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- Homework Statement
- ## r ={ \sqrt{c^2} + {x} } ##

- Relevant Equations
- ## r ={ \sqrt{c^2} + {x} } ##

If I wanted to remove c from the square root, ## r ={ \sqrt{c^2} + {x} } ## would this be correct ## r = \sqrt { {c} + {x} } {c} ## ?

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In summary, if you want to remove c from the square root, you need to use the following expression: ##r = \sqrt { {c} + {x} } {c}##.

- #1

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- Homework Statement
- ## r ={ \sqrt{c^2} + {x} } ##

- Relevant Equations
- ## r ={ \sqrt{c^2} + {x} } ##

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- #2

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if ##c## is a real number it is straightforward to remove the square root from ##c^2##...

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

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Assuming ##c,x\in\mathbb{R}##:

It is ##\sqrt{c^2}=|c|## where ##|c|## the absolute value of c.

On the other expression , ##\sqrt{c+xc}=\sqrt{c(x+1)}=\sqrt{c}\sqrt{x+1}## only if we are given that c,x+1 are positive. As you can see in this case there is no way to remove c from the square root.

It is ##\sqrt{c^2}=|c|## where ##|c|## the absolute value of c.

On the other expression , ##\sqrt{c+xc}=\sqrt{c(x+1)}=\sqrt{c}\sqrt{x+1}## only if we are given that c,x+1 are positive. As you can see in this case there is no way to remove c from the square root.

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okay thanks

- #6

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What are you really trying to do?rgtr said:How can I remove a single c from the square root?

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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)} } ## . Then ## \frac {T_B} {T_A} = \frac { \sqrt {( v^2 - c^2)} } {c} ## The problem I then encounter is I have equal amount of the c variables. So I figure I have mutliply both sides by ## \frac {1} {c^2} ## Is this correct? Or is there a better way? Or are my calculations wrong?

Also why do I divide instead of - TB and TA?

I managed to reduce it to ## \frac {T_B} {T_A} = \frac {( 2h)(c)} { \sqrt {(2h)( v^2 - c^2)} } ## . Then ## \frac {T_B} {T_A} = \frac { \sqrt {( v^2 - c^2)} } {c} ## The problem I then encounter is I have equal amount of the c variables. So I figure I have mutliply both sides by ## \frac {1} {c^2} ## Is this correct? Or is there a better way? Or are my calculations wrong?

Also why do I divide instead of - TB and TA?

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- #8

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You are solving for ##T_B##, ##T_A##, ##h##, ##H## or ##v##?rgtr said:I am trying to solve ##T_B = \frac {2h} {\sqrt {v^2 - c^2} } ## ,##T_A = \frac {2H} {c} ##

i.e. what are your knowns and that are your unknowns?

Also, can you back up and tell us what you are really trying to do instead of starting in the middle with partially simplified equations that may or may not be correct?

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##r = (\sqrt (c))(\sqrt (c)) + x##

##r^2 = (\sqrt (c))(\sqrt (c))^2 + x^2 + 2x(\sqrt(c^2))##

##r^2 = (\sqrt (c^2))(\sqrt (c^2)) + x^2 + 2x(\sqrt(c^2))##

##r^2 = c(\sqrt (c^2)) + x^2 + 2x(\sqrt(c^2))##

That seems to be the only way to forcefully produce a single c out of the square root.

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Are the ##h,H## the same thing?

- #12

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You have several problems (errors) here. In addition to Algebra issues, it looks like there must be errors with the physics. Assuming that ##v^2<c^2## you have the square root of a negative quantity.rgtr said:

I managed to reduce it to ## \frac {T_B} {T_A} = \frac {( 2h)(c)} { \sqrt {(2h)( v^2 - c^2)} } ## . Then ## \frac {T_B} {T_A} = \frac { \sqrt {( v^2 - c^2)} } {c} ## The problem I then encounter is I have equal amount of the c variables. So I figure I have mutliply both sides by ## \frac {1} {c^2} ## Is this correct? Or is there a better way? Or are my calculations wrong?

Also why do I divide instead of - TB and TA?

So, please answer @jbriggs444 questions much more thoroughly. What are you really trying to do?

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This makes sense as: in reality this can be simply written as ##r= c + x## . Squaring this equation would essentially give the same answer.Ameyzing73 said:

##r = (\sqrt (c))(\sqrt (c)) + x##

##r^2 = (\sqrt (c))(\sqrt (c))^2 + x^2 + 2x(\sqrt(c^2))##

##r^2 = (\sqrt (c^2))(\sqrt (c^2)) + x^2 + 2x(\sqrt(c^2))##

##r^2 = c(\sqrt (c^2)) + x^2 + 2x(\sqrt(c^2))##

That seems to be the only way to forcefully produce a single c out of the square root.

- #14

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Neither of your posts are correct.Ameyzing73 said:This makes sense as: in reality this can be simply written as ##r= c + x## . Squaring this equation would essentially give the same answer.

