Do the Large Hadron Collider take earth's rotational speed?

[howabout1337]

so according to wiki: protons have a Lorentz factor of about 7,500 and move at about 0.999999991 c, or about 3 metres per second slower than the speed of light (c)

If we consider speed of Earth's rotation or speed at which two galaxies approach each other (collision or otherwise)?

 

[wotanub]

Could you rephrase? I don't understand what you are asking.

 

[DrChinese]

If we consider speed of Earth's rotation or speed at which two galaxies approach each other (collision or otherwise)?

Welcome to PhysicsForums, howabout1337!

We really have no way to determine either the Earth's absolute velocity or absolute direction of travel, net of all of the gravitational effects of celestial bodies. We could be traveling a few percent of the speed of light, for all we know. So you might deduce from those statements that such does not matter.

In fact, according to relativity, the laws of physics are independent of any absolute reference frame. So the Earth's movement, per se, is not a factor in calculating the results. (As far as anyone knows, there is no absolute reference frame anyway.)

So when we say it is moving at 99.9999% of c, we mean relative to the observer.

 

[phinds]

so according to wiki: protons have a Lorentz factor of about 7,500 and move at about 0.999999991 c, or about 3 metres per second slower than the speed of light (c)

If we consider speed of Earth's rotation or speed at which two galaxies approach each other (collision or otherwise)?

As DrChinese implied, all motion is relative. You, as you are reading this, are both absolutely motionless from a frame of reference of the chair you are sitting in AND moving at 0.999999991 c from a frame of reference of that particle you mentioned.

Also, you may not be aware, speed "addition" in relativity is not simple arithmetic, it is algebraic and the sum NEVER exceeds c.

 

[howabout1337]

I don't have a background in physics. So excuse my ignorance.

So far that I know about velocity of a particle is it's a measure of distance per unit time. If a particle is at a position at one time, and another position at another time, we take the difference of the position and the time, thus we have the speed.

Now, the particle on the LHC is moving at a certain speed, assuming that at one point, it is moving parallel to that of the Earth's rotation. Wouldn't you agree that the position has changed, not only by the velocity at which it travels, but also at the velocity of the Earth's rotation? As in, the Earth has carried it a certain distance, as well as it's own velocity.

 

[phinds]

I don't have a background in physics. So excuse my ignorance.

So far that I know about velocity of a particle is it's a measure of distance per unit time. If a particle is at a position at one time, and another position at another time, we take the difference of the position and the time, thus we have the speed.

Now, the particle on the LHC is moving at a certain speed, assuming that at one point, it is moving parallel to that of the Earth's rotation. Wouldn't you agree that the position has changed, not only by the velocity at which it travels, but also at the velocity of the Earth's rotation? As in, the Earth has carried it a certain distance, as well as it's own velocity.

Yes, the movement of the Earth does change the resultant velocity vector but

(1) the amount the Earth moves while a particle moves in the LHC is utterly trivial
(2) relativistic velocities do NOT add as "v1+v2" (Google relativistic velocity addition if you want to know more)

 

[russ_watters]

Wouldn't you agree that the position has changed, not only by the velocity at which it travels, but also at the velocity of the Earth's rotation?

Not if the LHC is the reference frame.

The problem here is lack of definition of reference frame: the LHC is the reference frame so the particle speed is measured relative to it.

 

[phinds]

The problem here is lack of definition of reference frame: the LHC is the reference frame so the particle speed is measured relative to it.

Good point.

I was unthinkingly creating a reference frame which was neither the Earth nor the LHC to talk about relativistic velocity addition.

 

[krash661]

We really have no way to determine either the Earth's absolute velocity or absolute direction of travel, net of all of the gravitational effects of celestial bodies. We could be traveling a few percent of the speed of light, for all we know. So you might deduce from those statements that such does not matter.

i once came across this,

 

[phinds]

i once came across this,

An amusing presentation, although flawed.

1) He keeps adding vectors as thought they were scalers, but I understand why he doesn't want to add that complication just because it would make his presentation factual instead of wrong. he is not orders of magnitude wrong.

2) He mentions a "fixed point in the universe" which is also seriously wrong, but it is somewhat understandable what he means and he does point out that this has a difficulty, though the doesn't say what it is.

