What is Physical Significance of Lorentz Transforms ?

[mitesh9]

What is Physical Significance of "Lorentz Transforms"?

Can anyone explain me what is physical significance of "Lorentz Transforms"?

That means, apart from the math, how can we understand "how and why" length contraction or time dilation actually happen? and what is speed of light doing in the lorentz transform equations? Why is it not speed of sound or speed of any other wave but speed of light? I think these are too many!

Regards,

Mitesh

 

[robphy]

The Lorentz Transformation encodes
the principle of relativity (no preference for any inertial observer) together with the constancy of the speed of light.

 

[mitesh9]

Thank you sir,

But I was expecting some simpler (and yet detailed) answer, Like a thought experiment.

Say a rod of length L is moving in +x direction with a constant velocity v from us, the observer (and thus we and the rod are in different inertial frames moving with a relative velocity v). Now how would we derive Lorentz transform equations and incorporate the speed of light in it?

Regards,

Mitesh

 

[lbrits]

I like the old parable of the two mirrors placed inside a moving train, one on the floor and one directly above it on the ceiling. A laser is then shone in such a way that light bounces from the bottom mirror to the top mirror and then back down again. One asks, what is the path traveled by the beam of light.

For someone standing besides the tracks, the beam appears to move a greater distance, because the train is obviously moving forward. However, if one demands that light travel at the same speed for both observers (as both are equally entitled to solve maxwell's equations in their own frame), then one must conclude that the train has become shorter!

 

[jtbell]

One normally derives the Lorentz transformation by assuming that the speed of light is the same in any inertial reference frame.

 

[mitesh9]

I think I have understood somewhat, But if the train has shortened for the observer on the track, has it shortened as a whole, or just the distance between the two point (which will be different for an outside observer) on the cieling (or flore) is contracted so that the points coincide and the distance traveled by light in top-bottom-top two way journey remain same for both observers? In first case, the whole train should be shrunk to zero length, to coincide the two points exactly, while for second case, the contraction should be [(height of train)/c]*speed of train, but this is not Lorentz transform!

 

[lbrits]

You can put mirrors everywhere if you like... the conclusion would be that the whole train has to shrink, but only in the direction of motion.

 

[mitesh9]

Ok, Let's say, there are two mirrors (A, B) in a train, separated by a distance of c/2, and the speed of the train is v, so for an observer in the train, the distance traveled for light (in the direction of the train for A-B-A path) will be 2*c/2=c. For an out side observer, the distance traveled would be, c/2+v (for AtoB)+c/2-v (for BtoA)=c. Then why would we need the Lorentz transforms? Am I missing something?

 

[mitesh9]

(reformatted)
Let's say, there are two 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 traveled 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 traveled would be,
(c/2) + v + (c/2) - v = c.
(c/2) + v for distance A to B; (c/2) - v for distance B to A
(this is not velocity addition; c is just a number ca. equal to 300000)
Note that, without considering relativity (the principle of, SR or GR), the distance traveled by light for both the observers is c. The time taken is also c/c=1s. the laws of physics are same for both the observers. Then why would we need the Lorentz transforms? Am I missing something?

 

[lbrits]

(reformatted)
...stuff

I guess the point is that one mirror is directly above the other (one ceiling, one floor). Hell, you could just have one mirror on the ceiling and a laser on the floor. The height of the ceiling is d. According to the passenger, the distance traveled in one leg of the trip is d in a time t'.

According to the person on the ground, the distance traveled in one leg is $$\sqrt{d^2 + v^2 t^2}$$. Experimentally (or by Einsteinian reasoning) we find that the speed of light is the same in all reference frames. Thus
$$\frac{d}{t'} = \frac{ \sqrt{d^2 + v^2 t^2} } { t} = c$$,
which gives
$$t = \frac{d}{c} \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = t' \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
Thus we have shown that there is time dilation. Once you've established this, you can do further thought experiments to see what happens to lengths. Basically in order for the speed to remain the same, both the distance and the time have to change in such a way that the two effects cancel out. But It's not a simple ratio, because of the relativity of simultaneity.

 

[Antenna Guy]

According to the person on the ground, the distance traveled in one leg is $$\sqrt{d^2 + v^2 t^2}$$. Experimentally (or by Einsteinian reasoning) we find that the speed of light is the same in all reference frames.

Does the observer on the ground infer that the laser in the inertial frame is not pointed at the mirror (also in the inertial frame)?

Regards,

Bill

 

[robphy]

Thank you sir,

But I was expecting some simpler (and yet detailed) answer, Like a thought experiment.

