Find Relativistic Momentum Equation for a Moving Reference Frame?

[Ascendant0]

Homework Statement

Suppose that a mass $m$ has momentum $p$ and energy $E$, as measured in a frame $S$. Use the relations $p = m(dr)/(dt_0)$ and $E = mc^2(dt)/(dt_0)$ and the known transformation of dr and dt to find the values of $p'$ and $E'$ as measured in a second frame $S'$ traveling with speed $v$ along $O_x$

Relevant Equations

Lorentz Transformation for velocity: $u_x' = (u_x-v)/(1-(u_xv/c^2)$
Possibly time? $t = \gamma t_0$

I figured since $dr/dt$ is simply the velocity of the target mass, the velocity $u_x$ would simply have to be changed by the Lorentz transformation. Since the rest mass doesn't change, I think this should be as simple as taking the Lorentz transformation for velocity, and substituting the equation for $u_x'$ listed in the "relevant equations" section above, and multiply that by $m$ to get $mu' = p'$, so:

$p' = \gamma mu_x' = \gamma m(u_x-v)/(1-(u_xv/c^2)$

But, I can't find anything online about a conversion like this for some reason, and I'm not all that confident it would be that simple, but I'm not sure how else to look at this other than the velocity should be converted via the Lorentz transformation.

As far as $E'$, I'm not sure how to do that one. I don't see how I would add in some sort of Lorentz transformation into it in order to take into consideration the new velocity, unless I were to include it in the $\gamma$

 

[PeroK]

I suspect the problem is to find $E', p'$ in terms of $E, p$. I assume that $t_0$ is the proper time of the particle.

 

[Ascendant0]

I suspect the problem is to find $E', p'$ in terms of $E, p$. I assume that $t_0$ is the proper time of the particle.

Yes, that's all correct. The problem tells us to use the two relations there, $$p = m(dr)/(dt_0) $$
and $$ E = mc^2 (dt)/(dt_0) $$to find them. So going by that, that is what led me to the equation above for $p'$, but again, I'm not all that confident in that answer.

And I don't know how to go about solving for $E'$, so I'm lost on this one. I know the $mc^2$ won't change, since mass is not changing, and of course the constant $c$ isn't going to change. But, if I were to change the equation to $t$ instead of $t_0$ in that equation to convert to $E'$, I'd just end up changing it to $E =\gamma mc^2$, but then I don't know what $v$ would be to take into consideration both the $v$ of $S'$, as well as the $v_0$ in the momentum of the mass relative to the original frame. I feel like I'm not fully wrapping my brain around all the relations here in various frames

 

[PeroK]

Why not differentiate the Lorentz Transformations for the time and position of the particle with respect to its proper time?

 

[Ascendant0]

Why not differentiate the Lorentz Transformations for the time and position of the particle with respect to its proper time?

Thank you. I did this earlier today after class, but I wanted to post on here, because I'm not confident I simplified it enough, but not sure how else to simplify it

So, I took $p_x' = m(dr_x') = m \gamma (dx - vdt)/(dt_0) = \gamma p_x - \gamma mv(dt)/(dt_0)$

To finish with: $p_x' = \gamma [p_x - mv (dt)/(dt_0) ]$

I don't see any way to simplify it further, as the $mv$ is relativistic, so I can't break that down to $p$. I don't like leaving it with the $(dt)/(dt_0)$ though, but is there anything else I can do to make this simpler?

Additionally, I get where the $\gamma p_x$ comes from in the first part, that makes complete sense to me. But, I'm not sure where the second part of the equation comes from? I would think it would have to do with the energy $E$, but I don't see any way to change what I see there into $E$?

 

[TSny]

I don't see any way to change what I see there into $E$?

Homework Statement: … Use the relations $p = m(dr)/(dt_0)$ and $E = mc^2(dt)/(dt_0)$

 

[Ascendant0]

Omg, I can't believe I missed that. Thank you.

I finished the rest of my homework problems, and I'm trying to finish up the second part, finding the equation for $E'$. I believe it might be simpler than I originally thought...

Since $(dt)/(dt_0) = \gamma$, I can just take the $E = \gamma mc^2$ and take the derivative of that with respect to v (from $\gamma$), right?

With that, I'm getting:

$E' = -mc^2/((1-v^2/{c^2})^{3/2})$

The only thing that's throwing me off there is the negative sign I get from the derivative, but I'm thinking I would just make it positive since it will of course be a positive energy $E$?

