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What happens in the restframe with lightsource?

by faen
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faen
#1
Mar3-12, 07:16 AM
P: 140
So I was reading the proof of time dilation. It explains how the observer in a rest frame observes a longer path of the light than the observer in the moving frame.

However, I made a drawing of a slightly different experiment, where I can't predict the result of it. I was hoping somebody could help me out. In the picture below, the person in frame of reference A is moving close to the light speed. In the rest frame an observer is sending light between two vertical points as shown in the picture. The person in the rest frame would see that the light goes between the two points in time t determined by the given distance d and velocity c. Now, I am wondering what would the person in frame of reference A observe? If time is really slower for him, he would have to observe a shorter path with the same speed. However as far as I can see, the c and d variables is the same for the people in both frame of references giving equal time? I hope somebody can help me out with my confusion here, thanks a lot :)

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Fredrik
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Mar3-12, 07:55 AM
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In A's rest frame, the mirrors are moving, so the light going from the top mirror to the bottom is going "diagonally". The distance between the mirrors is the same in both frames. Light going from the top mirror to the bottom travels a distance d=ct in B's frame, and ct' in A's frame (where t' is the time measured by A). The pythagoran theorem tells us that ##(ct')^2=(ct)^2+(vt')^2##. Solve this for t, and you get
$$t=\frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma t'.$$.
faen
#3
Mar3-12, 08:32 AM
P: 140
Quote Quote by Fredrik View Post
In A's rest frame, the mirrors are moving, so the light going from the top mirror to the bottom is going "diagonally". The distance between the mirrors is the same in both frames. Light going from the top mirror to the bottom travels a distance d=ct in B's frame, and ct' in A's frame (where t' is the time measured by A). The pythagoran theorem tells us that ##(ct')^2=(ct)^2+(vt')^2##. Solve this for t, and you get
$$t=\frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma t'.$$.
Yes I thought about this too. But the result seems strange to me, because it now predicts that time goes slower in the rest frame and not in the moving frame, just because we changed the location of the experiment.

For example if we had done the experiment in frame A instead of B, t and t' would change place in the equation predicting opposite results, as such:

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

How is this possible?

Fredrik
#4
Mar3-12, 09:10 AM
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What happens in the restframe with lightsource?

It's possible because t and t' mean different things in the two equalities. Consider two inertial coordinate systems S and S' with a common origin. Denote the velocity of S' in S by v. We assume that v>0. Denote the event at the origin by O. Let E be an event on the time axis of S, and denote the time coordinate of E in S by t. Now the time dilation formula tells us that t'=γt. Here t' is the time coordinate of E in S'. Note that E is not on the time axis of S'.

But when you use the time dilation formula to go from S' to S instead, t' is the time coordinate of an event F on the time axis of S', and t is the time coordinate of F in S. So we're not just doing the calculation we did before "in reverse". It's a calculation that involves a different event.

This is what I said about the apparent contradiction in another thread:
Quote Quote by Fredrik View Post
"B's clock is slow relative to A" appears to contradict "A's clock is slow relative to B". To understand the problem here, it's essential that you understand that these statements are actually defined to mean something different from what they appear to be saying. What they actually mean is this:
"The coordinate system associated with A's motion assigns time coordinates to B's world line that increase faster along B's world line than the numbers displayed by B's clock"

"The coordinate system associated with B's motion assigns time coordinates to A's world line that increase faster along A's world line than the numbers displayed by A's clock"
ghwellsjr
#5
Mar3-12, 09:34 AM
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Quote Quote by faen View Post
Yes I thought about this too. But the result seems strange to me, because it now predicts that time goes slower in the rest frame and not in the moving frame, just because we changed the location of the experiment.

For example if we had done the experiment in frame A instead of B, t and t' would change place in the equation predicting opposite results, as such:

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

How is this possible?
According to Special Relativity, time does not go slower in any rest frame you choose, it only goes slower for objects/observers/clocks that are moving in any rest frame you choose. It's no different than saying that the rocket is moving in the ground's rest frame and the ground is moving in the rocket's rest frame. How is that possible?

