The twins struggle with Minkowski

[michel123456]

The speed of sound is not invariant. The "doppler effect" with light have nothing to do with waves going faster than others and getting closer to each other eventually joigning together. Waves of light go at the same speed no matter the speed of the source, that's what I have been told.

[Sisyphus]

The speed of sound is not invariant. The "doppler effect" with light have nothing to do with waves going faster than others and getting closer to each other eventually joigning together. Waves of light go at the same speed no matter the speed of the source, that's what I have been told.

 

The speed of sound isn't dependent on the speed of the source, either. Only the medium it's traveling through. If you're only looking at one frame of reference (and you are), then the situation is the same.

 

If you listen to a plane traveling at mach 0.5 away from you, then returning at the same speed, you'll hear it receding for 3 times as long as you hear it approaching.

[Janus]

The speed of sound is not invariant. The "doppler effect" with light have nothing to do with waves going faster than others and getting closer to each other eventually joigning together.

Neither does the Doppler effect for sound. If I'm standing next to the road and there is no wind, the speed of sound from the ambulance doesn't change, what changes is the distance between you and the ambulance, and the time it takes each successive "dee doo" to reach you.

 

Example: The ambulance starts its siren when it is 340 m away, travels towards you at 34 m/s and emits a "dee doo" once a second. Sound travels at 340 m/s, so you hear the fisrst "dee doo" in one second. At this same time, the ambulance emits its second "dee doo" but since it has traveled 34 meters closer to you since it emitted it last "dee doo", the sound only takes 0.9sec to reach you. So, you hear the first "dee doo" at one sec, and 0.9 sec later at 1.9 sec you hear the second. One tenth of a sec later, the ambulance emits a third "dee doo", by which time it is another 34 meters closer so the sound onlt takes 0.8 sec to travel the distance and arrives at 2.8 sec or 0.9 sec after the second "dee doo" In other words you hear the "dee doo"s 0.9 sec apart even though they were emitted 1 sec apart. Not only that, since the distance between you and the ambulance changed over the course of the emission of the 'dee doo's, each part of each sound wave was emitted at different distances, and you will hear the pitch of the "dee doo's as being higher. All without any chnge in the speed of sound you here.

Waves of light go at the same speed no matter the speed of the source, that's what I have been told.

 

 

Which just means that since the speed of light doesn't change for you, you can always treat it as if you are stationary and the other object is moving, unlike with sound, where you have to consider the speed of the receiver relative to the medium. This however does not change the fact that the distance between you is changing.

[swansont]

The speed of sound is not invariant. The "doppler effect" with light have nothing to do with waves going faster than others and getting closer to each other eventually joigning together. Waves of light go at the same speed no matter the speed of the source, that's what I have been told.

 

This isn't an issue of relativity. You're only dealing with one frame of reference, and the speed of sound in that frame is a constant. The effects are analogous.

 

I ment all observations done from FOR of Earth.

 

To show I am not completed stubborn.

 

If at the U-turn point, a flash light is emitted, I can imagine a spaceship tracking behind the flash, and arriving at Earth a few instant after the flash arrival.

But that is not exactly what happens.

In fact the spaceship flashes constantly.

The spaceship constantly follows constant flashes.

I must have wood in my head.

 

Yes, I know this is all measured from earth. But let's say there is a planet at the turnaround point (25 light-days away), in the same FOR as the earth, i.e. they are at rest with respect to each other. It sends a signal out signifying that it's January 1st. When does earth get the signal, and what does the earth observer set his calendar to be?

[DrRocket]

So we are looking in our telescopes, and after 50 days, we observe the spaceship making the U-turn.

 

.

.

.

 

Where am I wrong?

 

 

You are wrong all over the place,

 

Pay attention to what swansont, Sysiphus and Janus are telling you.

 

Correcting your mistakes is beginning to resemble "wack a mole". You really do need to read a book that will give you a complete and consistent treatment of relativity. Rindler's Essential Relativiity, Special, General and Cosmological might be appropriate. His Introduction to Special Relativity is also pretty good.

 

I like Naber's The Geometry of Minkowski Spacetime but you might find it a bit too heavy on mathematical abstraction.

