Speed of Light always remains constant ? think again.

[Anomalous]

[SOLVED] Speed of Light always remains constant ? think again.

Take 3 points P1, P2 and D

Point D is at a fantastic distance from P1 P2.

Point P1 is very near to P2 and both are at equal distance from D.

A Rocket R2 is waiting at P2

Rocket R1 is approaching at tremendous velocity at point P1 and towards D.

At the instance when the rocket R1 reaches P1, both the rockets simultaneously fire a powerful laser beam ( may be a few meters in length ) towards D.

Few hours after firing the laser, R2 starts accelerating towards D and quickly showing its raw power, matches the speed and starts flying with R1.

Now, will the speed of already fired lasers that haven't yet reached D with respect to R2 decrease ? now that it has come closer to the R1 hours after firing the laser, and while both the lasers together are approaching D.

 

[Meir Achuz]

The speed of each laser pulse is c.
Whatever the rockets do after firing is irrelevant.
The frequency of the pulse from R1 will be increased by the relatilvistic Doppler effect.

 

[russ_watters]

Like you, Anamalous, most scientists before Einstein assumed speed was relative (including the speed of light) and time was constant. As it turns out, they were wrong: time is relative and speed (of light) is constant.

How do we know? We measure both the (relative) rate of the passage of time and the speed of light on a virtually constant basis.

 

[Anomalous]

1) The speed of each laser pulse is c.

2) Whatever the rockets do after firing is irrelevant.

3) The frequency of the pulse from R1 will be increased by the relatilvistic Doppler effect.

1) LOL, obviously.

2) Why ?

3) But what about R2. And won't U care if R2 had achived 99% speed of light before catching up with R1 ?

 

[Anomalous]

...Now, will the speed of already fired lasers that haven't yet reached D with respect to R2 decrease ? now that it has come closer to the R1 hours after firing the laser, and while both the lasers together are approaching D.

Its clear here that R2 covered the same distance that R1 did after firing the lasers yet laser fired by R2 is not going to over take laser fired by R1 and hence comapred to R1 speed of light for R2 was less when it was accelerating near R1.

I know that I am wrong here , but I don't know how, and until then for me anyone who doesnot explain that to me is wrong.

 

[Crosson]

Now, will the speed of already fired lasers that haven't yet reached D with respect to R2 decrease ?

The reason that no one can explain the flaw in your reasoning is that you show a complete ignorance of the special theory of relativity. Read a book about SR, look at the reasons why the speed of light is a constant from all points of view at all times, then see if you still have this question.

 

[learningphysics]

Its clear here that R2 covered the same distance that R1 did after firing the lasers yet laser fired by R2 is not going to over take laser fired by R1 and hence comapred to R1 speed of light for R2 was less when it was accelerating near R1.

I know that I am wrong here , but I don't know how, and until then for me anyone who doesnot explain that to me is wrong.

Yes, according to R2, the speed of light is not constant because R2 is accelerating. But when R2 stays in a particular inertial frame it will measure the same speed of light as anyone else in an inertial frame.

Speed of light is the same in all inertial frames of reference.

In an non inertial frame the speed of light can be different, even 0.

Here's a specific example from Rindler's Introduction to Special relativity.
Take the example of a spaceship undergoing constant proper acceleration a (in it's own instantaneous frame it is always accelerating at a). Suppose that when the ship starts off from rest, a photon is emitted at a distance c^2/a behind the ship just as it leaves. Now from the point of view of the ship, the photon remains at that same distance behind it at all times. In other words that relative to the ship, the speed of the photon is 0.

The idea is that in a particular inertial frame, we will always measure the speed of light in vacuum to be c.

 

[dextercioby]

Yes, according to R2, the speed of light is not constant because R2 is accelerating. But when R2 stays in a particular inertial frame it will measure the same speed of light as anyone else in an inertial frame.

Speed of light is the same in all inertial frames of reference.

In an non inertial frame the speed of light can be different, even 0.

Here's a specific example from Rindler's Introduction to Special relativity.
Take the example of a spaceship undergoing constant proper acceleration a (in it's own instantaneous frame it is always accelerating at a). Suppose that when the ship starts off from rest, a photon is emitted at a distance c^2/a behind the ship just as it leaves. Now from the point of view of the ship, the photon remains at that same distance behind it at all times. In other words that relative to the ship, the speed of the photon is 0.

