Ok so i have the final test in my physics class tomorrow and my brain has just been freezing up all day. I can't even make this work, so now I'm asking for your help.

From this equation

I need to get what gamma equals. My own attempts so far has failed since gamma can't be 1 or lower for the 2nd equation to work.

This is what I've done so far, and it's not correct just thought I'd share what I've tried:

I know that's not right since that would mean that

So could someone PLEASE help me get gamma out of that equation. I know its really fucking simple but I feel like a dumbass.

(if you wanna test it out m = 9,1094*10^-31, c = 3*10^8 and E = 3,98*10^-17)

crate

United States2474 Posts

May 20 2010 22:42 GMT

On May 21 2010 07:38 Julmust wrote:
This is what I've done so far, and it's not correct just thought I'd share what I've tried:

I know that's not right since that would mean that

Your second equation here does not follow from your first. You messed up the algebra.

You have g = (E + mc^2)/(mc^2)
So g = E/(mc^2) + 1

Assault_1

Canada1950 Posts

May 20 2010 22:42 GMT

gamma=(E+mc^2)/(mc^2)

how do u get (E+mc^2)/(mc^2) = E

o_o

(E+mc^2)/(mc^2) = (E/(mc^2)) + 1

Impervious

Canada4211 Posts

May 20 2010 22:43 GMT

E/mc^2 + mc^2/mc^2 = gamma is your mistake, I believe.

tYsopz

Norway215 Posts

May 20 2010 22:44 GMT

What you got here is the formula for "total energy" and "mass energy". Total energy = mass energy + kinetic energy. As a result, kinetic energy = total energy - mass energy, which is what you are doing here.

E = ymc^2-mc^2
=>
E = mc^2(y-1)
=>
y = E/mc^2+1

y being gamma, idk how to make the symbol in windows.

dont do what the above poster did, you can use that formula as well to get the answer, but that's not how the "official" formula is, the relativistic formula for kinetic energy is: Ek = (y-1)m^2)
If you don't get to the official formula, chances are your answer will be deemed incorrect.

rushz0rz

Canada5300 Posts

May 20 2010 22:54 GMT

go here: http://demonstrations.wolfram.com/FundamentalLawOfFractions/

crate

United States2474 Posts

May 20 2010 22:55 GMT

As an aside, I'm not sure where your first equation for E is coming from.

Relativistic energy is E^2 = (pc)^2 + (mc^2)^2; this does not simplify to E = (gamma)mc^2 - mc^2 unless I'm missing something.

Julmust

Sweden4867 Posts

May 20 2010 22:55 GMT

thanks for the help guys, gonna need lots of luck tomorrow (or rather later today) if this is gonna work out.

Jonoman92

United States9104 Posts

May 20 2010 22:57 GMT

Factor out the mc^2:
E=mc^2(y-1)

Divide both sides:
E/(mc^2)=y-1

Add one to both sides:
(E/(mc^2))+1=y

tYsopz

Norway215 Posts

May 20 2010 23:00 GMT

#10

Formula for total relativistic energy is:
Etot = ymc^2
Formula for relativistic kinetic energy is:
Ek = (y-1)mc^2
Formula for mass energy / ("hvileenergi" in norwegian)
E0 = mc^2

E0 + Ek = Etot
if that wasn't clear.

tYsopz

Norway215 Posts

May 20 2010 23:07 GMT

#11

And crate, it does:
E^2 = (pc)^2 + (mc^2)^2
=>
E^2 = (γmc^2)^2 + (mc^2)^2
=>
E = (γmc^2) + (mc^2)
=>
E=(γ-1)mc^2

crate

United States2474 Posts

May 20 2010 23:08 GMT

#12

On May 21 2010 08:07 tYsopz wrote:
E^2 = (γmc^2)^2 + (mc^2)^2
=>
E = (γmc^2) + (mc^2)

This is wrong.
a^2 = b^2 + c^2
a != b + c

edit: your last equation also doesn't follow, it should be (γ+1)(mc^2) instead of (γ-1)(mc^2) as you've written.

If the original equation is referring to relativistic kinetic energy as E it's correct I think, but I dislike calling that E instead of E_k or T since it's misleading.

tYsopz

Norway215 Posts

May 20 2010 23:09 GMT

#13

that is awkward, yeah i was a bit too quick ><

Edit:
you know how it is when you know what it is supposed to look like :p

You sure about that formula, never seen it?

