Johnny5

Dapthar, your discussion of epsilon/delta proofs was extremely good. I did not have time to inspect it carefully, but when I want to, it is something that will hold my attention. I could see your careful use of logic. Regards

No.

This is false. I don't understand your statement. Space is three dimensional, in the sense that at most three infinite straight lines can meet at a point, and be mutually perpendicular. Not more than three. Time has no spatial extent, and hence it is not a dimension in any geometrical sense. This I do not understand.

I am aware of this... Lorentz contraction appears. In other words, you cannot have one without the other. One more thing, in my approach, I am going to reach a statement which is true, regardless of whether or not time dilates. So I am going to isolate the point in the derivation at which the postulate of relativity is invoked. It will be clear, its been a few years since i've ran through the derivation.

Here is what I am going to do. I am going to derive the time dilation formula, using the light clock example. Then we can all see what assumptions go into the conclusion. Measuring the speed of light In order to measure the speed of something, we need to be able to measure the distance it travels, and the time of travel. If the object travels non-uniformly (meaning it speeds up and slows down during the measurement) then all we can compute is the average speed it has. But there is one special case of motion, called by Galileo uniform motion, and in this case the average speed is equal to the instantaneous speed. What Galileo meant by "uniform motion" is that at each successive moment in time, the object has the same speed it previously had. In other words, the speed of the object is constant, for the duration of the measurement process. Let the measurement take place in deep space, far from any significant gravitational fields, and let the measurement process take place in an inertial frame of reference. The measurement is going to be of the speed of a photon, emitted by a laser. Let there be a mirror located a distance D away from the laser. The distance D is measured by a ruler which is at rest in the measurement frame. For the sake of definiteness, let the unit of length be the meter. Here is how the measurement is going to be made. A photon is going to be fired at the mirror, hit the mirror, and be reflected back. The time of travel of this photon is going to be measured by a single clock at rest in the measurement frame, located next to the laser. So we can know two measurements with almost complete certainty, or if you prefer with almost complete accuracy. The total distance travelled by the photon is 2D (since the photon goes back and forth), and the total time of travel is measured by a clock at rest in the measurement frame. Since the experiment is taking place in the vacuum of space, the photon does not encounter any impedance as it moves through the vacuum. Postulate 1: The speed of the photon during this experiment is constant. After the experiment is over, the experimenter will have two numbers. One will correspond to the time of travel, lets say in units of seconds. Let Dt denote the amount of time in this frame (frame where laser is at rest). Now, the other number will correspond to the distance travelled by the photon in this frame, in the amount of time as measured by this clock. The instantaneous speed of the photon is: v1 = 2D / Dt Let the experiment have been done. Let the computation yield the following value: v1 = 299792458 meters/second So an event took place in that frame, which started at the moment the photon exited the laser, and ended the moment the photon returned to the starting point. The time of this event, in this frame is Dt Now, consider the exact same event, as viewed in a frame in which the laser gun is moving from left to right, at a constant speed V. Since the laser gun (and mirror) are moving to the right at constant speed V, the photon traverses a triangular path in this frame, specifically an isosceles triangle is traced out. The altitude of the triangle is D. However, the total distance traveled by the photon is the sum of the lengths of the sides of the isosceles triangle. In this frame, there is a ruler which is at rest, which also has units of meters, and two clocks which are at rest. Let us stipulate that the clocks are synchronized, they are of identical construction to the clock which is at rest in the frame in which the laser is at rest. Let us call these two clocks Clock A, and Clock B, and let us refer to the clock that is attached to the laser as Clock L. Now, at the moment the photon is fired out of the laser, let Clock A and Clock L coincide. At the moment the photon has returned round trip, let Clock B and clock L coincide. Let the reading on clock A at the moment the photon was fired, be denoted by T1. Let the reading on clock B at the moment the photon went round trip be denoted by T2. Since Clock A and Clock B are synchronized and tick at the same rate, the total time of travel in this frame is T2-T1 Definition: Dt` = T2-T1 We can break the isosceles triangle down into two right triangles. One of the legs of which has length D. The other leg has the following length: V Dt` / 2 Convince yourself of this, ask questions if you don't understand. Now, using the Pythagorean theorem, the hypotenuse H of this right triangle is related to the lengths of the legs as follows: H2 = D2 + (V Dt` / 2 )2 So the speed of the photon in this frame is given by: V2 = 2H/ Dt` Since H2 = D2 + (V Dt` / 2 )2 it follows that: 4H2 = 4 [ D2 + (V Dt` / 2 )2 ] Therefore: (2H) 2 = 4 [ D2 + (V Dt` )2/4 ] Therefore: (2H) 2 = (2D)2 + (V Dt` )2 Taking the square root of both sides we have: 2H = [ (2D)2 + (V Dt` )2 ]1/2 Therefore we have: V2 = [ (2D)2 + (V Dt` )2 ]1/2 / Dt` V2 is the speed of light in this frame, and V1 is the speed of light in the