# (split) Relativity on a plane

Right. It can't display two different readings at the same time. But that doesn't mean the force on it isn't different in two different frames.

For another example: a ruler can't display two different readings at the same time. Yet if you take that ruler and measure the length of an object in two different frames, you'll get different answers.

All of that is true, however Tach is claiming something different. He says that there are two different forces on the scale in the same frame. His analysis is given in post #85.

All of that is true, however Tach is claiming something different. He says that there are two different forces on the scale in the same frame. His analysis is given in post #85.

Are you now denying that your analysis in post #85 claims there should be two different forces on the scale in one single reference frame?

Are you now denying that your analysis in post #85 claims there should be two different forces on the scale in one single reference frame?

It does? What in :

Tach said:
In the frame attached to A, the force applied to A as measured by A is :

$$\vec{F_A}=F \vec{u_y}$$

In the frame attached to B, the force applied to B as measured by B is :

$$\vec{F_B}=F \vec{u_y}$$

do you fail to understand?

Do you have an elementary English reading and comprehension problem? Because if you do, I am not a medical doctor, I can't help you.

Neddy Bate said:
Are you now denying that your analysis in post #85 claims there should be two different forces on the scale in one single reference frame?

It does? What in :

In the frame attached to A, the force applied to A as measured by A is :

$$\vec{F_A}=F \vec{u_y}$$

In the frame attached to B, the force applied to B as measured by B is :

$$\vec{F_B}=F \vec{u_y}$$

do you fail to understand?

Do you have an elementary English reading and comprehension problem? Because if you do, I am not a medical doctor, I can't help you.

But those are not in the same frame. Notice how one says, "frame attached to A", and the other says, "frame attached to B"? Let's stick to one frame, and see what your analysis in post #85 says:

In the frame attached to A, the force applied to A as measured by A is :
$$\vec{F_A}=F \vec{u_y}$$

In the frame attached to A, the force applied to B as measured by A is :
$$\vec{F_{AB}}=F/\gamma \vec{u_y}$$

D'oh! You have two different forces acting on the scale in the same frame!! How can you forget that was your claim? You must have a really, really bad memory!!! Maybe it's time for a stroll down Memory Lane:

In your post #128 you explained that there would be two different forces, because, "Why would the forces be the same?"
http://www.sciforums.com/showpost.php?p=2769341&postcount=128
think about the fact that A doesn't move wrt his scale whilst B moves (at speed v)? Why would the forces be the same?

Then in your post #133 you changed up and decided that there could be no forces at all. Your reasoning was, "How does B act on the scale at A? Through telepathic waves?"
http://www.sciforums.com/showpost.php?p=2769368&postcount=133
Scale A is comoving with ball A.
B is moving wrt A and the scale A.
How does B act on the scale at A? Through telepathic waves?

What a laugh you are!!!

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But those are not in the same frame. Notice how one says, "frame attached to A", and the other says, "frame attached to B"?

Of course they are not in the same frame, A and B are in motion wrt each other.

Let's stick to one frame, and see what your analysis in post #85 says

D'oh! You have two different forces acting on the scale in the same frame!!

This is what happens when you refuse to learn relativity, like Motor Daddy.

$$\vec{F_{AB}$$ and $$\vec{F_{BA}$$ are NOT what the balance measures, they are relativistic forces, they represent what A measures about B and what B measures about A because A and B are in motion wrt each other. 50 posts and you are still incapable of learning, just because you refuse to pick up a book. It is much more fun posting BS, like Motor Daddy, eh?

How can you forget that was your claim?

It is not "my" claim, it is what SR teaches you. Unless you are a retard who refuses to crack open a book.

$$\vec{F_{AB}$$ and $$\vec{F_{BA}$$ are NOT what the balance measures, they are relativistic forces, they represent what A measures about B and what B measures about A because A and B are in motion wrt each other.

If $$\vec{F_{AB}$$ and $$\vec{F_{BA}$$ are not what the balance measures, then let's not worry about them for now.

I would like to know the two forces (one upward, one downward) which act on the balance when ball B rolls over the red scale:

Notice that this drawing takes place entirely in the reference frame of ball A.
That is, ball A never moves in this frame. Let's stick to that reference frame.
Now, please tell me, do you know how to calculate the forces on the balance?

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If $$\vec{F_{AB}$$ and $$\vec{F_{BA}$$ are not what the balance measures, then let's not worry about them for now.

I would like to know the two forces (one upward, one downward) which act on the balance when ball B rolls over the red scale:

Notice that this drawing takes place entirely in the reference frame of ball A.
That is, ball A never moves in this frame. Let's stick to that reference frame.
Now, please tell me, do you know how to calculate the forces on the balance?

Of course I do but it is time that you hit the books and you learned how to do it yourself. You can start with something simple, like drawing the vectors representing the forces on the diagram.

Of course I do but it is time that you hit the books and you learned how to do it yourself. You can start with something simple, like drawing the vectors representing the forces on the diagram.

Yes, "of course" you know how to do it. You just don't want to do it right now, because you don't want to show off your skills too much.

Here, I'll save you some time. Since you already said that these forces do not act on the scale:
$$\vec{F_{AB}}=F/\gamma \vec{u_y}$$
$$\vec{F_{BA}}=F/\gamma \vec{u_y}$$

Then that leaves these forces acting on the scale:
$$\vec{F_A}=F \vec{u_y}$$
$$\vec{F_B}=F \vec{u_y}$$

where $$F=G \frac{m^2}{4r^2}$$

So, in other words, you have the force of attraction between the moving balls as identical to the force of attraction when the balls are stationary. Thanks a bunch for all of your help.

