# Relativistic Mass

#### misogynisticfeminist

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

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#### marlon

misogynisticfeminist said:
From what i heard, my friend told me relatvistic mass has just been recently proven wrong. Is that right????
No this is not right.

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

marlon

#### pmb_phy

misogynisticfeminist said:
From what i heard, my friend told me relatvistic mass has just been recently proven wrong. Is that right????
No. Absolultely not. Where did he get that idea from??

Pete

#### Janus

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

#### marlon

Thanks Janus for getting rid of these crappy posts

marlon

#### Tom Mattson

Staff Emeritus
Gold Member
misogynisticfeminist said:
From what i heard, my friend told me relatvistic mass has just been recently proven wrong. Is that right????
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).

pmb_phy said:
Where did he get that idea from??
Maybe his name is DW. :rofl:

#### pmb_phy

Tom Mattson said:
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).
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.
Maybe his name is DW. :rofl:

Pete

#### dextercioby

Homework Helper
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.

#### Tom Mattson

Staff Emeritus
Gold Member
pmb_phy said:
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').
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.

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#### Tom Mattson

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

#### dextercioby

Homework Helper
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.

#### pmb_phy

dextercioby said:
There's no need to introduce RELATIVISTIC MASS (a rather confusing concept,
What pat of it are you confused about?
...at least to the unfamiliar with subtleties of SR/GR),when u have already energy...
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

#### chroot

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

- Warren

#### dextercioby

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

Daniel.

#### dextercioby

Homework Helper
Okay,Warren,i'll back off.

Daniel.

#### pmb_phy

Tom Mattson said:
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.
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

#### Tom Mattson

Staff Emeritus
Gold Member
pmb_phy said:
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.
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.

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