The idea is that in a particular inertial frame, we will always measure the speed of light in vacuum to be c.

In its (sic!) "own instantaneous frame" it's (sic!) not moving...

Daniel.

P.S.Quote Rindler I'm sure that,if he's (sic!) that Rindler,he wouldn't have made that mistake.

 

[learningphysics]

In its (sic!) "own instantaneous frame" it's (sic!) not moving...

Daniel.

P.S.Quote Rindler I'm sure that,if he's (sic!) that Rindler,he wouldn't have made that mistake.

Yes it is not moving. velocity = 0. But acceleration does not = 0. So can't I say it is accelerating?

Hmmm... maybe I should have said that in the inertial frame where the ship's velocity is 0, its acceleration is a (proper acceleration).

Here's how Rindler's writes it:
"If we define the proper acceleration \alpha of P as that which is measured in P's instantaneous rest frame"

 

[Integral]

Now, will the speed of already fired lasers that haven't yet reached D with respect to R2 decrease ? now that it has come closer to the R1 hours after firing the laser, and while both the lasers together are approaching D.

This is a pretty ambiguous statement. I would assume the lasers are in the different ships so will therefore be traveling with the ship. What is the point of this? How does this have anything to do with the initial pulses fired by the different lasers?

I assume that the initial pulses were fired when the moving ship was by passing the stationary ship so, according to Maxwell's equations the pulses will be traveling in the same (or nearly so) direction at the same speed. Let us assume that they are traveling side by side. Now if we further assume that the lasers are of the same type (frequency) then when the pulse arrive at point D the observer there will be able to tell which ship was moving at the time the pulse was initialized as it will be blue shifted by an amount related to the speed of the ship. Since the pulses of light have traveled the same distance and since they were initiated at the same time they will arrive at D at the same time.

It is not clear to me what your proposed mechanism is that the speeding up of the stationary ship will magically effect the laser pulse it previously initiated?

 

[dextercioby]

It's unbelieveble.In its rest frame the ship's not moving,which means it has 0 velocity and 0 acceleration.He must refer to a comoving frame.

Daniel.

 

[DaveC426913]

Now, will the speed of already fired lasers that haven't yet reached D with respect to R2 decrease ? now that it has come closer to the R1 hours after firing the laser, and while both the lasers together are approaching D.

I'm following your thought experiment all the way up to this last paragraph. It is nonsensical. Can you restate and elaborate on your thinking?

 

[robphy]

It's unbelieveble.In its rest frame the ship's not moving,which means it has 0 velocity and 0 acceleration.He must refer to a comoving frame.

Daniel.

If the accelerating observer carries an accelerometer, he will notice that it is not in its "center" position. His 4-acceleration is a nonzero spacelike vector.

 

[pervect]

My $.02 on the accelerating spaceship. An accelerating spaceship can set up a coordinate system (which is not a frame) by using the coordinates (x,t) of an instantaneously co-moving observer.

Such a coordinate system is strictly limited in it's possible size. The coordinate system does not and cannot map every coordinate pair (xi, ti) of an inertial coordinate system to a coordinate pair (xa, ta) of the accelerated observer. This is covered in detail in MTW's section on "accelerated obsevers". There are two difficulties which prevent the coordinate system from covering all of space-time: the fact that geodesics allow multiple coordiantes for some events, and the existence of an event horizon.

The event horizon associated with the coordinate system of an accelerated observer is known as the "Rindler horizon". In the coordinate system of an accelerated obsever, a light beam exactly at the Rindler horizon will have a constant value for the distance coordinate, much as an outgoing light beam exactly at the event horizon of a Schwarzschild black hole will have a constant value for the 'R' coordiante.

The fact that the rate of change of the coordinate with respect to time is zero could be called a "zero velocity", I suppose. However, if you put any inertial observer moving at any rate at the same point in space-time, he will still find that the speed of light is equal to 'c' in his local coordinate system.

 

[Anomalous]

... 1) I would assume the lasers are in the different ships so will therefore be traveling with the ship.

2) What is the point of this?