3) He keeps adding "directions of motion" as though they don't combine and ends up having movement be in "7 different directions" which is going to be a bit tough in a 3-spacial-dimension universe, although again it's pretty understandable what he means (which is something like "COMPLICATED 3D motion which I choose to call 7-directional motion")

 

[krash661]

ok.
i'm interested in this at times,
so do you have any resources of such ?

edit-
or at lease have your own calculation ?

 

[phinds]

ok.
i'm interested in this at times,
so do you have any resources of such ?

edit-
or at lease have your own calculation ?

I can only suggest that you study basic cosmology.

 

[krash661]

can you show the correct calculation from your prospective?
at lease ?

edit-
i also came across this,
" Scalar (mathematics), a quantity that can multiply vectors in the context of vector spaces
Scalar (physics), a quantity which is independent of specific classes of coordinate systems
Vectors are quantities that are fully described by both a magnitude and a direction."

and also, clarify which type of vector you are referring to

 

[krash661]

Hmm.

 

[mfb]

Vectors are elements of vector spaces. There are no different types of vectors, just different types of vector spaces. I assume Phinds means a Minkowski space, as this is the framework of special relativity.
Somehow I doubt that this helped, but it is a precise answer to your last question.
All calculations can be done in this space and with Lorentz transformations.

 

[krash661]

There are no different types of vectors,

ok, so all these vectors below are all " vector spaces "?

http://en.wikipedia.org/wiki/Vector_(mathematics_and_physics )

 

[phinds]

can you show the correct calculation from your prospective?
at lease ?

No, the calculations would be incredibly complicated and a HUGE waste of time. You should focus on the concepts.

 

[mfb]

ok, so all these vectors below are all " vector spaces "?

http://en.wikipedia.org/wiki/Vector_(mathematics_and_physics)

They are all elements of vector spaces, sure. Even the wikipedia page you linked states that:

Many special instances of the general definition of vector as an element of a vector space are listed below.

(I fixed the link in the quote)

 

[Dale]

Now, the particle on the LHC is moving at a certain speed, assuming that at one point, it is moving parallel to that of the Earth's rotation. Wouldn't you agree that the position has changed, not only by the velocity at which it travels, but also at the velocity of the Earth's rotation? As in, the Earth has carried it a certain distance, as well as it's own velocity.

As phinds mentioned, velocity doesn't add that way. If you work out the math you find that a Lorentz factor of 7500 is c-2.664822 m/s. That is the speed of the particles in the frame where CERN is at rest. In the frame where the center of the Earth is at rest CERN is moving at 465 m/s (or it would if CERN were located at the equator). In that frame the forward moving particles are moving at c-2.66481 m/s and the backward moving particles are moving at -c+2.66483 m/s.

 

[howabout1337]

As phinds mentioned, velocity doesn't add that way. If you work out the math you find that a Lorentz factor of 7500 is c-2.664822 m/s. That is the speed of the particles in the frame where CERN is at rest. In the frame where the center of the Earth is at rest CERN is moving at 465 m/s (or it would if CERN were located at the equator). In that frame the forward moving particles are moving at c-2.66481 m/s and the backward moving particles are moving at -c+2.66483 m/s.

so what is the reference to the speed of the particle that it is given that speed? If the reference of speed of light is that of a moving object/particle, then we can assume that speed of light can be different? We just have to know how fast everything "stationary" is moving.

On that note, if I am riding a bicycle inside a ship that is as big is our solar system, from one end to another, and the ship is moving parallel to me, both of us traveling at Lorentz factor of 7500 , how would that play into the equation? I have always thought speed is a position in one space, over a time, another space. That would be my speed. I have been wrong my whole life.

 

[phinds]

I have been wrong my whole life.

Yes but since you have probably thought about only non-relativistic speeds your error has been negligible.

 

[Dale]

so what is the reference to the speed of the particle that it is given that speed?

Sorry, I cannot understand what you are asking here.

On that note, if I am riding a bicycle inside a ship that is as big is our solar system, from one end to another, and the ship is moving parallel to me, both of us traveling at Lorentz factor of 7500 , how would that play into the equation?

