Doppler Effect Of Gravitational Field

[OceanBreeze]

I am done with you crank, go bother someone else.

 

I did notice your red stamp in this thread.

 

I didn't remove it. :)

[ralfcis]

Victor I initially thought you were using some kind of new equations but your math makes no sense.

 

You use gammas and not velocities? Y_a  = Y_b =2.3

 

At first you say "The answer is your add the gamma's to find the speed "

 

then you multiply the gammas contradicting what you said

 

Y_a * Y_b = 5.26. 

 

Neither is the correct answer .9945. Tony is right.

 

Then you try to hide you don't know how to calculate the third scenario by giving the same answer of .9945 and finally admit trig is necessary but don't show how to use it. From what you're inferring, the answer would be sqrt(.9² + .9²) = .9 rt2 = 1.27 which would be wrong. The answer .9945 is also wrong. So instead of smoke bombing everyone, can you provide a correct answer or not? Maybe that lying moron exchemist can help you out.

Edited March 18, 2020 by ralfcis

[Vmedvil2]

Victor I initially thought you were using some kind of new equations but your math makes no sense.

 

You use gammas and not velocities? Y_a  = Y_b =2.3

 

At first you say "The answer is your add the gamma's to find the speed "

 

then you multiply the gammas contradicting what you said

 

Y_a * Y_b = 5.26. 

 

Neither is the correct answer .9945. Tony is right.

 

Then you try to hide you don't know how to calculate the third scenario by giving the same answer of .9945 and finally admit trig is necessary but don't show how to use it. From what you're inferring, the answer would be sqrt(.9² + .9²) = .9 rt2 = 1.27 which would be wrong. The answer .9945 is also wrong. So instead of smoke bombing everyone, can you provide a correct answer or not? Maybe that lying moron exchemist can help you out.

A) you didn't use the gamma multiplication formula you cannot just multiply them directly, you must use a formula for multiplying gammas(https://en.wikipedia.org/wiki/Velocity-addition_formula#Standard_configuration).

 

B ) No, the formula is valid for gammas in the Pythagorean theorem, I derived that when calculating multiple dimensions such as (x,y,z) axis that is the proper formula for calculating special relativity in multiple dimensions(https://en.wikipedia.org/wiki/Minkowski_space).

 

C) It is assumed you know this stuff as I didn't explain it in detail and left out stuff that I didn't think would cause you any trouble it seems I was wrong, I suppose I either over-estimated your intelligence or under-estimated your guile.

 

D) What did you expect am I supposed to hold everyone's hand the disagrees with relativity, I just call them a crank and move on, basically I did not feel like having to teach a class on "How to do relativity the right way".

 

Edited March 18, 2020 by VictorMedvil

[ralfcis]

A) Still don't see the gamma multiplication version of the velocity combo law.  First you say addition and now it's a relativistic multiplication of gammas. I'm not familiar with that form. Show it please. Here's a form I like: DSR_w =DSR_v *DSR_u where w is the velocity combination of u and v velocities using multiplication of Doppler Shift Ratio formulas. 

 

:cool: This is your go-to smoke bomb. There is no need to go to 3 spatial dimensions for the examples given so don't try to convince me there's a special answer there.

 

C) I'm just asking a question. Do you have an answer or are you just going to get testy about it.

 

D) Your little video is just more smoke. Can you answer the question or not? Just give me a numerical answer for the 3rd scenario if you don't want to reveal your magician's tricks. Don't post your light cone graphic as you usually do, that is also irrelevant.

Edited March 18, 2020 by ralfcis

[Vmedvil2]

A) Still don't see the gamma multiplication version of the velocity combo law.  First you say addition and now it's a relativistic multiplication of gammas. I'm not familiar with that form. Show it please.

 

:cool: This is your go-to smoke bomb. There is no need to go to 3 spatial dimensions for the examples given so don't try to convince me there's a special answer there.

 

C) I'm just asking a question. Do you have an answer or are you just going to get testy about it.

 

D) Your little video is just more smoke. Can you answer the question or not? Just give me a numerical answer for the 3rd scenario if you don't want to reveal your magician's tricks. Don't post your light cone graphic as you usually do, that is also irrelevant.

