of SR: The speed of a photon in any inertial reference frame is c=c`= 299792458 meters per second. (Keep in mind that in the derivation both F1, AND F2 are inertial reference frames. F1 is inertial by stipulation, and since F2 isn't accelerating relative to F1 and F1 is inertial, it follows that F2 is inertial.

Here is what is being perfectly proved:

Theorem(contingent knowledge): If the fundamental postulate of SR is true then (\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}})

Now, at this point reason alone cannot yet determine whether or not the fundamental postulate of SR is true. More reasoning will have to be done.

In order to prove that the fundamental postulate is false, you now have to prove the following theorem:

Theorem(contingent knowledge): If the fundamental postulate of SR is true then NOT( \Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}} ).

At that point you will have proven the following:


There is at least one statement X, such that:

Theorem(contingent knowledge): If the fundamental postulate of SR is true then (X and not X).

At this point, you can now prove the following:

Theorem(Absolute knowledge): The fundamental postulate of SR is false.


The main purpose of this excercise, is so that we can both have the same exact thought experiment in mind simultaneously, using the exact same letters to denote the variables and constants which are crucial to the time dilation formula.

Kind regards,

StarThrower

P.S. Here is a possible enstasis(enstasis is a greek word and means objection). For the appearance of mathematical terms like postulate, lemma, enstasis see Heath's translation of Euclid.

Enstasis:

Clock B is at a fixed location in F2, and the photon gun only passes near to clock B once during the event.


Solution: Let there be a third clock (clock C) which is also at rest in F2. Let the photon leave the photon gun at the precise moment when the apparatus passes by clock B, and let the photon return to the gun at the precise moment that the apparatus passes by clock C. Thus, observers at rest in F2 will know the clock readings of clocks B,C at the moment the photon gun passed them. Let clocks B,C have been synchronized. Thus, the time \Delta t^\prime can be computed by subtracting the reading of clock B from the reading of clock C.

Additionally, let it be the case that all three clocks are of identical construction, therefore they all tick in the same unit (let the unit of time on these clocks be the second). Since all clocks are of identical construction, we can be certain that they have the same rest rate (the rate they tick when they are at rest with respect to one another). Thus, if any two of these three clocks are at rest with respect to one another and have been synchronized, then they will remain synchronous.

In order to synchronize clocks B,C, we can imagine that a rigid ruler connects them, and that we have located the midpoint, and then a spherical light pulse goes off there, and travels to both clocks, setting them to zero simultaneously in reference frame F2. They therefore remain in sync forever in reference frame F2.

Incidentally, here is a link to Euclid's proof of the Pythagorean Theorem

Euclids Proof Of The Pythagorean Theorem (http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI47.html)

The above site is excellent.

There should be no objection to the use of the Pythagorean theorem in the derivation, because it is certainly true of rulers at rest with respect to each other in an inertial frame. My favorite proof of the Pythagorean theorem can be seen at the following site:

Simple Proof Of Pythagorean Theorem (http://mathforum.org/isaac/problems/pythagthm.html)

[Hurkyl]

I have seen that derivation, but I would normally take a totally different approach to the proof that just looks at what sort of coordinate transformations preserve straight lines and lines with coordiante velocity c.

[StarThrower]

I have seen that derivation, but I would normally take a totally different approach to the proof that just looks at what sort of coordinate transformations preserve straight lines and lines with coordiante velocity c.

I don't doubt you have seen it, but this is the derivation I want us to discuss. I have my reasons, humor me. I know it's just a bit of algebra, but it will help if we both are certain the other guy is using the exact same symbol we are, to denote the exact same thing.

Kind regards,

The Star

[Hurkyl]

For the record, I'd like to point out that you have assumed D is the same in both frames. I imagine (based on the rest of your post) you intended that the F2 be moving in a direction perpendicular to the ruler, though you never stated as such.


Anyways, I'll agree with you up until


S^2 = D^2 + \left(\frac{v \Delta t'}{2}\right)^2
.

because from this point, we have:


\begin{equation*}\begin{split}
S &= \frac{1}{2} c \Delta t' \\
D &= \frac{1}{2} c \Delta t \\
\left(\frac{1}{2} c \Delta t'\right)^2 &=
\left(\frac{1}{2} c \Delta t\right)^2 + \left(\frac{v \Delta t'}{2}\right)^2 \\
c^2 \Delta t'^2 &= c^2 \Delta t^2 + v^2 \Delta t'^2 \\
\Delta t^2 &= \left( 1 - \left(\frac{v}{c}\right)^2\right) \Delta t'^2 \\
\Delta t &= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t'
\end{split}\end{equation*}

[StarThrower]

For the record, I'd like to point out that you have assumed D is the same in both frames. I imagine (based on the rest of your post) you intended that the F2 be moving in a direction perpendicular to the ruler, though you never stated as such.


Anyways, I'll agree with you up until


S^2 = D^2 + \left(\frac{v \Delta t'}{2}\right)^2
.

because from this point, we have:


\begin{equation*}\begin{split}
S &= \frac{1}{2} c \Delta t' \\
D &= \frac{1}{2} c \Delta t \\
\left(\frac{1}{2} c \Delta t'\right)^2 &=
\left(\frac{1}{2} c \Delta t\right)^2 + \left(\frac{v \Delta t'}{2}\right)^2 \\
c^2 \Delta t'^2 &= c^2 \Delta t^2 + v^2 \Delta t'^2 \\
\Delta t^2 &= \left( 1 - \left(\frac{v}{c}\right)^2\right) \Delta t'^2 \\
\Delta t &= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t'
\end{split}\end{equation*}


Interesting.

[Severian596]

I'm definitely riveted.

[StarThrower]

Do you agree that if I can now show that:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}

SR is finally overthrown?

Kind regards,

The Star

[Severian596]

Wait a tick...you just lost me. You want to prove that one of special relativity's formulas is correct, and then determine that SR is overthrown because its formula is correct?

I'm thinking out loud. This formula comes about because of the "hypothesis" that c' = c no matter what reference frame you're in. So you want to prove that the formula is correct, but also assert that c' != c (please excuse the programming convention of NOT EQUAL)?

[StarThrower]

Wait a tick...you just lost me. You want to prove that one of special relativity's formulas is correct, and then determine that SR is overthrown because its formula is correct?

I'm thinking out loud. This formula comes about because of the "hypothesis" that c' = c no matter what reference frame you're in. So you want to prove that the formula is correct, but also assert that c' != c (please excuse the programming convention of NOT EQUAL)?

Hurkyl proved this:

If the fundamental postulate of SR is true then:

\Delta t^\prime = \frac{\Delta t}{\sqrt{1-v^2/c^2}}

When a theory self-contradicts, you cannot finish reasoning after the derivation of a single implication. In other words, when a theory self-contradicts, you can derive two implications as follows:

Let T denote a theory of physics.
Thus, T is the conjunction of many statements. Suppose that theory T is the result of conjoining exactly 8 uncertain statements, with 1000 certain statements. Consider just that part of the theory, which is uncertain.

Denote the uncertain part of a theory of physics using the following symbol:

\Im

Thus,

\Im = S_1 \wedge S_2 \wedge S_3 \wedge ... S_8

\wedge = AND

Let

\rightarrow = if...then

So when the uncertain part of a theory of physics is wrong, you can derive two implicative statements from the theory of the following form:

\Im \rightarrow X
\Im \rightarrow notX

In other words, you can find a statement X, such that:

\Im \rightarrow (X \wedge notX)

At this point in your reasoning, you will know that there is an error in the theory, although you won't know which of the 8 statements is false, all you will know is that at least one of the 8 statements is false.

Now, consider a theory which is based upon only one uncertain statement. For example, the theory of relativity certainly uses all statements of algebra, but such statements are true by stipulation.

Let us define the theory of relativity to be a theory with just one uncertain postulate, namely the following postulate:

Fundamental postulate SR: The speed of a photon in any inertial reference frame is c.

SR will be overthrown if we can now find a statement X, such that:

If the fundamental postulate of SR is true then X
If the fundamental postulate of SR is true then not X.

Hurkly just derived the following implication:

If the fundamental postulate of SR is true then:

\Delta t^\prime = \frac{\Delta t}{\sqrt{1-v^2/c^2}}

Suppose that I can now derive the following implication:

If the fundamental postulate of SR is true then:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}

It will thus follow that:

If the fundamental postulate of SR is true then:

\frac{1}{\sqrt{1-v^2/c^2}} = \sqrt{1-v^2/c^2}

From which it will follow that:

1 = 1 - v^2/c^2

From which it will follow that V=0.