The equation is quite straightforward:
$$w=\frac{u+v}{1+uv/c^2}$$

A Lorentz factor of 7500 is c - 2.66482 m/s. So plug that into the above for u and v and you get that w is c - 11.84 nm/s.

 

[howabout1337]

Sorry, I cannot understand what you are asking here.

I mean, what is the reference of the moving object in the first place? Since for all we know, everything is moving. We always reference things to speed of light, but we have to refer speed of light to something else too don't we? If I am moving in one direction, the speed of of light is moving with me in the same direction, would that mean speed can be faster? Since all I am doing is observe, and I don't really know my "true" speed.

The equation is quite straightforward:
$$w=\frac{u+v}{1+uv/c^2}$$

A Lorentz factor of 7500 is c - 2.66482 m/s. So plug that into the above for u and v and you get that w is c - 11.84 nm/s.

What will the observer see when they look at the ship and when they look at me? How fast will it take me to travel to the end of this ship? So unintuitive, for me at least.

 

[phinds]

I mean, what is the reference of the moving object in the first place? Since for all we know, everything is moving. We always reference things to speed of light,

No, we most emphatically do NOT.

... but we have to refer speed of light to something else too don't we?

No, the speed of light is constant in all inertial frames.

If I am moving in one direction, the speed of of light is moving with me in the same direction, would that mean speed can be faster? Since all I am doing is observe, and I don't really know my "true" speed.

Regardless of your speed, any light beams reaching you will be moving at c.

EDIT: and by the way, there IS no such thing as a "true" speed. All speed is relative to something (except for light which travels at c relative to everything)

 

[Nugatory]

The equation is quite straightforward:
$$w=\frac{u+v}{1+uv/c^2}$$

A Lorentz factor of 7500 is c - 2.66482 m/s. So plug that into the above for u and v and you get that w is c - 11.84 nm/s.

What will the observer see when they look at the ship and when they look at me? How fast will it take me to travel to the end of this ship? So unintuitive, for me at least.

You are riding your bicycle at speed u=c-2.66482 m/s relative to the spaceship. The time it takes you to travel from one end of the ship to the other is just the distance divided by your speed, no different than if you're riding your bicycle at 20 km/h between two towns 40 kilometers apart. The movement of the ship is of no more interest to you than the Earth's movement about the sun when you're riding the bicycle on earth.

Some external observer sees that ship zooming by at speed v=c-2.66482 m/s (although we could just as easily say that the ship is at rest and the observer is moving at that speed in the opposite direction).

How fast are you in the bicycle moving relative to the external observer? The answer is not w=u+v as you'd expect from your experience with speeds that are small compared with the speed of light, it's given by the formula that DaleSpam posted: w = c - 11.84 nm/s.

Next steps: Google around for "time dilation", "length contraction", "relativity of simultaneity", and "Lorentz transforms". They all play together to make this counterintuitive picture internally consistent and in agreement with experiment.

 

[Dale]

I mean, what is the reference of the moving object in the first place?

The reference frame of any object is, by definition, the frame where its velocity is 0.

We always reference things to speed of light

On the contrary, we never do that. We always reference things to some specified object with mass.

If I am moving in one direction, the speed of of light is moving with me in the same direction, would that mean speed can be faster?

Use the formula I provided above, substitute in u=c and simplify. What do you get? How does it depend on v?

 

[howabout1337]

The reference frame of any object is, by definition, the frame where its velocity is 0.

Wow thanks. You guys are explaining a lot things that I never even thought of before.

Ok, this is where I get confused regarding this:

We use another object as a frame of reference, and we assume that the frame has no velocity. If that object that we are referencing are actually moving, then use another object that has zero velocity to reference it. How do we know that something "stationary" is not actually moving on a bigger scare. For example, on earth, my mouse is not moving, but on a larger scale, it is (speed of Earth's rotation and speed of Earth going around the sun), relatively. If I take the speed of my "stationary" mouse as a reference to speed of light, can we assume that there may be an error somewhere? We know that speed is maximum for massless particles, that is known for sure. Why not use those as reference and work our way back?

 

[phinds]

Wow thanks. You guys are explaining a lot things that I never even thought of before.

