# Time dilation and basic math

1. Oct 9, 2006

### petm1

Time is a tool we use to measure motion, it is a dimension not in the same sense as length, width, or height, but rather it is used to show movement within the three named dimensions. Einstein used two light clocks, photons traveling between mirrors, in his ideal thought experiment. One clock stayed on earth with the first twin, the other clock left earth at the speed of light with the second twin. The earth bound twin sees his brother’s clock as slowing down, not because of a change in time rather a change in the timing between the two clocks. The earth bound twin sees the photons from his brothers clock first when the wave hits him after being emitted and second when the wave reaches him after the photons change directions at the top of the clock. The earth bound twin’s clock shows more time has elapsed between these sightings as his brother speeds away. Einstein called this difference between the clocks time dilation and proposed that only the second was affected. Using basic math to keep a ratio constant whenever you change the divisor; you must also change the dividend by the same amount. i.e. the expression 1/1, if you dilate the bottom number to a two it would be 1/2 which changes the overall amount of the ratio so to keep the ratio equivalent you must dilate the top number by the same amount which in our example would give you 2/2. To make Einstein's theory consistent with the math I learned as a boy if you dilate the second (in the ratio meters per second) you must also dilate the unit used to measure the distance traveled (meter) so that each clock would appear to each twin as still working correctly relative to themselves. By increasing the length of our unit of measurement we will be keeping the speed of light constant at 300,000k per second for each of the twins. By dilating the second only you are just showing the difference between the twins clocks but you are not showing why each twin still sees themselves as being unchanged. Time doesn’t decrease or increase with speed motion does, because clocks are how we measure motion then is stands to reason that it is our clocks and how they measure motion that increases or decreases with velocity.

2. Oct 9, 2006

### pess5

Yes.

In Lorentz's theory, he assumed an unchanging aether. The body's length and time had to contract in an unchanging space.

Per Einstein, a body never changes in and of itself per itself (ie the proper frame vantage). What changes is not the body or clock, but rather the way in which we measure the very dimensions of space & time themselves.

Lorentz published his paper on relativity some 6 months (or so) before Einstein published his 1905 Special Theory. Einstein says he never saw Lorentz's paper at that time, nor the famous Michelson/Morley interferometer null result. After Lorentz read it and thought about it, he ditched his theory and embraced Einstein's theory. He favored it because Einstein's method of attributing distortions to space & time itself was superior to Lorentz's belief that the electromagnetic field in atomic structure drove the distortions. Einstein's method showed that if space & time possessed a certain symmetry, then light can be invariant c while all use the very same mechanics.

pess

3. Oct 10, 2006

### Staff: Mentor

It might be helpful if you would define the word "clock" for us...

4. Oct 10, 2006

### petm1

Define the word clock; A device used for measuring and indicating time. In my example I used a light clock, but I believe anything with repeative motion would work. At this "time" we use the periodic oscillations of a cesium atom, about 9,192,631,770 hertz to represent a second.

5. Oct 10, 2006

### Staff: Mentor

Ok, so how does that definition fit with the piece I quoted above? It seems like you are saying that it is just clocks and not time itself that is relative.

6. Oct 10, 2006

### MeJennifer

Indeed.

In relativity it is not just clocks and lengths that are relative it is more general than that, all duration and distance intervals are relative. Only the combined interval in the form of: $$\inline {\sqrt { \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2 }}$$ is invariant, at least in special relativity since it assumes a flat space.

An alternative way of writing it is in tensor format:
$$ds^2= \eta_{\mu \nu} dx^\mu dx^\nu$$

$$\eta_{\mu \nu} = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{array} \right)$$

Last edited: Oct 10, 2006
7. Oct 10, 2006

### petm1

"Per Einstein, a body never changes in and of itself per itself (ie the proper frame vantage). What changes is not the body or clock, but rather the way in which we measure the very dimensions of space & time themselves."

This is what my original post was dealing with, why when dealing with time dilation does each observer still see themselves as being stationary?
I know that while sitting in front of this computer everything around me in my room appears to be standing still, but I also know that this 125 cubic meter space occupies about 3,750,000 cubic meters of space per second, i.e. using the rotation of the earth around itself, the sun, and the milky way galaxy. Now if I were the twin moving away at .9c in this same room, we would be occupying about 37,500,000,000 cubic meters of space per second, Add time dilation of about 2.5 and we would be occupying about 93,750,000,000 cubic meters. I don't know about you but by dilating time only I would think that I would notice not only the longer duration as compared to myself but also by the changes to the speed of light. I don't think that Einstein gets to break the basic rules of math, if you dilate time with an increase in your speed then you must dilate lenght at the same rate to keep everything relative around yourself, the meter and the second are directly related and to keep everything relative you must keep them proportional.