But not (v=0) by stipulation, since F1 and F2 are in relative motion.

Thus, it will follow that:

If c=c` then (v=0 and not (v=0)).

It will therefore follow that:

not (c=c`)

And c is the speed of light in inertial reference frame F1, and c` is the speed of light in inertial reference frame F2.

Thus, it is not the case that the speed of light is 299792458 meters per second in all inertial reference frames. In other words, the fundamental postulate of the theory of special relativity is false.

So, what I said was, that if I can now derive the second implication above, then the theory of SR is wrong.

Kind regards,

The Star

[Severian596]

You made it clear. I understood it that way, but wanted to make sure I was not confusing the prime character...the two formulae are very similar.

So following soon will be your proof of the proposed equation?

[Severian596]

This is strange though. This equation:


\Delta t = \frac{\Delta t'}{\sqrt{1-v^2/c^2}}


is true, and is used to calculate the dilation of the clock at rest with respect to the clock in motion...so I want to make sure you're not switching frames. You will attempt to show that relative to the same clock in the same reference frame, both equations are true?

[Pergatory]

Hurkly just derived the following implication:

If the fundamental postulate of SR is true then:

\Delta t^\prime = \frac{\Delta t}{\sqrt{1-v^2/c2}}

Suppose that I can now derive the following implication:

If the fundamental postulate of SR is true then:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c2}}

It will thus follow that:

If the fundamental postulate of SR is true then:

\frac{1}{\sqrt{1-v^2/c^2}} = \sqrt{1-v^2/c^2}


Could you please explain to me how you arrived at the last quoted conclusion?

[Severian596]

Could you please explain to me how you arrived at this last quoted conclusion?

Yes, please. You cannot substitute 1 for both \Delta t and \Delta t'...if that was in fact what you did. This would imply absolute time, and you can't use this implication in an attempt to prove just that.

[StarThrower]

This is strange though. This equation:


\Delta t = \frac{\Delta t'}{\sqrt{1-v^2/c^2}}


is true, and is used to calculate the dilation of the clock at rest with respect to the clock in motion...so I want to make sure you're not switching frames. You will attempt to show that relative to the same clock in the same reference frame, both equations are true?

Yes, be patient, I am awaiting Hurkyl's response.

Kind regards,

The Star

[StarThrower]

Could you please explain to me how you arrived at the last quoted conclusion?

I didn't arrive at that conclusion yet. That conclusion is contingent upon me showing within the framework of this problem that:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}

Kind regards,

The Star

[Hurkyl]

If &Delta;t is the time interval measured by clock A, and &Delta;t' is the elapsed time in the coordinate frame F2, then

&Delta;t = &radic;(1 - (v/c)^2) &Delta;t'

(sanity check: clock A is indeed running slower than coordinate time)

If you could also prove that

&Delta;t' = &radic;(1 - (v/c)^2) &Delta;t

for the same &Delta;t', &Delta;t, and v, then there would indeed be a contradiction in SR.


Furthermore, if one could build a model of SR in some mathematical theory (say... by using linear algebra to construct Minowski geometry), then you will also have proven that mathematical theory to be inconsistent.

[jdavel]

StarThrower, I finally see the source of your confusion.

You correctly say that the postulates of SR imply two equations,

Equation 1:

\Delta t^\prime = \frac{\Delta t}{\sqrt{1-v^2/c^2}}

and Equation 2:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}

If these weren't both true then there would be something fundamentally different between moving to the left and moving to the right. The postulates say there is no such difference.

But then you mistakenly say that these equations together imply a third equation,

Equation 3:

\frac{1}{\sqrt{1-v^2/c^2}} = \sqrt{1-v^2/c^2}

This simply (and obviously) isn't true.

The reason you can't get to Eq 3 from Eqs 1 & 2 is that the variables delta-t and delta-t' don't represent the same thing in Eq 1 that they do in Eq 2. So the ratio, delta-t/delta-t' in Eq 1 isn't the same as the ratio delta-t/delta-t' in Eq 2. In fact, they are exactly the reciprocals of eachother! So your Eq 3 should read:

\frac{1}{\sqrt{1-v^2/c^2}} = \frac{1}{\sqrt{1-v^2/c^2}}

which is true, but not very interesting, and certainly no threat to the validity of SR!

[Severian596]

I agree. The only way I can see StarThrower coming to the incorrect equation is by misinterpreting SR.

This is easy for him to fall into considering the following:
* He supports absolute time
* He supports absolute space (and therefore believes we live in Euclidian space)
* He adheres to classic Newtonian physics
* He believes he's brilliant and therefore justifies the previous three behaviors

Now I'm not saying he's ABSOLUTELY crazy because, well, that would be taking the same fanatic standpoint he's taking. I don't have all the answers, but I invest confidence in many of our past geniuses who advanced theory beyond Pythagoras and Euclid's day. I don't assume I have the luxury of being one of them myself...

[StarThrower]

I agree. The only way I can see StarThrower coming to the incorrect equation is by misinterpreting SR.

This is easy for him to fall into considering the following:
* He supports absolute time
* He supports absolute space (and therefore believes we live in Euclidian space)
* He adheres to classic Newtonian physics
* He believes he's brilliant and therefore justifies the previous three behaviors

Now I'm not saying he's ABSOLUTELY crazy because, well, that would be taking the same fanatic standpoint he's taking. I don't have all the answers, but I invest confidence in many of our past geniuses who advanced theory beyond Pythagoras and Euclid's day. I don't assume I have the luxury of being one of them myself...

Remember, if I can derive any contradiction whatsoever, then logically I can draw any conclusion whatsoever.

Look back at the equation Hurkyl arrived at:

For the record, I'd like to point out that you have assumed D is the same in both frames. I imagine (based on the rest of your post) you intended that the F2 be moving in a direction perpendicular to the ruler, though you never stated as such.


Anyways, I'll agree with you up until


S^2 = D^2 + \left(\frac{v \Delta t'}{2}\right)^2
.

because from this point, we have:


\begin{equation*}\begin{split}
S &= \frac{1}{2} c \Delta t' \\
D &= \frac{1}{2} c \Delta t \\
\left(\frac{1}{2} c \Delta t'\right)^2 &=
\left(\frac{1}{2} c \Delta t\right)^2 + \left(\frac{v \Delta t'}{2}\right)^2 \\
c^2 \Delta t'^2 &= c^2 \Delta t^2 + v^2 \Delta t'^2 \\
\Delta t^2 &= \left( 1 - \left(\frac{v}{c}\right)^2\right) \Delta t'^2 \\
\Delta t &= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t'
\end{split}\end{equation*}


This equation is valid for v=c is it not?

[Pergatory]

SR states that if we are in an inertial reference frame, v<c

[StarThrower]

SR states that if we are in an inertial reference frame, v<c

No, actually you just stated that... kind of ad hoc, if you know what I mean.

Kind regards,

The Star

[Severian596]

This equation

\Delta t &= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t'

is valid for v=c. Because of the flop you're no longer dividing by zero. The original equation was this:

\Delta t' &= \frac{\Delta t}{\sqrt{1 - \left( \frac{v}{c} \right)^2}}

This equation derives the time observed by a clock A at rest, of a clock B in motion, and denotes the observed time as \Delta t'. Now if we look at the first equation again:

\Delta t &= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t'

This equation derives the time observed by clock B in motion, of a clock A at rest, and denotes the observed time as \Delta t, with respect to clock A's reference frame. This equation is in fact valid if clock B is traveling at v=c because the amount of time that B will observe passing on A is zero!!

This equation tells us that photons do not observe time passing on any clocks because they travel at v=c.

EDIT:

I want to stress that the frame of reference for this equation did not change, therefore the symbols' meanings did not change. Thus there is no contradiction involved...all we did is multiply by the numerator. It doesn't support any contradiction (X and not X, for example) of special relativity. Deriving B's observation of A with respect to A's reference frame is completely valid, though a bit confusing if you don't keep the meaning straight.

[StarThrower]

This equation

\Delta t &= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t'

is valid for v=c. Because of the flop you're no longer dividing by zero. The original equation was this:

\Delta t' &= \frac{\Delta t}{\sqrt{1 - \left( \frac{v}{c} \right)^2}}

This equation derives the time observed by a clock A at rest, of a clock B in motion, and denotes the observed time as \Delta t'. Now if we look at the first equation again:

\Delta t &= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t'

This equatin derives the time observed by clock B in motion, of a clock A at rest, and denotes the observed time as \Delta t, with respect to clock A's reference frame. This equation is in fact valid if clock B is traveling at v=c because the amount of time that B will observe passing on A is zero!!

