# Is is not that Simple? (about Lorentz transforms)

1. May 6, 2008

### mitesh9

I posted a simple thread regarding Lorentz transforms, but it has been locked for "staff review". May be because I used some "Net-Speak" conventions while writing some responses (which has earned me infraction points, never mind, my mistake). However, one of my questions remained unanswered there, and so this post...

Let's say, there are two mirrors in a train (mirrors A & B, separated by a distance of c/2 km, perpendicular to the motion of the train) so that they are arranged as "Last Bogy - A - B - Engine", and the speed of the train is v km/s.

So for an observer in the train, the distance travelled for light (with speed c km/s in the direction of the train for A-B-A path) will be 2*c/2=c. For an out side observer, the distance travelled would be,
(c/2) + a + (c/2) - a = c.
(c/2) + a for distance A to B; (c/2) - a for distance B to A
(this is not velocity addition; c is just a number ca. equal to 300000)

Note that, without considering the principle of relativity (or LR or SR for that matter), the distance travelled by light for both the observers is c (no length contraction). The time taken is also c/c=1s (no time dilation). the laws of physics are same for both the observers. Then why would we need the Lorentz transforms? Do correct me if I'm missing something?

For more clarification, for an outside observer, when the light starts traveling from mirror A to B, it will travel in the same direction as the train itself, so the mirror B (which is first destination of the light ray) will also move away from the light with the speed v (speed of the train), and hence, the light will have to travel more distance then just distance between the mirrors, which will amount to c/2 + a. however, when the ray reflects back from B to A, the mirror B will be approaching the wave of light (as the light is traveling in opposite direction from the motion of the train), so the wave will travel the distance c/2 - a.

Last edited: May 6, 2008
2. May 6, 2008

### Staff: Mentor

Not sure what you're doing here:
(1) If the distance between the two mirrors is D according to train observers, then to the track observers the distance will be $D/\gamma$.
(2) If the time, according to track observers, for the light to travel from A to B is t1, then the distance traveled by the light will be $D/\gamma + v t_1$. For the return trip, the time is different (call it t2) and the distance traveled by the light will be $D/\gamma - v t_2$. The total distance isn't 2D or even $2D/\gamma$.

3. May 6, 2008

### mitesh9

Sure sir,

But that is assuming we know Lorentz transforms, however, as you can see, we started assuming no prior knowledge of Lorentz transforms, yet we reach the same conclusion that the lows of physics are same for both observers, inside and outside the train. And if this is the case, why do we need to use the Lorentz transforms in the first place?

4. May 6, 2008

### yuiop

Yes, what you are missing is what happens when you consider a one way measurement of the speed of light. Your example considers the two way speed of light and a number of self cancelations hide some issues. For a one way speed of light measurement (ignoring length contraction) the distance travelled by the light is c/2+a according to an observer external to the train, taking a time of (1/2+a/c) seconds and for the measurement in the opposite direction, the light travels a distance of c/2-a in a time of (1/2-a/c) seconds. Without using Lorentz transformations, how do you propose to explain why the observers measure the time for light to travel from the back to the front, to be the same as the time for light to travel front to the back?

Another reason we need Lorentz transformations, despite the laws of physics being the same in every inertial referfrence frame, is that "the laws of physics" that are consistent are not neccesarily what we classically assume them to be if we did not know the Lorentz transforms. For example the classical equation for kinetic energy is 1/2mv^2 but it turns out that this classical equation does not work at relativistic speeds and the "real" law for kinetic energy is (m*gamma-m)c^2. It can also be seen that if we only make measurements in the reference frame of the object then the kinetic energy of any object is zero which is not very useful. Sometimes we do not have the luxury of being able to make measurements in the rest frame of the object, such as when observing distant stars, so we have to use Lorentz transformations to calculate what is happening in the rest frame of the object. You might also note that we do not have the luxury of being able to use two way measurements when observing stars as it it is not practical to send light signals to a star and wait for the signal to come back. As I hinted before, your non-Lorentz method simply does not work for one way measurements.

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5. May 6, 2008

### atom888

I suggest you forget about this problem. There probably more than three changing parameter we must consider like x,y,z,s... as time, space, speed, observing speed and so on... Mathematically we can't visualize things more than 3 dimensions. To my understanding, I just see that as I move faster, I'm space warping(in a continuous manner), but the one who observe me will see that I'm traveling the whole distance like normal. So if somewhere in between, he freeze the picture and say, "hey, i see you here right now with this time". I say "yeah, i am in the exact location as you see, but my clock read different".

