High School Does everything in space have or inherit inertia?

[DeckSmeck]

Including light?

 

[phinds]

Does everything in space have or inherit inertia? Including light?

Why do you make any distinction between things in space vs things not in space?

What is your understanding of the definition of inertia?

 

[DeckSmeck]

I understand that everything is in space. I might mean momentum. I can translate my understanding like this. If an object is moving very quickly and a ray of light is emitted at what appears to be perpendicular to the path of the object, does the ray of light stay straight or does it bend over distance. Is the direction of the light particles/waves moving diagonally within the beam with respect to having the same inertia/momentum as the source therby keeping the ray straight?

 

[phinds]

I understand that everything is in space. I might mean momentum. I can translate my understanding like this. If an object is moving very quickly and a ray of light is emitted at what appears to be perpendicular to the path of the object, does the ray of light stay straight or does it bend over distance. Is the direction of the light particles/waves moving diagonally within the beam with respect to having the same inertia/momentum as the source therby keeping the ray straight?

A ray of light moves straight in the direction in which it was emitted regardless of the motion of the emitter (unless its path gets "curved" along a space-time geodesic by a gravity well, but that has nothing to do with the emitter).

 

[phinds]

I understand that everything is in space.

Do you feel that you are in space right now? I don't think *I* am in space ("space" meaning the void away from solid bodies)

 

[PeroK]

I understand that everything is in space. I might mean momentum. I can translate my understanding like this. If an object is moving very quickly and a ray of light is emitted at what appears to be perpendicular to the path of the object, does the ray of light stay straight or does it bend over distance. Is the direction of the light particles/waves moving diagonally within the beam with respect to having the same inertia/momentum as the source therby keeping the ray straight?

I think you are confusing yourself with different terminology and concepts. In particular, when you try to imagine a scenario or experiment in more than one reference frame.

To look at a problem, you first need to specify a reference frame. A lot of the time this is so obvious that you don't know you are doing it: analysing a tennis match (from the reference frame of the court); traffic on a highway (from the reference frame of the highway) etc.

To analyse the scenario you want to describe, you first need to specify a reference frame. This could be: the reference frame of a spaceship. Or, it could be a reference frame in which the spaceship is moving very fast.

All of the conceptual difficulties with Special Relativity, for example, come from how measurements in one reference frame compare to measurements in another. So, as soon as you start talking about light and "fast moving" objects, you must be clear and precise about who is doing the measuring - or, in other words, clearly specifiy your reference frame.

 

[DeckSmeck]

Do you feel that you are in space right now? I don't think *I* am in space ("space" meaning the void away from solid bodies)

Not really. I do have to adjust my intuition to account for being on an object in space if I am going to understand astronomy though. I am not trying to do precise calculations or measurements. I am only trying to understand how things move.

 

[DeckSmeck]

I think you are confusing yourself with different terminology and concepts. In particular, when you try to imagine a scenario or experiment in more than one reference frame.

To look at a problem, you first need to specify a reference frame. A lot of the time this is so obvious that you don't know you are doing it: analysing a tennis match (from the reference frame of the court); traffic on a highway (from the reference frame of the highway) etc.

To analyse the scenario you want to describe, you first need to specify a reference frame. This could be: the reference frame of a spaceship. Or, it could be a reference frame in which the spaceship is moving very fast.

All of the conceptual difficulties with Special Relativity, for example, come from how measurements in one reference frame compare to measurements in another. So, as soon as you start talking about light and "fast moving" objects, you must be clear and precise about who is doing the measuring - or, in other words, clearly specifiy your reference frame.

Ok. I need to think about it. My first frame of reference is from the origin of the beam which is moving, supposing I could know if it was going straight from my reference point.

 

[PeroK]

Ok. I need to think about it. My first frame of reference is from the origin of the beam which is moving, supposing I could know if it was going straight from my reference point.

Light travels in a straight line (unless it's affected by gravity). It doesn't move in a curve regardless of the motion of its source.

 

[DeckSmeck]

Light travels in a straight line (unless it's affected by gravity). It doesn't move in a curve regardless of the motion of its source.

Ok. But does the whole beam move sideways then or are we talking about the particles.

Here is what it seems like it should be to me.

Given that there are no additional forces other than direction and speed for simplicity:

Light particles have the same momentum as the source when emitted. So if a beam is emitted perpendicular to the source's direction of travel, the light particles travel away perpendicular to the source and in the direction that the source is moving. That keeps the beam straight and makes sense. But is it true?

 

[PeroK]

Ok. But does the whole beam move sideways then or are we talking about the particles.

Here is what it seems like it should be to me.

Given that there are no additional forces other than direction and speed for simplicity:

Light particles have the same momentum as the source when emitted. So if a beam is emitted perpendicular to the source's direction of travel, the light particles travel away perpendicular to the source and in the direction that the source is moving. That keeps the beam straight and makes sense. But is it true?

