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digi99
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I was calculating there a time dilation in frame A for B in frame B compared to a light wave, formula time dilation (lesser time) = V/C (or t' = t. (1 - V/C) or x' = x . (1 - V/C)) (I considered your own movement V. t is subtracted from the same light wave in frame A and B, the end effect is it will be smaller for B and distances too of course).
None of the observers need to do any calculations to get the time dilation - all they have to do is look at each other's clocks.the time dilation calculated by A is a factor γ bigger
Digi99 is trying to provide a simple way to illustrate time dilation which simply means a clock running slower the faster an observer moves. If you look at posts #1 and #6 on his link on the first post of this thread you will see the clearest explanation. He starts with a monochrome light source and two observers who have "special clocks" that can count the wave cycles of the light coming from the light source during a period of one second. The first observer's "special clock" counts out the same number of cycles as the light source is emitting in one second which he calls t. The second observer is moving at speed v away from the light source. He will count t(1-v/c) cycles coming from the light source. Thus, a very simple way to show that a moving observer's "special clock" runs slower than a stationary observer's "special clock". Note that at v=0 the "special clock" runs at the regular rate. At v=c the "special clock" comes to a standstill.digi99 said:I think totally I gave enough information, but I know I got very few feedback (only from Ghwellsjr but ended in a kind of doppler effect what I did not mean).
ghwellsjr said:Digi99 is trying to provide a simple way to illustrate time dilation which simply means a clock running slower the faster an observer moves. If you look at posts #1 and #6 on his link on the first post of this thread you will see the clearest explanation. He starts with a monochrome light source and two observers who have "special clocks" that can count the wave cycles of the light coming from the light source during a period of one second. The first observer's "special clock" counts out the same number of cycles as the light source is emitting in one second which he calls t. The second observer is moving at speed v away from the light source. He will count t(1-v/c) cycles coming from the light source. Thus, a very simple way to show that a moving observer's "special clock" runs slower than a stationary observer's "special clock". Note that at v=0 the "special clock" runs at the regular rate. At v=c the "special clock" comes to a standstill.
OK, this formula is incorrect. The correct formula is [itex]t'=\gamma(t-vx/c^2)[/itex]. See here.digi99 said:It's for me (mathematics thinking) a big question you don't see the relation t'=t(1-V/C) (before Lorentz) and after Lorentz t'=γ.t(1-V/C). I tried to explain how I come to t'=t(1-V/C) by thinking in the Newton way.
DaleSpam said:OK, this formula is incorrect. The correct formula is [itex]t'=\gamma(t-vx/c^2)[/itex]. See here.
OK, so since you are describing the light then using the notation I described above this should be written: [itex]t'_c=\gamma t_c(1-v/c)[/itex] which can be derived from the Lorentz transform as followsdigi99 said:The expression t(1-V/C) and x(1-V/C) are for the meassured light wave length (passing light wave) in frame A (how A it sees/calculated for B). So x is positioned on the light line and time t.
Huh? Please use the notation I suggested, or propose your own clear notation and I will use it. But I cannot tell if you intend these to be general coordinate transformations or if they are the coordinates of some specific worldline such as the worldline of the light pulse.digi99 said:In frame B you must take the coordinates of the transformed light line and so you must take t' and x'. Lorentz : x' = γ . (x - v.t) = γ . (x - v/c . t . c) = γ . (x - v/c . x) = γ . x . (1 - v/c). And t' = γ . t . (1 - v/c). Pfff...
DaleSpam said:OK, so since you are describing the light then using the notation I described above this should be written: [itex]t'_c=\gamma t_c(1-v/c)[/itex] which can be derived from the Lorentz transform as follows
By the second postulate [itex]t_c c=x_c[/itex]
By the Lorentz transform for the light wave [itex]t_c'=\gamma(t_c-vx_c/c^2)[/itex]
So by substitution [itex]t_c'=\gamma(t_c-v(t_c c)/c^2) = \gamma t_c(1-v/c)[/itex]
Um - no: light waves pass me just as fast when I move as when I don't.if you move, the light wave is slower passing you
In my notation [itex]t_c[/itex] is the time of an arbitrary event on the worldline of the light pulse in A's reference frame.digi99 said:So what you wrote here is what I suggested (you did it in the right way). Tc is the same time for the moving object.
Sure, if you are willing to accept the Lorentz transform then time dilation is very simple and doesn't require any drawings to explain.digi99 said:So from now on you can explain time dilation in a very simple way everybody in the world could understand, no magic anymore with complex drawings.