3) How does this have anything to do with the initial pulses fired by the different lasers?

4) I assume that the initial pulses were fired when the moving ship was by passing the stationary ship so, according to Maxwell's equations the pulses will be traveling in the same (or nearly so) direction at the same speed. Let us assume that they are traveling side by side. Now if we further assume that the lasers are of the same type (frequency) then when the pulse arrive at point D the observer there will be able to tell which ship was moving at the time the pulse was initialized as it will be blue shifted by an amount related to the speed of the ship. Since the pulses of light have traveled the same distance and since they were initiated at the same time they will arrive at D at the same time.

5) It is not clear to me what your proposed mechanism is that the speeding up of the stationary ship will magically effect the laser pulse it previously initiated?

1) right.

2) It shows that moving of R1 won't change speed of light ( according to established evidence ) when the lasers were fired and hence both Lasers will always travell together.

3) That they were fired by rockets with a speed difference.

4) Bingo.

5) U tell me, I am a Rookee.

 

[Anomalous]

I'm following your thought experiment all the way up to this last paragraph. It is nonsensical. Can you restate and elaborate on your thinking?

Can U ask which part didnt make sense to U sir.

 

[arildno]

Anomalous:
That the speed of light is constant for different observers is a postulate; the maths used in relativity uses this as one of its axioms, if you like (it is a necessarily true statement within relativity theory).

Thus, if you disagree to that, you've basically to complete one of two tasks:
1) Prove that relativity is mathematically inconsistent in some damaging way.
You've not done that.

2) Provide empirical results which must be interpreted as meaning that the speed of light cannot be regarded as constant.
You've not done that, either.

 

[quantumdude]

Anomalous: Now, will the speed of already fired lasers that haven't yet reached D with respect to R2 decrease ? now that it has come closer to the R1 hours after firing the laser, and while both the lasers together are approaching D.

Meir Achuz: The speed of each laser pulse is c.

Anomalous: LOL, obviously.

Well what exactly are you looking for? Are you looking for someone to do the calculation for you in both SR and Galilean relativity? Are you looking for someone to convince you that the SR prediction is correct? Are you looking for someone to do the experiment?

You asked a very cut-and-dry question, and you have been given an answer. You aren't going to get anywhere by snickering at the respondents. If you want something else, then ask for it.

 

[Crosson]

Whatever man, U have all the powers, but that's not the answer to my question, Can U please tell Mr. Corsson to answer my question now that he has read so many books, I mean what will inspire me to read now that I know that she can't answer me.

I would advise you to read "A brief history of time" by stephen hawking, to understand more about the constancy of the speed of light.

In a nutshell, the distance traveled by light, and the time taken, are different from different points of view (called frames of reference). But, the speed is always the same from everyones point of view.

 

[learningphysics]

I think it's an important distinction to make... that the speed of light is constant in all inertial frames, but not from all points of view.

 

[Maaneli]

Speed of light constancy

Take 3 points P1, P2 and D

Point D is at a fantastic distance from P1 P2.

Point P1 is very near to P2 and both are at equal distance from D.

A Rocket R2 is waiting at P2

Rocket R1 is approaching at tremendous velocity at point P1 and towards D.

At the instance when the rocket R1 reaches P1, both the rockets simultaneously fire a powerful laser beam ( may be a few meters in length ) towards D.

Few hours after firing the laser, R2 starts accelerating towards D and quickly showing its raw power, matches the speed and starts flying with R1.

Now, will the speed of already fired lasers that haven't yet reached D with respect to R2 decrease ? now that it has come closer to the R1 hours after firing the laser, and while both the lasers together are approaching D.

Actually, under certain conditions, the signal velocity of a photon can minutely exceed c, but not for the reasons you propose. See references:

http://www.npl.washington.edu/AV/altvw43.html

http://www-com.physik.hu-berlin.de/~scharnh/cite16.htm


Maaneli Derakhshani

 

[Anomalous]

...
You've not done that, either.

Well , I don't have the capacity to do that as I am not from a maths profession, I am just an amature, How can U expect something like that from me ? I did my best and no one told me why its wrong and I am happy of it. As long as I am happy that's what matters the most.

