Relativity paradox

[Motor Daddy]

gravitational acceleration, please open your own thread , you seem to be trying to learn 9-th grade physics, don't pollute this thread

"Gravitational acceleration"?

So, I cut the wires at the same time, the rod lands flat on the floor, and my watch says it takes 1 second from the time the wires are cut to the time the rod is flat on the floor. Are you saying my watch is wrong, that it really doesn't take one second??

 

[Tach]

"Gravitational acceleration"?

So, I cut the wires at the same time, the rod lands flat on the floor, and my watch says it takes 1 second from the time the wires are cut to the time the rod is flat on the floor. Are you saying my watch is wrong, that it really doesn't take one second??

If you plug in your numbers in the formula I gave you, you will get 1s. You can stop embarrassing yourself now by showing your crass ignorance, go away.

 

[Motor Daddy]

If you plug in your numbers in the formula I gave you, you will get 1s. You can stop embarrassing yourself now by showing your crass ignorance, go away.

Tach, I will embarrass the shit out of you if you persist in your BS. The formula you gave me is not needed, the time is only known when you MEASURE IT! Once the time is measured you can not change it, it is set in stone!

 

[Tach]

Tach, I will embarrass the shit out of you if you persist in your BS. The formula you gave me is not needed, the time is only known when you MEASURE IT! Once the time is measured you can not change it, it is set in stone!

Go away.

 

[Motor Daddy]

Go away.

I think I'd rather watch Pete rip you a new one, he's doing a fine job.

 

[Janus58]

Err, let's try again, with math this time:

$$t'=\gamma(t-vx/c^2)$$


$$dt'=\gamma(dt-vdx/c^2)$$

$$dt=0$$

$$dt'=\gamma(-vdx/c^2)$$

$$dx=x_{right}-x_{left} >0$$

Therefore:

$$t'_{right}-t'_{left}=dt' <0$$

i.e. the right end falls first. There is also a physical explanation for this: in the car frame , the two ends fall simultaneously, this is why, the platform observer must see the end farthest from him (the right end) fall first, not the other way around.

You appear to be a good person, genuinely interested in finding a solution. We can work together to plug the holes in your current solution. The biggest ones are in your attempt at creating a relativistic theory of elasticity.
I have an even bigger objection: the attempt at solving this problem using SR (and RoS, to make things even worse) is ill founded, the presence of the gravitational field makes this problem a candidate for GR, not SR.

Sorry, but no. What you just showed is that if the releasing of the end were simultaneous in the platform frame, then the right end will release first in the train frame. This is because from the train frame, the platform is moving right to left.

The observer's position on the platform has no bearing on this. he could be far ahead of the train ( and thus closer to the right end) or far behind the train (and closer to the left end).

Again, it is easy to show why it is the left end that drops first.

Consider the following animation taken from the train frame:

Here we see a flash of light emitted from the midpoint between two clocks and starts each clock upon arrival. The flash reaches each clock at the same time, and they start in sync.

Now consider things from the platform frame, where the clocks are moving left to right:

The light is emitted from a point midway between the clocks and expands outward at c. We can see the left clock meets up with the flash first and starts running before the flash reaches the right clock and starts it running.

Now before anyone says anything, these animations do not take time dilation or length contraction into account. If they did, then the second animation would show the clocks closer together and squashed, and they would run slower when running. But neither of these change the fact the one clock starts before the other and that is the left clock that does so.

 

[eram]

Tach, I will embarrass the shit out of you if you persist in your BS. The formula you gave me is not needed, the time is only known when you MEASURE IT! Once the time is measured you can not change it, it is set in stone!

This thread is leaving me feeling "weirded out".

 

[Tach]

duplicate post, deleted.

 

[Tach]

Sorry, but no.

Please see post 136 for the solution.

 

[Motor Daddy]

Now before anyone says anything, these animations do not take time dilation or length contraction into account. If they did, then the second animation would show the clocks closer together and squashed, and they would run slower when running. But neither of these change the fact the one clock starts before the other and that is the left clock that does so.

How much time elapses until the light reaches the clocks?

