Is Removing c from the Square Root in This Equation Correct?

[rgtr]

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 $r = \sqrt { {c} + {x} } {c}$ ?

 

[Delta2]

No they these two expressions for $r$ are not necessarily equal. For example c=2, x=1, the first gives $r=2+1=3$ while the second $r=\sqrt{2+2}=2$.

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

 

[rgtr]

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

 

[Delta2]

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.

 

[rgtr]

okay thanks

 

[jbriggs444]

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

What are you really trying to do?

 

[rgtr]

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?

 

[jbriggs444]

I am trying to solve $T_B = \frac {2h} {\sqrt {v^2 - c^2} }$ ,$T_A = \frac {2H} {c}$

You are solving for $T_B$, $T_A$, $h$, $H$ or $v$?

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?

 

[rgtr]

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.

 

[Ameyzing73]

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

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

 

[Delta2]

Are the $h,H$ the same thing?

 

[SammyS]

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?

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.

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

 

[Ameyzing73]

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

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

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

 

[SammyS]

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

Neither of your posts are correct.

Please, just let @jbriggs444 make a reply.

 

[rgtr]

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

 

[Delta2]

you mean this $$\frac{\frac{1}{1}}{\frac{1}{1}}$$?

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

 

[haruspex]

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?

First, as noted, you probably mean $\sqrt{c^2-v^2}$.
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?

 

[rgtr]

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?

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 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.

 

[haruspex]

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 am trying to solve $T_B = \frac {2h} {\sqrt {v^2 - c^2} }$ ,$T_A = \frac {2h} {c}$

I see you now have the more reasonable $c^2-v^2$ so that becomes
$T_B = \frac {2h} {\sqrt {c^2 - v^2} }$ ,$T_A = \frac {2h} {c}$

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

 

[rgtr]

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.

 

[haruspex]

I can also cancels out the c's.

This is where you lose me. What do you mean that you can cancel the c's?
You can do $\frac c{\sqrt{c^2-v^2}}=\frac 1{\sqrt{1-(\frac vc)^2}}$.

 

[rgtr]

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?

 

[haruspex]

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:
$=\frac 1{\sqrt{\frac 1{c^2}(c^2-v^2)}}$
$=\frac 1{\sqrt{1-(\frac vc)^2}}$.

 

[berkeman]

RSOTE = right side of the equation.

Please use the more standard "RHS" and "LHS" for designating the two sides of an equation. Thank you kindly.

 

[Delta2]

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?

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.

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

 

[berkeman]

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?

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.