This equation tells us that photons do not observe time passing on any clocks because they travel at v=c.

NO!

You are disobeying mathematical rules, namely you are dividing by zero in order to put the square root quantity in a denominator.

You absolutely are not permitted to do that. However, Hurkyl's formula

\Delta t &= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t'

certainly is valid.

Kind regards,

The Star

[Severian596]

NO!

You are disobeying mathematical rules, namely you are dividing by zero in order to put the square root quantity in a denominator.

You absolutely are not permitted to do that. However, Hurkyl's formula

\Delta t &= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t'

certainly is valid.

Kind regards,

The Star

Wait wait...first read my edit to my post just to make sure you get my meaning, then with a little less zeal and a little more annunciation tell me how I disobeyed mathematical rules. I'll examine my own post while you type.

[StarThrower]

EDIT:

I want to stress that the frame of reference for this equation did not change, therefore the symbols' meanings did not change. Thus there is no contradiction involved...all we did is multiply by the numerator. It doesn't support any contradiction (X and not X, for example) of special relativity. Deriving B's observation of A with respect to A's reference frame is completely valid, though a bit confusing if you don't keep the meaning straight.


Yes I know this. Let me state for the record:

Clock A is attached to the photon gun, and time measured by it is being denoted by \Delta t

Clock B was a clock moving at arbitrary speed v, relative to clock A. The Hurkelian result was then derived, where \Delta t^\prime is the time of the same event, but measured by clock B.

Up to this point you are good, so what are you trying to do now?

Kind regards,

The Star

[StarThrower]

Wait wait...first read my edit to my post just to make sure you get my meaning, then with a little less zeal and a little more annunciation tell me how I disobeyed mathematical rules. I'll examine my own post while you type.

I UNDERSTOOD YOU PERFECTLY.

The symbols meaning didn't change... of course the symbols meaning cannot change, that would be cheating just to derive a contradiction. Read what I wrote.

Kind regards,

The Star

[Severian596]

Yes I know this. Let me state for the record:

Clock A is attached to the photon gun, and time measured by it is being denoted by \Delta t

Clock B was a clock moving at arbitrary speed v, relative to clock A. The Hurkelian result was then derived, where \Delta t^\prime is the time of the same event, but measured by clock B.

Up to this point you are good, so what are you trying to do now?

Kind regards,

The Star

Please be careful of your terminology, because Delta t' is NOT an event, it's a measure of change in time between any two defined events. Please don't refer to it as the "time of the same event" unless you've first defined an event (ct,x,y,z) about which to speak. Furthermore the delta t of an event is not a delta at all, it's just t...


Up to this point you are good, so what are you trying to do now?


I'm trying to show you that in my opinion
1) v=c is valid for this equation, and
2) this fact does not contradict special relativity

Why? Well we set up two equations for A's rest frame, right? One solves for Delta t and the other solves for Delta t prime. All with respect to A. If we go ahead and solve each of them for v=c we get the following results:

\Delta t = 0
\Delta t' = Undefined (division by zero)

You'll grant me the right to manipulate the equation BEFORE plugging in values, right? So I didn't divide by zero. But why oh why don't these results spell the downfall of Special Relativity??

[StarThrower]

Please be careful of your terminology, because Delta t' is NOT an event, it's a measure of change in time between any two defined events. Please don't refer to it as the "time of the same event" unless you've first defined an event (ct,x,y,z) about which to speak. Furthermore the delta t of an event is not a delta at all, it's just t...

I didn't say Delta t` is an event.

Let me define the 'event' I am talking about. The event I am talking about begins at the moment in time when clock A coincides with clock B, and ends later, when clock A coincides with clock C.

Clock B and clock C are in an inertial reference frame, and are at rest with respect to each other. Additionally, they have been synchronized. At the moment in time when clock A and clock B coincide, clock B reads x, and clock C also reads x (because clock C is synchronized with clock B by stipulation). Later, clock A coincides with clock C, and at this moment in time, clock C reads Y. Thus, the total time of the event in the inertial reference frame in which clock B and clock C are at rest, is equal to Y-X. In other words:

\Delta t^\prime = Y-X

The above quantity is the "time of the event" in reference frame F2.

I have now defined the 'event' I am talking about.

Kind regards,

The Star

P.S. I know why some verbal confusion has arisen.

You refer to a single moment in time as an 'event'. That is the standard terminology yes. But there is another word for this, which is used in philosophical circles, and that word is 'state'.

You represent a state using the following symbolism: (ct,x,y,z)

But you have no precise notion of the "time of an event" which can only be defined using two states, one which occurred before the other, in universal time.

Consider two distinct states. Using your notation we have:

State 1 = S1 = (ct1,x1,y1,z1)
State 2 = S2 = (ct2,x2,y2,z2)

Let S1 occur before S2. I am calling [S1,S2] an event, the amount of time of this event is given by your (t2-t1).

Regards,

The Star

[Severian596]

I thought maybe you'd answer me, or try to tell me that it does in fact lead to a contradiction in SR.

This situation (where v=c) just doesn't make sense. Distance = rate*time. No matter what your rate, if time is zero, distance is zero. But for v=c, no matter what amount of time you try to assert has passed at clock A denoted by \Delta t, clock B traveling at v=c will measure \Delta t = 0, which is why we cannot define \Delta t' with respect to A. We just wouldn't understand it! No matter how much time we experience traveling with clock A, clock B never measures any change in our clock.

Your initial argument that two photons traveling parallel to each other are traveling relative to each other at velocity zero is classic newtonian velocity addition. This is an incorrect application of newtonian physics, just like the application of newtonian physics to explain the "almost insignificant" change in Mercury's perihelion did not apply (but was precisely explained by Einstein's equations). So, if I can prove that one of your assumptions is incorrect, I can call your entire conclusion incorrect.

I won't mention that you assume that we're in Euclidian space (not proven), and that space is the same for a photon as it is for us (not proven), AND that you assume time is absolute (which I'm sure you know carries implications of variable c and the ether, as well as absolute space, which you've admitted you agree with).

I don't know why I bother. Think what you like, but your thoughts aren't as original as you believe they are.

[Severian596]

The event I am talking about begins at the moment...and ends later...

I say again, events don't have a duration. They have 4 coordinates,

(time, x,y,z)

Anyway I'll try to make it through the rest of your post now...though right off the bat you show me that you are in fact confusing the heck out of me by calling an event a duration of time.

*sigh*

[Severian596]

I didn't say Delta t` is an event.

Let me define the 'event' I am talking about. The event I am talking about begins at the moment in time when clock A coincides with clock B, and ends later, when clock A coincides with clock C.

Clock B and clock C are in an inertial reference frame, and are at rest with respect to each other. Additionally, they have been synchronized. At the moment in time when clock A and clock B coincide, clock B reads x, and clock C also reads x (because clock C is synchronized with clock B by stipulation). Later, clock A coincides with clock C, and at this moment in time, clock C reads Y. Thus, the total time of the event in the inertial reference frame in which clock B and clock C are at rest, is equal to Y-X. In other words:

\Delta t^\prime = Y-X

The above quantity is the "time of the event" in reference frame F2.

I have now defined the 'event' I am talking about.

Kind regards,

The Star

Alright, this is convoluted at best, and a diagram may help or it may not, but

The event I am talking about begins at the moment in time when clock A coincides with clock B, and ends later, when clock A coincides with clock C.

I guess I'll assume that, in order to synchronize all three clocks a light source is placed in the center of a Circle (or sphere) in an extremly local spacial area with respect to clocks A, B, and C. A pulse of light is emitted from the light source, and the clocks begin ticking the moment the light reaches them. They are equidistant from the source and not in relative motion to each other or the light source, and therefore we should be confident that they are synchronized. So, in this reference frame F1 (notorious label by now) the lines of simultaneity for clocks A, B, and C are parallel to their spacial axis x with respect to their time axis t.


Clock B and clock C are in an inertial reference frame, and ...


Whoa! When did they change reference frames? You just synchronized them for crying out loud! So we must conclude that now clocks B and C are in a different reference frame F2 than A, then some force must have acted upon them for them. So an acceleration took place. Bad for SR....very bad...you're getting out of special relativity's jurisdiction.


Additionally, they [clocks B and C] have been synchronized.


As long as they've had the same force applied to them and are now in the same reference frame I assume this was not necessary. I'll assume also that you're just reiterating the fact.