Yes but I am lazy, it's like uhhhhhhh man, this is stuff you should understand by now and even if I give you a answer you will somehow say its wrong by your understanding of relativity as you think relativity is wrong, that's the truth. Well, the truth is I know relativity is correct as I have used it for many years with correct results, I don't think there is a set of words or calculations that will convince you thus I am wasting my time, that's the way I feel about this. Either take the blue pill or red pill, the blue pill is that relativity is correct and you believe me, the red pill is you believe whatever you want and go about your life, it doesn't matter to me as it doesn't affect me. I look at that picture of the light cone and see all the equations of relativity lined up in perfect harmony however for you it seems it doesn't represent the same. My suggestion is take a Modern physics course in college man, I hated college too however I think that would help fix the problems that you have with relativity. 

 

-Real talk from Victor Medvil

Edited March 18, 2020 by VictorMedvil

[ralfcis]

Ok so you're just as much of a math fraud as all the other math frauds on this forum. Just man up to your ignorance. I don't think relativity is wrong, I think Einstein's theory is wrong but my answers don't disagree with his, only my math method does. Two .9c ships leaving earth at right angles to each other, what is their relative velocity to each other? How hard could this be for a non-math fraud? Get 006, P8ly and exchemist in on this but not Popeye because he would just give you guys the answer.

Edited March 18, 2020 by ralfcis

[Vmedvil2]

Ok so you're just as much of a math fraud as all the other math frauds on this forum. Just man up to your ignorance. I don't think relativity is wrong, I think Einstein's theory is wrong but my answers don't disagree with his, only my math method does. Two .9c ships leaving earth at right angles to each other, what is their relative velocity to each other? How hard could this be for a non-math fraud? Get 006, P8ly and exchemist in on this but not Popeye because he would just give you guys the answer.

Okay, fine it is γ_x = 1/(1- V_x²/C²)^1/2    γ_y = 1/(1- V_y²/C²)^1/2  then γ_x² + γ_y² = γ_f²   Then reverse the equation 

γ_f = 1/(1- V_f²/C²)^1/2 and solve for V_f  Then you will arrive at the exact solution to that problem, plug and chug algebra style.

Edited March 18, 2020 by VictorMedvil

[ralfcis]

So I make a lot of mistakes with arithmetic but the answer I get is .9955c.  This number happens to be greater than the .9945c of them separating at 180 degrees. Your answer can't be correct as the speed addition at 180 degrees must be less than at 90 degrees because the vector is smaller. I don't understand how you can be using gammas to calculate anything about velocity combination. γ_x² + γ_y² = γ_f² means nothing. Even if you used velocities instead of gammas, it would still mean nothing. Try googling this question, I found an example. See if you can decipher it and apply it to this problem. If you're going to call people cranks, make sure you're not one yourself. I'm not saying Tony isn't a crank and has any intention of learning he's wrong but I'm hoping you're different.

 

http://electron6.phys.utk.edu/PhysicsProblems/Mechanics/8-Relativity/velocity_addition.html

[Vmedvil2]

So I make a lot of mistakes with arithmetic but the answer I get is .9955c.  This number happens to be greater than the .9945c of them separating at 180 degrees. Your answer can't be correct as the speed addition at 180 degrees must be less than at 90 degrees because the vector is smaller. I don't understand how you can be using gammas to calculate anything about velocity combination. γ_x² + γ_y² = γ_f² means nothing. Even if you used velocities instead of gammas, it would still mean nothing. Try googling this question, I found an example. See if you can decipher it and apply it to this problem. If you're going to call people cranks, make sure you're not one yourself. I'm not saying Tony isn't a crank and has any intention of learning he's wrong but I'm hoping you're different.

 

http://electron6.phys.utk.edu/PhysicsProblems/Mechanics/8-Relativity/velocity_addition.html

"( c )  |u'| = (u₁'² + u₂'² + u₃'²)^½ = c.

This is a postulate of special relativity."

 

γ_x² + γ_y² +γ_z² = γ_f²

 

 

That is the Dimensional field of special relativity(https://en.wikipedia.org/wiki/Higher_local_field).

Edited March 18, 2020 by VictorMedvil

[ralfcis]

I don't know this notation but I'm pretty sure u is a velocity and you're not using the formula correctly because your answer is wrong. Wiki is useless if you don't understand what you're flashing. Show me a formula that gets you the right answer and I'll translate it into a simple algebraic notation I understand. I'm not into equation porn. I've given you a link to an example, don't make me use your crank stamp on you.