No, [itex]v_c=v_c'=c[/itex]. The light wave passes at the same speed in every framedigi99 said:If you see a passing light wave as (relative) time, if you move, the light wave is slower passing you so time is going slower (time dilation).
This is the first mention here about counting cycles. I thought we were describing a brief pulse of light. If you are talking about a continuous source of coherent light, then is this source at rest in A's frame or B's frame?digi99 said:You have to compare it to the original size of the light wave when standing still, the finally effect will be when moving that the light wave you see wil being smaller, just as time do. It relates to the counting cycles of a light wave (the total length of the passing light wave).
DaleSpam, Please reread my previous post:DaleSpam said:In my notation [itex]t_c[/itex] is the time of an arbitrary event on the worldline of the light pulse in A's reference frame.
Sure, if you are willing to accept the Lorentz transform then time dilation is very simple and doesn't require any drawings to explain.
No, [itex]v_c=v_c'=c[/itex]. The light wave passes at the same speed in every frame
This is the first mention here about counting cycles. I thought we were describing a brief pulse of light. If you are talking about a continuous source of coherent light, then is this source at rest in A's frame or B's frame?
We will need to modify our notation. I suggest that we replace the subscript c with a subscript number indicating which cycle of the light wave is referenced.
ghwellsjr said:Digi99 is trying to provide a simple way to illustrate time dilation which simply means a clock running slower the faster an observer moves. If you look at posts #1 and #6 on his link on the first post of this thread you will see the clearest explanation. He starts with a monochrome light source and two observers who have "special clocks" that can count the wave cycles of the light coming from the light source during a period of one second. The first observer's "special clock" counts out the same number of cycles as the light source is emitting in one second which he calls t. The second observer is moving at speed v away from the light source. He will count t(1-v/c) cycles coming from the light source. Thus, a very simple way to show that a moving observer's "special clock" runs slower than a stationary observer's "special clock". Note that at v=0 the "special clock" runs at the regular rate. At v=c the "special clock" comes to a standstill.
ghwellsjr said:You are describing a situation in which B moves away from A, correct? After awhile, if B turns around and moves toward A, does your idea still work?
There is nothing in there about the length of the light. If you want to talk about the length of a pulse of light then you will need two separate worldlines, one for the front of the pulse and one for the back. I would recommend using a subscript 0 to indicate the front of the pulse and a subscript 1 to indicate the back of the pulse. Also, without loss of generality you can set your unit of time and distance such that the duration and length of the pulse is 1.digi99 said:Pure Lorentz but expressed in total length of the passing light waves
That doesn't really matter. The wavelength and the frequency are inevitably linked, so if you change the wavelength then you must change the frequency. It is Doppler whether you count periods or measure wavelengths.digi99 said:(in fact I am not counting periods, that's a practice problem maybe, but I considered only the length of the passed light signal
The difference in both length and frequency is due to Doppler shift, time dilation is a small part of the Doppler shift.digi99 said:everything is going smaller, the total length and the periods if you like, the difference is because of the time dilation.
So far, the formulas seem to have little to do with what you are saying. Can you express your formulas in the notation I have suggested for clarity? Where is the length of the light pulse?digi99 said:And the formulas are right, so I don't tell nonsence and it has nothing to do with Doppler (see the light waves on a distance). What I tell you is not important, look only to the formulas.
DaleSpam said:There is nothing in there about the length of the light. If you want to talk about the length of a pulse of light then you will need two separate worldlines, one for the front of the pulse and one for the back. I would recommend using a subscript 0 to indicate the front of the pulse and a subscript 1 to indicate the back of the pulse. Also, without loss of generality you can set your unit of time and distance such that the duration and length of the pulse is 1.
That doesn't really matter. The wavelength and the frequency are inevitably linked, so if you change the wavelength then you must change the frequency. It is Doppler whether you count periods or measure wavelengths.
The difference in both length and frequency is due to Doppler shift, time dilation is a small part of the Doppler shift.
So far, the formulas seem to have little to do with what you are saying. Can you express your formulas in the notation I have suggested for clarity? Where is the length of the light pulse?
Whether you call it a pulse, beam, wave, or signal is not really important. What is important is the math. Since it has a length it must have a front and a back which means you need two worldlines to describe it, not just one.digi99 said:I am not talking about pulse light, I am only using just 1 long light wave (= light signal) with speed C. Suppose that light wave is already left hours ago before, and see that light wave from coordinates 0,0 as it arrives on the moment B starts moving. I don't know the name for such light wave ... maybe a beam ..