At the moment in time when clock A and clock B coincide, clock B reads ...


Okay that's it. Now you're proposing a situation where special relativity does not apply. Because A and B are in different reference frames their lines of simultaneity are NOT parallel to each other. A sees B's lines of simultaneity at an angle, and B sees A's lines of simultaneity at an angle. Therefore there exists no "moment in time" that A and B coincide unless both A and B are in uniform motion with respect to some OTHER reference frame F3. This other reference frame (with respect to which A and B must have been synchronized before hand, then had equal forces applied to each of them to put them in different reference frames than F3 itself) could possibly measure "coinciding" times on A and B, but ONLY RELATIVE TO ITSELF.

Therefore because A and B will never coincide with respect to each other, the rest of your statements never occur.

[StarThrower]

I say again, events don't have a duration. They have 4 coordinates,

(time, x,y,z)

Anyway I'll try to make it through the rest of your post now...though right off the bat you show me that you are in fact confusing the heck out of me by calling an event a duration of time.

*sigh*

See my PS which I added to my previous post. It was/is:

P.S. I know why some verbal confusion has arisen.

You refer to a single moment in time as an 'event'. That is the standard terminology yes. But there is another word for this, which is used in philosophical circles, and that word is 'state'.

You represent a state using the following symbolism: (ct,x,y,z)

But you have no precise notion of the "time of an event" which can only be defined using two states, one of which must occur before the other, in universal time.

Consider two distinct states. Using your notation we have:

State 1 = S1 = (ct1,x1,y1,z1)
State 2 = S2 = (ct2,x2,y2,z2)

Let S1 occur before S2. I am calling [S1,S2] an event, the amount of time of this event is given by your (t2-t1).

Regards,

The Star

[Severian596]

Consider two distinct states. Using your notation we have:

State 1 = S1 = (ct1,x1,y1,z1)
State 2 = S2 = (ct2,x2,y2,z2)

Let S1 occur before S2. I am calling [S1,S2] an event, the amount of time of this event is given by your (t2-t1).


Thank you for this. I don't participate in philosophical circles, and we're not in one right now. Your 'event' is actually a duration, I'm sure you know that. Read my previous post to see why your proposed system is invalid...you make assumptions, leave out details, and change frames of reference willy nilly. You know better than that.

Oh and remember that your "Universal Time" you referred to when defining the philosopher's circle 'event' is actually just another frame of reference which is subject to relativity itself. In my post, that "Universal Time" could be in terms of F3, but that means you're not presenting all the necessary facts. And remember that there is no preferred frame of reference, and if there is one (in your opinion) how do YOU personally define that?

[StarThrower]

I say again, events don't have a duration. They have 4 coordinates,

(time, x,y,z)

Anyway I'll try to make it through the rest of your post now...though right off the bat you show me that you are in fact confusing the heck out of me by calling an event a duration of time.

*sigh*

I apologize for that, though I knew it would happen. I simply prefer the word 'state' to the word 'event' when I am discussing points of time, or moments in time, and I like to assign quantities of time to something that I wish to call 'events'.

You use the word 'event' to mean what I call 'state', and I use the word 'event' for something you don't have a word for.

Kind regards,

The Star

(P.S. It is easier for you to change than me.)

EDIT: Correction, you do have a word for it... duration. However, duration isn't a noun, the way an event is. Philosophers would have a field day with you. :smile:

To familiarize yourself with the philosophical use of the terms 'state' and 'event', peruse the following reputable site which discusses temporal logic.

Temporal Logic (http://plato.stanford.edu/entries/logic-temporal/)

Keep in mind that these guys are just dying to talk with physicists.

[Severian596]

duration isn't a noun, the way an event is.

Duration is a noun...and I'm not going to debate about it. Here's dictionary.com (http://dictionary.reference.com/search?q=duration)'s definition (beware an annoying popup).

[matt grime]

And linguists would have a field day with you. Not to mention psychologists, given your egoism.

[StarThrower]

Duration is a noun...and I'm not going to debate about it. Here's dictionary.com (http://dictionary.reference.com/search?q=duration)'s definition (beware an annoying popup).


Absolutely no point in arguing semantics. I can switch back and forth between your terminology and mine, I am intelligent enough for that. I presume you have the same capability?

Kind regards,

The Star

[StarThrower]

And linguists would have a field day with you. Not to mention psychologists, given your egoism.

In the time it takes them to comprehend me, I will have died. Hence, I conclude their efforts would be in vain. Better they should become physicists.

Kind regards,

The Universe

[Severian596]

To suppliment my post about your incorrect use of Special Relativity, peruse this answer (http://van.hep.uiuc.edu/van/qa/section/New_and_Exciting_Physics/Relativity/20021019150128.htm) to the twin paradox. The paradox is not a paradox at all. The individual cites his source and mentions that General Relativity is needed to correctly answer the paradox because Special Relativity does not supply a means to deal with acceleration and/or a preferred frame.

I admire your apparent understanding of physics (both classical and electromagnetic), and your mathematical ability. I'm just a beginning student to the topic(s) yet I spot your error so easily. Why do you persist in thinking you can prove an error in SR where SR does not apply?

[StarThrower]

To suppliment my post about your incorrect use of Special Relativity, peruse this answer (http://van.hep.uiuc.edu/van/qa/section/New_and_Exciting_Physics/Relativity/20021019150128.htm) to the twin paradox. The paradox is not a paradox at all. The individual cites his source and mentions that General Relativity is needed to correctly answer the paradox because Special Relativity does not supply a means to deal with acceleration and/or a preferred frame.

I admire your apparent understanding of physics (both classical and electromagnetic), and your mathematical ability. I'm just a beginning student to the topic(s) yet I spot your error so easily. Why do you persist in thinking you can prove an error in SR where SR does not apply?

Firstly, I didn't make the error someone accused me of, which was to reverse the meanings of \Delta t and \Delta t^\prime

I understand which quantity of time is being measured by which clock. That wasn't/isn't the contradiction I am reaching for. As for your new question about the twin paradox, let me read the article at the site first, and then I will come back and give you my comments. As for my persistence in thinking I have spotted an error in SR... the reason for that is because I have. I don't really believe I am the first to ponder it either. Nonetheless, the error to me seems obvious, yet it hasn't caught on to mainstream physics yet. And that is what I really cannot understand.

If the relative speed in this problem is given by v=c, then we have two photons moving at right angles to each other. The clocks B,C are thus in a photonic frame (see other threads on this). The conclusion I reach, is that if the fundamental postulate of the theory of special relativity is true, then photons cannot move relative to each other. That cannot possibly be true, based upon simple facts that come from sensory perception.

Now let me give this site a look, and I will come back and tell you what I think.

Kind regards,

StarThrower


EDIT: Correction, the clocks B,C are in a photonic reference frame (reference frame in which photons are at rest)

[matt grime]

So, you're an expert, a genius able to see to the very inherent contradictory nature of relativity, and you've not come across the twin paradox?

[Severian596]

That cannot possibly be true, based upon simple facts that come from sensory perception.

This is the biggest pitfall you could've fallen into. You cannot argue anything based "upon simple facts that come from sensory perception," especially mathematics. Sensory perception is not an axiom. In fact that this phrase even escaped your keyboard surprised the heck out of me. How atrocious.

[Severian596]

If the relative speed in this problem is given by v=c, then we have two photons moving at right angles to each other. The clocks A,B are thus in photonic frames (see other threads on this).

Note that this example should be stricken from the record unless StarThrower is willing to better qualify "this problem" and exactly what A and B have to do with any frames, much less "photonic" frames.

[StarThrower]

So, you're an expert, a genius able to see to the very inherent contradictory nature of relativity, and you've not come across the twin paradox?

Of course I have, I am merely looking at this site for the first time... then I will give my comments. Immediately upon looking at it I thought to express the population as follows:

P(t) = P_0 e^r^t

Give me a moment here, I'm reading.

Regards,

Star

[Severian596]

Nonetheless, the error to me seems obvious, yet it hasn't caught on to mainstream physics yet. And that is what I really cannot understand.

You need to understand relativity better. It never caught on because actual physicists (who know more on this topic than I do) evaluated it, pondered it, and dismissed it. Why? Because they know more than you.

[StarThrower]

This is the biggest pitfall you could've fallen into. You cannot argue anything based "upon simple facts that come from sensory perception," especially mathematics. Sensory perception is not an axiom. In fact that this phrase even escaped your keyboard surprised the heck out of me. How atrocious.

My apologies, allow me to be more professional.