 

[Anomalous]

removed non-content remarks

Well what exactly are you looking for? Are you looking for someone to do the calculation for you in both SR and Galilean relativity? Are you looking for someone to convince you that the SR prediction is correct? Are you looking for someone to do the experiment?

You asked a very cut-and-dry question, and you have been given an answer. You aren't going to get anywhere by snickering at the respondents. If you want something else, then ask for it.

What I am asking is that , without knowledge of all that super science, What seems so obvious to me seems so wrong to others here.

In my thought experiment it seems impossible that speed of light should remain at C for R2 as it covered vast distance after firing the lasers, hence while it was in acceleration its impossible that speed of light of the fired lasers was C at that time, Thats all I am saying.

 

[wisp]

Now, will the speed of already fired lasers that haven't yet reached D with respect to R2 decrease ? now that it has come closer to the R1 hours after firing the laser, and while both the lasers together are approaching D.

Anomalous

In a non-Einstein world you could say that R2 measures both light pulses traveling at speed c towards D.
If R2 moves towards D at speed V. Then R2 would measure the light pulses speed as u=lambda(c-V). Since R2 is moving the measured results are relative ones, because R2 experiences time dilation (lambda). In absolute time (which R2 cannot experience because of motion) the approach speed is simply c-V.
However, these my views and they do not agree with those of Einstein.

 

[learningphysics]

I was interested in finding out what the accelerating observer measures as the speed of light in this particular situation.
Can you check my calculations below?

Suppose that in some inertial frame S, we have two ships and a photon.

The equation of motion of ship 1 is: x = 0 (0<=t<=1)
x = (1/2)(0.4c)(t-1)^2 (t>1) (in other words acceleration = 0.4c)

The equation of motion of ship 2 is: x = 0.1ct

The equation of motion of the photon is: x = ct

We see that at t=1... ship 1's rest frame is this particular inertial frame (since its velocity at t=1 is 0).
At t=1, the distance between ship 1 and the photon is c.

The two ships meet at t = 2.
The spacetime coordinates of this event are x=0.2c, t=2

At t=2 the velocity of ship 1 is 0.4c.

Use lorentz transformation to transform the coordinates of this event to a frame S' moving at 0.4c relative to frame S :
x&#039; = \gamma(x-vt) = \gamma (0.2c - 0.4c(2))
x&#039;= -0.6c\gamma

t&#039; = \gamma (t-vx/c^2) = \gamma (2 - 0.4c(0.2c)/c^2)) = 1.92\gamma

The equation of motion of the photon transforms in this frame to simply:
x' = c t'

So at t&#039; = 1.92 \gamma, x&#039; = 1.92 c \gamma

So when ship 1 catches up with ship 2, it will measure a distance of 1.92 c \gamma + 0.6 c \gamma =2.52c \gamma between itself and the photon.

So the change in distance that ship 1 sees is 2.52c \gamma - c

Now we need to find the time that passes for ship 1 from the moment it starts accelerating to the moment it catches up with ship 2.

d\tau = \sqrt{1-v^2/c^2}dt
d\tau = \sqrt{1 - (0.4c(t-1))^2/c^2} dt
d\tau = \sqrt{1 - 0.16(t-1)^2}dt

If we integrate both sides from t=1 to t=2 we get:
\delta \tau = 0.97265 s

So the average speed measured by ship 1 from the moment it starts accelerating up to when it catches ship 2 is:

\frac{2.52c \gamma - c}{0.97265} = 1.799c

So for the accelerating observer the speed of light is on average greater than c(during the acceleration) ? Is the above correct?

 

[Crosson]

This attitude is going to make a common man like me to dismiss these technocrats as bluffs.


If they know something greate then they should be able to Rationaly explain it to a common man

Alright, I except your challenge.

Suppose their is a maximum speed in the universe c (not having to do with light).

I see a bird flying across the sky (to the left) at this maximum speed c. You are running past me in the oposite direction (to the right), at the close to the maximum speed 0.99*c.

Question: What is the speed of the bird from your point of view?

Well, if you don't think to hard, you will add the velocities and get 1.99c. But I told you to suppose c was the maximum speed! The correct answer is that you see the bird traveling at c.

The point of this is to show that IF there is a maximum speed THEN anything traveling at that speed from one point of view must travel at that speed from everyones point of view.