 

[Fednis48]

I am glad that you are attempting to formalize a little the dumbed down version of the problem. Since you are so intent on solving the dumbed down version instead of the realistic one, I'll solve this version for you.
First off, there are THREE frames of reference that you must consider, not TWO:
$$S"=$$ the frame of the train car
$$S'=$$ the frame of the falling rod
$$S=$$ the frame of the platform

There are two speeds that we need to consider:

$$v'=(0,-u)$$ the speed between $$S'$$ and $$S"$$
$$v=(V,0)$$ the speed between $$S$$ and $$S'$$

This is a classical problem of ______.
The angle made by the rod with the floor of the car , in the frame of the car $$S"$$ is $$\theta"$$.
The angle made by the rod with the platform , in the frame of the platform is $$\theta$$.
Form the description of Thomas precession:

$$tan(\theta)=\frac{v'}{v \gamma(v)}$$
$$tan(\theta")=\frac{v'\gamma(v')}{v}$$

So:

$$\tan(\theta")=tan(\theta) \gamma(v)\gamma(v')$$

Since $$\tan(\theta")=0$$ it follows $$tan(\theta)=0$$.
In other words:

$$\theta"=0 => \theta=0$$

QED.

Thanks for the notes on Thomas precession - they're very clear! Unfortunately, I don't think this is correct. The $$\theta$$ in Thomas precession refers to the direction of motion of the double-Lorentz-transformed reference frame relative to the original reference frame. In our case, that makes it the direction of motion of the bar in the platform frame. This is not the same as the angle the bar itself makes with the ground. It's not clear to me how one would include the angle of the bar in Thomas precession (maybe divide the bar up into a string of point particles and treat each separately?), but this isn't the right way.

In counterpoint, I'll try to formalize something people have been posting for a while now. Let's say for argument that if the bar forms an angle $$\phi$$ with the floor in the platform frame,

$$\phi=0$$

when it hits the floor like you've been saying. As you've acknowledged, since the bar doesn't "know" in advance where the floor is, this solution must be independent of height, that is:

$$\phi(h)=0$$

for all h. The bar is falling straight down, so h is a one-to-one function of t (to be precise, let's say t the time since the second wire was cut in the platform frame). We can just as easily write $$\phi$$ as a function of t, and take the limit as t goes to zero:

$$\lim_{t \to 0^+}\phi(t)=0$$

This means that $$\phi$$ must be zero (ie. the bar must be parallel to the floor) at the moment of release. This in turn means that, between when the first wire is cut and the second is cut, the bar must hover motionless in place, because any rotation during that time would make $$\phi\neq0$$. This is absurd; a metal bar pinned to an inertial reference frame at only one end will rotate under gravity, whether we use GR or SR. The puzzle, then, is not whether the two ends of the bar hit at different times, but how the two ends can hit at different times without deforming the final shape of the bar. I'm inclined to buy Janus' explanation for this.

 

[Tach]

Thanks for the notes on Thomas precession - they're very clear! Unfortunately, I don't think this is correct. The $$\theta$$ in Thomas precession refers to the direction of motion of the double-Lorentz-transformed reference frame relative to the original reference frame. In our case, that makes it the direction of motion of the bar in the platform frame. This is not the same as the angle the bar itself makes with the ground.

The two directions are one and the same. You need to look at the first picture in the Smoot class notes (page 1).

In counterpoint, I'll try to formalize something people have been posting for a while now. Let's say for argument that if the bar forms an angle $$\phi$$ with the floor in the platform frame,

$$\phi=0$$

when it hits the floor like you've been saying. As you've acknowledged, since the bar doesn't "know" in advance where the floor is, this solution must be independent of height, that is:

$$\phi(h)=0$$

for all h. The bar is falling straight down, so h is a one-to-one function of t (to be precise, let's say t the time since the second wire was cut in the platform frame). We can just as easily write $$\phi$$ as a function of t, and take the limit as t goes to zero:

$$\lim_{t \to 0^+}\phi(t)=0$$

This means that $$\phi$$ must be zero (ie. the bar must be parallel to the floor) at the moment of release.

IN THE FRAME OF THE CAR. You need to be able to specify frames.

The puzzle, then, is not whether the two ends of the bar hit at different times, but how the two ends can hit at different times without deforming the final shape of the bar. I'm inclined to buy Janus' explanation for this.