Then that leaves these forces acting on the scale:
$$\vec{F_A}=F \vec{u_y}$$
$$\vec{F_B}=F \vec{u_y}$$

where $$F=G \frac{m^2}{4r^2}$$

So, in other words, you have the force of attraction between the moving balls as identical to the force of attraction when the balls are stationary. Thanks a bunch for all of your help.

Nope, you can stop rolling your eyes, it is dead wrong. Draw the force $$F=G \frac{m^2}{4r^2}$$ as a vector and you might figure out your mistake. Or maybe not.

Again, I don't think you can apply the force law $$F=Gm_1m_2/r^2$$ in Relativity. The boosted object has an increased inertial mass as seen from the rest frame, yet if you naively use this as the mass in Newton's law, I believe you get a contradiction when switching frames. Hence you need GR to model a gravitational force.

Again, I don't think you can apply the force law $$F=Gm_1m_2/r^2$$ in Relativity. The boosted object has an increased inertial mass as seen from the rest frame, yet if you naively use this as the mass in Newton's law, I believe you get a contradiction when switching frames. Hence you need GR to model a gravitational force.

Yes, of course but we'll let him do his own nonsense stuff, in the process he might learn about central forces. Besides, he's abandoned any attempts at learning relativity (see above), he's just struggling with the problem as viewed in Newtonian mechanics.

Again, I don't think you can apply the force law $$F=Gm_1m_2/r^2$$ in Relativity. The boosted object has an increased inertial mass as seen from the rest frame, yet if you naively use this as the mass in Newton's law, I believe you get a contradiction when switching frames. Hence you need GR to model a gravitational force.

It may very well require GR to solve this problem. Unlike Tach, I don't pretend to know the answer. I was hoping someone who actually knew what they were doing might contribute something.

Earlier in the thread, I speculated that the force measured by the scale when the balls are relatively moving and aligned vertically might be:

$$F'_{y} = \gamma F_{y} = \gamma G \frac{m_1 m_2}{(2r)^2}$$

Where $$F_{y}$$ is the force between the balls when $$v=0$$:

$$F_{y} = G \frac{m_1 m_2}{(2r)^2}$$

And:
$$2r$$
is the distance between the centers of the balls when they are aligned vertically

And:
$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

Yes, "of course" you know how to do it. You just don't want to do it right now, because you don't want to show off your skills too much.

Here, I'll save you some time. Since you already said that these forces do not act on the scale:
$$\vec{F_{AB}}=F/\gamma \vec{u_y}$$
$$\vec{F_{BA}}=F/\gamma \vec{u_y}$$

Then that leaves these forces acting on the scale:
$$\vec{F_A}=F \vec{u_y}$$
$$\vec{F_B}=F \vec{u_y}$$

where $$F=G \frac{m^2}{4r^2}$$

So, in other words, you have the force of attraction between the moving balls as identical to the force of attraction when the balls are stationary. Thanks a bunch for all of your help.

Nope, you can stop rolling your eyes, it is dead wrong. Draw the force $$F=G \frac{m^2}{4r^2}$$ as a vector and you might figure out your mistake. Or maybe not.

My mistake? Those are your equations.

My mistake? Those are your equations.

The equations don't apply for the scenario you drew in post 147. You'll have to figure the equations on your own.

The equations don't apply for the scenario you drew in post 147. You'll have to figure the equations on your own.

Post 147 is a picture of the two ball scenario that we've been talking about all along. I'm glad to see you finally admit that your equations do not solve that problem. Might I ask what problem you thought we were talking about when you submitted those equations as your solution?

Post 147 is a picture of the two ball scenario that we've been talking about all along.

Maybe you have been talking. You did not manage to communicate it to anyone else. Proof is that you can't get the forces right even if after you dropped any pretense of solving the relativistic case.

I'm glad to see you finally admit that your equations do not solve that problem. Might I ask what problem you thought we were talking about when you submitted those equations as your solution?

You can read the solution in post 40 and you can figure out for yourself. It is quite clear.

Maybe you have been talking. You did not manage to communicate it to anyone else. Proof is that you can't get the forces right even if after you dropped any pretense of solving the relativistic case.

Alphanumeric understood it. Cpt. Bork understood it, too. Everything was explicitly defined back in post 74. The drawing matches that post exactly.

You can read the solution in post 40 and you can figure out for yourself. It is quite clear.

Your four equations were presented as a solution to the ball problem in post 85. They did not come from post 40, which has to do with torques on the plane. You are all cornfused, as usual.

Alphanumeric understood it. Cpt. Bork understood it, too. Everything was explicitly defined back in post 74. The drawing matches that post exactly.
Your four equations were presented as a solution to the ball problem in post 85. They did not come from post 40, which has to do with torques on the plane. You are all cornfused, as usual.

We were discussing relativistic transformation of force when you butted in only to find out much later that not only you don't know how force transforms in relativity but that you also believe that " The Lorentz transforms I know transform coordinates, not forces. ". This monster blunder from you post 132 pretty much sealed it , so from this point on you downgraded the problem to a non-relativistic one and you responded by producing the silly drawing from post 136. When asked to draw the forces, you waffled again and you have been waffling ever since waiting for someone else to solve your nonsensical exercise. So, you are stuck, tough!