Please, just let @jbriggs444 make a reply.

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

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Write click on the latex pic and choose Display as Tex Commands

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First, as noted, you probably mean ##\sqrt{c^2-v^2}##.rgtr said:

I managed to reduce it to ## \frac {T_B} {T_A} = \frac {( 2h)(c)} { \sqrt {(2h)( v^2 - c^2)} } ## . Then ## \frac {T_B} {T_A} = \frac { \sqrt {( v^2 - c^2)} } {c} ## The problem I then encounter is I have equal amount of the c variables. So I figure I have mutliply both sides by ## \frac {1} {c^2} ## Is this correct? Or is there a better way? Or are my calculations wrong?

Also why do I divide instead of - TB and TA?

Next, you have gone wrong in simplifying ## \frac {T_B} {T_A}##. In your first step, ##\frac {( 2h)(c)} { \sqrt {(2h)( v^2 - c^2)} } ## , I guess it is a typo that the 2h in the denominator is inside the square root. But your final ## \frac { \sqrt {( v^2 - c^2)} } {c} ## is inverted.

I don’t understand what you mean by "The problem I then encounter is I have equal amount of the c variables." Where were you hoping to go from there?

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What I mean is that ## \frac {c} {c } = 1 ## .haruspex said:I don’t understand what you mean by "The problem I then encounter is I have equal amount of the c variables." Where were you hoping to go from there?

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 algebra rule 4 the right hand side becomes the left side by using c. https://algebrarules.com/

## \frac {T_B} {T_A} = \frac { (c)(1)} {\sqrt { (c^2 - v^2)} } ##

part 4

The c -v should be square rooted but due to formating issues I left it I am in a rush.

## \frac {T_B} {T_A} = \frac{\frac{1}{c -v}}{\frac{1} {c} }##

part 5

I left the 1 at the bottom due to troubles with formatting.

## \frac {T_B} {T_A} = \frac{ \frac {1} {\sqrt {c^2-v^2}} } {\frac{c} {1} } ##

I fixed it finally.

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rgtr said:part 1

## \frac {T_B} {T_A} = \frac{\frac{2h}{c}}{\frac{ 2h} {c^2-v^2}} ##

## ##

You've lost a square root, and it seems inverted. In post #7 you wrote

I see you now have the more reasonable ##c^2-v^2## so that becomesrgtr said:I am trying to solve ##T_B = \frac {2h} {\sqrt {v^2 - c^2} } ## ,##T_A = \frac {2h} {c} ##

##T_B = \frac {2h} {\sqrt {c^2 - v^2} } ## ,##T_A = \frac {2h} {c} ##

From that I get ## \frac {T_B} {T_A} =\frac{\frac{2h}{\sqrt{c^2-v^2}}}{\frac{ 2h} {c}} =\frac c{\sqrt{c^2-v^2}}##.

## ##

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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.

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This is where you lose me. What do you mean that you can cancel the c's?rgtr said:I can also cancels out the c's.

You can do ##\frac c{\sqrt{c^2-v^2}}=\frac 1{\sqrt{1-(\frac vc)^2}}##.

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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 relativity I have a stationary light clock and a moving light clock. Why did I divide them and not minus the two frames?

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 relativity I have a stationary light clock and a moving light clock. Why did I divide them and not minus the two frames?

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rgtr said: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.

Dividing top and bottom by c:

##\frac c{\sqrt{c^2-v^2}}=\frac 1{\frac 1c\sqrt{c^2-v^2}}##

Moving the 1/c inside the square root:

##=\frac 1{\sqrt{\frac 1{c^2}(c^2-v^2)}}##

##=\frac 1{\sqrt{1-(\frac vc)^2}}##.

- #24

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Please use the more standard "RHS" and "LHS" for designating the two sides of an equation. Thank you kindly.rgtr said:RSOTE = right side of the equation.

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Because special relativity laws imply ##\frac{T_B}{T_A}=\gamma## and not ##T_B-T_A=\gamma##? Anyway post this question in the relativity forum so that the experts there can give more detailed and in depth answers.rgtr said:I just have 1 question in special relativity I have a stationary light clock and a moving light clock. Why did I divide them and not minus the two frames?

EDIT: You could 've taken ##T_B-T_A## and end up with ##T_B-T_A=(\gamma -1)T_A##

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rgtr said:I guess it is kind of self explained. Thanks I think the title needs a change.|

I just have 1 question in special relativity I have a stationary light clock and a moving light clock. Why did I divide them and not minus the two frames?

Delta2 said:Anyway post this question in the relativity forum so that the experts there can give more detailed and in depth answers.

No, @rgtr please send me a PM (private message -- click my avatar and "Start a Conversation") to explain your question. I will help you formulate it better and get it ready for a new thread start in the appropriate forum. Thank you.

This thread is closed for now.

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