Ok, this is where I get confused regarding this:

We use another object as a frame of reference, and we assume that the frame has no velocity. If that object that we are referencing are actually moving, then use another object that has zero velocity to reference it. How do we know that something "stationary" is not actually moving on a bigger scare. For example, on earth, my mouse is not moving, but on a larger scale, it is (speed of Earth's rotation and speed of Earth going around the sun), relatively. If I take the speed of my "stationary" mouse as a reference to speed of light, can we assume that there may be an error somewhere? We know that speed is maximum for massless particles, that is known for sure. Why not use those as reference and work our way back?

You misunderstand completely. There IS no absolute motion. ALL motion has to be relative to something. You and your chair may be motionless relative to each other but relative to an accelerated particle, you are both moving at almost the speed of light.

 

[Dale]

If that object that we are referencing are actually moving

There is no such thing as "actually moving". Movement is relative, so it is not moving relative to itself, or it may be moving relative to something else. The phrase "actually moving" has no meaning.

 

[russ_watters]

How do we know that something "stationary" is not actually moving on a bigger scale. For example, on earth, my mouse is not moving, but on a larger scale, it is (speed of Earth's rotation and speed of Earth going around the sun), relatively...

We don't know, but more importantly, we don't care. It doesn't matter because the laws of physics work the same for all such reference frames, so those "other motions" don't matter. They are undetectable internally and have no effect on how experiments work in your lab's frame. That is The Principle of Relativity, the first postulate of SR.

 

[Nugatory]

How do we know that something "stationary" is not actually moving on a bigger scale?

We don't.
It is meaningless to to talk about something "actually moving" or "actually at rest" - anything we say about motion is always relative to something else.

In everyday life and speech we tend to be a bit careless and say things like "The car is moving at 100 kilometers per hour", but if you think about it for a moment you'll see that this means "The car is moving at 100 kilometers per hour relative to the surface of the earth". Dig deep enough into any statement about speeds, and you'll find that there's never something that you can call the "real speed" or "actual speed".

 

[howabout1337]

OK, I think I am getting there.

Say two cars are traveling, one faster than the other. We'll call it car A and car B. Car A is faster than car B, not that much, but faster. When the person in car B looks at car A, he's going to say "Oh, he's traveling at his speed minus my speed". Me, the person who is "stationary", measures their speed to my speed, and say "hey they are traveling at their speed minus my speed". Obviously the person in car B looks at the person in car A's speed differently than I do. The person in car B calls me and say "hey that guy in car A is traveling at this speed", and I argue with him saying "he's not that speed, he's THIS speed". Obviously I didn't measure the speed between myself and the guy in car B.

Now, someone behind me shines a laser pointer right in between all of us.

Now, the guy is car A cannot see the guy in car B, or me, or the guy behind me, and he performs experiment on speed of light from the laser pointer, and he's going to say "hey this is speed of light". The guy in car B is going to say, "look at that dumb guy in car A, he doesn't know he's moving, let me measure the REAL speed of light", and he measures the speed of light relative to himself, "he will say, look this is the REAL speed of light". On the other hand, I am thinking to myself, "both of these guys don't know what they are talking about, They both think they are stationary. The B guy think he's stationary and making fun of that guy in car A"

How do I know that the guy behind me is not saying "haha, so all of these guys don't know that they are on a moving train?"

 

[HallsofIvy]

OK, I think I am getting there.

Say two cars are traveling, one faster than the other.

Relative to some point on, say, the ground?

We'll call it car A and car B. Car A is faster than car B, not that much, but faster. When the person in car B looks at car A, he's going to say "Oh, he's traveling at his speed minus my speed".

Relative to car A, yes.

Me, the person who is "stationary", measures their speed to my speed, and say "hey they are traveling at their speed minus my speed".

Having said you were "stationary", what do you mean by "my speed"?

Obviously the person in car B looks at the person in car A's speed differently than I do. The person in car B calls me and say "hey that guy in car A is traveling at this speed", and I argue with him saying "he's not that speed, he's THIS speed". Obviously I didn't measure the speed between myself and the guy in car B.

I don't know what you mean by "this speed" or "that speed". Let's put specific numbers to this:
You, who are "stationary", standing on the side of the road have speed 0 relative to the road. Car B has speed 80 kph relative to the road and car A has speed 100 kph relative to the road.