other frame. Since the other frame is an inertial reference frame, and this frame is moving uniformly relative to that, this too is an inertial reference frame. Recall: v1 = 2D / Dt Therefore: (V1)2 = (2D / Dt) 2 Squaring both sides of V2, we obtain: (V2)2 = [ (2D)2 + (V Dt` )2 ] / (Dt`)2 (V1Dt) 2 = (2D) 2 (V2)2 = [ (V1Dt) 2 + (V Dt` )2 ] / (Dt`)2 So we reach the following unconditionally true statement: (V2Dt`)2 = (V1Dt) 2 + (V Dt` )2 It involves the speed of light V1 in the rest frame of the laser. It involves the speed of light V2 in a reference frame in which the laser is moving at a constant relative speed V. It involves the time measurement Dt of this event E in the rest frame of the laser. It involves the time measurement Dt`of the exact same event, in a frame in which the laser is moving at constant speed V. At this point in the argument, we can either choose to invoke the fundamental postulate of the theory of special relativity, or not. Fundamental postulate SR: The speed of a photon in any inertial reference frame is 299792458 m/s. If we invoke the postulate, then V1=V2. Let us invoke the postulate, so that we can have shown how the postulate of SR leads to the time dilation formula. Assumption 1: V1=V2 Therefore: (V1Dt`)2 = (V1Dt) 2 + (V Dt` )2 Since we have assumed that there is a unique value for the speed of light in any inertial reference frame, let us set aside the symbol c, for this speed. Therefore we can write: (cDt`)2 = (cDt) 2 + (V Dt` )2 Since c^2 is nonzero, we can divide both sides of the statement above by c^2 to obtain: (Dt`)2 = (Dt) 2 + (V/c Dt` )2 From which it follows that: (Dt`)2 - (V/c Dt` )2 = (Dt) 2 From which it follows that: (Dt`) 2 [1 - (V/c)2 ] = (Dt) 2 Presuming that the quantity [1 - (V/c)2 ] is nonzero, we can divide both sides of the equation above by it, to obtain: (Dt`) 2 = (Dt) 2/ [1 - (V/c)2 ] Now, take the square root of both sides of the statement above, to obtain the following statement which will have the same truth value as the previous one: Dt` = Dt/ [1 - (V/c)2 ]1/2 The previous formula is the well known time dilation formula of SR. We can now remove the assumption, so that we have proven that the following statement is true: Theorem: If the speed of a photon is the same in any inertial reference frame then Dt` = Dt/ [1 - (V/c)2 ]1/2 Where the symbols have the meaning used in the derivation. QED Notice that if we assume that v1=v2 in the derivation above, we do not arrive at the result that: Dt` = Dt Delta t is the time of the event E in the rest frame of the laser, and Delta t` is the time of the same event, in a frame in which the laser is moving at a constant relative speed V. End of Derivation 1 So we have finally derived the time dilation formula from the fundamental postulate of the theory of special relativity. The next goal, is to show that if the fundamental postulate of the theory of SR is true, then the Lorentz contraction formula must be true. Let the length of the ruler in the frame in which the ruler isn't moving be denoted by L0. Regardless of the postulate of SR, the laser traversed the entire length of the ruler in the following amount of time in the rest frame of the ruler: Dt` And the laser was traveling at speed V in this frame. So, the total distance traveled by the laser in this frame is: V Dt` Which is just the length of the ruler in this frame, therefore: L0 = V Dt` Now, let us re-assume that the fundamental postulate of SR is true. Therefore: Dt` = Dt/ [1 - (V/c)2 ]1/2 Therefore: L0 = V Dt/ [1 - (V/c)2 ]1/2 Therefore: L0 [1 - (V/c)2 ]1/2 = V Dt Delta t is the time of event E in the rest frame of the laser, and V is the speed of any point on the ruler, in the rest frame of the laser. Let L denote the total distance traveled by any point on the ruler in the rest frame of the laser, during event E. Therefore we have: V Dt = L From which it follows that: L = L0 [1 - (V/c)2 ]1/2 The previous statement is called the Lorentz-Fitzgerald length contraction formula. We can now remove the assumption, so that we have proven that the following statement is true: Theorem: If the speed of a photon is the same in any inertial reference frame then L = L0 [1 - (V/c)2 ]1/2 Where the symbols get their meaning from the experiment. QED End of derivation 2 We can combine derivation one and two, as follows: Theorem: If the speed of a photon is the same in any inertial reference frame then (Dt` = Dt/ [1 - (V/c)2 ]1/2 AND L = L0 [1 - (V/c)2 ]1/2) QED Now think about how to interpret the length-contraction formula. At the beginning of the event, the point 0 on the ruler coincided with the laser. At the end of the event, the point L0 coincided with the laser. So, in the rest frame of the laser, the total distance traveled by the point 0 on the ruler was L, which is less than L0 Therefore, if the postulate of SR is true then the ruler is shorter in the rest frame of the laser. Keep in mind, that the length of the ruler L in this frame, is to be measured by a second ruler, which is at rest in the laser frame, and has the same units as the first ruler. Note that nothing in this derivation proved that the postulate of SR is true. However, we did succeed in proving the following fact: If either the length contraction formula or time dilation formula are false then the postulate of SR is false. Therefore, we can logically falsify SR, if either formula (LCF or TDF) leads to a contradiction. Let us make just one assumption. Assumption: The Length contraction formula is true. We will now encounter problems with simultaneity. Under the given assumption, it is provable that: X before Y and Y before X Where X,Y are different moments in time. Keep in mind that a moment in time doesn't last for any amount of time; consequently, the time dilation formula does not apply to moments in time, rather the time dilation formula applies to amounts of time. I will not prove this here. Regards