6. May 6, 2008

### mitesh9

Well, I think u r missing something too, buddy!
Surely, the outside observer measures two different times for two different journeys of the light way, but we must not forget that they are two "different" journeys after all. And thus, as distances for both journeys are different (by virtue of the train to be moving), the observer must notice two different distances. What is it that does not allow the observer to notice two different times for two different distances, especially when he considers the speed of light to be invariant?
Surely, it must not have anything to do with the principle of relativity. The reason is, even for one sided journey (say A to B), the observer notices more time for a longer journey, while for the opposite direction (B to A), the observer notices less time for a shorter journey.

As already has been explained before, the observer observes what he should, and, while he is completely aware about the relative velocity of the train (as is prerequisite if he wish to exercise the Lorentz transforms), he is bound to get two different results, which will not puzzle or surprise him, but will allow him to calculate the difference between the mirrors, and every other detail of the situation, which is accessible to the observer in the train.

I would humbly refuse to take this non-Lorentz method as "My Method", instead, let it be considered as simple logic. Yes, you are probably right that this simple logic may not be useful to describe the complexities of complex systems, however, may I bring your attention to a more important fact, that, as we have already seen in the second post, whenever a situation like this (as has been described by OP) comes, we fetch the lorentz transformation equations and our answers differ considerably using Lorentz transforms and Simple Logic. That means, one of them have to be wrong. I am unable to see, what is wrong in the Simple Logic? and if nothing, then why should the Lorentz transformations should be used at all?

Well, my dear friend, "I suggest we remember to keep the spirit of the discussion alive". You see, we are not getting any suitable answer for "why should we use lorentz transforms at all?", yet, If we are asked this question in examination paper, we will use them, without knowing what we are doing. That is no good. Even if we use space co-ordinates (x, y, z), we will arrive at same conclusion. And if we can satisfy all we need without considering space-time continuum, we must forget about it. We will not use sword to accomplish a task, achievable by a needle, especially, when we have doubts about what sword is and how it works.

Regards,

Mitesh

7. May 6, 2008

### yuiop

OK..I know I shouldn't take the bait..but here goes...

All the above is correct for the OUTSIDE observer on the track. Yes he notices that the light takes longer to travel the greater distance from the the back of the moving train to the front of the moving train and vice versa and he he will get a constant result for the speed of light in either direction.
I should have made it clear that I was talking about the observers on the train so the above quote would be better expressed as "Without using Lorentz transformations, how do you propose to explain why the observers (on the train) measure the time for light to travel from the back to the front, to be the same as the time for light to travel front to the back?" While the observers on the track measure a longer time for the light to go from the back to the front and a shorter time from the front to the back, the observers on the train measure the time to be the same in either direction. That is what you have to explain. That is the conclusion you must come to, if you believe any of the hundreds of experiments that have ben carried out and confirm time dilation and Special Relativity. If you simply say you do not believe any of the experiments, that the scientists are liars or careless in their analysis of the experiments you will not be entertained for very long on this forum

P.S. You, too should make it clear which observers you are talking about.

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8. May 6, 2008

### Staff: Mentor

I thought I showed that your steps were incorrect (even if you ignore Lorentz contraction). What is your reasoning? What are you assuming? What are you ignoring? Are you assuming that the speed of light is invariant, yet ignoring some of its consequences? On what basis are you concluding that the laws of physics are the same?

9. May 6, 2008

### kamerling

you can't make the galilean transform work with a constant speed of light and a universal time.
I'll call the distance between the front and back mirror d and consider the event as seen from the ground. (using c/2 for this is confusing)
on the leg where the light is catching up with the train, the position of the light
would be ct, and the position of the front mirror d + vt. They'll meet when ct = d + vt. this means the time that this takes is: t = d/(c-v) the distance the light traveled in that time is cd/(c-v). the distance of the return leg is cd/(c+v). if you add them you get

$$d_{ground} = \frac {2 d_{train} c^2} {c^2 - v^2}$$

A constant speed of light and absolute time are contradictory. The fact that you can derive some results without hitting the contradiction doesn't prove it is not contradictory.
In fact between us we have proved that it is contradictory by getting 2 different results for the distance that the light traveled.

