# We can't determine who is moving?

1. Oct 27, 2009

### yoelhalb

here I will ask on linear motion
if in linear motion both can claim that the other is moving and
because the one that is moving it clock will go slower so
both claim that the other ones clock is going slower so when they will
meet (for instance the motion is toward each other) who will actually be younger.

even more
consider
A <-------------------> B ------------------> c
when A and B are any objects moving apart with linear motion of any
speed and c is the light going to the right of both.
because both are linear so both claim that they are at rest and the
other one move and his clock is getting slower.
so let see how it's work A claims that he is at rest and the clock of
B gets slower so he still sees light constant, that's fine.
but B claims that he is at rest so A will move so according to B then
Ac (the speed of c according to A) will be more then the speed of
light and the movement of A (from B's point of view) makes the situation worse because its time
slows down so it is getting even much more then the speed of light.
and because that cant be so its clear that A is at rest and B is
moving.
so there is a way to say who is moving even with linear motion

2. Oct 27, 2009

### JesseM

No, A and B will both measure the speed of light to have the same value of c. In Newtonian physics it's true that if A was moving at speed u relative to B, and A measured an object moving at speed v relative to himself in the same direction, then in B's frame the object would be moving at speed u + v, but in relativity this formula no longer works, you must instead use the relativistic velocity addition formula (u + v)/(1 + uv/c^2). You can see that if the object is a light ray so v=c (i.e. A measured it to move at speed c in his rest frame), then the speed in B's rest frame would be (u + c)/(1 + uc/c^2) = (u + c)/(1 + u/c) = (u + c)/(1/c)*(c + u) = 1/(1/c) = c.

Keep in mind that each is using their own rulers and synchronized clocks to measure speed--speed is distance/time, so if I have a set of synchronized clocks attached to different markings on a ruler which is at rest to me, then if the object passes marking #1 when the clock there reads T1, and then later it passes marking #2 a distance L away from marking #1 when the clock at marking #2 reads T2, then in my frame the speed of this object must be L/(T2 - T1). But if B is using her own ruler and synchronized clocks at rest relative to herself, then in A's frame B's ruler will be shrunk by a factor of $$\sqrt{1 - v^2/c^2}$$ due to length contraction, the length of each tick of B's clocks will be expanded by $$1/\sqrt{1 - v^2/c^2}$$ due to time dilation, and two clocks which are synchronized and a distance of L apart in B's frame will be measured to be out-of-sync by vL/c^2 in A's frame due to the relativity of simultaneity. If you take all these effects you can see why, even if you calculate everything from A's perspective, it makes sense that B, using these distorted rulers and clocks ('distorted' only in A's frame of course), will measure the speed of a light beam to be c. Here's a numerical example I came up with a while ago:

Say there's a ruler that's 50 light-seconds long in its own rest frame, moving at 0.6c in my frame. In this case the relativistic gamma-factor (which determines the amount of length contraction and time dilation) is 1.25, so in my frame its length is 50/1.25 = 40 light seconds long. At the front and back of the ruler are clocks which are synchronized in the ruler's rest frame; because of the relativity of simultaneity, this means that in my frame they are out-of-sync, with the front clock's time being behind the back clock's time by vx/c^2 = (0.6c)(50 light-seconds)/c^2 = 30 seconds.

Now, when the back end of the moving ruler is lined up with the 0-light-seconds mark of my own ruler (with my own ruler at rest relative to me), I set up a light flash at that position. Let's say at this moment the clock at the back of the moving ruler reads a time of 0 seconds, and since the clock at the front is always behind it by 30 seconds in my frame, then in my frame the clock at the front must read -30 seconds at that moment. 100 seconds later in my frame, the back end will have moved (100 seconds)*(0.6c) = 60 light-seconds along my ruler, and since the ruler is 40 light-seconds long in my frame, this means the front end will be lined up with the 100-light-seconds mark on my ruler. Since 100 seconds have passed, if the light beam is moving at c in my frame it must have moved 100 light-seconds in that time, so it will also be at the 100-light-seconds mark on my ruler, just having caught up with the front end of the moving ruler.

