Question on four momentum squared

[bonbon22]

Homework Statement

Why does the four momentum squared lead to simply m^2*c^2

Relevant Equations

P (mc,mVx,mVy,mVz) <---- components of four vector where v is velocity in x y z directions
mc*mc - (three vector multiplied and added with corresponding parts)
ϒ = gamma factor = lorentz factor = (1/1-v[SUP]2[/SUP]/c[SUP]2[/SUP])[SUP]1/2[/SUP]
google lorentz factor if it looks confusing.
P ⋅ P/[p][p] =cos(θ) <----- three scalar product rule a level maths stuff

So i have taken a beginner course on relativity, first year physics student. I am confused as to why four momentum squared simply gives
m²* c²*ϒ² -(three vector multiplied and added with corresponding parts) *ϒ²
so as the three vector part which is being subtracted, is the same as - (P ⋅ P) *ϒ², a normal three dot product, which is the same as the - [p][p]cos(θ)*ϒ²
,but what would cos(θ) be ? If it is zero then we have an actual value for [p][p]cos(θ) = [p][p] which then i don't get how
m²* c²*ϒ² - [p][p]cos(θ)*ϒ² = is somehow simply m²* c²*ϒ²?

finally in many questions i have come across they seem to ignore fully the *ϒ²
(1/1-v²/c²)^1/2 factor when doing conservation of four momentum problems. Where P1 +P1 =P3
where P1, P2,P3 are four momentums with the four components. Squaring both sides i get P1² + P2² + P1*P2 = P3² which ends up being m1²c² +m2²c² plus four vector product = m3²c²[/SUP][/SUP]

 

[Ibix]

Your notation is somewhat hard to read, but you do not appear to me to have quoted the four momentum correctly. It is ##P=(\gamma mc,\gamma m\vec v)##, where ##m## is the rest mass and ##\vec v## is shorthand for the three spatial components of velocity. The answer you are looking for follows directly.

 

[bonbon22]

Your notation is somewhat hard to read, but you do not appear to me to have quoted the four momentum correctly. It is ##P=(\gamma mc,\gamma m\vec v)##, where ##m## is the rest mass and ##\vec v## is shorthand for the three spatial components of velocity. The answer you are looking for follows directly.

Exactly thank you that does look better I am new to using the symbols my mistake.

So squaring a four momentum vector with itself gives (##\gamma mc## )² subtracted with the ##\gamma##²## m ##²##\vec v## *##\vec v##

where the ##\vec v## *##\vec v## can be written as the [##\vec v##][##\vec v##]cos(θ)

P² = (##\gamma mc## )² - ##\gamma##²## m ##²##\vec v## *##\vec v##
or
P² = (##\gamma mc## )² - ##\gamma##²## m ##²[##\vec v##][##\vec v##]cos(θ)

im guessing as the the two four vectors are in the same direction the angle is zero the cos(θ) is then 1 simply so i don't simply get an answer of (##\gamma mc## )²

that was my question how did they arrive at the answer of (##\gamma mc## )².

 

[Ibix]

im guessing as the the two four vectors are in the same direction the angle is zero the cos(θ) is then 1 simply so i don't simply get an answer of (γmc\gamma mc )2

that was my question how did they arrive at the answer of (γmc\gamma mc )2.

You are correct that the ##\theta## is the angle between the vectors, and the angle between a vector and itself is, indeed, zero.

I'm not sure what you are expecting to be equal to ##(\gamma mc) ^2##. The modulus of the four momentum is just ##(mc) ^2##. The modulus of the four momentum (and any four vector) is invariant, and cannot depend on a frame-dependant quantity like ##\gamma##.

 

[PeroK]

Exactly thank you that does look better I am new to using the symbols my mistake.

So squaring a four momentum vector with itself gives (##\gamma mc## )² subtracted with the ##\gamma##²## m ##²##\vec v## *##\vec v##

where the ##\vec v## *##\vec v## can be written as the [##\vec v##][##\vec v##]cos(θ)

P² = (##\gamma mc## )² - ##\gamma##²## m ##²##\vec v## *##\vec v##
or
P² = (##\gamma mc## )² - ##\gamma##²## m ##²[##\vec v##][##\vec v##]cos(θ)

im guessing as the the two four vectors are in the same direction the angle is zero the cos(θ) is then 1 simply so i don't simply get an answer of (##\gamma mc## )²

that was my question how did they arrive at the answer of (##\gamma mc## )².

