# Attention Paid To Accelerating Reference Frames Overthrows SR

1. Mar 25, 2004

### 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.

2. Mar 25, 2004

### 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?

Last edited: Mar 25, 2004
3. Mar 25, 2004

### Chen

4. Mar 25, 2004

### 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.

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.

5. Mar 25, 2004

### StarThrower

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

6. Mar 25, 2004

### Staff: Mentor

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.
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.

Last edited: Mar 25, 2004
7. Mar 25, 2004

### Severian596

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 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.

8. Mar 25, 2004

### 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:

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.

Last edited: Mar 25, 2004
9. Mar 25, 2004

### Severian596

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.

10. Mar 31, 2004

### StarThrower

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

11. Apr 2, 2004

### DrMatrix

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.
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.

12. Apr 2, 2004

### DrMatrix

This is allowed because the two clocks are next to each other
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.

13. Apr 2, 2004

### Hurkyl

Staff Emeritus
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:

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

Last edited: Apr 2, 2004
14. Apr 3, 2004

### StarThrower

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

15. Apr 3, 2004

### StarThrower

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

Last edited: Apr 3, 2004
16. Apr 3, 2004

### StarThrower

Do you want me to show you how Hurkyl?

17. Apr 3, 2004

### DrMatrix

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.

18. Apr 3, 2004

### StarThrower

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

19. Apr 3, 2004

### StarThrower

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

Last edited: Apr 3, 2004
20. Apr 3, 2004

### Hurkyl

Staff Emeritus
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.