I could have been clearer. Please feel free to ask for clarification when necessary.
A(x_A, y_A,t) is the set of events that satisfy the equation given:
$$y_A = x_A \tan(\theta_A)$$
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Angle between the orientation of a moving object and its velocity
- Thread starter Pete
- Start date
A(x_A, y_A,t) is the set of events that satisfy the equation given:
$$y_A = x_A \tan(\theta_A)$$
(x_A, y_A) is the set of points that satisfy the equation given:
$$y_A = x_A \tan(\theta_A)$$
So, they are just points like P. Therefore , there is no reason to transform any differently from the way P transforms.
I don't think you understand my ideas at all.Yes, it showed that your ideas about colinearity are unphysical.
Post 4 certainly doesn't address them.
If the questions are so elementary, why are you avoiding them?So, try to understand the math and please stop harassing with the same elementary questions.
I think you'd want to label it $$A(\theta_{A},x_{A},y_{A})$$ because it's the parameter which defines your line also. If you want to pass through the point $$(x_{A},y_{A})$$ with gradient $$\tan(\theta_{A})$$ then $$A(\theta_{A},x_{A},y_{A})$$ is the set of points which satisfy $$y-y_{A} = \tan \theta_{A}(x-x_{A})$$. I think that's what you mean.A(x_A, y_A,t) is the set of events that satisfy the equation given:
$$y_A = x_A \tan(\theta_A)$$
/edit
Opps, just seen your intercept is 0, in which case you actually just label it by the angle, since you wish to define the set $$A(\theta_{A})$$ by those points satisfying $$y - x \tan\theta_{A} = 0$$.
It's clear from the pictures but we wouldn't want Tach to have a reason to complain, now would we?
Perhaps I have misunderstood.This is not what I claim, and pete knows it.
If the angle between a surface and its velocity is zero in a given reference frame, do you or do you not claim that the angle between the surface and its velocity is zero in all reference frames?
I don't think you understand my ideas at all.
Post 4 certainly doesn't address them.
Of course it does, it shows that your conclusions are unphysical, they are contradicting the fundamentals of light aberration.
Perhaps I have misunderstood.
If the angle between a surface and its velocity is zero in a given reference frame, do you or do you not claim that the angle between the surface and its velocity is zero in all reference frames?
Yes, this is precisely what I claim and no, this is not what AN used in his counter-example. He used a surface that moves in the direction of its normal. This is why his counter-example is not relevant to the discussion.
That was my counter example. And the original counter example I gave in your mirrored wheel thread was another one, just rotated 90 degrees!This is not what I claim, and pete knows it. It is a mirror that has a zero angle between its velocity and its tangent. So, your counter-example is irrelevant.
Did you even read what I said?
That was my counter example. And the original counter example I gave in your mirrored wheel thread was another one, just rotated 90 degrees!
Did you even read what I said?
Yes, I do, I just explained a second time why your counter-example shows that you did not understand my claim. Perhaps I should let pete explain to you what the claim is and why your counter-example is irrelevant.
I think you'd want to label it $$A(\theta_{A},x_{A},y_{A})$$ because it's the parameter which defines your line also. If you want to pass through the point $$(x_{A},y_{A})$$ with gradient $$\tan(\theta_{A})$$ then $$A(\theta_{A},x_{A},y_{A})$$ is the set of points which satisfy $$y-y_{A} = \tan \theta_{A}(x-x_{A})$$. I think that's what you mean.
Thanks. It made sense to me while I was writing it, but I can see how it is bad.
Tach, my apologies for the confusion.
Correction:
A is the surface defined by:
$$y = x \ tan(\theta_A)$$
Also:
The worldline of P is defined by:
$$\begin{align}
x &= t |\vec{V_p}| \cos(\theta_P) \\
y &= t |\vec{V_p}| \sin(\theta_P)
\end{align}$$
I've fixed in the OP as well.
The mirror's normal is orthogonal to its motion in frame S, satisfying your initial condition of moving parallel to it's tangent. Do a boost in the direction of the normal and you end up with a situation where the velocity in the new frame is not orthogonal to the mirror's normal. It's a relevant counter example.
Explain precisely why this doesn't apply. No "Ask Pete!" or excuses, actually reply.
Explain precisely why this doesn't apply. No "Ask Pete!" or excuses, actually reply.
The mirror's normal is orthogonal to its motion in frame S, satisfying your initial condition of moving parallel to it's tangent.
Let's try again: the velocity and the tangent are colinear. Like in the case of a spinning wheel.
Do a boost in the direction of the normal and you end up with a situation where the velocity in the new frame is not orthogonal to the mirror's normal.
