Proof of the Fundamental Theorem of Calculus

  • #51


So finally my summary.

The formula I found in my first topic (already a problem in many forums) (1 - V/C) . t before Lorentz and γ . (1 - V/C) . t after Lorentz was correct, for a light wave. The meaning lies in the relativistic Doppler effect, but it does not mean it can be have a second explanation. You could think in frame A is the prediction for the time dilation (1 - V/C) . t, in reality it's a factor γ bigger. The formula Lorentz came with, is dificult to show in little pieces with a formula like mine, otherwise you could found 1 / γ too, with the math integration function. So even if you see in many little pieces (1 - ΔV/C) . Δt you will not find 1 / γ. So for an easier explanation for the time dilation, you may not use it in this way, but you may use the traveled path of the light wave 1 / γ. That the length of a piece of lightwave would be γ has not a special meaning, because it is the length of a moving object, not measured while standing still. You could say the coordinates c (as used by DaleSpam) go after transformation to γ . c. But the path of a traveled lightwave c . t goes to c . t', so the lightwave will going smaller in size like all other objects in frame B (B is moving in frame A), t' = 1 / γ . t and x' = 1 / γ . x.

When something moves it will not going smaller to 1 / γ (and lower time) for us, but for the moving object itselves in his rest frame. But we calculate that length and time dilation for corrections, the moving object itselves in this own rest frame is not aware of this changes. Some thinks, a time must be corrected (my confusion in the beginning) by Lorentz, yes that's
true but not the time we meassure in our frame for the moving object, but the time of the moving object in his own rest frame.

Now my reformulation to explain time dilation easier for others (I think) on my website or in fact exactly what it is (maybe still difficult, but a confusing drawing does not make it more simple). My goal is to understand (general) relativity and all related topics (parallel universa etc.) but to give it through to others in a simple way (e.g. when I die my knowledge is gone, what's the point of knowing that difficult knowlegde only for myself). Time dilation is the very basic term to understand first.

In my explanation on my website I will first start about light and relation (relative) time and distance etc. before I explain next:

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First an explanation for a normal object, different than light.

Consider a passing train. If you are standing still, count the meters train passing with a special device developed for it for e.g. about 5 seconds. Now walk with a constant speed in the driving direction of a train and count with that device again for the same period, so 5 seconds in this case. Because you are moving you will count lesser meters (in fact your own
movement lesser).

Light presents time and distance in the most exact way by nature. At any moment by nature traveled distance / time = always the lightspeed c, so just nature, nothing to understand only to accept. So light is the most exact clock existing, so the length of the path of a traveled light wave can be recalculated to time passed. Every clock can meassure time, but all clocks needs corrections, light is the onliest exact clock by nature (in fact it defines relative time).

Consider a passing light wave in mind as the passing (relative) time. If you are standing still, after the same 5 seconds as for the train, the passing light wave is in distance c . 5 meters further. Now walk with the same constant speed in the direction of the light wave, the light wave seems to go slower for you in mind but by nature at any moment the speed must be
the same c, so this will only be possible when time and traveled distance of the light wave is going smaller for you (light wave is going smaller). Compared to standing still, the light wave has traveled in the same 5 seconds, lesser distance while moving. So the lesser distance represents lesser time. So compared to standing still, the 5 seconds while moving takes lesser time. So while moving time is going slower, 1 second moving goes slower than 1 second standing still. A clock represents the same time, so a moving clock goes slower. The difference in time (expressed in time while standing still), is called time dilation. It's only to accept nature, nothing more. But you are not aware of the slowing going time. Like the light wave is going smaller, everything is going smaller for you while moving in the same ration. Again just nature. In physics by formula is the passing time t' while moving equal 1 / γ . t (t is the passing time while standing still, γ is a calculated constant factor but 1 / γ is always lesser 1).

Maybe not exactly true, but easy to remember :

See a passing light wave as the (relative) time passing, if you move in the direction of that light wave it goes slower for you, so (relative) time goes slower.

----------------------------------

Ok, some will say, turn the direction of the train. You may not compare this (and in fact, you use not the most difficult explanation) because a train has not the same speed at any moment while you are moving, that's something extra's. Some will say, turn the direction of the light wave, that will be exactly the same but more difficult to understand, a movement gives a time dilation, but the direction of a movement is not important only the positive value in meters of your movement, time can't be negative (maybe a time dilation can when a speed can be greater than c somewhere proved in the future, in that case you will be going faster older I guess but not going back in time).
 
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  • #52


DaleSpam said:
In post 32 I did both the total length of the path (Doppler formula) and the length of the light wave (γ). Is that clear? I am not sure if you are confused or not.

I was writing my explanation on my website as in the earlier answer but better (pictures and scales for distances/times) and I was thinking about the γ length and want it to be cleared. Yes still confused about how to use Lorentz, but it does not change my explanation.

