# Short Lorentz derivation problem …

1. Mar 12, 2013

### digi99

I had 1 year ago a little discussion with DaleSpam about the short Lorentz derivation (triangle) if a photon alone can be seen as a light beam.

1) The long Lorentz derivation seems me correct, so no discussion about that
2) For the short Lorentz derivation with the triangle (c.t , v.t and c.t’) I had my doubts, but who am I, so maybe correct too
3) Now I have because of 2 the next problem, an equal idea, maybe I just don’t see it:

Consider a light source with speed V along a straight line and the source emitted only one photon as in 2).

Start-------t,A--------source with speed V------------------------t’,B----------------------photon

t = time t in inertial system s1
A = distance between startpoint and source (having speed V)
t’ = time t’ in inertial system s2 (rest system of source)
B = distance between source and photon

It’s like the triangle from 2 but in flat form (photon horizontally traveling in stead of vertically in s2).

The observer in s1 sees the photon from the start point in his/her inertial system s1 with speed C and time t.

An observer In system s2 sees the photon with time t’ and C.

So an observer in s1 may write: c.t (= A + B) = v.t (= A) + c.t’ (= B) or (c-v).t = c.t’, this gives not Lorentz but t’ = (1 – v/c).t

What I am doing wrong here (very curious) ?

Last edited: Mar 12, 2013
2. Mar 12, 2013

### PAllen

Well for one, is a foton something you sit on? (or is that a futon?). Sorry, but grammar and spelling help.

Anyway, the following is just wrong:

c.t (= A + B) = v.t (= A) + c.t’ (= B)

You are mixing quantities from different frames in an invalid way. What would be true in s1 is the tautological:

ct = vt + (c-v)t

The quantitiy ct' has no direct meaning in s1.

3. Mar 12, 2013

### digi99

Thanks for your answer PAllen, yes I mean a photon (foton is Dutch), sorry.

What you explain I thought 1 year ago also about the triangle, what is the difference?

Last edited: Mar 12, 2013
4. Mar 12, 2013

### PAllen

I don't know what you mean by triangle. Try to explain a question clearly and succinctly, and it will more likely be answered.

5. Mar 12, 2013

### digi99

The Wiki triangle

I thought this was well known.

I citate myself :

"The much known triangle how Lorentz can be derived in a more simple way (see wiki). If a light source is moving horizontally which let bounce 1 vibrating photon (pulse, direction vibration is horizontally) vertically between 2 mirrors (with speed of light C), we as standing still see that passing photon as two legs of a triangle (compare with a passing translucent train where somebody lets bounce a ball and we look to it from a station)."

I hope you know now which triangle I mean .. $(c.t)^2 = (v.t)^2 + (c.t')^2$

Last edited: Mar 12, 2013
6. Mar 13, 2013

### PAllen

No, I am not actually familiar with this derivation. Wiki didn't exist when I studied relativity, and no book I own mentions it. It is clearly not necessary to understand relativity.

I still don't know what specific question you are asking. Referring to 'the triangle' does not tell me at all what exactly you are asking.

[Edit: I just looked up Lorentz transform in wiki and do not see what triangle you are referring to. What would be helpful is something like:

- Here is a link to a simple derivation of Lorentz transform: .<link>, with section information if link doesn't go to right section
- I want to change the configuration used in the derivation as follows
- How would the derivation proceed? I tried .... but it leads to a wrong answer...

I already explained that an equation you wrote is wrong. What I cannot do is tell you what you should do instead without a better defined question from you.]

Last edited: Mar 13, 2013
7. Mar 13, 2013

### PAllen

I am going to make a guess about what your issue is.

For a photon going perpendicular to the relative motion between s1 and s2, as seen in s2, you can use the fact that distances orthogonal to s1-s2 motion is the same in both frames (there are physical arguments you can use to motivate this so you don't have to arbitrarily assume it). For distances parallel to the motion this is not so. You must figure in length contraction, which what you are ultimately trying to derive. Thus there is probably no approach using photons parallel to the relative motion that corresponds well to a derivation relying on the agreement on orthogonal distances.

8. Mar 13, 2013

### Fredrik

Staff Emeritus
The triangle argument is a simple way to derive the time dilation formula from the assumptions a) the speed of light is invariant, b) two inertial observers agree about distances that are perpendicular to the line of motion. It goes roughly like this:

Consider a spaceship that's moving with velocity v relative to an observer outside the ship. When the ship passes the observer, it fires a laser inside the ship that in the ship's rest frame is aimed in the direction perpendicular to the line of motion. I imagine a rocket going to the right, and a laser fired from the side that's up in the picture, to the side that's down in the picture.

Let's say that the laser hits the opposite wall a time t0 later in the ship's rest frame, and let's call the time between the two events (emission and hitting the wall) in the external observer's rest frame t. From the external observer's point of view, the ship has moved a distance vt when the laser hits the wall, so the path of the laser is not perpendicular to the line of motion. The path of the laser and the path of the ship can respectively be viewed as the hypothenuse and the short leg of a right triangle. The length of the hypothenuse is ct. The length of the short leg is vt. And since both observers will agree about the width of the ship (because it's perpendicular to the line of motion), the long leg of the triangle is ct0.

Now we can use the pythagorean theorem to find the relationship between t and t0.
\begin{align}(ct)^2 &=(vt)^2+(ct_0)^2\\
(c^2-v^2)t^2 &=c^2t_0^2\\
t^2 &=\frac{c^2t_0^2}{c^2-v^2}\\
t^2 &=\frac{t_0^2}{1-\frac{v^2}{c^2}}\\
t &=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}.
\end{align}

9. Mar 14, 2013

### digi99

Thank you very much for your answer PAllen en for the example Fredrik, I understand what you say (but still difficult, reason of my questions to be sure how you think), both says that both observers agreed about the same distance (no length contraction), even it are 2 frames.

