I am having trouble solving this equation

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  • #1
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Homework Statement:

I am having trouble solving this equation?

Relevant Equations:

## \frac { {1 - \frac {v^2} {c^2}}} {\sqrt {1 - \frac {v^2} {c^2}}} ##
## \frac {\sqrt {1 - \frac {v^2} {c^2}}} { {1 - \frac {v^2} {c^2}}} ## I got the equation wrong this is the correct equation.
 

Answers and Replies

  • #2
berkeman
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Homework Statement:: I am having trouble solving this equation?
Relevant Equations:: ## \frac { {1 - \frac {v^2} {c^2}}} {\sqrt {1 - \frac {v^2} {c^2}}} ##

## \frac {\sqrt {1 - \frac {v^2} {c^2}}} { {1 - \frac {v^2} {c^2}}} ## I got the equation wrong this is the correct equation.
That's an expression, not an equation (an equation has an "=" sign). Are you asked to simplify the expression? Subject to any conditions?
 
  • #3
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Homework Statement:: I am having trouble solving this equation?
Relevant Equations:: ## \frac { {1 - \frac {v^2} {c^2}}} {\sqrt {1 - \frac {v^2} {c^2}}} ##

## \frac {\sqrt {1 - \frac {v^2} {c^2}}} { {1 - \frac {v^2} {c^2}}} ## I got the equation wrong this is the correct equation.
As @berkeman said, this is an expression, not an equation. You can't "solve" an expression. Most likely the instructions for the problem ask you to simplify this expression.

Hint:
$$\frac{\sqrt x}{x} = \frac 1 {\sqrt x}$$
 
  • #4
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is it possible from the expression

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

to get

## \frac 1 {\sqrt {1 - \frac {v^2} {c^2}}} ##
 
  • #5
phinds
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is it possible from the expression

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

to get

## \frac 1 {\sqrt {1 - \frac {v^2} {c^2}}} ##
Have you made any attempt to do so? Our goal is to help you learn how to do things, no to do them for you.

EDIT: also, did you read post #3?
 
  • #6
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I don't know where to start.
 
  • #7
berkeman
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Homework Statement:: I am having trouble solving this equation?
Relevant Equations:: ## \frac { {1 - \frac {v^2} {c^2}}} {\sqrt {1 - \frac {v^2} {c^2}}} ##

## \frac {\sqrt {1 - \frac {v^2} {c^2}}} { {1 - \frac {v^2} {c^2}}} ## I got the equation wrong this is the correct equation.
I don't know where to start.
But you haven't responded to our questions about what you are supposed to do. You've already been shown how to simplify the expression some, do you understand that? And unless you post the full text of the problem statement that you were given, there's not much more that we can do. Did your problem statement say "solve this equation" even though it is just an expression? If it said to solve it, did it say what to "solve" it for?
 
  • #8
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Sorry my fault. Its not a problem statement. I Just made up this example. My goal was to take the

## x_A = gamma (x_B - vt_B) ## and get ## x_B = gamma (x_A - vt_A) ##

To get this I was told to combine ## x_A = gamma (x_b - vt_b) ## , ## t_A = gamma (t_B - \frac{ vx} {c^2}) ##

Maybe this is bad advice. Eventually I get ## \frac {\sqrt {1 - \frac {v^2} {c^2}}} { {1 - \frac {v^2} {c^2}}} ## along with some other terms.

This is only 2 dimensions of a Lorentz transform.
 
  • #9
phinds
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You have not answered the question " You've already been shown how to simplify the expression some, do you understand that? " You HAVE been shown. DO you understand that? If not, just say so, but please make some effort.
 
  • #10
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OK sorry. Is it okay if i post an image from imgur? Or in the future do you prefer latex math?

Would something like this image work?
 
Last edited:
  • #11
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Is it okay if i post an image from imgur?
It might be OK if your writing was legible.
Or in the future do you prefer latex math?
Yes, very much
Would something like this image work?
No, not at all. It looks like you are saying that ##\sqrt{1 - \frac{v^2}{c^2}} = 1##, which would be true only if v = 0.
I'm having a hard time understanding how you can be studying Lorentz transforms, but are unable to simplify an expression that is at roughly the 9th grade algebra level.

$$\frac{\sqrt x} x = \frac{\sqrt x}{(\sqrt x)^2}$$
I'm hoping you can simplify that. If not, you really should brush up on plain old algebra.
 
  • #12
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I am a little rusty and have forgot a lot of stuff. Is that Okay?

I always thought that if you had a variable, an example could be, ##x## and wanted to get the other side of the equation and there is nothing there you just add a ##1## to the other side. Ex ##x = 1##
 
Last edited:
  • #13
phinds
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I always thought that if you had a variable, an example could be, ##x## and wanted to get the other side of the equation and there is nothing there you just add a ##1## to the other side. Ex ##x = 1##
OK, you thought something completely wrong. That happens. Now reread post #11 and act on it.
 
  • #14
PeroK
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is it possible from the expression

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

to get

## \frac 1 {\sqrt {1 - \frac {v^2} {c^2}}} ##
I don't want to put you off, but undergraduate level physics is going to be tough/impossible if you don't see that those two expressions are obviously equal. It's not a question of whether it is possible to get from one to the other, but you should see getting from one to the other as an obvious single algebraic step.

I am a little rusty and have forgot a lot of stuff. Is that Okay?

I always thought that if you had a variable, an example could be, ##x## and wanted to get the other side of the equation and there is nothing there you just add a ##1## to the other side. Ex ##x = 1##
You need some serious mathematics revision before you can tackle SR. You're not just a little rusty.
 
  • #15
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You need some serious mathematics revision before you can tackle SR. You're not just a little rusty.
Amen to that!
 

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