Prove Speed Any Particle with Mass >0 is Always < c

[knowLittle]

Homework Statement

Use:
$\beta =\dfrac {\overline {u}} {c}=\dfrac {\overline {p}c} {E}\&\&E^{2}=\left( pc^{2}\right) +\left( mc^{2}\right) ^{2}$

Homework Equations

The Attempt at a Solution

Solving the second equation for E, I get:
$\left( \dfrac {E u} {c}\right) ^{2}+m^{2}c^{4}>0$
$E^{2}u^{2}>-m^{2}c^{6}$

$\left( p\dfrac {c^{2}} {u}\right) ^{2}u^{2}>-m^{2}c^{6}$
p^2 > -m^2 c^2

$\gamma ^{2}m^{2}v^{2}$ > -m^2 c^2
v^2 >-c^2 +v^2
v^2<c^2 -v^2
√2 v<c
v < (c) / √2, and because √2 << c
v<c

Is this correct? I don't think it is, but I cannot think of any other way.

 

[darkxponent]

You din't proved v<c!. You proved sqrt(2)*v<c, which is certainly not true.

There are many arguments that prove this. The best i remember is the Force method.

Hint: What is the equation of Force?

 

[knowLittle]

I am required to use these equations:

Homework Statement

Use:
$\beta =\dfrac {\overline {u}} {c}=\dfrac {\overline {p}c} {E}\&\&E^{2}=\left( pc^{2}\right) +\left( mc^{2}\right) ^{2}$

 

[darkxponent]

I am required to use these equations:

As i told you there are many methods. These equations might get used in some or other. But you don't need these equations in force method. First derive the equation of Force. Use F = dp/dt.

 

[knowLittle]

I don't know how to proceed :/
dp= gamma (m*v_2 - m*v_1 )
dt= gamma(d t_0), by time dilation formula

F=ma=dp/dt
m du/dt = gamma( (m*v_2 - m*v_1 ) )/dt

 

[Ibix]

Your first approach isn't far off. Try not dumping terms just because it gets you an inequality. See where it takes you.

 

[vela]

Your second equation as written is incorrect. It should be
$$E^2 = (pc)^2 + (mc^2)^2.$$ Just use this equation to eliminate E in the first equation and show that $1/\beta > 1$.

 

[knowLittle]

Ok, thanks. I will follow what vela and lbix suggested.

Vela which second equation are you referring to?

 

[darkxponent]

I don't know how to proceed :/
dp= gamma (m*v_2 - m*v_1 )
dt= gamma(d t_0), by time dilation formula

F=ma=dp/dt
m du/dt = gamma( (m*v_2 - m*v_1 ) )/dt

These are not correct.
dp=d(gamma*mv).
dt =dt.(you are at rest not moving. Time dilation is for moving frame)

Now F= dp/dt, remember gamma is a function of v, not a constant, write the whole function and then diffrentiate dp/dt.

You can go with your first approach. Remeber the constraint given in the problem. It says something about the mass the object!


As i said to you 'the best i remember is force method' because only the force method gives you physical understanding of why the speed of any object can't exceed 'c'. The method you are using is all about solving equations mathematically, it won't give you physical understanding.

 

[Ibix]

Vela was observing a missing square in $E^2=p^2c^2+m^2c^4$ in your first post. I ignored it because the rest of your post implies the correct form.

 

[darkxponent]

I don't know how to proceed :/
dp= gamma (m*v_2 - m*v_1 )
dt= gamma(d t_0), by time dilation formula

F=ma=dp/dt
m du/dt = gamma( (m*v_2 - m*v_1 ) )/dt

These are not correct.
dp=d(gamma*mv).
dt =dt.(you are at rest not moving. Time dilation is for moving frame)

Now F= dp/dt, remember gamma is a function of v, not a constant, write the whole function and then diffrentiate dp/dt.

You can go with your first approach. Remeber the constraint given in the problem. It says something about the mass the object!


As i said to you 'the best i remember is force method' because only the force method gives you physical understanding of why the speed of any object can't exceed 'c'. The method you are using is all about solving equations mathematically, it won't give you physical understanding.

 

[knowLittle]

I think I found the proof.
$E^{2}-E^{2}\dfrac {u^{2}} {c^{2}}=\left( mc^{2}\right) ^{2}$

$E^{2}\left(1-\dfrac{u^{2}}{c^{2}}\right)=m^{2}c^{4}$

$\dfrac{E^{2}}{c^{6}}\left(c^{2}-u^{2}\right)=m^{2}$

$\dfrac{E}{c^{3}}\left(\left(c-u\right)\left(c+u\right)\right)^{\dfrac{1}{2}} =m$

If m>0, then u<c.

Is this a correct proof?

 

[darkxponent]

That is correct. Good work!

 

[knowLittle]

How do I make this a robust proof?

 

[darkxponent]

How do I make this a robust proof?

I don't know what you mean by robust proof. But if you are looking for the proof which gives some physical understanding, then derive the equation for force as i told you earlier.