So you say because of length contraction in other directions you can't use distances from 2 frames in a derivation seen from one observer or the other.

Is this because you have to be sure in Physics, the observer from s1 have to do with length contraction (we know the Lorentz factor), but you are not 100% sure what the observer in s2 sees (of course he sees a meter stick not length contracted, but you can never say the length of his/her meter stick in s1 is exactly the same as in s2) ?

What is the meaning from c . t' for the observer in s2 (the Lorentz factor is known) ? I think that you will answer, that distance is observed by s1 but because of length contraction it is not a real distance (only observed), and you can't say the real distance for s2 is c . t . 1/γ because t is not known in s2 as for s1 ..

10. Mar 14, 2013

### PAllen

The quantity ct' in S2 indeed corresponds to the quantity ct/γ in s1. However, the bigger issue is that with colinear motion (where there are differences in clock rates, ruler lengths, and simultaneity between s2 and s1; whereas in orthogonal direction, there are no differences in rulers or simultaneity), the equation you expect to be true if you want to use t' is:

ct = vt + γ(c-v)t'

This can be understood by s2 clock is slow per s1 by γ, and the separation rate between the photon and s2 origin is c-v per s1. Solving this for t', you get:

t/γ = t'

but this is a meaningless derivation because you had to assume everything you were hoping to derive. You need to grasp the point I made several posts back: there is no satisfactory analog of the triangle derivation using colinear photons, because there are no measurements in s2 that s1 will agree with in this set up. The triangle derivation exploits the fact that there s2 and s1 agree on distances orthogonal to their relative motion.

11. Mar 15, 2013

### digi99

This answer was very helpfull PAllen, thanks, this is not misty anymore for me .. the triangle now also fully understood ..

12. Apr 14, 2013

### digi99

One thing I don't understand about the above triangle argument.

If you use other speeds in the triangle in stead of C (so V1, V2 and V) and you use $t' = 1/γ . t$ you will find, this is only possible when V1=C and V2=C ?

Is time dilation not always equal perpendicular on the direction of motion for an inertial system with speed V and a time t ?

13. Apr 15, 2013

### Fredrik

Staff Emeritus
The triangle argument obviously only makes sense if the c in the calculation is an invariant speed. You can't just replace it with some other speed. And if you do, even though it doesn't make sense, then how could the calculation lead to something that involves c, when there's no c in the calculation?

I don't understand the last question. Time dilation is independent of the direction of motion. It only depends on the speed.

14. Apr 15, 2013

### digi99

I am very glad with this example, because after 1.5 years to be busy with this subject (between in of course, and it takes much time) I have the idea that I miss something in the time dilation concept and still see things wrong. That is the difference when you folllow a study and can talk about it. And sometimes you have the feeling, now I don't understand it at all anymore. You need than a little something to be on track again. That's very nice about this forum.

Maybe I see that little misunderstanding with this example.

If you use the Lorentz transformation (speed of light involved) for this triangle with other speeds than C, you can calculate the time dilation perpendicular on the motion of the inertial system, indeed the same in all directions, and perpendicular on the motion you have no to do with length contraction, so you may involve this perpendicular distance $v2. t'$ in a derivation seen from the other frame where the observer is standing still (there is the speed for this motion seen as v1 on the hypotenusa).

1) Why is this not valid : $(v1 . t)^2 = (v . t)^2 + (v2 . t')^2$ ?

2) Not direct related to this problem, but maybe I see time as registered in a clock also too simple, for me it is just a motion to register the rotation of the Earth, true or is there something extra I forget what is a important part of a clock ?

15. Apr 15, 2013

### Fredrik

Staff Emeritus
I still don't understand what you're asking. The point of the triangle argument is to derive a small part of what we can get from the Lorentz transformation, without actually using the Lorentz transformation. It's a way to show how the existence of an invariant speed leads to time dilation by a factor of gamma. So I don't understand why you're talking about doing Lorentz transformations now. Also, there are only two people in this scenario, one in the rocket and one outside of it. The velocity of the rocket in the coordinate system that's comoving with the guy who isn't in the rocket, is v. So why would we consider any other velocities? I mean, what other coordinate system should we transform to and why?

16. Apr 15, 2013

### digi99

Just as PAllen wrote, you may do this short derivation with two frames only because there is no length contraction perpendicular on the moving coordinate system for the photon.

So in my equal case I see an observator in frame 1 where a rocket is going on the hypotenusa and de x-axes velocity is v.

Than I see a frame2 moving along the x-axes with speed v, so perpendicular on the direction of frame2 you can see the rocket in this frame on the y axes and x'=0. You know already from the other short derivation what the time t' is.

Maybe I may use this situation mathematically but the solution is that this situation is only valid for v1=v2=c. And that looks weird because the travelled vertical distance of the rocket is seen the same from both frames (no length contraction) at any moment. But if you take a snapshot from this situation this seems not to be possible (only for c)!

Last edited: Apr 15, 2013
17. Apr 16, 2013

### digi99

Ok now I understand this.

An observer in frame 2 sees the rocket lower on the y' axes and an observer in frame 1 sees the rocket higher.

There is no length contraction in frame 2 on the y' axes.

The triangle argument is only true when the speed is c, and just as PAllen wrote, on that moment both sees the same traveling path for the photon on the y or y' axes. So only for the speed c the triangle is a closed figure and you may use pythagoras.

So only for the photon and the fact there is no length contraction on y' both oberservers see the same at the same moment, not in other situations (still no length contraction on y' but it goes slower, you see eventually a maxima on the y' axes later as observer in frame 2).