But it does help to consider that in the ground's rest frame, the clock on the rocket is ticking slower and it is moving away from the clock on the ground. So between each tick of the rocket clock, as it observes the ground clock, it has moved farther away and so the image of the ground clock's ticks have farther to travel resulting in a longer time compared to its own ticks. It turns out that the ratio of the ground clock's tick interval compared to the rocket clock's tick interval as determined by the rocket is the same as the ratio of the rocket clock's tick interval compared to the ground clock's tick as determined by the ground. Work it out, you'll see that this is true.
faen
#6
Mar3-12, 12:22 PM
P: 140
Quote Quote by ghwellsjr View Post
According to Special Relativity, time does not go slower in any rest frame you choose, it only goes slower for objects/observers/clocks that are moving in any rest frame you choose. It's no different than saying that the rocket is moving in the ground's rest frame and the ground is moving in the rocket's rest frame. How is that possible?

But it does help to consider that in the ground's rest frame, the clock on the rocket is ticking slower and it is moving away from the clock on the ground. So between each tick of the rocket clock, as it observes the ground clock, it has moved farther away and so the image of the ground clock's ticks have farther to travel resulting in a longer time compared to its own ticks. It turns out that the ratio of the ground clock's tick interval compared to the rocket clock's tick interval as determined by the rocket is the same as the ratio of the rocket clock's tick interval compared to the ground clock's tick as determined by the ground. Work it out, you'll see that this is true.
That was a surprising result. I think I more or less understand it now. Thanks a lot for clearing it up for me, to both of you. I have one more question though.

If a person travels close to the light speed, will he become older slower than the observer not traveling or is this a myth then?
Goodison_Lad
#7
Mar3-12, 12:24 PM
P: 59
I think the confusion might lie in the use of the word ‘really’. If ‘time is really slower for him’, in an absolute sense, there’d be a problem because it would imply that time in the ‘rest’ frame was running faster than his own. Time is only running more slowly in A’s frame as observed from the ‘rest’ frame on the ground. As far as A is concerned, his time is just fine: to him, it is time in the ground’s frame that is running slower than his own. The slowing of time isn’t an absolute thing in the ordinary sense - it is a conclusion reached by one observer on time in the other frame.

It’s best to avoid saying time ‘really’ is running more slowly on the ship because this might suggest that the situation is one-sided. It seems to imply that the ship is ‘really’ moving, and the ground is ‘really’ stationary. Relativity doesn’t allow us to say this – instead, all we can assert is that the ship and the ground are moving relative to each other.

To observer A, the light clock on the ground is moving past him, so as far as he’s concerned the light which leaves the source and reaches the mirror hasn’t just covered distance d in the vertical direction – it must appear to follow a diagonal path, which, therefore, is longer than d. Since, as you rightly say, all observers measure light to travel at velocity c, A must conclude that the light has had to take longer to go from the source to the mirror than if the clock were stationary in his own frame. So A concludes, from observing B’s light clock, that time must be running slower in B’s frame than his own.

So why does observer B say that time in A’s frame is running slower? Suppose A had an identical light clock on his ship. He would observe the light moving a vertical distance d, in his frame, from the source to the mirror. But B would observe, because of the relative motion of the ship, the light tracing a diagonal path. This path would be longer than d, leading B to conclude that, since the light, traveling at velocity c, takes longer to cover this diagonal distance than it would take to travel the just the vertical distance d, time on the ship must be running slower than time on the ground.

It’s always tempting to think of something ‘moving’, but a more accurate expression (albeit a clumsier one) would be something like ‘moving relative to a certain frame of reference’. In special relativity, this helps keep at bay the completely natural, though wrong, notion that something is ‘really’ moving and that something else is ‘really’ still.
Goodison_Lad
#8
Mar3-12, 12:32 PM
P: 59
Sorry – didn’t see the last posts – I should pay more attention.

Somebody traveling at close to light speed (or any speed relative to us) would appear to age more slowly than us. But that’s only as measured by us. To the person traveling, it is we who are doing the moving relative to them, so they would see us as ageing more slowly than them.

Again, the situation is perfectly symmetric (so long as nobody changes velocity). Both are right.
John232
#9
Mar3-12, 03:27 PM
P: 249
How is it possible that the length contraction equation is inverted in relation to the Lorenzt transformation as compared to the time dialation equation? I think I recall haveing to take the inverse of the result in order to actually get a smaller value for the length of time that the object seen to travel at a velocity would experience. For instance, you can't have x/z and (x)(z) and get smaller values for both when z is the same value.

