B Understanding Centre of Mass in Stationary Objects

[Ross B]

are 2 objects in the same frame if their centre of mass is stationary wrt each other?

 

[PeterDonis]

Objects aren't "in" one frame vs. another. Any reference frame will describe all objects in the area being described; different frames will just assign different coordinates to them.

 

[Ross B]

u say "different frames" isn't it implicit in that statement one frame vs another?

 

[Ross B]

if one frame cannot be distinguished from another how do you know when to use a diff corrd system?

 

[PeterDonis]

u say "different frames" isn't it implicit in that statement one frame vs another?

Yes, but I didn't say different objects will be in different frames. I said different frames will assign different coordinates to the same object.

if one frame cannot be distinguished from another

How did you get that out of what I said?

how do you know when to use a diff corrd system?

The choice of coordinates is a matter of convenience--you choose coordinates to make whatever calculation you want to do as simple and easy as possible.

How much background do you have in physics? What textbooks have you read?

 

[Ross B]

the book Serway 4th ed

 

[PeterDonis]

the book Serway 4th ed

What does it say about reference frames?

 

[Herman Trivilino]

I find this so confusing arent 2 objects in the same FOR if they are not moving relative to each other?

If two objects are not moving relative to each other then there are frames of reference in which they are both at rest. We usually refer to any of these frames of reference as a rest frame of those objects. Some will go so far as to say that the objects are "in" these frames of reference, but as has already been pointed out to you, those objects have a location in any frame of reference, regardless of their state of motion in those frames.

In the other thread that you started you were told that it takes 8 minutes for light to travel from Earth to the sun, but that such a claim was frame-dependent. That is, in a frame of reference in which both Earth and the sun are at rest, that transit time is indeed 8 minutes. But in a frame of reference in which they are moving that time might be less than 8 minutes. But that speed would have to be a significant fraction of the speed of light for the time to differ from 8 minutes by any significant amount.

 

[Ross B]

so going back to my original post. If u are standing on the sun and you point a laser at where u think the Earth is will the laser miss the Earth as it is pointed at a false image which is 8 minutes behind where the Earth actually is ?

 

[Ross B]

there is no facility to edit yr post, like FB, so if u think of a bit u want to add us have to do an entirely new post?

 

[SlowThinker]

so going back to my original post. If u are standing on the sun and you point a laser at where u think the Earth is will the laser miss the Earth as it is pointed at a false image which is 8 minutes behind where the Earth actually is ?

If you point the laser at the image of Earth that you see (you being in the Sun), you miss Earth by 16 minutes, or about 30000 km.
If you point the laser to where you "think" the Earth is now, you miss by 8 minutes, or 15000 km.
Of course you can watch the Earth move and point the laser in front of Earth to hit it, just like shooters do when they want to hit a moving object.

Posts can usually be edited for a while after being sent, but the feature might unlock only after you make a certain number of posts.

 

[Herman Trivilino]

so going back to my original post. If u are standing on the sun and you point a laser at where u think the Earth is will the laser miss the Earth as it is pointed at a false image which is 8 minutes behind where the Earth actually is ?

There's nothing false about the image. Hold your hand as far away from your eyes as you can and look at it. You see it as it was about 3 ns ago, because it takes light about 3 ns to travel the length of your arm. Everything you see is a consequence of a real image forming on your retina.

If you lead Earth by 8 minutes when you aim, you'll still miss by 8 minutes because that's how long it will take for the laser beam to reach Earth, so as @SlowThinker points out, it's a total of 16 minutes.

Is this the situation you were pondering when you asked your original question? If so, a lot of frustration could have been avoided if you'd have mentioned that. Usually when people ask me a question I try to figure out why they're asking, because that helps me form a more meaningful answer. When we're really stumped on something we often don't know how to formulate the right question.

 

[PeterDonis]

there is no facility to edit yr post, like FB, so if u think of a bit u want to add us have to do an entirely new post?

There is a fairly short window of time where you can edit a post. However, if you find yourself continually making one sentence posts and then making more one sentence posts shortly afterward, you should consider taking more time to think through your posts before making them at all. You are much more likely to get useful responses if you take the time to post fully formed questions that can stand on their own, instead of half-baked ones that you find yourself continually needing to add to.

 

[Ross B]

There's nothing false about the image. Hold your hand as far away from your eyes as you can and look at it. You see it as it was about 3 ns ago, because it takes light about 3 ns to travel the length of your arm. Everything you see is a consequence of a real image forming on your retina.

If you lead Earth by 8 minutes when you aim, you'll still miss by 8 minutes because that's how long it will take for the laser beam to reach Earth, so as @SlowThinker points out, it's a total of 16 minutes.