Consider the following experiment performed by an ancient caveman:

He started a fire for the women to sit, directly on his left, so he didnt have to listen to them squalk, and then started a fire directly in front of him, for the men to sit around. He noticed that not only was the front of his body warm, but so was the left side of his body too. He concluded that photons can move relative to each other. :biggrin:

P.S. Or was this a thought experiment? Geeze, I forget. Maybe he just thought he was warm???


Regards,

Star

[StarThrower]

You need to understand relativity better. It never caught on because actual physicists (who know more on this topic than I do) evaluated it, pondered it, and dismissed it. Why? Because they know more than you.

They cannot make me believe that which is false. They can confuse me about it for awhile, but they can never make me believe it.

Regards,

StarThrower

[Severian596]

He concluded that photons can move relative to each other.

This line should be changed to read "He concluded that photons can move relative to himself." I can only mildly apprecite your candor.

Imagine that you have two trains A and B that leave the station C like this:

A
^
|
|
|
|
C----------->B

And they just so happen to be traveling at the speed of light. They'll never observe each other, so they'll never determine how fast they're traveling relative to each other.

[Severian596]

And...I shall exit stage left, as I should've done from square one.

Goodbye, StarThrower

[StarThrower]

To suppliment my post about your incorrect use of Special Relativity, peruse this answer (http://van.hep.uiuc.edu/van/qa/section/New_and_Exciting_Physics/Relativity/20021019150128.htm) to the twin paradox. The paradox is not a paradox at all. The individual cites his source and mentions that General Relativity is needed to correctly answer the paradox because Special Relativity does not supply a means to deal with acceleration and/or a preferred frame.



Ok I've looked at the site. The first thing to say, is that I understood Dolan's question.

If you have ever been to the carnival, there is a ride where you are spun around really fast, and then the floor drops out, but you don't fall to the earth. In this ride, you are pinned to the wall by a force which has arisen because you are in uniform circular motion.

Principle of equivalence: It is impossible to tell if you are at rest on the surface of a body exerting a gravitational force F on you, or accelerating uniformly through space.

Dolan asks about the following experiment.

You have two vials of some radioactive substance, with a precisely known half life. For example, suppose that the half life is one hour. The samples are identical, in the sense that they contain the same initial population. So for example, suppose that P0 = 1,000. Thus, if the vials are left at rest, then in one hour, there will be (approximately) 500 radioactive particles left in both samples.

Now, perform this experiment, but this time put one of the samples in a device like the carnival ride described before.

The particles in the centrifuge are obviously subjected to a force which the particles in the vial on the bench are not subjected to.

Dolan's question then, is whether or not the experiment has ever been done, and whether or not any time dilation was observed.

[StarThrower]

This line should be changed to read "He concluded that photons can move relative to himself." I can only mildly apprecite your candor.

Imagine that you have two trains A and B that leave the station C like this:

A
^
|
|
|
|
C----------->B

And they just so happen to be traveling at the speed of light. They'll never observe each other, so they'll never determine how fast they're traveling relative to each other.

Umm but they are moving relative to each other, even if they cannot measure the relative speed.

[Severian596]

Umm but they are moving relative to each other, even if they cannot measure the relative speed.

That's it! That's your flaw. Why should train A or B assume anything based on YOUR measurements. They will never detect each other's presence, so therefore they are NOT moving relative to each other. YOU may determine that they're moving at X speed relative to each other, or Y speed, or whatever speed, but YOU have the luxury of experiencing time and space in a way completely removed from what either of these trains must experience. To them, YOUR claim that they are moving relative to each other is completely preposterous! No matter how long they vigilantly wait for information that lets them draw a conclusion about their motion RELATIVE TO EACH OTHER it will never come.

There IS no absolute preferred frame, and that means YOUR FRAME is not the divine frame from which to ordain all conclusions about the relative motion of any two bodies, or the nature of space and time.

Believe whatever you like. I'm not wasting another moment on you. I've wasted enough.

[StarThrower]

That's it! That's your flaw. Why should train A or B assume anything based on YOUR measurements. They will never detect each other's presence, so therefore they are NOT moving relative to each other. YOU may determine that they're moving at X speed relative to each other, or Y speed, or whatever speed, but YOU have the luxury of experiencing time and space in a way completely removed from what either of these trains must experience. To them, YOUR claim that they are moving relative to each other is completely preposterous! No matter how long they vigilantly wait for information that lets them draw a conclusion about their motion RELATIVE TO EACH OTHER it will never come.

There IS no absolute preferred frame, and that means YOUR FRAME is not the divine frame from which to ordain all conclusions about the relative motion of any two bodies, or the nature of space and time.

Believe whatever you like. I'm not wasting another moment on you. I've wasted enough.

I thought you exited stage left.

Self-contradiction? Oh now you are totally coming unglued huh. I find a flaw in the theory of SR, and you come apart.

[StarThrower]

That's it! That's your flaw. Why should train A or B assume anything based on YOUR measurements. They will never detect each other's presence, so therefore they are NOT moving relative to each other. YOU may determine that they're moving at X speed relative to each other, or Y speed, or whatever speed, but YOU have the luxury of experiencing time and space in a way completely removed from what either of these trains must experience. To them, YOUR claim that they are moving relative to each other is completely preposterous! No matter how long they vigilantly wait for information that lets them draw a conclusion about their motion RELATIVE TO EACH OTHER it will never come.

There IS no absolute preferred frame, and that means YOUR FRAME is not the divine frame from which to ordain all conclusions about the relative motion of any two bodies, or the nature of space and time.

Believe whatever you like. I'm not wasting another moment on you. I've wasted enough.

Let them be tied together with a very long string, with a lot of slack in it. Eventually, that string is going to break, and they will 'know' they are/were in relative motion. It will break at the moment in Universal time at which the distance between them exceeds the string length.

Regards,

The Star

[DrMatrix]

moment in Universal time If you're going to assume SR (or GR), you must drop the notion of Universal time. Otherwise, you will run into a contradiction. :wink:

[Severian596]

Let them be tied together with a very long string, with a lot of slack in it. Eventually, that string is going to break, and they will 'know' they are/were in relative motion. It will break at the moment in Universal time at which the distance between them exceeds the string length.

Regards,

The Star

Ignoring the impossibility of a string (which has mass) "keeping up" with the photon travelling at c, the compression wave triggered by the snapping string would travel at roughly the speed of sound, just as any compression wave. For example if you and your friend stand with thighs against a table and he bumps it, how much time passes before you feel the table move? The speed is not infinite, and in fact travels at the speed of sound through the medium of the solid table. The photon would never learn of the "universal time" at which the string snapped.

I will discuss the theory of SR with you no further, but am happy to educate you on some basic physical phenomenon.

[Hurkyl]

This equation is valid for v=c is it not?

The equation is well-defined, if that's what you mean. In the spirit of the discussion thus far, I'd have to say we currently have no reason to think that the equation would be correct in SR; v=c violates the hypotheses of our derivation.

[StarThrower]

The equation is well-defined, if that's what you mean. In the spirit of the discussion thus far, I'd have to say we currently have no reason to think that the equation would be correct in SR; v=c violates the hypotheses of our derivation.

Well, we have the derivation before us, and there was no reason to forbid v=c in the derivation. v = \infty certainly must be forbidden, but not v=c. The reason v cannot be infinite, is because as the relative speed v increases, the angle between the sides of the isosceles triangle approaches 180 degrees, at which point the distance D would have to equal zero. But D is nonzero by stipulation. In fact, the only stipulation made about the relative speed, was that it be nonzero (so that we have a triangle in F2).

As we perform algebraic steps, we certainly must be careful not to divide by zero, or WE make an error.

The next thing to say, is that you certainly were correct about my assumption about D, namely that there is no length contraction in a direction perpendicular to the velocity. You also correctly saw that the relative velocity vector was perpendicular to the photon's velocity vector in the lab frame. As for D not contracting, that is unobjectionable as far as I can see, and as for the relative velocity vector, the direction is just one of the stipulations of the experiment.

But, I am sorry to say, v=c is not forbidden by nature, nor was it forbidden in this wonderfully simple derivation of the time dilation formula. Special relativity has an insurmountable problem to contend with, which is namely that photons cannot move relative to one another.

The logic finishes off this way:

Premise 1: If c=c` then (photons can't move relative to one another).
Premise 2: Photons can move relative to one another.
Conclusion: Not (c=c`)

The truth of premise 1 is known deductively, you in fact derived the result for us.

The truth of premise two is known inductively, through our common sensory perception.