It has made me wonder if the real time dialation equation could look identical to the length contraction equation. I think the distance ct' would have to be assigned to the vertical distance. This would assume that the object in question measures their time by seeing the light travel straight up and down this vertical distance. Their time would then have to slow down in order for them to measure the same speed for light makeing this trip giving a smaller value for t to account for the smaller distance traveled. Then the distance the object traveled would have to be vt, this says that the observer watching the object travel used their own time to determine this as how they measure time. Then the other side of the triangle ct would be how an observer at rest measured the photon traveling at na angle using their own time as they measure it at rest. This would be a longer distance traveled and then would require a longer duration of time for the photon to be seen to travel this distance for an observer at rest. So in effect all you would be doing is comparing the relation of two sides of a triangle from one frame of reference to another side of a triangle that exist in another frame of reference. The amount of time t' would have to "really" go slower in order for them to measure the same speed of light when it is seen to travel a shorter distance.
Fredrik
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Mar3-12, 09:03 PM
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Quote Quote by faen View Post
That was a surprising result. I think I more or less understand it now. Thanks a lot for clearing it up for me, to both of you. I have one more question though.

If a person travels close to the light speed, will he become older slower than the observer not traveling or is this a myth then?
As Goodison_Lad said, they are both aging slowly in the other one's rest frame. (In A's frame, B is aging slower. In B's frame, A is aging slower). There's no contradiction here, because of what I said in my previous post. If one of them turns around so that they eventually meet again, the one who turned around will be younger. This result can't be calculated with the time dilation formula alone, because there are at least three inertial frames involved.
Fredrik
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Mar3-12, 09:11 PM
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Quote Quote by John232 View Post
How is it possible that the length contraction equation is inverted in relation to the Lorenzt transformation as compared to the time dialation equation?
The time dilation formula tells you the difference between the t' coordinates that another observer would assign to two events on the t axis of the inertial coordinate system in which you are at rest. The reason why the Lorentz contraction formula is different is that it doesn't tell us the difference between the x' coordinates that another observer would assign to two events on our x axis. Instead, it tells us the difference between the x' coordinates he would assign to two simultaneous (in his rest frame) events on the world lines of the endpoints of an object that's at rest relative to you. So the two events he's comparing are on one of his simultaneity lines, not yours. If he had compared two events on one of your simultaneity lines, the formula would have had the same factor as the time dilation formula.
John232
#12
Mar4-12, 12:25 PM
P: 249
Quote Quote by Fredrik View Post
In A's rest frame, the mirrors are moving, so the light going from the top mirror to the bottom is going "diagonally". The distance between the mirrors is the same in both frames. Light going from the top mirror to the bottom travels a distance d=ct in B's frame, and ct' in A's frame (where t' is the time measured by A). The pythagoran theorem tells us that ##(ct')^2=(ct)^2+(vt')^2##. Solve this for t, and you get
$$t=\frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma t'.$$.
In this equation you got the wrong relation to t and the Lorentz factor. t'=yt not t=yt'. Then x'=x/y, where y is the lorentz factor. It would seem that the assigning of t' can be counter intuitive and gives different relations for t and x to the lorentz factor, just to be clear about what I was saying before. I think I meant to say lorentz factor instead, before.

But, then say you where doing the proof on a black board and explaining how these variables are supposed to be assigned, how do you explain that the time variables are set up to be the opposite of what they "should" be? It looks like you have the original equation set up correctly, but then it was worked wrong or was a typo. I think the way you had it set up should give the correct relation to the lorentz factor. But then why would you measure the distance an object traveled as vt' when you are at rest using the moveing object dialated time and then measure ct' as the hypotenus as an object see's light travel at a greater distance at an angle again using the objects that is traveling in motions time, and then saying that it is compared in relation to the pythagrean to your own time as a light beam travels straight up and down to you at rest?