Yes, this is the fine point of the argument between me an pete, boosts do not preserve orthogonality but they preserve colinearity. See the calculations in post 4.
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Of course it does, it shows that your conclusions are unphysical, they are contradicting the fundamentals of light aberration.
No. That's a different scenario.
Please, stop avoiding these simple questions. I really need to understand what you're thinking, or we'll never get this resolved:
An observer in a train watches the ground passing by, and notes that the ground's velocity is parallel to the ground itself.
Do you agree?
An observer in an elevator watches the ground falling away, and notes that the ground's velocity is perpendicular to the ground.
Do you agree?
Yes. And then you do a boost. In the case of the flat mirror I choose the direction orthogonal to the velocity of the mirror. In the wheel's case it doesn't matter since it's rotationally symmetric and I can always find a piece of the wheel which fits the same conditions (ie the piece in the same direction from the axle as the boost velocity vector).Let's try again: the velocity and the tangent are colinear. Like in the case of a spinning wheel.
Let's call the velocity in the original frame (v,0). Let's say the flat mirror is on the y=0 line (but we can still tell it's moving in the x direction, despite being infinite in extent). The parametrisation of the mirror is then (s,0) for $$s \in \mathbb{R}$$. I boost by (0,v'). What's the parametrisation of the mirror in the new frame? Does it have any time dependent term in the y' component? Yes or no?
For example, if this were Galilean then a point in the original frame would be written as $$X(t) = (x_{0}+vt,0)$$ and in the new frame $$X'(t) = (x_{0}+vt,v't)$$. Thus $$\mathbf{e}_{y'} \cdot \dot{X'}(t) \neq 0$$. it is moving in the y' direction. The Lorentz version is less pleasant but the result $$\mathbf{e}_{y'} \cdot \dot{X'}(t) \neq 0$$ is still true.
If you disagree please show the full transformation.
Yes. And then you do a boost. In the case of the flat mirror I choose the direction orthogonal to the velocity of the mirror. In the wheel's case it doesn't matter since it's rotationally symmetric and I can always find a piece of the wheel which fits the same conditions (ie the piece in the same direction from the axle as the boost velocity vector).
Let's call the velocity in the original frame (v,0). Let's say the flat mirror is on the y=0 line (but we can still tell it's moving in the x direction, despite being infinite in extent). The parametrisation of the mirror is then (s,0) for $$s \in \mathbb{R}$$. I boost by (0,v'). What's the parametrisation of the mirror in the new frame?
I have already answered this hundreds of posts ago in the "rolling whee" thread.
Does it have any time dependent term in the y' component? Yes or no?
Yes, it does. Also answered in the same thread. pete and I are way past this, so, instead of going over stuff that we have aklready covered, please let us finish what we are doing.
For example, if this were Galilean then a point in the original frame would be written as $$X(t) = (x_{0}+vt,0)$$ and in the new frame $$X'(t) = (x_{0}+vt,v't)$$. Thus $$\mathbf{e}_{y'} \cdot \dot{X'}(t) \neq 0$$. it is moving in the y' direction.
If you disagree please show the full transformation.
Already answered this hundreds of posts ago. The full Lorentz transform, it is not very difficult.
The Lorentz version is less pleasant but the result $$\mathbf{e}_{y'} \cdot \dot{X'}(t) \neq 0$$ is still true.
I have already addressed this issue with you long ago, the y-component is irrelevant, what is relevant is that the velocity in frame S' is colinear with the tangent to the wheel. This is precisely the point where pete and I have arrived. So, rather than having to explain this over to you, why don't you let us continue the debate?
No. That's a different scenario.
All I need is ONE scenario to prove your mathematical model is unphysical. I have provided that scenario.
There is no point in discussing your scenarios since your math is invalidated by physics.
Tach, we're not way past it at all.
You were wrong hundreds of posts ago, you're wrong now.
Well, your math is wrong, resulting into results that contradict the physics but you will never admit it.
All I need is ONE scenario to prove your mathematical model is unphysical. I have provided that scenario.
There is no point in discussing your scenarios since your math is invalidated by physics.
- The mathematical model in question is special relativity.
- Your persistent avoidance of simple questions is telling.
- You should try intellectual honesty. It would look good on you.
Easy to say, but where's the proof?Well, your math is wrong
- The mathematical model in question is special relativity.
No, ir is your (mis)interpretation of special relativity
[*]Your persistent avoidance of simple questions is telling.
First off, you never addressed my contentions, you simply pretended that they don't exist. Tell you what, address my contentions and I will answer your questions. I know that this will prove too much for you, since it renders your theory as unphysical.
[*]You should try intellectual honesty. It would look good on you.
Mirror....mirror.
Easy to say, but where's the proof?
What do you think post 5 is?