xyzt

I am done correcting your errors. Bye!

It is called "beating the strawman to death". We should let him do it, I corrected all his errors but to no avail. What I explained to you. Not what you continue to erroneously claim.

Your posts are total nonsense.

This doesn't belong in the main forum.

Nope [math]dx'[/math] is the displacement in the frame where I also calculated [math]dy',F'_x,F'_y,dW'[/math]. You have to come to grips with the fact that you have no clue when it comes to SR.

No, I am not. I am simply trying to teach you relativity. But you seem unwilling and unable to learn. You would much rather cling to your misinterpretations of SR than learn SR. Not when [math]dx=0[/math]. One gets [math]dx'=dx/\gamma[/math] when one imposes [math]dt'=0[/math] in [math]dx'=\gamma(dx-Vdt)[/math] and [math]dt'=\gamma(dt-Vdx/c^2)=0[/math]. This is NOT the case here. The case here is [math]dx=0[/math] and basic math teaches you that [math]dx'=-\gamma V dt[/math] as I tried to teach you. Sorry I failed, there is nothing more that I can do for you.

The transformation of force with the minus sign is derived from the Lorentz transform: [math]x'=\gamma(x-Vt)[/math], so: [math]dx'=\gamma(dx-Vdt)=-\gamma V dt[/math] Now try calculating [math]dW'=F'_xdx'+F'_ydy'[/math]. If you do it correctly you should be getting [math]dW'=\gamma dW[/math]. So, once again, all you are finding is your mistakes in understanding SR.

yep

Good, you are starting to learn. Took you a few days to reproduce the formula that I have shown you a few days ago. You got the sign wrong because you have the first formula wrong: [math]F'_x=F_x+\frac{u_yV}{c^2+\frac{u_xV}{c^2}}F_y+\frac{u_zV}{c^2+\frac{u_xV}{c^2}}F_z[/math] I'll have Ranch with the above. You keep making the mistakes, I keep correcting them. The only "mistakes" that you have found in SR are your own.

He knows he's wrong, he will just never admit it. ...which is exactly what I explained to you in an earlier post: [math]dW'=F'_ydy'+F'_xdx'=F_y(dy \sqrt{1-V^2/c^2}+\gamma V dt \frac{u_yV}{c^2})=F_y dy(\frac{1}{\gamma}+\gamma \frac{V^2}{c^2})=F_ydy \gamma=\gamma dW > dW [/math] There is a [math]F'_ydy'[/math] component and a [math]F'_xdx'[/math] component. You keep denying the [math]F'_xdx'[/math] component because you don't know relativity. ....in frame S ONLY. In frame S', the acceleration is no longer perpendicular on the speed V, there is also a parallel component that shows up. This is the cause of the component [math]F'_x \ne 0[/math] that shows up in frame S'.

[math]L'=L \sqrt{1-(v/c)^2}[/math] When [math]v=0.76c[/math], [math]\sqrt{1-(v/c)^2}=0.66[/math]

Relativity teaches you that your claim (like most , if not all of your other claims) is false. You keep piling false claims on top of false claims. [math]F_x=0[/math] [math]u_x=0[/math] [math]F'_x=F_x+\frac{u_yV}{c^2+\frac{u_xV}{c^2}}F_y=0+\frac{u_yV}{c^2}F_y=\frac{u_yV}{c^2}F_y[/math] See how easy is to disprove your falsities? The guy is a mechanical engineer. I wonder how he got his degree. Yep Nope, I'll let you figure out your mistakes by yourself on this one. No, it doesn't, [math]dW'=\gamma F_y dy=\gamma dW[/math]. I'll let you figure out your errors by yourself. You can't do physics and you demonstrated that you can't do basic math either. Ahh, I missed this glaring error as well. You see, force in SR is NOT [math]\vec{F}=m \vec{a}[/math], it is defined as [math]\vec{F}=\frac{d}{dt}(\gamma(u) m \vec{u})[/math]. Try to do the Lorentz transform , you will see your errors. Or , most likely, you will add more errors on top of the existent errors.

This makes absolutely no sense.

The level of math knowledge is a very good indicator of physics knowledge. This is why I always ask for a mathematical formulations of the individual's prose claims.

The above is valid only for uniform motion (constant velocity). It is not valid for general cases. I asked for math to see your level of knowledge. Thank you for obliging.

The above sounds like a classical word salad. Can you post the math that goes with it or is it just that, a word salad?

No, it doesn't. You have been at this for quite a while.

If they are side by side and there is no relative motion, the two objects are at rest in the same reference frame. Do not try (again) to apply the above to light beams, it doesn't apply.

Who is Trout? Anyways, you clearly don't understand the subject matter, yet you persist in posting.

The part that you seem to be missing is that the foundation of time measurement in atomic (Cs,Ru) clocks has nothing to do with motion, the hyperfine transition is a quantum effect related to spin flipping. This effect is a quantum effect, therefore cannot be viewed in reductionist fashion as "motion".

If, in frame S you have* [math]dW=F_ydy[/math], then in frame S', moving in the x direction with speed [math]V[/math], you will have: [math]F'_y=F_y \sqrt{1-V^2/c^2}=\frac{F_y}{\gamma}[/math] [math]F'_x=\frac{u_yV}{c^2}F_y[/math] [math]dy'=dy[/math] [math]dx'=\gamma(dx+Vdt)=\gamma V dt[/math] so [math]dW'=F'_ydy'+F'_xdx'=F_y(dy \sqrt{1-V^2/c^2}+\gamma V dt \frac{u_yV}{c^2})=F_y dy(\frac{1}{\gamma}+\gamma \frac{V^2}{c^2})=F_ydy \gamma=\gamma dW > dW [/math] So? Work, is frame variant, as explained multiple times to you. What I see that having failed to prove your case for "work", you graduated to your other "disproofs" of SR. I have already explained to you that you cannot disprove SR by using SR, all you can do is to demonstrate your inability to understand and learn SR. Have fun! ----------------------------------------------------------------------------------------------------------------------------------------- * This is not how mechanical work is described in relativity. I explained that to you in a previous post but it fell on deaf years.

This kind of persistence in posting the SAME errors AFTER you have ALREADY been corrected is exactly what made me post what I posted.

I read your "paper", it is the same mistakes you are posting in the forum. Yet, you fail to transform [math]ds[/math] when passing from one frame to the other. You cling to your incorrect claim that [math]ds[/math] is frame-invariant. SR teaches you that it ISN'T. Err, wrong again, the general formula is: [math]\mathbf{F} = \gamma(\mathbf{v})^3 m_0 \, \mathbf{a}_\parallel + \gamma(\mathbf{v}) m_0 \, \mathbf{a}_\perp [/math] Only when the component of the acceleration perpendicular on the velocity [math]\mathbf{v}[/math] is zero does the formula reduce to what you wrote. In general case, it doesn't. I am going to tell you one last time : you CANNOT USE SR in order to "disprove" SR.

Good, you have learned something. dE represents WORK. By definition. So, there is something new that you need to learn. This is incorrect, you are misapplying relativity in order to "prove it wrong". I have shown you how to do the calculations correctly, you just repeated the correct calculations at the beginning of this post, now you are reverting to your initial errors. False. Relativity teaches you that length (S) is NOT "constant in both frames". Length is frame variant. You need to LEARN relativity before you try to "disprove" it. What you are really disproving is your MISCONCEPTIONS about relativity, not relativity itself.

I do not know why the moderators don't let me post negs, you definitely deserve it for the above.