What is the gamma factor in a particle's rest frame?

[dyn]

Hi.
If the 4-momentum in SR is given by p = m γ(v) ( 1 , v ) then in the rest frame of a particle the 4-momentum is ( m , 0 , 0 , 0 ) using c=1 units. This uses the fact that because v = 0 then γ(v) = 1.
I'm confused about this last sentence. The gamma factor is used for the relative velocity between 2 inertial frames but in the rest frame the particle is at rest relative to every other inertial frame so how can we specify a gamma factor ?

 

[sweet springs]

v is velocity of a body in your IFR. v=0 means the body is at rest in your IFR.

 

[PeroK]

Hi.
If the 4-momentum in SR is given by p = m γ(v) ( 1 , v ) then in the rest frame of a particle the 4-momentum is ( m , 0 , 0 , 0 ) using c=1 units. This uses the fact that because v = 0 then γ(v) = 1.
I'm confused about this last sentence. The gamma factor is used for the relative velocity between 2 inertial frames but in the rest frame the particle is at rest relative to every other inertial frame so how can we specify a gamma factor ?

I think you may be confusing the gamma factor that connects two IRF's with the gamma factor of a particle. In fact, if you have a particle and two IRF's then there are three gamma factors.

Suppose the frames are $S$ and $S'$ and the relative velocity between them is $v$. Then we have a gamma factor $\gamma_v$ that determines the Lorentz Transformation between the frames.

If we then have a particle moving with velocity $u$ in frame $S$ and velocity $u'$ in frame $S'$, then the particle has a different gamma factor in each frame and the energy-momentum of that particle in those frames is given by:

$p^\mu = m \gamma (1, \vec{u}) \$ and $p'^\mu = m \gamma' (1, \vec{u'})$

Where $\gamma$ is the gamma factor of the particle in frame $S$ and $\gamma'$ is the gamma factor of the particle in frame $S'$.

There is a useful relationship between these gamma factors, which is:

$\gamma' = \gamma_v \gamma (1 - vu_x)$

(Assuming, as usual, that $S'$ is moving with velocity $v$ in the +ve x-direction in frame $S$.)

It's a good exercise to derive that formula.

 

[Dale]

This uses the fact that because v = 0 then γ(v) = 1.
I'm confused about this last sentence.

I am not sure how this can be confusing.
$\gamma(v)=1/\sqrt{1-v^2/c^2}$ so clearly $\gamma(0)=1$

$\gamma(v)$ is just a function. It does show up in the Lorentz transform, but it is still just a function which can be evaluated for any v in its domain.

 

[Sorcerer]

Hi.
If the 4-momentum in SR is given by p = m γ(v) ( 1 , v ) then in the rest frame of a particle the 4-momentum is ( m , 0 , 0 , 0 ) using c=1 units. This uses the fact that because v = 0 then γ(v) = 1.
I'm confused about this last sentence. The gamma factor is used for the relative velocity between 2 inertial frames but in the rest frame the particle is at rest relative to every other inertial frame so how can we specify a gamma factor ?

I think what confuses you is that when considering the speed v between frames, at least TWO observers MUST be considered. But when considering the gamma factor of a particle in your frame, at least ONE observer must be considered. If you are at rest and the particle moves at speed u, it’s gamma factor FOR YOUR FRAME is γ_u. (Of course you can combine the two situations, like in your original example).Useful information:

https://courses.lumenlearning.com/physics/chapter/28-5-relativistic-momentum/

As PeroK pointed out, in your example there are THREE gamma factors: a gamma factor relating the two reference frames (γ_v), but there are also gamma factors for the particles themselves in each frame (γ_u and γ_u’).The gamma factor of the particle in your rest frame is 1, as you said.

 

[dyn]

I think what confuses you is that when considering the speed v between frames, at least TWO observers MUST be considered. But when considering the gamma factor of a particle in your frame, at least ONE observer must be considered. If you are at rest and the particle moves at speed u, it’s gamma factor FOR YOUR FRAME is γ_u. (Of course you can combine the two situations, like in your original example)
.

Yes this is what is confusing me. For questions involving 4-momentum , the gamma factor uses the velocity of the particle in my frame but that is the only frame being used and I am the only observer. Can the second frame be considered as the rest frame of the particle with the particle at its origin ? Does it have to be at the origin ? In which case the velocity of the particle in my frame is the relative velocity between these 2 frames .

 

[Dale]

Yes this is what is confusing me. For questions involving 4-momentum , the gamma factor uses the velocity of the particle in my frame but that is the only frame being used and I am the only observer.

Again, that is one context in which the gamma factor arises, but it is just a function that can arise in other contexts. In this context v is just the velocity of the particle in a given frame.

 

[Ibix]

Yes this is what is confusing me. For questions involving 4-momentum , the gamma factor uses the velocity of the particle in my frame but that is the only frame being used and I am the only observer. Can the second frame be considered as the rest frame of the particle with the particle at its origin ? Does it have to be at the origin ? In which case the velocity of the particle in my frame is the relative velocity between these 2 frames .

Let me ask you a question: if we always wrote the Lorentz transforms as$$\begin{eqnarray*}
t'&=&\frac{t-vx/c^2}{\sqrt{1-v^2/c^2}}\\
x'&=&\frac{x-vt}{\sqrt{1-v^2/c^2}}
\end{eqnarray*}$$and always wrote the particle momentum as$$p=\frac{mv}{\sqrt{1-v^2/c^2}}$$then would the appearance of that same square root term in both bother you? You'd just use the appropriate $v$ and carry on. But since we've given it its own symbol you seem to be getting confused. Don't. $\gamma$ is simply an expression that comes up often enough that we got fed up of writing out that square root every time.

Another thing to ask yourself - $v$ appears in the Lorentz transforms and in the expression for momentum. Why aren't you worried about $v$ in the same way you are worried about $\gamma$?

Basically, you are confusing yourself, I think. There is no mystery - $\gamma$ isn't solely associated with Lorentz transforms. It appears in many formulae in special relativity. Generally it's obvious which velocity you should use because there's only one physically relevant velocity which, in the case of a particle's momentum, is the velocity of the particle. What else could matter?

 

[Herman Trivilino]

The gamma factor is used for the relative velocity between 2 inertial frames but in the rest frame the particle is at rest relative to every other inertial frame so how can we specify a gamma factor ?

The relative velocity between two inertial frames can be zero! In that case $\gamma=1$.