# 4-momentum basic problem

1. Aug 27, 2008

### Newtons Balls

1. The problem statement, all variables and given/known data
A particle has 4-momentum:
P$$^{u}$$ = c(2$$\sqrt{2}$$,1,0,-1)
Where c denotes the speed of light.
Calculate the particle's rest mass m, its energy E, its speed v and its kinetic energy T.

2. Relevant equations

Well the relativistic relation between E, m and p is:
$$m^{2}$$ = $$E^{2}$$ - $$p^{2}$$
3. The attempt at a solution

I know this is meant to be really easy but this 4-momentum stuff is confusing me rather allot :/

The notation I've seen in my textbook indicates the 4th component should be the temporal component. The textbook also seems to indicate that the temporal component is equal to mc and E/c. This would give negative energy and negative mass, which I don't think is correct- am I misunderstanding what the textbook is saying?

After doing a bit of reading I've seen things saying that the 4-momentum inner producted with itself is related to the rest mass of the particle like this:
$$-|p|^{2}=m^{2}c^{2}$$

1) is this relationship correct?
2) I haven't actually managed to find anything explicitly saying what the inner product in Mikowski space is, would it be the standard dot product still?
3) if the relationship is correct do I use all four components in this dot product or just the first three?

I'm a bit rubbish at this stuff, I'm trying to teach myself the course from a textbook and whilst its quite good at explaining the kinematic aspects I'm having trouble following it when it gets onto mass/energy relationships...

Thanks

Last edited: Aug 27, 2008
2. Aug 27, 2008

### tiny-tim

i see no c …

Hi Newtons Balls!
erm … what c?
Yes … some books use the 4th, and some the 1st.

This question means the 1st.

But you seem to have copied it wrong anyway …

do you mean c(√2,1,0,-1)?

3. Aug 27, 2008

### Newtons Balls

yes, sorry I meant c in front of the bracket, not the 2. I'll edit that now!

Thanks for clarifying the notation :)

4. Aug 27, 2008

### Newtons Balls

If the 1st part of the 4-momentum in this case is the temporal component, is the other part of the textbook correct in saying this part is = E/C and mc? and does this m refer to proper mass or relativistic mass? its not at all clear :/

If it does refer to proper mass then this would mean the proper mass is 2$$\sqrt{2}$$. However if you inner product the entire 4-momentum with itself as I read elsewhere, which should result in $$-m^{2}$$$$c^{2}$$, this leads to a mass of $$\sqrt{10}$$. I'm guessing one of the masses is relativistic? Also I'm not sure what the negative sign is about ^_^

This would all be wrong, of course, if the dot product is not an inner product of Mikowski space...

5. Aug 27, 2008

### tiny-tim

(btw, you don't have to keep writing tex … you can put = and - in the middle of a tex expression)

This is the correct equation.

Why didn't you use it?
However did you get √10?

$$(t,x,y,z)^2 = t^2 - x^2 - y^2 - z^2$$

Start again.

6. Aug 27, 2008

### Newtons Balls

Thanks.

The reason I didn't use it yet is I don't know E yet.
My textbook did mention that the temporal component is = E/c, is this correct?
If that is true then $$E=2\sqrt{2}c^{2}$$
and I can go on and (hopefully) do the rest...

7. Aug 27, 2008

### tiny-tim

… the clue's in the name …

The reason why it's called the 4-momentum is because that's what it is … three bits are the momentum, p, and the other bit is the energy, E.

So E = 2√2 c and p= (c,0,-c).

8. Aug 27, 2008

### Newtons Balls

Thanks.

So $$E=2\sqrt{2}c, p=\sqrt{2}c, m=\sqrt{6}c$$?

I really hope I've finally got this :P

9. Aug 27, 2008

### tiny-tim

That's fine.

Now what are the speed and the KE?

10. Aug 27, 2008

### Newtons Balls

Well for v:
$$p=mv\gamma$$?

Unfortunately that doesn't rearrange reasonably... is that equation wrong or am I being an idiot with my re-arranging? ^_^

11. Aug 27, 2008

### tiny-tim

Good.

And similarly, E = … ?

12. Aug 27, 2008

### Newtons Balls

$$E=c^{2}m\gamma$$?

Or you mean T?

$$T=E - c^{2}m$$?

I assume you mean use the E equation to work out v, then minus that from E to work out T?

13. Aug 27, 2008

### tiny-tim

That's right! So v = … ?
hmm … I was actually expecting just $$E\ =\ m\gamma$$ …

I wonder whether I'm using the wrong units?

14. Aug 27, 2008

### Newtons Balls

Hmm. No I think you're right about not needing the $$c^{2}$$. I keep referring to this textbook and it doesn't use natural units. :(

15. Aug 27, 2008

### Newtons Balls

Anyway, just did the calculation quickly and I get 0.5c, which sounds reasonable.
And I got T=$$\sqrt8-\sqrt6$$c.. whatever that is. Sound right?

16. Aug 27, 2008

### tiny-tim

hmm … not a good idea to use different books …

Relativity is notoriously a subject where there are different conventions.

Stick to whatever your professors use, or you'll get confused and risk misunderstanding questions in your exams.
I don't like the sound of that …

you're supposed to say something like $$p\ =\ mv\,\gamma\ \ E\ =\ m\,\gamma$$

so v = p/E,

which in this case is √2/2√2, = 0.5c (uh … where did that c come from? )

oh … are you sure the original question wasn't (√2c,1,0,-1)?

That would look much better dimensionally.

17. Aug 27, 2008

### Newtons Balls

Its definitely $$c(2\sqrt2,1,0,-1)$$

And as for the textbook, it is the recommended text for the module. I haven't sat the module which explains my nubbishness. Long story short I have to teach myself this module over the summer for next year :( The textbook isn't very good for self learning though it seems. Kinematics and all the other stuff seemed to work okay, but I can't get my head around this 4-momentum/energy/mass related stuff.

So...as for this velocity thing: you're saying its just... 0.5 with no c? that does sound a little odd ^_^

18. Aug 27, 2008

### tiny-tim

No … it must have a c in it.

Which is why, on thinking about it, I expected the 4-momentum to have a c in the t position.

I learnt relativity with units so that c = 1, so I never had to bother with factors of c.

Clearly some of the equations I'm familiar with have an extra c or c2 with your units.

19. Aug 27, 2008

### Newtons Balls

*groan*

I've just been trying to sort out the units.

Doesn't the initial relativistic equation mean P and E must be dimensionally the same?

$$m^{2} = E^{2} - p^{2}$$

Then how can the velocity be E/p, as that would make it dimensionless?

20. Aug 27, 2008

### tiny-tim

You're supposed to be supplying us with the equations!

$$m^{2} = E^{2} - p^{2}$$ looks to me like a c = 1 equation.

It doesn't seem to fit the units in the rest of your problem.

btw, in space-time, velocity is dimensionless (space and time have the same dimensions) with c = 1 units.