It turns out that in a universe with a maximum speed, space (distances) and time (durations) must be relative to your point of view! Spacetime is a cone, rather then a flat sheet.

 

[SHtRO]

To make the answer simple for whomever is listening (if you are listening and you don't understand already...)...

Variations in velocity of the source of light affect WAVELENGTH (and hence FREQUENCY), they do not affect the SPEED of light. The SPEED of LIGHT is CONSTANT. That is the whole point of relativity. You cannot understand this unless you understand simple wave concepts such as FREQUENCY and WAVELENGTH. Light is not sound, if it is traveling through a medium, we have not yet detected that medium. All OBSERVERS "see" the SPEED of LIGHT as c (~126,000 mps) but they observe different frequencies depending upon gravity and the velocity of the light source RELATIVE to the OBSERVER.

This was all demonstrated quite nicely by Albert Einstein in the early 20th century and is only scientifically disputed by a few (rather strange, IMHO) scientists. There are plenty of SPECIAL RELATIVITY experiments demonstrating these concepts and they are quite nicely represented by the GENERAL THEORY OF RELATIVITY which states that E=mc^2.

 

[El Hombre Invisible]

Yes! Tell us your theories!

BTW, SHtRO: E=mc^2 was a conclusion of SR, not GR. And one of the observations that also needs to be explained by alternative theories is why the Universe is expanding as we go forward in time (hence gets smaller and smaller the further back in time you go).

 

[Nereid]

Hmm, let's see now, Anomalous is using a computer and the internet to post onto Physics Forums. These things don't grow on trees, so how did they come about? The methods used to make Anomalous' life interesting and rewarding (computers, the internet; the medicines he may have taken to keep healthy, the food he enjoys eating, and so on) are the same used to determine that the speed of light is constant.

May we now expect Anomalous to foreswear use of all things which come from the application of a method he doesn't believe in? Perhaps there's no need to ban him ... his own sense of morality, fair-play, and honesty will compel him to leave us (after all, bearing the shame of realising he's naught more than a cynical hypocrite may be more intolerable to him than seeing the reflection of his 'tude mirrored in others' eyes?).

 

[Daminc]

That was almost poetic Nereid

 

[Janus]

Mnetor note:

I just spent a considerable amount of time cleaning this thread of off topic comments. Let's try and keep it this way.

 

[DaveC426913]

Can U ask which part didnt make sense to U sir.

OK, the above also doesn't make sense. I can figure out what you meant to say, but Jeez, that's the problem with your initial question too. It is confusing to read, particularly the second, run-on sentence. Just rewrite it, using shorter sentences and filling in details of your thinking.

 

[DaveC426913]

The redshift blue shift thing that's all due to difference in speed of light; its just that I believe that when number of protons hit us at a faster rate, we get a higher frequency and viceversa. Naa, there is no ether.

No. Red light and blue light travel at the same speed.

Are you looking for answsers, or are you developing your own hypotheses?

 

[O Great One]

It's o.k Anomalous. If you spend your time thinking about SR, I'm sure you can come up with paradoxes. How about the pole-in-the-barn paradox with a twist? The solution given is based on relativity of simultaneity. In the barn's rest frame the doors close at the same time, where as in the pole's rest frame the doors close at different times. Now, let's say that there is a grenade in the back of the barn and there is fishing line running from the pin in the grenade to one of the doors and more fishing line running from the rest of the grenade to the other door. Now, if both doors close then the line is pulled tight and *KABOOM*. But, if only one door closes then the grenade does not explode. So, how can the barn simultaneously be exploded and not exploded?

 

[Phobos]

No I was talking about what's out there, What is obvious dosent need to be doubted.

When it comes to Relativity, common sense (what is 'obvious' to us) often fails. Our common sense is developed from our everyday experience here on Earth where everything we interact with is moving slowly relative to us. In Relativity, you're dealing with extremes (e.g., speed of light, gravitational fields of massive objects) that we just don't think about in our "common" experience.

 

[Gokul43201]

Since this (what follows) is the claim you are making, let's, for the sake of science and sanity, address it.

Its clear here that R2 covered the same distance that R1 did after firing the lasers

Yes, this follows your statement that R2 "starts flying with R1". However, you claim that speeds and positions can be matched after some period of positive acceleration. This is not possible.

