Jump to content
Sign in to follow this  
Companiero

Time dilation determined by light direction?!

Recommended Posts

Hi! My second post apparently, also in the form of a question. (if u havent noticed the thread sister of this one)

 

This time about time dilation. It's a common experiment to prove time dilation with the light reflecting a mirror, observed from a reference frame moving with velosity near c (vertical light beam path vs. horizontal movement), thus creating the right angled triangle. And I understand this.

However, what would be the effects on time dilation if the movement of the observer relative to the mirrors was in the same direction as the light beam path? And what if it was the opposite direction?

I forward the notion, that this way time dilation cant be proven. In fact in the two cases, we get two different values for the time: that it passes faster and in the second case slower than time in the reference frame. Consequently, time dilation depends on the direction of the beam of light and the observer, which is absurd (it is supposed to be solely dependent on the relative velocity between the two). Where am I mistaken?

 

Any help would be more than welcomed. Also, links to articles about this problem could be very helpful.

 

NOte: I prefer to get only reasonable replies, and not the Matrix-Oracle type of prophesies, which not even the author understands what they mean.

Share this post


Link to post
Share on other sites

Well for one thing, time dilation has been proven. It must be taken into account when programming and using GPS satellites. They move fast enough that there is an ever so slight effect. If we didn't compensate for this effect, GPS systems would be innacurate by several hundred meters or more.

 

Otherwise, I'm not really sure the point you are making. So I'm just going to say something that may help- remember that c is contasnt. Moving into a light beam, moving at right angles to it, moving away from the light beam, it always seems to be moving at the same speed. I dont know if this has any relevance at all to what you said.

Share this post


Link to post
Share on other sites

Maybe it would help if you explained the experiment a bit better as I wouldn´t be so sure that everyone knows it. For example: Where is time measured at all? Or even more general: What and how do you measure?

 

Another thing I´d like to mention: I don´t know how the others feel about your note you put on the end of your posts but to me it sounds pretty arrogant (are you an ongoing physicist, perhaps?). Maybe you should skip that and learn to tell usefull statements from crap. An ability that´s usefull quite often, anyways.

Share this post


Link to post
Share on other sites

If you place the path of light along the path of travel, it makes no difference which direction the mirror is from the light source, the return time is the same in both cases.

 

In the case where the mirror is placed in front of the light, the time it takes to reach the mirror is equal to :

 

[math]T_1 = \frac{x_1}{c-V}[/math]

 

the time it takes for the light to return from the mirror is

 

[math]T_2 = \frac{x_1}{c+V}[/math]

 

In the case where the mirror is placed behind the light source, these two times are reversed. (both of these cases are as measured by someone that the mirror and light are moving with resepect to.

 

In both cases, the total time for the light to travel to and back from the mirror is:

 

[math]T_t = \frac{x_1}{c-V}+\frac{x_1}{c+V} = x_1 \left(\frac{1}{c-V}+\frac{1}{c+V}\right)[/math]

[math]= \frac{2x_1c}{c^2-v^2} = \frac{2x_1}{c\left(1-\frac{v^2}{c^2}\right)}[/math]

 

x1 is the distance from the from light to mirror as measured by this same observer. But this distance is length contracted from the distance as measured by the frame of the light source and mirror, which we will call x0

 

The tranformation between these two measurements is

 

[math]x_1 = x_0\sqrt{1-\frac{v^2}{c^2}}[/math]

 

Thus

[math]T_t =\frac{2x_0\sqrt{1-\frac{v^2}{c^2}}}{c\left(1-\frac{v^2}{c^2}\right)}[/math]

[math]=\frac{2x_0}{c\sqrt{1-\frac{v^2}{c^2}}}[/math]

 

If t0 is the time time it takes in the mirror/ lightsource frame for the light to return, then

 

[math]t_0 = \frac{2x_0}{c}[/math]

 

and

 

[math]T_t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}[/math]

 

Which is the time dilation formula.

 

So time dilation holds no matter what the orientation of the light and mirror.

Share this post


Link to post
Share on other sites

Thank you Janus for the lovely response.

I anticipated your reply would be the way you wrote. ;)

However, what I don’t get here is why is it important for us to consider the light path all the way until it reaches back the source. Why not just from the source to the mirror (from point A to point B) in which case, we get two different times (t1 and t2, in your example) for the cases with opposite light direction, as opposed to the time we measure from the mirror (“static”) reference frame (to/2, in your example)? Thank you again for your help.

I don´t know how the others feel about your note you put on the end of your posts but to me it sounds pretty arrogant (are you an ongoing physicist, perhaps?).