 

Someone else might have better recommendations.

[michel123456]

Sorry, I have to apologize.

 

If at the U-turn point, a flash light is emitted, I can imagine a spaceship tracking behind the flash, and arriving at Earth a few instant after the flash arrival.

 

After a second thought, I cannot imagine that either. I completely forgot that the spaceship is under uniform linear motion. IOW it is at rest in his own FOR. It cannot "track behind its own image".

 

The only moment it can gain against its own image is during acceleration, that point we have negligate on diagram but explains the difference in the slope of the spaceship.

 

Am I wrong there too?

 

But even then, this "gain" will be lost when reducing speed at arrival. I suppose intuitively, due to symmetry.

Edited March 2, 2011 by michel123456

[Janus]

Sorry, I have to aplogize.

 

 

 

After a second thought, I cannot imagine that either. I completely forgot that the spaceship is under uniform linear motion. IOW it is at rest in his own FOR. It cannot "track behind its own image".

 

The only moment it can gain against its own image is during acceleration, that point we have negligate on diagram but explains the difference in the slope of the spaceship.

 

Am I wrong there too?

 

But even thain, this "gain" will be lost when reducing speed at arrival. I suppose intuitively, due to symmetry.

 

You're doing an improper frame switch. So far, this whole thread has been about the Earth frame. In fact, in your very first post you said:

 

We don't care about what is the observation of the traveler, we care only to represent what is observed from Earth's FOR.

 

And in the Earth frame, the ship does chase after its own light.

 

 

Besides, even if you switch to the traveler frame, you will find that the light from the u-turn will reach the Earth 25 yrs ahead of the traveler by the Earth's clock. You just have to take into effect that in the traveler FOR, the Earth is rushing to meet the light at 0.5c, that the earth clock runs slower and that the Earth clock reads later than 50 yrs right after the u-turn is made(Relativity of Simultaneity).

[michel123456]

(...)You just have to take into effect that in the traveler FOR, the Earth is rushing to meet the light at 0.5c, (...)

 

That is very Newtonian. The speed of light minus the speed of the Earth.

[swansont]

That is very Newtonian. The speed of light minus the speed of the Earth.

 

The speed of light is measured with respect to your own frame. When measured against another frame, you don't get c, nor do you expect to.

[michel123456]

The speed of light is measured with respect to your own frame. When measured against another frame, you don't get c, nor do you expect to.

 

The light that comes from all stars of the universe don't reach the Earth at C ?

[Sisyphus]

The light that comes from all stars of the universe don't reach the Earth at C ?

 

Of course it does. Where do you think it's implied that it doesn't?

[michel123456]

I don't understand your post #34

 

The speed of light is measured with respect to your own frame. When measured against another frame, you don't get c, nor do you expect to.

 

Interrogation point bolded.

[swansont]

The light that comes from all stars of the universe don't reach the Earth at C ?

 

The speed of light from the stars is measured in our reference frame; we are at rest. So of course you get c. What you aren't doing is measuring the speed of the light by comparing it to a moving object, i.e. in a different frame of reference.

[michel123456]

Now I don't understand your post #38

 

The speed of light from the stars is measured in our reference frame; we are at rest. So of course you get c. What you aren't doing is measuring the speed of the light by comparing it to a moving object, i.e. in a different frame of reference.

 

Aren't the stars around us moving relatively to us?

[swansont]

Now I don't understand your post #38

 

 

 

Aren't the stars around us moving relatively to us?

 

Yes, the stars are moving, but we aren't measuring the speed of light with respect to them.

[michel123456]

What are we doing when measuring the speed of light coming from the stars?

[Sisyphus]

In every frame of reference, light is moving at C. Therefore, in a given frame of reference, it is moving at C relative to something which is at rest in that frame. In that same frame, it is not moving at C relative to something which is not at rest.

[swansont]

What are we doing when measuring the speed of light coming from the stars?

 

We are measuring it with respect to things at rest in our frame.

[michel123456]

O.K. now I understand what you ment in post #34. Thank you both Swansont & Sisyphus.