The conclusion is sequitur, via the natural deduction known as modus tollens.


Kind regards,

The Star Thrower

[Hurkyl]

But, I am sorry to say, there is no reason to forbid v=c in this wonderfully simple derivation of the time dilation formula.

Sure there is.

For instance, you assert:

Now, consider this event as viewed from a reference frame F2 which is moving at speed v relative to F1. In this frame, the path of the photon is not a straight line, but rather the path is triangular

Tell me, how long are the three sides to the triangle? For the sake of argument, let's say D is 1 meter.


There is a second clock (clock B) at rest in F2, and let the time it takes the photon to move from the photon gun to the mirror and back to the photon gun in F2 be measured by clock B to be \Delta t'

If, say, \Delta t happened to be 1 nanosecond, what is \Delta t'?


In order to synchronize clocks B,C, we can imagine that a rigid ruler connects them, and that we have located the midpoint, and then a spherical light pulse goes off there, and travels to both clocks, setting them to zero simultaneously in reference frame F2. They therefore remain in sync forever in reference frame F2.

How long is the ruler? How long does it take to synchronize the clocks?

[StarThrower]

Tell me, how long are the three sides to the triangle? For the sake of argument, let's say D is 1 meter.

The answer comes from Euclidean geometry of course.

We have a right triangle, the hypotenuse of which is S. The base of which is 1/2 v delta t`, and the height of which is D. You want to let D=1 meter.

Additionally, S = 1/2 c` delta t`

Thus

(\frac{c^\prime \Delta t^\prime}{2} )^2 = 1 + (\frac{v \Delta t^\prime}{2} )^2

Give me a relative speed v in meters per second, and I can finish.

Regards,

Star

[Hurkyl]

Give me a relative speed v in meters per second, and I can finish.

You wanted to set v = c; use that.

[StarThrower]

You wanted to set v = c; use that.

Give me a numerical value please Mr Hurkyl, since the speed of light depends upon inertial reference frame. :smile:

Regards,

Star

[Hurkyl]

I thought you were trying to derive a contradiction within SR, you know, where the speed of light doesn't depend upon inertial reference frame.

[StarThrower]

I thought you were trying to derive a contradiction within SR, you know, where the speed of light doesn't depend upon inertial reference frame.

I thought we just figured out that SR is wrong. Therefore, the correct answer to your current question comes right out of Euclid.

Don't you realize that photons can move relative to one another????


I thought we were clear on at least that much.

Kind regards,

StarThrower

[Hurkyl]

I'm still critiquing your proof; we most certainly have not agreed SR is wrong.

You still require the assumption that inertial reference frames can travel at c. The derivation of the time dilation formula does not work if v >= c; if you would carry out the exercises I described, the reason would be clear.

In the end, this will amount to a proof that inertial reference frames cannot have relative velocity c in SR. Until you can prove inertial reference frames can have relative velocity c in SR, you have not proven SR internally inconsistent.

[StarThrower]

Don't go so fast.

We have a right triangle, the hypotenuse of which is S. The base of which is 1/2 v delta t`, and the height of which is D. You want to let D=1 meter.

Additionally, S = 1/2 c` delta t`

Thus

(\frac{c^\prime \Delta t^\prime}{2} )^2 = 1 + (\frac{v \Delta t^\prime}{2} )^2

In the lab frame, there is an experiment going on, and the experimenters there came up with a value of c = 299792458 meters per second for the speed of light in that frame. The following equation holds:

c = \frac{2D}{\Delta t}

Additionally:

c^\prime = \frac{2S}{\Delta t^\prime}

If my memory serves me.

So do you want me to use v = 299792458 meters per second?


P.S. Got to go to work now, be back tomorrow. I will do the things you ask, you will see that SR contradicts.

Kind regards,

Star

[Hurkyl]

Yes, yes I would.

[Hurkyl]

Shall we make a clarification on the definition of "inertial reference frame?"

In particular, shall we declare that if an event (or state, if you prefer) occurs in one inertial reference frame, then it must occur in all inertial reference frames? Or shall we permit inertial reference frames that violate this condition?

[StarThrower]

Shall we make a clarification on the definition of "inertial reference frame?"

In particular, shall we declare that if an event (or state, if you prefer) occurs in one inertial reference frame, then it must occur in all inertial reference frames? Or shall we permit inertial reference frames that violate this condition?

What do you mean?

Regards,

Star

[StarThrower]

I am not sure where Hurkyl is going with the numerical examples, but here is the logic:

In the problem that we are analyzing, we have a right triangle, the hypotenuse of which is S. The base of which is 1/2 v delta t`, and the height of which is D.

Additionally, S = 1/2 c` delta t`

Thus

(\frac{c^\prime \Delta t^\prime}{2} )^2 = D^2 + (\frac{v \Delta t^\prime}{2} )^2

In the lab frame, there is an experiment going on, and the experimenters there came up with a value of c = 299792458 meters per second for the speed of light in that frame. The following equation holds:

c = \frac{2D}{\Delta t}

Additionally:

c^\prime = \frac{2S}{\Delta t^\prime}

Using this information, Hurkyl was able to derive for us the following result:

\Delta t &= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t\'\;

No one has a problem with the formula above, and c is substitutable for v in the formula. When v=c, the formula predicts that delta t is equal to zero, which means that no time passes in the lab frame, and that leads to an explicit contradiction. So, here is the logic up to this point in the argument:

If c=c` and v = c then (x and not x), for some statement x.

At that point, we have the following absolute fact (independent of relativity), which I am sure Hurkyl will agree to:

Not [ c=c` and v = c]

where c,c`, and v are defined as stipulated in this problem.

Hence, we can all use demorgan's theorem of logic to arrive at:

not (c=c`) OR not (v=c)

In order to let v=c in this problem, let two photons have been emitted from the photon gun at right angles to one another. Thus, clock B is moving away from the photon gun at speed 299792458 meters per second.

There are THREE frames you can analyze the relative motion of the photons in. The frame where the gun is at rest, the frame where photon 1 is at rest, and the frame where photon 2 is at rest.

I will save everyone a lot of time.

In order to see where the error in the special theory of relativity is, consider the relative speed of the two photons. In a reference frame in which photon 1 is at rest, photon two has speed c2. In a reference frame in which photon 2 is at rest, photon one has speed c1.

Since speed is relative, we have c2=c1. It now follows that:

\Delta t = \Delta t^\prime

The problem in the theory of special relativity should now be visible to all.

Kind regards,

The Star

[Severian596]

What do you mean?

Regards,

Star


An event is something that happens at a position in the inertial frame at a given time in that frame. Thus an event is located by three position coordinates and one time in each inertial frame. Of course, the coordinates of the events will be different in different inertial frames moving relative to one another at constant velocity, but the event itself is absolute. Examples of events are birth, death, explosions, detectors going off, etc. In other words, if you are born in one inertial frame, you are also born in any other inertial frame - the time and location of your birth may be different in different frames, but the fact of your birth is absolute.


And there you go

[StarThrower]

Shall we make a clarification on the definition of "inertial reference frame?"

In particular, shall we declare that if an event (or state, if you prefer) occurs in one inertial reference frame, then it must occur in all inertial reference frames? Or shall we permit inertial reference frames that violate this condition?

No reference frame can violate that condition (assuming I understand you correctly).

Regards,

StarThrower

[Hurkyl]

The point of my numerical examples was to demonstrate that the conditions of the thought experiment fail in the case where v = c.

IF the photon does indeed traverse a triangular path (or even a degenerate one) from the gun to the mirror and back to the gun, then we can form the equation:

(\frac{c \Delta t^\prime}{2} )^2 = D^2 + (\frac{c \Delta t^\prime}{2} )^2

which simplifies to

0 = D^2

Which is clearly false.

Our assumption that, within SR, there exists a reference frame with velocity c relative to F1 in which the photon under analysis traverses this triangular path is fase.

So only does the derivation of the time dilation formula fail, we also have that F2 fails to be an inertial reference frame, because it is impossible for it to observe the three events (or states, if you prefer) where the photon leaves the gun, strikes the mirror, and returns to the gun.

[StarThrower]

The point of my numerical examples was to demonstrate that the conditions of the thought experiment fail in the case where v = c.

IF the photon does indeed traverse a triangular path from the gun to the mirror and back to the gun, then we can form the equation:

(\frac{c \Delta t^\prime}{2} )^2 = D^2 + (\frac{c \Delta t^\prime}{2} )^2

which simplifies to

0 = D^2

Which is clearly false.