 

[vela]

Omg, I can't believe I missed that. Thank you.

I finished the rest of my homework problems, and I'm trying to finish up the second part, finding the equation for $E'$. I believe it might be simpler than I originally thought...

Since $(dt)/(dt_0) = \gamma$, I can just take the $E = \gamma mc^2$ and take the derivative of that with respect to v (from $\gamma$), right?

No. Why do you think differentiating $E$ with respect to $v$ would give you the energy measured in $S'$? You want to use the same approach you did for $p'$.

The only thing that's throwing me off there is the negative sign I get from the derivative, but I'm thinking I would just make it positive since it will of course be a positive energy $E$?

You can't erase a sign just because it's inconvenient. A wrong sign is typically an indication you made a mistake which you should find and correct.

 

[Ascendant0]

No. Why do you think differentiating $E$ with respect to $v$ would give you the energy measured in $S'$? You want to use the same approach you did for $p'$.


You can't erase a sign just because it's inconvenient. A wrong sign is typically an indication you made a mistake which you should find and correct.

I figured it was wrong, but I'm struggling, and people complain on here if they feel you aren't making an effort, which I am, I'm just stuck on this one.

With $p$, I saw how the $m(dr'/dt_0)$ could be used, but with energy, I'm not seeing any derivative I could use for energy. Any derivative I have on my formulas sheet relating to energy is also the derivative of E, like $dE = (dp)/(dt) udt$, which obviously wouldn't do me any good and isn't even one of the equations included here.

If I took $E = mc^2(dt)/(dt_0)$ and changed $dt$ to $dt'$, then it's basically going to be $t' = t/\gamma$, along with $1/t_0 = \gamma/t$, and they'd just cancel out, which of course doesn't do me any good (unless they don't cancel out and there's a difference between the two that I'm not seeing). But anything I manipulate with the equations they gave, I can't see a way to take a derivative that would give me $E'$

Based off the other equations, I know it's going to look something like:

$E' = \gamma(E - [something])$, just no clue what the [something] is, nor how to derive the equation

 

[vela]

What are the Lorentz transformation equations you have?

 

[Ascendant0]

What are the Lorentz transformation equations you have?

The ones for position:

$x' = \gamma (x - vt)$ and the inverse

Velocity:

$u'= (u_x-v)/(1-(u_xv/c^2)$ and the inverse

... and I actually think I found it just now. I kept looking at $t = \gamma t'$, which I had grouped with the aforementioned, and I forgot on the back of my reference sheet, I have:

$t' = \gamma (t-vx/c^2)$

Ok, I know what to do now. I have to redo my formula sheet so I can include that near the other ones. Our professor didn't cover it until the day after the other ones (only gave us the $t = \gamma t'$ with the other transformations the first day), so I didn't have it grouped with them.

I'm not 100% on it, but I think I know the right direction now. I'm going to try to work this out real quick now before I head to bed.

 

[Ascendant0]

What are the Lorentz transformation equations you have?

So I believe it's:

$E' = \gamma^2mc^2(dt-dv(x)/(c^2))$, and then just simplify this as I'm able, right?

 

[vela]

So I believe it's:

$E' = \gamma^2mc^2(dt-dv(x)/(c^2))$, and then just simplify this as I'm able, right?

If you check the units, you should notice the righthand side has the wrong units, so it can't be correct. It's better if you show your work instead of just posting a result and asking if it's right.

 

[Ascendant0]

If you check the units, you should notice the righthand side has the wrong units, so it can't be correct. It's better if you show your work instead of just posting a result and asking if it's right.

Thank you for pointing that out. I finally came back to finish this since it's due tomorrow. I think I have it now

So, I took $E = mc^2 (dt)/(dt_0)$ and $t' = \gamma (t' - vx'/(c^2)$, then:

$E' = mc^2 (dt')/(dt_0) = mc^2 \gamma ((dt)/(dt_0) - (v/c^2) (dx/dt_0) = \gamma (E - vp)$

So $E' = \gamma (E - vp)$

That seems right looking at the other equations, and it is also consistent units-wise. I had expected it to be " $\gamma (E - [something])$", and it being $vp$ make sense

Thank you for pointing out the units (and all your previous help of course). That's something that I have to remember to check on things like this