I think their should be a better answer than if you convert the change in time to actual time you end up haveing to take the inverse of the change in time that in turns gives you a smaller value for time for in object in motion, and then how does the act of accouting for the change in time affect how variables are even assigned? Or is accouting for the change in time even necessary in this algebra problem? I think it would be funny if Einstein goofed on setting up this equation and misassigned the time variables. I think if you exchanged the time variables and then considered them as the actual times, then the answer you would get wouldn't be all that much different than if you did a conversion that takes the inverse of the answer that would in turn give lower values for the objects time that was in motion. The only difference would be the difference you would get in an answer by switching the vairiables of an equation and then takeing the inverse of it, but I don't think doing that gives you the same answer.

I would hate to be right about this because if there was something wrong with this equation it would mean that we would all have to start modern physics over. Then again they don't really use this in particle physics anyways...
Fredrik
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Mar4-12, 12:52 PM
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Quote Quote by John232 View Post
In this equation you got the wrong relation to t and the Lorentz factor. t'=yt not t=yt'.
Good catch. I seem to have exchanged the meaning of t and t' somewhere. When t and t' are defined as I described, the result should be t'=γt.
Tantalos
#14
Mar4-12, 01:24 PM
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Quote Quote by Fredrik View Post
As Goodison_Lad said, they are both aging slowly in the other one's rest frame. (In A's frame, B is aging slower. In B's frame, A is aging slower). There's no contradiction here, because of what I said in my previous post. If one of them turns around so that they eventually meet again, the one who turned around will be younger. This result can't be calculated with the time dilation formula alone, because there are at least three inertial frames involved.
And what if both decide to return and then meet halfway?

I think my aging process will not change if someone runs away from me.
John232
#15
Mar4-12, 01:46 PM
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Quote Quote by Fredrik View Post
Good catch. I seem to have exchanged the meaning of t and t' somewhere. When t and t' are defined as I described, the result should be t'=γt.
No problem, I have wrestled with this idea for some time. I still haven't been able to get an answer that resovles the confusion of this issue. I have thought about trying to publish my own theory of special relativity that only switchs the time variables and doesn't consider the time as being a change in time, and I don't know how far into the mathmatics I would have to go into other theories that depend on this equation. I am sure it would find a lot of opposition since it would mean that everything we know about modern physics could be slightly wrong, but their is a growing opionion that relativity may still need some work done on it inoder to obtain grand unification and that may conclude that relativity as we know is not the final answer. Or, that a more accurate theory could get rid of dark matter or something of the like, I think it could be that something that physics has been looking for...
Fredrik
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Mar4-12, 01:49 PM
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Quote Quote by Tantalos View Post
And what if both decide to return and then meet halfway?
Then they'll be the same age when they meet.
faen
#17
Mar4-12, 02:58 PM
P: 140
I tried to do some research about the time travel thing. As far as I've understood while the person is in the space ship he turns around, accellerates, and he will feel that he is in a gravitational field. Being in a gravitational field, causes time to go slower. Thus for him time will go slower for him and faster on earth, while observers from earth predict the same, that time goes slower for him and faster for them. However this accelleration lasts only a very short amount of time compared to the whole trip. Therefor he will only feel as if he is in a gravitational field slowing his time only for as long as he accellerates. After he is finished accellerating both observers will experience that the other persons clock is slower, right?

So is it only the time during the accelleration that the person in the rockets time is really going relatively slower, or his time is really going relatively slower the whole trip?

Also, if gravity is curvature in space-time, how does such curvature provide matter with energy just by being curves?
Tantalos
#18
Mar5-12, 10:19 AM
P: 46
Quote Quote by Fredrik View Post
As Goodison_Lad said, they are both aging slowly in the other one's rest frame. (In A's frame, B is aging slower. In B's frame, A is aging slower). There's no contradiction here, because of what I said in my previous post. If one of them turns around so that they eventually meet again, the one who turned around will be younger. This result can't be calculated with the time dilation formula alone, because there are at least three inertial frames involved.
Quote Quote by Fredrik View Post
Then they'll be the same age when they meet.
This is against logic. According to you if B travels first and then A goes that same distance with the same speed as B to meet him then A will be younger.

And what if they decide to travel both in one space ship the same distance and speed? Logically they will be of the same age, but with the previous result their age depends on whether they travel alone or with a company.


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