# Relativistic Mass

1. Feb 24, 2005

### misogynisticfeminist

From what i heard, my friend told me relatvistic mass has just been recently proven wrong. Is that right????

2. Feb 24, 2005

### marlon

No this is not right.

How would one prove relativistic mass to be wrong ???

marlon

3. Feb 24, 2005

### pmb_phy

No. Absolultely not. Where did he get that idea from??

Pete

4. Feb 24, 2005

### Janus

Staff Emeritus
Attention:
Tempest99 is just a new user name for Physicsguru. Accordingly, I have deleted all of his posts and any posts referencing them.

5. Feb 24, 2005

### marlon

Thanks Janus for getting rid of these crappy posts

marlon

6. Feb 24, 2005

### Tom Mattson

Staff Emeritus
You can't prove it wrong, because it's a definition. Take the expressions for momentum and total energy in natural units:

$$p=\gamma m_0 v$$
$$E=\gamma m_0$$

If you want to define mass as the (Lorentz-invariant) norm of the 4-momentum, then $m_0$ is that mass. It doesn't change with velocity. On the other hand, there is nothing stopping you from associating $\gamma m_0$ together and defining that product as $m$. This $m$ is called the relativistic mass, and it is a function of velocity.

So, the only way to disprove relativistic mass is to disprove the associative property of real numbers under multiplication (it can't be done).

Maybe his name is DW. :rofl:

7. Feb 24, 2005

### pmb_phy

I'll have to disagree with you on this point. Relativistic mass is defined such that the quantity mv is always conserved in collision (wherein one then defines the product mv as 'momentum'). Had nature been such that this is never measured in the lab then it would follow that the definition is incorrect in that it can't fit into what nature had in mind.

Pete

8. Feb 24, 2005

### dextercioby

Tom,since "c" is absolute (and NOT BECAUSE IT IS CONVENTIONALLY TAKEN AS "+1"),one should not speak about RELATIVISTIC MASS,but about ENERGY...

Daniel.

9. Feb 24, 2005

### Tom Mattson

Staff Emeritus
Is that not what I said? If you define $\gamma m_0$ as $m$, then $p=mv$ and $E=m$ (in natural units). These quantites are conserved.

Last edited: Feb 24, 2005
10. Feb 24, 2005

### Tom Mattson

Staff Emeritus
I'm not sure I get what you are saying. $E$ and $m$ are only identical in natural units. In the SI system, they have different dimensions.

11. Feb 24, 2005

### dextercioby

There's no need to introduce RELATIVISTIC MASS (a rather confusing concept,at least to the unfamiliar with subtleties of SR/GR),when u have already energy...

Daniel.

12. Feb 24, 2005

### pmb_phy

What pat of it are you confused about?
Relativistic mass is defined differently than energy. Under certain cercumstances one is proportional to the other, I.e. E = -mc^2. It appers to me that you're asserting that the "m" in m = E/c^2 is identical to the "m" in m = p/v. If so then the assertion is incorrect since, in the general, E/c^2 does not equal m = p/v.

Pete

13. Feb 24, 2005

### chroot

Staff Emeritus
Admin note: Let's not make the same old arguments about relativistic mass all over again. Otherwise, this thread will be locked.

- Warren

14. Feb 24, 2005

### dextercioby

Okay,neglect gravitational potential in that square root.Are u claiming that the "m"-s are different in SR...??

Daniel.

15. Feb 24, 2005

### dextercioby

Okay,Warren,i'll back off.

Daniel.

16. Feb 24, 2005

### pmb_phy

I was saying that relativistic mass is defined such that mv is a conserved quantity. What you posted implied that the definition cannot be wrong. What I posted was why a definition can be wrong. Its a fine point though and not worth worrying about.

Pete

17. Feb 24, 2005

### Tom Mattson

Staff Emeritus
I get you. What I should have said was:

Taking SR for granted, relativistic mass cannot be disproven.

Naturally, relativistic mass would be falsified if SR were falsified.

Last edited: Feb 24, 2005