Since 100 seconds passed in my frame, this means 100/1.25 = 80 seconds have passed on the clocks at the front and back of the moving ruler. Since the clock at the back read 0 seconds when the flash was set off, it now reads 80 seconds; and since the clock at the front read -30 seconds, it now reads 50 seconds. And remember, the ruler was 50 light-seconds long in its own rest frame! So in its frame, where the clock at the front is synchronized with the clock at the back, the light flash was set off at the back when the clock there read 0 seconds, and the light beam passed the clock at the front when its time read 50 seconds, so since the ruler is 50-light-seconds long, the beam must have been moving at 50 light-seconds/50 seconds = c as well! So you can see that everything works out--if I measure distances and times with rulers and clocks at rest in my frame, I conclude the light beam moved at 1 c, and if a moving observer measures distance and times with rulers and clocks at rest in his frame, he also concludes the same light beam moved at 1 c.

If you want to also consider what happens if, after reaching the front end of the moving ruler at 100 seconds in my frame, the light then bounces back towards the back in the opposite direction towards the back end, then at 125 seconds in my frame the light will be at a position of 75 light-seconds on my ruler, and the back end of the moving ruler will be at that position as well. Since 125 seconds have passed in my frame, 125/1.25 = 100 seconds will have passed on the clock at the back of the moving ruler. Now remember that on the clock at the front read 50 seconds when the light reached it, and the ruler is 50 light-seconds long in its own rest frame, so an observer on the moving ruler will have measured the light to take an additional 50 seconds to travel the 50 light-seconds from front end to back end.

3. Oct 27, 2009

### yoelhalb

Thanks
But what about my first question who is younger.
Also will your furmulas also work for the following
c1 <----- A <------> B <----> C <----> D ---> c2
when the uppercase letters are objects moving away from each othes with a speed the closest to the speed of light (because each one claims that he is at rest and he may have from both sides something traveling close to the speed of light) and c1 and c2 are light
Will in that case the speed of light remain constant and also no one of the objects violate the speed of light.
take in mind that you can have such objects infinitly

4. Oct 27, 2009

### JesseM

If they are moving towards each other, you mean? In this case it will depend on the details of their relative ages at earlier points on the trip, how fast they are moving towards each other, etc. For example, suppose in my frame both of them are moving at 0.6c towards one another, and at some moment they both simultaneously (according to my frame's definition of simultaneity) pass planets which are 30 light years apart, so each planet is 15 light years away from the point where they will meet. In my frame it will take each one 15/0.6c = 25 years to reach the midpoint where they meet, and during this time each one's aging is slowed down by a factor of $$\sqrt{1 - 0.6^2}$$ = 0.8 due to time dilation, so they will each age 25*0.8 = 20 years between passing their respective planets and meeting one another. So, who is older will just depend on how old they each were when they passed their planets. For example, if A was 32 when he passed his planet and B was 30 when she passed hers, then when they meet A will be 52 and B will be 50.

You can analyze the same situation from the perspective of either one's frame, and there will be different conclusions about how rapidly each one was aging (and in other frames the events of their passing their planets will not have been simultaneous), but they will end up with the same conclusion about how old each one is when they meet. For example, in A's frame, at the moment A is 32 and passing his planet, B is older than 30 and has passed her planet a while ago--in A's frame the event of B passing her planet happened 22.5 years earlier than the event of A passing his own (this can be derived using the Lorentz transformation--if we set the event of A passing his planet to happen at the origin at x=x'=0 and t=t'=0, then if we know the event of B passing her planet happened at x=30, t=0 in the frame of the observer who sees A moving at 0.6c, then in A's own frame this event must happen at t' = gamma*(t - vx/c^2) = 1.25*(-0.6*30) = -22.5). Using the velocity addition formula, in A's frame B must be moving at a velocity of (0.6c + 0.6c)/(1 + 0.6*0.6) = 0.88235c, so B's aging must be slow by a factor of $$\sqrt{1 - 0.88235^2}$$ = 0.4706, so in the 22.5 years between B passing her planet and A passing his, B will have aged 22.5*0.4706 = 10.59 years, and thus be aged 30 + 10.59 = 40.59 years at the moment A is passing his planet at age 32. Then for A it takes another 20 years to reach the midpoint, during which time B has aged 20*0.4706 = 9.41 more years, so B will be aged 40.59 + 9.41 = 50 years when they meet at the midpoint, exactly the same as was predicted in the other frame.