I've seen any number of people who would disagree with you (about the definition of derivative). Do you know what they are talking about?

There is nothing sure about this. This is perhaps the most complex problem in the history of theoretical physics. Well I am. So either go off and familiarize yourself with it, or have me explain it to you. Either or. (Somehow I think you're pulling my leg though, but no matter...) You will use the Pythagorean theorem in the derivation.

Ok, we are getting nowhere very fast. Is there any error in the definition of the derivative?

Are you familiar with the light clock derivation of the time dilation formula?

Well then we might not have any kind of disagreement here. Answer some questions please. Is space continuous? Is time continuous? Is motion continuous? Is angle measure continuous? Thank you.

There is such a high degree of symmetry in the problem you should expect the clocks to tick at the same rate. You are not exploiting that fact that each clock is in an inertial frame. I believe I added that neither is being subjected to any external force. and I don't know what you meant by a virtual frame. In A's rest frame, the relative speed is v1. In B's rest frame, the relative speed is v2. As long as they use the same units of length, and time, and the same definition of relative speed, it must be the case that: v1=v2. Actually, that fact has nothing whatsoever to do with measurements. It's a fact independently of how they measure v, or even if they can measure v. It's just that if they use the same units, their computations will yield the same answer.

I'm not sure what you mean by "indistinguishable" here. We can establish their position as follows: There is one and only one place in the universe, which is the center of mass of the universe. Let a three dimensional inertial reference frame have been set up there, with its origin permanently being the center of inertia of the universe. So we can use the center of mass of the universe to establish their position. The universe might not be homogenous. I understand the time dilation formula, I really wish you would attempt to derive it.

And this is what I am saying leads to a contradiction. Interesting article by the way, I thought it was very well written. Wikipedia seems to be getting better.

Well I don't know what you want from me. The question you asked is, "how do you define something to be infinitessimally small?" I don't see that you can. If you know better, then tell me. I thought you were going to use the limit concept to do exactly that.

Certainly.

Yes, its the famous definition of the limit, i know that. Well run through your own reasoning again for me. As far as how do I define something which is infinitessimally small, the answer is almost trivial. Take the difference between the quantity over two consecutive moments in time.

Yes. In either frame, when they compute the speed of the other using rulers and clocks at rest, they must come up with the same relative speed v.

How did you arrive at the conclusion that the time shown on each would be different?

They would mean nothing that I have ever pondered. Let me state my position. Time is measured by clocks. Length is measured by rulers. With that in mind, imaginary time is meaningless. With that in mind, imaginary length is meaningless.

Is this the reasoning you're talking about here: [math]\lim_{x\to c} f(x) = l \Leftrightarrow \forall \epsilon > 0 \exists \delta > 0 \text{ such that } |x-c| < \delta \Rightarrow | f(x) - l | < \epsilon[/math]. ?

What do you mean that dt cannot be defined by itself. Explain that right there, that.

I would answer this way. Depends on the object. Your question is conceptual. Suppose the object was a particle, with no internal parts. Then there is nothing which can exert a force upon this particle, so that it is a free particle. Therefore, in any inertial reference frame, either this particle is at rest, or moving in a straight line at a constant speed. So there would be no gravity, no gravitational force. On the other hand, suppose that all the matter in the universe, which currently exists, wherever it is, used to be concentrated into a single body. Gravity would make the shape spherical rather quickly (and I know this is true without needing a theory of gravity, I can see what it does to moons, planets, suns, etc. ). And if it was already spherical, then all the parts would pull on all the other parts. So I would say that in the case where the one object isn't a particle, that gravity would shape that object. Regards PS: As for whether or not every single particle constantly gravitationally interacts with every other particle, I have to say I don't know. In fact, I am not even sure zero-dimensional particles exist.