Here's what you said :

Few hours after firing the laser, R2 starts accelerating towards D and quickly showing its raw power, matches the speed and starts flying with R1.

I assume that all speeds spoken of are measured by an observer at rest in the rest frame of P1, P2 and D. Even simple, non-relativistic kinematics will show you that if R2's speed is monotonically increasing, and R1's is constant, then if R2 covers the same distance as R1 in less (or equal) time than R1, then R2's final velocity will greatly exceed R1's. The only way they can match positions and speeds is if R2 slows down as it approaches R1.

Does this new knowledge change anything ? If not, let's proceed.

yet laser fired by R2 is not going to over take laser fired by R1

There are several points to address here :

This is the first statement you are making about things necessarily traveling at or near the speed of light, so you must first tell us what framework you are working under. Clearly, you are not basing this statement on the framework of SR (or are you ?). Are you using the "Galilean framework", where the speed of the light beam at P1 is v+c ? Or are you using some other system of thought for understanding the kinematics of fast moving objects ?

Opening this clause with 'yet' means that what follows is related to the preceding clause (talking about "raw power" and "matching speeds"). So, this indicates that you expect the motions of the rockets, subsequent to firing the lasers to affect the possibility of one beam overtaking the other. The only way this might be true, in any system of thought, is if the kinematics of the beams (with respect to an observer at rest in this frame) is coupled forever to the kinematics of the source that emitted it - some kind of "entanglement" is suggested. Alternatively, you might be suggesting that the kinematics of the beams is being described with respect to one of the two rockets, but there is no such mention in the clause. So, this clause merely leaves the reader saying "uhhh?"

Nevertheless, let's accept it and analyze what it might be saying. You then use the word(s) "over take" (by which I'll assume you mean the single word 'overtake'), which is not well-defined. Does A have to lag behind B in order to 'overtake' B ? If A and B are traveling side-by-side and A speeds up, does it 'overtake' B ? Either way, you are claiming that the beam from P2 does not get ahead of the beam from P1 (with respect to the observer at rest ?). There is no suggested reason why this must be true, given that we have no idea what system of science is being used in this discussion. In both Galilean and SR frameworks, this statement will indeed be true, and in fact, in both frameworks it would be true independent of the subsequent dynamics of the rockets. For now let's accept it as true, and proceed.

and hence comapred to R1 speed of light for R2 was less when it was accelerating near R1.

Now where did this come from ? Starting off with "and hence", indicates that what follows must be derived from what preceded. Where is the logical connect ? If it is just our failing to connect the dots, do show us how they are connected.

This is like saying : "Today is a warm day, and hence Chopin never played piano. Prove me wrong."

Anyway, there are other issues to sort out with this clause too.

You talk of the "speed of light for R2". So, you seem to suggest that speeds are observer dependent (though you haven't mention how an observer measures the speed of light). So why, until this point, have to never mentioned the frame in which various speeds that you talk of are measured ? Let's ignore that too for now.

You are saying that, at some point of time, R2 was "accelerating near R1". I assume that by "accelerating" you are referring to a positive acceleration with respect to the direction vector pointing from P2 to D (as measured by the observer at rest). As I explained earlier, monotonic acceleration of R2 can not result in the final specified condition of matching speeds and positions. The only way R2 can be positively accelerating beside R1 with the same instantaneous velocity as R1, is if it had already overtaken R1, then slowed down to a speed below R1's (still maintaining a lead over R1) and then subsequently sped up again, allowing R1 to catch up with it (rather than the other way round). But there is absolutely no mention that this the the second time the rockets will be side by side, or that it is R1 that comes up to R2 from behind. In fact, the opposite is suggested.

 

[wisp]

... let's, for the sake of science and sanity, address it.
...

A simplified approach.

Since the light fired from both rockets travels at the same speed c (the motion of the light's source doesn't affect the speed of light - proven fact), we could simplify matters by letting point D emit light towards R1 and R2, and look at it in terms of Doppler shift.

The stationary rocket will record the light traveling towards it as having an observed frequency the same as the source, Fobs = Fsource.

