can't_think_of_a_name 0 Posted October 25 (edited) https://en.wikipedia.org/wiki/Lorentz_factor Just scroll a tiny bit to find what I am referring to. Edited October 25 by can't_think_of_a_name 0 Share this post Link to post Share on other sites

swansont 7365 Posted October 25 Could you be more specific? I don't see a use of the reciprocal for speed. I see a reciprocal for gamma mentioned, which they call alpha. Is that what you're referring to? The only mention of "reciprocal" is in this context. 0 Share this post Link to post Share on other sites

Endy0816 447 Posted October 25 (edited) Going based off the table in the numerical values section; one is v/c and the other the multiplicative inverse or reciprocal of the lorentz factor. v is always less than c, so for v/c you end up with: 0 ≤ β < 1 Now for the reciprocal of the lorentz factor you're doing the equivalent of finding the length of one side of a square with an area equal to the shaded section below. Sqrt(1^2 - β^2) Edited October 25 by Endy0816 0 Share this post Link to post Share on other sites

can't_think_of_a_name 0 Posted October 25 6 hours ago, swansont said: Could you be more specific? I don't see a use of the reciprocal for speed. I see a reciprocal for gamma mentioned, which they call alpha. Is that what you're referring to? The only mention of "reciprocal" is in this context. I start with gamma. The part that confuses me is that I want velocity = , from gamma . I get v = c/gamma.(I could have made a math mistake.) It never says I need B = v/c. I know B isn't the correct symbol I am just using it here. 0 Share this post Link to post Share on other sites

swansont 7365 Posted October 25 48 minutes ago, can't_think_of_a_name said: I start with gamma. The part that confuses me is that I want velocity = , from gamma . I get v = c/gamma.(I could have made a math mistake.) It never says I need B = v/c. I know B isn't the correct symbol I am just using it here. Invert gamma (which is alpha) and square it. Rearrange v^2/c^2 = 1 - alpha^2 Take a square root 0 Share this post Link to post Share on other sites

can't_think_of_a_name 0 Posted October 25 I know I got v = gamma/c. Like stated earlier why does b= v/c give a different answer. I was solving a problem problem 3 A. I used V = c/gamma but they want b = c/v. This confuses me why does one works and not the other? I guess I really didn't explain this well. In my course they never explain the difference. If the link doesn't work I will post the question. From here https://d3c33hcgiwev3.cloudfront.net/_d37cb29a797de375eb7866e695098a79_Wk7_problemsetsolutions.pdf?Expires=1603756800&Signature=SUmoIwTy2VAIs1CfSGO~F7C3BD7lIJKyMQJuedeO4RUqiCeo9HTLJk50r~oKjI6pAFoaSG5p-Pu3FRHuPiNizcGdd6EdCj4Eer1tU4BqbIBsdzW0WjLXxR8E~-5gZx3LDteO6L4ruR80eOQi1EzUdBj8Z9lnXJ6kQuj1fHpJGik_&Key-Pair-Id=APKAJLTNE6QMUY6HBC5A 0 Share this post Link to post Share on other sites

can't_think_of_a_name 0 Posted October 26 I meant v = c/ gamma 0 Share this post Link to post Share on other sites

joigus 281 Posted October 26 6 hours ago, can't_think_of_a_name said: I meant v = c/ gamma No. It's, \[\gamma^{2}\left(1-\beta^{2}\right)=1\] Gamma is a number always bigger than one. Beta is a number always less than one (in absolute value.) The absolute value of beta determines gamma. 0 Share this post Link to post Share on other sites

swansont 7365 Posted October 26 9 hours ago, can't_think_of_a_name said: I meant v = c/ gamma What speed does this apply to? I don't recognize this as a valid equation. 0 Share this post Link to post Share on other sites

can't_think_of_a_name 0 Posted October 30 I apologize for asking basic question I guess when I was originally in school either I didn't pay attention or wasn't taught properly. I assume the former but I can't be certain. I appreciate the help on basic questions. why is this wrong? Gamma = 1 / √ (1) - v^2 / c^2 = 1 / Gamma = √ 1 - v^2 c^2/ c^2 = c / Gamma = √ 1 - v^2 +v^2 = c + v / gamma = √ 1 = switch the c + v to v + c = v + c / gamma = √ 1= v + c / gamma = √ 1 = v + c gamma / gamma = √ 1 (gamma^2) = v + c -c = √ gamma^2 = v = √ gamma^2 - c or 1 / Gamma - c^2 = √ 1 - v^2 / c^2 - c^2 = 1 / Gamma - c^2 = √ 1 - v^2 = 1 + v / Gamma - c = √ 1 - v^2 + v^2 = 1 +v / Gamma - c = √ 1 =switch the 1+ v to v + 1 = v + 1 / gamma - c = √ 1= v + 1 gamma / gamma - c = √ 1 (gamma) = (v + 1) -c / - c = √ gamma -c^2 = - (v+1) = √ gamma -c^2 = -v -1 = gamma -c^2 = -v = √ gamma^2 - c^2 + 1^2 = - v / - = √ gamma^2 - c^2 + 1^2/ - = v = √ gamma^2 - c^2 + 1^2 / - = v = √ -gamma^2 + c^2 - 1^2 0 Share this post Link to post Share on other sites