 

[knowLittle]

I don't know the exact definition of a robust proof either, but I have heard it so many times in textbooks that I know there's a difference between a weak proof and a robust proof.

I will try you approach, darkxponent.

Can anyone tell me, if this proof is robust?

Thank you all for your help. :)

 

[jeppetrost]

I can see a much more simple proof, which might seem more 'robust' (and which has been mentioned before).

Look at your second equation. If m>0 then E > (pc). Now your first equation is v/c = (pc)/E.

What does that tell you about v/c ?

 

[knowLittle]

If E>pc, then pc/E <1.
Then, the ratio of v/c <1 .

For v/c <1, c>v or v<c.

:>
I am not sure what robust means, but I am ok with this one too.

 

[vela]

I think I found the proof.
$E^{2}-E^{2}\dfrac {u^{2}} {c^{2}}=\left( mc^{2}\right) ^{2}$

$E^{2}\left(1-\dfrac{u^{2}}{c^{2}}\right)=m^{2}c^{4}$

$\dfrac{E^{2}}{c^{6}}\left(c^{2}-u^{2}\right)=m^{2}$

$\dfrac{E}{c^{3}}\left(\left(c-u\right)\left(c+u\right)\right)^{\dfrac{1}{2}} =m$

If m>0, then u<c.

Is this a correct proof?

It wouldn't be a very good proof, even if it were correct. Ideally, you end up with an inequality that is what you're trying to show, i.e. $|u|<c$. Instead, you have a relatively big leap of logic. How exactly does m>0 imply u<c? You should consider that when you take square root, there's another possible solution:
$$\dfrac{E}{c^{3}}\left(\left(c-u\right)\left(c+u\right)\right)^{\dfrac{1}{2}} = -m.$$ This ruins your (already flawed) argument. According to your logic, since the righthand side is negative, u isn't necessarily less than c, right? Can you even justify taking the square root? By assuming you can take the square root of the lefthand side, you've implicitly assumed what you're trying to prove.

Go back to the second line in your proof. If you can somehow justify that $E \ne 0$, you can divide both sides by $E^2$. The righthand side will be $\frac{m^2c^4}{E^2} = \left(\frac{mc^2}{E}\right)^2$. What can you conclude about the sign of that term?

 

[darkxponent]

If E>pc, then pc/E <1.
Then, the ratio of v/c <1 .

For v/c <1, c>v or v<c.

:>
I am not sure what robust means, but I am ok with this one too.

Good to see. So proofs are raining from everywhere. As i told there are many arguments. You already got two of them. I don't know how many of these proofs are. Both of these are more or less similar(solving inequalities). If you are looking for the so called robust proof, try the Force method. I can show it to you(as you already have done yours), but i would like you to do it yourself.

 

[darkxponent]

It wouldn't be a very good proof, even if it were correct. Ideally, you end up with an inequality that is what you're trying to show, i.e. $|u|<c$. Instead, you have a relatively big leap of logic. How exactly does m>0 imply u<c? You should consider that when you take square root, there's another possible solution:
$$\dfrac{E}{c^{3}}\left(\left(c-u\right)\left(c+u\right)\right)^{\dfrac{1}{2}} = -m.$$ This ruins your (already flawed) argument. According to your logic, since the righthand side is negative, u isn't necessarily less than c, right? Can you even justify taking the square root? By assuming you can take the square root of the lefthand side, you've implicitly assumed what you're trying to prove.

Go back to the second line in your proof. If you can somehow justify that $E \ne 0$, you can divide both sides by $E^2$. The righthand side will be $\frac{m^2c^4}{E^2} = \left(\frac{mc^2}{E}\right)^2$. What can you conclude about the sign of that term?

I have to disagree with you here. After taking the squareroot, taking only positive m is okay because negative m makes no sence(as squareroot of real numbers is never negative).

I somehow agree that taking root was not required as the proof was done in the second-last equation itself.

I agree that this is not a good proof because this doesn't give any physical understanding. That is why i recommended the Force method.

PS: Correct me if i am wrong.

 

[knowLittle]

I agree with darkexponent's question.

Don't Physicist "just" discard answers that don't make physical sense?

I will try your approach too, vela.

 

[vela]

Your proof was not a good proof you never actually explicitly show the result you're trying to prove. Instead, you rely on the reader to fill in the steps to reach the conclusion |u|<c when it's not obvious. If you can't write down the final steps, it's a good sign you're cheating (proof-wise, that is) and you need to rethink the approach.

I pointed out some holes in your logic. You're looking for a robust proof, right? That means you need impeccable logic. You can't just throw stuff out or sweep it under the rug because it breaks your argument.

To reiterate: Your claim that m>0 implies u<c is wrong, but let's assume that step is okay. Your proof still has a problem because you can't arbitrarily throw out the -m solution. Note that's not the same thing as saying m<0, so claiming it's not physical isn't going to fly.

 

[knowLittle]

Yes, I am looking for a robust proof.

Thank you for your advice. I will consider it and re-work the solution.