When the rocket moves relative to the source (D) at speed V. It will record a frequency of Fobs= Fsource*squareroot[(1+V/c)/(1-V/c)]

The bottom line is: both SR and an alternative approach using c+V will give the same answers for the Doppler frequencey shift. And so you can't use this to show relativity is wrong, which is frustrating, but that's the way it is.

 

[c_helm@yahoo.com]

Take 3 points P1, P2 and D

Point D is at a fantastic distance from P1 P2.

Point P1 is very near to P2 and both are at equal distance from D.

A Rocket R2 is waiting at P2

Rocket R1 is approaching at tremendous velocity at point P1 and towards D.

At the instance when the rocket R1 reaches P1, both the rockets simultaneously fire a powerful laser beam ( may be a few meters in length ) towards D.

Few hours after firing the laser, R2 starts accelerating towards D and quickly showing its raw power, matches the speed and starts flying with R1.

Now, will the speed of already fired lasers that haven't yet reached D with respect to R2 decrease ? now that it has come closer to the R1 hours after firing the laser, and while both the lasers together are approaching D.

I am a rookie at best, so please forgive me if this question seems simplistic or its answer painfully obvious. But...

If R1 is in motion (at a tremendous velocity) as it approaches P1, and R2 is waiting, therefore at rest, at P2: would the relative gravity have more effect on the light leaving R2? Is it still accepted that light "bends" if enough gravitational force is exerted? So, if there is enough gravity to pull at R2's laser, but R1's momentum is enough to overcome it, will R2's beam actually have a bit farther to travel since it will be approaching D as a curve-ball? If so, then the speed is constant between the two, but one must travel farther, arriving later.

Am I way off base? Be gentle.

 

[EnumaElish]

Not even a rookie, a beginner at best

Anomalous,

I may be totally off the wall here. But bear with me. (I will especially appreciate any comments pointing to any errors in my posting.)

R2 has been speeding up to catch up with R1, so its crew have been measuring time at a slower pace. At the point where R2 catches up with R1, its clock no longer matches with R1's. R2's clock has been slowed down relative to R1's clock.

Had that not happened, R1 and R2 would have measured different speeds of light. But because it happens, they do not.

Put differently, since both crews take a measurement and observe the same speed of light, their notions of time must be different.

I hope this is correct and useful.

Take 3 points P1, P2 and D

Point D is at a fantastic distance from P1 P2.

Point P1 is very near to P2 and both are at equal distance from D.

A Rocket R2 is waiting at P2

Rocket R1 is approaching at tremendous velocity at point P1 and towards D.

At the instance when the rocket R1 reaches P1, both the rockets simultaneously fire a powerful laser beam ( may be a few meters in length ) towards D.

Few hours after firing the laser, R2 starts accelerating towards D and quickly showing its raw power, matches the speed and starts flying with R1.

Now, will the speed of already fired lasers that haven't yet reached D with respect to R2 decrease ? now that it has come closer to the R1 hours after firing the laser, and while both the lasers together are approaching D.

 

[clj4]

The bottom line is: both SR and an alternative approach using c+V will give the same answers for the Doppler frequencey shift. And so you can't use this to show relativity is wrong, which is frustrating, but that's the way it is.

This is patently wrong. SR and the ballistic ('c+v') theory give different answers for the Ives-Stilwell experiment. SR gives the correct result (confirmed by experiment), the 'c+v' gives the wrong one.

 

[LostInSpaceTime]

What I am asking is that , without knowledge of all that super science, What seems so obvious to me seems so wrong to others here.

In my thought experiment it seems impossible that speed of light should remain at C for R2 as it covered vast distance after firing the lasers, hence while it was in acceleration its impossible that speed of light of the fired lasers was C at that time, Thats all I am saying.

Not sure but I think you might be confused by the whole if your going 60km/hr and you throw a ball out the car window in the same direction at 15 km/hr the the balls total speed is 75km/h scenario.

Basically (from what i understand) light doesn't work that way. If you shoot a light beam out the window as you are going almost the speed of light the light beam still only travels c. I know this is an old thread but it would appear that's the source of confusion. (Either that or I've totally missed the point to his question)

 

[pervect]

This thread is over two years old. Let's let it rest in peace, and if you have questions, start a new thread, hopefully one of higher quality.