8. Oct 10, 2006

### pess5

When one is inertial, one cannot FEEL his own inertia. One is weightless. Since we cannot FEEL our inertia, we assume ourselves stationary. Imagine a half dozen inertial folks whizzing by each other in some flat local spacetime. There is no reason that one observer is preferred as stationary over the other. As inertial observers, they are all equally obliged.

x/t=c=X/T. X>x and T>t. The frame X,T is dilated wrt the frame x/t. Einstein breaks no rules. I'm not exactly sure what you are trying to say here? It sounds like you disagree with him in the 1st half, and agree in the last half. Are you saying that Einstein got it wrong, or that you agree with Einstein?

pess

9. Oct 11, 2006

### petm1

I agree with Einstein all the way, but I don't think he took it far enough. Einstein showed that the "proper frame vantage" to a outside observer will change due to velocity via time dilation, yet to the observer within the pfv everything will appear to be unchanged. If I were to increase my velocity to .9c having felt the change and knowing that I was going that fast compared to the observer I left behind Knowing that my second had dilated about two and a half times yet keeping my meter stick the same I would see a difference in the speed of light. To keep all things proportional my meter stick would have to dilate with my second. Think of it another way, my second may be longer than my outside observer due to time dilation but the distance that a photon has to travel in that second would still be 300,000k meters relative to me otherwise I would "see" a change in the speed of light.

10. Oct 11, 2006

### JesseM

The mathematics of relativity is such that different observers already measure the speed of light to be the same due to a combination of time dilation, length contraction, and the different definitions of simultaneity in each frame (if one observer's clocks are in sync in his frame, they will be out of sync in the frame of another observer moving relative to him). You can check the math for yourself to see no changes are needed to relativity to ensure it works out. If you like, I provided a simple numerical example showing how observers in motion using their own clocks and rulers will measure the speed of light to be the same in my first post on this thread.

11. Oct 11, 2006

### petm1

I am not talking about how I would see another clock or how I would measure the speed of light, I am talking about why each of the frames will always see themselves as, for a lack of a better word, normal. Everyone uses the lorentz transformation on the second with no problem thinking that the twins will age different, Einstein and lorentz disagreed about length contraction being a real or a visible phenomenon. All I am saying is you must use the same lorentz transformation on the meter to see your reality remain unchanged and the ratio of meters per second to remain proportional.

12. Oct 11, 2006

### JesseM

But that's just one of the two postulates of relativity--that the laws of physics will look the same in every inertial frame, so that if you have two sealed compartments in motion relative to each other, there won't be any difference in what observers inside the two compartments will see. If time dilated but rulers didn't shrink, then this wouldn't work, and you can derive both time dilation and length contraction from this postulate plus the postulate that the speed of light is the same in every frame.
I still don't understand what you're arguing--you do realize the lorentz transformation is already used on the meter in relativity, right? I'm pretty sure there was no disagreement between Einstein and Lorentz on this point, although there may have been some disagreement on what both length contraction and time dilation meant--whether there was an objective answer to which of two clocks was really ticking slower, or which of two rulers was really shrunk.

13. Oct 12, 2006

### petm1

Yes I do understand that the lorentz transformation is already used on the meter but it is used outside our frame to account for what you would "see", to the people inside our frame you see no changes we don't "see, feel or notice the contraction. If the other frame is moving away from our frame you would not see the constriction at all for it only showes constriction in the direction of travel. The laws of physics will look the same in every inertial frame. This is what I am questioning why? If you use the lorentz transformations to dilate the second and constrict the meter then you are changing the ratio of meters per second and therefore the speed of light in our home frame.

14. Oct 12, 2006

### JesseM

Yes, and the same is true of time dilation.
You would see contraction along whatever direction the object was travelling relative to you--why do you say "you would not see the constriction at all"?
So you're just asking how every frame can measure the speed of light the same if their distances and times are different? Keep in mind that the relativity of simultaneity also comes into play here--each frame would measure the speed of a light beam by using a pair of clocks which are at rest and synchronized in that frame, and then if the beam passed the first clock when it read t1 and passed the second clock when it read t2, and the distance between them according to rulers at rest in this frame is d, then the speed of the beam would be (t2 - t1)/d. But two clocks which are synchronized in one frame will be measured to be out-of-sync in every other frame (if two clocks are a distance x apart in their own rest frame, and synchronized in that frame, then in another frame which sees them moving at speed v along the axis joining the clocks, the clock at the back will have its time vx/c^2 ahead of the time on the clock at the front), so that's a third part of how each frame's measurements are different, besides the facts that their rulers measure length differently and their clocks measure time differently.