Remember that ##\gamma^2 = \frac{1}{1- v^2/c^2}##.

Or, sometimes more useful is:

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

 

[bonbon22]

Remember that ##\gamma^2 = \frac{1}{1- v^2/c^2}##.

Or, sometimes more useful is:

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

i see, but how does that link in with the problem?

 

[Orodruin]

i see, but how does that link in with the problem?

It simplifies the expression to ##m^2 c^2##, which is the correct result.

 

[bonbon22]

You are correct that the ##\theta## is the angle between the vectors, and the angle between a vector and itself is, indeed, zero.

I'm not sure what you are expecting to be equal to ##(\gamma mc) ^2##. The modulus of the four momentum is just ##(mc) ^2##. The modulus of the four momentum (and any four vector) is invariant, and cannot depend on a frame-dependant quantity like ##\gamma##.

I see, i understand so in this case do we just ignore the gamma factor? Can we only ignore this factor when gamma is equal to 1 for that to be true wouldn't the velocity be equal to C?

 

[Orodruin]

I see, i understand so in this case do we just ignore the gamma factor?

No. But cos(0)=1.

 

[PeroK]

I see, i understand so in this case do we just ignore the gamma factor? Can we only ignore this factor when gamma is equal to 1 for that to be true wouldn't the velocity be equal to C?

##\gamma = 1## when ##v = 0##.

No one is ignoring the gamma factor. Have you heard of algebra or the concept of simplifying an expression?

What level of mathematics have you studied?

 

[Ibix]

I see, i understand so in this case do we just ignore the gamma factor? Can we only ignore this factor when gamma is equal to 1 for that to be true wouldn't the velocity be equal to C?

No. You wrote:

P² = (##\gamma mc## )² - ##\gamma##²## m ##²##\vec v## *##\vec v##

(incidentally, the * should be a dot) and we've just agreed that this is ##(\gamma mc)^2-\gamma^2 m^2v^2## because the dot product of a 3-vector with itself is just its length squared. Getting to the correct result is just algebra from here (hint: ##\gamma=1/\sqrt{1-v^2/c^2}=c/\sqrt{c^2-v^2}##). As @PeroK asks, how much algebra do you know?

 

[bonbon22]

##\gamma = 1## when ##v = 0##.

No one is ignoring the gamma factor. Have you heard of algebra or the concept of simplifying an expression?

What level of mathematics have you studied?

Why are you tryn diss me mate you feelin alright. I understand the expression but why would you write v=o when you square a four momentum. If you read my post you would know what level I am studying at.

 

[Ibix]

Why are you tryn diss me mate you feelin alright. I understand the expression but why would you write v=o when you square a four momentum.

You wouldn't. Nobody is doing this, as has been noted several times.

As has also been said a couple of times, all you need to do is work through the algebra and you will find the answer you are expecting. How far have you got with this?

 

[bonbon22]

You wouldn't. Nobody is doing this, as has been noted several times.

As has also been said a couple of times, all you need to do is work through the algebra and you will find the answer you are expecting. How far have you got with this?

I just realized my mistake apologies, for making v = to c and actually solving the problem cheers you two. @Ibix @PeroK

 

[Ibix]

I just realized my mistake apologies, for making v = to c and actually solving the problem cheers you two. @Ibix @PeroK

...um, you don't need to make ##v## equal to ##c##. The ##\gamma## drops out for any ##v##.

 

[bonbon22]

...um, you don't need to make ##v## equal to ##c##. The ##\gamma## drops out for any ##v##.

no i understand that replying to what @PeroK said i wrongly made v = to c to make gamma = to 1 by mistake. Wait why does it drop out for any v.

 

[vela]

Wait why does it drop out for any v.

Based on what you've learned in this thread, why don't you post what you have so far? I don't think it's clear to any of us here what you're getting stuck on.