10. May 6, 2008

### yuiop

Doc Al raised an important point in observation (2) of his post (that I missed too). You should compare it carefully to the equation you gave in post#1 and it will show you where you are going wrong.

11. May 6, 2008

### mitesh9

Thank you, but (though I don't understand it exactly), there was no bait, I suppose!

Well, for the observer in the train, everything is simple, 'coz that is his own frame of reference, and mirrors or train or everything in the train, including the source are stationary for him. The only thing in motion for him is the outside observer. If he asks the outside observer, he will say, "my time reading for both way journey is same, how come your times are different?" The outside observer will reply, "that's because you are moving with a relative velocity of v with respect to me, however, if I conduct the same experiment on the track here, you will observe that time for both journeys is different, yet it would be same for me ".
In all above posts, an extensive description is given for outside observer only.

Well, I don't think there is any need to be so much harsh on this. Nor do I wont to prove (or propose) any new theory, or disprove relativity. With all due respect to all the scientists, I can not agree (or disagree) with them, unless I understand the matter. I (or anybody for that matter) must not worship scientists, just because they are scientists! You would have noticed that there are more than 100 scientists who claim to have found loop-holes, but we don't believe them (yet we can see that their imagination is better than that). They don't raise problem like the mirror situation above, but deal with variety of complexities, and yet we call them crackpots (at least on PF). I have seen so many discussion on PF dealing with problems of very high complexities (certainly out of my reach), so it is obvious that I would rest assured to get answers of such simple queries. Yet you see, that is not happening.
And yes, you are right about being entertained on this forum. I have seen that you question the SR or GR and you are sure to be banned today or tomorrow, but I don't think that's the spirit of the game.

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12. May 6, 2008

### paw

To understand why we need SR and the Lorentz transform you need to set aside your 'thought experiment' for a moment. We need SR and the LT to explain certain observed phenomena. For example, muons are created when cosmic rays collide with atoms in the Earths upper atmosphere. Even though the muons are travelling at nearly c they have a half life short enough that they should not be observed at the Earths surface. However they are observed.

You need a theory which not only explains that observation but that doesn't contradict other observations and that reduces to the classical case for non relativistic speeds. That theory is SR and includes the LT. I challenge you to explain muon detection at the Earths surface any other way. SR (and by extension, GR) is the only theory to do so successfully.

Now to your 'thought experiment'. c is a velocity. You are confusing the issue by using it as a distance. The observer on the train sees the light pulse travel from A to B in 2a/c seconds. In this case 2*150000m/300000ms-1 = 1s. Classically, the track observer sees the light pulse take a/(c+v) for the trip in the direction of the train and a/(c-v) for the return trip. For v << c this would reduce to 2a/c seconds; the same as the train observer. However, for v-->c this would approach a/2c for A to B and a/0 for B to A. In other words your 'thought experiment' is logically inconsistant. You NEED to use the LT to get a result that is not undefined in the case that v-->c.

13. May 6, 2008

### mitesh9

Well, Sir, It is the answers of the questions you asked, that I'm searching for here. The reason how I conclude how the "laws of physics are same in train and ground" is, that with invariant speed of light (which is standard, against which everything else is matched and manipulated in SR) in both frames, the laws of physics will be same as well. Also, You will notice that, your explanation in second post, point (2), is contradicted by kamerling, exactly as u said it.

correct, and thus for reverse path the distance would be d - vt, we add them to get 2d, which is for ground observer, and yet equal to the observer in the train.
How is this further calculation necessary? In fact, the further calculation is misleading. As u can see, it contradicts Doc Al's post and gives contradicting results.
On the contrary, I have assumed the speed of light to be constant, and derived that time is also constant. Two different results for two different distances, If I may correct you, and it is the case for Lorentz transforms as well, for instance cd/(c+v) and cd/(c-v).

I think I should repeat, that why must we use Lorentz transforms, if the current situation of mirrors can be correctly explained without it? and if the situation is wrong, where does it go wrong? Saying that you will get different results using Lorentz transforms is, as we all should agree, not a proper answer.

14. May 6, 2008

### Staff: Mentor

Wrong. My post and kamerling's post are entirely consistent.