That's sort of complicating a simple matter. It's really all about geometry.

Let's assume that an object is moving along my x-axis. A spaceship say. That spaceshift could describe any direction as its x-axis, but let's say it chooses the same x-axis as me.. If it emits a beam of light (or anything) in the y-direction (according to me), then that won't be along its y-axis. If we assume the ship is moving to the right, then in the ship's reference frame the beam is directed backwards at an angle.

Likewise, if the beam is fired in the y-direction according to the ship, then the beam will be moving forwards at an angle in my reference frame.

It's just geometry, really.

 

[DeckSmeck]

Rather some vector between perpendicular and the direction the source is traveling. I think that is how it is described. Vector? Sorry, I am such a layman.

 

[PeroK]

Rather some vector between perpendicular and the direction the source is traveling. I think that is how it is described. Vector? Sorry, I am such a layman.

You could try this:

https://www.khanacademy.org/science...ocity-time/v/introduction-to-reference-frames

 

[DeckSmeck]

That's sort of complicating a simple matter. It's really all about geometry.

Let's assume that an object is moving along my x-axis. A spaceship say. That spaceshift could describe any direction as its x-axis, but let's say it chooses the same x-axis as me.. If it emits a beam of light (or anything) in the y-direction (according to me), then that won't be along its y-axis. If we assume the ship is moving to the right, then in the ship's reference frame the beam is directed backwards at an angle.

Likewise, if the beam is fired in the y-direction according to the ship, then the beam will be moving forwards at an angle in my reference frame.

It's just geometry, really.

I have to think about this. It seems more complicated because now we have two ships. Why can't it be one ship traveling sideways along any axis with the headlights on. I just want to know if the light particles have the ship's momentum acting on it.

 

[PeroK]

I have to think about this. It seems more complicated because now we have two ships. Why can't it be one ship traveling sideways along any axis with the headlights on. I just want to know if the light particles have the ship's momentum acting on it.

With light there is an important difference between speed and momentum. The light emitted by any source has the same speed, independent of the motion of the source.

This is the fundamental difference between Newtonian physics and Relativity.

If we stick to one dimension, then light emitted to the front of the source will have speed $c$; and light emitted to the rear of the source will also have speed $c$.

Where light differs from other things is that that is the same in any reference frame. The ship itself, for example, will measure light traveling at $c$ in both the forward and backward directions. And, in a reference frame where the ship is moving, the speed will still be $c$ in both directions.

This is very difference from classical physics, where the speed of the light would be different in different reference frames. For example if the light was traveling at speed $c$ in the ship's frame, then it would be traveling at $c + v$ and $c-v$ in a reference frame where the ship was moving at speed $v$.

This is the fundamental difference between Newtonian physics and Relativity.

The momentum of light depends on its frequency and wavelength; not on its speed, which is universally constant.

 

[DeckSmeck]

With light there is an important difference between speed and momentum. The light emitted by any source has the same speed, independent of the motion of the source.

This is the fundamental difference between Newtonian physics and Relativity.

If we stick to one dimension, then light emitted to the front of the source will have speed $c$; and light emitted to the rear of the source will also have speed $c$.

Where light differs from other things is that that is the same in any reference frame. The ship itself, for example, will measure light traveling at $c$ in both the forward and backward directions. And, in a reference frame where the ship is moving, the speed will still be $c$ in both directions.

This is very difference from classical physics, where the speed of the light would be different in different reference frames. For example if the light was traveling at speed $c$ in the ship's frame, then it would be traveling at $c + v$ and $c-v$ in a reference frame where the ship was moving at speed $v$.

This is the fundamental difference between Newtonian physics and Relativity.

The momentum of light depends on its frequency and wavelength; not on its speed, which is universally constant.

Ok. What about sideways though? If a laser travels away at $c$ from the ship long enough to make a one mile long beam, is the beam moving with and away from the ship as it travels? I might lack the ability to ask this question right, I think.

 

[DeckSmeck]

I don't know how a photon is oriented with respect to forward and rear, or if it has an orientation, but if it did, is the photon traveling something like

Forward speed + inherited speed of source = $c$?

 

[phinds]

... is the photon traveling something like

Forward speed + inherited speed of source = $c$?

Absolutely not. The speed of light is c REGARDLESS of the speed of the emitter. The speed of the emitter is completely irrelevant.

 

[DeckSmeck]

Absolutely not. The speed of light is c REGARDLESS of the speed of the emitter. The speed of the emitter is completely irrelevant.

Let $c$ be constant.

Let $p$ be a unique proton.

Let v1 be velocity of $p$ emitted in the direction of a laser.

Let v2 be velocity of $p$ in the direction of the source.

$p$(v1) + $p$(v2) = $c$

 

[phinds]

Let $c$ be constant.