Now:

Relative to you, you have speed 0, car A has speed 100 kph and car B has speed 80 kph.
(Exactly the same as "relative to the road", of course.)

Relative to the person in car A, you (and the road) have speed -100 kph, car A has speed 0, and car B has speed -20 kph.

Relative to the person in car B, you (and the road) have speed -80 kph, car A has speed 20 kph and car B has speed 0.

Now, someone behind me shines a laser pointer right in between all of us.

Now, the guy is car A cannot see the guy in car B, or me, or the guy behind me, and he performs experiment on speed of light from the laser pointer, and he's going to say "hey this is speed of light". The guy in car B is going to say, "look at that dumb guy in car A, he doesn't know he's moving, let me measure the REAL speed of light", and he measures the speed of light relative to himself, "he will say, look this is the REAL speed of light". On the other hand, I am thinking to myself, "both of these guys don't know what they are talking about, They both think they are stationary. The B guy think he's stationary and making fun of that guy in car A"

How do I know that the guy behind me is not saying "haha, so all of these guys don't know that they are on a moving train?"

I don't know why any of those guys would think anyone was dumb. They should understand that the speed of light would be measured to be the same at any speed.

Look at it this way: according to classical, Newtonian, Physics, we can add speeds: u+ v.

If I am standing at the side of the road, you approach me on the back of a truck going 60 kph (relative to me) and I throw a ball to you at 70 kph (relative to me) classical physics would say the ball will be going toward you at 60+ 70= 130 kpg relative to you.

Similarly, if I shine a light at you with speed "c", relative to me, classical physics says you would see it coming toward you with speed 60+ c.

Now, relativity says you must use the formula $(u+ v)(1+ uv/c^2)$.
In the first example, that would be $(60+ 70)/(1+ (60)(70)/c^2)= 130/(1+ 4200/300000^2)$= 129.999999 kph. The difference is, of course, so small it isn't noticed.

In the second example we would have $(60+ c)/(1+ 60c/c^2)= (60+ c)/(1+ 60/c)$.
Multiplying both numerator and denonator by c we gave [itex]c(60+ c)/[c(1+ 60/c)]= c(60+ c)/(60+ c)= c[itex]. That is, the speed of light is the same relative to both of us.

 

[Nugatory]

Say two cars are traveling, one faster than the other. We'll call it car A and car B. Car A is faster than car B, not that much, but faster. When the person in car B looks at car A, he's going to say "Oh, he's traveling at his speed minus my speed which is zero".

Almost... I've added a few very important words above. The key here is that as far as person B is concerned, he is the stationary one and I and the ground that I am standing on are moving backwards at some speed.

Now, someone behind me shines a laser pointer right in between all of us.

Now, the guy is car A cannot see the guy in car B, or me, or the guy behind me, and he performs experiment on speed of light from the laser pointer, and he's going to say "hey this is speed of light". The guy in car B is going to say, "look at that dumb guy in car A, he doesn't know he's moving, let me measure the REAL speed of light", and he measures the speed of light relative to himself, "he will say, look this is the REAL speed of light". On the other hand, I am thinking to myself, "both of these guys don't know what they are talking about, They both think they are stationary. The B guy think he's stationary and making fun of that guy in car A"

And the joke is on all of them, because they ALL measure the exact same value for the speed of light.

How do I know that the guy behind me is not saying "haha, so all of these guys don't know that they are on a moving train?"

Exactly... You don't. In fact, there's somebody on a spaceship flying by somewhere saying "haha, so all those guys don't know that they're on a moving planet". The Earth is a lot bigger than a train, but it's the exact same problem otherwise.

 

[howabout1337]

Now, relativity says you must use the formula $(u+ v)/(1+ uv/c^2)$.

What is this formula called? How can I derive it?

Also, I can't wrap my head around each person's position relative to the tip of the light. What is causing them to slow down as they reach the speed of light? Is there some kind of drag like a feather dropped in the air (there is a maximum velocity on it)? Who is the one slowing down relative to each other?

By that fact, can our total change in position be greater than c? As in, can our total change in position be greater than maximum speed c? For example, can my change in position be 4 seconds*c in 2 seconds? No wormhole or shortcuts, just compounding speed (even in relativistic term).