Edit: also, my formula for relativistic momentum is wrong.

crate

United States2474 Posts

May 20 2010 23:14 GMT

#14

On May 21 2010 08:09 tYsopz wrote:
that is awkward, yeah i was a bit too quick ><

Edit:
you know how it is when you know what it is supposed to look like :p

You sure about that formula, never seen it?

Relativistic total energy is E = (pc)^2 + (mc^2)^2, yes. p is relativistic momentum: p = γmv

http://en.wikipedia.org/wiki/Mass_in_special_relativity#The_relativistic_energy-momentum_equation

tYsopz

Norway215 Posts

May 20 2010 23:17 GMT

#15

Since the formula is correct, you should be able to get to E = mc^2(y-1) from E^2 = (pc)^2 + (mc^2)^2

unless the fact it is 4 hours past my bedtime is putting a stopper to my ability to think.

Chill

Calgary25989 Posts

May 20 2010 23:36 GMT

#16

i think your problem is basic math, not physics.

Edit: To be clearer, the request doesn't even make sense. How are we going to just "get rid" of a variable?

HeavOnEarth

United States7087 Posts

May 20 2010 23:43 GMT

#17

Um,.. all i can say is good luck taking calculus(and harder physics), if you aren't comfortable with simple fractions.
@ Chill , he meant "out of the equation", instead of "get out"
So like,
"help me get gamma, out of that equation."
Also -insert obligatory pout about homework threads in a very biased manner-

Julmust

Sweden4867 Posts

May 21 2010 00:02 GMT

#18

On May 21 2010 08:36 Chill wrote:
i think your problem is basic math, not physics.

my problem was that my brain wasn't functioning correctly, I solved the problem (which, as it shows, had nothing to do with this equation

) and now it's back online.

tYsopz

Norway215 Posts

May 21 2010 00:04 GMT

#19

E^2=(pc)^2 + (mc^2)^2
=>
E^2 = (γmvc)^2 + m^2c^4
=>
E^2 = γ^2m^2v^2c^2 + m^2c^4
=>
E^2 = (1/(1-(v^2/c^2)))m^2v^2c^2 + m^2c^4
=>
E^2 = (1/((c^2-v^2)/c^2)))*m^2v^2c^2 + m^2c^4
=>
E^2 = (c^2/(c^2-v^2)*m^2v^2c^2 + m^2c^4
=>
E^2 = m^2c^4(v^2/(c^2-v^2)+1)
=>
E^2 = m^2c^4(v^2+c^2-v^2)/(c^2-v^2)
=>
E^2 = m^2c^4(c^2/(c^2-v^2))
=>
E^2 = m^2c^4(1/(c^2-v^2)/c^2)
=>
E^2 = m^2c^4(1/(1-v^2/c^2))
=>
E^2 = m^2c^4γ^2
=>
E = γmc^2

Uggh, should have used my scanner.

Darby.mcg

United States16 Posts

May 21 2010 00:29 GMT

#20

I'm going to college next year, will these sorts of things become useful in the near or far future if I'm going into physics?

@tYsopz: I'm going to assume that y^2 = 1/(1-v^2/c^2)

FiBsTeR

United States415 Posts

May 21 2010 00:34 GMT

#21

On May 21 2010 09:29 Darby.mcg wrote:
I'm going to college next year, will these sorts of things become useful in the near or far future if I'm going into physics?

It depends on the structure of the course, but I think if you can do the math, you can do the physics. But the converse might not be true... yeah I'm a math major.

tYsopz

Norway215 Posts

May 21 2010 00:41 GMT

#22

The math/algebra in physics isn't too hard, but there are certain things you just need to be aware of or you won't be able to transform equations. For example you have to know that 1/(1-v^2/c^2) = c^2/(c^2-v^2). It is easy to see, but you have to know you need to do it, else you might do countless other transformations that won't really get you closer to the equation you want. There's a lot of problems just like this in theory of relativity that you need to be familiar with, since they potentially can take a loooot of time if you are unlucky with how you approach them. Actually ran into a problem at my exam that i didn't have time to complete, and it might have cost me the A+.

(I'm of course referring to the transformation of E^2=(pc)^2 + (mc^2)^2 => E = γmc^2)

and yeah y^2 = 1/(1-v^2/c^2), seeing as y = 1/(sqrt(1-v^2/c^2)) (lorentz factor)

Darby.mcg

United States16 Posts

May 21 2010 02:12 GMT

#23

Ok, thanks, I'm currently one of the best people in all of my math science courses, and love it so far, the only thing I'm sad is that I couldn't go higher, my HS doesn't offer AP classes. That's what college is supposed to be for though! hopefully it wont totally suck.

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