Is this the situation you were pondering when you asked your original question? If so, a lot of frustration could have been avoided if you'd have mentioned that. Usually when people ask me a question I try to figure out why they're asking, because that helps me form a more meaningful answer. When we're really stumped on something we often don't know how to formulate the right question.

I just used the term false image to differentiate between it and where the Earth actually is placed in space

so to be clear if I point the laser at the image of the Earth that is formed on my retina I will miss it. If I lead the image of the Earth that is formed on my retina by 16 minutes I should get a reflection ...yeh?

 

[SlowThinker]

If I lead the image of the Earth that is formed on my retina by 16 minutes I should get a reflection ...yeh?

Sounds correct. Why do you have doubts, and why are you asking this in a Relativity section? Is this the first step to something else?

 

[Ross B]

as an outside observer of our galaxy (the Milky Way) is the Earth in the same frame as the galaxy?

 

[PeterDonis]

as an outside observer of our galaxy (the Milky Way) is the Earth in the same frame as the galaxy?

All objects are in all frames. See post #2.

I suspect that answer won't remove the question in your mind, but if so, that's because "in the same frame" is the wrong way to express whatever you're actually trying to ask. I strongly suggest taking some time to work through what your textbook says about reference frames and learn the appropriate language for describing what you're thinking. It will pay big dividends in the long run.

 

[sdkfz]

as an outside observer of our galaxy (the Milky Way) is the Earth in the same frame as the galaxy?

Peter Donis answered that in post #2. Think of it this way: if you are standing "still" on the side of the road and a car drives past, does it vanish? No. It's not stationary with respect to you, but there it is, and it's position and speed can be described relative to a frame in which you are still.

Having said that, our Galaxy as a whole is not made of parts that are all stationary with respect to each other. e.g. our solar system is moving up and down relative to the Galactic plane.

Essentially, the answer to your question is "yes" but it's not "yes" due to what I suspect you think yes means.

 

[Ross B]

so to be clear if I point the laser at the image of the Earth that is formed on my retina I will miss it. If I lead the image of the Earth that is formed on my retina by 16 minutes I should get a reflection ...yeh?

In reference to this comment above. If I point the laser at the image on my retina (the laser misses and there will be no reflection). If I note that angle and then gradually rotate the angle of the laser until I get a reflection. If I measure that angle, light is C ...can I measure the velocity that the Earth is moving around the sun?

 

[PeterDonis]

if I point the laser at the image of the Earth that is formed on my retina I will miss it

Yes.

If I lead the image of the Earth that is formed on my retina by 16 minutes I should get a reflection ...yeh?

Yes.

If I note that angle and then gradually rotate the angle of the laser until I get a reflection.

It takes 16 minutes for light to make the round trip, so you won't be able to do this unless you rotate the laser very slowly--waiting for 16 minutes each time you move it a little bit, to see if there's a reflection, before moving it again. But during that time the Earth's image that you see is moving too, so you're not really gaining anything.

 

[Ross B]

Yes.
Yes.
It takes 16 minutes for light to make the round trip, so you won't be able to do this unless you rotate the laser very slowly--waiting for 16 minutes each time you move it a little bit, to see if there's a reflection, before moving it again. But during that time the Earth's image that you see is moving too, so you're not really gaining anything.

I could just point it at the retina image each time, note the angle, rotate it (with a slight increment) note the angle and repeat ...that would work I think...until eventually I got a reflection ...it would be weird tho cos the reflection would be from, what would appear to be, empty space ...does that mean the laws of physics are not working in this frame as empty space is reflecting a laser ? and space that appears full (the retina image) is not reflecting a laser?

 

[Ibix]

it would be weird tho cos the reflection would be from, what would appear to be, empty space

What? No. Why would you think that?

You lead the Earth's apparent position by sixteen minutes when you send out your laser pulse. The pulse return time is sixteen minutes. Where will the Earth appear to be when the pulse returns?

 

[Ross B]

What? No. Why would you think that?

You lead the Earth's apparent position by sixteen minutes when you send out your laser pulse. The pulse return time is sixteen minutes. Where will the Earth appear to be when the pulse returns?

I was told the following list of facts

1 if I fired the laser at where the Earth appeared to be, according to my retina, the laser would miss, and there would be no reflection, as the laser would miss. So I am firing the laser at what appears to be a solid object and I get no reflection.

bizarre measurement 1

2 if I fired the laser leading where the Earth appeared to be, according to my retina, by about 16 minutes (ie empty space) then the laser would reflect off where the Earth actually was so I would detect a reflection. From empty space!

bizarre measurement 2

 

[Ibix]

1 if I fired the laser at where the Earth appeared to be, according to my retina, the laser would miss, and there would be no reflection, as the laser would miss...bizarre measurement 1

If you don't lead a moving target, you miss it. This is not bizarre.