But Janus' explanation has been shown to contain multiple errors.

 

[Fednis48]

The two directions are one and the same. You need to look at the first picture in the Smoot class notes (page 1).

No. The Smoot notes just treat one reference frame that's moving with respect to another. The orientation of the bar in a reference frame is a separate and independent variable from the direction of its motion in that frame. In section 2, the Smoot notes even use the expressions for $$\theta$$ to examine the motion of a point particle, which has no orientation. If nothing else, you maintain that $$\theta=0$$ in the platform frame; if the bar's $$\theta$$ of orientation and its $$\theta$$ of motional direction are the same, this would imply that the bar is moving horizontally, which is wrong because it's falling.

IN THE FRAME OF THE CAR. You need to be able to specify frames.

Look back at my post; I did specify that $$\phi$$, t, and h were all in the platform frame.

But Janus' explanation has been shown to contain multiple errors.

Yeah, it probably does need some polish. But it makes qualitative sense. I'm not enough of a relativity specialist to iron out the details myself, so I'm inclined to think Janus' explanation is qualitatively right, pending counterexample.

 

[Tach]

No. The Smoot notes just treat one reference frame that's moving with respect to another. The orientation of the bar in a reference frame is a separate and independent variable from the direction of its motion in that frame. In section 2, the Smoot notes even use the expressions for $$\theta$$ to examine the motion of a point particle, which has no orientation. If nothing else, you maintain that $$\theta=0$$ in the platform frame;

$$S$$ is the car frame, can't you read the definitions in post 136? $$\theta=0$$ on the CAR FRAME. Can't you read the definitions in post 136?

if the bar's $$\theta$$ of orientation and its $$\theta$$ of motional direction are the same, this would imply that the bar is moving horizontally, which is wrong because it's falling.

$$\theta$$ is the angle with the x-axis, "falling" means $$\theta=0$$.

I'm not enough of a relativity specialist to iron out the details myself,

The classical cop-out.

 

[Fednis48]

$$S$$ is the car frame, can't you read the definitions in post 136? $$\theta=0$$ on the CAR FRAME. Can't you read the definitions in post 136?

...seriously? From post 136:

$$S=$$ the frame of the platform

And I know that wasn't a typo on your part, because for the Smoot notes to work, $$S''$$ has to be moving both horizontally and vertically with respect to $$S$$. This means either $$S$$ is the platform frame and $$S''$$ is the bar frame or vice-versa. In either case, $$S'$$ has to be the car frame.

$$\theta$$ is the angle with the x-axis, "falling" means $$\theta=0$$.

Ok, but the bar isn't falling vertically in the platform frame, either. It's falling down while moving horizontally, which is definitely a nonzero $$\theta$$ no matter how you define it.

 

[Tach]

...seriously? From post 136:

And I know that wasn't a typo on your part, because for the Smoot notes to work, $$S''$$ has to be moving both horizontally and vertically with respect to $$S$$. This means either $$S$$ is the platform frame and $$S''$$ is the bar frame or vice-versa. In either case, $$S'$$ has to be the car frame.

You realize that motion is relative, don't you? A

Ok, but the bar isn't falling vertically in the platform frame, either.

It is falling vertically in the car frame, meaning that the car frame is moving vertically in the rod frame. Motion is relative, you realize that?
The car frame is also moving horizontally in the platform frame.

You realize that motion is relative and this is all that counts, don't you?

Since you have som much difficulty with 3 frames of reference, let's try the exercise with only two frames, like here. This is a pretty good page, it tells you that , if the slope of a moving rod is $$tan(\theta)$$ in one frame, it will be $$\gamma tan(\theta)$$ in a frame boosted in the x direction:

$$tan(\theta')=\gamma tan(\theta)$$. This results into a "Thomas precession" (rotation) EXCEPT for the trivial case when $$\theta=0$$ (rod transported perpendicular on the y axis). In that case :
$$\theta'=\theta=0$$ (no precession).

 

[Pete]

Wrong, we aren't talking about the shape of the rod in the train car frame as viewed from the platform frame.
We are talking about the shape of the rod in the platform frame as measured in the platform frame vs. the shape of the rod in the train car frame as measured in the train car frame. We are talking about a comparison of the proper shapes of the rod.

