Understanding Centre of Mass in Stationary Objects

[Ross B]

What do you think?

as you may be aware in dealing with stuff I tend not to guess as weird dang often happens...my guess is yes it will still lag

Can a spaceship longer than a barn, fit within the barn?

see what I mean

 

[Ibix]

Why guess? Why not work it out? Where is the Earth when it emits a pulse of light? Where is it when the pulse arrives at you? Which direction would you have to point a telescope to catch that light?

It's easiest to work in the frame where the Earth and the Sun are at rest, but you can use another frame if you like.

 

[Ross B]

Why guess? Why not work it out? Where is the Earth when it emits a pulse of light? Where is it when the pulse arrives at you? Which direction would you have to point a telescope to catch that light?

It's easiest to work in the frame where the Earth and the Sun are at rest, but you can use another frame if you like.

Well I thought i had figured it out but I get a weird result so I must be doing something wrong

so I have Earth sun distance = 150 million km
lest says the lag angle is 20 degrees
the speed of light is C

It appears to me I can calculate W

 

[Ibix]

You are working in a frame where the rocket is not at rest, which I advised against since it makes the maths harder.

However, you can do it if you wish. You are correct that light will appear to come in at an angle viewed from this frame; however you are not viewing from this frame. You are viewing from the Sun which is moving, so you need to correct for angular aberration to determine what you'd actually measure. Imagine a straight tube across your rocket joining Earth and Sun. Go back to your diagram and work out if the light pulse is ever outside this tube. That should tell you all you need to know.

 

[Ross B]

You are working in a frame where the rocket is not at rest, which I advised against since it makes the maths harder.

However, you can do it if you wish. You are correct that light will appear to come in at an angle viewed from this frame; however you are not viewing from this frame. You are viewing from the Sun which is moving, so you need to correct for angular aberration to determine what you'd actually measure. Imagine a straight tube across your rocket joining Earth and Sun. Go back to your diagram and work out if the light pulse is ever outside this tube. That should tell you all you need to know.

sorry I don't understand so it is possible to calc W given the above?

 

[Ibix]

What did I say the last three times you asked?

Here's the easy way to do it: work in the frame where the Sun and Earth are at rest. Can it possibly be lagging? Why would that answer change just because there happens to be someone moving relative to the Sun and Earth watching them?

 

[Ross B]

I this right
I can work out x from the angles
I can work out L from the angles
T1 - T2 = 2L/C
W = 2x/(T1-T2)
W = 2x/(2L/C)

 

[Ibix]

No, you cannot measure w. And it's the fourth time I've told you that.

What you are doing here is standing somewhere outside the rocket and watching it go past. You are measuring the speed of the rocket relative to you. You can do that - but that's not the question you asked. You wanted to know if you could measure the speed of the rocket from within. No you cannot, because inside the rocket you do not measure the 20 degree angle. Here's why not:

The Sun is on the left. The Earth is on the right. There is a straight tube connecting the two. There is a red pulse of light that flies from the Sun to the Earth and back. You can join the red dots to recover your first diagram if you want. But at all times, the red dot is directly between the Earth and the Sun. So the light always comes directly from where the Earth is.

Your diagram fails to account for the fact that the detectors attached to the Sun are moving too. So the angle you are marking as 20 degrees, they see as zero.

 

[Ross B]

No, you cannot measure w. And it's the fourth time I've told you that.

What you are doing here is standing somewhere outside the rocket and watching it go past. You are measuring the speed of the rocket relative to you. You can do that - but that's not the question you asked. You wanted to know if you could measure the speed of the rocket from within. No you cannot, because inside the rocket you do not measure the 20 degree angle. Here's why not:
View attachment 208426
The Sun is on the left. The Earth is on the right. There is a straight tube connecting the two. There is a red pulse of light that flies from the Sun to the Earth and back. You can join the red dots to recover your first diagram if you want. But at all times, the red dot is directly between the Earth and the Sun. So the light always comes directly from where the Earth is.

Your diagram fails to account for the fact that the detectors attached to the Sun are moving too. So the angle you are marking as 20 degrees, they see as zero.

but I thought u said before in the rocket scenario that the image of the sun would lag the actual sun ?

 

[Ibix]

but I thought u said before in the rocket scenario that the image of the sun would lag the actual sun ?

I've just reviewed all my posts in this thread and I cannot see that I said that anywhere.

 

[Ross B]

I think your wrong I think this is the proper pic

mauve is the image of the sun

black is the sun actual

 

[Ibix]

So... despite there being a tube physically connecting the Earth to the Sun in this setup, you are telling me the Sun isn't at the other end of the tube?