Janus 1077 Posted October 30 46 minutes ago, can't_think_of_a_name said: I apologize for asking basic question I guess when I was originally in school either I didn't pay attention or wasn't taught properly. I assume the former but I can't be certain. I appreciate the help on basic questions. why is this wrong? Gamma = 1 / √ (1) - v^2 / c^2 = 1 / Gamma = √ 1 - v^2 c^2/ c^2 = c / Gamma = √ 1 - v^2 +v^2 = c + v / gamma = √ 1 = switch the c + v to v + c = v + c / gamma = √ 1= v + c / gamma = √ 1 = v + c gamma / gamma = √ 1 (gamma^2) = v + c -c = √ gamma^2 = v = √ gamma^2 - c or 1 / Gamma - c^2 = √ 1 - v^2 / c^2 - c^2 = 1 / Gamma - c^2 = √ 1 - v^2 = 1 + v / Gamma - c = √ 1 - v^2 + v^2 = 1 +v / Gamma - c = √ 1 =switch the 1+ v to v + 1 = v + 1 / gamma - c = √ 1= v + 1 gamma / gamma - c = √ 1 (gamma) = (v + 1) -c / - c = √ gamma -c^2 = - (v+1) = √ gamma -c^2 = -v -1 = gamma -c^2 = -v = √ gamma^2 - c^2 + 1^2 = - v / - = √ gamma^2 - c^2 + 1^2/ - = v = √ gamma^2 - c^2 + 1^2 / - = v = √ -gamma^2 + c^2 - 1^2 To start off. 1 / Gamma = √ 1 - v^2 c^2/ c^2 = c / Gamma = √ 1 - v^2 +v^2 You have to multiply all the factors under the radical by c^2 if you want to move c to the other side of the equation This leaves c/Gamma =√ (c^2 - v^2) And since c^2 ≠ 1+v^2 you can't get to where you got. And at the end, your answer is not a multiple of c, so that right there should have been a tip-off that you did something wrong along the way. To solve for v from 1 / Gamma = √ 1 - v^2/c^2 You first square both sides: 1/Gamma^2 = 1- v^2/c^2 (you square both the 1 and gamma, but since 1^2 = 1...) v^2/c^2 = 1-1/gamma^2 v^2 = c^2(1-1/gamma^2) take the square root of both sides: v= c√(1-1/gamma^2) Thus if v = 0.6c Then Gamma = 1/√(1- 0.6c^2/c^2) = 1.25 and v = c√(1-1/1.25^2) = 0.6c 1 Share this post Link to post Share on other sites

can't_think_of_a_name 0 Posted October 30 (edited) 1 / Gamma = √ (1 - v^2) c^2/ c^2 is the same as 1 / Gamma = √ (c^2 - v^2) c^2/ c^2 shouldn't this give c / Gamma = √ 1 - v^2 if I do it my way. Or is that just not how algebra works? Edited October 30 by can't_think_of_a_name 0 Share this post Link to post Share on other sites

Janus 1077 Posted October 30 (edited) 1-v^2/c^2 is not the same as (1-v^2)/c^2 So for example, again using v= 0.6c 1- (0.6c)^2/c^2 = 1- 0.6^2 = 0.64 but (1-(0.6c)^2)/c^2 = 1/c^2- 0.6^2 = 1/c^2 - 0.36 1-v^2/c^2 = (1-v^2)/c^2 is like saying (1-1/2) = (1-1)/2 but solving the left side gives 0.5 and solving the right side gives 0 Edited October 30 by Janus 0 Share this post Link to post Share on other sites

can't_think_of_a_name 0 Posted October 30 Hopefully the last question about basic math. Does √1 = 1^-1. Also I can go √1^2 = 1. Are there any other ways to remove a square root? 0 Share this post Link to post Share on other sites

can't_think_of_a_name 0 Posted October 31 I also have one more question if I have c+t =g. If I want to move g to the left of the = sign. I don't care how pointless this is. I go c+t-g = 0? 0 Share this post Link to post Share on other sites

joigus 281 Posted October 31 8 hours ago, can't_think_of_a_name said: Hopefully the last question about basic math. Does √1 = 1^-1. Also I can go √1^2 = 1. Are there any other ways to remove a square root? Number 1 is a bit misleading, because \( \sqrt{1} = 1 \). Generally there are two basic ways of undoing a square root. One is squaring a root; e.g., \[\sqrt{a}=2\] which gives, \[a=4\] and the other is the one you suggest --rooting a square--, but with that one you must be careful: \[\sqrt{a^{2}}=4\] which gives, \[a=\pm4\] Another possible way to get square roots out of the way is to remember that sums times differences give differences of squares. As in, \[\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=\sqrt{a}^{2}-\sqrt{b}^{2}=a-b\] You can prove quite amazing identities with this: \[\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}+\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}=\frac{2a}{a-b}\] \[\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}=-\frac{2b}{a-b}\] There's almost no end to fun with square roots! 3 hours ago, can't_think_of_a_name said: I also have one more question if I have c+t =g. If I want to move g to the left of the = sign. I don't care how pointless this is. I go c+t-g = 0? Exactly. 1 Share this post Link to post Share on other sites