If you want to see an example of how these three factors come together to make it so that each observer measures the same speed for a given light beam, then look at my first post on this thread which I linked to earlier.

15. Oct 12, 2006

### petm1

"You would see contraction along whatever direction the object was travelling relative to you--why do you say "you would not see the constriction at all"?"

I said that when an object is moving away from you would be seeing its backside, the side opposite its direction of travel.

"So you're just asking how every frame can measure the speed of light the same if their distances and times are different?"

No, I am saying the reason that all observers see themselves as normal is their meters per second is proportional to themselves, because basic math states that to keep a ratio equivalent what ever you do to a second (time dilation) happens to your meter (length dilation), if your time dilates by a factor of 2.5 at .9c then your meter dilates by 2.5 times.

16. Oct 12, 2006

### yogi

In SR each observer will see the clock in the relatively moving frame run slower and the lengths contracted - but SR does not assert that clocks in the other frame actually run slower or that lengths actually contract - what is predicted and observed is that clocks put in motion relative to a frame which remains inertial will accumulate less time than clocks which have not undergone a change in motion. An actual objective age difference is always the result of some acceleration - whether it be a one way experiment such as a high speed laboratory particle or a round trip twin thought experiment - but the clocks themselves are not affected by the change in motion - they still tick at one second per second in their own frame - but because of the invariance of the spacetime interval, in the fame that has been accelerated, the combined spatial distance squared (vt) + the temporal distance squared (ct') will always equal the square of the temporal distance (ct) in the fame which has remained inertial.

Last edited: Oct 12, 2006
17. Oct 13, 2006

### petm1

"In SR each observer will see the clock in the relatively moving frame run slower and the lengths contracted"

The key word here is "see", via light waves or if you will a stream of photons. A photon which we symbolized by planck's constant, measured in Joule-second, times frequency, measured in hertz (cycles per second).

"but SR does not assert that clocks in the other frame actually run slower or that lengths actually contract"

This is the part we are discussing now.

"what is predicted and observed is that clocks put in motion relative to a frame which remains inertial will accumulate less time than clocks which have not undergone a change in motion."

Yes we all "see" this. To change a frames motion you have to change its energy level, or what I see as a change in its "volume" of space occupied over time.

"they still tick at one second per second in their own frame"

Yes all observers see their own clocks as still ticking one second per second, or they are staying relative to their clocks. To put it another way each observer is taking their second and then taking the length that light has traveled in that second dividing by 300,000k and that is the measure of their meter. This works out because meters per second stay relative to themselves, this would not work if the meter did not change or was contracted and the second did. Even the size of their photons stays relative, but the size of their photons are still tied with their dilated second. Which brings us back to basic math, to stay relative to your own frame your second must stay proportional to your meter.

"but because of the invariance of the spacetime interval, in the fame that has been accelerated, the combined spatial distance squared (vt) + the temporal distance squared (ct') will always equal the square of the temporal distance (ct) in the fame which has remained inertial."

I have no problem with this statement, I am talking about what is happening with the frame that is moving.

18. Oct 13, 2006

Staff Emeritus
Nothing is happening to it. For the people in it it's a rest frame and they see physics just as if they were absolutely at rest. This is "Galilean relativity" because Galileo demonstrated this point with the example of experiments on shore versus the same experiments done on a smoothly moving ship.

What changes is the interactions between the two frames. In these simple cases the interaction is only in emitting and perceiving light, but in the experiments hooked up to colliders like Tevatron and LHC, and going back to the original cyclotron, the interactions are much richer, and relativity has been confirmed in careful experiments for generations.

19. Oct 13, 2006

### petm1

"Nothing is happening to it. For the people in it it's a rest frame and they see physics just as if they were absolutely at rest. This is "Galilean relativity" because Galileo demonstrated this point with the example of experiments on shore versus the same experiments done on a smoothly moving ship."

Galileo never used time dilation.

"What changes is the interactions between the two frames. In these simple cases the interaction is only in emitting and perceiving light, but in the experiments hooked up to colliders like Tevatron and LHC, and going back to the original cyclotron, the interactions are much richer, and relativity has been confirmed in careful experiments for generations."