 

But the only direct measurement we can make is in our own FOR, from light moving at C hitting some measurement device. We cannot directly measure the speed of light relative to another FOR. The only thing we know for sure is that the other observer in its own FOR will also find the same value for C.

 

What you propose, if I understand well, is that from our FOR here on Earth, we have to consider that the spaceship approaching us at 0,5 C under constant linear motion is chasing its own image.

 

We can imagine the same situation in reverse, saying that for some supposed alien planet, the Earth is chasing its own image because the Earth approaches the alien planet at 0,5 C.

 

Did I make any mistake in the above statements?

Edited March 2, 2011 by michel123456

[Iggy]

What you propose, if I understand well, is that from our FOR here on Earth, we have to consider that the spaceship approaching us at 0,5 C under constant linear motion is chasing its own image.

 

We can imagine the same situation in reverse, saying that for some supposed alien planet, the Earth is chasing its own image because the Earth approaches the alien planet at 0,5 C.

 

Did I make any mistake in the above statements?

 

As I understand them, they are exactly correct.

 

A good example, because it is observable, is the frame of a muon which is created by cosmic rays in the upper atmosphere and moves toward the earth's surface at 98% the speed of light. In our earth frame light moving from the upper atmosphere to the lower atmosphere moves at 1c and muons moving from the upper to lower atmosphere move at .98c. They are chasing light, as you put it, in our frame.

 

In the muon's frame, the muon itself is at rest. Light moves from the surface of the earth towards it at 1c and the earth moves towards it at .98c. So, I agree, the earth is chasing its own light in the muon's frame.

[michel123456]

Here I have to be really careful. Maybe I misinterpreted something.

 

I suppose that in order to agree in the above statements, you must have made a substraction of speeds, stating for example that between 0.98C and 1C, there is a 0.02% gap, representing the muons (or Earth) chasing their own image. This gap is not observable inside a specific FOR, but observing from one FOR the other.

 

Is it correct to make a simple substraction of speeds?

Why not use the relativistic formula for composition of velocities?

[Iggy]

Here I have to be really careful. Maybe I misinterpreted something.

 

I suppose that in order to agree in the above statements, you must have made a substraction of speeds, stating for example that between 0.98C and 1C, there is a 0.02% gap, representing the muons (or Earth) chasing their own image.

 

That's right.

 

If one thing in my frame is going c and something else in my frame is going .98c then the difference in the speeds according to my frame is c-.98c.

 

This gap is not observable inside a specific FOR, but observing from one FOR the other.

 

I'm not sure what you mean by that.

 

Is it correct to make a simple substraction of speeds?

Why not use the relativistic formula for composition of velocities?

 

Yes, because we are not switching frames you don't have to use the relativistic formula. The relativistic addition formula is when you ask "A is going u in my FOR and B is going v in the FOR of A -- how fast is B going in my frame?"

 

We are saying "A is going u in my frame and B is going v in my frame -- how much faster is B going than A in my frame?". For this question you don't need the relativistic addition formula.

 

In case I explain badly, here is a website explaining the difference:

 

How can that be right?

 

Naively the relativistic formula for adding velocities does not seem to make sense. This is due to a misunderstanding of the question which can easily be confused with the following one: Suppose the object B above is an experimenter who has set up a reference frame consisting of a marked ruler with clocks positioned at measured intervals along it. He has synchronised the clocks carefully by sending light signals along the line taking into account the time taken for the signals to travel the measured distances. He now observes the objects A and C which he sees coming towards him from opposite directions. By watching the times they pass the clocks at measured distances he can calculate the speeds they are moving towards him. Sure enough he finds that A is moving at a speed v and C is moving at speed u. What will B observe as the speed at which the two objects are coming together? It is not difficult to see that the answer must be u+v whether or not the problem is treated relativistically. In this sense velocities add according to ordinary vector addition.

 

But that was a different question from the one asked before. Originally we wanted to know the speed of C as measured relative to A not the speed at which B observes them moving together. This is different because the rulers and clocks set up by B do not measure distances and times correctly in the reference from of A where the clocks do not even show the same time. To go from the reference frame of A to the reference frame of B you need to apply a Lorentz transformation on co-ordinates as follows (taking the x-axis parallel to the direction of travel):

 

http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/velocity.html

[michel123456]

Sorry in advance for the long post.

