Proof of the Fundamental Theorem of Calculus

[digi99]

Do you follow this so far?

Yes DaleSpam, I can follow it and I am very curious to the result ...

 

[Dale]

OK, so you want to find
L_c'=|x_1'-x_0'|

From the Lorentz transform:
t'=\gamma(t-vx) and x'=\gamma(x-vt)
So in B's frame we have
(x_0',t_0')=\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',t_1')=\gamma(x_1-v t_1, t_1-v x_1)=\gamma((t-1)-v t,t-v (t-1))

So now
L_c'=|x_1'-x_0'|=|\gamma((t-1-vt)-(t-vt))|=\gamma

 

[digi99]

Today/tomorrow I can finish my topic, because I wanted to finish it in the good way and so spent my time to it. So I was reading everything again in my notes, and I found many mistakes (that I have learned now, order your notes otherwise it will be a mess, I did it all in a hurry between my work which asks a lot of time from me too, and besides I am a starter with mathematics knowledge from a long time ago and not much physics, but I learn quick).

In my old topic (see link to that old topic in first answer in this topic) I wrote once that I was wrong and came with another formula (something different but right) and that was the end. But my statement was/is still there (see later).

Than I came in discussion with DaleSpam.

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.

And that I misinterpret probably. Lorentz is meant to calculate time and distance how the moving object B it experiences (like a person) in rest (but moving in frame A). So you may use that formula only to calculate time and distance when you are in rest. That found factor 1/γ (in case of constant speed) you may use than for corrections in time and distances in the objects rest frame. What I did was now to use the transformed coordinates also for other movements in frame A for frame B. In fact you may only transform coordinates for other moving objects with the factor 1/γ. So I came again in the wrong way on the wrong formulas as first found in my old topic.

So I thought now knowing what is right, maybe I can find a formula now which relates 1/γ and (1 - V/C). Yes I found, so you can see the relation between the prediction from A of the time dilation in frame A and the really time dilation in frame B. That's is just as I thought a linear relation.

The formula is generally in itex (yes I am willing to learn Simon, it's not the whole effect what I wanted, later maybe) :

t'=t . 1/γ . 1/\sqrt{1+v^2/c^2} . (1-v/c)

And of course for length's/distances too.

So new times and distances are expressed in a way I thought in frame A (prediction time dilation). And in fact in diagrams like prof. Sander Bais is working like that too in thinking (seeing a light wave as just something else).

Generally you may I think always say when you move compared to somebody in rest, times goes slower compared to a passing light wave. When you move, you move a specified distance in frame A, at the same time the light wave travels a specified distance (started together with you, mover). So in that time you traveled also a distance, when you arrive has the light wave lesser distance traveled seen by you, so lesser time than in frame A at the same time. This can only give a time dilation (light wave was a clock too). A light wave cannot travel the same distance seen from A and seen from B because B has moved.

If you change the direction of the light wave does not matter in this thinking, only the positive distance you moved is important (not the direction for calculating time).

I wrote (digi99): The length of the light wave L_c = X_c - 0 = X_c.

No, it isn't.

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). Or must we not talk about length but just over the length of a traveled path (maybe is meassuring length for an object something else as for light itself) ?

DaleSpam, I wait first for an answer on this new situation. Finally you will see a shorter light wave (and so related time), this must be visible finally in a diagram too ...

 

[digi99]

Stupid, formula is wrong, rest story is correct, maybe found a new one.

 

[Dale]

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).

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]

So later maybe I find one after reading Sander Bais.

For now the most closes formula's are from my old topic.

ΔXb_light = ΔXa_light . (1 - V/C).
ΔTb_light = ΔTa_light . (1 - V/C)

After Lorentz:

ΔXb_light = γ . (ΔXa_light - V/C . ΔXb_seen_from_a) (ΔXb_seen_from_a = V . ΔTa_light = V . T)

ΔTb_light = γ . ΔTa_light . (1 - V2/C2)

Expressed in X' and T' for the moving object B :

T' = γ . (T - V/C . X/C)

 

[digi99]

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.

Ok understand, but I (we) have to see the distance of the traveled light wave in a diagram.
That's my statement. And you know that will be shorter always (in some way related to your movement, in fact I say nothing special or not understandable I think).

 

[Dale]

Ok understand, but I (we) have to see the distance of the traveled light wave in a diagram.
That's my statement. And you know that will be shorter always (in some way related to your movement, in fact I say nothing special or not understandable I think).