We know that t' = 1/γ . t for how far the light wave would traveled after Lorentz. But if I look to the transformation for the light wave I found γ . (1 -V/C) . t and this is not the same time. If I try to transform a speedline e.g. x(t) = 5 . t, I find other transform formules for 1 second of time, so in the transformed points when standing still is t' = 1/γ . t but not in the other points (between 2 seconds).

I am in doubt now if you may Lorentz for other points except standing still, and if yes which meaning has it ? Why are the times for the light wave different ?

I thought that a time dilation in one frame B, is for all the times compared to frame A (where B is moving) ? So a car's speed of 50 km / hour is still 50 km / hour in the other frame only distances and times are shorter but the ration is the same as for light is ... so confused again ...
 
  • #53


digi99 said:
I am in doubt now if you may Lorentz for other points except standing still, and if yes which meaning has it ?
You may use the Lorentz transform for any coordinates. The meaning is always the same, the Lorentz transforms the coordinates from one frame into the other. There is no restriction on which objects or which coordinates may be Lorentz transformed.

For example, in units where c=1 suppose you have a particle traveling in the x direction at .2 c starting at x=1. Then the particle's worldline would be represented by the parametric function:
f(t)=(t,.2t+1)

If you wanted to find its worldline in any other frame you would use the Lorentz transform to get the coordinates in that new frame.
 
  • #54


DaleSpam said:
You may use the Lorentz transform for any coordinates. The meaning is always the same, the Lorentz transforms the coordinates from one frame into the other. There is no restriction on which objects or which coordinates may be Lorentz transformed.

Hi DaleSpam, thanks again, I am back for a little while, it's because you read at most about 1/γ.t relations.

1) Ok in points (t, v.t), (t,v.t+1), (t,v.t+2) etc. in frame B you have always the relation with 1/γ.Δt for B standing still in frame B (in fact all points can be described in this way). For other movements in frame A (maybe logically) you have other time dilations in frame B, even time dilations compared to B standing still (if you subtract the 1/γ factor) because the relation between t and x is different (not a t, v.t relation) ?

2) We know that t' = 1/γ . t for B standing still in his frame B, so for him a light wave from frame A would traveled c.t' in his frame B (c.t in frame A from the start point of moving B in frame A, that's (0,0)). But if I look to the transformation for the light wave I (you too) found γ . (1 -V/C) . t and this is not the same time. Why are the times for the light wave different or what's different how one come to this different times ?
 
  • #55


digi99 said:
1) Ok in points (t, v.t), (t,v.t+1), (t,v.t+2) etc. in frame B you have always the relation with 1/γ.Δt for B standing still in frame B (in fact all points can be described in this way). For other movements in frame A (maybe logically) you have other time dilations in frame B, even time dilations compared to B standing still (if you subtract the 1/γ factor) because the relation between t and x is different (not a t, v.t relation) ?
The Lorentz transform is the relation that applies in all cases. The other, simplified relations only apply in certain circumstances (e.g. the clock is at rest in one of the frames).

digi99 said:
2) We know that t' = 1/γ . t for B standing still in his frame B, so for him a light wave from frame A would traveled c.t' in his frame B (c.t in frame A from the start point of moving B in frame A, that's (0,0)). But if I look to the transformation for the light wave I (you too) found γ . (1 -V/C) . t and this is not the same time. Why are the times for the light wave different or what's different how one come to this different times ?
Can you answer this question yourself? Think about the circumstances required for the simplified time dilation formula and whether or not it can ever apply to light.
 
  • #56


DaleSpam said:
The Lorentz transform is the relation that applies in all cases. The other, simplified relations only apply in certain circumstances (e.g. the clock is at rest in one of the frames).

Can you answer this question yourself? Think about the circumstances required for the simplified time dilation formula and whether or not it can ever apply to light.

DaleSpam said:
Can you answer this question yourself? Think about the circumstances required for the simplified time dilation formula and whether or not it can ever apply to light.

Thanks again DaleSpam. The reason that my topic had take such a long time is, that many physics students don't know how to use Lorentz (for light), so it's not that easy I guess. Finally you confirmed as first the found formula γ . (1 - V/C) . t.

But I think I get it now.

E.g. when you calculate the time included dilation for B standing still, that time is valid for all movements from frame A in frame B, how B it sees. So in frame A you use time t for all movements, in frame B for B standing still you take t' = 1/γ . t for all movements (so how B its experiences when standing still).

If you transform another speedline in frame A to frame B, you get the time in frame B for that speedline, how all other movements are seen in that speedline (e.g. could be a moving person C other than B). If you want to see the time when C is in rest, you have to consider it's rest frame (of course not possible for light).

Ok now I understand c.t' for the light wave, it's the time how B it sees/experiences. So the formula γ . (1 - V/C) . t is not valid how B it sees when standing still.

Now finally the transformed light wave ... hmm ... not that easy ...