How is it false?

I believe you arrived at the following formula, which I will just refer to as the Hurkelian result:

\Delta t &= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t\'\;

So when v=c the previous equation leads us to the following equation:

\Delta t = 0 {\Delta t^\prime}

From which we can see that the left hand side must be zero. On the other hand, Delta t` doesn't have to be zero, and the equation is still true.

Did you mean something else?

Regards,

Star

[Hurkyl]

You're seriously asking me why D2 = 0 is false?

[StarThrower]

You're seriously asking me why D2 = 0 is false?

Yeah, I want to see your reasoning.

Here is one line of reasoning:

c = \frac{2D}{\Delta t}

Hence:

c \Delta t= 2D

So if delta t is equal to zero (and it is if v=c) then D=0.

But if you have a different line of reasoning, I want to see it. The reason I assumed you did, is because you wrote a formula which involved delta t` and D, and then implied the result that D=0 followed from that formula.

Kind regards,

StarThrower

[Hurkyl]

There's a whole grab-bag of reasons why we conclude D = 0. The problem is that D doesn't have to be zero in the thought experiment.


Or, I suppose we can conclude that there is no distance between mirrors and photon guns instead of there being no inertial reference frames with relative velocity c.

[StarThrower]

There's a whole grab-bag of reasons why we conclude D = 0. You presumed D was not zero, though, in the thought experiment, did you not?


Yes, D is nonzero by stipulation. We have a big experiment set up in our lab, involving a laser gun, and a distant mirror. We are absolutely certain that not (D=0) in reality. But here we are investigating theory. And as we are investigating theory, we have come across something odd. We have come across the result that:

(\Delta t = 0) \wedge (D=0)

(the wedge symbol ^ means AND)

We now quickly ask ourselves how we arrived at this obviously incorrect result.

Let us briefly tabulate all unclosed assumptions:

1. c=c`
2. v=c

Thus, we are here:

If c=c` and v=c then (D=0 and not (D=0)).

(Keep in mind, that at the beginning of this reasoning event, we stipulated as true, that not (D=0). Then, during this reasoning event, we opened the scope of several assumptions, which allowed us to conclude that D=0. We can now close the scope of all the assumptions, to arrive at a statement which is true, and not true by assumption. Just plain old true.)

Then we use reductio ad absurdum to quickly get here:

not (c=c` and v=c )

Then DeMorgan's law of conjunction gets us here:

not (c=c`) OR not (v=c )

And here is where I wanted to be, watching the world wake up to history.

Kind regards,

The Star Thrower

[Hurkyl]

not (c=c`) OR not (v=c )

If this is all you're trying to prove, we could have gotten here much sooner, and in a much less sloppy manner. Oddly enough, if you've paid attention, you'd've noticed we've been telling you not(v=c), so you wouldn't have wasted all this time proving the above statement which we already take to be true.

[StarThrower]

If this is all you're trying to prove, we could have gotten here much sooner, and in a much less sloppy manner. Oddly enough, if you've paid attention, you'd've noticed we've been telling you not(v=c), so you wouldn't have wasted all this time proving the above statement which we already take to be true.

You have been telling me that not (v=c) is true???

No, that is not what we currently can be said to know epistemologically speaking. We do not know that not (v=c). What we know is this:


Current Knowledge: not (c=c`) OR not (v=c)

P.S. This is not all I am trying to prove, I will prove not (c=c`). Additionally, the sloppy manner you speak of has tremendous pedagogical value.

[Hurkyl]

This whole thought experiment is supposed to be done in the context of SR, remember? And since SR => not(v=c), it is thus true that not(v=c).

[StarThrower]

This whole thought experiment is supposed to be done in the context of SR, remember? And since SR => not(v=c), it is thus true that not(v=c).

I guess I should repeat my point, it is a logico/deductive one.

A theory of physics is a conjunction of many statements. If any one of the statements is false, the whole theory is internally inconsistent.

The special theory of relativity is composed of many statements obviously. Some of which a reasoning agent is certain is true, and some of which a reasoning agent is not certain of the truth value of.

Suppose a theory of physics T consists of exactly 8 statements whose truth value we are uncertain of. We can define the portion of the theory which is uncertain to be the conjunction of these 8 statements. The rest of the theory we are certain of, and we cannot be certain of that which is false, hence the rest of the theory is consistent.

\Im = S_1 \wedge S_2 \wedge S_3 ... \wedge S_8

Now, if the uncertain portion of this theory leads us to a contradiction, then we know that at least one of the eight statements is false, though we will not yet know which one.

Let us define the special theory of relativity to be a theory of physics which consists of exactly one uncertain postulate, namely its fundamental postulate:

The Fundamental Postulate Of The Special Theory Of Relativity

The speed of a photon in any inertial reference frame is 299792458 meters per second.

Now, I introduced a do-able experiment, designed to measure the speed of light in two inertial reference frames moving relative to each other at a constant speed v. (This is exactly the 'thought experiment' to consider, in order to falsify SR)

Now, we reached the following true statement:

not [ c = c` and v=c ]

Where c,c`, and v were defined for this experiment.

The previous statement is logically equivalent to the following statement:

If c=c` then not (v=c)

But the case where v=c is where two photons are fired at right angles to one another, and this experiment is do-able in reality.

You have a dilemma on your hands, I do not.

Kind regards,

StarThrower

[Pergatory]

This whole thought experiment is supposed to be done in the context of SR, remember? And since SR => not(v=c), it is thus true that not(v=c).

Thank you!

StarThrower, despite all the math that has been thrown around you have only restated the same thing over and over again which is that you believe inertial reference frames can have v=c. Unless you address this SPECIFICALLY, you are not making progress, all you have done so far is tell us why SR is invalid if v=c.

[Hurkyl]

But the case where v=c is where two photons are fired at right angles to one another, and this experiment is do-able in reality.

Sure, we can fire two photons at right angles. However, this experiment which is doable in reality is not the thought experiment we were discussing. Where are the clocks attached by a rod? Where is the inertial frame of reference? Where is the triangular photon path?

[StarThrower]

The case where v=c is do-able. The theory of SR says that the experiment is not do-able. Therefore, SR has been falsified.

Kind regards,

StarThrower

Clock B is a photon, and clock C is a photon. Under SR's assumption, they move at the same speed relative to the photon gun. Hence, the distance between them isn't changing, in other words photon B isn't moving relative to photon C. So it is as if photon B and photon C are attached by a rigid rod according to SR. That having been said, the rest frame of the photon gun was stipulated to be an inertial reference frame. So there is at least one inertial reference frame. Now, consider the do-able experiment.

Two photons are simultaneously emitted from the origin, at right angles in inertial reference frame F1.

To be clear, photon A travels in the j^ direction (direction of increasing y coordinates), and photon B travels in the -i^ direction (direction of decreasing x coordinates.

The photon gun is located at the origin of F1 obviously, and there is a clock at rest there, clock A, which ticks in units of seconds. Now, 299792458 meters in the j^ direction, there is a mirror stationed. Every time a photon has been fired at that mirror, it took two seconds to return to the photon gun, as measured by clock A.

Similarly, let there be a 2nd mirror stationed 299792458 meters in the -i^ direction.

Under the operating assumption of SR, when we fire the two photons, they should return to the photon gun simultaneously.

Clearly, the two photons are in relative motion during the experiment. I now ask a very simple question.

What is the speed of photon A relative to photon B?

The result which you must inevitably get, is that:

\Delta t = \Delta t^\prime

Kind regards,

StarThrower

[Hurkyl]

The case where v=c is do-able.

On what grounds do you assert that?

[StarThrower]

On what grounds do you assert that?

I've done the experiment.

Regards,

StarThrower

[Hurkyl]

Excellent! What was your experimental setup? What were the lengths of the sides of the triangle? What was the the reading on the clocks at each event of interest? What was the experimental error?

And what bearing does the experiment have on the internal consistency of SR?

[Severian596]

I'm also curious on how he attained an inertial reference frame during the experiment...

[StarThrower]

Excellent! What was your experimental setup? What were the lengths of the sides of the triangle? What was the the reading on the clocks at each event of interest? What was the experimental error?

And what bearing does the experiment have on the internal consistency of SR?

The distance to the mirror was about 20 feet or so, and an oscilloscope was used to make a time calculation. The laser used was helium neon. As for all the juicy details, I don't remember them now. But, you don't need them anyways, to check the internal consistency of SR. All you need to do, is compute the relative speed of two photons fired at right angles to one another simultaneously.