True statement, it is not just the interaction in emitting and perceiving light, it is happening to the frames themselves. I do not see my own time dilation, not because it is not there, it is because my meter stays proportional to my second at all times. It is basic math if you want to change the divisor, yet keep the ratio equivelent, you must change the dividend by the same amount.

Has any one in all of these experiments looked for a size increase in their particle. Light waves increase as they move. Every experiment I have read about or seen pictures of, the small particle appears to be bigger. I am not talking about any thing that would make us see anything different, yet when you do the basic math of keeping the meter proportional with your time and its time dilation, you get rid of most of the paradoxes related to SR. Twins paradox, i.e. when the moving twin's meter dilates big enough so his clock face measures the same size as the earth's orbit then his minute equals the earth bound twin's year, you can show this to childern and have them understand. Ehrenfest paradox, points the way, because he thought that the notion of rigidity is not generallly compatible with special relativity.

Last edited: Oct 13, 2006
20. Oct 13, 2006

Staff Emeritus
The bolded statement is false. If your dilation were real in your frame, it would have some definite value. But any observer in the universe might see you, and have any speed relative to you that is physically possible. And EVERY ONE of those observers would see your dilation as a DIFFERENT value, based on that relative speed. So your dilation could have any positive value at all! And all at once. Therefore the dilation does not belong to you or to your frame; it belongs to each relation you have with any other frame. Only by pertaining to the relationship does it become single-valued.

21. Oct 13, 2006

### JesseM

No, the time dilation equation doesn't tell you how fast you will actually see a moving clock ticking using light waves--what you see is affected by Doppler shift as well, which is a consequence of the fact that the light from each successive tick on the clock has a shorter or longer distance to travel to reach your eyes than the last one, depending on whether the clock is moving towards you or away from you. Because of this, a clock moving towards you will actually appear to tick faster than your own clock, not slower. Time dilation is what you're left with when you factor out the effects of signal delays. One way to do this would be to make only local measurements using a set of clocks at different locations, which are at rest and "synchronized" in my rest frame (although 'synchronized' in my frame means they're out-of-sync in other frames, because of disagreements about simultaneity). For example, if I have two synchronized clocks A and B at different locations, and a moving clock C passes by A when A reads t=1 second and C also reads t=1 second at the moment it passes, then later C passes by B when B reads t=3 seconds while C only reads t=2 seconds at the moment it passes, then I can say that C is ticking at half the speed of A and B in my rest frame.
SR would say it's meaningless to say clocks "actually" run slower or that they "actually" run at the same rate unless you specify slower or the same relative to some other clock or set of clocks. There are no absolute, non-relative truths about the rate of clock ticking in relativity.
But they could also define their meter in some other way, like 1.89 * 10^10 times the radius of the first orbit of a hydrogen atom which is at rest in their frame. Likewise, you don't have to define time in terms of light speed, you can also define a second as 9192631770 oscillations of a cesium-133 atom at rest in your frame. No matter what physical definition of distance and time they use, they will still find that the distance/time for light is unchanged (at least if they are measuring the two-way speed of light...the one-way speed of light also depends on what clock synchronization procedure they use).
What do you mean by "to stay relative to your own frame"? Suppose we lived in a universe where the laws of physics were different, where clocks slowed down depending on their speed relative to some preferred frame (say, the frame of the ether) while rulers did not shrink. Certainly if this were the case, observers in different frames would not measure the same speed of light if they used their own rulers and clocks. But in terms of their perceptions, their clocks would still appear to tick at the normal rate (since the rate of their brain processes changes along with the clocks), and their rulers would still appear to measure length normally.

Last edited: Oct 13, 2006
22. Oct 13, 2006

### petm1

I said; "I do not see my own time dilation, not because it is not there, it is because my meter stays proportional to my second at all times."

Every moving frame has time dilation, we may only note it with when viewing another frame and it may be that the time dilation is a product of both frames relative velocity but that does not make my statement false.

23. Oct 13, 2006

### petm1

I must admit, That I have been thinking that the results of any time dilation computation was credited to the other observer's frame, but I think it works out better knowing that we see the total time dilation effect of both observers frames while looking at the other observer. If you knew your own velocity then you could discount that portion, like we do for Doppler.

24. Oct 13, 2006

### JesseM

Our own velocity relative to what? Are you trying to argue that we have an absolute velocity, and likewise that there is an absolute truth about how much time dilation we are experiencing?

25. Oct 13, 2006

### petm1

We compare our motion with everything we see and nothing that is at absolute rest and no I don't think that we can know our absolute velocity.

By the way today is the first time that I have tried to use the quote option on this board, up to now I have been putting all quotes between the "." symbol.