 

This below is from your link:

 

"Suppose an object A is moving with a velocity v relative to an object B, and B is moving with a velocity u (in the same direction) relative to an object C. What is the velocity of A relative to C?

 

 

In non-relativistic mechanics the velocities are simply added and the answer is that A is moving with a velocity w = u+v relative to C. But in special relativity the velocities must be combined using the formula

 

[math] w=\frac{u+v}{1+uv/c^2} [/math]

 

 

Now lets label the diagram in correspondance with the muons example:

 

C=the earth's Frame Of Reference

B= muon's FOR

u=0.98c

w=c

 

Actually, from theory, we also know that [math]v=c[/math]. So we know almost everything.

 

What we don't know is the gap between w and v. Following your precedent post, I suppose you are stating that the gap as observed in Earth's FOR is [math]w-u[/math] that is [math]c-0.98c=.02c[/math](1)

 

Let's first make a simple check of the relativistic formula (just in order to make sure it is applicable here, people with knowledge can bypass this and go to point 2 directly)

 

1.Let's make the check.

with:

 

[math]u=0.98c[/math]

[math]w=c[/math]

 

inputting in

 

[math] w=\frac{u+v}{1+uv/c^2} [/math]

 

we get

 

[math] c=\frac{0.98c+v}{1+0.98cv/c^2} [/math]

 

[math] c(1+0.98v/c)=0.98c+v [/math]

 

[math] c+0.98v=0.98c+v [/math]

 

[math] c-0.98c=v-0.98v [/math]

 

[math] 0.02c=0.02v [/math]

 

[math] c=v [/math]

 

Which was the expected result. The formula says that v, which is the gap in muon's FOR, is equal to the Speed Of Light.

 

2. But that was not the question. The question is what is the gap in Earth's FOR.

 

Then we have to make the exact reverse calculations, labelling the same diagram as seen from the FOR of the muon: now the Earth is moving.

 

We have the same graph

 

 

with new labelling

C=the muon's Frame Of Reference

B= Earth's FOR

u=0.98c

 

There is no need to redo the calculations, see point 1. The result will be the same, since the input values are the same. The result is [math]v=c[/math]

 

Which means that a scientist riding upon a muon can calculate that the gap in Earth's FOR is the Speed Of Light.

 

3.If we take as granted the result coming from simple substraction 0,02c see (1), and input this in the equations as seen from the muon's FOR, we have

 

[math]u=0,98c[/math]

 

[math]v=0,02c[/math]

 

which gives

 

[math]w=\frac{c}{1+(0.98c 0,02c)/c^2}[/math]

 

[math]w=\frac{c}{1,0196}[/math]

 

which is not an expected result. Theory states that [math]w=c[/math] in Earth's FOR.

 

Michel

[swansont]

The velocity addition equation works for objects in a single frame of reference. You are mixing frames, so the formula does not apply. The solution is simply this: v = d/t. How far does each object move in a time t, relative to the observer. The relative speed is the difference in the distance traveled divided by that time. Which just turns out to be the difference in their speeds. If a rocket travels at c/2, then an observer will see the light outpacing it by c/2. If the rocket goes in the other direction, the separation occurs at 3c/2.

[Iggy]

3.If we take as granted the result coming from simple substraction 0,02c see (1), and input this in the equations as seen from the muon's FOR, we have

 

[math]u=0,98c[/math]

 

[math]v=0,02c[/math]

 

which gives

 

[math]w=\frac{c}{1+(0.98c 0,02c)/c^2}[/math]

 

[math]w=\frac{c}{1,0196}[/math]

 

which is not an expected result. Theory states that [math]w=c[/math] in Earth's FOR.

 

Michel

 

Use u+v=w if u, v and w are all measured by C.

 

Use the relativistic formula if u and w are measured by C and v is measured by B.

 

They are solutions to different problems so you can't substitute the solution of one into the other.

Edited March 6, 2011 by Iggy