Say a rod of length L is moving in +x direction with a constant velocity v from us, the observer (and thus we and the rod are in different inertial frames moving with a relative velocity v). Now how would we derive Lorentz transform equations and incorporate the speed of light in it?

Regards,

Mitesh

In my opinion, there is an unnecessary importance [probably due to historical emphasis and textbook emphasis] placed on "time dilation" and "length contraction", which are merely side effects of more fundamental things like the invariance of the interval and the Lorentz Transformations themselves.

Probably the quickest, least mathematically sophisticated, and arguably more physically-direct approach to the Lorentz Transformations is the Bondi k-calculus (which uses the Doppler effect). [One reason for this is that one is essentially using the eigenvectors of the Lorentz boost Transformation.]

Here's a diagram

$$<br /> \]<br /> <br /> \unitlength 1mm<br /> \begin{picture}(55,90)(0,0)<br /> \linethickness{0.3mm}<br /> \put(20,10){\line(0,1){80}}<br /> \linethickness{0.3mm}<br /> \multiput(20,90)(0.12,-0.12){250}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(20,30)(0.12,0.12){250}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(20,10)(0.12,0.2){250}{\line(0,1){0.2}}<br /> \put(15,30){\makebox(0,0)[cc]{t}}<br /> \put(14,60){\makebox(0,0)[cc]{\gamma kt}}<br /> \put(20,60){\circle*{2}}<br /> \put(15,90){\makebox(0,0)[cc]{k^2t}}<br /> <br /> \put(55,60){\makebox(0,0)[cc]{kt}}<br /> <br /> \end{picture}<br /> \[<br />$$

I don't have time right now to fully develop this to directly answer your question, but here are some earlier threads to get you started [if you are interested].
https://www.physicsforums.com/showthread.php?p=934989#post934989
https://www.physicsforums.com/showthread.php?t=117439

 

[mitesh9]

@Ibrits

Dear Sir, May I remind u that u just asserted "u may put the mirrors where ever u want", based upon which I formatted my question, and now u r changing ur statement! It is more of an escapism on ur part, or may be u r being generous ant replying me just for the sake of replying me (of course with an intension of maintaining my enthusiasm), however, It is adding to my confusion, and not helping any bit. (My apologies if my comments are inappropriate).

@robphy

I'm going through the prescribed posts, meanwhile, do consider my above posted question (which I believe is valid, even if I take the mirrors to be perpendicular to the direction of motion, as the length contraction is in the direction of motion). Note that, in above described situation, the speed of light, the distance between the mirrors and the time to travel the distance, all are equal for both observer in relative motion, then why should we need Lorentz transformations (or any relativity at all)?

 

[lbrits]

Sorry to have confused you. I said you can place the mirrors wherever along the train you want, meaning that the effect is independent on where you place the mirrors. I did say that they had to be above one another. It's hard to convey these things without diagrams. May I suggest you try a book on special relativity. Pretty much anyone will cover this well

Also, go easy on the sms-speak.

 

[mitesh9]

@Ibrits
I offer my apologies again, but u c, relativity is not simple to understand, and that's what all the texts convey. The recent one I used was "perspectives of modern physics" by Arthur Baiser.

@robphy
Dear Sir,
As suggested, I went through the posts, however, I don't see how are they related to my query? Again, here my query is

Let's say, there are two mirrors in a train (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 traveled 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 traveled would be,
(c/2) + v + (c/2) - v = c.
(c/2) + v for distance A to B; (c/2) - v 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 traveled 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?

 

[atom888]

(c/2) + v for distance A to B; (c/2) - v for distance B to A
(QUOTE]



why u have -v on the second term? The train going in 1 direction. The final answer shoudl be c+2v where 2v is some non converted form of distance.

 

[mitesh9]

I think not,
We are considering two way travel of the light between mirrors A and B in a Path A-B-A. So, when a ray starts from A to be, due to velocity of train being v, from the point of view of an outside observer, the mirror will move with the train and thus, the light has to move an extra distance v, however, in return journey, the mirror B is moving with velocity v towards the light wave, thus it will have to cover the distance (c/2)-v. And hence, for at outsider point of view as well, the distance covered by light will be same as that for an observer inside the train (i.e. c only).

 

[atom888]

Oh i see, you put the mirrior side-way. I thought you put it up and down.

your watch won't read 1 sec. It will read 1 sec + some change by the time light complete the cycle. Why? because you see the train travel at v, it actual speed is v+some change. Or u leave the train out of this and do dilution on the light. Eitherway you have to account for the train (hence v) dilution.

 

[Integral]

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Integral