# Light V/S non EM wave.

1. Jan 23, 2008

### sheetalbhat

hey..while looking into the consequences of the invariance of light....we see that time needs to be slowed down in order to incorporate it in all the appropriate condtns..normally every body( in most of the books) gives an example.......if we are in a space ship(moving horizontally with a speed of v) and we are having a pair of mirrors(perpendicularly opposite, and kept vertical with reference to ship)..then the time taken by the light to emanate from one mirror and come back after reflection is lesser than the time noted by the stationery observer(located outside the ship) who is seeing the oblique path of light inside the ship from outside.....all this is fine..even the maths is fine..and we get the desired result..but the key thing to which we attribute everything here is that..since we are dealing with light(EM wave), it has to be the same velocity no matter what the frame of reference is......now..
the question is ...if i replace the two mirrors with a sound making radiator and keep a signal reflector just exactly the mirrors were, i see the same effect again...that is the time dilation between the two observers....

though this experiment is exhibited to us to show that invariance of EM wave velocity leads to lapse of time..but it is working for non EM waves too...

i hope u got the question rightly....

Last edited: Jan 23, 2008
2. Jan 23, 2008

### Fredrik

Staff Emeritus
The purpose of that thought experiment (involving light) is to give you a triangle with sides ct, ct' and vt. (The hypothenuse is ct). When you apply the Pythagorean theorem and solve for t or t', you get the relativistic time dilation formula. The reason this doesn't work with sound is that "c" wouldn't be the same in both places.

Edit: I didn't read your thought experiment carefully enough. I thought this was about another thought experiment.

Edit 2: I read your post again, and now I think this is the thought experiment I had in mind, except that in my version we only consider light going from the location of one mirror to the location of the other (instead of bouncing back and forth between the two locations). That's really all we need. The mirrors are irrelevant.

Last edited: Jan 23, 2008
3. Jan 23, 2008

### sheetalbhat

hey...u said that...."The reason this doesn't work with sound is that "c" wouldn't be the same in both places."
i didn't get u here...if the air inside the ship and the air outside is the same(refractive index is same)...then, lets say, as the sound emanates from the source, the stationary observer and the person inside the ship both come to know.after that, when sound comes back after reflection, they again get to know..so time dilation will again be there..because, in the Pythagoras triangle..the base will vt, hyp will be Ut, U being the speed of sound in that particular air, and third side being Ut'......i am saying this because the statinary observer will calculate the time as if wave is goin straight and will get the time as 2L/U.......so the perpendicular(third side ) will be Ut'.....so it will again come the same..........please explain...

4. Jan 23, 2008

### Fredrik

Staff Emeritus
The hypothenuse isn't Ut (because the velocity isn't U). You didn't consider the fact that the air is moving relative to the observer who measures the time t (the "stationary" observer).

I agree that the other two sides are vt and Ut'.

5. Jan 23, 2008

### sheetalbhat

hey..i am sorry..but ....here i have understood some points and i would like to mention it..then u plz tell me....whether i am right or wrong............................if the wave into consideration is a non EM wave, then the hyp will be the resultant velocity of the velocity of the ship and the velocity of the wave itself......and in that way, if we calculate the time for the events from both the observers....it will come out to be same........and if the wave is and EM wave..then the hyp is not the resultant but only C still...due to this.....time has to dilate............

am i right......

6. Jan 23, 2008

### pervect

Staff Emeritus
No, you don't see the same effect, because, unlike the speed of light, the speed of sound is not a universal constant, and not the same for all observers.

The speed of sound in a steel bar has one value for an observer at rest with respect to the bar, and another value for an observer moving relative to the bar.

7. Jan 23, 2008

### Fredrik

Staff Emeritus
Yes, I think you understand it now. When we're dealing with light, the hypothenuse is ct and the other sides vt and ct'. This gives us

$$c^2t^2=v^2t^2+c^2t'^2$$

Solve for t' and we get

$$t'=t\sqrt{1-\frac{v^2}{c^2}}$$

When we're dealing with sound, the hypothenuse is bt and the other sides are vt and ct'. (Here c is the speed of sound in air, and b the speed of sound in the moving air).

This gives us

$$b^2t^2=v^2t^2+c^2t'^2$$

Solve for t' and we get

$$t'=t\sqrt{\frac{b^2-v^2}{c^2}}$$

We see that t=t' if and only if b2=v2+c2, and we know that this is exactly what b2 must be, because b is the magnitude of the sum of two perpendicular vectors, one with magnitude c and the other with magnitude v.

8. Jan 23, 2008

### sheetalbhat

hey...fredrik...u got it in the perfect way........i think what i was missing was the change of velocity of sound with respect to moving air( vehicle)though it being a very obvious thing....i couldn't catch it.......thanx a lot.......

will trouble u again...after some time with some more doubts....bye...yaar....

9. Jan 23, 2008

### sheetalbhat

hey once more....let me put it this way....just now we concluded that for non Em waves there won't be any dliation in time for the event concerned above.....now if we remove all the components of the experiment...that is to say....only space ship is there....inside there is nothing( i mean no mirrors and all)....it is moving at the velocity of of v....then we say normally that....if i am moving in this space ship at a velocity approachable to velocity of light then, my clock time will run slower compared to a stationary oberver and after a considerable time i can very well say to that stationary observer that i am younger than you.....
here we calculate the slowing down of the time....by the formula (deduced probably from the above experiment)....which u have mentioned in one of ur replies.....
my question is ...how do we physically justify this dilation in this case..mathematically it is ok....but...u got my point....how do we know it that the formula is universally applicable in all conditions of constant velocity....i mean to say...this formula is a sort of special case.....plz explain...........

10. Jan 23, 2008

### Staff: Mentor

In physics, like all sciences, you generate your theories with appropriate assumptions, math, logic, and predictions. You then devise experiments to test the accuracy of your theory's predictions. Physical justification is obtained experimentally. The best website that I know of about the experimental verification of relativity is: http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

For your particular question on time dilation, pay attention to the experiments on muon decay. Fast-moving muons literally have a longer lifespan then their stationary twins.

11. Jan 24, 2008

### Fredrik

Staff Emeritus
How do you know that if you drop a rock, it will fall to the ground? You don't really. Every rock you drop is a special case. You can't really know that any formula is "universally applicable". You just assume that a certain formula is valid and call that your "theory". Then you do experiments to try to prove that theory wrong.

If it bothers you that this derivation used light that moves in a specific direction, you can find other derivations that doesn't use this trick. For example, you can derive the Lorentz transformation the way I did in this thread and use it to derive the time dilation formula.