Yes, as long as you do not confuse length and distance traveled you can calculate either, in fact, I have done both above. The distance traveled is given by the Doppler formula:
(x_0',t_0')=\sqrt{\frac{1-v}{1+v}}(t,t)

 

[digi99]

Yes, as long as you do not confuse length and distance traveled you can calculate either, in fact, I have done both above. The distance traveled is given by the Doppler formula:
(x_0',t_0')=\sqrt{\frac{1-v}{1+v}}(t,t)

I forgot to thank you for your answer (math). I think there is an error in your formula (the same as you thought some time ago, you forgot to divide by c^2 for the time transformation, I am never sure so I say it only).

But the situation has changed now, my formula is not valid anymore compared with after Lorentz. It can now be having another meaning.

The distance of light is c.t in frame A and c.t' in frame B used in the derivation of Lorentz, you can't say after that derivation, that way of thinking is not valid anymore. So the distance traveled by the light wave is c.t in frame A and c.t' in frame B (just a fact). You willl find the same result in a diagram. And don't think how do I meassure that, maybe you get the doppler effect in that way.

Compared to a train, A sees x meters train passing (a special device counts the meters), B is moving y from A into the driving direction of the train, and he counts x-y meters train (with that special device in front of his eyes). The same is valid for a passing light wave presenting time by distance (his movement from B gives a time dilation). As I think time is a equal term like distance, distance / time = always the lightspeed.

The meaning of (1 - V/C) can be different explained, maybe is it valid in each little piece during the movement of B.

1/y = \sqrt{(1-v/c)(1+v/c)} and you see a relation with (1 - v/c), so a lineair relation is not to expect.

But if you say in each little piece of movement Δb_light = Δa_light . (1 - V/C), maybe you find by the math integration 1/γ. My math knowledge is from 35 years ago (as I studied it), sometimes I use formulas for my work but the more difficult things are gone (an integration, no it's gone, maybe if I try hard but now I don't see the point anymore).

 

[Dale]

I forgot to thank you for your answer (math). I think there is an error in your formula (the same as you thought some time ago, you forgot to divide by c^2 for the time transformation, I am never sure so I say it only).

You are correct, there are many factors of c or c² missing, but it is not a mistake. I was explicitly using units where c=1 specifically so that I could neglect those factors, that is why I made such a point of my units so many times. If you want to use units where c is not 1 then you can analyze the units of each expression to determine where to put the factors of c. For example, in the expression 1-v, 1 is unitless and v has units of speed, so you need to divide v by c to get a unitless number that you can subtract from 1.

The distance of light is c.t in frame A and c.t' in frame B used in the derivation of Lorentz, you can't say after that derivation, that way of thinking is not valid anymore. So the distance traveled by the light wave is c.t in frame A and c.t' in frame B (just a fact).

Yes, this is the second postulate, the invariance of c.

As I think time is a equal term like distance, distance / time = always the lightspeed.

Yes, in all frames.

The meaning of (1 - V/C) can be different explained, maybe is it valid in each little piece during the movement of B.

1/y = \sqrt{(1-v/c)(1+v/c)} and you see a relation with (1 - v/c), so a lineair relation is not to expect.

All fine.

But if you say in each little piece of movement Δb_light = Δa_light . (1 - V/C)

This equation is incorrect. I derived the correct one above.

 

[digi99]

OK, so you want to find
L_c'=|x_1'-x_0'|

From the Lorentz transform:
t'=\gamma(t-vx) and x'=\gamma(x-vt)
So in B's frame we have
(x_0',t_0')=\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',t_1')=\gamma(x_1-v t_1, t_1-v x_1)=\gamma((t-1)-v t,t-v (t-1))

So now
L_c'=|x_1'-x_0'|=|\gamma((t-1-vt)-(t-vt))|=\gamma

Thanks DaleSpam, I understand now (I had not look enough). You just transformed the light line from frame A to frame B and indeed that formula of my is involved now (very good). One thing I don't understand is, you find γ as length while the length contraction is 1/γ. Can you explain that ?

Later today I come back on my statement, my subject after many weeks (to close it and going to use for 100% on my website, I will formulate again), but you have explained the formula ... but is also logic that it is valid in my thinking, only in a very little piece of movement and never as one total piece ... because nature is correcting immediately because of the constant speed C ...

 

[Dale]

One thing I don't understand is, you find ³ as length while the length contraction is 1/³. Can you explain that ?

The length contraction formula does not apply. Length contraction only applies when the object is at rest in one of the frames. A pulse of light is not at rest in any frame.

This is a very common mistake and is the reason I recommend against using the length contraction and time dilation formulas. They automatically fall out of the Lorentz transforms when appropriate, and using the Lorentz transforms prevents accidentally using the length contraction and time dilation formulas when you shouldn't.

 

[digi99]

The length contraction formula does not apply. Length contraction only applies when the object is at rest in one of the frames. A pulse of light is not at rest in any frame.

This is a very common mistake and is the reason I recommend against using the length contraction and time dilation formulas. They automatically fall out of the Lorentz transforms when appropriate, and using the Lorentz transforms prevents accidentally using the length contraction and time dilation formulas when you shouldn't.