Edited March 18, 2020 by ralfcis

[Vmedvil2]

I don't know this notation but I'm pretty sure u is a velocity and you're not using the formula correctly because your answer is wrong. Wiki is useless if you don't understand what you're flashing. Show me a formula that gets you the right answer and I'll translate it into a simple algebraic notation I understand. I'm not into equation porn. I've given you a link to an example, don't make me use your crank stamp on you.

Fine i'll put it in a different form.

 

X = (1- V_x²/C²)^1/2    Y = (1- V_y²/C²)^1/2  then X² + Y² = r²  , r = (1- V_f²/C²)^1/2 

 

x² + y² +z² = r²

 

 

Now look that up, that is the invariant metric.

Edited March 18, 2020 by VictorMedvil

[ralfcis]

Are you just taking wild guesses now? How is this older form:

 

 γ_x = 1/(1- V_x²/C²)^1/2    γ_y = 1/(1- V_y²/C²)^1/2  then γ_x² + γ_y² = γ_f²   Then reverse the equation 

γ_f = 1/(1- V_f²/C²)^1/2

 

Any different from your new form:

 

X = (1- V_x²/C²)^1/2    Y = (1- V_y²/C²)^1/2  then x² + y² = r²  , r = (1- V_f²/C²)^1/2 

 

x² + y² +z² = r² 

 

You've never done math before have you. You can't read the link I gave you, can you. It's time to give up Victor. Maybe one of the other math frauds can step up. Take your time. Just translate the link I gave you into the much simpler problem of two ships leaving earth at .9c at right angles, what is their relativistic velocity? Tony needs an answer. I doubt exchemist can do simple algebra but I'd love to see him give it the old Cambridge try.

Edited March 18, 2020 by ralfcis

[Vmedvil2]

Are you just taking wild guesses now? How is this older form:

 

 γ_x = 1/(1- V_x²/C²)^1/2    γ_y = 1/(1- V_y²/C²)^1/2  then γ_x² + γ_y² = γ_f²   Then reverse the equation 

γ_f = 1/(1- V_f²/C²)^1/2

 

Any different from your new form:

 

X = (1- V_x²/C²)^1/2    Y = (1- V_y²/C²)^1/2  then x² + y² = r²  , r = (1- V_f²/C²)^1/2 

 

x² + y² +z² = r² 

 

You've never done math before have you. You can't read the link I gave you, can you. It's time to give up Victor. Maybe one of the other math frauds can step up. Take your time. Just translate the link I gave you into the much simpler problem of two ships leaving earth at .9c at right angles, what is their relativistic velocity? Tony needs an answer.

And this shows how uneducated you really are about special relativity or you would know exactly what I just did. Everyone that sees that will know the who the true math fraud is.

Edited March 18, 2020 by VictorMedvil

[ralfcis]

What's your numerical answer then Victor. Changing Y_x to 1/X and the rest of your superficial lipstick isn't going to get you a correct answer. Show me.

Edited March 18, 2020 by ralfcis

[TonyYuan2020]

And this shows how uneducated you really are about special relativity or you would know exactly what I just did. Everyone that sees that will know the who the true math fraud is.

The whole story is in it. You can see it from the back. I've basically demonstrated that special relativity is a mathematical game.

If I'm wrong, please point it out for me.

http://www.sciforums.com/threads/a-story-about-special-relativity-who-can-explain-it%EF%BC%9F.162954/

 

w = (u+v)/(1+uv/c^2) is key

Edited March 18, 2020 by TonyYuan2020

[Vmedvil2]

What's your numerical answer then Victor. Changing Y_x to 1/X and the rest of your superficial lipstick isn't going to get you a correct answer. Show me.

Haha, I don't wanna 

[ralfcis]

Victor, baby, the formula was in your Wiki link. Please don't compliment me any more on my progress through relativity as you clearly have no grasp of even your own wiki articles. It's embarrassing for me to be promoted by people who know nothing. I'll just give you and the rest of the effort impaired and intellectually challenged on here the formula:

 

w = sqrt (u²+v²+2uv cos@ - (vu sin@/c)²) / (1+vu cos@/c²)

 

@= 90 degrees so cos@=0 and sin@=1

 

u=v=.9c

 

answer w=.9817c which is the relativistically combined u=v=.9c leaving at right angles from earth.

 

See that answer is less than .9945c as it should be.

 

You had absolutely no clue Victor yet you lack the integrity to admit it. Do you feel embarrassed?

Edited March 18, 2020 by ralfcis