The proper shape of the rod is the shape of the rod measured in the rod's rest frame.
The disagreement is over the shape of the rod measured in the train and platform rest frames.

Actually you have been disagreeing on that by virtue of disagreeing on the angle of fall. You may not realize or admit it but this is the case.

Prove it.

I have, several times.

Prove it.

This is a classical problem of Thomas precession.

...

$$tan(\theta)=\frac{v'}{v \gamma(v)}$$
...
$$\tan(\theta)=0$$

This implies $$v'=0$$, so there must be a mistake somewhere.

It seems to me that the Smoot notes analyse the angle of the displacement over time of a moving object, rather than the instantaneous angle of a rod.

I'll do a proper derivation later today.

 

[Fednis48]

You realize that motion is relative, don't you? A

...

It is falling vertically in the car frame, meaning that the car frame is moving vertically in the rod frame. Motion is relative, you realize that?
The car frame is also moving horizontally in the platform frame.

You realize that motion is relative and this is all that counts, don't you?

I do know that motion is relative, but I don't really get your point here. Yes the bar is falling vertically in the car frame, yes the car is moving vertically in the bar frame, and yes the bar is parallel to the floor in both frames. In post 151, I showed that if you say the bar is parallel to the ground in the platform frame, you'll reach absurd conclusions. What, if anything, was incorrect about that calculation?

Since you have som much difficulty with 3 frames of reference, let's try the exercise with only two frames, like (url). This is a pretty good page, it tells you that , if the slope of a moving rod is $$tan(\theta)$$ in one frame, it will be $$\gamma tan(\theta)$$ in a frame boosted in the x direction:

$$tan(\theta')=\gamma tan(\theta)$$. This results into a "Thomas precession" (rotation) EXCEPT for the trivial case when $$\theta=0$$ (rod transported perpendicular on the y axis). In that case :
$$\theta'=\theta=0$$ (no precession).

I looked through those notes, and I actually couldn't find the formula you cited. Where in the notes is it? The closest formula I could find was

$$\phi=\gamma\tan^{-1}(\gamma \tan \theta)$$

but relates the rod's positional vector to its orientation, rather than its orientations in two different frames, so it doesn't answer the question at hand. One thing I did notice was that the very first figure of the notes shows a vector that is tilted in a rest frame but horizontal in an x-boosted frame. This is exactly what I'm trying to argue: the rod is horizontal in the car frame (and in its own frame), but that does not mean it's horizontal in the platform frame.

 

[Motor Daddy]

Don't forget, Tach, the test of the rod falling in the train frame needs to match the top 20! In the train frame there is a specific elapsed time that it takes for that rod to travel, and a specific distance that it travels. I hate to break it to you, but that is gonna leave a mark!

 

[Undefined]

we aren't talking about the shape of the rod in the train car frame as viewed from the platform frame.
We are talking about the shape of the rod in the platform frame as measured in the platform frame vs. the shape of the rod in the train car frame as measured in the train car frame. We are talking about a comparison of the proper shapes of the rod.

Thankyou for before praising my naive reasoning and understanding. But now I am confused by what you say above. I naively understand when we "measure" the rod within its proper frame it is a "proper measurement". I also naively understand when we look from the embankment it is "relative measurement" using an observer's image of the rod from a distance. You use and switch from one kind of measurement to the other kind in your explanation as if they were *equivalently* based on information available to both frames in the same form? That is what confuses me. One measurement (proper) is a measurement based on "inside information" in proper frame, but the other is not a measurement proper, it is a relative view isn't it, based on "remote information" in different frame? A relative view is not really a "measurement" is it, because we have to use some theory on it to turn the image we "relatively view" into a "rod proper frame measurement" *equivalency* don't we? The constant switching and equating of "proper measurement" process with "theoretically adjusted relative view" process confuses me every time I try to naively understand what you mean. My head is spinning because I try to understand but I can't if you are not clearer which is what. Will you explain for me what is which when you refer to the different frame observations: is it a "measurement" or is it a "theoretically adjusted view" or whatever?