 

[Ross B]

referring to your pic if my eye was at b as the photons are arriving at my eye from position C the sun would appear to be at position c but the actual sun would be at position D

you are confusing coherent light with non coherent light

 

[Ibix]

It would appear at D, which you can tell because all of the light you can see from it is coming down the tube.

This really is easier to understand in the frame where the rocket isn't moving. It's instantly clear you can't measure a lag in that frame. Then how could you possibly be measuring a lag in any other frame? You can't get two different results just because there happens to be somebody watching you who is moving with respect to you.

 

[Ross B]

It would appear at D, which you can tell because all of the light you can see from it is coming down the tube.

This really is easier to understand in the frame where the rocket isn't moving. It's instantly clear you can't measure a lag in that frame. Then how could you possibly be measuring a lag in any other frame? You can't get two different results just because there happens to be somebody watching you who is moving with respect to you.

we can't go any further as I think it would appear at C.

If your eye was facing D the light from C would come in the corner of your eye, not straight on, the light would strike your eye at an angle

If light reflected off an object at C, there is no way the object would appear to be at d, that would mean the laws of physics were not working properly

 

[Ibix]

we can't go any further as I think it would appear at C.

If your eye was facing D the light from C would come in the corner of your eye, not straight on, the light would strike your eye at an angle

No, because (in the frame you insist on working in) your eye is moving sideways as the light enters. That's what my diagram shows. The pipe is a really long eye, and the laser pulse travels right down the centre of it the whole way.

Feel free to try it. Stand beside a road bouncing a basketball straight up and down. According to the driver of a passing car you're doing 30mph and the ball appears to be moving diagonally. Does it make any difference to you? According to you, you'd have to adjust the angle you're bouncing the ball just because the driver was watching you.

 

[Ross B]

No, because (in the frame you insist on working in) your eye is moving sideways as the light enters. That's what my diagram shows. The pipe is a really long eye, and the laser pulse travels right down the centre of it the whole way.

Feel free to try it. Stand beside a road bouncing a basketball straight up and down. According to the driver of a passing car you're doing 30mph and the ball appears to be moving diagonally. Does it make any difference to you? According to you, you'd have to adjust the angle you're bouncing the ball just because the driver was watching you.

that is not the same scenario tho as the driver is not at rest wrt to the ball and the bouncer. In my scenario the observer, laser, earth, sun and frame are all at rest wrt one another

the angle the light enters yr eye is NOT a function of W, but is a function of the width of the frame.

As the light is entering the left of yr eye, your eye moving to the right actually makes the angle of entry more rakeish

surely if light bounced off an object at C, the object would have to appears at C

empty the spaceship, fix an object at c, speed the spaceship up to W, bounce a light off the object at C, will it appear at C?

 

[Ibix]

that is not the same scenario tho as the driver is not at rest wrt to the ball and the bouncer. In my scenario the observer, earth, sun and frame are all at rest wrt one another

Then why are you drawing them in a moving frame?

 

[Ross B]

Then why are you drawing them in a moving frame?

empty the spaceship, fix an object at c wrt my eye, speed the spaceship up to W, put a strobe at C, bounce a light off the object at C, will I perceive it to be at C?

 

[Ibix]

What have I said the last five times you've asked this question?

I think I'm going to duck out of this thread now. I've told you the answer; I've drawn a diagram demonstrating the answer; and I've suggested an experiment you can do to test the answer. I don't think there's anything more I can offer.

 

[Ross B]

What have I said the last five times you've asked this question?

I think I'm going to duck out of this thread now. I've told you the answer; I've drawn a diagram demonstrating the answer; and I've suggested an experiment you can do to test the answer. I don't think there's anything more I can offer.

I agree ...I don't agree with you...I also proposed an experiment so you can test yr answer

 

[Ross B]

What have I said the last five times you've asked this question?

I think I'm going to duck out of this thread now. I've told you the answer; I've drawn a diagram demonstrating the answer; and I've suggested an experiment you can do to test the answer. I don't think there's anything more I can offer.

a further problem u have is the tubes would not appear straight, they would appear bent

 

[fresh_42]

This thread is going nowhere.

a) It's a re-opening of a closed thread which was already an exception we've made.
b) The answer has been given several times.
c) All participants who actually know the answer and tried to explain it, gave up.

As virtual means on the internet are restricted, the only remaining advice is to read a book about special relativity. Here's the first entry I received from google.com

https://web.stanford.edu/~oas/SI/SRGR/notes/srHarris.pdf

which starts right away by explaining frames aka coordinate systems. However, there are many. many more, such that one of them should be appropriate.

Sorry, that we couldn't solve this one to the satisfaction of the OP, but these cases occur.

Thread closed.