As someone correctly pointed out, SR says the relative speed should be c. So under the single assumption of SR, the triangle in the rest frame of the photon gun is an equilateral triangle. Thus, the angle is 60, when it is 90 in reality. Thus, the theory of relativity ends in contradiction. It is internally inconsistent.


Kind regards,

StarThrower

P.S. Of course you do need to prove that the photon's frame is inertial. :smile:

[Hurkyl]

Er, so the oscillosope was moving at light speed?

And what definition are you using for "relative velocity of photons" and what bearing does it have on the validity of SR?

[Hurkyl]

You seem to have lost all of your attention to detail. Don't you find it telling that you get sloppy every time you try to make your conclusion that SR is inconsistent?

[StarThrower]

Er, so the oscillosope was moving at light speed?

And what definition are you using for "relative velocity of photons" and what bearing does it have on the validity of SR?

The experiment was designed to simply measure the speed of light. We couldn't even get 299792458 exactly. There was only one mirror, there didn't need to be two. There was only one clock, and it was at rest in the lab frame.

But of course you are just sidestepping the issue.

Two photons are emitted from the origin of inertial reference frame F1, at right angles. What is their relative speed?

Assume the fundamental postulate of SR is correct.

Thus, each photon is moving at speed 299792458 meters per second relative to the photon gun.

Finish the analysis, tell me what the relative speed of these two photons is, assuming SR is correct.

Kind regards,

The Star

[Hurkyl]

But of course you are just sidestepping the issue.

You claimed to have carried out the thought experiment we were describing, with v = c. Whether or not I'm sidestepping, you are being intentionally misleading, at best.


Two photons are emitted from the origin of inertial reference frame F1, at right angles. What is their relative speed?

The only definition of relative speed of which I'm aware does not apply to this situation. Would you care to provide one?

[StarThrower]

You seem to have lost all of your attention to detail. Don't you find it telling that you get sloppy every time you try to make your conclusion that SR is inconsistent?

You know what they say... you can lead a horse to water, but you can't make him drink.

And no I'm not getting sloppy. I have led you to where I wanted to. Consider two photons fired at right angles to one another. Assume SR is correct. What is their relative speed?

Regards,

Star

[StarThrower]

The only definition of relative speed I'm aware does not apply to this situation. Would you care to provide one?

Speed in an inertial frame = distance traveled in inertial frame divided by time of travel in inertial frame

Regards,

StarThrower

[Hurkyl]

Ok, so that's speed in an inertial frame. What's relative speed?

[StarThrower]

Ok, so that's speed in an inertial frame. What's relative speed?

Suppose that two objects are not moving relative to each other.

The distance between them in any frame is unchanging.

In this case, the relative velocity v is zero.

Case II: not (v=0)

Suppose that two objects are in relative motion. Thus, the distance between them is changing.

Choose one object to be at rest, at the origin of its own frame of reference, say F1. Since the two objects are in relative motion, the distance between them is changing. Thus, the coordinates of object 2 are changing in F1.

Let the position of object 2 at the current moment in time, in F1 be denoted by (x,y,z).

Define the position vector of object 2 in F1 as follows:

\vec{r} = xi + yj + zk

Let there be an infinite number of clocks in F1, one at every point. Let all the clocks be synchronized.

Let (x1,y1,z1) be the position of object 2 in F1, the moment when all clocks read 1. Let (x2,y2,z2) be the position of object 2 in F1, the moment when all clocks read 2, and so on.

The relative speed of object 1 to object 2 can be defined as follows:

|\vec{V}| = \frac{|\vec{r2} - \vec{r1}|}{2-1}

Or we can define time of travel using arbitrary clock readings. Let object 2 be located at (x1,y1,z1) when all clocks read A, and later, let it be located at position (x2,y2,z2) when all clocks read B. The clocks increase in their readings, tick at the same rate (by hypothesis), hence B>A. The relative speed of object 2 to object one is the magnitude of the objects velocity vector which is:

\vec{V} = \frac{\vec{r2} - \vec{r1}}{B-A}

Kind regards,

StarThrower

Keep in mind, that the above definition is only useful when the relative speed is constant in time. If the objects are accelerating relative to each other, then the denominator must be vanishingly small.

[Hurkyl]

Suppose that two objects are in relative motion. Thus, the distance between them is changing.

According to some reference frame, I presume you mean?


Choose one object to be at rest, at the origin of its own frame of reference, say F1.

This doesn't apply to photons. Would you care to provide a definition of relative speed that does work for photons?

[StarThrower]

Choose one object to be at rest, at the origin of its own frame of reference, say F1.


This doesn't apply to photons. Would you care to provide a definition of relative speed that does work for photons?


Yes it does.

Kind Regards,

StarThrower

P.S. Certainly you can analyze the motion in a reference frame in which the photon is at rest. Whether or not that frame happens to be inertial is a totally separate issue.

[Tom Mattson]

So this is what this thread has boiled down to?

OK, fine.

Yes it does.


No, it doesn't.

There is no reference frame in which the photon is at rest.

[Hurkyl]

Yes it does.

This definition requires one of the objects in question to have a rest frame. Since photons don't have rest frames, this definition does not work.

[StarThrower]

So this is what this thread has boiled down to?

OK, fine.



No, it doesn't.

There is no reference frame in which the photon is at rest.

Incorrect Tom. You can choose any object in the universe you wish, and fix a rectangular coordinate system to it.

What you are concerned about is whether or not such a reference frame is an inertial reference frame, well I have news for you... if the photon isn't being subjected to a force, then it is.

Kind regards,

The Star

[Tom Mattson]

Incorrect Tom.


:rolleyes:


You can choose any object in the universe you wish, and fix a rectangular coordinate system to it. Makes sense to place the object at the origin of the system/reference frame. Then you get to the concept of "worldline" but I'm not going there.


Whatever. If you fix a rectangular system to a photon, then that rectangular coordinate system will be moving at the speed of light in anyone's frame.


What you are concerned about is whether or not such a reference frame is an inertial reference frame, well I have news for you... it is.


If you consider an inertial reference frame to be one that can always be brought to rest by a change of coordinates, then I have news for you: A photon cannot be placed at the origin of such a frame.

[StarThrower]

:

If you consider an inertial reference frame to be one that can always be brought to rest by a change of coordinates, then I have news for you: A photon cannot be placed at the origin of such a frame.

Notice I changed that post slightly. It now reads:

If the photon isn't being subjected to an outside force, then that photon is in an inertial reference frame.

Kind regards,

StarThrower

[Tom Mattson]

Notice I changed that post slightly. It now reads:

If the photon isn't being subjected to an outside force, then that photon is in an inertial reference frame.


So? What does that have to do with the fact that you cannot consider the photon as the origin of a stationary frame?

[StarThrower]

This definition requires one of the objects in question to have a rest frame. Since photons don't have rest frames, this definition does not work.

Well then Hurkyl, that's really what it all boils down to huh?

A photon placed at the origin of a frame is at rest in a frame. It's in a rest frame, regardless of whether or not that frame is inertial. I mean we can view life from a photon's point of view.

It all boils down to this:

Is a photon which isn't being subjected to a force in an inertial reference frame?

The answer of course is yes.

Kind regards,

StarThrower

[Hurkyl]

A photon placed at the origin of a frame is at rest in a frame.

Too bad that the time axis of such a coordinate chart is a null vector, and thus can't be considered a frame.


Is a photon which isn't being subjected to a force in an inertial reference frame?

The answer of course is yes.

Why of course?

[Hurkyl]

And no I'm not getting sloppy.

Earlier on, you insisted on long, detailed proofs, even of elementary facts (though, admittedly, you were making tons of implicit assumptions).

Now, your assertions are backed up with phrases like "of course it's yes".

That's what I call getting sloppy. You were only interested in proving the things with which you know I'll agree; once we get to points of debate, you have lost all interest in proof.

[russ_watters]

Well, what a nice thread. Scrolling back a little, I found this gem:
I did not "assume" the speed of light depends upon inertial frame, in fact, I assumed the exact opposite (that it doesn't), and subsequently arrived at a contradiction. Translation: I assume SR to be wrong, therefore SR is wrong.

Hmm....speaking of contradictions. If you consider an inertial reference frame to be one that can always be brought to rest by a change of coordinates, then I have news for you: A photon cannot be placed at the origin of such a frame. We've discussed this one too.

Essentially guys, StarThrower bases his beliefs here on several basic assumptions that are contrary to accepted physics. I honestly don't know how to fix that: like he said, "you can lead a horse to water..."