Recent content by rgtr

Sorry I was multitasking while posting the first time.## VBLA = VBLT + VTA ## = ## VBLT = VBLA - VTA ## When using the example ## VBLT = VBLA - VTA## velocity addition on the right hand of the equal sign of the equal sign, does the right hand of the equal sign always need to equal the...

Sorry I took a break from physics for a while I got busy and just have a followup/ confirmation question. Where ball 1 or B_1 is the left ball. Also obviously Alice is on the ground seeing the train move at velocity v in the right direction. ## VPA = VTA -(VB_1A) ## , when using velocity...

I think I meant to say horizontal component. Saying vertical component was just a mistype. Whoops.

In this picture it shows a light clock. Let's use the moving light clock example. Am I essentially calculating the b component of moving clock. Assume the moving frame is the B frame. Assume the stationary frame is the A frame https://simple.wikipedia.org/wiki/Light_clock Or essentially the b...

Thanks for the excellent explanation it was very clear now. Question for anyone. In this picture it shows a light clock. Let's use the moving light clock example. Am I essentially calculating the b component of moving clock. Assume the moving frame is the B frame. Assume the stationary frame is...

Why does it need to be a function of v? And what do you mean by a function of v?

In special relativity I can get ## \gamma ## , ## \frac {T_B}{T_A}=\gamma ## Why do I not go ##{T_B} - {T_A} = \gamma## ? ##T_B = \frac {2H} {c^2 - v^2}## . ## T_B ## is the moving light clock. ## T_A = \frac {2H} {c^2} ## . ##T_A ## is the stationary light clockI assume LaTeX doesn't work...

RSOTE = right side of the equation. Okay the RSOTE you just mentioned is confusing me. Can you explain it? The confusing part is the c's. ## \frac c { \sqrt{ { c } } } = 1 ## I guess it is kind of self explained. Thanks I think the title needs a change.| I just have 1 question in special...

Ya the squre root was just a typo. My question is am I on the right track? I can also cancels out the c's.

What I mean is that ## \frac {c} {c } = 1 ## . Both H's are the same.part 1 ## \frac {T_B} {T_A} = \frac{\frac{2h}{c^2-v^2}}{\frac{ 2h} {c}} ## part 2 Then I performed algebra rules 10 https://algebrarules.com/## \frac {T_B} {T_A} = \frac {(2h)(c^2- v^2)}{(2h)(c} ## part 3 Then I use...

In LaTeX how do I go 1/1/1/1/?

I am trying to get gamma. so iow's ## \frac {T_B} {T_A } = gamma## I would have to do the calculations. I guess I am asking am I am on the right track? Sure I will show my math. Let me make a new post.

I am trying to solve ##T_B = \frac {2h} {\sqrt {v^2 - c^2} } ## ,##T_A = \frac {2h} {c} ## Then I need to divide TB by TA. This is in reference to special relativity deriving time dilation. I managed to reduce it to ## \frac {T_B} {T_A} = \frac {( 2h)(c)} { \sqrt {(2h)( v^2 - c^2)} } ## ...

okay thanks

How can I remove a single c from the square root?