Sorry if that came across as rude. But to tell you the truth, I consider arrogant being someone who acts like they know it all and try to act like “too wise to even be understood”. I cannot tell you the amount of wrong information I have read on this forum, that can only mislead people who want to learn.

I’m not proffesional physicist, but I happen to be a student who is state champion in physics for the past three years, and I consider myself to understand this subject pretty much, except for this god forsaken relativity. :P I never pretend I know something unless I really do, and prefer to ask questions than give answers. But I do recognize an informed person when I see one. (i think its better if this paragraph isnt quoted for the sake of staying on topic ;))

Share this post


Link to post
Share on other sites
Thank you Janus for the lovely response.

I anticipated your reply would be the way you wrote. ;)

However' date=' what I don’t get here is why is it important for us to consider the light path all the way until it reaches back the source. Why not just from the source to the mirror (from point A to point B) in which case, we get two different times (t1 and t2, in your example) for the cases with opposite light direction, as opposed to the time we measure from the mirror (“static”) reference frame (to/2, in your example)? Thank you again for your help.

 

[/quote']

 

The object of the light clock is two measure the time rate at a single point as measured from two different frames. Otherwise you run into Relativity of Simultaneity issues. Events that are separated by space that are simultaneous in one frame, are not simultaneous in the other.

 

Example, assume that you have two mirrors, one ahead and one trailing with the lightsource placed exactly between. in the frame of the mirrors and light source, the light reaches both mirrors at the same time. But from a frame which has a relative velocity to the mirrors, the light reaches one mirror before the other.

 

The only way to get both frames to agree that two events happened at the same time is to have them happen at the same place. Thus if we start our clock when the light leaves and stop it when it returns, both frames will agree that that was when the clock stopped and started.

Share this post


Link to post
Share on other sites

However' date=' what I don’t get here is why is it important for us to consider the light path all the way until it reaches back the source. Why not just from the source to the mirror (from point A to point B) in which case, we get two different times (t1 and t2, in your example) for the cases with opposite light direction, as opposed to the time we measure from the mirror (“static”) reference frame (to/2, in your example)?[/quote']

 

That's true, the equations for the invariability of the speed of light always contain c+V and c-V. Yet it's not supposed to matter (according to relativity). It's been proposed by opponents of relativity that this is a key to unravelling the mystery. There are in fact two different times. Why? Because of the addition of velocities (I know, I know, relativity doesn't allow for those). When you take into account c+v and c-V you end up with an AVERAGE over the entire trip. A one way trip is impossible to detect because the information from the light beam hitting the mirror in the first place has to reach the sender somehow (rebound trip - t2). So when you work out the numbers, you come up with an average. And what's the average?? C.

 

Please note - this is my opinion and I do not claim to be a physicist. I have a strong mathematical background (Aerospace engineer) but my mind works best with concrete ideas. That's probably why I'm an engineer.

Share this post


Link to post
Share on other sites

Maybe I understand Companiero's original question perfectly. Companiero: perhaps you make the most common mistake of them all... perhaps you forget that the distortions of SR are THREEfold, entailing length contraction, time dilation plus time dissynchronicity. You can't compute anything at all under relativity without working in that third element.

 

I offer my web site for plain elucidation of it all, with diagrams. The appended diagrams illustrate how the three distortions conspire to make all seem relativistic.

Share this post


Link to post
Share on other sites

Thank you Janus for explaining that to me. It was obvious, but I guess I didnt notice it.

 

lightSword, your site doesnt show up.

 

However, I believe the original purpose of this thread has been exhausted, and can now be closed/deleted (whatever you guys do with useless threads).

Share this post


Link to post
Share on other sites

i have a quick questino on time dilation.....

 

doesnt one have to incorporate the speed of the earth going through the universe/orbiting the sun? if people were to be launched into space and perhaps land on anohter planet, they would be traveling at an entirely differnt speed than that of those on earth, and though it may not be c, it should cause some sort of descrepincy in the passage time of the space traveler versus the passage of time of someone on earth.

Share this post


Link to post
Share on other sites

I get a chuckle from your phrase, "and though it may not be c". Why should it be c? why should it even be close to c?? Relativity proclaims that it could never be c. If you're talking about landing on another planet in our own galaxy, then its velocity with respect to Earth would definitely be very small compared to c, and so the clock discrepancy effect would be miniscule, virtually negligible. Yes, there would be some tiny discrepancy, but realize that it's not Earth's velocity "through the universe" (as you say) that matters, because relativity permits no such absolute measurement to be defined. All that pertains in such relativistic computations is the velocity of those travelers with respect to Earth.

 

With relativity, you can only compare one observatory to another... there's no "looking at the bigger picture".

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.