2 if I fired the laser leading where the Earth appeared to be, according to my retina, by about 16 minutes (ie empty space) then the laser would reflect of the Earth so I would detect a reflection. From empty space!

Why do you think the reflection would come from empty space? If you aim at the point where Earth will appear to be in sixteen minutes, where will Earth appear to be when the reflection comes back sixteen minutes later?

 

[SlowThinker]

bizarre measurement

It may be better to imagine a short laser flash rather than a beam, at least for now.
If you flash the laser, you won't see it reflected back for 16 minutes. You seem to not notice this.
By that time, the image of Earth will be at the place where you see the reflected flash.

If you continuously point 16 minutes ahead of the Earth you see, you will see the bright spot always going with Earth, but the light you see was created 16 minutes ago.

 

[Ross B]

lets say I now put the Earth and sun in a rocket. The Rocket is stationary wrt to me.The Rocket will travel along the x axis. The Rocket is 150 million km wide. I glue the Earth to the wall parallel to the x-axis (wall 1). I glue the sun on the other wall at a spot that is perpendicular to the earth. This wall is also parrallell to the x-axis (wall2). The distance from Earth to sun is 150 million Km

I now sit on the sun and fire the rocket which accelerates until it establishes a steady velocity of W. I do not know what W is

I start incrementally firing my laser , note the angles, repeat, until I get a reflection.

I know Earth sun distance and C and the angles - from that can I calculate W?

 

[Ibix]

No.

You seem to be bouncing all over the place. This thread isn't remotely about centre of mass, and it isn't even about the Earth in orbit any more, which at least was the topic of your other thread. Is there something you are trying to understand?

 

[Ross B]

No.

You seem to be bouncing all over the place. This thread isn't remotely about centre of mass, and it isn't even about the Earth in orbit any more, which at least was the topic of your other thread. Is there something you are trying to understand?

why not?

I have three knowns and 1 unknown?

 

[Ibix]

You will learn more if you read the responses carefully.

No.

My point is that this:

yes given the above can I calc w?

is pretty much unrelated to what you were asking before. Why are you asking these questions? What is it that links these questions together? If there isn't anything you'd be better starting separate threads, since then people won't be trying to second guess what the link is. If there is a theme, it would help if you'd say what it is.

 

[Ross B]

well hang on before I did ask if I knew the lead angle, Earth sun distance and C could I determine the rotational velocity of the Earth ...u must have missed that post

Instead of second guessing why don't you just answer the post ...why are u second guessing ? Isnt that just a waste of energy

there is a theme the common factors are the speed of light, the earth, the sun, the Earth sun distance, leading an lagging images, a laser, and laser angles

 

[Ibix]

I didn't miss it. You're asking two completely different questions. Yes, you can measure the speed of the Earth relative to the Sun. You cannot measure its speed relative to some arbitrarily chosen fixed point (which is what you now want to do), not without involving that fixed point somehow.

In short, you cannot ever measure the speed of something. You can only measure its speed relative to your measurement apparatus. If you'd left your apparatus behind when you fired your rocket, then you could measure the speed of the rocket. Or if you'd left a marker behind you could measure the speed of the marker from the rocket. But you cannot do it from inside the rocket because "speed" without the "relative to a specified thing" is meaningless.

 

[Ross B]

I didn't miss it. You're asking two completely different questions. Yes, you can measure the speed of the Earth relative to the Sun. You cannot measure its speed relative to some arbitrarily chosen fixed point (which is what you now want to do), not without involving that fixed point somehow.

In short, you cannot ever measure the speed of something. You can only measure its speed relative to your measurement apparatus. If you'd left your apparatus behind when you fired your rocket, then you could measure the speed of the rocket. Or if you'd left a marker behind you could measure the speed of the marker from the rocket. But you cannot do it from inside the rocket because "speed" without the "relative to a specified thing" is meaningless.

Im not sure what yr answer is , is that a yes I can measure W or no I can't measure W ?

 

[Ibix]

Im not sure what yr answer is , is that a yes I can measure W or no I can't measure W ?

I've answered this question twice already. Will you read my answer if I write it out a third time?

 

[Ross B]

I've answered this question twice already. Will you read my answer if I write it out a third time?

OK so that is a NO

so in the space rocket scenario - the earth, sun, laser and me are all stationary wrt each other. The rocket is speeding along at W along the x axis. Will the image of the Earth lag where it actually is?

 

[Ibix]

What do you think?

 

[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