X and x' are coordinates, not lengths. A length is the difference between two coordinates, i.e. in my suggested notation something like [itex]L_c=x_1-x_0[/itex]digi99 said:The coordinate x and x' represent the total length of the passing light wave
Any signal may be decomposed into a sum of sine waves using the Fourier transform. Even if your signal is not repetitive you still have frequencies and wavelengths and Doppler shift.digi99 said:(maybe you think the wave length, as related with some cycles, no I mean the length of the traveled path of the passing light signal), so what have I to do with cycles and frequency in this case ?
Yes, there have been a number of communication problems and now some math problems, which is why I have requested that you be clear in your notation. I don't understand why you are so reluctant to do so.digi99 said:Would this be a communication problem, that when I say total length of the light wave, you think to the wave length (some cycles) ? (I mean the total length of the traveled path of the passing light wave between 0,0 and 0,x)
DaleSpam said:Whether you call it a pulse, beam, wave, or signal is not really important. What is important is the math. Since it has a length it must have a front and a back which means you need two worldlines to describe it, not just one.
X and x' are coordinates, not lengths. A length is the difference between two coordinates, i.e. [itex]L_c=x_1-x_0[/itex]
Simon Bridge said:@digi99: in order to communicate effectively it is required that there is a common language. This is a subject which has been under active study now for a long time, so there is a language that has already been developed for it. If you refuse to use that language you will be unable to communicate your ideas. The others have been trying to get you to use that language with the special notation: I urge you to adopt it.
Please bear in mind that this is such a well studied field that it is very unlikely that you have come up with anything not thought of before. Try to listen to what people here are trying to tell you because they are genuinely trying to help you avoid some quite common pitfalls that you seem determined to jump down into. (Specifically, but not restricted to, the idea that a particular reference frame is "really stationary" and everyone else just thinks they are stationary when they are "in fact" moving.)
The closest I can figure is that you believe you have come up with a method of teaching about time dilation that is simpler than the history-tested approaches commonly used to date. The trouble is that the simplifications involve ignoring quite a large chunk of relativity... which is what we want to teach. By concentrating on light pulses and paths like that you introduce ideas which will lead to worse confusions later.
So, instead, we require the students to do some hard work at the start, to make more important concepts easier to learn later. We also have to be careful not to leave too much hanging loose for pseudoscience and crackpots to take advantage of.
Thanks, this is helpful. I will walk through it step-by-step with you. Again, for simplicity I am using units where c=1.digi99 said:Ok DaleSpam, I understand and shall give the situation again with the right notation.
Are you familiar with the parametric equation of a line? We can write this worldline as: [itex](x_c,t_c)=(t,t)[/itex]. Do you see how this equation contains both the event [itex](0,0)[/itex] and [itex](x_c,t_c)[/itex] and it constrains it so that it has the correct speed to be light [itex]dx_c/dt=1=c[/itex]?digi99 said:I have in frame A a light wave following the x-as from (0,0) to (X_c, T_c),
Similarly, for B we have the worldline [itex](x_b,t_b)=(v_b t,t)[/itex].digi99 said:I have B moving from (0,0) to (X_b,T_b) with speed V_b.
No, it isn't. When we measure the length of something we measure the distance between the front and back at the same time. For instance, my car is about 4 m long. When I am driving at 100 kph if I took your approach and measured the the length of the car at t=0 and t=T_b=1h then I would get that my car is 100 km long. This is clearly not correct. I need to measure the position of the front and the back at the same time in order to get the length.digi99 said:The length of the light wave L_c = X_c - 0 = X_c.
DaleSpam said:Do you follow this so far?
DaleSpam said:You can use the Lorentz transform to change coordinates for any object or path. It is not limited to light nor to paths of objects which are stationary in one of the frames.
DaleSpam said:I wrote (digi99): The length of the light wave L_c = X_c - 0 = X_c.
No, it isn't.
The length of something is not at all the same as the distance traveled. If I drive my car for an hour the distance traveled by my car is 100 km, but the length of my car is still just 4 m. The two concepts are completely different.digi99 said:This I don't understand, the traveled path of a light wave is c.t in the derivation for Lorentz too. In my eyes is the traveled path also c.t, so L_c = X_c (started at origin (0,0)). The traveled path is equal to the length of the passing light wave (was a beam).