Prove v<c for particles with m>0 using two identities

[Ascendant0]

Homework Statement

Using the two identities:


$\beta = u/c = pc/E$ ... and ... $E^2 = (pc)^2 + (mc^2)^2$


prove that the speed of any particle with $m > 0$ is always less than $c$

Relevant Equations

$\beta = u/c = pc/E$ ... and ... $E^2 = (pc)^2 + (mc^2)^2$

I can't figure out how to prove this using only those two identities? I mean in general, I could prove it easy when using relativistic equations, and showing that if $v = c$, the denominator becomes 0, and if $v>c$, the denominator becomes an imaginary number (a negative square root). But, with only having those two eqs above, I don't see how to show it?

 

[Ibix]

What have you tried so far?

 

[PeroK]

It might help to use units where $c = 1$. That simplifies things. And, once you see it that way, you can see how it works with $c \ne 1$.

 

[kuruman]

I cannot be more specific without giving everything away. Can you show that $\beta<1$ using the two given equations?

 

[Ascendant0]

What have you tried so far?

I figured this problem out myself earlier today, but thanks for asking

I always do try to do these problems on my own, but I feel like trying to type out all of the stuff I write in LaTeX would be an impractical use of my already limited time, especially because I had about a half hour of scribbles trying to figure this one out, lol. I just couldn't find any limiting factor that would require $v$ to be less than $c$, until I split the problem up and started making comparisons between $mc^2$ and $pc$. As soon as I did that, I saw where I could take $\beta$ and combine it with the $E^2$ eq to show that $pc^2/E^2 < 1$, and then of course $v^2/c^2 <1$, and I feel like it's blatantly obvious from there, lol.

Would it make more sense to simply post a screenshot of my hand-written work when asking questions like this so people can know where I'm at on the problem and what I've tried already?

 

[Ascendant0]

I cannot be more specific without giving everything away. Can you show that $\beta<1$ using the two given equations?

Thanks, and that's exactly what I did earlier today. I covered it in my last response. My problem was that I wasn't splitting up the $E^2$ equation and relating the two values on the right initially, so I couldn't for the life of me see why $v$ would have to be less than $c$. As soon as I was thinking that having mass means $E^2$ had to be greater than $(pc)^2$ and related that to $\beta$, then it was obvious where to go from there. That's when I got the $1 > (pc)^2$ while I was driving home with the kids, and nearly locked up the breaks to write it down to make absolutely sure it didn't fall out of my head before I wrote it down, lol.

 

[Ibix]

Would it make more sense to simply post a screenshot of my hand-written work when asking questions like this so people can know where I'm at on the problem and what I've tried already?

Forum rules (see here, which is linked from the global guidelines) require you to show evidence of effort before we can help. Apart from anything else, organising your thoughts to explain what you've done may lead you to an insight into what you missed.

Whether a photo of your hand written work is enough is doubtful. The rules say that it is not, and I personally don't usually try to decipher a photo of someone's handwriting - if you can't be bothered to make it easily readable, I can't be bothered to read it. But I have seen such threads answered.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

Would it make more sense to simply post a screenshot of my hand-written work

No, because images are not acceptable ways to post that kind of content here. You need to post text as text and equations as LaTeX equations.

so people can know where I'm at on the problem and what I've tried already?

As @Ibix has pointed out, the PF rules already require you to show what you have already tried. You just need to do it as described above.

Since you have solved the problem, this thread will remain closed. One final note, however: if you think someone's post violates the PF rules, and you use the Report button, please don't also respond in the thread with basically the same thing you said in your report. (You should not respond in the thread in such cases anyway; please just use the Report button.) That just means we moderators now have to deal with two posts instead of one.