Finally if we consider things from the perspective of B's frame, by symmetry we can see that in this frame it must be A who passed his planet 22.5 years earlier than B passed hers, and A who only aged 10.59 years in the 22.5 years between A passing his planet and B passing hers, so A must be 32 + 10.59 = 42.59 years at the moment B is passing her planet at age 30. Then for B it takes another 20 years to reach the midpoint, during which time A has aged 20*0.4706 = 9.41 more years, so A will be aged 42.59 + 9.41 years = 52 years when they meet at the midpoint.
Yes, you can use the velocity addition formula here too. For example, suppose that C measures D to be moving the right at 0.9c, and B measures C to be moving to the right at 0.9c, and A measures B to be moving to the right at 0.9c. Then the velocity addition formula tells us that in B's frame, D is moving at (0.9c + 0.9c)/(1 + 0.9*0.9) = 0.994475c. Now if we know that in B's frame D is moving at 0.994475c, and we know that in A's frame B is moving at 0.9c, then this is just another example where we have two frames and two velocities so we can use the velocity addition formula again, this time concluding that in A's frame D is moving at (0.9c + 0.994475c)/(1 + 0.9*0.994475) = 0.9997c. No matter how many times you combine velocities smaller than c in the velocity addition formula you'll never get a velocity greater than or equal to c (and if you plug in a velocity of exactly c in the formula, you'll always get a velocity of c back).

5. Oct 27, 2009

### yoelhalb

I would appreciate if you can explain me why the rotation of the earth on its axis is not considered linear motion and why for example the orbit of the earth around the sun can be considered.
(My argument that rotation is not considered is that if not then we can claim that all the stars circle us every day obviously faster then the speed of light, and the argument for the earth around the sun is a quate that I say in the name of Einstein and his book the evoulotion of physics page 160 that the question of wheter the earth or the sun is in center is a relativistic question.)
By the way (as I saw that you are very familiar with the subject and very nice) I have written a paper to answer the question of what is time and what causes it to flow and I want to publish it but I want it first to be reviewd by someone that understands the object and hear its opinion on that, would you like to do it?
Thanks.

6. Oct 27, 2009

### JesseM

Are you asking this question in the context of general relativity, where spacetime around the Earth is curved by its mass, or in the context of special relativity where we could treat the Earth as a sphere of negligible mass rotating in flat spacetime? In an SR context, rotation is obviously not linear because it would involve traveling in a circle which means the direction is constantly changing as seen by an inertial observer (and the person traveling in a circle on the surface of the Earth would feel G-forces due to acceleration, in this case the so-called "centrifugal force"). In the curved spacetime of GR, things are more complicated because no coordinate system covering a large region of curved spacetime can be considered "inertial", and no paths through curved spacetime would normally be referred to as "linear" (as an analogy, consider paths on a curved 2D surface like the surface of the sphere--can you draw any path on the surface of a sphere that you'd call 'linear'?)
It is only in inertial coordinate systems that nothing is allowed to travel faster than c, in non-inertial coordinate systems, whether in SR or GR, objects can have coordinate velocities greater than c (although in a GR context we can still say that if we zoom in on an arbitrarily small region of curved spacetime, a coordinate system in that tiny region can be "locally inertial" and the laws of physics observed in that system will be arbitrarily close to those seen in inertial frames in SR--this is the http://www.aei.mpg.de/einsteinOnline/en/spotlights/equivalence_principle/index.html [Broken]--so in this sense nothing can locally travel faster than c in GR).
He was probably talking about GR rather than SR, where a principle called "diffeomorphism invariance" allows you to use absolutely any coordinate system and the Einstein field equation relating curvature to mass will still apply--see http://www.aei.mpg.de/einsteinOnline/en/spotlights/background_independence/index.html [Broken] for more details. In an SR context, if we treated the Sun and Earth as having negligible mass, and the Sun as moving inertially while the Earth circled around it, then all inertial reference frames would agree the Earth was accelerating while the Sun was not.
If it's not too long I'll take a look at it and give you some comments, sure.

Last edited by a moderator: May 4, 2017
7. Oct 28, 2009

### yoelhalb

I am sorry for the long delay.
But of course I dont want to post the paper on this form how can we arrange that?
Thanks anyway

8. Oct 28, 2009

### yoelhalb

I saw it in relation to the question wheter the earth or the sun is in center and the queote should say that way can use whatever coordinate system we want

9. Oct 28, 2009

### hamster143

There's no actual, absolute "younger" if your objects are in different points of spacetime. The question is ill-formed. That is one of the essential parts of SR.