Very interesting, what you say sounds logically, light is never at rest. Ok in length gives it another value than expected, possible it's light and speed c, the important factor, but it's path is c.t' = c. 1/γ . t. When you consider this as length e.g. on level t, so c.t is the length in frame A, is than the length γ . c . t or 1/γ . c . t, that's confusing, something I don't see right ? Light waves can not be bigger than other objects while moving ?

What I said about little pieces is not true too, because the speed is constant V, but maybe when you consider little pieces ΔV.t in (1 - V/C) you will find 1/γ or γ with math integration. Is this ever tried (I think so but without results) ?

 

[digi99]

That's what I told him and that's what Simon Bridge told him, although it's not normal Doppler because he's basing the time duration on the stationary frame instead of the moving observer's frame.

I even pointed out that if the moving observer changes direction and returns to the stationary observer, both their "special clocks" will end up with the same "time" on them instead of what should be happening according to the Twin Paradox.

But he still thinks its a better way to illustrate time dilation even though he realizes that it only "works" in one direction and even though it only "works" correctly at v=0 and v=c.

You were right from the beginning with the Doppler effect, I don't get it but I go to read again (besides I had wrong transformations), because it will always there I guess in the rest frame of a moving object. But I come with a conclusion I will place on my website, later I will make a new formulation. But maybe I understand because if something moves, you are in the "middle" of a light wave, is the reason probably. But if you consider a light wave on distance from both frames, also Doppler ?

 

[digi99]

OK, so you want to find
L_c'=|x_1'-x_0'|

From the Lorentz transform:
t'=\gamma(t-vx) and x'=\gamma(x-vt)
So in B's frame we have
(x_0',t_0')=\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',t_1')=\gamma(x_1-v t_1, t_1-v x_1)=\gamma((t-1)-v t,t-v (t-1))

So now
L_c'=|x_1'-x_0'|=|\gamma((t-1-vt)-(t-vt))|=\gamma

Hi DaleSpam,

I want it let it go, while reading my books I could judge it again with more knowledge, but it confuses me how you use Lorentz, I still not get the picture how to use it, that was all wrong how I did it, but now I see, you use Lorentz how I did ?

I was/am thinking now (see an answer from me) that you may use Lorentz for transforming the coordinates for the moving object only, so you could see how B sees time (and distances of course) but all his coordinates are transformed to x' = 0 for any time (t' = 1/γ.t).

But I understand while looking to your solution, you may transform all coordinates from frame A to frame B with the Lorentz formula, so you see every point how B it sees at rest. So other speedlines you can transform in that way, so you see that speedline how B it sees. But transforming other points does not give t' = 1/γ . t anymore, this is my confusion now ?
After many weeks I have at least learn how to use Lorentz (cannot be that difficult, so I don't make such errors a next time).

Besides 1-V/C is a wrong thought of me because even in little pieces is never t' = (1 - V/C) . t at any moment. It's always the very know triangle relation between c.t, c.t' and v.t, but difficult to see mathematically if as well B and a light wave are all on the x-axes in the same direction. But maybe after reading my books.

Besides I thought today, looking to the length contraction, near the light speed is all very small comparing to standing still, so a foton with light speed is never visible for us (mass 0).

Depending on your answers concerning Lorentz I have to check what I was doing wrong in the Lorentz transformations where I found expressions with (1 - V/C) because I did it in the same way as you did.

 

[digi99]

That's what I told him and that's what Simon Bridge told him, although it's not normal Doppler because he's basing the time duration on the stationary frame instead of the moving observer's frame.

I even pointed out that if the moving observer changes direction and returns to the stationary observer, both their "special clocks" will end up with the same "time" on them instead of what should be happening according to the Twin Paradox.

But he still thinks its a better way to illustrate time dilation even though he realizes that it only "works" in one direction and even though it only "works" correctly at v=0 and v=c.

You were right from the beginning with the Doppler effect, I don't get it but I go to read again (besides I had wrong transformations), because it will always there I guess in the rest frame of a moving object. But I come with a conclusion I will place on my website, later I will make a new formulation. But maybe I understand because if something moves, you are in the "middle" of a light wave, is the reason probably. But if you consider a light wave on distance from both frames, also Doppler ?

I am a starter so understanding goes a little bit slower for me. Now I understand that my formula (question was has it a meaning) is the Doppler effect, in case it would be there, so not always is in another frame.