Let $p$ be a unique proton.

Let v1 be velocity of $p$ emitted in the direction of a laser.

Let v2 be velocity of $p$ in the direction of the source.

$p$(v1) + $p$(v2) = $c$

I ASSUME that you mean photon, not proton. If so, then if I understand your scenario correctly, you have v1 = c and v2 = c and your equation is nonsense (or it should be v1 + v2 = 0)

EDIT: depending on how this gets interpreted, v2 = -c, not v2 = c

In any case, you REALLY need to stop trying to use the speed of a photon as anything other than c.

 

[DeckSmeck]

Let $c$ be constant.

Let $p$ be a unique proton.

Let v1 be velocity of $p$ emitted in the direction of a laser.

Let v2 be velocity of $p$ in the direction of the source.

$p$(v1) + $p$(v2) = $c$

Or is it

v1(p) + v2(p) = c
?

 

[DeckSmeck]

I ASSUME that you mean photon, not proton. If so, then if I understand your scenario correctly, you have v1 = c and v2 = c and your equation is nonsense (or it should be v1 + v2 = 0)

EDIT: depending on how this gets interpreted, v2 = -c, not v2 = c

In any case, you REALLY need to stop trying to use the speed of a photon as anything other than c.

Yes. Photon.

Yes v1 + v2. Does the p go in the parenthesis. Trying to say v1 of p + v2 of p = $c$.

 

[phinds]

Yes. Photon.

Yes v1 + v2. Does the p go in the parenthesis. Trying to say v1 of p + v2 of p = $c$.

This makes no sense. You have said v1 is the speed of a photon. That's c. So if your equation is correct, the v1 + v2 = c is the same as saying that c + v2 = c, or v2 = 0, which I assume is not what you intend.

AGAIN, you have to stop thinking that the speed of a photon is anything but c. If you keep just saying the same wrong thing over and over, that does not make it right, but it will likely get this thread shut down since it's pointless to discuss it further if you persist.

 

[DeckSmeck]

I ASSUME that you mean photon, not proton. If so, then if I understand your scenario correctly, you have v1 = c and v2 = c and your equation is nonsense (or it should be v1 + v2 = 0)

EDIT: depending on how this gets interpreted, v2 = -c, not v2 = c

In any case, you REALLY need to stop trying to use the speed of a photon as anything other than c.

How then do I describe the adjustment to the direction of the photon by the speed of the source? Even if it w

This makes no sense. You have said v1 is the speed of a photon. That's c. So if your equation is correct, the v1 + v2 = c is the same as saying that c + v2 = c, or v2 = 0, which I assume is not what you intend.

AGAIN, you have to stop thinking that the speed of a photon is anything but c. If you keep just saying the same wrong thing over and over, that does not make it right, but it will likely get this thread shut down since it's pointless to discuss it further if you persist.

Relax. I just have to learn more math to talk to you. I can't speak your language and you can't understand my question.

 

[phinds]

How then do I describe the adjustment to the direction of the photon by the speed of the source?

As you have been told over and over the speed of the source is irrelevant. If I am sitting on Earth and you go past me in a spaceship traveling at .5c relative to me and you turn on a flashlight, the beam of light moves away from you at c. I also experience it as traveling at c. If your speed were .9c, then the beam would move away from you at c and I also agree that it is moving at c. If your speed were .1c it would all be the same. The speed of the source does not matter.

 

[DeckSmeck]

As you have been told over and over the speed of the source is irrelevant. If I am sitting on Earth and you go past me in a spaceship traveling at .5c relative to me and you turn on a flashlight, the beam of light moves away from you at c. I also experience it as traveling at c. If your speed were .9c, then the beam would move away from you at c and I also agree that it is moving at c. If your speed were .1c it would all be the same. The speed of the source does not matter.

Let's talk about a rock instead. I shoot a rock off my spaceship directly to the right. How do I calculate the resulting vector of the rock if the final speed is 100. What does the equation look like.

 

[phinds]

Let's talk about a rock instead. I shoot a rock off my spaceship directly to the right. How do I calculate the resulting vector of the rock if the final speed is 100. What does the equation look like.

Rocks do not act the way light acts, so that exercise is pointless to bring up in a discussion of light. Perhaps that's your problem You think that a rock and a photon act the same way. They don't.

 

[DeckSmeck]

Rocks do not act the way light acts, so that exercise is pointless to bring up in a discussion of light. Perhaps that's your problem You think that a rock and a photon act the same way. They don't.

Well help me with the rock first then. What would the equation look like on a 2d plane?

 

[phinds]

Well help me with the rock first then. What would the equation look like on a 2d plane?

Rocks do not act the way light acts, so that exercise is pointless to bring up in a discussion of light.

 

[DeckSmeck]

Rocks do not act the way light acts, so that exercise is pointless to bring up in a discussion of light.

I can fix that.