Wrong again. Please read more carefully. (The times for each leg of the round-trip path are not equal.)

Still wrong. There's no contradiction with my post.

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15. May 6, 2008

### mitesh9

@Doc Al
You should notice sir, that, Kamerling has deduced D=2D/gamma, where as in 2nd point of your first post, you have said that "... The total distance isn't 2D or even 2D/gamma", and that is not consistency!

further, you said "The times for each leg of the round-trip path are not equal", of course they are not, and we are considering that difference. The point is, the speed of the train being uniform, when the mirrors and the train are going in same direction, at the speed the mirror was receding the train, in reverse trip, with the same speed the other mirror will approach the ray, which cancels out, leaving the total distance covered 2d. (or c as in my original post)

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16. May 6, 2008

### mitesh9

Well, I think that makes sense. But one question to all, We put c + v and c - v terms here, is it not velocity addition, which is I think is prohibited in SR. Well, yes, I can see that this is not done directly, but through some math that reveal this in 3rd or 4th step, yet, can we not give it a thought?

By the way, Let me have some time to understand it a bit more clearly. The same points were made by Doc Al (using Lorentz transforms), Kamerling (using galilian transforms) and paw made who made above quoted response. Some different and interesting points from Kev and atom888 are also acknowledged.

Well, we have seen that answers of such trivial things are also not so obvious (I'm still not convinced though, may I get it in the end)!

17. May 6, 2008

### Ich

That's plain wrong. Try to answer this:
How long does a light ray travel to catch up with a mirror, receding with velocity v, with a head start of 300000 km?

18. May 6, 2008

### mitesh9

Well, I already said I got that point in response to paw's post above...

Yet, how come anything other then photon travel at that speed??

Edit: My mistake again...Sorry, I now understood it correctly (poor english probably.

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19. May 6, 2008

### yuiop

Sorry, If I sounded harsh. I was just a friendly warning, that some of the moderators on PF can be harsh if you go down certain roads

This is one place you are going wrong. You are adding (d+vt)+(d-vt) to get 2d which would be true if t was the same in both directions, but it is not. With the position of the back being represented by (a) and the one at the front be represented by (b) then the equation should be $$(d+vt_{AB})+(d-vt_{BA})$$ which is not equal to 2d. Lets flesh the example out a bit with a numerical example with v = 0.8c and the the distance d=AB in the rest frame = 0.5 light seconds so that the two way trip ABA for light should take one second in the rest frame of the train.

The time for light to catch go from the back to the front of the train according to the track observer (ignoring length contraction) is $$t_{AB} = d/(c-v) = 0.5/0.2 = 2.5$$ seconds and the time to go in the opposite direction is $$t_{BA} = d/(c+v) = 0.5/1.2 = 0.2777$$ seconds. Add then up and you get a total round trip time of 2.7777 seconds and not the the 1 second you assumed. At the very least you must conclude that the clock at the back of the train is running at a different speed to the clocks of the observers alongsde the railway track. Now at 0.8c the gamma factor $1/\sqrt{1-v^2/c^2}$ is 0.6. If we assume the clock on the train is running slower than the track clocks by this factor due to time dilation, then the rear train clock should have registered 2.7777*0.6 = 1.666 seconds which is still too long. Now if we reduce that time by same gamma facteor of 0.6 again to allow for the length contraction of the train we get required 1.666*0.6 = 1 second that the rear train clock should read for the round trip time of the light according to the train observer. Without the Lorentz transforms there is no explanation why the clock onboard the train records one second for the round trip time of the light signal.

[EDIT]Perhaps I should explain where the (c-v) in the equations comes from. When the light goes from the back to the front of the train of length d the train advances a distance v*t while the light is traveling. We can say that the distance the light travels is c*t and the distance the front of the train travels is v*t. The total distance is c*t = d + v*t By re-arranging the equation we get t=d/(c-v) this does not imply that the light signal is travelling at anything other than c ;)

The total distance travelled by the light signal in the track frame is d/(1-v/c)+d/(1+v/c) = 2d/(1-v^2/c^2) = 2d*gamma^2

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20. May 6, 2008

### Staff: Mentor

You might want to reread kamerling's post. He deduced that the roundtrip distance = 2Dc^2/(c^2 - v^2). That's not 2D/gamma!

Wrong again: It doesn't "cancel out". Did you read my earlier post?