 

[digi99]

OK, so you want to find
L_c'=|x_1'-x_0'|

From the Lorentz transform:
t'=\gamma(t-vx) and x'=\gamma(x-vt)
So in B's frame we have
(x_0',t_0')=\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',t_1')=\gamma(x_1-v t_1, t_1-v x_1)=\gamma((t-1)-v t,t-v (t-1))

So now
L_c'=|x_1'-x_0'|=|\gamma((t-1-vt)-(t-vt))|=\gamma

I learned from your Star Track series, never give up because I am still not satisfied with this topic (who is right or not is not important, but I don't get it).

See answer #18 in this topic. I agreed already with Ghwellsjr that I came to the Doppler effect when I was combining some formulas in math. Exact the same formulas you did found now and with formulas I thought valid.

But in my old topic I came with a kind of basic formulas too in the very beginning so I thought the relation is there between 1 - V/C before and after Lorentz. But I thought later the way I found them is not right, so I came to the last formulas close to it (#36 in this topic, a summarization of my original topic, the formulas where I found the Doppler effect too, not in frequencies but just as you in transformed coordinates of a light wave).

In the wrong formulas I thought they are not right because the relation 1/γ was not found, like you proved with the length Lc. So it is possible, that Lorentz is not valid for a light wave itself (speed c) but you used the Lorentz formulas also for transformations of a light wave. Very very confusing all.

Additional: Ok I understand the Doppler effect now, it's of course everywhere in the rest frame of the moving object, because everything moves in that frame with the same speed. Clear. But so there is a relation between 1 - V/C before and after Lorentz (I suppose now it may be used for transform all the linespeeds included light itselves, maybe all moving objects in the rest frame of the moving object has a length factor γ, only objects in rest 1/γ).

BUT I am talking in my topic about the total length of the path of a light wave (not the wave length, language confusion) and that must be before Lorentz c.t and after c.t' (and t' = 1/γ . t and not γ . t) and that I don't see back in the diagram. Funny but my statement is still there what I said (except time dilation calculated before Lorentz, that's be seen as the Doppler effect but you may explain that how you like, the path of the light wave is going shorter to explain the constant speed, the path (distance) presents time, so time will be lesser, or time dilation explained) ...

 

[digi99]

Last answer #47 changed ...

 

[Dale]

Light waves can not be bigger than other objects while moving ?

Why not? The thing to do here is not to simply apply the length contraction formula, but to derive a general formula that would work for v=0 and v=c. Some objects will contract some will expand, all depending on the various velocities. It would be a tedious exercise.

But maybe I understand because if something moves, you are in the "middle" of a light wave, is the reason probably. But if you consider a light wave on distance from both frames, also Doppler ?

Light from a source that is coming towards you is blueshifted and light from a source that is going away from you is redshifted.

But I understand while looking to your solution, you may transform all coordinates from frame A to frame B with the Lorentz formula, so you see every point how B it sees at rest. So other speedlines you can transform in that way, so you see that speedline how B it sees. But transforming other points does not give t' = 1/γ . t anymore, this is my confusion now ?

This is correct. You may transform all coordinates from A to B using the Lorentz formula. You are correct that not all points will give t'=t/γ, which is why it is important to use the whole formula, and not just part.

So it is possible, that Lorentz is not valid for a light wave itself (speed c) but you used the Lorentz formulas also for transformations of a light wave. Very very confusing all.

You can certainly use the Lorentz transform for a light wave. Light waves are governed by Maxwell's equations, and Maxwell's equations are invariant under the Lorentz transform.

What you cannot do is use a Lorentz transform to get a frame where a light wave is at rest. In all frames, the light wave will travel at c.

BUT I am talking in my topic about the total length of the path of a light wave (not the wave length, language confusion) and that must be before Lorentz c.t and after c.t' (and t' = 1/γ . t and not γ . t) and that I don't see back in the diagram.

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.

 

[digi99]

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.

Yes indeed DaleSpam it is clear now, if you fill in c.t there you will find t' = 1/γ .t so the light wave has traveled path c.t'. Of course is length from something that is moving different than in rest.

Thanks very much, thanks to you I understand Lorentz now fully. Also good for others to read I guess because without study you don't find this easily in popular books or internet.

Later I come with a reformulation again in a try to explain time dilation in a few sentences without complex drawings most people will only confuse (most are no Beta people).

So the meaning of the formula 1 - V/C is the Doppler effect, while you move (v.t) the frequency of the light wave will change, but the traveled path is still t. Only after Lorentz it will be t'. So I may not talk about a time dilation seen in frame A, it's just the Doppler effect.

So I go to explain it something different but in somehow equal words with the path of a light wave (the most exact clock) as main role ..

 

[digi99]

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:

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

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).

 

[digi99]

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 ...

 

[Dale]

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.

 

[digi99]

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 ?

 

[Dale]

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).

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.

 

[digi99]

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.

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 ?

 

[digi99]

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 ...

 

[digi99]

Previous answer changed ...

 

[digi99]

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).

 

[Dale]

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:

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.