I guess the time that the light wave experiences (γ . (1 - V/C) . t), so we (in frame A or in frame B) can calculate it's time dilation etc. That time dilation gives the relativistic Doppler effect (an effect caused by time dilation). Now you see the length γ in picture too, because (1 - V/C) . t presents a piece of lightwave in frame A (cut off by V/C), that length is γ bigger in frame B. But compared to its original size (without V/C) it will be smaller too because γ . (1 - V/C) < 1 (see my old topic, with examples in calculations).

So your own movement makes the traveled path of the light wave smaller, also when B is standing still, in that case I showed already too that your own movement is involved in the formulas (but in fact already to seen in the Lorentz transformation formulas for x).

Is this all right (this could be the final end of this or my old topic) ?

So B experiences the relativistic Doppler effect but he sees the traveled path of the light wave different than the light wave itselves ?
 
  • #57


DaleSpam said:
Can you answer this question yourself? Think about the circumstances required for the simplified time dilation formula and whether or not it can ever apply to light.

Fine, I can still edit this topic (forget the previous answer DaleSpam).

So in fact I mean, B standing still do not experience the relativistic Doppler effect (like we in frame A for the same light wave when standing still, just another time 1/γ . t).

We in frame A experience the relativistic Doppler effect of a light wave in the (for us) moving frame B and can calulate it's time dilation (that's why the calculated t' of the light wave is different than for B standing still in frame B). The formula γ . (1 - V/C) . t is for the light wave in frame A for a moving light source with speed V (my confusion all the time I think, I considered the whole rest frame B only for B itselves). It's still the same light wave of course, but B travels with the lightwave with the same speed V, so no Doppler effect for B.

All transformations of movements from frame A shows the coordinates (time, x, y, z) when the whole frame moves with a speed V, if B is a person/object it shows the time too when B is standing still, but shows of course also the coordinates and time of other objects.

Let me know if I am right now, so this topic can be finished ...
 
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  • #58


Previous answer changed ...
 
  • #59


DaleSpam said:
OK, so you want to find
L_c&#039;=|x_1&#039;-x_0&#039;|

From the Lorentz transform:
t&#039;=\gamma(t-vx) and x&#039;=\gamma(x-vt)
So in B's frame we have
(x_0&#039;,t_0&#039;)=\gamma(x_0-v t_0, t_0-v x_0)=\gamma(t-v t,t-v t) =
\frac{1}{\sqrt{1-v^2}}(t(1-v),t(1-v)) = \sqrt{\frac{1-v}{1+v}}(t,t)
Which is essentially just the relativistic Doppler formula.

Similarly we have
(x_1&#039;,t_1&#039;)=\gamma(x_1-v t_1, t_1-v x_1)=\gamma((t-1)-v t,t-v (t-1))

So now
L_c&#039;=|x_1&#039;-x_0&#039;|=|\gamma((t-1-vt)-(t-vt))|=\gamma

So after thinking a while I stay at my last answer (you are probably with Christmas holidays). My found and your found formula is for a moving light source with speed V (and gives the relativistic Doppler effect).

My prediction (1 - V/C) . t was for the time B could expected more or less when standing still and there was not direct a relation after Lorentz other than already mentioned some answers ago.

So you said my formula was wrong (was only a prediction), because it is the relativitic Doppler effect. But you are talking (I was too in my old topic, the confusion) now about a moving light source and that's different. So thinking in only the movement of B, the prediction saids it's time will be slower and that time is indeed slower after Lorentz (and some of relation with (1 - V/C)). That's all. Because I am a starter this took all a long time (but understandable, it's not easy and can be confusing sometimes).
 
  • #60


Hi digi99, sorry about the delay in responding. I have to read your posts many times to get to the point where I think that I understand what you are saying. It seems OK to me except for one small detail:
digi99 said:
It's still the same light wave of course, but B travels with the lightwave with the same speed V, so no Doppler effect for B.
I would not say that B travels with the lightwave since the light wave travels at c and B does not. I would say that B travels with the source of the lightwave. The v used in the Doppler shift formula is the relative velocity between the source and the detector. B detects no shift because B travels with the same speed v of the source.

If you want to talk about the speed wrt the lightwave, that is always c.
 
  • #61


DaleSpam said:
It seems OK to me except for one small detail:I would not say that B travels with the lightwave since the light wave travels at c and B does not. I would say that B travels with the source of the lightwave. The v used in the Doppler shift formula is the relative velocity between the source and the detector. B detects no shift because B travels with the same speed v of the source.

If you want to talk about the speed wrt the lightwave, that is always c.

Hi DaleSpam. As usual you are right, one have to carefully compose his/her sentences in physics. Excellent, this topic is now fully analysed. I have learned a lot from you, I will read my books much easier now because of the basic understandings of time dilation (just nature). Slowly I will place links in my other 2 topics to this topic as the final end, one per day :smile:
 

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